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 Text-book of theoretical naval architect 
 
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TEXTBOOK 
 
 OF 
 
 THEORETICAL NAVAL ARCHITECTU 
 
TEXT-BOOK 
 
 OF 
 
 THEORETICAL NAVAL 
 ARCHITECTURE 
 
 E. L. ATTWOOD 
 
 ASSISTANT CONSTRUCTOR, ROYAL NAVY 
 
 MEMBER OF THE INSTITUTION OP NAVAI- ARCHITECTS 
 
 LECTUREK ON NAVAL ARCHITECTURE AT THE WEST HAM MUNICIPAI- 
 
 TECHNICAL INSTITUTE 
 
 WITH 114 DIAGRAMS 
 
 LONGMANS, GREEN, AND CO. 
 
 39 PATERNOSTER ROW, LONDON 
 
 NEW YORK AND BOMBAY 
 
 1899 
 
 .i U rights reserved 
 
 % 
 
PREFACE 
 
 This book has been prepared in order to provide students and 
 draughtsmen engaged in Shipbuilders' and Naval Architects' 
 drawing offices with a text-book which should explain the 
 calculations which continually have to be performed. It is 
 intended, also, that the work, and more especially its later 
 portions, shall serve as a text-book for the theoretical portion 
 of the examinations of the Science and Art Department in 
 Naval Architecture. It has not been found possible to include 
 all the subjects given in the Honours portion of the syllabus, 
 such as advanced stability work, the rolling of ships, the vibra- 
 tion of ships, etc. These subjects will be found fully treated 
 in one or other of the books given in the list on page 292. 
 
 A special feature of the book is the large number of 
 examples given in the text and at the ends of the chapters. 
 By means of these examples, the student is able to test his 
 grasp of the principles and processes given in the text. It is 
 hoped that these examples, many of which have been taken 
 from actual drawing ofiSce calculations, will form a valuable 
 feature of the book. 
 
 Particulars are given throughout the work and on page 292 
 as to the books that should be consulted for fuller treatment of 
 the subjects dealt with. 
 
 In the Appendix are given the syllabus and specimen 
 questions of the examination in Naval Architecture conducted 
 
VI Preface. 
 
 by the Science and Art Department. These are given by the 
 permission of the Controller of Her Majesty's Stationery Office. 
 I have to thank Mr. A. W. Johns, Instructor in Naval 
 Architecture at the Royal Naval College, Greenwich, for 
 reading through the proofs and for sundry suggestions. I also 
 wish to express my indebtedness to Sir W. H. White, K.C.B., 
 F.R.S., Assistant Controller and Director of Naval Construction 
 of the Royal Navy, for the interest he has shown and the 
 encouragement he has given me during the progress of the 
 book. 
 
 E. L. ATTWOOD. 
 
 LONDOX, 
 
 February, 1899. 
 
CONTENTS 
 
 CHAPTER PACK 
 
 I. Areas, Volumes, Weights, Displacement, etc. . . i 
 
 II. Moments, Centre of Gravity, Centre of Buoyancy, 
 
 Displacement Table, Planimeter, etc. , . 43 
 
 III. Conditions of Equilibrium, Transverse Metacentre, 
 
 Moment of Inertia, Transverse BM, Inclining 
 Experiment, Metacentric Height, etc. ... 86 
 
 IV. Longitudinal Metacentre, Longitudinal BM, 
 
 Change of Trim . . 132 
 
 V. Statical Stability, Curves of Stability, Calcula- 
 tions for Curves of Stability, Integrator, Dy- 
 namical Stability ... . .158 
 
 VI. Calculation of Weights and Strength of Butt 
 
 Connections. Strains experienced by Ships . . 188 
 
 VII. Horse-power, Effective and Indicated— Resistance 
 of Ships — Coefficients of Speed — Law of Corre- 
 sponding Speeds 214 
 
 Appendix .... . . . . 245 
 
 Index . . ... 293 
 
TEXT-BOOK 
 
 OF 
 
 THEORETICAL NAVAL ARCHITECTURE 
 
 CHAPTER I. 
 
 AREAS, VOLUMES, WEIGHTS, DISPLACEMENT, ETC. 
 
 Areas of Plane Figures. 
 
 A Rectangle. — This is a four-sided figure having its opposite 
 sides parallel to one another and all its angles right angles. 
 Such a figure is shown in d. C 
 
 Fig. I. Its area is the pro- 
 duct of the length and the 
 breadth, or AB X BC. Thus 
 a rectangular plate 6 feet 
 long and 3 feet broad will 
 contain^- 
 
 A. 
 
 6 X 3 = 18 square feet ^'°- '• 
 
 and if of such a thickness as to weigh 1 2\ lbs. per square foot, 
 will weigh — 
 
 18 X 12^ = 225 lbs. 
 
 A Square. — This is a particular case of the above, the 
 length being equal to the breadth. Thus a square hatch of 
 3^^ feet side will have an area of— 
 
 02 -^02 — 2-^2 4 
 
 = 12^ square feet 
 
 irtt 
 
Theoretical Naval Architecture. 
 
 A Triangle.— This is a figure contained by three straight 
 hnes, as ABC in Fig. 2. From the vertex C drop a perpen- 
 dicular on to the base AB 
 (or AB produced, if neces- 
 sary). Then the area is 
 given by half the product 
 of the base into the height, 
 or — 
 
 i(AB X CD) 
 
 If we draw through the 
 apex C a line parallel to 
 the base AB, any triangle 
 having its apex on this line, 
 and having AB for its base, will be equal in area to the 
 triangle ABC. If more convenient, we can consider either A 
 or B as the apex, and BC or AC accordingly as the base. 
 
 Thus a triangle of base s| feet and perpendicular drawn 
 from the apex 2,\ feet, will have for its area — 
 
 = 63^ square feet 
 
 If this triangle be the form of a plate weighing 20 lbs. to 
 the square foot, the weight of the plate will be — 
 ff X 20 = 123! lbs. 
 
 A Trapezoid. — This is a figure formed of four straight 
 
 lines, of which two only are 
 parallel. Fig. 3 gives such a 
 figure, ABCD. 
 
 If the lengths of the parallel 
 sides AB and CD are a and b 
 respectively, and h is the per- 
 pendicular distance between 
 them, the area of the trapezoid 
 is given by — 
 
 \(a ^b)-Kh 
 
 or one-half the sum of the parallel sides multiplied by the 
 perpendicular distance between them. 
 
 f<Jl 
 
 Fig. 
 
 B. 
 
Areas, Volumes, Weights, Displacement, etc. 3 
 
 Example. — An armour plate is of the form of a trapezoid with parallel 
 sides 8' 3" and 8' 9" long, and their distance apart 12 feet. Find its 
 weight if 6 inches thick, the material of the armour plate weighing 490 lbs. 
 per cubic foot. 
 
 First we must find the area, which is given by — 
 
 ( 8' 3" + 8' 9" '\ 
 
 I ^ ) X 12 square feet = 1^ x 12 
 
 = 102 square feet 
 
 The plate being 6 inches thick = \ foot, the cubical contents of the 
 plate will be — 
 
 102 X J = 51 cubic feet 
 
 The weight will therefore be — 
 
 51 X 490 
 
 SI X 490 lbs. =^-^^^ 
 
 = II"IS tons 
 
 A Trapezium is a quadrilateral or four-sided figure of 
 which no two sides are parallel. 
 
 Such a figure is ABCD (Fig. 4). Its area may be found 
 by drawing a diagonal BD 
 and adding together the 
 areas of the triangles ABD, 
 BDC. These both have the 
 same base, BD, Therefore 
 from A and C drop per- 
 pendiculars AE and CF on 
 to BD. Then the area of 
 the trapezium is given by — 
 
 i(AE + CF) X BD 
 
 Example. — Draw a trapezium F'g- 4- 
 
 on scale J inch = i foot, where 
 
 four sides taken in order are 6, 5, 6, and 10 feet respectively, and the 
 diagonal from the starting-point 10 feet. Find its area in square feet. 
 
 Alts. 40 sq. feet. 
 
 A Circle. — This is a figure all points of whose boundary 
 are equally distant from a fixed point within it called the centre. 
 The boundary is called its circumference^ and any line from the 
 centre to the circumference is called a radius. Any line passing 
 through the centre and with its ends on the circumference 
 is called a diameter. 
 
4 Theoretical Naval Architecture. 
 
 The ratio between the circumference of a circle and its 
 diameter is called tr^ and it = 3'i4i6, or nearly -^ 
 
 Thus the length of a thin wire forming the circumference 
 of a circle of diameter 5 feet is given by — 
 
 IT X 5 = 5 X 3"i4i6 feet 
 = 157080 feet 
 or using tt = -^, the circumference = 5 x -,- 
 = iA^= i5f feet nearly 
 
 The circumference of a mast 2' 6" in diameter is given by — 
 
 2i X ^ feet = I X ^ 
 
 = ^ = jf feet nearly 
 
 The area of a circle of diameter d is given by — 
 '^^^ (d' = d^ d) 
 
 Thus a solid pillar 4 inches in diameter has a sectional 
 area of — 
 
 4 --7-X4 
 
 = I z-f square inches nearly 
 
 A hollow pillar 5 inches external diameter and \ inch thick 
 will have a sectional area obtained by subtracting the area of 
 a circle 4^ inches diameter from the area of a circle 5 inches 
 diameter 
 
 =r-f)-(^) 
 
 — 3' 7 3 square inches 
 The same result may be obtained by taking a mean 
 diameter of the ring, finding its circumference, and multiplying 
 by the breadth of the ring. 
 
 Mean diameter = 4f inches 
 Circumference = ■— x ^ inches 
 
 Area = (^ X ■^) X ^ square inches 
 = 373 square inches as before 
 
 ' This is the Greek letter /«', and is always used to denote 3'i4i6, or ^ 
 nearly ; that is, the ratio borne by the circumference of a circle to its 
 diameter. 
 
A)'eas, Vohnnes, Weights, Displacement, etc. 5 
 
 Trapezoidal Rule.— We have already seen (p. 2) that 
 the area of a trapezoid, as ABCD, Fig. 5, is given by 
 |(AD + BC)AB, or caUing AD, EC, and AB y^, y^, and A 
 respectively the area is given by — 
 
 If, now, we have two trapezoids joined together, as in 
 
 B. 
 
 Fig. s. 
 
 Fig. 6, having BE = AB, the area of the added part will be 
 given by — 
 
 My^+ysVi 
 
 The area of the whole figure is given by- — 
 
 iCjfi +y2)^ + Uj^ -Vy^h = \h{y-, 4- 2y^ +.!'„) 
 If we took a third trapezoid and joined on in a similar 
 manner, the area of the whole figure would be given by 
 
 ^h{y, + 2y^ + 273 + J'4) = h [J^^ + J2 + ^/s ) 
 
 Trapezoidal rule for finding the area of a curvilinear figure, 
 as ABCD, Fig. 7. 
 
 Divide the base AB into a convenient number of equal 
 parts, as AE, EG, etc., each of length equal to h, say. Set up 
 perpendiculars to the base, as EF, GH, etc. If we join DF, 
 FH, etc., by straight lines, shown dotted, the area required 
 will very nearly equal the sum of the areas of the trapezoids 
 ADFE, EFHG, etc. Or using the lengths y^, y^, etc., as 
 indicated in the figure — 
 
 (yi+ji 
 
 Area 
 
 = ,{• 
 
 +y2+ys+y4+yn+ya 
 
6 Theoretical Naval Architecture. 
 
 In the case of the area shown in Fig. 7, the area will be 
 somewhat greater than that given by this rule. If the curve, 
 however, bent towards the base line, the actual area would be 
 somewhat less than that given by this rule. In any case, the 
 closer the perpendiculars are taken together the less will be 
 the error involved by using this rule. Putting this rule into 
 words, we have — 
 
 To find the area of a curvilinear figure, as ABCD, Fig. 7, 
 by means of the trapezoidal rule, divide the base into any con- 
 venient number of equal parts, and erect perpendiculars to the base 
 meeting the curve ; then to the half -sum of the first and last of 
 these add the sum of all the intermediate ones ; the result multi- 
 plied by the common distance apart will give the area required. 
 
 The perpendiculars to the base AB, as AD, EF, are termed 
 " ordinates" and any measurement along the base from a given 
 starting-point is termed an '^abscissa." Thus the point P on 
 the curve has an ordinate OP and an abscissa AO when 
 referred to the point A as origin. 
 
 Simpson's First Rule.^ — This rule assumes that the 
 curved line DC, forming one boundary of the curvilinear area 
 ABCD, Fig. 8, is a portion of a curve known as a parabola 
 of the second order."^ In practice it is found that the results 
 given by its application to ordinary curves are very accurate, 
 
 ' It is usual to call these rules Simpson's rules, but the first rule 
 was given before Simpson's time by James Stirling, in his " Methodus 
 Differentialis," published in 1730. 
 
 "^ A "parabola of the second order" is one whose equation referred 
 to co-ordinate axes is of the formj = a„ + a,x + a.^pc'-, where «„, ij„ aj are 
 constants. 
 
Areas, Volumes, Weights, Displacement, etc. 7 
 
 and it is this rule that is most extensively used in finding the 
 areas of curvilinear figures required in ship calculations. 
 
 Let ABCD, Fig. 8, be a figure bounded on one side by 
 the curved line DC, which, as 
 stated above, is assumed to be 
 a parabola of the second order. 
 AB is the base, and AD and 
 BC are end ordinates perpen- 
 dicular to the base. 
 
 Bisect AB in E, and draw 
 EF perpendicular to AB, meet- 
 ing the curve in F. Then the 
 area is given by — 
 
 ^i^ 
 
 S 
 
 tJ! 
 
 h. 
 
 E. 
 
 Fig. 8. 
 
 iAE(AD + 4EF -t- BC) 
 
 or using y-^, y^, y.-, to represent the ordinates, h the common 
 interval between them — 
 
 Area = -(ji + 4j/2-t-j3) 
 
 Now, a long curvilinear area ^ may be divided up into a 
 number of portions similar to the above, to each of which the 
 above rule will apply. Thus the area of the portion GHNM 
 of the area Fig. 7 will be given by — 
 
 -U'3 + 4^4+^5) 
 
 and the portion MNCB will have an area given by— 
 A, 
 
 Therefore the total area will be, supposing all the ordinates 
 are a common distance h apart — 
 
 -{y\ + Ay 2 + 2j3 + 4ji'4 + ^y^ + Ay^ +7?) 
 
 Ordinates, as GH, MN, which divide the figure into the 
 elementary areas are termed " dividing ordinates." 
 
 Ordinates between these, as EF, KL, OP, are termed 
 
 " intermediate ordinates" 
 
 ' The curve is supposed continuous. If the curvature changes abruptly 
 at any point, this point must be at a dividing ordinate. 
 
Theoretical Naval Architecture. 
 
 Notice that the area must have an even number of intervals, 
 or, what is the same thing, an odd number of ordinates, for 
 Simpson's first rule to be applicable. 
 
 Therefore, putting Simpson's first rule into words, we 
 have — 
 
 Divide the base into a convenient even number of equal parts, 
 and erect ordinates meeting the curve. Then to the sum of the end 
 ordinates add four times the even ordinates and twice the odd 
 ordinates. The sum thus obtained, multiplied by one-thii'd the 
 common distance apart of the ordinates, will give the area. 
 
 Approximate Proof of Simpson's First Rule. — The 
 truth of Simpson's first rule may be vmderstood by the following 
 approximate proof : ' — 
 
 Let DFC, Fig. 9, be a curved line on the base AB, and 
 with end ordinates AD, EC perpendicular to AB. Divide AB 
 equally in E, and draw the ordinate EF perpendicular to AB. 
 Then with the ordinary notation — 
 
 h 
 Area = -(;'i + 4J2 +^3) 
 
 by Simpson's 
 
 first rule. 
 
 Now divide AB into three equal 
 parts by the points G and H. 
 Draw perpendiculars GJ and 
 HK to the base AB. At F 
 draw a tangent to the curve, 
 meeting GJ and HK in J and 
 K. Join DJ and KC. Now, 
 it is evident that the area we 
 want is very nearly equal to 
 the area ADJKCB. This 
 will be found by adding to- 
 gether the areas of the trape- 
 zoids ADJG, GJKH, HKCB. 
 
 Area of ADJG = i(AD -f GJ)AG 
 „ GJKH = i(GJ -f HK)GH 
 „ HKCB = i(HK -f BC)HB 
 
 ' Another proof will be found on p. 73. 
 be found in appendix A, p. 245. 
 
 The mathematical proof will 
 
Areas, Volumes, Weights, Displacement, etc. g 
 
 Now, AG = GH = HB = ^AB = |AE, therefore the total 
 
 area is- 
 
 i(^)(AD + 2GJ + 2HK + BC) 
 
 Now, AE = h, and GJ + HK = 2EF (this may be seen at 
 once by measuring with a strip of paper), therefore the total, 
 area is — 
 
 ^(AD + 4EF + BC) = -iy, + Ay, +y,) 
 
 which is the same as that given by Simpson's first rule. 
 
 Application of Simpson's First 'RvXs.— Example.— A curvi- 
 linear area has ordinates at a common distance apart' of 2 feet, the lengths 
 being 1-45, 2-65, 4-35, 6-45, 8-50, 10-40, and 11-85 feet respectively. 
 Find the area of the figure in square feet. 
 
 In finding the area of such a curvilinear figure by means of Simpson's 
 first rule, the work is arranged as follows : — 
 
 Number of 
 
 Length of 
 
 Simpson's 
 
 Functions of 
 
 ordinate. 
 
 ordinate. 
 
 multipliers.' 
 
 ordinates. 
 
 I 
 
 I -45 
 
 I 
 
 I "45 
 
 2 
 
 2-6s 
 
 4 
 
 10-60 
 
 3 
 
 4-35 
 
 2 
 
 8-70 
 
 4 
 
 6-4S 
 
 4 
 
 25-80 
 
 5 
 
 8-50 
 
 2 
 
 17-00 
 
 6 
 
 10-40 
 
 4 
 
 41-60 
 
 7 
 
 11-85 
 
 I 
 
 11-85 
 
 Common interval = 2 feet 
 \ common interval = f feet 
 
 area = 117 x J : 
 
 117-00 sum of functions 
 
 I = 78 square feet 
 
 The length of this curvilinear figure is 1 2 feet, and it has 
 
 been divided into an even number of intervals, viz. 6, 2 feet 
 
 apart, giving an odd number of ordinates, viz. 7. We are 
 
 consequently able to apply Simpson's first rule to finding its 
 
 area. Four columns are used. In the first column are placed 
 
 the numbers of the ordinates, starting from one end of the 
 
 figure. In the second column are placed, in the proper order, 
 
 the lengths of the ordinates corresponding to the numbers in 
 
 the first column. These lengths are expressed in feet and 
 
 ' Sometimes the multipliers used are half these, viz. i, 2, i, 2, i, 2, J, 
 and the result at the end is multiplied by two-thirds the common interval. 
 
lO 
 
 Theoretical Naval Architecture. 
 
 decimals of a foot, and are best measured off with a decimal 
 scale. If a scale showing feet and inches is used, then the 
 inches should be converted into decimals of a foot ; thus, 
 6' 9" = 675', and 6' 3^" = 6-3'. In the next column are placed 
 Simpson's multipliers in their proper order and opposite their 
 corresponding ordinates. The order may be remembered by 
 combining together the multipliers for the elementary area first 
 considered — 
 
 I 4 I 
 
 I 4 I 
 
 ori42424i 
 The last column contains the product of the length of the 
 ordinate and its multipUer given in the third column. These 
 are termed the '■'■functions of ordinates." The sum of the 
 figures in the last column is termed the " sum of functions of 
 ordinates." This has to be multiplied by one-third the common 
 interval, or in this case f. The area then is given by — 
 117 X f = 78 square feet 
 Simpson's Second Rule. — This rule assumes that the 
 curved line DC, forming one boundary of the curvilinear area 
 
 H. 
 
 
 ^ 
 
 C 
 
 ''^"""^ 
 
 
 
 
 0. 
 
 X 
 
 ^ 
 
 
 ^ 
 
 
 f^ 
 
 fj. 
 
 7v. 
 
 
 7v. 
 
 
 7^ 
 
 
 A. E. F B 
 
 Fig. 10. 
 
 ABCD, Fig. 10, is a portion of a curve known as " a parabola 
 
 of tJie third order" ^ 
 
 Let ABCD, Fig. 10, be a figure bounded on one side by 
 
 the curved line DC, which, as stated above, is assumed to be 
 
 ' A "parabola of the third order" is one whose equation referred to 
 co-ordinate axes is of the form y ■=■ a^-\- a\X -t- a^x' -f- ^3^;', where a,,, u^, 
 !((;, «3 are constants. 
 
Areas, Volumes, Weights, Displacement, etc. ii 
 
 " a parabola of the third order" AB is the base, and AD and 
 BC are end ordinates perpendicular to the base. Divide the 
 base AB into three equal parts by points E and F, and draw 
 EG, FH perpendicular to AB, meeting the curve in G and H 
 respectively. Then the area is given by — 
 
 |AE(AD + 3EG + 3FH + BC) 
 or, using 71, y^, y^, y^ to represent the ordinates, and h the 
 common interval between them — 
 
 Area = \h{yy + zy^ + 3^3 + ^4) 
 Now, a long curvilinear area^ may be divided into a 
 number of portions similar to the above, to each of which the 
 above rule will apply. Thus the area of the portion KLCB in 
 Fig. 7 will be given by — 
 
 Consequently the total area of ABCD, Fig. 7, will be, 
 supposing all the ordinates are a common distance h apart — 
 \Kyx + 3J'2 + Zy^ + 2^4 + 3^5 + 3_y8 + y,) 
 
 The ordinate KL is termed a " dividing oi'dinafe^' and the 
 others, EF, GH, MN, OP, are termed " intermediate ordinates'.' 
 This rule may be approximately proved by a process similar to 
 that adopted on p. 8 for the first rule. 
 
 Application of Simpson's Second Riile. — Example. — A cur- 
 vilinear area has ordinates at a common distance apart of 2 feet, the 
 lengths being i'4S, 2'65, 4'3S, 6'45, 8-50, I0'40, and ii'85 feet respectively. 
 Find the area of the figure in square feet by the use of Simpson's second rule. 
 
 In finding the area of such a curvilinear figure by means of Simpson's 
 second rule, the work is arranged as follows : — 
 
 Number of 
 
 Length of 
 
 Simpson's 
 
 Functions of 
 
 ordinate. 
 
 ordinate. 
 
 multipliers. 
 
 ordinates. 
 
 I 
 
 I-4S 
 
 I 
 
 I -45 
 
 2 
 
 2-65 
 
 3 
 
 7-95 
 
 3 
 
 4-3S 
 
 3 
 
 13-05 
 
 4 
 
 6 '45 
 
 2 
 
 12-90 
 
 S 
 
 8-50 
 
 3 
 
 25-50 
 
 6 
 
 io"40 
 
 3 
 
 31-20 
 
 7 
 
 11-85, 
 
 I 
 
 n-85 
 
 103-90 sum of functions 
 Common interval = 2 feet 
 I common interval = | z= f 
 
 103-9 X I = 77-925 square feet 
 
 ' See footnote on p. 7. 
 
12 Theoretical Naval Ai'chitecture. 
 
 This curvilinear area is the same as already taken for an 
 example of the application of Simpson's first rule. It will be 
 noticed that the number of intervals is 6 or a tnuUiple of 3. 
 We are consequently able to apply Simpson's second rule to 
 finding the area. The columns are arranged as in the previous 
 case, the multipliers used being those for the second rule. 
 The order may be remembered by combining together the 
 multipliers for the elementary area with three intervals first 
 considered — 
 
 I 3 3 I 
 
 I 3 3 I 
 
 or I 3 3 2 3 3 I 
 For nine intervals the multipliers would be i, 3, 3, 2, 3, 3, 
 
 2, 3. 3> I- 
 
 The sum of the functions of ordinates has in this case to be 
 multiplied by f the common interval, or f x 2 = f, and con- 
 sequently the area is — 
 
 io3'9 X f = 77'925 square feet 
 
 It will be noticed how nearly the areas as obtained by the 
 two rules agree. In practice the first rule is used in nearly all 
 cases, because it is much simpler than the second rule and 
 quite as accurate. It sometimes happens, however, that we 
 only have four ordinates to deal with, and in this case Simp- 
 son's second rule must be used. 
 
 To find the Area of a Portion of a Curvilinear Area 
 contained between Two Consecutive Ordinates. — Such 
 a portion is AEFD, Fig. 8. In order to obtain this area, we 
 require the three ordinates to the curve ^ijVajj. The curve 
 DFC is assumed to be, as in Simpson's first rule, a parabola 
 of the second order. Using the ordinary notation, we 
 have — 
 
 Area of ADFE = 1^^(571 -f Sv^ -;',) 
 or, putting this into words^ — 
 
 To eight times the middle ordinate add five times tfie near end 
 ordinate and subtract tlie far end ordinate ; multiply the remainder 
 by Yi the common intei~val: the result will be the area. 
 
 Thus, if the ordinates of the curve in Fig. 8 be 8-5, 10-4, 
 
Areas, Volumes, Weights, Displacement, etc. 13 
 
 ii*85 feet, and 2 feet apart, the area of AEFD will be given 
 by- 
 
 iV X 2(5 X 8's + 8 X 10-4 - 11-85) = 18-97 square feet 
 Similarly the area of EBCF will be given by — 
 
 B. 
 
 •Jj X 2(5 X 11-85 + 8 X 10-4 - 8-5) = 22-32 square feet 
 
 giving a total area of the whole figure as 41-29 square feet. 
 
 Obtaining this area by means of Simpson's first rule, we 
 should obtain 41-3 square feet. 
 
 This rule is sometimes known as the "five-eight " rule. 
 
 Subdivided Intervals. — When the curvature of a line 
 forming a boundary of an area, as Fig. 11, is very sharp, it is 
 found that the distance apart of ordinates, as used for the 
 straighter part of the curve, does not give a sufficiently accurate 
 result. In such a case, ordinates 
 are drawn at a sub-multiple of 
 the ordinary distance apart of 
 the main ordinates. 
 
 Take ABC, a quadrant of a 
 circle (Fig. 11), and draw the 
 three ordinates 72, jCs, jCj a dis- 
 tance h apart. Then we should 
 get the area approximately by 
 putting the ordinates through 
 Simpson's first rule. Now, the 
 curve EFC is very sharp, and 
 the result obtained is very far 
 from being an accurate one, 
 ordinates y', y" 
 given by — 
 
 S 
 
 S 
 
 f*> 
 
 c. 
 
 Fig. 
 
 Now put in the intermediate 
 Then the area of the portion DEC will be 
 
 {\){y. + Ay' + ^y* + Af+y^) 
 
 or we may write this— (js — o at end) 
 
 ^h{\y,+ 2y+y,+ 2y" + h.) 
 The area of the portion ABED is given by— 
 ^h{y, + Ayi-\-y-i) 
 
14 
 
 Theoretical Naval Architecture. 
 
 or the area of the whole figure — 
 
 \Kyx + ^y^ + ^\y, + ^y' + J'4 + 2/ + \y^ 
 
 Thus the multipliers for ordinates one-half the ordinary distance 
 apart are \, 2, \, and for ordinates one-quarter the ordinary 
 distance apart are \, i, i i, \. Thus we diminish the 
 multiplier of each ordinate of a set of subdivided intervals m 
 the same proportion as the intervals are subdivided. Each 
 ordinate is then multiplied by its proper multiplier found in 
 this way, and the sum of the products multiplied by | or f the 
 whole interval according as the first or second rule is used. 
 An exercise on the use and necessity for subdivided intervals 
 will be found on p. 41. 
 
 Algebraic Expression for the Area of a Figure 
 bounded by a Plane Curve.— It is often convenient to be 
 able to express in a short form the area of a plane curvilinear 
 figure. 
 
 In Fig. 12, let ABCD be a strip cut off by the ordinates 
 
 AB, CD, a distance t^x apart, 
 
 l^x being supposed small. 
 
 Then the area of this strip is 
 
 very nearly— 
 
 y X ^x 
 where y is the length of the 
 ordinate AB. If now we 
 imagine the strip to become 
 indefinitely narrow, the small 
 triangularpiece BDE will dis- 
 appear, and calling dx the 
 
 Fig. 
 
 breadth of the strip, its area will be — 
 
 y y. dx 
 
 The area of the whole curvilinear figure would be found if 
 we added together the areas of all such strips, and this could 
 be written — 
 
 iy .dx 
 where the symbol / may be regarded as indicating the sum 
 of all such strips as y . dx. We have already found that 
 
Areas, Volumes, Weights, Displacement, etc. 15 
 
 Simpson's rules enable us to find the areas of such figures, 
 so we may ],ook upon the expression for the area — 
 
 ^y .dx 
 
 as meaning that, to find the area of a figure, we take the 
 length of the ordinate y at convenient intervals, and put them 
 through Simpson's multipliers. The result, multiplied by \ or 
 f the common interval, as the case may be, will give the area. 
 A proper understanding of the above will be found of great 
 service in dealing with moments in the next chapter. 
 
 To find the Area of a Figure bounded by a Plane 
 Curve and Two Badii. — Let OAB, Fig. 13, be such a figure, 
 OA, OB being the 
 bounding radii. 
 
 Take two points 
 very close together on 
 the curve PP'; join OP, 
 OP', and let OP = r 
 and the small angle 
 POP' = A^ in circular 
 measure.^ Then OP 
 = OP' = r very nearly, 
 and the area of the 
 elementary portion 
 
 being the length of PP', 
 and regarding OPP' as 
 a triangle. If now we consider OP, OP' to become in- 
 definitely close together, and consequently the angle POP' 
 indefinitely small = dQ say, any error in regarding POP' as a 
 triangle will disappear, and we shall have — 
 
 Area POP' =^.de 
 2 
 
 and the whole area AOB is the sum of all such areas which 
 
 can be drawn between OA and OB, or — 
 
 Fig. 13. 
 
 / 
 
 '-.dd 
 
 ' See p. 86. 
 
i6 
 
 Theoretical Naval Architecture. 
 
 Now, this exactly corresponds to the algebraic expression 
 for the area of an ordinary plane curvilinear area, viz. — 
 
 y . dx (see p. 15) 
 
 y corresponding to and dx corresponding to 
 
 Therefore 
 
 divide the angle between the bounding radii into an even 
 number of equal angular intervals by means of radii. Measure 
 these radii, and treat their half-squares as ordinates of a curve 
 by Simpson's first rule, multiplying the addition by \ the 
 common angular interval in circular measure. Simpson's second 
 rule may be used in a similar manner. 
 
 The circular measure of an angW^ is the number of degrees 
 
 it contains multiplied by — , or o'oi745. Thus the circular 
 
 180 
 
 measure of — 
 
 90 
 
 3-^4x6 ^ ^.^^^s 
 
 2 2 
 
 and the circular measure of 15° is 0-26175. 
 
 Example. — To find the area of a figure bounded by a plane curve and 
 two radii 90° apart, the lengths of radii 15° apart being o, 2"6, 5-2, 7-8, lO'S, 
 13-1. 157- 
 
 Angle from 
 
 Length of 
 
 Square of 
 
 Simpson's 
 
 Functions of 
 
 first radius. 
 
 radius. 
 
 length. 
 
 multipUers. 
 I 
 
 squares. 
 O'O 
 
 0° 
 
 O'O 
 
 0-0 
 
 K 
 
 2-6 
 
 6-8 
 
 4 
 
 27-2 
 
 30° 
 
 S"2 
 
 27-0 
 
 2 
 
 54-0 
 
 is: 
 
 7-8 
 
 6o-8 
 
 4 
 
 243-2 
 
 60° 
 
 10-5 
 
 II0'2 
 
 2 
 
 220-4 
 
 75° 
 
 13-1 
 
 171-6 
 
 4 
 
 686-4 
 
 90° 
 
 157 
 
 246-S 
 
 I 
 
 246-5 
 
 1477-7 sum of 
 functions 
 Circular measure of 15° = 0-26175 
 
 .'. area = 1477-7 X i X 0-26175 X \ 
 = 64 j square feet nearly 
 The process is exactly the same as in Simpson's rule for a plane area 
 with equidistant ordinates. To save labour, the squares of the radii are put 
 through the proper multipliers, the multiplication by J being performed at 
 the end. 
 
 See also p. 86. 
 
Areas, Volumes, Weights, Displacement, etc. 17 
 
 Measurement of Volumes. 
 
 The Capacity or Volume of a Rectangular Block 
 
 is the product of the length, breadth, and depth, or, in other 
 words, the area of one face multiplied by the thickness. All 
 these dimensions must be expressed in the same units. Thus 
 the volume of an armour plate 12 feet long, 8^ feet wide, and 
 18 inches thick, is given by — 
 
 12 X 8i X i| = 12 X \^ X I = ^^ = i48i cubic feet 
 
 The Volume of a Solid of Constant Section is the 
 area of its section multiplied by its length. Thus, a pipe 2 feet 
 
 in diameter and 100 feet long has a section of — = -^ square 
 
 feet, and a volume of -7- X 100 = -^^ = 314! cubic feet. 
 
 A hollow pillar / 6" long, 5 inches external diameter, and 
 \ inch thick, has a sectional area of — 
 
 373 square inches 
 
 or 3_Z3 square feet 
 144 
 
 and the volume of material of which it is composed is — 
 
 ^373) 
 V 144 / 
 
 ^ 15 ^ i8;65 
 
 44/2 96 
 
 = o"i95 cubic foot 
 
 Volume of a Sphere. — This is given by ^ . d^, where d 
 
 6 
 
 is the diameter. Thus the volume of a ball 3 inches in dia- 
 meter is given by — 
 
 77 22 X 27 
 
 -7. 27 = 
 
 6 ' 42 
 
 = i4y cubic inches 
 
 Volume of a Pyramid. — This is a solid having a base 
 in the shape of a polygon, and a point called its vertex not in 
 the same plane as the base. The vertex is joined by straight 
 lines to all points on the boundary of the base. Its volume is 
 given by the product of the area of the base and one-third the 
 
 C 
 
Theoretical Naval Architecture. 
 
 perpendicular distance of the vertex from the base, A cone is 
 a particular case of the pyramid having for its base a figure 
 with a continuous curve, and a right circular cone is a cone 
 having for its base a circle and its vertex immediately over the 
 centre of the base. 
 
 To find the Volume of a Solid bounded by a 
 Curved Surface. — ^The volumes of such bodies as this are 
 continually required in ship calculation work, the most 
 important case being the volume of the under-water portion 
 of a vessel. In this case, the volume is bounded on one side 
 by a plane surface, the water-plane of the vessel. Volumes 
 of compartments are frequently required, such as those for con- 
 taining fresh water or coal-bunkers. The body is divided by 
 a series of planes spaced equally apart. The area of each 
 section is obtained by means of one of the rules already 
 explained. These areas are treated as the ordinates of a new 
 curve, which may be run in, with ordinates the spacing of the 
 planes apart. It is often desirable to draw this curve with 
 areas as ordinates as in Fig. 14, because, if the surface is a fair 
 
 CURVE 
 
 -IT^FI- 
 
 Fic. 14. 
 
 surface, the curve of areas should be a fair curve, and should 
 run evenly through all the spots ; any inaccuracy may then be 
 detected. The area of the curve of areas is then obtained by 
 one of Simpson's rules as convenient, and this area will re- 
 present the cubical contents of the body. 
 
 Example. — A coal-bunker has sections 17' 6" apart, and the areas of 
 these sections are 98, 123, 137, 135, 122 square feet respectively. Find the 
 
Areas, Volumes, Weights, Displacement, etc. 19 
 
 volume of the bunker and the number of tons of coal it will hold, taking 
 44 cubic feet of coal to weigh \ ton. 
 
 
 Simpson's 
 
 Functions of 
 
 
 multipliers. 
 
 areas. 
 
 98 
 
 I 
 
 98 
 
 123 
 
 4 
 
 492 
 
 137 
 
 2 
 
 274 
 
 135 
 
 4 
 
 540 
 
 122 
 
 I 
 
 122 
 
 1526 sum of functions 
 
 \ common interval = 3 X 17.2 = ^^ 
 
 .". volume = 1526 X -^ cubic feet 
 = 8902 cubic feet, 
 and the bunker will hold Sf^ = 202 tons 
 
 The under-water portion of a ship is symmetrical about the 
 fore-and-aft middle line plane, so that only one-half need be 
 calculated. We may divide the volume in two ways — 
 
 1. By equidistant planes parallel to the load water-plane. 
 
 2. By equidistant planes perpendicular to the middle-line 
 plane and to the load water-plane. 
 
 The volume as obtained by both methods should be the 
 same, and they are used to check each other. 
 
 Examples. — i. The under-water portion of a vessel is divided by vertical 
 sections 10 feet apart of the following areas : o'3, 227, 48'8, 73'2, 88-4, 
 82'8, 587, 26-2, 3'9 square feet. Find the volume in cubic feet. (The 
 curve of sectional areas is given in Fig. 15.) 
 
 CuRyE OF Sectiqn/m. Areas. 
 
 The number of ordinates being odd, Simpson's first rule can be applied 
 as indicated in the following calculation : — 
 
20 
 
 Theoretical Naval Architecture. 
 
 Number of 
 
 Area of 
 
 Simpson's 
 
 Function of 
 
 section. 
 
 section. 
 
 multipliers. 
 
 area. 
 
 j 
 
 I 
 
 o'3 
 
 I 
 
 0-3 
 
 2 
 
 227 
 
 4 
 
 90-8 
 
 3 
 
 48'8 
 
 2 
 
 97-6 
 
 4 
 
 73 '2 
 
 4 
 
 292-8 
 
 
 88-4 
 
 2 
 
 176-8 
 
 6 
 
 82-8 
 
 4 
 
 33 1 "2 
 
 7 
 
 587 
 
 2 
 
 117-4 
 
 8 
 
 26-2 
 
 4 
 
 io4"8 
 
 9 
 
 3-9 
 
 1 
 
 3'9 
 
 1215 -6 sum of functions 
 
 \ common interval = ^ 
 
 .". volume = 1215-6 X -'3°- 
 
 = 4052 cubic feet 
 
 2. The under- water portion of the above vessel is divided by planes 
 parallel to the load water-plane and I J feet apart of the following areas : 
 944, 795, 605, 396, 231, 120, 68, 25, 8 square feet. Find the volume in 
 cubic feet. 
 
 The number of areas being odd, Simpson's first rule can be applied, as 
 indicated in the following calculation : — 
 
 Number of 
 
 Area of 
 
 Simpson's 
 
 Function of 
 
 water-line. 
 
 water-plane. 
 
 multipliers. 
 
 area. 
 
 I 
 
 944 
 
 I 
 
 944 
 
 2 
 
 795 
 
 4 
 
 3180 
 
 3 
 
 605 
 
 2 
 
 1210 
 
 4 
 
 396 
 
 4 
 
 1584 
 
 5 
 
 231 
 
 2 
 
 462 
 
 6 
 
 120 
 
 4 
 
 480 
 
 7 
 
 68 
 
 2 
 
 136 
 
 8 
 
 25 
 
 4 
 
 100 
 
 9 
 
 8 
 
 I 
 
 8 
 
 8104 sum of functions 
 
 \ common interval = J x | 
 
 .'. volume = 8104 X J 
 
 = 4052 cubic feet 
 
 which is the same result as was obtained above by taking the areas of 
 vertical sections and putting them through Simpson's rule. 
 
 In practice this volume is found by means of a " displace- 
 ment sheet," or by the " planimeter." An explanation of 
 these will be found in Chapter II. 
 
Areas, Volumes, Weights, Displacement, etc. 21 
 
 Displacement. — The amount of water displaced or put 
 aside by a vessel afloat is termed her " displacement:' This 
 may be reckoned as a volume, when it is expressed in cubic 
 feet, or as a weight, when it is expressed in tons. It is usual 
 to take salt water to weigh 64 lbs. per cubic foot, and conse- 
 quently -^f^ = 35 cubic feet of salt water will weigh one 
 ton. Fresh water, on the other hand, is regarded as weighing 
 ea^lbs. per cubic foot, or 36 cubic feet to the ton. The 
 volume displacement is therefore 35 or 36 times the weight dis- 
 placement, according as we are dealing with salt or fresh water. 
 
 If a vessel is floating in equilibritim in still water, the weight 
 of water s/ie displaces must exactly equal the weight of t/ie vessel 
 herself with every thing she has on board. 
 
 That this must be true may be understood from the follow- 
 ing illustrations — 
 
 I. Take a large basin and stand it in a dish (see Fig. 16). 
 
 ^ZZX 
 
 n r 
 
 -r-— r 
 
 I ; 
 
 V ^-- -" ~7 
 
 Fig. 16. 
 
 Just fill the basin to the brim with water. Now carefully place 
 a smaller basin into the water. It will be found that some of 
 the water in the large basin will be displaced, and water will 
 spill over the edge of the large basin into the dish below. 
 It is evident that the water displaced by the basin is equal 
 in amount to the water that has been caught by the dish, and 
 if this water be weighed it will be found, if the experiment be 
 conducted accurately, that the small basin is equal in weight to 
 the water in the dish, that is, to the water it has displaced. 
 
 2. Consider a vessel floating in equilibrium in still water, and 
 imagine, if it were possible, that the water is solidified, main- 
 taining the same level, and therefore the same density. If now 
 we lift the vessel out, we shall have a cavity left behind which 
 
22 
 
 Theoretical Naval Architecture. 
 
 will be exactly of the form of the under-water portion of the 
 ship, as Fig. 17. Now let the cavity be filled up with water. 
 The amount of water we pour in will evidently be equal to the 
 volume of displacement of the vessel. Now suppose that the 
 solidified water outside again becomes water. The water we 
 have poured in will remain where it is, and will be supported 
 by the water surrounding it. The support given, first to the 
 vessel and now to the water we have poured in, by the sur- 
 
 Water 
 
 Surface. 
 
 Fig. 17. 
 
 rounding water must be the same, since the condition of the 
 outside water is the same. Consequently, it follows that the 
 weight of the vessel must equal the weight of water poured 
 in to fill the cavity, or, in other words, the weight of the 
 vessel is equal to the weight of water displaced. 
 
 If the vessel whose displacement has been calculated on P 
 is floating at her L.W.P. in salt water, her total weight will be — 
 
 4052 -^ 35 = "S'S tons 
 If she floated at the same L.W.P. in fresh water, her total weight 
 would be — 
 
 4052 -7-36 = ii2-| tons 
 
 It will be at once seen that this property of floating bodies 
 is of very great assistance to us in dealing with ships. For, to 
 find the weight of a ship floating at a given line, we do not 
 need to estimate the weight of the ship, but we calculate out 
 from the drawings the displacement in tons up to the given line, 
 and this must equal the total weight of the ship. 
 
 Curve of Displacement. — The calculation given on p. 20 
 gives the displacement of the vessel up to the load-water plane, 
 but the draught of a ship continually varies owing to different 
 weights of cargo, coal, stores, etc., on board, and it is desirable 
 
Areas, Volumes, Weights, Displacement, etc. 23 
 
 to have a means of determining quickly the displacement at 
 any given draught. From the rules we have already investi- 
 gated, the displacement in tons can be calculated up to each 
 water-plane in succession. If we set down a scale of mean 
 draughts, and set off perpendiculars to this scale at the places 
 where each water-plane comes, and on these set off on a con- 
 venient scale the displacement we have found up to that water- 
 plane, then we should have a number of spots through which we 
 shall be able to pass a fair curve if the calculations are correct. 
 
 — ^ Scale for Displacement. — r— 
 
 FIG. 18. 
 
 A curve obtained in this way is termed a " curve of displacement" 
 and at any given mean draught we can measure the displace- 
 ment of the vessel at that draught, and consequently know at 
 once the total weight of the vessel with everything she has on 
 board. This will not be quite accurate if the vessel is floating 
 at a water-plane not parallel to the designed load water-plane. 
 Fig. 18 gives a "curve of displacement" for a vessel, and the 
 following calculation shows in detail the method of obtaining 
 the information necessary to construct it. 
 
24 
 
 Theoretical Naval Architecture. 
 
 The areas of a vessel's water-planes, two feet apart, are as 
 follows : — 
 
 L.W.L. ... 
 
 . . . 7800 square feet. 
 
 2W.L. ... 
 
 - 7450 
 
 3W.L. ... 
 
 ... 6960 
 
 4W.L. ... 
 
 ... 6290 
 
 SW.L. ... 
 
 ... 5460 
 
 6 W.L. ... 
 
 ... 4320 
 
 7W.L. ... 
 
 ... 2610 „ 
 
 iraught to the 
 
 L.W.L. is 1 4' 0", and the displace 
 
 ment below the lowest W.L. is 7 1 tons. 
 To find the displacement to the L.W.L. 
 
 Number of 
 
 Area of 
 
 Simpson's 
 
 Function of 
 
 W.L. 
 
 water-plane. 
 
 multipliers. 
 
 area. 
 
 1 
 
 7800 
 
 I 
 
 7,800 
 
 2 
 
 7450 
 
 4 
 
 29,800 
 
 3 
 
 6960 
 
 2 
 
 13,920 
 
 4 
 
 6290 
 
 4 
 
 25,160 
 
 5 
 
 5460 
 
 2 
 
 10,920 
 
 6 
 
 4320 
 
 4 
 
 17,280 
 
 7 
 
 2610 
 
 I 
 
 2,610 
 
 107,490 
 
 \ common interval = \y~ 
 
 :. displacement in cubic feet = 107,490 X f 
 and displacement in tons, salt water = 107,490 X -ToT 
 
 = 2047 tons without the 
 appendage 
 
 Next we require the displacement up to No. 2 W.L., and 
 we subtract from the total the displacement of the layer between 
 I and 2, which is found by using ^& five-eight rule as follows : — 
 
 Number of 
 W.L. 
 
 Area of 
 water-plane. 
 
 Simpson's 
 multipliers. 
 
 Function of 
 area. 
 
 I 
 
 2 
 
 3 
 
 7800 
 
 7450 
 6960 
 
 5 
 
 8 
 — I 
 
 39,000 
 
 59,600 
 
 -6,960 
 
 91,640 
 
1 
 
 3S 
 
 Areas, Volumes, Weights, Displacement, etc. 
 
 Displacement in tons between ) _ ^ „ 
 
 No. I and No. 2 W.L.'s ( ~ ^''4° ^ tV X 
 
 = 436 tons nearly 
 .". the displacement up to No. 2 ) 
 
 W.L. is 2047 - 436 j = '^" *°"^ y^l\!cLO^xt the 
 
 appendage 
 
 The displacement between i and 3 W.L.'s can be found by 
 putting the areas of i, 2 and 3 W.L.'s through Simpson's first 
 rule, the result being 848 tons nearly. 
 
 .". the displacement up to No. 3 ) 
 
 W.L. is 2047 - 848 f = "99 tons without the 
 
 appendage 
 
 The displacement up to No. 4 W.L. can be obtained by 
 putting the areas of 4, 5, 6, and 7 W.L.'s through Simpson's 
 second rule, the result being — 
 
 819 tons without the appendage 
 
 The displacement up to No. 5 W.L. can be obtained by 
 putting the areas of 5, 6, and 7 W.L.'s through Simpson's first 
 rule, the result being — 
 
 482 tons without the appendage 
 
 The displacement up to No. 6 W.L. can be obtained by 
 means of the five-eight rale, the result being — 
 
 201 tons without the appendage 
 
 Collecting the above results together, and adding in the 
 appendage below No. 7 W.L., we have — 
 
 jlacement up 
 
 to L. W.L. ... 
 
 ... 21 18 tons 
 
 
 2 W.L. ... 
 
 ... 1682 „ 
 
 
 3 W.L. ... 
 
 ... 1270 ,, 
 
 
 4 W.L. ... 
 
 ... 890 „ 
 
 
 S W.L. ... 
 
 •■• SS3 ., 
 
 
 6 W.L. ... 
 
 ... 272 ,, 
 
 
 7 W.L. ... 
 
 71 „ 
 
 These displacements, set out at the corresponding draughts, 
 are shown in Fig. 18, and the fair curve drawn through forms 
 the "curve of displacement" of the vessel. It is usual to com- 
 plete the curve as indicated right down to the keel, although 
 
26 
 
 Theoretical Naval Architecture. 
 
 the ship could never float at a less draught than that given by 
 the weight of her structure alone, or when she was launched. 
 
 Tons per Inch Immersion. — It is frequently necessary 
 to know how much a vessel will sink, when floating at a given 
 water-line, if certain known weights are placed on board, or 
 how much she will rise if certain known weights are removed. 
 Since the total displacement of the vessel must equal the weight 
 of the vessel herself, the extra displacement caused by putting 
 a weight on board must equal this weight. If A is the area 
 
 ■ Tons Per Inch Immersion 
 
 Fig. 19, 
 
 of a given water-plane in square feet, then the displacement 
 of a layer i foot thick at this water-plane, supposing the vessel 
 parallel-sided in its neighbourhood, is — 
 
 A cubic feet 
 
 A 
 
 or — tons m salt water 
 35 
 
 For a layer i inch thick only, the displacement is— 
 
 A 
 
 35 X. 12 
 
 tons 
 
Areas, Vohimes, Weights, Displacement, etc. 27 
 
 and this must be the number of tons we must place on board 
 in order to sink the vessel i inch, or the number of tons we 
 must take out in order to lighten the vessel i inch. This is 
 termed the " tons per inch immersion " at the given water-line. 
 This assumes that the vessel is parallel-sided at the water-line 
 for the depth of i inch up and i inch down, which may, for all 
 practical purposes, be taken as the case. If, then, we obtain 
 the tons per inch immersion at successive water-planes parallel 
 to the load water-plane, we shall be able to construct a " ctirve 
 0/ tons per inch immersion" in the same way in which the curve 
 of displacement was constructed. Such a curve is shown in 
 Fig. 19, constructed for the same vessel for which the displace- 
 ment curve was calculated. By setting up any mean draught, 
 say 1 1 feet, we can measm-e off the " tons per inch immersion," 
 supposing the vessel is floating parallel to the load water-plane ; 
 in this case it is \']\ tons. Suppose this ship is floating at a 
 mean draught of 1 1 feet, and we wish to know how much she 
 will lighten by burning 100 tons of coal. We find, as above, 
 the tons per inch to be 17^, and the decrease in draught is 
 therefore — 
 
 100 -4- i7-|- = sf inches nearly 
 
 Curve of Areas of Midship Section. — This curve is 
 usually plotted off on the same drawing as the displacement 
 curve and the curve of tons per inch immersion. The ordi- 
 nates of the immersed part of the midship section being known, 
 we can calculate its area up to each of the water-planes in 
 exactly the same way as the displacement has been calculated. 
 These areas are set out on a convenient scale at the respective 
 mean draughts, and a line drawn through the points thus 
 obtained. If the calculations are correct, this should be a fair 
 curve, and is known as " the curve of areas of midship section." 
 By means of this curve we are able to determine the area of 
 the midship section up to any given mean draught. 
 
 Fig. 20 gives the curve of areas of midship section for the 
 vessel for which we have already determined the displacement 
 curve and the curve of tons per inch immersion. 
 
 Coefficient of Fineness of Midship Section.— If we 
 
28 Theoretical Naval Architecture. 
 
 draw a rectangle with depth equal to the draught of water at 
 the midship section to top of keel, and breadth equal to the 
 
 Aheas of Midi Sec: — Sq: Ft; < 
 
 extreme breadth at the midship section, we shall obtain what 
 may be termed the circumscribing rectangle of the immersed 
 midship section. The area of the immersed midship section 
 will be less than the area of this rectangle, and the ratio — 
 
 area of immersed midship se ction 
 area of its circumscribing rectangle 
 
 is termed the coefficient of fineness of midship section. 
 
 Example.— The midship section of a vessel is 68 feet broad at its 
 broadest part, and the draught of water is 26 feet. The area of the immersed 
 midship section is 1584 square feet. Find the coefficient of fineness of the 
 midship section. 
 
 Area of circumscribing rectangle = 68 x 26 
 
 = 1768 square feet 
 .". coefficient = \f^ = 0-895 
 
 If a vessel of similar form to the above has a breadth at 
 
Areas, Volumes, Weights, Displacement, etc. 29 
 
 the midship section of 59' 6" and a draught of 22' 9", the area 
 of its immersed midship section will be — 
 
 59^ X 22f X 0-895 — 1213 square feet 
 
 The value of the midship section coefficient varies in 
 ordinary ships from about 0*85 to 0-95, the latter value being 
 for a section with very flat bottom. 
 
 Coefficient of Fineness of Water-plane. — This is 
 the ratio between the area of the water-plane and its circum • 
 scribing rectangle. 
 
 The value of this coefficient for the load waiter-plane may 
 be taken as follows : — 
 
 For ships with fine ends 07 
 
 For ships of ordinary form 0-75 
 
 For ships with bluff ends 0'85 
 
 Block Coefficient of Fineness of Displacement. — 
 
 This is the ratio of the volume of displacement to the volume 
 of a block having the same length between perpendiculars, 
 extreme breadth, and mean draught as the vessel. The 
 draught should be taken from the top of keel. 
 
 Thus a vessel is 380 feet long, 75 feet broad, with 27' 6" 
 mean draught, and 14,150 tons displacement. What is its 
 block coefficient of fineness or displacement ? 
 
 Volume of displacement = 14,150 X 35 cubic feet 
 Volume of circumscribing solid = 380 X 75 X 27-^ cubic feet 
 .•. coefficient of fineness of I _ 14150 X 35 
 displacement 1 ~ 380 X 75 X 27I 
 
 = 0-63 
 
 This coefficient gives a very good indication of the fineness 
 of the underwater portion of a vessel, and can be calculated 
 and tabulated for vessels with known speeds. Then, if in the 
 early stages of a design we have the desired dimensions given, 
 with the speed required, we can select the coefficient of fineness 
 which appears most suitable for the vessel, and so determine 
 very quickly the displacement that can be obtained under the 
 conditions given. 
 
30 Theoretical Naval Architecture. 
 
 Example. — A vessel has to be 400 feet long, 42 feet beam, 17 feet draught, 
 and I3i knots speed. What would be the probable displacement 1 
 
 From available data, it would appear that a block coefficient of fineness 
 of 0'625 would be desirable. Consequently the displacement would be — 
 
 (400 X 42 X 17 X 0-625) -^ 35 tons = 5100 tons about 
 
 The following may be taken as average values of the block 
 coefficient of fineness of displacement in various types of 
 ships : — 
 
 Recent battleships ... ... ■6o-'6s 
 
 Recent fast cruisers 'So-'SS 
 
 Fast mail steamers ... ... 'SO-'JS 
 
 Ordinary steamships ... ... 'SS-'^S 
 
 Cargo steamers ... ... 'Ss-'So 
 
 Sailing vessels ... ■65-75 
 
 Steam-yachts ■35-'4S 
 
 Frismatic Coefficient of Fineness of Displace- 
 ment. — This coefficient is often used as a criterion of the 
 fineness of the underwater portion of a vessel. It is the ratio 
 between the volume of displacement and the vohime of a 
 prismatic solid the same length between perpendiculars as the 
 vessel, and having a constant cross-section equal in area to the 
 immersed midship section. 
 
 Example. — A vessel is 300 feet long, 2100 tons displacement, and has 
 the area of her immersed midship section 425 square feet. What is her 
 prismatic coefficient of fineness ? 
 
 Volume of displacement = 2100 x 35 cubic feet 
 Volume of prismatic solid = 300 x 425 ,, 
 
 .'. coefficient = — 
 
 300 X 425 
 
 = 0-577 
 
 Difference in Draught of Water when floating 
 in Sea Water and when floating in River Water. — 
 
 Sea water is denser than river water ; that is to say, a given 
 volume of sea water — say a cubic foot — weighs more than the 
 same volume of river water. In consequence of this, a vessel, 
 on passing from the river to the sea, if she maintains the same 
 weight, will rise in the water, and have a greater freeboard 
 than when she started. Sea water weighs 64 lbs. to the cubic 
 foot, and the water in a river such as the Thames may be 
 taken as weighing 63 lbs. to the cubic foot. In Fig. 21, let 
 the right-hand portion represent the ship floating in river water, 
 
Areas, Volumes, Weights, Displacement, etc. 31 
 
 and the left-hand portion represent the ship floating in salt 
 water. The difiference between the two water-planes will be 
 the amount the ship will rise on passing into sea water. 
 
 Let W = the weight of the ship in tons ; 
 
 T = the tons per inch immersion at the water-line 
 
 W'L' in salt water ; 
 X = the difference in draught between the water-lines 
 WL, W'L' in inches. 
 Then the volume of displacement — 
 
 m river water = 
 
 in sea water = 
 
 the volume of the layer = 
 
 W X 2240 
 W X 2240 
 W X 2240 W X 2240 
 
 63 
 
 W X 2240 
 63 X 64 
 
 64 
 
 Now, the volume of the layer also = * X T x ^- ; there- 
 
 fore we have- 
 
 i^ X A X 3 4." 
 
 W X 2240 
 
 e* 63 X 64 
 
 or if = -T— =i inches 
 63 1 
 
32 
 
 Theoretical Naval Architecture. 
 
 This may be put in another way. A ship, if floating in salt 
 water, will weigh ^ less than if floating to the same water-line 
 in river water. Thus, if W is the weight of the ship floating at a 
 given hne in salt water, her weight if floating at the same line 
 in fresh water is — 
 
 gVW less 
 
 and this must be the weight of the layer of displacement 
 between the salt-water line and the river-water line for a given 
 weight W of the ship. If T be the tons per inch for salt water, 
 the tons per inch for fresh water will be ffT. Therefore the 
 difference in draught will be — 
 
 W 
 
 : inches, as above 
 
 64 " • 84 ^ 6^T 
 
 Sinkage caused by a Central Compartment of a 
 Vessel being open to the Sea. — Take the simple case of a 
 box-shaped vessel, ABCD, Fig. 22, floating at the water-line WL. 
 
 w 
 
 M 
 
 
 N. 
 
 U 
 
 
 
 
 
 i^^i— 
 
 
 w 
 
 K 
 
 ^H 
 
 P 
 
 L 
 
 
 
 D. 
 
 1 
 
 r 
 
 H. 
 
 
 -*. 
 
 Fig. 
 
 This vessel has two water-tight athwartship bulkheads in the 
 middle portion, EF and GH. A hole is made in the bottom or 
 side below water somewhere between these bulkheads. We 
 will take a definite case, and work it out in detail to illustrate 
 the principles involved in such a problem. 
 
 Length of box-shaped vessel 
 
 ... 100 feet. 
 
 Breadth 
 
 ... 20 „ 
 
 Depth 
 
 20 „ 
 
 Draught ,, „ 
 
 10 ., 
 
 Distance of bulkheads apart 
 
 20 „ 
 
 If the vessel is assumed to be floating in salt water, 
 weight must be — 
 
 100 X 20 X 10 
 
 Its 
 
 35 
 
 = --3-5— tons 
 
Areas, Volumes, Weights, Displacement, etc. 33 
 
 Now, this weight remains the same after the bilging as 
 before, but the buoyancy has been diminished by the opening 
 of the compartment KPHF to the sea. This lost buoyancy 
 must be made up by the vessel sinking in the water until the 
 volume of displacement is the same as it originally was. 
 Suppose W'L' to be the new water-line, then the new volume of 
 displacement is given by the addition of the volumes of W'MFD 
 and NL'CH, or, calling d the new draught of water in feet — 
 
 (40 X 20 X (/)+ (40 X 20 x^) = i6ooi^ cubic feet 
 
 The original volume of displacement was — 
 
 100 X 20 X 10= 20,000 cubic feet 
 .". 1600 d = 20,000 
 
 .-. d=\^=i2' 6" 
 
 that is, the new draught of water is 12' 6", or the vessel will 
 sink a distance of 2' 6". 
 
 The problem may be looked at from another point of view. 
 The lost buoyancy is 20 X 20 x 10 cubic feet = 4000 cubic 
 feet; this has to be made up by the volumes W'MKW and 
 NL'LP, or the area of the intact water-plane multiplied by 
 the increase in draught. Calling x the increase in draught, we 
 shall have — 
 
 80 X 20 X .» = 4000 
 
 4000 I r gj 
 
 = 2' 6" 
 
 which is the same result as was obtained above. 
 
 If the bilged compartment contains stores, etc., the amount 
 of water which enters from the sea will be less than if the com- 
 partment were quite empty. The volume of the lost displace- 
 ment will then be given by the volume of the compartment up 
 to the original water-line less the volume occupied by the 
 stores. 
 
 Thus, suppose the compartment bilged in the above 
 example to contain coal, stowed so that 44 cubic feet of it will 
 weigh one ton, the weight of the solid coal being taken at 
 80 lbs. to the cubic foot. 
 
 D 
 
34 Theoretical Naval Architecture. 
 
 I cubic foot of coal, if solid, weighs 80 lbs. 
 
 I „ „ as stowed „ ^^ = 51 lbs. 
 
 Therefore in every cubic foot of the compartment there is — 
 
 fi cubic feet solid coal 
 
 If „ space into which water will find its way 
 
 The lost buoyancy is therefore — 
 
 If X 4000 =1450 cubic feet 
 
 The area of the intact water-plane will also be affected in 
 the same way ; the portion of the water-plane between the bulk- 
 heads will contribute — 
 
 fi X 20 X 20 = 255 square feet to the area 
 
 The area of the intact waterplane is therefore — 
 
 1600 -|- 25s = 1855 square feet 
 
 The sinkage in feet is therefore — 
 
 Hff =078, or 9-36 inches 
 
 In the case of a ship the same principles apply, supposing 
 the compartment to be a central one, and we have — 
 
 Sinkage of vessel 1 _ volume of lost buoyancy in cubic feet 
 in feet I area of intact water-plane in square feet 
 
 In the case of a compartment bilged which is not in the 
 middle of the length, change of the trim occurs. The method 
 of calculating this for any given case will be dealt with in 
 Chapter IV. 
 
 In the above example, if the transverse bulkheads EF and 
 GH had stopped just below the new water-line W'L', it is 
 evident that the water would flow over their tops, and the 
 vessel would sink. But if the tops were connected by a water- 
 tight flat, the water would then be confined to the space, and 
 the vessel would remain afloat. 
 
Areas, Volumes, Weights, Displacement, etc. 35 
 
 Velocity of Inflow of Water into a Vessel on 
 Bilging.— 
 
 Let A = area of the hole in square feet ; 
 
 d = the distance the centre of the hole below the 
 
 surface in feet ; 
 V = initial rate of inflow of the water in feet per 
 second. 
 
 Then v = S^/d nearly 
 and consequently the volume of water ) ^ _ 
 
 passing through the hole per second j = ^ ^^ x A cub. ft. 
 
 Thus, if a hole 2 square feet in area, 4 feet below the water- 
 line, were made in the side of a vessel, the amount of water, 
 approximately, that would flow into the vessel would be as 
 follows : — 
 
 Cubic feet per second = 8 x ^4 X 2 
 
 = 32 
 Cubic feet per minute = 32 x 60 
 
 Tons of water per minute = 
 
 35 
 = 54-85 
 
 Weights of Materials. — The following table gives 
 average weights which may be used in calculating the weights 
 of materials employed in shipbuilding : — 
 
 Steel 
 
 490 lbs 
 
 per cubic foot 
 
 Wrought iron 
 
 480 
 
 
 
 Cast iron 
 
 445 
 
 
 
 Copper 
 
 550 
 
 
 
 Brass 
 
 530 
 
 
 
 Zinc 
 
 44S 
 
 
 
 Gunmetal 
 
 .. 528 
 
 
 
 Lead 
 
 712 
 
 
 
 Elm (English) 
 
 35 
 
 
 
 „ (Canadian) 
 
 45 
 
 
 
 Fir (Dantzic) 
 
 36 
 
 
 
 Greenheart 
 
 72 
 
 
 
 Mahogany 
 
 .. 40-48 
 
 
 
 „ (for boats) 
 
 35 
 
 
 
 Oak (English) 
 
 52 
 
 
 
 „ (African) 
 
 62 
 
 
 
 Pine (Pitch) 
 
 40 
 
 
 
 „ (red) 
 
 36 
 
 
 
 ,, (yellow) 
 
 30 
 
 
 
 Teak 
 
 5° 
 
 
 
36 
 
 Theoretical Naval Architecture. 
 
 It follows, from the weights per cubic foot of iron and 
 steel given above, that an iron plate i inch thick weighs 40 lbs. 
 per square foot, and a steel plate i inch thick weighs 40-8 lbs. 
 per square foot. 
 
 " The weight per square foot may be obtained for other 
 thicknesses from these values, and we have the following : — 
 
 Thickness in 
 inches. 
 
 Weight per square foot in 
 pounds. 
 
 Iron. 
 
 Steel. 
 
 1 
 
 ! 
 
 I 
 
 10 
 
 15 
 20 
 
 25 
 30 
 
 35 
 40 
 
 10-2 
 
 15-3 
 20-4 
 
 25-5 
 30-6 
 
 357 
 40-8 
 
 It is convenient to have the weight of steel per square foot 
 when specified in one-twentieths of an inch, as is the case in 
 Lloyd's rules — 
 
 Thickness in 
 inches. 
 
 Weight per 
 square foot 
 in pounds. 
 
 Thickness in 
 inches. 
 
 Weight per 
 square foot 
 in pounds. 
 
 1 
 SO 
 
 2-04 
 
 4-08 
 
 6-12 
 
 8-16 
 
 10-20 
 
 12-24 
 
 14-28 
 
 16-32 
 
 18-36 
 
 20-40 
 
 i: 
 
 1=1 
 
 2:^4^ 
 26-52 
 28-56 
 30-60 
 32-64 
 34-68 
 36-72 
 38-76 
 40-80 
 
Areas, Volumes, Weights, Displacement, etc. 37 
 
 Examples to Chapter I. 
 
 What is its weight if its 
 Ans. 95 lbs. 
 
 Fig. 
 
 1. A plate has the form shown in Fig. 23 
 weight per square foot is 10 lbs. ? 
 
 2. The material of an 
 armour plate weighs 490 lbs. 
 a cubic foot. A certain 
 plate is ordered 400 lbs. per 
 square foot : what is its 
 thickness ? 
 
 Ans. 9'8 inches. 
 
 3. Steel armour plates, 
 as in the previous question, 
 are ordered 400 lbs. per 
 square foot instead of 10 
 inches thick. What is the 
 saving of weight per lOo 
 square feet of surface of this 
 armour ? 
 
 Ans. 833 lbs., or o"37 ton. 
 
 4. An iron plate is of the dimensions shown in Fig. 24. What is its area ? 
 If two lightening holes 2' 3" in diameter are cut in it, what will its 
 
 area then be ? . 
 
 Afis. 33I square feet ; 1 I 
 
 25 '8 square feet. 
 
 5. A hollow pillar is 4 inches 
 external diameter and | inch 
 thick. What is its sectional 
 area, and what would be the 
 weight in pounds of 10 feet of 
 this pillar if made of wrought 
 iron ? 
 
 Ans. 4-27 square inches ; 
 142 lbs. 
 
 6. A steel plate is of the 
 form and dimensions shown in Fig. 23. What is its weight ? (A steel plate 
 I inch thick weighs 25'5 lbs. per square foot.) 
 
 Afis. 1267 lbs. 
 
 k-- 
 
 
 Fig. 
 
 
 I 
 
 Jl:: 
 
 ?-ol. 
 
 ^ THICK. 
 
 14-6. 
 
 A 
 
 Fig. 23. 
 
 7. A wrought-iron armour plate is 15' 3" long, 3' 6" wide, and4i inches 
 thick. Calculate its weight in tons. 
 
 Ams. 4-29 tons. 
 
38 Theoretical Naval Architecture. 
 
 8. A solid pillar of iron of circular section is 6' lo" long and 2\ inches 
 in diameter. What is its weight ? 
 
 Ans. 90J lbs. 
 
 9. A Dantzic fir deck plank is 22 feet long and 4 inches thick, and 
 tapers in width from 9 inches at one end to 6 inches at the other. What is 
 its weight ? 
 
 Ans. 165 lbs. 
 
 10. A solid pillar of iron is 7' 3" long and 2i inches diameter. What 
 is its weight ? 
 
 Ans. 143 lbs. 
 
 11. The total area of the deck plan of a vessel is 4500 square feet. 
 What would be the surface of deck plank to be worked, if there are— 
 
 4 hatchways, each 4' X 2j' 
 
 2 „ ,, 10' X 6' 
 and two circular skylights, each 4 feet in diameter, over which no plank is 
 to be laid ? 
 
 Ans. 43r4'86 square feet. 
 
 12. A pipe is 6 inches diameter inside. How many cubic feet of water 
 will a length of 100 feet of this pipe contain ? 
 
 Ans. 19 '6 cubic feel. 
 
 13. A mast 90 feet in length and 3 feet external diameter, is composed of 
 20 lb. plating worked flush-jointed on three T-bars, each 5" X 3" X isJlbs. 
 per foot. Estimate the weight, omitting straps, and rivet heads. 
 
 Ans. 9 J tons nearly. 
 
 14. A curve has the following ordinates, I ' 4" apart : io'86, I3'S3, I4'S8, 
 15-05, I5'24, IS'28, I5'22 feet respectively. Draw this curve, and find 
 its area^ 
 
 (1) By Simpson's first rule ; 
 
 (2) By Simpson's second rule. 
 
 Ans. (i) ii6'o7 square feet; (2) Ii6'03 square feet. 
 
 15. The semi-ordinates in feet of a vessel's midship section, starting 
 from the load water-line, are 26'6, 26'8, 26'8, 26'4, 25'4, 23'4, and i8'5 feet 
 respectively, the ordinates being 3 feet apart. Below the lowest ordinate 
 there is an area for one side of the section of 24'6 square feet. Find the 
 area of the midship section, using — 
 
 ( 1 ) Simpson's first rule ; 
 
 (2) Simpson's second rule. 
 
 Ans. (i) 961 square feet ; (2) 9607 square feet. 
 
 16. The internal dimensions of a tank for holding fresh water are 
 8' o" X 3' 6" X 2' 6". How many tons of water will it contain ? 
 
 Ans. I '94. 
 
 17. The .^a^-ordinates of a deck plan in feet are respectively ij, 5§, 
 loi. 13!) I4i) I4i> 12J, 9, and 3J, and the length of the plan is 128 feet. 
 
 Find the area of the deck plan in square yards. 
 
 Ans. 296. 
 
 18. Referring to the previous question, find the area in square feet of the 
 portion of the plan between the ordinates i J and 5|. 
 
 Ans. 1067. 
 
 19. The half-ordinates of the midship section of a vessel are 22'3, 22'2, 
 217, 20'6, I7'2, I3'2, and 8 feet in length respectively. The common 
 interval between consecutive ordinates is 3 feet between the first and fifth 
 ordinates, and i' 6" between the fifth and seventh. Calculate the total area 
 of the section in square feet. 
 
 Ans. 585 '2 square feet. 
 
Areas, Volumes, Weights, Displacement, etc. 39 
 
 20. Obtain the total area included between the first and fourth ordinates 
 of the section given in the preceding question. 
 
 A71S. 392 '8 square feet. 
 
 21. The semi-ordinates of the load water-plane of a vessel are 0'2, 3'6, 
 7'4, 10, II, 107, 9'3, 65, and 2 feet respectively, and they are 15 feet 
 apart. What is the area of the load water-plane ? 
 
 Ans. 1808 square feet. 
 
 22. Referring to the previous question, what weight must be taken out 
 of the vessel to lighten her 3^ inches ? 
 
 What additional immersion would result by placing 5 tons on board ? 
 
 Ans. 15 tons ; i '16 inch. 
 
 23. The " tons per inch immersion " of a vessel when floating in salt 
 water at a certain water -plane is 44'S. What is the area of this plane ? 
 
 Ans. 18,690 square feet. 
 
 24. A curvilinear area has ordinates 3 feet apart of length 97, lO'O, and 
 1 3 '3 feet respectively. Find — 
 
 (i) The area between the first and second ordinates. 
 
 (2) The area between the second and third ordinates. 
 
 (3) Check the addition of these results by finding the area of the whole 
 
 figure by Simpson's first rule. 
 
 25. Assuming the truth of the five-eight rule for finding the area between 
 two consecutive ordinates of a curve, prove the truth of the rule known as 
 Simpson's first rule. 
 
 26. A curvilinear area has the following ordinates at equidistant intervals 
 of 18 feet : 6-20, 1380, 2r90, 26-40, 22-35, 14' 7°, and 7-35 feet. 
 Assuming that Simpson's first rule is correct, find the percentage of error 
 that would be involved by using — 
 
 (1) The trapezoidal rule ; 
 
 (2) Simpson's second rule. 
 
 Ans. (l) 1-2 per cent. ; (2) 0-4 per cent. 
 
 27. A compartment for containing fresh water has a mean section of 
 the form shown in Fig. 26. The length 
 of the compartment is 1 2 feet. How many 
 tons of water willit contain? 
 
 Ans. 17 tons. 
 
 28. A compartment 20 feet long, 20 
 feet broad, and SJ feet deep, has to be 
 lined with teak 3 inches in thickness. 
 Estimate the amount of teak required in 
 cubic feet, and in tons. 
 
 Ans. 365 cubic feet; 8-1 tons. 
 
 29. The areas of the water-line sec- 
 tions of a vessel in square feet are re- 
 spectively 2000, 2000, 1600, 1250, and 
 300. The common interval between them 
 is l§ foot. Find the displacement of the 
 vessel in tons in salt water, neglecting the 
 small portion below the lowest water-line 
 section. 
 
 Ans. 264I tons. 
 
 30. A series of areas, 17' 6" apart, contain 0-94, 2-08, 3-74, 5-33, 8-27, 
 12-14, 16-96, 21-82, 24-68, 24-66, 22-56, 17-90, 12-66, 8-40, 5-69, 3-73, 
 2-61, 2-06, o square feet respectively.. Find the volume of which the above 
 are the sectional areas. 
 
 Alls. 3429 cubic feet. 
 
 8-8. 
 
 Fig. 26. 
 
40 Theoretical Naval Architecture. 
 
 31. Show how to estimate the change in the mean draught of a vessel in 
 going from salt to river water, and vice versd. 
 
 A vessel floats at a certain draught in river water, and when floating in 
 sea water without any change in lading, it is found that an addition of 1 75 
 tons is required to bring the vessel to the same draught as in river water. 
 What is the displacement after the addition of the weight named ? 
 
 Ans. 11,200 tons. 
 
 32. The vertical sections of a vessel 10 feet apart have the following 
 areas : 10, 50, 60, 70, 50, 40, 20 square feet. Find the volume of displace- 
 ment, and the displacement in tons in salt and fresh water. 
 
 Ans. 2966 cubic feet ; 84' 7 tons, 82"4 tons. 
 
 33. A cylinder is 500 feet long, 20 feet diameter, and floats with the 
 axis in the water-line. Find its weight when floating thus in salt water. 
 What weight should be taken out in order that the cylinder should float 
 with its axis in the surface if placed into fresh water ? 
 
 Ans. 2244 tons ; 62 tons. 
 
 34. A vessel is 500 feet long, 60 feet broad, and floats at a mean draught 
 of 25 feet when in salt water. Make an approximation to her draught 
 when she passes into river water. (Coefficient of displacement, 0'5 ; coefficient 
 ofL.W.P., 0-6.) 
 
 Ans. 25' 4". 
 
 35. A piece of teak is 20 feet long, 4J inches thick, and its breadth 
 tapers from 12 inches at one end to 9 inches at the other end. What is its 
 weight, and how many cubic feet of water would it displace if placed into 
 fresh water (36 cubic feet to the ton) ? 
 
 Ans. 328 lbs. ; 5i cubic feet nearly. 
 
 36. The area of a water-plane is 5443 square feet. Find the tons per 
 inch immersion. Supposing 40 tons placed on board, how much would the 
 vessel sink ? 
 
 State any slight error that may be involved in any assumption made. If 
 40 tons were taken out, would the vessel rise the same amount ? What 
 further information would you require to give a more accurate answer ? 
 
 Ans. 1 2 -96 tons; 3- 1 inches nearly. 
 
 37. Bilge keels are to be fitted to a ship whose tons per inch are 48. 
 The estimated weight of the bilge keels is 36 tons, and the volume they 
 occupy is 840 cubic feet. What will be the increase of draught due to 
 fitting these bilge keels ? 
 
 Afis. J inch. 
 
 38. The tons per inch of a vessel at water-lines 2 feet apart are 19 '45, 
 18-51, 17-25, 15-6, 1355, 10-87, and 6-52, the lowest water-line being 18 
 inches above the underside of flat keel. Draw the curve of tons per inch 
 immersion to scale, and estimate the number of tons necessary to sink the 
 vessel from a draught of 12 feet to a draught of 13' 6". 
 
 Ans. 344 tons. 
 
 39. The steamship Umbria is 500 feet long, 57 feet broad, 22' 6" 
 draught, 9860 tons displacement, 11 50 square feet area of immersed midship 
 section. Find — 
 
 (i) Block coefficient of displacement. 
 
 (2) Prismatic ,, ,, 
 
 (3) Midship-section coefficient. 
 
 /«^. (1)0-538; (2)0-6; (3)0-896. 
 
 40. 1 he steamship Orient is 445 (eet long, 46 feet broad, 21' 4i'' draught 
 mean ; the midship section coefficient is 0-919, the blocjc coefficient of dis- 
 placement is 0-621. Find — 
 
Areas, Volumes, Weights, Displacement, etc. 41 
 
 (1) Displacement in tons. 
 
 (2) Area of immersed midship section. 
 
 (3) Prismatic coefficient of displacement. 
 
 Ans. (i) 7763 tons ; (2) 904 square feet ; (3) 0'675. 
 
 41. A vessel is 144 feet long, 22' 6" broad, 9 feet draught ; displacement, 
 334 tons salt water ; area of midship section, 124 square feet. Find — 
 
 (1) Block coefficient of displacement. 
 
 (2) Prismatic ,, „ 
 
 (3) Midship-section coefficient. 
 
 Ans. (i) 0-4; (2) o'6ss ; (2) 0'6i2. 
 
 42. Find the displacement in tons in salt water, area of the immersed 
 midship section, prismatic coefficient of displacement, having given the 
 following particulars : Length, i68 feet ; breadth, 25 feet ; draught, 10' 6" j 
 midship-section coefficient, 087 ; block coefficient of displacement, 0'S95. 
 
 Ans. 750 tons ; 228'5 square feet ; 0"585. 
 
 43. A vessel in the form of a box, 100 feet long, lo feet broad, and 20 
 feet depth, floats at a draught of S feet. Find the draught if a central 
 compartment 10 feet long is bilged below water. 
 
 . ^'"- 5' ^^"• 
 
 44. In a given ship, pillars in the hold can be either solid iron 4! inches 
 
 diameter, or hollow iron 6 inches diameter and half inch thick. Find the 
 saving in weight for every 100 feet length of these pillars, if hollow pillars 
 are adopted instead of solid, neglecting the effect of the solid heads and 
 heels of the hollow pillars. 
 
 Ans. I '35 ton. 
 
 45. What is the solid contents of a tree whose girth (circumference) is 
 60 inches, and length is 18 feet ? 
 
 Ans. 35'8 cubic feet nearly. 
 
 46. A portion of a cylindrical steel stern shaft casing is I2| feet long, 
 I J inch thick, and its external diameter is 14 inches. Find its weight in 
 pounds. 
 
 Ans. 2170 lbs. 
 
 47. A floating body has a water-plane whose semi-ordinates 25 feet 
 apart are 0-3, 8, 12, 10, 2 feet respectively, and every square station is in 
 the form of a circle with its centre in the water-plane. Find the volume of 
 displacement (ir = -^?). 
 
 Ans. 12,414 cubic feet. 
 
 48. A quadrant of 16 feet radius is divided by means of ordinates parallel 
 to one radius, and the following distances away : 4, 8, 10, 12, 13, 14, 1$ 
 feet respectively. The lengths of these ordinates are found to be 1 5 '49, 
 I3'86, 12-49, lO'SS, 9"33, 775, and S'57 feet respectively. Find — 
 
 (i) The exact area to two places of decimals. 
 
 (2) The area by using only ordinates 4 feet apart. 
 
 (3) The area by using also the half-ordinates. 
 
 (4) The area by using all the ordinates given above. 
 
 (5) The area as accurately as it is possible, supposing the ordinate I2'49 
 
 had not been given. 
 Ans. (I) 201-06 ; (2) 197-33 ; (3) I997S 5 (4) 200-59 ; (5) 200-50. 
 
 49. A cylindrical vessel Jo ^et long and 16 feet diameter floats at a 
 constant draught of 12 feet in salt water. Using the information given in 
 the previous question, find the displacement in tons. 
 
 Ans. 231 tons nearly. 
 
 50. A bunker 24 feet long has a mean section of the form of a trapezoid, 
 with length of parallel sides 3 feet and 4-8 feet, and distance between them 
 10-5 feet. Find the number of tons of coal contained in the bunker, assuming 
 
42 Theoretical Naval Architecture. 
 
 I ton to occupy 43 cubic feet. If the parallel sides are perpendicular to 
 one of the other sides, and the side 4'8 feet long is at the top of the section, 
 where will the top of 17 tons of coal be, supposing it to be evenly 
 distributed ? 
 
 (This latter part should be done by a process of trial and error.) 
 
 Ans.-2i-?, tons ; 2' 3" below the top. 
 
 51. The sections of a ship are 20 feet apart. A coal-bunker extends 
 from 9 feet abaft No. 8 section to i foot abaft No. 15 section, the total 
 length of the bunker thus being 132 feet. The areas of sections of the 
 bunker at Nos. 8, 11, and 15 are found to be 126, 177, and 1 45 square 
 feet respectively. With this information given, estimate the capacity of the 
 bunker, assuming 44 cubic feet of coal to go to the ton. Stations numbered 
 from forward. 
 
 Alls. 495 tons. 
 
 52. The tons per inch immersion at water-lines 2 feet apart are l8'09, 
 i6"8o, I5"i5, I3'15, iO'49, and 6-48. The draught of water to the top 
 water-line is 11' 6", and below the lowest water-line there is a displacement 
 of 7S'3 tons. Find the displacement in tons, and construct a curve of 
 displacement. 
 
 Ans. 1 712 tons. 
 
 53. A tube 35 feet long, 16 feet diameter, closed at the ends, floats in 
 salt water with its axis in the surface. Find approximately the thickness of 
 the tube, supposed to be of iron, neglecting the weight of the ends. 
 
 Ans. 0*27 foot. 
 
 54. Find the floating power of a topmast, length 64 feet, mean diameter 
 21 inches, the wood of the topmast weighing 36 lbs. per cubic foot. 
 
 (The floating power of a spar is the weight it will sustain, and this is 
 the difference between its own weight and that of the water it displaces. 
 In constructing a raft, it has to be borne in mind that all the weight of 
 human beings is to be placed on it, and that a great quantity of provisions 
 and water may be safely carried under it. For instance, a cask of beef 
 slung beneath would be 116 lbs., above 300 lbs. See " Sailor's Pocket- 
 book," by Admiral Bedford.) 
 
 Ans. 4310 lbs. 
 
CHAPTER II. 
 
 MOMENTS, CENTRE OF GRAVITY, CENTRE OF BUOY- 
 ANCY, DISPLACEMENT TABLE, PLANIMETBK, ETC. 
 
 Principle of Moments.— The moment of a force about 
 any given line is the product of the force into the perpen- 
 dicular distance of its line of action from that Une. It may 
 also be regarded as the tendency to turn about the line. A 
 man pushes at the end of a capstan bar (as Fig. 27) with a 
 
 Fig. 27. 
 
 certain force. The tendency of the capstan to turn about its 
 axis is given by the force exerted by the man multiplied by 
 his distance from the centre of the capstan, and this is the 
 moment of the force about the axis. If P is the force exerted 
 by the man in pounds (see Fig. 27), and d is his distance from 
 the axis in feet, then — 
 
 The moment about the axis = V y. d foot-lbs. 
 
 The same moment can be obtained by a smaller force with 
 a larger leverage, or a larger force with a smaller leverage, and 
 the moment can be increased : — 
 
 (i) By increasing the force ; 
 
 (2) By increasing the distance of the force from the axis. 
 
44 
 
 Theoretical Naval Architecture. 
 
 If, in addition, there is another man helping the first man, 
 exerting a force of P' lbs. at a distance from the axis of d' 
 feet, the total moment about the axis is — 
 
 (P X ^ + (P'X rf') foot-lbs. 
 
 We must now distinguish between moments tending to turn 
 one way and those tending to turn in the opposite direction. 
 
 Thus, in the above case, we may take a rope being wound 
 on to the drum of the capstan, hauling a weight W lbs. If the 
 radius of the drum be « feet, then the rope tends to turn the 
 capstan in the opposite direction to the men, and the moment 
 about the axis is given by — 
 
 W X « foot-lbs. 
 
 If the weight is just balanced, then there is no tendency to 
 turn, and hence no moment about the axis of the capstan, and 
 leaving out of account all consideration of friction, we have — 
 
 (P X ^) + (P' X rf') = W X « 
 The most common forces we have to deal with are those 
 caused by gravity, or the attraction of bodies to the earth. This 
 is known as their weight, and the direction of these forces must 
 all be parallel at any given place. If we have a number of 
 weights, Wi, W2, and Wj, on a beam at A, B, and C (Fig. 28), 
 
 El 
 
 w 
 
 Fig. 28. 
 
 whose end is fixed at O, the moment of these weights about O 
 is given by — 
 
 (Wi X AO) + (W, X BO) + (W3 X CO) 
 This gives the tendency of the beam to turn about O, due to 
 
Moments, Centre of Gravity, Centre of Buoyancy, etc. 45 
 
 the weights Wj, Wg, and W3 placed upon it, and the beam must 
 be strong enough at O in order to resist this tendency, or, as 
 it is termed, the bending moment. Now, we can evidently 
 place a single weight W, equal to the sum of the weights 
 Wi, W2, and W5, at some point on the beam so that its moment 
 about O shall be the same as that due to the three weights. 
 If P be this point, then we must have — 
 
 W X OP = (Wi X OA) + (W2 X OB) + (W3 X OC) 
 or, since W = Wi + W2 + W3 
 
 OP - (Wi X OA) + (W, X OB) + (W3 X OC) 
 
 W, + W, + W3 
 
 Example.: — Four weights, 30, 40, 50, 60 lbs. respectively, are placed 
 on a beam fixed at one end, O, at distances from O of 3, 4, 5, 6 feet respec- 
 tively. Find the bending moment at O, and also the position of a single 
 weight equal to the four weights which will give the same bending moment. 
 
 Bending moment at O = (30 X 3) + (4° X 4) + (50 X S) + (60 x 6) 
 = 90 + 160 + 250 + 360 
 = 860 foot-lbs. 
 Total weight = 180 lbs. 
 /. position of single weight = f|§ = 4^ feet from O 
 
 Centre of Gravity. — The single weight W above, when 
 placed at P, has the same effect on the beam at O and at 
 any other point of the beam, as the three weights Wj, W2, and 
 W3. The point P is termed the centre of gravity of the weights 
 Wi, W2, and W3. Thus we may define the centre of gravity 
 of a number of weights as follows : — 
 
 The centre of gravity of a system of weights is that point 
 at which we may regard the whole system as being concentrated, 
 and at which the same effect is produced as by the original system 
 of weights. 
 
 This definition will apply to the case of a solid body, since 
 we may regard it as composed of a very large number of small 
 particles, each of which has a definite weight and occupies a 
 definite position. A homogeneous solid has the same density 
 throughout its volume ; and all the solids with which we have 
 to deal are taken as homogeneous unless otherwise specified. 
 
 It follows, from the above definition of the centre of 
 gravity, that if a body is suspended at its centre of gravity. 
 
46 
 
 Theoretical Naval Architecture. 
 
 it would be perfectly balanced and have no tendency to move 
 away from any position in which it might be placed. 
 
 To Find the Position of the Centre of Gravity 
 of a number of Weights lying in a Plane. — Two lines 
 are drawn in the plane at right angles, and the moment of the 
 system of weights is found successively about each of these 
 lines. The total weight being known, the distance of the 
 centre of gravity from each of these lines is found, and conse- 
 quently the position of the centre of gravity definitely fixed. 
 
 Fig. 29. 
 
 The following example will illustrate the principles in- 
 volved : Four weights, of 15, 3, 10, and 5 lbs. respectively, 
 are lying on a table in definite positions as shown in Fig. 29. 
 Find the position of the centre of gravity of these weights. 
 (If the legs of the table were removed, this would be the place 
 where we should attach a rope to the table in order that it 
 should remain horizontal, the weight of the table being 
 neglected.) 
 
Moments, Centre of Gravity, Centre of Buoyancy, etc. 47 
 
 Draw two lines, Ox, Oy, at right angles on the table in 
 any convenient position, and measure the distances of each of 
 the weights from Ox, Oy respectively: these distances are 
 indicated in the figure. The total weight is 33 lbs. The 
 moment of the weights about O* is — 
 
 (is X 7) + (3 X 3) + (10 X 5) + (5 X rs) = 171-5 foot-lbs. 
 
 The distanceof the centre of gravity from O^ = ~ — ^ = 5"2 feet 
 
 33 
 
 If we draw a line A A a distance of 5-2 feet from Ox, the 
 centre of gravity of the weights must be somewhere in the 
 line AA. 
 
 Similarly, we take moments about Oy, finding that the 
 moment is 150 foot-lbs., and the distance of the centre of 
 gravity from Oy is — 
 
 W = 4-25 feet 
 
 If we draw a line BB a distance of 4*55 feet from Oy, the 
 centre of gravity of the weights must be somewhere in the line 
 BB. The point G, where AA and BB meet, will be the centre 
 of gravity of the weights. 
 
 Centres of Gravity of Plane Areas. — A plane area has 
 length and breadth, but no thickness, and in order to give a 
 definite meaning to what is termed its centre of gravity, the 
 area is supposed to be the surface of a thin lamina or plate of 
 homogeneous material of uniform thickness. With this sup- 
 position, the centre of gravity of a plane area is that point at 
 which it can be suspended and remain in equilibrium. 
 
 Centres of Gravity of Plane Figures. 
 
 Circle. — The centre of gravity of a circle is obviously at 
 its centre. 
 
 Square and Rectangle. — The centre of gravity of 
 either of these figures is at the point where the diagonals 
 intersect. 
 
 Rhombus and Rhomboid. — The centre of gravity of 
 either of these figures is at the point where the diagonals 
 intersect. 
 
48 
 
 Theoretical Naval Architecture. 
 
 Triangle.— Take the triangle ABC, Fig. 30. Bisect any 
 two sides BC, AC in the points D and E. Join AD, BE. The 
 point G where these two lines intersect is the centre of gravity 
 
 of the triangle. It can be 
 proved that the point G is 
 situated so that DG is one-third 
 DA, and EG is one-third EB. 
 We therefore have the following 
 rules : — 
 
 I. Bisect any two sides of 
 the triangle, and join the points 
 thus obtained to the opposite angu- 
 lar points. The7i the point in 
 which these two lines intersect is the centre of gravity of the triangle. 
 2. Bisect any side of the triangle, and join the point thus 
 obtained with the opposite angular point. The centre of gravity 
 of the triangle will be on this line, and at a point at one-third its 
 length measured from the bisected side. 
 
 Trapezium. — Let ABCD, Fig. 31, be a trapezium. By 
 joining the corners A and C we can divide the figure into two 
 triangles, ADC, ABC. The centres of gravity, E and F, of 
 
 these triangles can be 
 found as indicated 
 above. Join EF. The 
 centre of gravity of the 
 whole figure must be 
 somewhere in the line 
 EF. Again, join the 
 corners D and B, thus 
 dividing the figure into 
 S C two triangles ADB, 
 
 ^"=- 3'- CDB. The centres of 
 
 gravity, H and K, of these triangles can be found. The 
 centre of gravity of the whole figure must be somewhere in the 
 line HK; therefore the point G, where the lines HK and EF 
 intersect, must be the centre of gravity of the trapezium. 
 
 The following is a more convenient method of finding the 
 centre of gravity of a trapezium. 
 
Moments, Centre of Gravity, Centre of Buoyancy, etc. 49 
 
 Let ABCD, Fig. 32, be a trapezium. Draw the diagonals 
 AC, BD, intersecting at E. In the figure CE is greater than 
 
 Fig. 32. 
 
 EA, and DE is greater than EB. Make CH = EA and DF 
 = EB. Join FH. Then the centre of gravity of the triangle 
 EFH will also be the centre of gravity of the trapezium ABCD. 
 
 (A useful exercise in drawing would be to take a trapezium 
 on a large scale and find its 
 centre of gravity by each of 
 the above methods. If the 
 drawing is accurately done, the 
 point should be in precisely 
 the same position as found by 
 each method.) 
 
 To find the Centre of 
 Gravity of a Plane Area 
 by Experiment. — Draw out 
 the area on a piece of card- 
 board or stiff paper, and cut 
 out the shape. Then suspend 
 the cardboard as indicated in 
 Fig. 33, a small weight, W, 
 being allowed to hang plumb. 
 A line drawn behind the string AW must pass through the 
 centre of gravity. Mark on the cardboard two points on the 
 string, as A and B, and join. Then the centre of gravity must 
 lie on AB. Now suspend the cardboard by another point, C, 
 
 £ 
 
 OW. 
 
 Fig. 33. 
 
so 
 
 Theoretical Naval Architecture. 
 
 as in Fig. 34, and draw the line CD immediately behind the 
 string of the plumb-bob W. Then also the centre of gravity 
 must lie on the line CD. Consequently it follows that the 
 
 point of intersection G of the 
 lines AB and CD must be the 
 centre of gravity of the given 
 area. 
 
 Example. — Set out the section of 
 a beam on a piece of stiff paper, and 
 find by experiment the position of its 
 centre of gravity, the beam being formed 
 of a bulb plate 9 inches deep and 
 J inch thick, having two angles on the 
 upper edge, each 3" X 3" X J". 
 
 Ans. 3 inches from the top. 
 
 Centres of Gravity of Solids 
 formed of Homogeneous 
 Material. 
 Sphere. — The centre of 
 gravity of a sphere is at its centre. 
 Cylinder. — The centre of 
 gravity of a cylinder is at one- 
 half its height from the base, on 
 the line joining the centres of 
 gravity of the ends. 
 Pyramid or Cone. — The centre of gravity of a pyramid 
 or cone is at one- fourth the height of the apex from the base, 
 on the line joining the centre of gravity of the base to the 
 apex. 
 
 Moment of an Area. 
 
 .The geometrical moment of a plane area relatively to a! 
 given axis, is the product of its area into the perpendicular 
 distance of its centre of gravity from the given axis. It fellows' 
 that the position of the centre of gravity is known relatively to 
 the given axis if we know the geometrical moment about the 
 axis and also the area, for the distance will be the moment 
 divided by the area. It is usual to speak of the moment of an 
 
Moments, Centre of Gravity, Centre of Buoyancy, etc. 5 1 
 
 area about a given axis when the geometrical moment is really 
 meant. 
 
 To find the Position of the Centre of Gravity of 
 a Curvilinear Area with respect to one of its Ordi- 
 nates.— Let AEDO, Fig. 35, be a plane curvilinear area, and 
 
 we wish to find its centre of gravity with respect to the end 
 ordinate, OA. To do this, we must first find the moment of 
 the total area about OA, and this divided by the area of the 
 figure itself will give the distance of the centre of gravity from 
 OA. Take any ordinate, PQ, a distance of x from OA, and 
 at PQ draw a strip A;« wide. Then the area of the strip is 
 V X A* very nearly, and the moment of the strip about OA is 
 {y X t^oc)x very nearly. 
 
 If now A* be made indefinitely small, the moment of the 
 strip about OA will be — 
 
 y .X .dx 
 
 Now, we can imagine the whole area divided up into such 
 strips, and if we added up the moments about OA of all such 
 strips, we should obtain the total moment about OA. Ther^pre, 
 using the notation we employed for finding the area of a plane' 
 curvilinear figure on p. 14, we shall have — 
 
 Moment of the total area about OA = jy .x.dx 
 
 The expression for the area is — 
 
 jy ■ dx 
 
^i 
 
 Theoretical Naval Architecture. 
 
 and this is of the same form as the expression for the moment. 
 Therefore, instead of y we put yx through Simpson's rule in 
 the ordinary way, and the result will be the moment about OA. 
 Set off on BC a length BF = BC X h, and on DE a length 
 DG = DE X 2h. Then draw through all such points a curve, 
 as OFG.* Any ordinate of this curve will give the ordinate of 
 the original curve at that point multiplied by its distance from 
 OA. The area of an elementary strip of this new curve will be 
 y .X .dx, and the total area of the new curve will he jy . x . dx, 
 or the moment of the original figure about OA. Therefore, to 
 find the moment of a curvilinear figure about an end ordinate, 
 we take each ordinate and multiply it by its distance from the 
 end ordinate. These products, put through Simpson's rule in 
 the ordinary way, will give the moment of the figure about the 
 end ordinate. This moment divided by the area will give the 
 distance of the centre of gravity of the area from the end 
 ordinate. 
 
 Example.— A. midship section has semi-ordinates, i' 6" apart, com- 
 mencing at the L.W.L., of length 8"6o, S'lo, 6'9S, 4-90, 275, I'So, 070 
 feet respectively. Find the area of the section and the distance of its C.G. 
 from the L.W.L. 
 
 Number of 
 
 Length of 
 
 Simpson's 
 
 Function of 
 
 _ Number of 
 
 intervals from 
 
 No. I ord. 
 
 Products for 
 
 ordinates. 
 
 ordinates. 
 
 multipliers. 
 
 ordinates. 
 
 moment. 
 
 I 
 
 8 -60 
 
 I 
 
 8 -60 
 
 
 
 0-0 
 
 2 
 
 8- 10 
 
 4 
 
 32-40 
 
 I 
 
 32-40 
 
 3 
 
 6-95 
 
 2 
 
 13-90 
 
 2 
 
 27-80 
 
 4 
 
 4-90 
 
 4 
 
 19-60 
 
 3 
 
 58-80 
 
 5 
 
 27s 
 
 2 
 
 5-50 
 
 4 
 
 22-00 
 
 6 
 
 1-50 
 
 4 
 
 6-00 
 
 S 
 
 30-00 
 
 7 
 
 070 
 
 I 
 
 0-70 
 
 6 
 
 4-20 
 
 86-70 
 
 Function of 
 moment 
 
 } I7S-' 
 
 The half-area will be given by 86-70 x (J X I -5) = 43-35 square feet 
 and the whole area is 86-70 „ 
 
 The arrangement above is adopted in order to save labour in 
 
 ' If the original curve is a parabola of the second order, this curve will 
 be one of the third order, and it can be proved that Simpson's first rule 
 will integrate a parabola of the third order. 
 
Moments, Centre of Gravity, Centre of Buoyancy, etc. 53 
 
 finding the moment of the area. In the fourth column we have 
 the functions of the ordinates, or the ordinates multiplied succes- 
 sively by their proper multipliers. In the fifth column is placed, 
 not the actual distance of each ordinate from the No. i ordi- 
 nate, but the number of intervals away, and the distance apart 
 is brought in at the end. In the sixth column the products of 
 the functions in column 4 and the multipliers in column 5 are 
 placed. It will be noticed that we have put the ordinates 
 through Simpson's multipliers first, and then multiplied by the 
 numbers in the fifth column after. This is the reverse to the 
 rule given in words above, which was put into that form in 
 order to bring out the principle involved more plainly. The 
 final result will, of course, be the same in either case, the 
 method adopted giving the result with the least amount of 
 labour, because column 4 is wanted for finding the area. The 
 sum of the products in column 6 will not be the moment 
 required, because it has to be multiplied as follows : First, by 
 one-third the common interval, and second, by the distance 
 apart of the ordinates. 
 
 I = 175-20 X (I X 4) X li 
 
 The moment of the half-area 
 about the L.W.L. 
 
 = i3i"4 
 
 and the distance of the C.G. of the half-area from the L.W.L. 
 
 is — 
 
 i3i'4 
 
 Moment -r area = = 3*03 feet 
 
 43-35 "* ^ 
 
 It will be noticed that we have multiplied both columns 
 4 and 6 by one-third the common interval, the distance of the 
 C.G. from No. i ordinate being obtained by — 
 
 i75'2°X (jx 1-5) X 1-5 
 8670 X (i X 1-5) 
 
 The expression \ X i'5 is common to both top and bottom, 
 and so can be cancelled out, and we have — 
 
 175-20 X 1-5 f , 
 
 86-70 = 3-°3 feet 
 
54 
 
 Theoretical Naval Architecture. 
 
 The position of the centre of gravity of the half-area with 
 regard to the L.W.L. is evidently the same as that of the whole 
 area. 
 
 When finding the centre of gravity of a large area, such as 
 a water-plane of a vessel, it is usual to take moments about 
 the middle ordinate. This considerably simplifies the work, 
 because the multipliers in column 5 are not so large. 
 
 Example. — The semi-ordinates of the load water-plane of a vessel 395 
 feet long are, commencing from forward, o, io'2, 20'o, 27'4, S^'i, 34'o> 
 33'8, 317, 27'6, 20-6, 9'4. Find the area and the distance of its C.G. 
 from the middle ordinate. 
 
 In addition to the above, there is an appendage abaft the last ordinate, 
 having an area of 153 square feet, and whose C.G. is 5'6 feet abaft the last 
 ordinate. Taking this appendage into account, find the area and the 
 position of the C.G. of the water-plane. 
 
 Number of 
 
 Length of 
 
 Simpson's 
 
 Function of 
 
 Number of 
 
 interval from 
 
 mid. ord. 
 
 Product for 
 
 ordinates. 
 
 ordinates. 
 
 multipliers. 
 
 ordinates. 
 
 moment. 
 
 I 
 
 O'O 
 
 I 
 
 O'O 
 
 s 
 
 O-Q 
 
 2 
 
 IO-2 
 
 4 
 
 40-8 
 
 4 
 
 163-2 
 
 3 
 
 20 'O 
 
 2 
 
 40-0 
 
 3 
 
 120-0 
 
 4 
 
 27-4 
 
 4 
 
 109-6 
 
 2 
 
 219-2 
 
 S 
 6 
 
 7 
 
 32-1 
 
 34'o 
 33-8 
 
 2 
 
 4 
 2 
 
 64-2 
 
 136-0 
 
 67-6 
 
 I 
 
 I 
 
 64-2 
 
 566-6 
 
 67-6 
 
 8 
 
 317 
 
 4 
 
 126-8 
 
 2 
 
 253-6 
 
 9 
 
 27-6 
 
 2 
 
 SS-2 
 
 3 
 
 165-6 
 
 10 
 
 20 '6 
 
 4 
 
 82-4 
 
 4 
 
 329-6 
 
 II 
 
 9-4 
 
 I 
 
 9'4 
 
 5 
 
 47 -0 
 
 732-0 
 
 The half-area will be given by — 
 
 732-0 X (J X 39-5) = 9638 square feet 
 
 863-4 
 
 The fifth column gives the number of intervals away from the middle 
 ordinate, and the products are obtained for the forward portion adding up 
 to 566-6, and they are obtained for the after portion adding up to 863-4, 
 This gives an excess aft of 863-4 — S66-6 = 296-8. The distance of the C.G. 
 abaft the middle ordinate is then given by — 
 
 ^96:L^?1S = ,6.01 feet 
 732-0 
 
 The area of both sides is 19,276 square feet. 
 
 The second part of the question takes into account an appendage abaft 
 No. II ordinate, having an area of 153 square feet. 
 
Moments, Centre of Gravity, Centre of Btcoyancy, etc. 5 5 
 
 The total area will then be — 
 
 19,276 4- 153 = 19.429 square feet 
 
 To find the position of the C.G. of the whole water-plane, we take 
 moments about No. 6 ordinate, the distance of the C.G. of the appendage 
 from it being — 
 
 i97'S + 5'6 = 203-1 feet 
 Moment of main area abaft No. 6 ordinate = 19,276 x i6-oi = 308,609 
 „ appendage „ „ = 153 x 203-1 = 31-074 
 
 .•. total moment abaft No. 6 ordinate = 308,609 + 31,074 
 
 = 339.683 
 and the distance of the centre of gravity \ _ 339683 _ . o j- 
 of the whole area abaft No. 6 ordinate J — 19429^ ~ '7 '4° leet 
 
 To find the Position of the Centre of Gravity of 
 a Curvilinear Area contained between Two Con- 
 secutive Ordinates with respect to the Near End 
 Ordinate. — The rule investigated in the previous paragraph 
 for finding the centre of gravity of an area about its end ordi- 
 nate fails when applied to such a case as the above. For 
 instance, try the following example : — 
 
 A curve has ordinates 10, 9, 7 feet long, 4 feet apart. To 
 find the position of the centre of gravity of the portion between 
 the two first ordinates with respect to the end ordinate. 
 
 Ordinates. 
 
 Simpson's 
 multipliers. 
 
 Functions. 
 
 Multipliers 
 for moment. 
 
 Products for 
 moment. 
 
 10 
 
 9 
 
 7 
 
 s 
 
 8 
 — I 
 
 SO 
 
 72 
 -7 
 
 
 I 
 2 
 
 
 
 72 
 -14 
 
 IIS 
 
 58 
 
 Centre of gravity from the end ordinate would be — 
 5 8x4 
 
 "5 
 
 ._2_ 
 '116 
 
 feet 
 
 Now this is evidently wrong, since the shape of the curve is 
 such that the centre of gravity ought to be slightly less than 
 2 feet from the end ordinate. 
 
 We must use the following rule : — 
 
 To ten times the middle ordinate add three times the near 
 end ordinate and siibtract tJie far end ordinate. Multiply the 
 
56 
 
 Theoretical Naval Architecture. 
 
 remainder by one-twenty-fourth the square of the common interval. 
 The product will be the moment about the end ordinate. 
 
 Using y-i, y^, y^, for the lengths of the ordinates, and h the 
 common interval, the moment of the portion between the 
 ordinates y^ and y^ about the ordinate yi is given by — 
 
 l^ 
 
 —izyi + 10^2 - yi) 
 
 24 
 
 We will now apply this rule to the case considered above. 
 
 Ordinates. 
 
 Area. 
 
 Moment. 
 
 Simpson's 
 multipliers. 
 
 Functions. 
 
 Simpson's 
 multipliers. 
 
 Functions. 
 
 10 
 9 
 
 7 
 
 — I 
 
 50 
 
 72 
 
 7 
 
 3 
 10 
 
 — I 
 
 30 
 90 
 
 -7 
 
 IIS "3 
 
 Moment = 113 X y| 
 Area = 115 X -5% 
 
 Therefore distance of the centre of gravity from the end ordi- 
 nate is — 
 
 "3 X il ^ 113 X 2 X 3 
 115 X^ 115x3 
 
 = HI = 1-965 feet 
 
 This result is what one might expect by considering the 
 shape of the curve. 
 
 To find the Position of the Centre of Gravity of a 
 Curvilinear Area with respect to its Base. — Let DABC, 
 
 Fig. 36, be a plain curvilinear area. We wish to find the 
 distance of its centre of gravity from the base DC. To do 
 this, we must first find the moment of the figure about DC and 
 divide it by the area. Take any ordinate PQ, and at PQ 
 draw a consecutive ordinate giving a strip ^x wide. Then the 
 area of the strip is — 
 
 y X ^x very nearly 
 
Moments, Centre of Gravity, Centre of Buoyancy, etc. 57 
 
 and regarding it as a rectangle, its centre of gravity is at a 
 distance of \y from the base. Therefore the moment of the 
 strip about the base is — 
 
 If now we consider the strip 
 to be indefinitely thin, its 
 moment about the base will 
 be— 
 
 \f.dx 
 
 and the moment of the total 
 area about the base must be 
 the sum of the moments of all such strips, or — 
 
 This expression for the moment is of the same form as that for 
 the area, viz. Jj . dx. Therefore, instead of y we put ky^ ' 
 through Simpson's rule in the ordinary way, and the result will 
 be the moment of the curve about DC. 
 
 Fig. 36. 
 
 Example. — An athwartship coal-bunker is 6 feet long in a fore-and-aft 
 direction. It is bounded at the sides by two longitudinal bulkheads 34 feet 
 apart, and by a horizontal line at the top. The bottom is formed by the 
 inner bottom of the ship, and is in the form of a curve having vertical 
 ordinates measured from the top of 12 '5, 15'0, i6'0, 16-3, i6"4, l6"3, i6'0, 
 I5'0, I2'S feet respectively, the first and last ordinates being on the bulk- 
 heads. Find — 
 
 (i) The number of tons of coal the bunker will hold. 
 
 (2) The distance of the centre of gravity of the coal from the top. 
 The inner bottom is symmetrical either side of the middle line, so we ■ 
 need only deal with one side. The work is arranged as follows :- — 
 
 
 Simpson's 
 
 Functions of 
 
 Squares of 
 
 Simpson's 
 
 Functions 
 
 
 multipliers. 
 
 ordinates. 
 
 ordinates. 
 
 multipliers. 
 
 of squares. 
 
 l6-4 
 
 I 
 
 16-4 
 
 269 
 
 I 
 
 269 
 
 i6-3 
 
 4 
 
 65-2 
 
 266 
 
 4 
 
 1064 
 
 16 -o 
 
 2 
 
 32-0 
 
 256 
 
 2 
 
 512 
 
 15-0 
 
 4 
 
 60 '0 
 
 225 
 
 4 
 
 900 
 
 12-5 
 
 I 
 
 12-5 
 
 J 56 
 
 I 
 
 156 
 
 Function of area l86'i 
 
 Function of moment 2901 
 
 ' This assumes that Simpson's first rule, which will most probably be 
 used, will correctly integrate a parabola of the fourth order, which can be 
 shown to be the case for all practical purposes. 
 
58 Theoretical Naval Architecture. 
 
 Common interval = 4'2S feet 
 Half-area of section = i86t X J X 4*25 square feet 
 
 4.'2^ X 2 X 6 
 
 Volume of bunker = i86"l X — — cubic feet 
 
 3 
 Number of tons of coal = l86-i X JJ 
 = 72 tons 
 
 Moment of half-area below top = 2901 x — X ^— = 
 
 23 
 
 And distance of C.G. from the top = 
 
 area 
 
 I 4-25 
 2901 x-x— ■" 
 
 ^^86-ix^'5 
 
 = 7'8 feet 
 
 In the first three columns we proceed in the ordinary way 
 for finding the area. In the fourth column is placed, not the 
 half-squares, but the squares of the ordinates in column i,the 
 multiplicatiorl by \ being brought in at the end. These 
 squares are then put through Simpson's multipliers, and the 
 addition of column 6 will give a function of the moment of 
 the area about the base. This multiplied by \ and by \ the 
 common interval gives the actual moment. This moment 
 divided by the area gives the distance of the centre of gravity 
 we want. It will be noticed that \ the common interval 
 comes in top and bottom, so that we divide the function of the 
 moment 2901 by the function of the area i86"i, and then 
 multiply by \ to get the distance of centre of gravity required. 
 
 It is not often required in practice to find the centre of 
 gravity of an area with respect to its base, because most of the 
 areas we have to deal with are symmetrical either side of a 
 centre line (as water-planes), but the problem sometimes occurs, 
 the question above being an example. 
 
 To find the Position of the Centre of Gravity of 
 an Area bounded by a Curve and Two Radii. — We 
 have already seen (p. 15) how to find the area of a figure such 
 as this. It is simply a step further to find the position of the 
 centre of gravity with reference to either of the bounding radii. 
 Let OAB, Fig. 13, be a figure bounded by a curve, AB, and 
 two bounding radii, OA, OB. Take any radius OP, the angle 
 BOP being called B, and the length of OP being called r. 
 
Moments, Centre of Gravity, Centre of Buoyancy, etc. 59 
 
 Draw a consecutive radius, OP' ; the angle POP' being indefi- 
 nitely small, we may call it dO. Using the assumptions we 
 have already employed in finding areas, the area POP' = 
 \7^.M, POP' being regarded as a triangle. The centre of 
 gravity of POP' is at g, and O^ = | r, and gm is drawn perpen- 
 dicular to OB, and gm = \r. sin B (see p. 87). 
 
 The moment of the area ] ,, , „^ 
 
 POP' about OB [ = (*' • '^^) X (t^' • ^"^ ^) 
 
 = \r' . sin ^ . dO 
 
 The moment of the whole figure about OB is the sum of 
 the moments of all such small areas as POP', or, using the 
 ordinary rotation — 
 
 i/;-^sin e.de 
 
 This is precisely similar in form to the expression we found 
 for the area of such a figure as the above (see p. 15), viz. — 
 
 \ii^.de 
 
 so that, instead of putting \ r^ through Simpson's rule, measuring 
 r at equidistant angular intervals, we put \r^ . sin ^ through 
 the rule in a similar way. This will be best illustrated by the 
 following example : — 
 
 Example. — Find the area and position of centre of gravity of a quadrant 
 of a circle with reference to one of its bounding radii, the radius being 
 10 feet. 
 
 We will divide tire quadrant by radii 15° apart, and thus be able to use 
 Simpson's first rule. 
 
 , 
 
 II 
 
 3 >• 
 
 Is 
 
 _3 
 
 "g.a 
 
 3 9. 
 
 s 
 
 5 
 
 u 
 
 
 1-1 
 
 
 
 I 
 
 10 
 
 100 
 
 I 
 
 100 
 
 1000 
 
 
 
 O'O 
 
 
 
 I 
 
 
 
 2 
 
 10 
 
 100 
 
 4 
 
 400 
 
 1000 
 
 15 
 
 0-258 
 
 258 
 
 4 
 
 1,032 
 
 ?, 
 
 lO 
 
 100 
 
 2 
 
 200 
 
 1000 
 
 30 
 
 o'5oo 
 
 500 
 
 2 
 
 1,000 
 
 4 
 
 10 
 
 100 
 
 4 
 
 400 
 
 1000 
 
 45 
 
 0707 
 
 707 
 
 4 
 
 2,828 
 
 S 
 
 10 
 
 100 
 
 2 
 
 200 
 
 1000 
 
 60 
 
 0-866 
 
 866 
 
 2 
 
 1. 732 
 
 6 
 
 10 
 
 100 
 
 4 
 
 400 
 
 1000 
 
 7S 
 
 0-965 
 
 965 
 
 4 
 
 3,860 
 
 7 
 
 10 
 
 100 
 
 I 
 
 100 
 
 1000 
 
 90 
 
 I -000 
 
 1000 
 
 I 
 
 1,000 
 
 Function of area 1800 
 
 Function of moment 11,452 
 
6o Theoretical Naval Architecture. 
 
 The circular measure of l8o° = ir = 3-1416 
 1,0^3:1416 
 
 12 
 
 area 
 
 II, 
 Moment -i- area = ■ 
 
 /x 3-1416 N 
 = i8ooxJx(^jX-^ j 
 
 = 78-54 square feet 
 
 I ( \ 3-1416 \ 
 Moment of area about the first radius =11,452 >< -^ I T^ j2 J 
 
 therefore distance of centre of gravity from the first radius is— 
 
 I /I 3-i4i6\ 
 
 I / I 31416 \ 
 1800 X - X ( - X Z 
 
 1 1452 X 2 , ^ 
 
 = o ., „ = 4'24 feet 
 1800 X 3 ^ ^ 
 
 The exact distance of the centre of gravity of a quadrant 
 
 from either of its bounding radii is — times the radius, and if 
 
 this is appHed to the above example, it will be found that the 
 result is correct to two places of decimals, and would have 
 been more correct if we had put in the values of the sines of 
 the angles to a larger number of decimal places. 
 
 Centre of Gravity of a Solid Body which is 
 bounded by a Curved Surface and a Plane. — In the 
 first chapter we saw that the finding the volume of such a solid 
 as this was similar in principle to the finding the area of a 
 plane curve, the only difference being that we substitute areas 
 for simple ordinates, and as a result get the volume required. 
 The operation of finding the centre of gravity of a volume in 
 relation to one of the dividing planes is precisely similar to the 
 operation of finding the centre of gravity of a curvilinear area 
 in relation to one of its ordinates. This will be illustrated by 
 the following example : — 
 
 Example. — A coal-bunker has sections 17' 6" apart, and the areas of 
 these sections, commencing from forward, are 98, 123, 137, 135, 122 square 
 feet respectively. Find the volume of the bunker, and the position of its 
 centre of gravity in a fore-and-aft direction. 
 
Moments, Centre of Gravity, Centre of Buoyancy, etc. 6i 
 
 Areas. 
 
 Simpson's 
 multipliers. 
 
 Functions of 
 areas. 
 
 Number of 
 
 intervals from 
 
 forward. 
 
 Products for 
 moments. 
 
 98 
 123 
 137 
 13s 
 122 
 
 I 
 
 4 
 2 
 
 4 
 
 I 
 
 98 
 492 
 274 
 540 
 
 122 
 
 
 I 
 2 
 
 3 
 
 4 
 
 
 
 492 
 
 548 
 
 1620 
 
 488 
 
 1526 
 
 3148 
 
 Volume = 1526 X J X 17 J = 8go2 cubic feet 
 moment = 3148 x 3 X 17I x 17J 
 . distance of centre of gravity \ _ 3148 X I7"5 
 from forward end / 1526 
 
 : 36-2 feet 
 
 It is always advisable to roughly check any result such 
 as this ; and if this habit is formed, it will often prevent mis- 
 takes being made. The total length of this bunker is 4 X 
 17' 6" = 70 feet, and the areas of the sections show that the 
 bunker is fuller aft than forward, and so, on the face of it, we 
 should expect the position of the centre of gravity to be some- 
 what abaft the middle of the length ; and this is shown to be 
 so by the result of the calculation. Also as regards the volume. 
 This must be less than the volume of a solid 70 feet long, and 
 having a constant section equal to the area of the middle 
 section of the bunker. The .volume of such a body would 
 be 70 X 137 = 9590 cubic feet. The volume, as found by 
 the calculation, is 8902 cubic feet, thus giving a coefiScient of 
 11^ = 0-93 nearly, which is a reasonable result to expect. 
 
 Centre of Buoyancy. — The centre of buoyancy of a 
 vessel is the centre of gravity of the underwater volume, or, 
 more simply, the centre of gravity of the displaced water. 
 This has nothing whatever to do with the centre of gravity of 
 the ship herself. The centre of buoyancy is determined solely 
 by the shape of the underwater portion of the ship. The 
 centre of gravity of the ship is determined by the distribution 
 of the weights forming the structure, and of all the weights she 
 has on board. Take the case of two sister ships built from 
 the same lines, and each carrying the same weight of cargo 
 and floating at the same water-line. The centre of buoyancy 
 
62 
 
 Theoretical Naval Architecture. 
 
 of each of these ships must necessarily be in the same position. 
 But suppose they are engaged in diiferent trades — the first, 
 say, carrying a cargo of steel rails and other heavy weights, 
 which are stowed low down. The second, we may suppose, 
 carries a cargo of homogeneous materials, and this has to be 
 stowed much higher than the cargo in the first vessel. It is 
 ■evident that the centre of gravity in the first vessel must be 
 much lower down than in the second, although as regards form 
 they are precisely similar. This distinction between the centre 
 of buoyancy and the centre of gravity is a very important 
 one, and should always be borne in mind. 
 
 To find the Position of the Centre of Buoyancy of 
 a Vessel in a Fore-and-aft Direction, having given 
 the Areas of Equidistant Transverse Sections. — The 
 following example will illustrate the principles involved : — 
 
 Example. — The underwater portion of a vessel is divided by transverse 
 sections lo feet apart of the following areas, commencing from forward : 0-2, 
 22-7, 48'8, 73-2, 88-4, 82-8, 58-7, 26-2, 3-9 square feet respectively. Find 
 the position of the centre of buoyancy relative to the middle section. 
 
 i Number of 
 ' station. 
 
 ] 
 
 Area of 
 section. 
 
 Simpson's 
 multipliers. 
 
 Functions of 
 area. 
 
 Number of 
 
 intervals 
 
 from middle. 
 
 Product for 
 moment. 
 
 1 I 
 1 ^ 
 
 \ 3 
 4 
 
 0-2 
 
 22-7 
 48-8 
 
 73-2 
 88-4 
 
 82-8 
 
 587 
 26-2 
 
 3-9 
 
 I 
 
 4 
 
 2 , 
 
 4 
 
 0-2 
 
 90-8 
 
 97-6 
 
 292-8 
 
 4 
 3 
 2 
 I 
 
 
 
 I 
 2 
 
 3 
 4 
 
 0-8 
 272-4 
 195-2 
 292-8 
 
 5 
 
 2 1 176-8 
 
 761-2 
 
 6 
 
 7 
 8 
 
 9 
 
 4 
 2 
 
 4 
 I 
 
 331 "2 
 117-4 
 104-8 
 
 3-9 
 
 331-2 
 234-8 
 
 314-4 
 15-6 
 
 Function of displacement 1215-5 
 
 Function of 
 moment 
 
 \ 896-1 
 
 Volume of displacement = 1215-5 x 3° 
 
 excess of products aft = 896-0 — 761-2 = 134-8 
 moment aft = 134-8 x 3' X 10 
 
 C.B. abaft middle 
 
 _ 134-8 X ^ X 10 
 
 ~ 1215-5 X 'i? 
 
 134-8 X 10 , , 
 
 _ J-?_.._ =1-11 feet 
 
 1215-5 
 
Moments, Centre of Gravity, Centre of Buoyancy, etc. 63 
 
 The centre of gravity of a plane area is fully determined 
 when we know its position relative to two lines in the plane, 
 which are generally taken at right angles to one another. The 
 centre of gravity of a volume is fully determined when we 
 know its position relative to three planes, which are generally 
 taken at right angles to one another. In the case of the under- 
 water volume of a ship, we need only calculate the position of 
 its centre of gravity relative to (i) the load water-plane, and 
 (2) an athwartship section (usually the section amidships), 
 because, the two sides of the ship being identical, the centre 
 of gravity of the displacement must lie in the middle-line 
 longitudinal plane of the ship. 
 
 Approximate Position of the Centre of Buoyancy. — 
 In vessels of ordinary form, it is found that the distance of the 
 centre of buoyancy below the L.W.L. varies from about ^ to ^ 
 of the mean moulded draught, the latter being the case in 
 vessels of full form. For yachts and vessels of unusual form, 
 such a rule as this cannot be employed. 
 
 Example. — A vessel 13' 3" mean draught has her C.B. 5'34 feet below 
 L.W.L. 
 
 Here the proportion of the draught is — 
 
 ^^ =0-403 = -— 
 
 1 3 '25 
 This is an example of a fine vessel. 
 
 Example. — A vessel 27' 6" mean draught has her C.B. 12-02 feet below 
 L.W.L. 
 
 Here the proportion of the draught is — 
 
 12-02 _ 875 
 27-5 ~ 20 
 
 This is an example of a fuller vessel than the first case. 
 
 Normand's Approximate Formula for the Distance 
 of the Centre of Buoyancy below the Load Water- 
 line.^ 
 
 Let V — volume of displacement up to the load-line in 
 cubic feet ; 
 A = the area of the load water-plane in square feet ; 
 d = the mean draught (to top of keel) in feet. 
 
 ' See a paper in Transactions of the Institution of Naval Architects, b/ 
 Mr. S. W. F. Morrish, M.I.N. A., in 1892. 
 
64 Theoretical Naval Architecture. 
 
 Then centre of buoyancy below L.W.L. = i I „ "f" a y 
 
 This rule gives exceedingly good results for vessels of 
 ordinary form. In the early stages of a design the above 
 particulars would be known as some of the elements of the 
 design, and so the vertical position of the centre of buoyancy 
 can be located very nearly indeed. In cases in which the 
 stability of the vessel has to be approximated to, it is important 
 to know where the C.B. is, as will be seen later when we are 
 dealing with the question of stabiUty. 
 
 The rule is based upon a very ingenious assumption, and 
 the proof is given in Appendix A, p. 249. 
 
 The Area of a Curve of Displaeement divided by 
 the Load Displacement gives the Distance of the 
 Centre of Buoyancy below the Load Water-line. — This 
 is an interesting property of the curve of displacement. A 
 demonstration of it will be found in Appendix A, p. 247. 
 
 Displacement Sheet. — We now proceed to investigate 
 the method that is very generally employed in practice to find 
 the displacement of a vessel, and also the position of its centre 
 of buoyancy both in a longitudinal and a vertical direction. 
 The calculation is performed on what is termed a " Displace- 
 ment Sheet" or ^"^ Displacement Tabled' and a specimen calcula- 
 tion is given at the end of the book for a single-screw tug of 
 the following dimensions : — 
 
 Length between perpendiculars 
 
 75' 
 
 0" 
 
 Breadth moulded 
 
 14' 
 
 6" 
 
 Depth moulded 
 
 8' 
 
 3" 
 
 Draught moulded forward 
 
 5' 
 
 S" 
 
 aft 
 
 6' 
 
 2" 
 
 ,, ,, mean 
 
 5' 
 
 <A 
 
 The sheer drawing of the vessel is given on Plate I. This 
 drawing consists of three portions — the body plan, the half- 
 breadth plan, and the sheer. The sheer plan shows the ship 
 in side elevation, the load water-line being horizontal, and the 
 keel, in this case, sloping down from forward to aft. The ship 
 is supposed cut by a number of transverse vertical planes, 
 which are shown in the sheer plan as straight lines, numbered 
 
Moments, Centre of Gravity, Centre of Buoyancy, etc. 65 
 
 I, 2, 3, etc. Now, each of these transverse sections of the ship 
 has a definite shape, and the form of each half-section to the 
 outside of frames is shown in the body-plan, the sections being 
 numbered as in the sheer. The sections of the forward end 
 form what is termed the ^^ fore-body" and those of the after 
 end the " after-body." Again, the ship may be supposed to be 
 cut by a series of equidistant horizontal planes, of which the 
 load water-plane is one. The shape of the curve traced on each 
 of these planes by the moulded surface of the ship is given in 
 the half-breadth plan, and the curves are numbered A, i, 2, 3, 
 etc., to agree with the corresponding lines in the sheer and 
 body plan. Each of these plans must agree with the other 
 two. Take a special station, for example. No. 4. The breadth 
 of the ship at No. 4 station at the level of No. 3 water-plane is 
 Od in the body-plan, but it is also given in the half-breadth plan 
 by Oa, and therefore Oa must exactly equal Od. The process 
 of making all such points correspond exactly is known as 
 "fairing." For full information as to the methods adopted in 
 fairing, the student is referred to the works on " Laying-off " 
 given below.^ For purposes of reference, the dimensions of 
 the vessel and other particulars are placed at the top of the 
 displacement sheet. The water-lines are arranged on the 
 sheer drawing with a view to this calculation, and in this case 
 are spaced at an equidistant spacing apart of i foot, with an 
 intermediate water-line between Nos. 5 and 6. The number 
 of water-lines is such that Simpson's first rule can be used, and 
 the multipliers are, commencing with the load water-plane— 
 
 I 4 2 4 i| 2 i 
 
 The close spacing near the bottom is very necessary to 
 ensure accuracy, as the curvature of the midship sections of the 
 vessel is very sharp as the bottom is approached, and, as we 
 saw on p. 13, Simpson's rules cannot accurately deal with areas 
 such as these unless intermediate ordinates are introduced. 
 Below No. 6 water-plane there is a volume the depth of which 
 increases as we go aft, and the sections of this volume are very 
 
 ' " Laying Off," by Mr, S, J. P, Thearle j " Laying Off," by Mr. T. H, 
 Watson, 
 
 F 
 
66 Theoretical Naval Architecture. 
 
 nearly triangles. This volume is dealt with separately on the 
 left-hand side of the table, and is termed an " appendage!' 
 
 In order to find the volume of displacement between water- 
 planes I and 6, we can first determine the areas of the water- 
 planes, and then put these areas through Simpson's rule. To 
 find the area of any of the water-planes, we must proceed in 
 the ordinary manner and divide its length by ordinates so that 
 Simpson's rule (preferably the first rule) can be used. In the 
 case before us, the length is from the after-edge of the stem to 
 the forward edge of the body post, viz. 71 feet, and this 
 length is divided into ten equal parts, giving ordinates to each 
 of the water-planes at a distance apart of 7'i feet. The dis- 
 placement-sheet is arranged so that we can put the lengths of 
 the semi-ordinates of the water-planes in the columns headed 
 respectively L.W.L., 2W.L., 3W.L., etc., the semi-ordinates at 
 the several stations being placed in the same line as the number 
 of ordinates given at the extreme left of the table. The lengths 
 of the semi-ordinates are shown in italics. Thus, for instance, 
 the lengths of the semi-ordinates of No. 3 W.L., as measured 
 off, are 0-05, 1-82, 4-05, 5-90, 6-90, 7-25, 7-04, 6-51, 5-35, 
 2'85, and o'os feet, commencing with the forward ordinate 
 No. I, and these are put down in italics ^ as shown beneath 
 the heading 3 W.L. in the table. The columns under the 
 heading of each W.L. are divided into two, the semi-ordinates 
 being placed in the first column. In the second column of each 
 water-line is placed the product obtained by multiplying the 
 semi-ordinate by the corresponding multiplier to find the area. 
 These multipUers are placed at column 2 at the left, opposite 
 the numbers of the ordinates. We have, therefore, under the 
 heading of each water-line what we have termed the "funciions 
 of 01-dinaies," and if these functions are added up, we shall 
 obtain what we have termed the "function of area." 
 
 Taking No. 3 W.L. as an instance, the '^function " of its 
 area is i44'io, and to convert this "function" into the actual 
 area, we must multiply by one-third the common interval to 
 complete Simpson's first rule, i.e. by ^ x 7*1 j and also by 2 
 
 * In practice, it is advisable to put down the lengths of the semi- 
 ordinates in some distinctive colour, such as red. 
 
Moments, Centre of Gravity, Centre of Buoyancy, etc. 67 
 
 to obtain the area of the water-plane on both sides of the ship. 
 We should thus obtain the area of No. 3 W.L. — 
 
 i44"io X -^ X 7'i X 2 = 682"o7 square feet 
 
 The functions of the area of each water-plane are placed at 
 the bottom of the columns, the figures being, starting with 
 the L.W.L., 16370, 155-36, i44"io, 12874, ioS"67, 87-27, 
 and 60-97 . To get the actual areas of each of the water-planes, 
 we should, as above, multiply each of these functions by |- x 
 7-1 X 2. Having the areas, we could proceed as on p. 20 to 
 find the volume of displacement between No. i and No. 6 
 water-lines, but we do not proceed quite in this way; we 
 put the "functions of areas'' through Simpson's rule, and 
 multiply afterwards by |- X 7-1 X 2, the same result being 
 obtained with much less work. Below the " functions of 
 areas " are placed the Simpson's multipliers, and the products 
 163-70, 621-44, etc., are obtained. These products added up 
 give i95i'83. This number is a function of the volume of 
 displacement, this volume being given by first multiplying it 
 by one-third the vertical interval, i.e. -g- X i ; and then by 
 i X 7-1 X 2, as seen above. The volume of displacement 
 between No. i W.L. and No. 6 W.L. is therefore — 
 
 i95r83 X (I X i) X (^ X 7-1) X 2 = 3079-5 cubic feet 
 
 and the displacement in 1 3079-5 
 
 / u ^ \i f = ^—- = 87-98 tons 
 tons (salt water) ^ I 35 ' ■' 
 
 We have thus found the displacement by dividing the 
 volume under water by a series of equidistant horizontal planes ; 
 but we could also find the displacement by dividing the under- 
 water volume by a series of equidistant vertical planes, as we 
 saw in Chapter I. This is done on the displacement sheet, 
 an excellent check being thus obtained on the accuracy of the 
 work. Take No. 4 section, for instance : its semi-ordinates, 
 commencing with the L.W.L., are 6-40, 6*24, 5-90, 5-32, 4-30, 
 3-40, and 2-25 feet. These ordinates are already put down 
 opposite No. 4 ordinate. If these are multiplied successively 
 by the multipliers, i, 4, 2, 4, i\, 2, ^ and the sum of the 
 
 ' Thirty-five cubic feet of salt water taken to weigh one ton. 
 
68 Theoretical Naval Architecture. 
 
 functions of ordinates taken, we shall obtain the "■function of 
 area" of No. 4 section between the L.W.L. and 6 W.L. This 
 is done in the table by placing the functions of ordinates 
 immediately below the corresponding ordinate, the multiplier 
 being given at the head of each column. We thus obtain a 
 series of horizontal rows, and these rows are added up, the 
 results being placed in the column headed " Function of areas." 
 Each of these functions multiplied by one-third the common 
 interval, i.e. -j X i, and then by 2 for both sides, will give 
 the areas of the transverse sections between the L.W.L. and 
 6 W. L. ; but, as before, this multiplication is left till the end 
 of the calculation. These functions of areas 'are put through 
 Simpson's multipliers, the products being placed in the column 
 headed "-^ Multiples of areas'' This column is added up, giving 
 the result i95i"83. To obtain the volume of displacement, 
 we multiply this by (|- X i) X 2 x (|^ X 7'i). It will be noticed 
 that we obtain the number 1951 "83 by using the horizontal 
 water-lines and the vertical sections ; and this must evidently 
 be the case, because the displacement by either method must 
 be the same. The correspondence of these additions forms 
 the check, spoken of above, of the accuracy of the work. We 
 thus have the result that the volume of displacement from 
 L.W.L. to 6 W.L. is 3079*5 cubic feet, and the displacement 
 in tons of this portion 87'98 tons in salt water. This is termed 
 the " Main solid," and forms by far the greater portion of the 
 displacement. 
 
 We now have to consider the portion we have left out below 
 No. 6 water-plane. Such a volume as this is termed an 
 " appendage." The sections of this appendage are given in the 
 body-plan at the several stations. The form of these sections 
 are traced off, and by the ordinary rules their areas are found 
 in square feet. We have, therefore, this volume divided by, a 
 series of equidistant planes the same as the main solid, and we 
 can put the areas of the sections through Simpson's rule and 
 obtain the volume. This calculation is done on the left-hand 
 side of the sheet, the areas being placed in column 3, and the . 
 functions of the areas in column 4. The addition of these 
 functions is 49*99, and this multiplied by -^ x 7'i gives the 
 
Moments, Centre of Gravity, Centre of Buoyancy, etc. 6g 
 
 volume of the appendage in cubic feet, viz. 118-3; and this 
 volume divided by 35 gives the number of tons the appendage 
 displaces in salt water, viz. 3-38 tons. The total displacement 
 is thus obtained by adding together the main solid and the 
 appendage, giving 91-36 tons in salt water. The displacement 
 in fresh water (36 cubic feet to the ton) would be 88-8 tons. 
 
 The sheer drawing for this vessel as given on Plate I. was 
 drawn to the frame line, i.e. to the moulded dimensions of the 
 ship ; but the actual ship is fuller than this, because of the outer 
 bottom plating, and this plating will contribute a small amount 
 to the displacement, but this is often neglected. Some sheer 
 drawings, on the other hand, are drawn so that the lines include 
 a mean thickness of plating outside the frame hne, and when 
 this is the case, the displacement sheet gives the actual dis- 
 placement, including the effect of the plating. For a sheathed 
 ship this is also true; in this latter case, the displacement 
 given by the sheathing would be too great to be neglected. 
 When the sheer drawing is drawn to the outside of sheathing, 
 or to a mean thickness of plating, it is evident that the ship 
 must be laid off on the mould loft floor, so that, when built, 
 she shall have the form given by the sheer drawing. 
 
 We now have to find the position of the centre of buoyancy 
 both in a fore-and-aft and in a vertical direction. (It must be 
 in the middle-line plane of the ship, since both sides are sym- 
 metrical.) Take first the fore-and-aft position. This is found 
 with reference to No. 6 station. The functions of the areas of 
 the sections are 0-55, 23-055, etc., and in the column headed 
 " Multiples of areas " we have these functions put through 
 Simpson's multipliers. We now multiply these multiples by 
 the number of intervals they respectively are from No. 6 station, 
 viz. 5, 4, etc., and thus obtain a column headed " Moments." 
 This column is added up for the fore body, giving 1505-43, and 
 for the after body, giving 1913-02, the difference being 407-59 
 in favour of the after body. To get the actual moment of the 
 volume abaft No. 6 station, we should multiply this difference 
 by (|- X i) for the vertical direction, (|- X 7-1) for the fore-and- 
 aft direction, and by 2 for both sides, and then by 7-1, since we 
 have only multiplied by the number of intervals away, and not 
 
70 Theoretical Naval Architecture. 
 
 by the actual distances, or 407-59 X (I- X i) X (i X 7"i) X 2 
 X 7' I. The volume, as we have seen above, is given by — 
 
 1951-83 X (i X i) X (i X 7-1) X 2 
 
 The distance of the centre of gravity of the main solid from 
 No. 6 station will be — 
 
 Moment 4- volume 
 
 But on putting this down we shall see that we can cancel out, 
 leaving us with — 
 
 407-59 X 7-1 or. 
 
 ^ > ^^ ' = 1-48 feet 
 1951-83 
 
 which is the distance of the centre of gravity of the main solid 
 abaft No. 6 station. The distance of the centre of gravity of 
 the appendage abaft No. 6 station is 4-0 feet ; the working is 
 shown on the left-hand side of the table, and requires no further 
 explanation. These results for the main solid and for the 
 appendage are combined together at the bottom ; the displace- 
 ment of each in tons is multiplied by the distance of its centre 
 of gravity abaft No. 6 station, giving the moments. The total 
 moment is i43'73, and the total displacement is 91-36 tons, 
 and this gives the centre of gravity of the total displacement, or 
 what we term the centre of buoyancy, C.B., 1-57 feet abaft No. 
 6 station. 
 
 Now we have to consider the vertical position of the 
 C.B., and this is determined with reference to the load water- 
 line. For the main solid the process is precisely similar 
 to that adopted for finding the horizontal position, with the 
 exception that we take our moments all below the load water- 
 plane, the number of intervals being small compared with the 
 horizontal intervals. We obtain, as indicated on the sheet, the 
 centre of gravity of the main solid at a distance of 2-21 feet 
 below the L.W.L. For the appendage, we proceed as shown 
 on the left-hand side of the sheet. When finding the areas of 
 the sections of the appendage, we spot off as nearly as possible 
 the centre of gravity of each section, and measure its distance 
 below No. 6 W.L. If the sections happen to be triangles, this will, 
 of course, be one-third the depth. These distances are placed in 
 
Moments, Centre of Gravity, Centre of Buoyancy, etc. 71 
 
 a column as shown, and the " functions of areas " are respec- 
 tively multiplied by them, e.g. for No. 4 station the function of 
 the area is 5*92, and this is multiplied by o'2 2, the distance 
 of the centre of gravity of the section of the appendage below 
 No. 6 W.L. We thus obtain a column which, added up, gives 
 a total of 1378. To get the actual moment, we only have to 
 multiply this by I- X 7'i. The volume of the appendage is 
 49'99 X (^ X 7'i). So that the distance of the centre of 
 gravity of the whole appendage below No. 6 W.L. is given 
 
 by moment H- volume, or = o'27 feet, and therefore the 
 
 49-99 
 
 centre of gravity of the appendage is 5*27 feet below the 
 L.W.L. The results for the main solid and for the appendage 
 are combined together in the table at the bottom, giving the 
 final position of the C.B. of the whole displacement as 2'32 
 feet below the L.W.L. 
 
 It will be of interest at this stage to test the two approxi- 
 mations that were given on p. 63 for the distance of the C.B. 
 below the L.W.L. The first was that this distance would be 
 from -^ to -^ of the mean draught to top of keel (z'.^. the mean 
 moulded draught). For this vessel the distance is 2"32 feet, 
 and the mean moulded draught is 5' 9-|", or 5-8 feet, and so 
 
 2'32 
 
 we have the ratio — ^q-, or exactly ^. The second approxi- 
 mation (Normand's), p. 63, was — 
 
 *(M) 
 
 All these are readily obtainable from the displacement sheet, 
 and if worked out its value is found to be 2*29 feet. This 
 agrees well with the actual result, 2-32 feet, the error being 
 3 in 232, or less than i-| per cent. 
 
 For large vessels a precisely similar displacement-sheet is 
 prepared, but it is usual to add in the effect of other appen- 
 dages besides that below the lowest W.L. A specimen calcu- 
 lation is shown below. In this case the sheer drawing was 
 made to include a mean thickness of plating. The appendages 
 are — 
 
72 
 
 Theoretical Naval Architecture. 
 
 Before fore perpendicular (ram bow). 
 
 Abaft after perpendicular. 
 
 Rudder. 
 
 Shaft-tubes, etc. (including propellers, shafts, swells, and 
 struts). 
 
 Bilge keels (if fitted). 
 The effect of these appendages, outside the naked hull, is to 
 increase the displacement by 6i'5 tons, and to throw the C.B. 
 aft from 8-44 feet to 8'88 feet abaft the middle ordinate. The 
 effect on the vertical position of the C.B. is of very small 
 amount. 
 
 Summary. 
 
 Item. 
 
 Displace- 
 ment 
 in tons. 
 
 
 
 From Middle Ordinate. 
 
 
 Forward, 
 
 Aft. 
 
 Lever. 
 
 Moment. 
 
 Lever. 
 
 Moment. 
 
 Lever. 
 
 Moment. 
 
 Main portion 
 Below 8 W.L. 
 Abaft A. P. 
 Before F.P. 
 Rudder 
 Shaft-tubes, etc. 
 
 I3,319'6 
 
 92 1 -0 
 
 34-1 
 
 I2'3 
 
 4-0 
 
 II-2 
 
 II'II 
 
 25-47 
 3-60 
 
 13-77 
 
 1 8 -60 
 1 6 -So 
 
 147,981 
 23,458 
 
 ^6^ 
 
 74 
 188 
 
 192-80 
 
 2,352 
 
 8-58 
 
 645 
 
 195-05 
 
 196-30 
 i49'oo 
 
 114,282 
 5,940 
 6,651 
 
 785 
 1,679 
 
 14,302-1 
 
 2,352 
 
 1 2 02 feet 
 below L. W.L. 
 
 I29,3.;7 
 2,352 
 
 ) 126,985 
 
 8-88 feet 
 aft. 
 
 The total displacement up to the L. W.L. is 14,302 tons. The centre 
 of buoyancy is 12-02 feet below the L.W.L. and 8-88 feet abaft the middle 
 ordinate. 
 
 Graphic or Geometrical Method of calculating 
 Displacement and Position of Centre of Buoyancy. — 
 
 There is one property of the curve, known as the ^^ parabola of 
 the second order" (see p. 6), that can be used in calculating 
 by a graphic method the area of a figure bounded by such a 
 
Moments, Centre of Gravity, Centre of Buoyancy, etc. 73 
 
 B. 
 
 curve. Let BFC, Fig. 37, be a curve bounding the figure 
 ABCD, and suppose the curve is a "parabola of the second 
 order!' Draw the ordinate EF 
 midway between AB and DC ; then 
 the following is a property of the 
 curve BFC : — the area of the seg- 
 ment BCF is given by two-thirds 
 the product of the deflection GF 
 and the base AD, or — 
 
 Area BCF = f X GF X AD 
 
 Make GH = | GF. Then- 
 
 Area BCF = GH X AD 
 
 E. 
 Fig. 37. 
 
 Now, the area of the trapezoid ABCD is given by 
 AD X EG, and consequently — 
 
 The area ADCFB = AD x EH > 
 
 Thus, if we have a long curvilinear area, we can divide it up 
 as for Simpson's first rule, and set off" on each of the inter- 
 mediate ordinates two-thirds the deflection of the curve above 
 or below the straight line joining the extremities of the dividing 
 ordinates. Then add together on a strip of paper all such 
 distances as EH right along, and the sum multiplied by the 
 
 ' This property may be used to prove the rule known as Simpson's 
 first rule. Call AB, EF, DC respectively ji, _j'j, jj. Then we have — 
 
 EG =^^L±ii3 and FG = y^ - EG 
 
 . FG =y, - 
 EH = EG + GH 
 
 _ / >! +J'3 \ _|_ /2j/, -j/i -yA 
 
 = iiyi + 4^2 + n) 
 
 and calling AE = /i, we have — 
 
 Area ADCFB = ^(y, + 4y, +y,) 
 which is the same expression as given by Simpson's first rule. 
 
74 
 
 Theoretical Naval Architecture. 
 
 distance apart of the dividing ordinates, as AD, will give the 
 area required. Thus in Fig. 38, AB is divided into equal parts 
 
 ^-^:::i 
 
 K 
 
 
 
 ^^^E 
 
 
 
 N;^ 
 
 
 
 ;^fe;. 
 
 G. F 
 
 Fig. 38. 
 
 L. 
 
 as shown. D and E are joined, also E and C ; MO is set 
 off = f HM, and NP is set off = f NK. Then- 
 Area ADEF = AF X GO 
 and area FECB = FB X LP 
 and the whole area ABCD = AF X (GO + LP) 
 
 ^^^e can represent the area ABCD by a length equal to 
 GO + LP on a convenient scale, if we remember that this length 
 has to be multiplied by AF to get the area. This principle can 
 be extended to finding the areas of longer figures, such as 
 water-planes, and we now proceed to show how the displacement 
 and centre of buoyancy of a ship can be determined by its use. 
 The assumption we made at starting is supposed to hold good 
 with all the curves we have to deal, i.e. that the portions 
 between the ordinates are supposed to be '■'■parabolas of tlu 
 second order!' This is also the assumption we make when 
 using Simpson's first rule for finding displacement in the ordi- 
 nary way. 
 
 Plate I. represents the ordinary sheer drawing of a vessel, 
 and the underwater portion is divided by the level water-planes 
 shown by the half-breadth plan. The areas of each of these 
 planes can be determined graphically as above described, the 
 area being represented by a certain length obtained by the 
 addition of all such lengths as GO, etc.. Fig. 38, the interval 
 
Moments, Centre of Gravity, Centre of Buoyancy, etc. 75 
 
 being constant for all the water-planes. Let AB, Fig. 39, be 
 set vertically to represent the extreme moulded draught of the 
 vessel. Draw BC at right angles to AB, to represent on a 
 convenient scale the area of the L.W.L. obtained as above. 
 Similarly, DE, FG are set out to represent on the same scale 
 the areas of water-planes 2 and 3, and so on for each water- 
 plane. A curve drawn through all such points as C, E, and G 
 
 L.W.L. 
 
 R. C. L. 
 
 
 
 B 
 
 s\ \e. 
 
 V 
 
 3^ 
 
 D. 
 
 
 > 
 
 V 
 
 F 
 
 X \ 
 
 Im. 
 
 ^ 
 
 
 \ 
 
 V \\\ 
 
 2.W.L. 
 
 3.W.L. 
 
 4.W.L. 
 
 5.W.L. 
 
 6.W.L. 
 
 Fig. 39. 
 
 will give a " curve of areas of water-planes'' Now, the area of this 
 curve up to the L.W.L. gives us the volume of displacement up 
 to the L.W.L., as we have seen in Chapter I., and we can readily 
 find the area of the figure ABCEG by the graphic method, and 
 this area will give us the displacement up to the L.W.L. 
 Similarly, the area of ADEG will give the displacement up to 
 2 W.L., and so on. Therefore set off BL to represent on a 
 convenient scale the area of the figure ABCE, DK on the 
 
^6 Theoretical Naval Architecture. 
 
 same scale to represent the area ADEG, and so on. Then 
 a curve drawn through all such points as L, K will give us a 
 " curve of displacement" and the ordinate of this curve at any 
 draught will give the displacement at that draught, BL being 
 the load displacement. 
 
 We now have to determine the distance of the centre 
 of buoyancy below the L.W.L., and to find this we must get 
 the moment of the displacement about the L.W.L. and 
 divide this by the volume of displacement below the L.W.L. 
 We now construct a curve, BPMA, such that the ordinate at 
 any draught represents the area of the water-plane at that 
 draught multiplied by the depth of the water-plane below the 
 L.W.L. Thus DP represents on a convenient scale the area 
 of No. 2 water-plane multiplied by DB, the distance below the 
 L.W.L. The ordinate of this curve at the L.W.L. must evi- 
 dently be zero. This curve is a curve of " moments of areas 
 of water-planes" about the L.W.L. The area of this curve up 
 to the L.W.L. will evidently be the moment of the load dis- 
 placement about the L.W.L., and thus the length BR is set out 
 to equal on a convenient scale the area of BPMA. Similarly, 
 DS is set out to represent, on the same scale, the area of 
 DPMA, and thus the moment of the displacement up to 2 W.L. 
 about the L.W.L. These areas are found graphically as in the 
 preceding cases. Thus a curve RSTA can be drawn in, and 
 BR -^ BL, or moment of load displacement about L.W.L. -f- 
 load displacement, gives us the depth of the centre of buoyancy 
 for the load displacement below the L.W.L. 
 
 Exactly the same course is pursued for finding the displace- 
 ment and the longitudinal position of the centre of buoyancy, 
 only in this case we use a curve of areas of transverse sections 
 instead of a curve of areas of water-planes, and we get the 
 moments of the transverse areas about the middle ordinate. 
 Fig. 40 gives the forms the various curves take for the fore 
 body. AA is the " curve of areas of transverse sections ; " BB 
 is the " curve of displacement " for the fore body, OB being the 
 displacement of the fore body. CC is the curve of " moments 
 of areas of transverse sections " about No. 6 ordinate ; DD is 
 the curve of " moment of displacement " about No. 6 ordinate. 
 
Moments, Centre of Gravity, Centre of Buoyancy, etc. yy 
 
 OD being the moment of the fore-body displacement about 
 No. 6 ordinate. Similar curves can be drawn for the after 
 body, and the difference of the moments of the fore and after 
 bodies divided by the load displacement will give the distance 
 
 Fig. 40, 
 
 of the centre of buoyancy forward or aft of No. 6 ordinate, as 
 the case may be. The total displacement must be the same as 
 found by the preceding method. 
 
 Method of finding Areas by Means of the Plani- 
 meter. — This instrument is frequently employed to find the 
 area of plane curvilinear figures, and thus the volume of dis- 
 placement of a vessel can be determined. One form of the 
 instrument is shown in diagram by Fig. 41. It is supported at 
 three places : first, by a weighted pin, which is fixed in position 
 by being pressed into the paper; second, by a wheel, which 
 actuates a circular horizontal disc, the wheel and disc both 
 being graduated ; and third, by a blunt pointer. The instru- 
 ment is placed on the drawing, the pin is fixed in a convenient 
 position, and the pointer is placed on a marked spot A on the 
 boundary of the curve of which the area is required. The 
 reading given by the wheel and disc is noted. On passing 
 round the boundary of the area with the pointer (the same way 
 as the hands of a clock) back to the starting-point, another 
 reading is obtained. The difference of the two readings is 
 proportional to the area of the figure, the multiplier required to 
 convert the difference into the area depending on the instru- 
 ment and on the scale to which the figure is drawn. Particu- 
 lars concerning the necessary multipliers are given with the 
 instrument ; but it is a good practice to pass round figures of 
 known area to get accustomed to its use. 
 
y8 Theoretical Naval Architecture. 
 
 By the use of the planimeter the vohime of displacement of 
 a vessel can very readily be determined. The body plan is 
 taken, and the L.W.L. is marked on. The pointer of the in- 
 strument is then passed round each section in turn, up to the 
 L.W.L., the readings being tabulated. If the differences of the 
 readings were each multiplied by the proper multiplier, we 
 should obtain the area of each of the transverse sections, and 
 so, by direct application of Simpson's rules, we should find the 
 
 Ayf Pointer. 
 
 required volume of displacement. Or we could put the actual 
 difference of readings through Simpson's multipliers, and 
 multiply at the end by the constant multiplier. 
 
 It is frequently the practice to shorten the process as 
 follows : The body-plan is arranged so that Simpson's first rale 
 will be used, i.e. an odd number of sections is employed. 
 The pointer is passed round the first and last sections, and 
 the reading is recorded. It is then passed round all the even 
 sections, 2, 4, 6, etc., and the reading is recorded. Finally, 
 it is passed round all the odd sections except the first and last, ' 
 viz. 3, 5, 7, etc., and the reading is put down. The differences 
 of the readings are found and put down in a column. The 
 
Moments, Centre of Gravity, Centre of Buoyancy, etc. yg 
 
 first difference is multiplied by i, the next difference is multi- 
 plied by 4, and the last by 2. The sum of these products is 
 then multiplied for Simpson's first rule, and then by the proper 
 multiplier for the instrument and scale used. The work can 
 conveniently be arranged thus : 
 
 Numbers of 
 sections. 
 
 Readings. 
 
 5,124 
 5.360 
 
 18,681 
 31.758 
 
 Differences 
 of readings. 
 
 Simpson's 
 multipliers. 
 
 Products, j 
 
 1 
 
 Initial reading 
 
 1, 21 
 
 2, 4. 6, 8, 10, 1 
 
 12, 14, 16, \ 
 18, 20 ... I 
 
 3, 5. 7, 9. II. 1 
 
 13. 15. 17. 
 19 ) 
 
 236 
 13.321 
 
 13.077 
 
 I 
 4 
 
 2 
 
 ! 
 236 
 53.284 
 
 26,154 
 
 79.674 
 
 The multiplier for the instrument and scale of the drawing 
 used and to complete the use of Simpson's first rule is -f ; so 
 that the volume of displacement is 79,674 X -f cubic feet, and 
 the displacement in tons is 79,674 X f X -^ = 2732 tons. 
 
 There are two things to be noticed in the use of the plani- 
 meter : first, it is not necessary to set the instrument to the 
 exact zero, which is somewhat troublesome to do ; and second, 
 the horizontal disc must be watched to see how many times 
 it makes the complete revolution, the complete revolution 
 meaning a reading of 10,000. 
 
 It is also possible to find the vertical position of the centre 
 of buoyancy by means of the planimeter. By the method 
 above described we can determine the displacement up to each 
 water-line in succession, and so draw in on a convenient scale 
 the ordinary curve of displacement. Now we can run round 
 this curve with the planimeter and find its area. This area 
 divided by the top ordinate {i.e. the load displacement) will 
 give the distance of the centre of buoyancy below the load-line 
 (see p. 64). 
 
 To find the centre of buoyancy in a fore-and-aft direction, 
 it is necessary to tabulate the differences for each section, 
 and treat these differences in precisely the same way as the 
 
8o 
 
 Theoretical Naval Architecture. 
 
 " functions of areas of vertical sections " are treated in the 
 ordinary displacement sheet. 
 
 Method of approximating to the Area of the 
 Wetted Surface ^ by " Kirk's " Analysis.— The ship is 
 assumed to be represented by a block model, shaped as 
 shown in Fig. 42, formed of a parallel middle body and a 
 
 Fig. 42. 
 
 tapered entrance and run which are taken as of equal length. 
 The depth of the model is equal to the mean draught, and the 
 length of the model is equal to the length of the vessel. The 
 breadth is not equal to the breadth of the vessel, but is equal 
 to area of immersed midship section -H mean draught. The 
 displacement of the model is made equal to that of the vessel. 
 We then have — 
 
 Volume of displace-1 _ -y 
 
 > — V , bay 
 
 ment ) 
 
 = AG X area of midship section 
 
 _ V 
 
 ~ area of midship section 
 
 .*. length of entrance! ,,,,,. V 
 
 \ = length of ship j — ^t-t^ : — 
 
 or run ) ° " area of midship section 
 
 B' X D 
 
 • The area of wetted surface can be closely approximated to by patting 
 a curve of girths (modified for the slope of the level lines, see p. 192) 
 through Simpson's rule. 
 
Moments, Centre of Gravity, Centre of Buoyancy, etc. 8 1 
 
 where L = length of ship ; 
 
 B' = breadth of model ; 
 
 D = mean draught. 
 Having found these particulars, the surface of the model 
 can be readily calculated. 
 
 Area of bottom = AG X B' 
 Area of both sides = 2(GH + 2AE) X mean draught 
 
 The surface of a model formed in this way approximates 
 very closely to the actual wetted surface of the vessel. It is 
 stated that in very fine ships the surface of the model exceeds 
 the actual wetted surface by about 8 per cent., for ordinary 
 steamers by about 3 per cent., and for full ships by 2 per cent. 
 
 By considering the above method, we may obtain an 
 approximate formula for the wetted surface — 
 
 Area of bottom = Y\ 
 
 Area of sides = 2L'D 
 
 where L' is the length along ADCB. Then — 
 
 V 
 Surface = 2L'D + — 
 
 This gives rather too great a result, as seen above ; and if 
 we take — 
 
 V 
 Surface = 2LD + r=- 
 
 we shall get the area of the wetted surface slightly in excess, 
 but this will allow for appendages, such as keels, etc. 
 
 Since V = /S . LBD, where k is the block coefficient of 
 displacement, we may write — 
 
 Surface = 2LD + /J . LB 
 
 Approximate Formulae for finding Wetted Surface. 
 
 — Mr. Denny gives the following formula for the area of 
 wetted surface : — 
 
82 Theoretical Naval Architecture. 
 
 V 
 ryLD + ^ 
 
 which is seen to be very nearly that obtained above. 
 
 Mr. Taylor, in his work on " Resistance and Propulsion 
 of Ships," gives the following formula : — 
 
 where W is the displacement in tons. 
 
 Approximate Method of determining the Mean 
 Wetted Girth of Ships, given by Mr. A. Blechynden, 
 M.I.N. A. {Transactions of Instittition of Naval Architects, 
 1888)— 
 
 Let M = midship wetted girth measured on midship section in 
 
 feet; 
 
 L = length between perpendiculars in feet ; 
 
 V = volume of displacement in cubic feet ; 
 
 S = area of midship section in square feet ; 
 
 D = moulded draught in feet ; 
 
 V 
 c = prismatic coefficient of fineness = j-^^t-f (see p. 30) 
 
 m = mean wetted girth in feet. 
 
 Then m = o'gS^M + 2(1 - c)D 
 
 Examples to Chapter II. 
 
 1. A ship lias the folic 
 
 20 tons 
 45 „ 
 15 .- 
 60 „ 
 
 40 » 
 
 30 .. 
 
 Show that these weights will have the same effect on the trim of the shijj 
 as a single weight of 210 tons placed I5f feet abaft amidships. 
 
 2. Six weights are placed on a drawing-board. The weights are 3, 4, 
 S, 6, 7, 8 lbs. respectively. Their respective distances from one edge are 
 5, 4J, 4, 3i, 3, 2 feet respectively, and from the edge at right angles, |, f, 
 I, i|, 2, 2I feet respectively. The drawing-board weighs 6 lbs., and is 
 6 feet long and 3 feet broad. Find the position where a single support 
 would need to be placed in order that the board should remain horizontal. 
 
 A71S. 3'27 feet from short edge, i -58 feet from long edge. 
 
 lowing 
 
 weights placed on 
 
 board : — 
 
 
 . 100 feet before 
 
 amidships 
 
 
 . 80 
 
 J) 
 
 
 • 40 
 
 ,, 
 
 
 50 feet abaft 
 
 )) 
 
 
 • 80 
 
 J) 
 
 
 .. no 
 
 »» 
 
Moments, Centre of Gravity, Centre of Buoyancy, etc. 83 
 
 3. An area bounded by a curve and a straight line is divided by ordinates 
 4 feet apart of the following lengths : o, 12-5, 14-3, 15-1, 15-5, 15-4, 14-8, 
 I4"0, o feet respectively. Find — 
 
 (1) Area in square feet. 
 
 (2) Position of centre of gravity relative to the first ordinate. 
 
 (3) Position of the centre of gravity relative to the base. 
 
 Ans. (I) 423 square feet; (2) 16-27 feet ; (3) 7-24 feet. 
 
 4. A triangle ABC has its base BC 15 feet long, and its height 25 
 feet. A line is drawn 10 feet from A parallel to the base, meeting AB and 
 AC in D and E. Find the distance of the centre of gravity of DBCE 
 from the apex. 
 
 Ans. 18-57 feet. 
 
 5. The semi-ordinates of a water-plane in feet, commencing from the 
 after end, are 5-2, 10-2, 14-4, 17-9, 20-6, 22-7, 24-3, 25-5, 26-2, 26-5, 
 26-6, 26-3, 25-4, 23-9, 21-8, 1-88, 15-4, 11-5, 7-2, 3-3, 2-2. The distance 
 apart is 15 feet. Find the area of the water-plane, and the position of the 
 centre of gravity in relation to the middle ordinate. 
 
 Ans. 11,176 square feet ; 10-15 feet abaft middle. 
 
 6. Find the area and transverse position of the centre of gravity of 
 "half" a water-line plane, the ordinates in feet being 0-5, 6, 12, 16, 12, 10, 
 and 0-5 respectively, the common interval being 15 feet. 
 
 Ans. 885 square feet ; 6-05 feet. 
 
 7. The areas of sections 17' 6" apart through a bunker, commencing 
 from forward, are 65, 98, 123, 137, 135, 122, 96 square feet respectively. 
 The length of bunker is 100 feet, and its fore end is i' 6" forward of the 
 section whose area is 65 square feet. Draw in a curve of sectional areas, 
 and obtain, by using convenient ordinates, the number of cubic feet in the 
 bunker, and the number of tons of coal it will contain, assuming that 43 cubic 
 feet of coal weigh i ton. Find also the position of the C.G. of the coal 
 relative to the after end of the bunker. 
 
 Ans. 272 tons ; 464 feet from the after end. 
 
 8. The tons per inch in salt water of a vessel at water-lines 3 feet 
 apart, commencing with the L.W.L., are 31-2, 30-0, 28-35, z6-2i, 23-38, 
 19-5, 12-9. Find the displacement in salt and fresh water and the position 
 of the C.B. below the L.W.L., neglecting the portion below the lowest 
 W.L. Draw in the tons per inch curve for salt water to a convenient scale, 
 and estimate from it the weight necessary to be taken out in order to lighten 
 the vessel 2' 3I" from the L.W.L. The mean draught is 20' 6". 
 
 Ans. S405 tons ; 5255 tons ; 8-01 feet ; 847 tons. 
 
 9. In the preceding question, calling the L.W.L. i, find the displacement 
 up to 2 W.L., 3 W.L., and 4 W.L., and draw in a curve of displacement 
 from the results you obtain, and check your answer to the latter part of the 
 question. 
 
 10. The tons per inch of a ship's displacement at water-lines 4 feet 
 apart, commencing at the L.W.L., are 44-3, 42-7, 40-5, 37-5, 33-3. Find 
 number of tons displacement, and the depth of C.B. below the top W.L. 
 
 Ans. 7670 tons j 7-6 feet. 
 
 11. The ship in the previous question has two water-tight transverse 
 bulkheads 38 feet apart amidship, and water-tight flats at 4 feet below and 
 3 feet above the normal L.W.L. If a hole is made in the side 2 feet 
 below the L.W.L., how much would the vessel sink, taking the breadth 
 of the L.W.L. amidships as 70 feet? Indicate the steps where, owing to 
 insufficient information, you are unable to obtain a perfectly accurate result. 
 
 Ans. 8 inches. 
 
 12. The areas of transverse sections of a coal-bunker 19 feet apart are 
 
84 Theoretical Naval Architecture. 
 
 respectively 63-2, 93-6, iivd, loS'S, 94-8 square feet, and the centres 
 of gravity of these sections are lO'S, ii-6, 12-2, 117, II-2 feet respectively 
 below the L.W.L. Find the number of tons of coal the bunker will hold, 
 and the vertical position of its centre of gravity (44 cubic feet of coal to the 
 ton). 
 
 Ans. I74"3 tons ; il"68 feet below L.W.L. 
 
 13. A vessel is 180 feet long, and the transverse sections from the load 
 water-line to the keel are semicircles. Find the longitudinal position of 
 the centre of buoyancy, the ordinates of the load water-plane being i, 5, 13, 
 15, 14, 12, and 10 feet respectively. 
 
 Ans. \Qi(>"i feet from the finer end. 
 
 14. Estimate the distance of the centre of buoyancy of a vessel below 
 the L.W.L., the vessel having 22' 6" mean moulded draught, block co- 
 efficient of displacement 0-55, coefficient of fineness of L.W.L. 07 (use 
 Normand's formula, p. 63). 
 
 Ans. 9-65 feet. 
 
 15. A vessel of 2210 tons displacement, 13' 6" draught, and area 
 of load water-plane 8160 square feet, has the C.B., calculated on the dis- 
 placement sheet, at a distance of 5-43 feet below the L.W.L. Check this 
 result. 
 
 16. The main portion of the displacement of a vessel has been calculated 
 and found to be 10,466 tons, and its centre of gravity is 10-48 feet below 
 the L.W.L., and 5 '85 feet abaft the middle ordinate. In addition to this, 
 there are the following appendages : — 
 
 tons. 
 Below lowest W.L. 263, 24-8 ft. below L.W.L., 4-4 ft. abaft mid. ord. 
 Forward ... ... 5, i2'o ,, ,, 202 ft. forward of mid. 
 
 ord. 
 
 Stern i6, 2'8 ,, ,, 201 ft. abaft mid. ord. 
 
 Rudder 16, 17-5 ,, ,, 200 ,, „ 
 
 Bilge keels ... 20, 20 „ ,, o ,, ,, 
 
 Shafting, etc. ... 18, 15 ,, ,, 140 „ „ 
 
 Find the total displacement and position of the centre of buoyancy. 
 
 Ans. 10,804 tons ; C.B. 6'5 abaft mid. ord., 1086 ft. below L.W.L. 
 
 17. The displacements of a vessel up to water-planes 4 feet apart 
 are 10,804, 8612, 6511, 4550, 2810, 1331, and 263 tons respectively. The 
 draught is 26 feet. Find the distance of the centre of buoyancy below the 
 load water-line. Would you call the above a fine or a full ship ? 
 
 Ans. io'9 feet nearly. 
 
 18. The load displacement of a ship is 5000 tons, and the centre of 
 buoyancy is 10 feet below the load water-line. In the light condition the 
 displacement of the ship is 2000 tons, and the centre of gravity of the layer 
 between the load and light lines is 6 feet below the load-line. Find the 
 vertical position of the centre of buoyancy below the light line in the light 
 condition. 
 
 Ans. 4 feet, assuming that the C.G. of the layer is at half its depth. 
 
 19. Ascertain the displacement and position of the centre of buoyancy 
 of a floating body of length 140 feet, depth 10 feet, the forward section 
 being a triangle 10 feet wide at the deck and with its apex at the keel, and 
 the after section a trapezoid 20 feet wide at the deck and 10 feet wide at 
 the keel, the sides of the vessel being plane surfaces ; draught of water 
 may be taken as 7 feet. 
 
 Ans. 238 tons ; 56-3 feet before after end, 3 feet below water-line. 
 
 20. Show by experiment or otherwise that the centre of gravity of a 
 
Moments, Centre of Gravity, Centre of Buoyancy, etc. 85 
 
 quadrant of a circle 3 inches radius is i '8 inches from the right angle of 
 the quadrant. 
 
 21. A floating body has a constant triangular section, vertex down- 
 wards, and has a constant draught of 12 feet in fresh water, the breadth at 
 the water-line being 24 feet. The keel just touches a quantity of mud of 
 specific gravity 2. The water-level now falls 6 feet. How far will the 
 body sink into the mud ? 
 
 Ans. 4 feet iij inches. 
 
CHAPTER III. 
 
 CONDITIONS OF EQUILIBRIUM, TRANSVERSE META- 
 CENTRE, MOMENT OF INERTIA, TRANSVERSE BM, 
 INCLINING EXPERIMENT, METACENTRIC HEIGHT, 
 ETC. 
 
 Trigonometry. — The student of this subject will find it a 
 distinct advantage, especially when dealuig with the question 
 of stability, if he has a knowledge of some of the elementary 
 portions of trigonometry. The following are some properties 
 which should be thoroughly grasped : — 
 
 Circular Measure of Angles. — The degree is the unit gene- 
 rally employed for the measurement of angles. A right angle 
 
 is divided into 90 equal 
 parts, and each of these 
 parts is termed a ^^ de- 
 gree" If two lines, as 
 OA, OB, Fig. 43> are 
 inclined to each other, 
 forming the angle AOB, 
 and we draw at any radius 
 OA an arc AB from the 
 centre O, cutting OA, 
 ^"=- «• OB in A and B, then 
 
 length of arc AB -^- radius OA is termed the circular measure 
 of the angle AOB. Or, putting it more shortly — 
 
 Circular measure = 
 
 arc 
 
 radius 
 
 The circular measure of four right \ _ circumference of a circle 
 angles, or 360 degrees I ~ radius 
 
Conditions of Equilibrium, Transverse Metacentre, etc. 87 
 
 The circular measure of a right angle 
 
 Since 360 degrees = ztt in circular measure, then the angle 
 whose circular measure is unity is — 
 
 — = 57-3 degrees 
 
 The circular measure of i degree is —^ = °'°i745) and 
 
 thus the circular measure of any angle is found by multiplying 
 the number of degrees in it by o'oi74S. 
 
 Trigonometrical Ratios^ 
 etc. — Let BOC, Fig. 44, be 
 any angle ; take any point P 
 in one of the sides OC, and 
 draw PM perpendicular to 
 OB. Call the angle BOC, 6? 
 PM is termed the perpen- 
 dicular. 
 OM is termed the base. 
 OP is termed the hypo- 
 tenuse. 
 
 BASE. M. 
 
 Fig. 44. 
 
 Then— 
 
 PM perpendicular . . „ . . . 
 
 -pr^r = , ^ = Sine 6, usually written sin 6 
 
 OP hypotenuse ^ 
 
 OM base . . „ .^^ ^ 
 
 -=r=r = , = cosine 6, usually written cos 
 
 OP hypotenuse 
 
 ^^ _ P^'^P " _ tannnt 6, usually written tan 6 
 
 OM base ^ 
 
 These ratios will have the same value wherever P is taken 
 on the line OC. 
 
 ' An aid to memory which is found of assistance by many in learning 
 these ratios is — 
 
 Sin perplexes hypocxiVes 
 Cos of base hypocxlsy. 
 ' eis3. Greek letter (theta) often used to denote an angle. 
 
88 
 
 Theoretical Naval Architecture. 
 
 We can write sin 6 = i- — - 
 hyp. 
 
 n base 
 
 cos 6 = 
 
 hyp. 
 
 and also tan — -. 
 
 cos 
 
 There are names for the inversions of the above ratios, 
 which it is not proposed to use in this work. 
 
 For small angles, the value of the angle Q in circular 
 measure is very nearly the same as the values of sin 6 and 
 tan 6. This will be seen by comparing the values of 6, sin 0, 
 and tan 6 for the following angles : — 
 
 Angle in 
 degrees. 
 
 Angle in 
 circular 
 measure. 
 
 Sin D. 
 
 TanO. 
 
 2 
 
 4 
 6 
 8 
 
 10 
 
 0-0349 
 0-0698 
 0-1047 
 0-1396 
 0-I745 
 
 0-0349 
 0-0697 
 0-1045 
 0-1392 
 0-1736 
 
 0-0349 
 0-0699 
 O-I051 
 0-1405 
 0-1763 
 
 Up to 10° they have the same values to two places of 
 decimals, and for smaller angles the agreement in value is 
 closer still. 
 
 Further information can be obtained by reference to a 
 pocket-book, as " Mackrow's " or " Moles worth's," or an 
 elementary text-book on trigonometry. 
 
 Conditions that must hold in the Case of a Vessel 
 floating freely, and at Rest in Still Water. — We 
 saw in Chapter I. that, for a vessel floating in still water, 
 the weight of the ship with everything she has on board must 
 equal the weight of the displaced water. To demonstrate this, 
 we imagined the cavity left by the ship when lifted out of the 
 water to be filled with water (see Fig. 17). Now, the upward 
 support of the surrounding water must exactly balance the weight 
 of the water poured in. This weight may be regarded as acting 
 downwards through its centre of gravity, or, as we now term 
 it, the centre of buoyancy. Consequently, the upward support 
 
Conditions of Equilibrium, Transverse Metacentre, etc. 89 
 
 of the water, or the buoyancy, must act through the centre of 
 buoyancy. All the horizontal pressures of the water on the 
 surface of the ship must evidently balance among themselves. 
 We therefore have the following forces acting upon the ship : — 
 
 (i) The weight acting downwards through the C.G. ; 
 
 (2) The upward support of the water, or, as it is termed, 
 the buoyancy, acting upwards through the C.B. ; 
 and for the ship to be at rest, these two forces must act in the 
 same line and counteract each other. Consequently, we also 
 have the following condition : — 
 
 Tlie centre of gravity of the ship, with everything she has on 
 board, must be in the same vertical line as the centre of buoyancy. 
 
 If a rope is pulled at both ends by two men exerting the 
 same strength, the rope will evidently remain stationary; and 
 this is the case with a ship floating freely and at rest in still 
 water. She will have no tendency to move of herself so long 
 as the C.G. and the C.B. are in the same vertical line. 
 
 Definition of Statical Stability. — The statical 
 stability of a vessel is the tendency she has to return to the 
 upright when inclined away from that position. It is evident 
 that under ordinary conditions of service a vessel cannot 
 always remain upright ; she is continually being forced away 
 from the upright by external forces, such as the action of 
 the wind and the waves. It is very important that the ship 
 shall have such qualities that these inclinations that are forced 
 upon her shall not affect her safety; and it is the object of the 
 present chapter to discuss how these qualities can be secured 
 and made the subject of calculation so far as small angles of 
 inclination are concerned. 
 
 A ship is said to be in stable equilibrium for a given direc- 
 tion of inclination if, on being slightly inclined in that direction 
 from her position of rest, she tends to return to that position. 
 
 A ship is said to be in unstable equilibrium for a given 
 direction of inclination if, on being slightly inclined in that 
 direction from her position of rest, she tends to move away 
 farther from that position. 
 
 A ship is said to be in neutral or indifferent equilibrium 
 for a given direction of inclination if, on being slightly inclined 
 
90 
 
 Theoretical Naval Architecture. 
 
 in that direction from her position of rest, she neither tends 
 to return to nor move farther from that position. 
 
 These three cases are represented- by the case of a heavy 
 sphere placed upon a horizontal table. 
 
 1. If the sphere is weighted so that its C.G. is out of the 
 centre, and the C.G. is vertically below the centre, it will be 
 in stable equilibrium. 
 
 2. If the same sphere is placed so that its C.G. is vertically 
 above the centre, it will be in unstable equilibrium. 
 
 3. If the sphere is formed of homogeneous material so that 
 its C.G. is at the centre, it will be in neutral or indifferent 
 equilibrium. 
 
 Transverse Metacentre. — We shall deal first with 
 transverse inclinations, because they are the more important, 
 and deal with inclinations in a longitudinal or fore-and-aft 
 direction in the next chapter. 
 
 Let Fig. 45 represent the section of a ship steadily inclined 
 
 Stable. 
 
 7 Yw. 
 
 Fig. 45. 
 
 at a small angle from the upright by some external force, 
 such as the wind. The vessel has the same weight before and 
 after the inclination, and consequently has the same volume 
 of displacement. We must assume that no weights on board 
 shift, and consequently the centre of gravity remains in the 
 same position in the ship. But although the total volume of 
 
Conditions of Eqtnlibrimn, Transverse Metacentre, etc. g i 
 
 displacement remains the same, the shape of this volume 
 changes, and consequently the centre of buoyancy will shift 
 from its original position. In the figure the ship is repre- 
 sented by the section, WAL being the immersed section 
 when upright, WL being the position of the water-line on 
 the ship. On being inclined, W'L' becomes the water-line, 
 and WAL' represents the immersed volume of the ship, which, 
 although different in shape, must have the same volume as the 
 original immersed volume WAL. 
 
 The wedge-shaped volume represented by WSW, which 
 has come out of the water, is termed the " emerged" or " out" 
 wedge. The wedge-shaped volume represented by LSL', 
 which has gone into the water, is termed the ^^ immersed" or 
 " in " wedge. Since the ship retains the same volume of 
 displacement, it follows that the volume of the emerged wedge 
 WSW is equal to the volume of the immersed wedge LSL'. 
 It is only for small angles of inclination that the point S, 
 where the water-lines intersect, falls on the middle line of the 
 vessel. For larger angles it moves further out, as shown in 
 Fig. 77. 
 
 Now consider the vessel inclined at a small angle from 
 the upright, as in Fig. 45. The new volume of displacement 
 WAL' has its centre of buoyancy in a certain position, say B'. 
 This position might be calculated from the drawings in the 
 same manner as we found the point B, the original centre of 
 buoyancy ; but we shall see shortly how to fix the position of 
 the point B' much more easily. 
 
 B' being the new centre of buoyancy, the upward force of 
 the buoyancy must act through B', while the weight of the ship 
 acts vertically down through G, the centre of gravity of the 
 ship. Suppose the vertical through B' cuts the middle line of 
 the ship in M ; then we shall have two equal forces acting on 
 the ship, viz. — 
 
 (i) Weight acting vertically down through the centre of 
 gravity. 
 
 (2) Buoyancy acting vertically up through the new centre 
 of buoyancy. 
 But they do not act in the same vertical line. Such a system 
 
92 
 
 Theoretical Naval Architecture. 
 
 of forces is termed a couple. Draw GZ perpendicular to tlie 
 vertical through B'. Then the equal forces act at a distance 
 from each other of GZ. This distance is termed the arm of 
 the couple, and the moment of the couple is W X GZ. On 
 looking at the figure, it is seen that the couple is tending to 
 take the ship back to the upright. If the relative positions 
 of G and M were such that the couple acted as in Fig. 46, 
 the couple would tend to take the ship farther away from the 
 upright ; and again, if G and M coincided, we should have the 
 forces acting in the same vertical line, and consequently no 
 
 couple at all, and the ship would have no tendency to move 
 either to the upright or away from it. 
 
 We see, therefore, that for a ship to be in stable equilibrium 
 for any direction of inclination, it is necessary that the meta- 
 centre be above the centre of gravity of the ship. We now 
 group together the three conditions which must be fulfilled in 
 order that a ship may float freely and at rest in stable equili- 
 brium — 
 
 (i) The weight of water displaced must equal the total 
 weight of the ship (see p. 21). 
 
Conditions of Equilibrium, Transverse Metacentre, etc. 93 
 
 (2) The centre of gravity of the ship must be in the same 
 vertical line as the centre of gravity of the displaced water 
 (centre of buoyancy) (see p. 89). 
 
 (3) The centre of gravity of the ship must be below the 
 metacentre. 
 
 For small transverse inclinations, M is termed the transverse 
 metacentre, which we may accordingly define as follows : — 
 
 For a given plane of flotation of a vessel in the upright 
 condition, let B be the centre of buoyancy, and BM the vertical 
 through it. Suppose the vessel inclined transversely through 
 a very small angle, retaining the same volume of displacement, 
 B' being the new centre of buoyancy, and B'M the vertical 
 through it, meeting BM in M. Then this point of intersection, 
 M, is termed the transverse metacentre. 
 
 There are two things in this definition that should be noted : 
 (1) the angle of inclination is supposed very small, and (2) the 
 volume of displacement remains the same. 
 
 It is found that, for all practical purposes, in ordinary ships 
 the point M does not change in position for inclinations up to 
 as large as 10° to 15°; but beyond this it takes up different 
 positions. 
 
 We may now say, with reference to a ship's initial stability 
 or stability in the upright condition — 
 
 (i) If G is below M, the ship is in stable equilibrium. 
 
 (2) If G is above M, the ship is in unstable equilibrium. 
 
 (3) If G coincides with M, the ship is in neutral or in- 
 different equilibrium. 
 
 We thus see how important the relative positions of the 
 centre of gravity and the transverse metacentre are as affecting 
 a ship's initial stability. The distance GM is termed the 
 transverse metacentric height, or, more generally, simply the 
 metacentric height. 
 
 We have seen that for small angles M remains practically 
 in a constant position, and consequently we may say GZ 
 = GM . sin 6 for angles up to 10° to 15°, say. GZ is the 
 arm of the couple, and so we can say that the moment of the 
 couple is — 
 
 W X GM . sin 51 
 
94 Theoretical Naval ArchitecHire. 
 
 If M is above G, this moment tends to right the ship, and 
 we may therefore say that the moment of statical stability at the 
 angle Q is — 
 
 W X GM . sin e 
 
 This is termed the metacentric method of determining a 
 vessel's stability. It can only be used at small angles of 
 inclination to the upright, viz. up to from lo to 15 degrees. 
 
 Example. — A vessel of 14,000 tons displacement has a metacentric 
 height of 3I feet. Then, if she, is steadily inclined at an angle of 10°, the 
 tendency she has to return to the upright, or, as we have termed it, the 
 moment of statical stability, is — 
 
 14,000 X 3*5 X sin 10° = 8506 foot-tons 
 
 We shall discuss later how the distance between G and M, 
 or the metacentric height, influences the behaviour of a ship, 
 and what its value should be in various cases ; we must now 
 investigate the methods which are employed by naval archi- 
 tects to determine the distance for any given ship. 
 
 There are two things to be found, viz. (i) the position of 
 G, the centre of gravity of the vessel ; (2) the position of M, 
 the transverse metacentre. 
 
 Now, G depends solely upon the vertical distribution of the 
 weights forming the structure and lading of the ship, and the 
 methods employed to find its position we shall deal with 
 separately ; but M depends solely upon the form of the ship, 
 and its position can be determined when the geometrical form 
 of the underwater portion of the ship is known. Before we 
 proceed with the investigation of the rules necessary to do this, 
 we must consider certain geometrical principles which have Jto 
 be employed. 
 
 Centre of Flotation. — If a floating body is slightly 
 inclined so as to maintain the same volume of displacement, 
 the new water-plane must pass through the centre of gravity of 
 the original water-plane. In order that the same volume of 
 displacement may be retained, the volume of the immersed 
 wedge SLLi, Fig 47, must equal the volume of the emerged 
 wedge SWWi. Call y an ordinate on the immersed side, and 
 y' an ordinate on the emerged side of the water-plane. Then 
 
Conditions of Equilibrium, Transverse Metacentre, etc. 95 
 
 the areas of the sections of the immersed and emerged wedges 
 are respectively (since LLi =y.d0, WWi = y . dO, dO being 
 the small angle of inclination) — 
 
 tv' . do, ^,{y'f . d6 
 
 and using the notation we have already employed — 
 
 Volume of immersed wedge = i/y .dO.dx 
 „ emerged „ =ij(yf.de.dx 
 
 and accordingly — 
 
 yf.de.dx = y(y)\de.dx 
 
 or ^Jf.dx =\Ky)\dx 
 
 But \^y . dx is the moment of the immersed portion of the 
 water-plane about the intersection, 
 and ^J(yy • dx is the moment of 
 the emerged portion of the water- 
 plane about the intersection (see 
 p. 57); therefore the moment of 
 one side of the water-plane about 
 the intersection is the same as the 
 moment of the other side, and 
 consequently the line of inter- 
 section passes through the centre 
 of gravity of the water-plane. 
 The centre of gravity of the water- 
 plane is termed the centre of flota- 
 tion. In whatever direction a 
 ship is inclined, transversely, 
 longitudinally, or in any interme- 
 diate direction, through a small 
 angle, the line of intersection of 
 the new water-plane with the 
 original water-plane must always 
 pass through the centre of flotation. For transverse inclinations 
 of a ship the line of intersection is the centre line of the water- 
 plane ; for longitudinal inclinations the fore-and-aft position of 
 the centre of flotation has to be calculated, as we shall see 
 when we deal with longitudinal inclinations. 
 
 ■ Plan.- — 
 
 Fig. 47, 
 
96 
 
 Theoretical Naval Architecture. 
 
 Shift of the Centre of Gravity of a Figure due to 
 the Shift of a Portion of the Figure.— Let ABCD, 
 Fig. 48, be a square with its centre at G ; this point will also 
 be its centre of gravity. Suppose one corner of the square EF 
 
 is taken away and placed 
 in the position FK, forming 
 a new figure, ADKHGE. 
 We wish to find the centre 
 of gravity of this new figure. 
 The centre of gravity of the 
 original figure was at G, 
 and a portion of it, EF, 
 with its centre of gravity at 
 g, has been shifted so that 
 its centre of gravity now is at ^. Then this important 
 principle holds good — 
 
 The centre of gravity of the figure will shift to G', such that 
 GG' is parallel to^, and if A be the original area of the square, 
 and a be the area shifted— 
 
 Fig. 48. 
 
 In this case, if 2 15 be a side of the square— 
 
 A = 4^^ 
 « = 3' _ 
 gg' = b^2 
 
 If- X b^2 
 Ab' 
 
 and therefore GG' = 
 
 = 0-353^ 
 
 In the same way, g^' being the horizontal shift of the centre 
 of gravity of the corner EF, the horizontal shift of the centre 
 of gravity of the whole area will be given by — 
 
 GG" 
 
 a X gt 
 
 In this case gg" = b 
 and therefore GG" = \b 
 
 The same principle applies to the shift of the centre of 
 gravity of a volume or a weight due to the shift of a portion of 
 
Conditions of Equilibrium, Transverse Metacentre, etc. 97 
 
 it. The small portion multiplied by its shift is equal to the 
 whole body multiplied by its shift, and the shifts are in parallel 
 directions. 
 
 The uses that are made of this will become more apparent 
 as we proceed, but the following examples will serve as illus- 
 trations : — 
 
 Example. — A vessel weighing W tons has a weight w tons on the deck. 
 This is shifted transversely across the deck a distance of d feet, as in Fig. 49. 
 Find the shift of the C.G. of the vessel both in direction and amount. 
 
 Fig. 49. 
 
 G will move to G' such that GG' will be parallel to the line joining 
 the original and final positions of the weight w ; 
 
 and GG' = 
 
 WAd 
 W 
 
 If OT = 70 tons, </ = 30 feet, W = 5000 tons, then— 
 
 GG' = 22Ji3o _ .^ ^ggj _ Q.^2 jgpt 
 5000 
 Example.— in a vessel of 4000 tons displacement, suppose 100 tons of 
 coal to be shifted so that its C.G. moves 18 feet transversely and 44 feet 
 vertically. Find the shift of the C.G. of the vessel. 
 
 , ^ 100 X 18 , 
 
 The C.G. will move horizontally an amount equal to —^ o 45 it. 
 
 , 100 X 4'S ,,,-- r. 
 
 and vertically an amount equal to —^^^= ° "^5 it. 
 Moment of Inertia.— We have dealt in Chapter II. with 
 
98 
 
 Theoretical Naval Architecture. 
 
 the moment of a force about a given point, and we defined it as 
 the product of the force and the perpendicular distance of its 
 Hne of action from the point; also the moment of an area 
 about a given axis as being the area multiphed by the distance 
 of its centre of gravity from the axis. We could find the 
 moment of a large area about a given axis by dividing it into 
 a number of small areas and summing up the moments of all 
 these small areas about the axis. In this we notice that the 
 area or force is multiphed simply by the distance. Now we 
 have to go a step further, and imagine that each small area is 
 multiplied by the square of its distance from a given axis. If 
 all such products are added together for an area, we should 
 obtain not the simple moment, but what may be termed the 
 
 Fig. 50. 
 
 moment of the second degree, or more often the moment oj 
 ineiiia of the area about the given axis.* We therefore define 
 the moment of inertia of an area about a given axis as 
 follows : — 
 
 ' This is the geometrical moment of inertia. Strictly speaking, moment 
 of inertia involves the mass of the body. We make here the same assump- 
 tion that we did in simple moments (p. 47), viz. that the area is the 
 surface of a very thin lamina or plate of homogeneous material of uniform 
 thickness. 
 
Conditions of Equilibrium, Transverse Metacentre, etc. 99 
 
 Itnagine the area divided info very small areas, and each such 
 small area multiplied by the square of its distance from the given 
 axis ; then, if all tliese products be added together, we shall obtain 
 the moment of inertia of the total area about the given axis. 
 
 Thus in Fig. 50, let 00 be the axis. Take a very small 
 area, calling it dK, distance y from the axis. Then the sum 
 of all such products as dh. X f, or (using the notation we have 
 employed) j;f .dK, will be the moment of inertia of the area 
 about the axis 00. 
 
 To determine this for any figure requires the application of 
 advanced mathematics, but the result for certain regular figures 
 are given below. 
 
 It is found that we can always express the moment of 
 inertia, often written I, of a plane area about a given axis by 
 the expression — 
 
 nKh^ 
 
 where A is the area of the figure ; 
 
 h is the depth of the figure perpendicular to the axis ; 
 ji is a coefficient depending on the shape of the figure 
 and the position of the axis. 
 
 First, when the axis is through the centre of gravity of 
 the figure parallel to the base, as in Figs. 51 and 52 — 
 
 I N 
 
 Fig. si. 
 
 Fig. 52. 
 
 for a circle n = ^,^o that I = ^M 
 for a rectangle « = iV> » ^ = i^A/? 
 for a triangle n = xj, 
 
 I ^:^Ah"' 
 
100 Theoretical Naval Architecture. 
 
 Second, when the axis is one of the sides — 
 
 for a rectangle n = \, so that I = \ti.}i' 
 for a triangle n =\, „ I = \hJf' 
 
 Example. — Two squares of side a axe joined to form a rectangle. The 
 I of each square about the common side is — 
 
 ]{a^)a' {a^ = area) 
 
 the I of both about the common side will be the sum of each taken 
 separately, or — 
 
 If, however, we took the whole figure and treated it as a rectangle, its I 
 about the common side would be — 
 
 ^\{2a^)(2a)' = la* (area = 2a^) 
 
 which is the same result as was obtained before. 
 
 To find the moment of inertia of a plane figure about an axis 
 parallel to and a given distance from an axis through its centre 
 of gravity. 
 
 Suppose the moment of inertia about the axis NN passing 
 through the centre of gravity of the figure (Fig. 53) is I„, the 
 
 area of the figure is A, and 
 00, the given axis, is parallel 
 to NN and a distance y 
 from it. Then the moment 
 of inertia (I) of the figure 
 about 00 is given by — 
 
 I = I„ + A/ 
 The moment of inertia of an 
 area about any axis is there- 
 fore determined by adding 
 to the moment of inertia of 
 the area about a parallel axis 
 through the centre of gravity, 
 the product of the area into 
 the square of the distance 
 between the two axes. We 
 
 Fig. S3. 
 
 see from this that the moment of inertia of a figure about an 
 axis through its own centre of gravity is always less than about 
 any other axis parallel to it. 
 
Conditions of Equilibrium, Transverse Metacentre, etc. loi 
 
 Example. — Having given the moment of inertia of tlie triangle in 
 Fig. 52 about the axis NN through the centre of gravity as tsA/^', find the 
 moment of inertia about the base parallel to NN. 
 
 Applying the above rule, we have — 
 
 I = VbAA' + A 
 
 iS)' 
 
 which agrees with the value given above for the moment of inertia of a 
 triangle about its base. 
 
 Example. — Find the moment of inertia of a triangle of area A and 
 height h about an axis through the vertex parallel to the base. 
 
 Ans. JA/i^. 
 Example. — A rectangle is 4 inches long and 3 inches broad. Compare 
 the ratio of its moment of inertia about an axis through the centre parallel 
 to the long and short sides respectively. 
 
 Ans. 9 : 16. 
 Example. — A square of 12 inches side has another symmetrical square 
 of half its area cut out of the centre. Compare the moments of inertia 
 about an axis through the centre parallel to one side of, the original 
 square, the square cut out, the remaining area. 
 
 Ans. As 4 : I : 3, the ratio of the areas being 4:2:2. 
 
 This last example illustrates the important fact that if an 
 area is distributed away from the centre of gravity, the moment 
 of inertia is very much greater than if the same area were 
 massed near the centre of gravity. 
 
 To find the Moment of Inertia of a Plane Cur- 
 vilinear Figure (as Fig. 36, p. 57) about its Base. — Take 
 a strip PQ of length y and breadth (indefinitely small) dx. 
 Then, if we regard PQ as a rectangle, its moment of inertia 
 about the base DC is — 
 
 \{y . dx)/' = ^y" .dx (y .dx = area) 
 
 and the moment of inertia of the whole figure about DC will 
 be the sum of all such expressions as this ; or— 
 
 that is, we put the third part of the cubes of the ordinates of the 
 curve through either of Simpson's rules. For the water-plane 
 of a ship (for which we usually require to find the moment of 
 inertia about the centre line), we must add :the moment of 
 inertia of both sides together : and, since these are symmetrical, 
 we have — 
 
 I = ^ jy^ .dx (y — semi-ordinate of water-plane) 
 
102 
 
 Theoretical Naval Architecture. 
 
 In finding the moment of inertia of a water-plane about the 
 centre line, the work is arranged as follows : — 
 
 Number of 
 
 Semi-ordinates 
 
 Cubes of 
 
 Simpson's 
 
 Functions of 
 
 ordinate. 
 
 of water-plane. 
 
 semi-ordinates. 
 
 multipliers. 
 
 cubes. 
 
 I 
 
 COS 
 
 
 
 I 
 
 — 
 
 2 
 
 4-65 
 
 lOI 
 
 4 
 
 404 
 
 3 
 
 10-05 
 
 IOI5 
 
 2 
 
 2,030 
 
 4 
 
 H'30 
 
 2924 
 
 4 
 
 11,696 
 
 5 
 
 1675 
 
 4699 
 
 2 
 
 9.398 
 
 6 
 
 1765 
 
 5498 
 
 4 
 
 21,992 
 
 7 
 
 17-40 
 
 5268 
 
 2 
 
 10,536 
 
 8 
 
 16 -20 
 
 4252 
 
 4 
 
 17,008 
 
 9 
 
 i3'SS 
 
 2488 
 
 2 
 
 4,976 
 
 lO 
 
 9-65 
 
 899 
 
 4 
 
 3,596 
 
 II 
 
 3-6S 
 
 49 
 
 I 
 
 49 
 
 81,685 
 
 Common interval =28 feet 
 
 Moment of inertia = 81,685 X f X ^ = 508,262 ^ 
 
 The semi-ordinates are placed in column 2, and the cubes 
 of these are placed in column 3. It is not necessary, in ordi- 
 nary cases, to put any decimal places in the cube ; the nearest 
 whole number is sufficient. It is best to take the cubes out 
 of Barlow's tables or out of a pocket-book, as " Mackrow," 
 since the labour of cubing the numbers is very great. These 
 cubes are put through Simpson's multipliers in the ordinary 
 way, giving column 5. The sum of the functions of cubes has 
 to be treated as follows : First there is the multiplier for Simp- 
 son's rule, viz. -^ X 28, and then the f of the expression 
 f /y . dx, which takes into account both sides. The multiplier, 
 therefore, is f X -^j and the sum of the numbers in column 5 
 multiplied by this will give the moment of inertia required. 
 
 Approximation to the Moment of Inertia of a 
 Ship's Water-plane about the Centre Line. — We have 
 seen that for certain regular figures we can express the moment 
 of inertia about an axis through the centre of gravity in the 
 form nKW', where ?z is a coefficient varying for each figure. 
 We can, in the same way, express the I of a water-plane area 
 
 ' This calculation for the L.W.P. is usually done on the displacement 
 sheet. 
 
Conditions of Equilibrium, Transverse Metacentre, etc. 103 
 
 about the centre line, but it is not convenient to use the area 
 as we have done above. We know that the area can be 
 expressed in the form — 
 
 /^ X L X B 
 where L is the extreme length ; 
 B „ „ breadth; 
 
 .^ is a coefficient of fineness ; 
 
 so that we can write — 
 
 where fi is a new coefficient that will vary for different shapes 
 of water-planes. If we can find what the values of the co- 
 efficient n are for ordinary water-planes, it would be very 
 useful in checking our calculation work. Taking the case of 
 a L.W.P. in the form of a rectangle, we should find that n = 
 o'o8, and for a L.W.P. in the form of two triangles, n = o'oz. 
 
 These are two extreme cases, and we should expect for 
 ordinary ships the value of the coefficient n would lie between 
 these values. This is found to be the case, and we may take 
 the following approximate values for the value of n in the 
 formula I = nlS& : — 
 
 For ships whose load water-planes are extremely fine . . . o"04 
 
 ,, ,, ,, „ moderately fine ... o'os 
 
 „ ,, ,, very full o'o6 
 
 For the water-plane whose moment of inertia we calcu- 
 lated above, we have, length 280 feet, breadth 35-3 feet, and 
 I = 508,262 in foot-units. Therefore the value of the coefficient 
 
 n is — 
 
 [;o8262 
 
 — ;r-== ; rr, = 0'04I 
 
 z8o X (35'3)' 
 Formula for finding the Distance of the Trans- 
 verse Metacentre above the Centre of Buoyancy 
 (BM). — We have already discussed in Chapter II. how the 
 position of the centre of buoyancy can be determined if the under- 
 water form of the ship is known, and now we proceed to discuss 
 how the distance BM is found. Knowing this, we are able to 
 fix the position of the transverse metacentre in the ship. 
 
104 Theoretical Naval Architecture. 
 
 Let Fig. 45, p. 90, represent a ship heeled over to a very 
 small angle 6 (much exaggerated in the figure). 
 
 B is the centre of buoyancy in the upright position when 
 
 floating at the water-line WL. 
 B' is the centre of buoyancy in the inclined position when 
 
 floating at the water-line W'L'. 
 V is the volume of either the immersed edge LSL' or the 
 
 emerged wedge WSW. 
 V is the total volume of displacement. 
 g is the centre of gravity of the emerged wedge. 
 ^ is the centre of gravity of the immersed wedge. 
 Then, using the principle given on p. 96, BB' will be parallel 
 to^/, and— 
 
 BB' = ^ 
 
 since the new displacement is formed by taking away the wedge 
 WSW from the original displacement, and putting it in the 
 position LSL'. 
 
 Now for the very small angle of inclination, we may say 
 that— 
 
 BB' . 
 BM = ^^" ^ 
 or BB' = BM sin 
 
 so that we can find BM if we can determine the value of 
 V X git since V, the volume of displacement, is known. 
 
 Let Fig. 54 be a section of the vessel; wl, 7t/l', the original 
 and new water-lines respectively, the angle of inclination being 
 very small. Then we may term wSw' the emerged triangle, 
 and ^l' the immersed triangle, being transverse sections of 
 the emerged and immersed wedges, and ww', II' being for all 
 practical purposes straight lines. If j be the half-breadth of 
 the water-line at this section, we can say ww' = W =y sin 6, 
 and the area of either of the triangles is — 
 
 \y Xy sinO = \f sin 6 
 
 Let a,d h& the centres of gravity of the triangles w^w', ISi' 
 respectively ; then we can say, seeing that 6 is very small, that 
 
Conditions of Equilibrium, Transverse Metacentre, etc. 105 
 
 ad = |_y, since the centre of gravity of a triangle is two-thirds 
 the height from the apex. The new immersed section being 
 regarded as formed by the transference of the triangle wSte/ 
 
 Fig. 54. 
 
 to the position occupied by the triangle /S/', the moment of 
 transference is — 
 
 (\f sin e)xiy= if sin 6 
 
 and for a very small length dx of the water-line the moment 
 will be — 
 
 fy sin 6 . dx 
 
 since the small volume is \y sin 6 . dx, and the shift of its 
 centre of gravity is \y. If now we summed all such expres- 
 sions as this for the whole length of the ship, we should get 
 the moment of the transference of the wedge, or z; X gg' . 
 Therefore we may say, using the ordinary notation — 
 
 ■v^g^ = /I/ sin e . dx 
 = f sin ^ ^y^ . dx 
 
 therefore we have — 
 
 BB' = BM sin B = 
 
 or BM = 
 
 V X g^ I sin Q // . dx 
 V ~ V 
 
 V 
 
 But the numerator of this expression is what we have found to 
 
io6 Theoretical Naval Architecture. 
 
 be the moment of inertia of a water-plane about its centre line, 
 y being a semi-ordinate ; therefore we can write — 
 
 We have seen, on p. loi, how the moment of inertia of a 
 water-plane is found for any given case, and knowing the 
 volume of displacement, we can then determine the distance 
 BM, and so, knowing the position of the C.B., fix the position 
 of the transverse metacentre in the ship. 
 
 Example. — k lighter is in the form of a box, 120 feet long, 30 feet 
 broad, and floats at a draught of 10 feet. Find its transverse BM. 
 
 In this case the water -plane is a rectangle 120' X 30', and we want its 
 I about the middle line. Using the formula for the I of a rectangle about 
 an axis through its centre parallel to a side, -^A/^", we have — 
 
 I = n X 3600 X 900 [h = 30) 
 
 = 270,000 
 V, the volume of displacement, = 120 x 30 X 10 = 36,000 
 
 .-. BM = ^7°,ooo^y.gfeet 
 36,000 
 
 Example. — A pontoon of 10 feet draught has a constant section in the 
 form of a trapezoid, breadth at the water-line 30 feet, breadth at base 
 20 feet, length 120 feet. Find the transverse BM. 
 
 Ans. 9 feet. 
 
 It will be noticed that the water-plane in this question is 
 
 the same as in the previous question, but the displacement being 
 
 less, the BM is greater. M is therefore higher in the ship for 
 
 two reasons. BM is greater and B is higher in the second case. 
 
 Example. — A raft is formed of two cylinders 5 feet in diameter, parallel 
 throughout their lengths, and 10 feet apart, centre to centre. The raft floats 
 with the axes of the cylinders in the surface. Find the transverse BM. 
 
 We shall find that the length does not affect the result, but we will 
 suppose the length is / feet. We may find the I of the water-plane in two 
 ways. It consists of two rectangles each /' X 5', and their centre lines 
 are lo feet apart. 
 
 1. The water-plane may be regarded as formed by cutting a rectangle 
 /' X 5' out of a rectangle /' x 15' ; 
 
 .-. I = W X IS) X 15^ - W X 5) X 5' 
 
 = tWi5' - t) 
 = af|a/ 
 
 this being about a fore-and-aft axis at the centre of the raft. 
 
 2. We may take the two rectangles separately, and find the I of each 
 about the centre line of the raft, which is 5 feet from the line through the 
 centre of each rectangle. Using the formula — 
 
 I = I„ + Ay 
 
 = 'mi 
 
Conditions of Equilibrium, Transverse Metacentre, etc. 107 
 
 and for both rectangles the moment of inertia will be twice this, or 3^2/^ as 
 obtained above. 
 
 We have to find the volume of displacement, which works out to ^l 
 cubic feet. The distance BM is therefore — 
 
 a?i2/-7-i'7f/=i3-8feet 
 
 Example. — A raft is formed of three cylinders, 5 feet in diameter, 
 parallel and symmetrical throughout their lengths, the breadth extreme 
 being 25 feet. The raft floats with the axes of the cylinders in the surface. 
 Find the transverse BM. 
 
 The moment of inertia of the water-plane of this raft is best found by 
 using the formula I = I„ + Aj/° for the two outside rectangles, and adding 
 it to I„, the moment of inertia of the centre rectangle about the middle line. 
 We therefore have for the whole water-plane I = ^i^l, where / = the 
 length ; and the volume of displacement being S|^/, the value of BM will be 
 35 feet. 
 
 Approximate Formula for the Height of the Trans- 
 verse Metacentre above the Centre of Buoyancy. — 
 
 The formula for BM is — 
 
 We have seen that we may express I as iiLW, where n is 
 a coefficient which varies for different shapes of water-planes, 
 but which will be the same for two ships whose water-planes 
 are similar. 
 
 We have also seen that we may express V as kLED, where 
 D is the mean moulded draft (to top of keel amidships), and k 
 is a coefficient which varies for different forms, but which will 
 be the same for two ships whose under-water forms are similar. 
 Therefore we may say — 
 
 « X L X B' 
 
 BM = 
 
 >6 X L X B X D 
 
 B^ 
 
 where a is a coefficient obtained from the coefficients n and k. 
 Sir William White, in the " Manual of Naval Architecture," 
 gives the value of a as being between o"o8 and o'l, a usual 
 value for merchant ships being o'og. The above formula 
 shows very clearly that the breadth is more effective than the 
 draught in determining what the value of BM is in any given 
 case. It will also be noticed that the length is not brought in. 
 
io8 
 
 Theoretical Naval Architecture. 
 
 The ship for which the moment of inertia of a water-plane 
 was calculated on p. 102, had a displacement of 1837 tons up 
 to that water-plane. The value of BM is therefore — 
 
 508262 
 1837 X 35 
 
 7 "9 1 feet 
 
 The breadth and mean draught were 35*3 and 13^ feet re- 
 spectively. Consequently the value of the coefficient a is 
 o"o84. 
 
 To prove that a Homogeneous Log of Timber of 
 Square Section and Specific Gravity 0*5 cannot float 
 in Presh Water with One of its Faces Horizontal.— 
 The log having a specific gravity of 0-5 will float, and will float 
 with half its substance immersed. The condition that it shall 
 float in stable equilibrium, as regards transverse inclination, in 
 any position is that the transverse metacentre shall be above 
 the centre of gravity. 
 
 Let the section be as indicated in Fig. 55, with side length 
 I 2a. And suppose the log 
 
 is placed in the water with 
 one side of this section 
 horizontal. Then the 
 draught-line will be at a 
 distance a from the bot- 
 tom, and the log, being 
 homogeneous, i.e. of the 
 same quality all through, 
 will have its C.G. in the 
 middle at G, at a distance 
 also of a from the bottom. 
 The centre of buoyancy 
 will be at a distance of 
 
 
 , 
 
 
 Ja. 
 
 
 
 
 
 
 
 i 
 
 
 
 
 
 i W. 
 
 
 G 
 
 
 L. 
 
 
 
 M. 
 
 
 — 
 
 
 
 ■^- 
 
 
 
 
 
 
 
 
 
 I 
 
 Flc. 55. 
 
 - from the bottom. The height of the transverse metacentre 
 above the centre of buoyancy is given by — 
 
 BM =Y 
 
Conditions of Eqnilibriwn, Transverse Metacentre, etc. 109 
 
 where I = moment of inertia of water-plane about a longi- 
 tudinal axis through its centre 
 V = volume of displacement in cubic feet. 
 
 Now, the water-plane of the log is a rectangle of length / 
 and breadth 2a, and therefore — 
 
 its I = Jg /. 2a{2df = ^-Jd' 
 and Y = 1 . 2a . a = 2lc^ 
 :. BM = ^laP -^ 2/fl2 = i« 
 But BG = i« 
 
 therefore the transverse metacentre is below the centre of 
 gravity, and consequently the log cannot float in the position 
 given. 
 
 If, now, the log be assumed floating with one corner down- 
 ward, it will be found by a precisely similar method that — 
 
 BG = o'47ia; 
 and BM = o'g4$a 
 
 Thus in this case the transverse metacentre is above the 
 centre of gravity, and consequently the log will float in stable 
 equilibrium. 
 
 It can also be shown by similar methods that the position 
 of stable equilibrium for all directions of inclination of a cube 
 composed of homogeneous material of specific gravity o'5 is 
 with one comer downwards. 
 
 Metacentric Diagram.— We have seen how the position 
 of the transverse metacentre can be determined for any given 
 ship floating at a definite water-line. It is often necessary, 
 however, to know the position of the metacentre when the ship 
 is floating at some different water-line ; as, for instance, when 
 coal or stores have been consumed, or when the ship is in a light 
 condition. It is usual to construct a diagram which will show 
 at once, for any given mean draught which the vessel may have, 
 the position of the transverse metacentre. Such a diagram is 
 shown in Fig. 56, and it is constructed in the following manner : 
 A line WiLj is drawn to represent the load water-line, and 
 parallel to it are drawn W3L2, W3L2, W4L4 to represent the 
 
no 
 
 Theoretical Naval Architecture. 
 
 water-lines Nos. 2, 3, and 4, which are used for calculating the 
 displacement, the proper distance apart, a convenient scale 
 being \ inch to i foot. A line L1L4 is drawn cutting these 
 level lines, and inclined to them at an angle of 45°. Through 
 the points of intersection Li, L2, L3, Lj, are drawn vertical lines 
 as shown. The ship is then supposed to float successively at 
 these water-lines, and the position of the centre of buoyancy 
 and the distance of the transverse metacentre above the C.B. 
 
 Fig. s6. 
 
 calculated for each case. The methods employed for finding 
 the position of the C.B. at the different water-lines have already 
 been dealt with in Chapter II. On the vertical lines are then 
 set down from the L.W.L. the respective distances of the 
 centres of buoyancy below the L.W.L. Thus LjBi is the 
 distance when floating at the L.W.L., and AB3 the distance 
 when floating at No. 3 W.L. In this way the points Bi, fij, 
 B3, B4 are obtained ; and if the calculations are correct, a fair 
 
Conditions of Eqinlibriwn, Transverse Metacentre, etc. 1 1 1 
 
 line can be drawn passing through all these spots as shown. 
 Such a curve is termed the curve of centres of buoyancy. It is 
 usually found to be rather a flat curve, being straight near the 
 load-liiie condition. The distance BM for each water-line is 
 then set up from Bi, Bj, B3, Bj respectively, giving the points 
 Ml, Ma, Ms, Mj. A curve can then be drawn through these 
 points, which is termed the curve of transverse meiacentres. 
 Now, suppose the ship is floating at some intermediate water- 
 line — say wl: through /, where wl cuts the 45° line, draw a 
 vertical cutting the curves of centres of buoyancy and meta- 
 centres in b and m respectively. Then m will be the position 
 of the transverse metacentre of the ship when floating at the 
 water-line wl. It will be noticed that we have supposed the 
 ship to float always with the water-plane parallel to the L.W.P. ; 
 that is to say, she does not alter trim. For water-planes not 
 parallel to the L.W.P. we take the mean draught {i.e. the 
 draughts at the fore-and-aft perpendiculars are added together 
 and divided by 2), and find the position of M on the meta- 
 centric diagram for the water-plane, parallel to the L.W.P., 
 corresponding to this mean draught. Unless the change of 
 trim is very considerable, this is found to be correct enough 
 for all practical purposes. Suppose, however, the ship trims 
 very much by the stern,^ owing to coal or stores forward being 
 consumed, the shape of her water-plane will be very different 
 from the shape it would have if she were floating at her normal 
 trim or parallel to the L.W.P. ; generally the water-plane will 
 be fuller under these circumstances, and the moment of inertia 
 will be greater, and consequently M higher in the ship, than 
 would be given on the metacentric diagram. When a ship 
 is inclined, an operation that will be described later, she 
 is frequently in an unfinished condition, and trims consider- 
 ably by the stern. It is necessary to know the position of 
 the transverse metacentre accurately for this condition, and 
 
 ' This would be the case in the following : A ship is designed to float 
 at a draught of 17 feet forward and 19 feet aft, or, as we say, 2 feet by the 
 stern. If her draught is, say, 16 feet forward and 20 feet aft, she will have 
 the same mean draught as designed, viz. 18 feet, but she will trim 2 feet 
 more by the stern. 
 
1 1 2 Theoretical Naval A rchitectnre. 
 
 consequently the metacentric diagram cannot be used, but a 
 separate calculation made for the water-plane at which the 
 vessel is floating. 
 
 On the metacentric diagram is placed also the position of 
 the centre of gravity of the ship under certain conditions. For 
 a merchant ship these conditions may vary considerably owing 
 to the nature of the cargo carried. There are two conditions 
 for which the C.G. may be readily determined, viz. the light 
 condition, and the condition when loaded to the load-line with 
 a homogeneous cargo. The light condition may be defined as 
 follows : No cargo, coal, stores, or any weights on board not 
 actually forming a part of the hull and machinery, but includ- 
 ing the water in boilers and condensers. The draught-lines for 
 the various conditions are put on the metacentric diagram, and 
 the position of the centre of gravity for each condition placed 
 in its proper vertical position. The various values for GM, the 
 metacentric height, are thus obtained : 
 
 On the left of the diagram are placed, at the various water- 
 lines, the mean draught, displacement, and tons per inch. 
 
 There are two forms of section for which it is instructive to 
 construct the metacentric diagram. 
 
 1. A floating body of constant rectangular section. 
 
 2. A floating body of constant triangular section, the apex 
 of the triangle being at the bottom. 
 
 I. For a body having a constant rectangular section, the 
 moment of inertia of the water-plane is the same for all 
 draughts, but the volume of displacement varies. Suppose the 
 rectangular box is 80 feet long, 8 feet broad, 9 feet deep. 
 Then the moment of inertia of the water-plane for all draughts 
 is — 
 
 ^(80 X 8) X 8^ = i^f^ 
 
 The volumes of displacement are as follows : — 
 
 Draught 6 inches V = 8ox8xf cubic feet 
 
 1 foot V = 8ox8 
 
 2 feet V = 8ox8x2 
 
 4 V = 80x8x4 
 
 7 , V = 8ox8X7 
 
 9 „ V =80x8x9 
 
Conditions of Equilibrium, Transverse Metacentre, etc. 1 1 3 
 and the values of BM are therefore as follows : — 
 
 Draught 6 inches 
 
 .. BM = 10-66 feet 
 
 „ I foot 
 
 .. BM= 5-33 „ 
 
 2 feet 
 
 .. BM= 2-66 „ 
 
 „ 4 .. 
 
 .. BM= 1-33 „ 
 
 .. 7 „ 
 
 .. BM= 076 ,, 
 
 ,. 9 M 
 
 .. BM = 059 „ 
 
 The centre of buoyancy is always at half-draught, so that 
 its locus or path will be a straight line,^ and if the values obtained 
 above are set off from the centres of buoyancy at the various 
 water-lines, we shall obtain the curve of transverse metacentres 
 as shown in Fig. 57 by the curve AA, the line BB being the 
 corresponding locus of the centres of buoyancy. 
 
 3-0, 
 
 Fig. 57. 
 
 2. For a floating body with a constant triangular section, the 
 locus of centres of buoyancy is also a straight line, because it is 
 always two-thirds the draught above the base.^ Suppose the 
 triangular section to be 10 feet broad at the top and 9 feet deep, 
 the length of the body being 120 feet. In this case we must 
 calculate the moment of inertia of each water-plane and the 
 volume of displacement up to each. The results are found to 
 be as follows : — 
 
 ' This may be seen by finding a few spots on this locus. 
 
 I 
 
114 Theoretical Naval Architecture. 
 
 Draught i foot 
 ,, 2 feet 
 
 !> 4 )! 
 
 BM = 0'20 feet 
 BM = 0-41 „ 
 BM = o-82 „ 
 BM = 1-23 „ 
 BM = 1-85 „ 
 
 9 .. 
 
 These values are set up from the respective centres of 
 buoyancy, and give the locus of transverse metacentres, which 
 is found to be a straight line, as shown by CC in Fig. 57, DD 
 being the locus of centres of buoyancy. 
 
 Approximation to Locus of Centres of Buoyancy 
 on the Metacentric Diagram. — We have seen (p. 63) how 
 the distance of the centre of buoyancy below the L.W.L. can 
 be approximately determined. The locus of centres of buoyancy , 
 in the metacentric diagram is, in most cases, very nearly straight 
 for the portion near the load-line, and if we could obtain easily 
 the direction the curve takes on leaving the position for the load 
 water-line, we should obtain a very close approximation to the 
 actual curve itself. It might be desirable to obtain such an 
 approximation in the early stages of a design, when it would 
 not be convenient to calculate the actual positions of the centre 
 of buoyancy, in order to accurately construct the curve. 
 
 Let Q be the angle the tangent to the curve of buoyancy at 
 the load condition makes with the horizontal, as in 
 Kg. 56; 
 A., the area of the load water-plane in square feet ; 
 V, the volume of displacement up to the load water-line 
 
 in cubic feet ; 
 h, the distance of the centre of buoyancy of the load 
 displacement below the load water-line in feet. 
 Then the direction of the tangent to the curve of buoyancy is 
 given by — 
 
 . Ah 
 tan B = -=^ 
 
 Each of the terms in the latter expression are known or can 
 be readily approximated to,^ and we can thus determine the in- 
 clination at which the curve of centres of buoyancy will start, 
 and this will closely follow the actual curve. 
 
 ' See Example 39, p. 131, for a further approximation. 
 
Conditions of Equilibrium, Transverse Metacentre, etc. 115 
 
 In a given case — 
 
 A = 7854 square feet 
 
 h = 5-45 feet 
 
 V = 2140 X 35 cubic feet 
 
 so that — 
 
 ^ 7854 X 5'45 
 
 tan B = ^—^^ ^-^ 
 
 2140 X 35 
 
 = 0-572' 
 
 Finding the Metacentric Height by Experiment. 
 Inclining Experiment. — We have been dealing up to the 
 present with the purely geometrical aspect of initial stability, 
 viz. the methods employed and the principles involved in 
 finding the position of the transverse metacentre. All that is 
 needed in order to determine this point is the form of the 
 underwater portion of the vessel. But in order to know any- 
 thing about the vessel's initial stability, we must also know the 
 vertical position of the centre of gravity of the ship, and it is to 
 determine this point that the inclining experiment is performed. 
 This is done as the vessel approaches completion, when 
 weights that have yet to go on board can be determined 
 together with their final positions. Weights are shifted trans- 
 versely across the deck, and by using the principle explained on 
 p. 97, we can tell at once the horizontal shift of the centre of 
 gravity of the ship herself due to this shift of the weights on 
 board. The weight of the ship can be determined by calculating 
 the displacement up to the water-line she floats at, during the 
 experiment. (An approximate method of determining this 
 displacement when the vessel floats out of her designed trim 
 
 ' The best way to set off this tangent is, not to find the angle 9 in 
 degrees and then set it off by means of a protractor, but to set off a 
 horizontal line of lo feet long (on a convenient scale), and from the end 
 set down a vertical line 572 feet long on the same scale. This will give 
 
 the inclination required, for tan 9 = ^ — '- — = 0'572. 
 
 This remark applies to any case in which an angle has to be set off very 
 accurately. A table of tangents is consulted and the tangent of the required 
 angle is found, and a similar process to the above is gone through. 
 
Ii6 
 
 Theoretical Naval Architecture. 
 
 will he found on p. 140.) Using the notation employfed on 
 p. 97, and illustrated by Fig. 49, we have — 
 
 GG' = 
 
 w X d 
 W 
 
 Now, unless prevented by external forces, it is evident that 
 the vessel must incline over to such an angle that the centre of 
 gravity G' and the centre of buoyancy B' are in the same verti- 
 cal line (see Fig. 58), and, the angle of inclination being small, 
 
 Fig. 58. 
 
 M will be the transverse metacentre. If now we call 6 the 
 angle of inclination to the upright, GM being the " metacentric 
 height "— 
 
 GG' 
 GM 
 GG' 
 
 tan^ 
 w X d 
 ^ W X tan ^ 
 
 using the value found above for GG'. The only term that we 
 do not yet know in this expression is tan 6, and this is found in 
 the following manner : At two or three convenient positions 
 
 tan l9 = 
 
 GM = 
 
Conditions of Equilibrium, Transverse Metacentre, etc. Wj 
 
 in the ship ^ (such as at bulkheads or down hatchways) plumb- 
 bobs are suspended from a point in the middle line of the ship, 
 and at a convenient distance from the point of suspension a 
 horizontal batten is fixed, with the centre line of the ship marked 
 on it, as shown by PQ in Fig. 58. Before the ship is inclined, 
 the plumb-hne should coincide, as nearly as possible, with the 
 centre-line of the ship — that is to say, the ship should be prac- 
 tically upright. When the ship is heeled over to the angle 6, 
 the plumb-line will also be inclined at the same angle, Q, to the 
 original vertical or centre line of the ship, and if / be the 
 distance of the horizontal batten below the point of suspension 
 in inches, and a the deviation of the plumb-line along the 
 batten, also in inches, the angle 6 is at once determined, for— 
 
 tan d = 
 so that we can write — 
 
 In practice it is convenient to check the results obtained by 
 dividing the weight w into four equal parts, placing two sets on 
 one side and two sets on the other side, arranged as in Fig. 59. 
 
 The experiment is then performed in the following order : — 
 
 (a) See if the ship is floating upright, in which case the 
 plumb-lines will coincide with the centre of the ship. 
 
 {b) The weight (i). Fig. 59, is shifted from port to star- 
 board on to the top of weight (3) through the distance d feet, 
 say, and the deviations of the plumb-lines are noted when the 
 ship settles down at a steady angle. 
 
 (c) The weight (2) is shifted from port to starboard on to 
 the top of weight (4) through the distance d ,feet, and the 
 deviations of the plumb-line noted. 
 
 {d) The weights (i) and (2) are replaced in their original 
 positions, when the vessel should again resume her upright 
 position. 
 
 ' If two positions are taken, one is forward and the other aft. If three 
 positions are taken, one is forward, one aft, and one amidships. 
 
ii8 Theoretical Naval Architecture. 
 
 {e) The weight (3) is moved from starboard to port, and 
 the deviations of the plumb-Unes noted. 
 
 (/■) The weight (4) is rhoved from starboard to port, and 
 the deviations of the plumb-Hnes noted. 
 
 With the above method of conducting the experiment,^ and 
 using two plumb-Hnes, we obtain eight readings, and if three 
 plumb-lines were used we should obtain twelve readings. It is 
 important that such checks should be obtained, as a single result 
 might be rendered quite incorrect, owing to the influence of the 
 hawsers, etc. A specimen experiment is given on p. 119, in 
 which two plumb-lines were used. The deviations obtained 
 
 
 rpf^ rsirti 
 
 Fig. 59- 
 
 are set out in detail, the mean deviation for a shift of 1 2\ tons 
 through 36 feet being 5^ inches, or the mean deviation for a 
 shift of 25 tons through 36 feet is loyg inches. 
 
 Precautions to be taken when performing an Inclining Experi- 
 ment.— K rough estimate should be made of the GM expected 
 at the time of the experiment ; the weight of ballast can then be 
 determined which will give an inclination of about 4° or 5° when 
 one-half is moved a known distance across the deck. The weight 
 of ballast thus found can then be got ready for the experiment. 
 
 K personal inspection should be made to see that all weights 
 likely to shift are efficiently secured, the ship cleared of all 
 
 ' There is a slight rise of G, the centre of gravity of the ship, in this 
 method ; but the error involved is inappteciable. 
 
Conditions of Equilibritim, Transverse Metacentre, etc. 1 19 
 
 free water, and boilers either emptied or run up quite full. 
 Any floating stages should be released or secured by very slack 
 painters. 
 
 If possible a fine day should be chosen, with the water calm 
 and little wind. All men not actually employed on the experi- 
 ment should be sent ashore. Saturday afternoon or a dinner 
 hour is found a convenient time, since then the majority of 
 the workmen eniployed finishing the ship are likely to be away. 
 
 The ship should be hauled head or stern on to the wind, 
 if any, and secured by hawsers at the bow and stern. When 
 taking the readings, these hawsers should be slacked out, so as 
 to ensure that they do not influence the reading. The ship 
 should be plumbed upright before commencing. 
 
 An account should be taken, with positions of all weights to 
 be placed on board to complete, of all weights to be removed, 
 such as yard plant, etc., and all weights that have to be shifted. 
 
 The following is a specimen report of an inclining experi- 
 ment : — . 
 
 Report on Inclining Experiment performed on 
 
 at . Density of water cubic feet to the ton. 
 
 Draught of water ... ... ... ... 16' 9" forward. 
 
 „ ,, 22' 10" aft. 
 
 Displacement in tons at this draught ... ... 5372 
 
 The wind was slight, and the ship was kept head to wind during the 
 experiment. Ballast used for inclining, 50 tons. Lengths of pendulums, 
 two in number, 15 feet. Shift of ballast across deck, 36 feet. 
 
 
 Deviation of pendulum in 15 feet. 
 
 Forward. 
 
 Aft. 
 
 Experiment I, 12J tons port to starboard 
 ,, 2, 12^ ,, ,, 
 
 Ballast replaced, zero checked 
 
 Experiment 3, 12 J tons starboard to port 
 ,, 4, 12J „ „ 
 
 54" 
 loi" 
 right 
 
 s 
 
 S*" 
 I0|" 
 right 
 
 St" 
 
 lOj" 
 
 The condition of the ship at the time of inclining is as defined below : 
 
 Bilges dry. 
 Water-tanks empty. 
 
I20 Theoretical Naval Architecture. 
 
 No water in boilers, feed-tanks, condensers, distillers, cisterns, etc. 
 
 Workmen on board, 66. 
 
 Tools on board, 5 tons. 
 
 Masts and spars complete. 
 
 No boats on board. 
 
 Bunkers full. 
 
 Anchors and cables, complete and stowed. 
 
 No provisions or stores on board. 
 
 Engineers' stores, half on board. 
 
 Hull complete. 
 
 The mean deviation in 15 feet for a shift of 25 tons through 36 feet is 
 lo/g inches = I0'3I2 inches. 
 
 ... GM = ^5 X 36 X 15 X 12 ^ ^^^^ 
 
 10-312 X 5372 
 
 The ship being in an incomplete condition at the time of 
 the inchning experiment, it was necessary to take an accurate 
 account of all weights that had to go on board to complete, 
 with their positions in the ship, together with an account of 
 all weights that had to be removed, with their positions. The 
 total weights were then obtained, together with the position of 
 their final centre of gravity, both in a longitudinal and vertical 
 direction. For the ship of which the inclining experiment is 
 given above, it was found that to fully complete her a total 
 weight of 595 tons had to be placed on board, having its 
 centre of gravity 11 feet before the midship ordinate, and 3*05 
 feet below the designed L.W.L. Also 63 tons of yard plant, 
 men, etc., had to be removed, with centre of gravity 14 feet 
 abaft the midship ordinate, and 15 feet above the designed 
 L.W.L. The centre of buoyancy of the ship at the experi- 
 mental water-line was io'8 feet abaft the midship ordinate, 
 and the transverse metacentre at this line was calculated at 
 3-14 feet above the designed L.W.L. 
 
 We may now calculate the final position of the centre of 
 gravity of the completed ship as follows, remembering that 
 in the experimental condition the centre of gravity must be 
 in the same vertical line as the centre of buoyancy. The 
 vertical position of G in the experimental condition is found 
 by subtracting the experimental GM, viz. 2*92 feet, from the 
 height of the metacentre above the L.W.L. as given above, 
 viz. 3" 1 4 feet. 
 
Conditions of Equilibrium, Transverse Metacentre, etc. 
 
 121 
 
 
 a 
 
 e5 
 
 Above 
 L.W.L. 
 
 Below 
 L.-W.L. 
 
 Abaft 
 amidships. 
 
 Before 
 amidships. 
 
 1^ 
 
 s 
 
 
 > 
 
 1 
 
 > 
 
 
 II 
 
 s 
 
 
 
 6545 
 654s 
 
 Weight of ship at timel 
 of experiment ...J 
 
 Weight to go on board"! 
 to complete / 
 
 Weight to be taken"! 
 from ship / 
 
 5372 
 
 595 
 
 5967 
 
 63 
 
 0'22 
 
 IS 
 
 1 182 
 1 182 
 
 945 
 
 3-05 
 
 1813 
 1813 
 
 10-8 
 
 I4'0 
 
 58,017 
 
 58,017 
 882 
 
 5904 
 
 237 
 
 I8I3 
 237 
 
 1576 
 
 S7.I35 
 6,545 
 
 50.590 
 
 6545 
 
 The final position of the centre of gravity of the ship is 
 therefore — 
 
 iffl = 0-266 feet below the L.W.L. 
 -Hii- = 8'S7 feet abaft amidships 
 
 the final displacement being 5904 tons. 
 
 The mean draught corresponding to the displacement can 
 be found by the methods we have already dealt with, and corre- 
 sponding to this draught, we can find on the metacentric 
 diagram the position of the transverse metacentre. In this 
 case the metacentre was 2*73 feet above the L.W.L., and 
 consequently the value of GM for the completed condition 
 was — 
 
 2"73 + 0*266 = 2*996 feet 
 
 or say, for all practical purposes, that the transverse meta- 
 centric height in the completed condition was 3 feet. 
 
 It is also possible to ascertain what the draughts forward 
 and aft will be in the completed condition, as we shall see in 
 the next chapter. 
 
 Values of GM, the " Metacentric Height." — We 
 have discussed in this chapter the methods adopted to find 
 for a given ship the value of the transverse metacentric height 
 
122 Theoretical Naval Architecture. 
 
 GM. This distance depends upon two things : the position of 
 G, the centre of gravity of the ship ; and the position of M, the 
 transverse metacentre. The first is dependent on the vertical 
 distribution of the weights forming the structure and lading of 
 the ship, and its position in the ship must vary with differences 
 in the disposition of the cargo carried. The transverse meta- 
 centre depends solely upon the form of the ship, and its 
 position can be completely determined for any given draught 
 of water when we have the sheer drawing of the vessel. There 
 are two steps to be taken in finding its position for any given 
 ship floating at a certain water-line. 
 
 1. We must find the vertical position of the centre of 
 buoyancy, the methods adopted being explained in Chapter II. 
 
 2. We then find the distance separating the centre of 
 buoyancy and the transverse metacentre, or BM, as explained 
 in the present chapter. 
 
 By this means we determine the position of M in the ship. 
 
 The methods of estimating the position of G, the centre 
 of gravity for a new ship, will be dealt with separately in 
 Chapter VI. ; but we have already seen how the position of G 
 can be determined for a given ship by means of the inclining 
 experiment. Having thus obtained the position of M and G in 
 the ship, we get the distance GM, or the metacentric height. 
 
 The following table gives the values of the metacentric height 
 in certain classes of ships. For fuller information reference 
 must be made to the works quoted at the end of the book. 
 
 Type of ship. 
 
 Values of GM. 
 
 Harbour vessels, as tugs, etc 
 
 Modern protected cruisers 
 
 Modern British battleships 
 
 Older central citadel armourclads 
 Shallow-draught gunboats for river service 
 Merchant steamers (varying according to \ 
 the nature and distribution of the cargo) / 
 Sailing-vessels ; 
 
 IS to i8 inches 
 
 2 to 2j feet 
 3i feet 
 
 4 to 8 feet 
 12 feet 
 
 1 to 3 feet 
 
 3 to 3J feet i 
 
 The amount of metacentric height given to a vessel is based 
 largely upon experience with successful ships. In order that 
 
Conditions of Eqtiilibrium, Transverse Metacentre, etc. 123 
 
 a vessel may be " stiff" that is, difBcult to incline by external 
 forces — ^as, for example, by the pressure of the wind on the 
 sails — the metacentric height must be large. This is seen by 
 reference to the expression for the moment of statical stability 
 at small angles of inclination from the upright, viz. — 
 
 W X GM sin 6 (see p. 94) 
 
 W being the weight of the ship in tons ; B being the angle of 
 inclination, supposed small. This, being the moment tending 
 to right the ship, is directly dependent on GM. A " craiik " 
 ship is a ship very easily inclined, and in such a ship the 
 metacentric height is small. For steadiness in a seaway the 
 metacentric height must be small. 
 
 There are thus two opposing conditions to fulfil — 
 
 1. The metacentric height GM must be enough to enable 
 the ship to resist inclination by external forces. This is espe- 
 cially the case in sailing-ships, in order that they may be able 
 to stand up under canvas without heeling too much. In the 
 case of the older battleships with short armour belts and 
 unprotected ends, sufficient metacentric height had to be pro- 
 vided to allow of the ends being riddled, and the consequent 
 reduction of the moment of inertia of the water-plane. 
 
 2. The metacentric height must be moderate enough (if 
 this can be done consistently with other conditions being 
 satisfied) to make the vessel steady in a seaway. A ship which 
 has a very large GM comes back to the upright very suddenly 
 after being inclined, and consequently a vessel with small 
 GM is much more comfortable at sea, and, in the case of a 
 man-of-war, affords a much steadier gun platform. 
 
 In the case of sailing-ships, a metacentric height of from 
 3 to 3-|- feet is provided under ordinary conditions of service, 
 in order to allow the vessel to stand up under her canvas. It is, 
 however, quite possible that, when loaded with homogeneous 
 cargoes, as wool, etc., this amount cannot be obtained, on 
 account of the centre of gravity of the cargo being high up in 
 the ship. In this case, it would be advisable to take in water 
 or other ballast in order to lower the centre of gravity, and 
 thus increase the metacentric height. 
 
124 Theoretical Naval Architecture. 
 
 In merchant steamers the conditions continually vary on 
 account of the varying nature and distribution of the cargo 
 carried, and it is probable that a GM of i foot should be the 
 minimum provided when carrying a homogeneous cargo (con- 
 sistently with satisfactory stability being obtained at large 
 inclinations). There are, however, cases on record of vessels 
 going long voyages with a metacentric height of less than 
 1 foot, and being reported as comfortable and seaworthy. Mr. 
 Denny (Transactions of the Institution of Naval Architects, 
 1896) mentioned a case of a merchant steamer, 320 feet long 
 (carrying a homogeneous cargo), which sailed habitually with 
 a metacentric height of o*6 of a foot, the captain reporting her 
 behaviour as admirable in a seaway, and in every way com- 
 fortable and safe. 
 
 Effect on Initial Stability due to the Presence of 
 Free Water in a Ship. — On reference to p. 118, where the 
 inclining experiment for obtaining the vertical position of the 
 centre of gravity of a ship is explained, it will be noticed that 
 special attention is drawn to the necessity for ascertaining 
 that no free water is allowed to remain in the ship while the 
 experiment is being performed. By free water is meant water 
 having a free surface. In the case of the boilers, for instance, 
 they should either be emptied or run up quite full. We now 
 proceed to ascertain the necessity for taking this precaution. 
 If a compartment, such as a ballast tank in the double bottom, 
 or a boiler, is run up quite full, it is evident that the water will 
 have precisely the same effect on the ship as if it were a solid 
 body having the same weight and position of its centre of 
 gravity as the water, and this can be allowed for with very 
 little difficulty. Suppose, however, that we have on board in a 
 compartment, such as a ballast tank in the double bottom, 
 a quantity of water, and the water does not completely fill the 
 tank, but has a free surface, as wl. Fig. 60.' If the ship is 
 heeled over to a small angle 6, the water in the tank must 
 adjust itself so that its surface id I' is parallel to the level water- 
 line W'L'. Let the volume of either of the small wedges wjzc/, 
 Isl' be Wo, and g, ^ the positions of their centres of gravity, b, b' 
 
 ' Fig. 60 is drawn out of proportion for the sake of clearness. 
 
Conditions of Equilibrium, Transverse Metacentre, etc. 125 
 
 being the centres of gravity of the whole volume of water in 
 the upright and inclined positions respectively. Then, if V„ 
 be the total volume of water in the tank, we have — 
 
 V„ X bb' = VoX gi" 
 andbb' = ^Xig' 
 and bb' is parallel to gg-'. Now, in precisely the same way as we 
 
 / 
 
 Fig. 60. 
 
 found the moment of transference of the wedges WSW, LSL', 
 in Fig. 45, we can find the moment of transference of the small 
 wedges wszef, IsP, viz. — 
 
 vo xgg' = ix e 
 
 where / is the moment of inertia of the free surface of the water 
 in the tank about a fore-and-aft axis through s ; and 6 is the 
 circular measure of the angle of inclination. 
 
 Substituting this value for Vf, X g^, we have — 
 
 ix 6 
 
 bb' = 
 
 V„ 
 
 Draw the new vertical through b', meeting the middle hne in m ; 
 then — 
 
 bbf = bin X e 
 
126 Theoretical Naval Architecture. 
 
 and consequently — 
 
 bin X 6 = -^^jT- 
 
 and im = zr^ 
 
 Now, if the water were solid its centre of gravity would be 
 at li both in the upright and inclined conditions, but the weight 
 of the water now acts through the point d' in the line i^m, and 
 its efifect on the ship is just the same as if it were a solid 
 weight concentrated at the point m. So that, although 6 is 
 the actual centre of gravity of the water, its effect on the ship, 
 when inclined through ever so small an angle, is the same as 
 though it were at the point m, and in consequence of this the 
 point m is termed the virtual centre of gravity of the water.^ 
 This may be made clearer by the following illustrations : — 
 
 1. Suppose that one instant the water is solid, with its 
 centre of gravity at b, and the following instant it became water. 
 Then, for small angles of inclination, its effect on the ship would 
 be the same as if we had raised its weight through a vertical 
 distance bm from its actual to its virtual centre of gravity. 
 
 2. Imagine a pendulum suspended at m, with its bob at b. 
 On the ship being inclined to the small angle B, the pendulum 
 will take up the position 7nV, and this corresponds exactly to 
 the action of the water. 
 
 We thus see that the centre of gravity of the ship cannot be 
 regarded as being at G, but as having risen to Go, and if W, be 
 
 the weight of water in tons =— (the water being supposed 
 
 salt), we have — 
 
 W X GG„ = Wo X bm 
 
 = -- X im 
 35 
 and therefore — 
 
 ' See a paper by Mr. W. Hok, at the Institution of Naval Architects, 
 189s, on "The Transverse Stability of Floating Vessels containing 
 Liquids, with Special Reference to Ships carrying Oil in Bulk." See also 
 a paper in the "Transactions of the Institution of Engineers and Ship- 
 builders in Scotland for 1889," by the late Professor Jenkins, on the 
 stability of vessels carrying oil in bulk. 
 
Conditions of Equilibrium, Transverse Metacentre, etc. 1 27 
 
 Vo 
 
 GG„ — ^ — y X btn (V = volume of displacement) 
 But we have seen that — 
 
 and therefore- 
 
 bm = :;rj- 
 
 
 The new moment of stabihty at the angle 6 is — 
 W X G„M X sin e = W X (GM - GG„) sin 6 
 = W X (gM -;^) sin 61 
 
 the metacentric height being reduced by the simple expres- 
 
 i 
 sion :^, We notice here that the amount of water does not 
 
 eifect the result, but only the moment of inertia of the free 
 surface. The necessity for the precaution of clearing all free 
 water out of a ship on inclining is now apparent. A small 
 quantity of water will have as much effect on the position of 
 the centre of gravity, and therefore on the trustworthiness of 
 the result obtained, as a large quantity of water, provided it 
 has the same form of free surface. If a small quantity of 
 water has a large free surface, it will have more effect than 
 a very large quantity of water having a smaller free surface. 
 
 Example. — A vessel has a compartment of the double bottom at the 
 middle line, 60 feet long and 30 feet broad, partially filled with salt water. 
 The total displacement is 9100 tons, and centre of gravity of the ship and 
 water is 0*26 feet below the water-line. Find the loss of metacentric 
 height due to the water having a free surface. 
 
 We have here given the position of the centre of gravity of the ship and 
 the water. The rise of this centre of gravity due to the mobility of the 
 water is, using the above notation — 
 
 i 
 
 V 
 
 and i — ^(60 X 30) X (30)^ 
 
 = S X (30)= ^ 
 
 Since the free surface is a rectangle 60 feet long and 30 broad 
 
 and V = 9100 X 35 cubic feet 
 
 therefore the loss in metacentric height = -^ — = 0-424 feet 
 
 ^ 9100 X 35 
 
Ans. 
 
 128 Theoretical Naval Architecture. 
 
 Examples to Chapter III. 
 
 1. Find the circular measure of S|°, ioJ°, isi°. 
 
 Ans. 0'09599; 0-17889; o-2748g. 
 
 2. Show that sin 10° is one-half per cent. less in value than the circular 
 measure of 10°, and that tan io° is one per cent, greater in value than the 
 circular measure of 10°. 
 
 3. A cylinder weighing 500 lbs., whose centre of gravity is 2 feet from 
 the axis, is placed on a smooth table and takes up a position of stable 
 equilibrium. It is rolled along parallel to itself through an angle of 60°. 
 What will be the tendency then to return to the original position ? 
 
 Ans. 866 foot-lbs. 
 
 4. Find the moment of inertia about the longest axis through the centre 
 of gravity, of a figure formed of a square of side la, having a semicircle at 
 each end. 
 
 5. Find the moment of inertia of a square of side 2a about a diagonal. 
 
 Ans. §a*. 
 
 6. A square has a similar square cut out of its centre such that the 
 moment of inertia (about a line through the centre parallel to one side) of 
 the small square and of the portion remaining is the same. What pro- 
 portion of the area of the original square is cut out ? 
 
 Ans. 071 nearly. 
 
 7. A vessel of rectangular cross-section throughout floats at a constant 
 draught of 10 feet, and has its centre of gravity in the load water-plane. 
 The successive half-ordinates of the load water-plane in feet are o'5, 6, 12, 
 16, 15, 9, o; and the common interval 20 feet. Find the transverse 
 metacentric height. 
 
 Ans. 8 inches. 
 
 8. A log of fir, specific gravity 0'5, is 12 feet long, and the section is 
 2 feet square. What is its transverse metacentric height when floating in 
 equilibrium in fresh water ? 
 
 Ans. o'47 foot. 
 
 9. The semi-ordinates of a water-plane 34 feet apart are 0-4, 137, 
 25-4, 32-1, 34-6, 35-0, 34-9, 34-2, 32-1, 23-9, 6-9 feet respectively. Find 
 its moment of inertia about the centre line. 
 
 Ans. 6,012,862. 
 
 10. The semi-ordinates of the load water-plane of a vessel are o, 3 '35, 
 6-41, 863, 9-93, 10-44, io'37, 9'94, 8-96, 7'i6, and 2-5 feet respectively. 
 These ordinates being 21 feet apart, find — • 
 
 (i) The tons per inch immersion. 
 
 (2) The distance between the centre of buoyancy and the transverse 
 metacentre, the load displacement being 484 tons. 
 
 Ans. (i) 7-73 tons; (2) 5-2 feet nearly. 
 
 11. The semi-ordinates, l6-6 feet apart, of a vessel's water-plane are 
 0-2, 2-3, 6-4, 9-9, 12-3, 13-5, 13-8, 13-7, 12-8, 10-6, 6-4, 1-9, 0-2 feet 
 respectively, and the displacement up to this water-plane is 220 tons. Find 
 the length of the transverse BM. 
 
 Ans. 20"6 feet. 
 
 12. A vessel of 613 tons displacement was inclined by moving 30 cwt. 
 of rivets across the deck through a distance of 22' 6" The end of a plumb- 
 
Conditions of Equilibrmin, Transverse Metaceutre, etc. 129 
 
 line 10 feet long moved through 2J inches. What was the metacentric 
 heiglit at the time of the experiment ? 
 
 Ans. 2 '93 feet. 
 
 13. The semi-oidinates of a ship's water-plane 35 feet apart are, com- 
 mencing from forward, 0-4, 7-12, 15'28, 21-88, 25-62, 26-9, 26-32, 24-42, 
 20-8, 15-15. 6-39 feet respectively. There is an after appendage of lib 
 square feet, with its centre of gravity 1 80 feet abaft the midship ordinate. 
 Find— 
 
 (1) The area of the water-plane. 
 
 (2) The tons per inch immersion. 
 
 (3) The distance of the centre of flotation abaft amidships. 
 
 (4) The position of the transverse metacentre above the L. W.L., taking 
 
 the displacement up to the above line as 5372 tons, and the 
 centre of buoyancy of this displacement 8-61 feet below the 
 L.W.L. 
 Ans. (l) 13,292 square feet ; (2) 31-6 tons; (3) 14-65 feet ; (4) 3-34 
 feet. 
 
 14. A ship displacing 9972 tons is inclined by moving 40 tons 54 feel 
 across the deck, and a mean deviation of 9J inches is obtained by pendulums 
 15 feet long. Find the metacentric height at the time of the operation. 
 
 Ans. 4-18 feet. 
 
 15. A ship weighing 10,333 tons was inclined by shifting 40 tons 52 
 feet across the deck. The tangent of the angle of inclination caused was 
 found to be 0-05. If the transverse metacentre was 4-75 feet above the 
 designed L. W.L., what was the position of the centre of gravity of the ship 
 at the time of the experiment ? 
 
 Ans. 0-73 foot above the L.W.L. 
 
 16. A vessel of 26 feet draught has the moment of inertia of the L. W.P. 
 about a longitudinal axis through its centre of gravity 6,500,000 in foot- 
 units. The area of the L.W.P. is 20,000 square feet, the volume of dis- 
 placement 400,000 cubic feet, and the centre of gravity of the ship may be 
 taken in the L.W.P. Approximate to the metacentric height. 
 
 Ans. sJ feet. 
 
 17. Prove the rule given on p. 60 for the distance of the centre of 
 
 gravity of a semicircle of radius a from the diameter, viz. —a, by finding 
 
 3"' 
 the transverse BM of a pontoon of circular section floating with its axis in 
 the surface of the water. 
 
 (M in this case is in the centre of section.) 
 
 18. Take a body shaped as in Kirk's analysis, p 80, of length 140 
 feet ; length of parallel middle body, 100 feet ; extreme breadth, 30 feet ; 
 draught, 12 feet. Find the transverse BM. 
 
 Am. 5-7 feet. 
 
 19. A vessel of 1792 tons displacement is inclined by shifting 5 tons 
 already on board transversely across the deck through 20 feet. The end 
 of a plumb-line 15 feet long moves through 5J inches. Determine the 
 metacentric height at the time of the experiment. 
 
 Ans. I -91 feet. 
 20. A vessel of displacement 1 722 tons is inclined by shifting 6 tons of 
 ballast across the deck through 22 j feet. A mean deviation of loj inches 
 is obtained with pendulums 15 feet long. The transverse metacentre is 
 1 5 "28 feet above the keel. Find the position of the centre of gravity of the 
 ship with reference to the keel. 
 
 Ans. 1 3 "95 feet. 
 
130 Theoretical Naval Architecture. 
 
 21. The ship in tlie previous question has 169 tons to go on board at 
 10 feet above Iceel, and 32 tons to come out at 20 feet above keel. Find 
 the metacentric height Vfhen completed, the transverse metacentre at the 
 disiJlacement of 1859 tons being I5'3 feet above keel. 
 
 Ans. I '8 feet. 
 
 22. A vessel of 7000 tons displacement has a weight of 30 tons moved 
 transversely across the deck through a distance of 50 feet, and a plumb-bob 
 hung down a hatchway shows a deviation of 12 inches in 15 feet. What 
 was the metacentric height at the time of the operation ? 
 
 Atis. 3 '2 1 feet. 
 
 23. A box is 200 feet long, 30 feet broad, and weighs 2000 tons. Find 
 the height of the transverse metacentre above the bottom when the box is 
 floating in salt water on an even keel. Ans. I2'26 feet. 
 
 24. Show that for a rectangular box floating at a uniform draught of d 
 feet, the breadth being 12 feet, the distance of the transverse metacentre 
 
 above the bottom is given by — ^ feet, and thus the transverse meta- 
 centre is in the water-line when the draught is 4'9 feet. 
 
 25. A floating body has a constant triangular section. If the breadth 
 at the water-line is nj2 times the draught, show that the curve of metacentres 
 in the metacentric diagram lies along the line drawn from zero draught at 
 45° to the horizontal, and therefore the metacentre is in the water-line for 
 all draughts. 
 
 26. A floating body has a square section with one side horizontal. 
 Show that the transverse metacentre lies above the centre of the square 
 so long as the draught does not much exceed 21 per cent, of the depth of 
 the square. Also show that as the draught gets beyond 21 per cent, of the 
 depth, the metacentre falls below the centre and remains below until 
 the draught reaches 79 per cent, of the depth ; it then rises again above 
 the centre of the square, and continues to rise as long as any part of the 
 square is out of the water. 
 
 (This may be done by constructing a metacentric diagram, or by using the 
 methods of algebra, in which case a quadratic equation has to be solved.) 
 
 27. Show that a square log of timber of 12 inches side, 10 feet long, and 
 weighing 320 lbs., must be loaded so that its centre of gravity is more than 
 1 inch below the centre in order that it may float with a side horizontal 
 in water of which 35 cubic feet weigh 1 ton. 
 
 28. A prismatic vessel is 70 feet long. The section is formed at the 
 lower part by an isosceles triangle, vertex downwards, the base being 20 
 feet, and the height 5 feet ; above this is a rectangle 20 feet wide and 5 f^ct 
 high. Construct to scale the metacentric diagram for all drafts. 
 
 29. A vessel's load water -plane is 380 feet long, and 75 feet broad, and 
 its moment of inertia in foot-units about the centre line works out to 
 8,000,000 about. State whether you consider this a. reasonable result to 
 obtain, the water-plane not being very fine. 
 
 B' 
 
 30. Find the value of the coefficient a in the formula BM = a^ 
 
 referred to on p. 107, for floating bodies having the following sections 
 throughout their length : — 
 
 (a) Rectangular cross-section. 
 
 (b) Triangular cross-section, vertex down. 
 
 (c) Vertical-sided for one half the draught, the lower half of the section 
 
 being in the form of a triangle. 
 
 Ans. {a) o'o8; (b) o-l6 ; (i:) O'll. 
 
Conditions of Equilibrium, Transverse Metacentre, etc. 131 
 
 For ordinary sliips the value of a will lie between the first and last of 
 these. 
 
 31. A lighter in the form of a box is 100 feet long, 20 feet broad, and 
 floats at a constant draught of 4 feet. The metacentric height when empty 
 is 6 feet. Two bulkheads are built lo feet from either end. Show that a 
 small quantity of water introduced into the central compartment will render 
 the lighter unstable in the upright condition. 
 
 32. At one time, in ships which were found to possess insufficient sta- 
 bility, girdling was secured to the ship in the neighbourhood of the water- 
 line. Indicate how far the stability would be influenced by this means. 
 
 33. A floating body has a constant triangular section. If the breadth 
 at the water-line is equal to the draught, show that the locus of metacentres 
 in the metacentric diagram makes an angle with the horizontal of about 40°. 
 
 34. A cylinder is placed into water with its axis vertical. Show that if 
 the centre of gravity is in the water-plane, the cylinder will float upright if 
 the radius -^ the draught is greater than ,J2. 
 
 35. In a wholly submerged body show that for stable equilibrium the 
 centre of gravity must lie below the centre of buoyancy. 
 
 36. A floating body has a constant triangular section, vertex down- 
 wards, and has a constant draught of 12 feet, the breadth at the water-line 
 being 24 feet. The keel just touches a quantity of mud, specific gravity 2. 
 The water-level now falls 6 feet : find the amount by which the metacentric 
 height is diminished due to this. 
 
 Aiu. 2\ feet about. 
 
 37. A floating body of circular section 6 feet in diameter has a meta- 
 centric height of I '27 feet. Show that the centre of buoyancy and centre 
 of gravity coincide, when the body is floating with the axis in the surface. 
 
 38. It is desired to increase the metacentric height of a vessel which is 
 being taken in hand for a complete overhaul. Discuss the three following 
 methods of doing this, assuming the ship has a metacentric diagram as in 
 Fig. 56, the extreme load draught being 15 feet : — 
 
 ( 1 ) Placing ballast in the bottom. 
 
 (2) Removing top weight. 
 
 (3) Placing a girdling round the ship in the neighbourhood of the 
 
 water-line. 
 
 39. Show that the angle 9 in Fig. 56 is between 29° and 30° for a 
 vessel whose coefficient of L.W.P. is 0'75, and whose block coefficient 
 of displacement is o'55. In any case, if these coefficients are denoted by 
 
 « and k respectively, show that tan fl = J -f- ^r approximately (use Nor- 
 
 mand's formula, p. 63). 
 
CHAPTER IV. 
 
 LONGITUDINAL METACENTRE, LONGITUDINAL BM, 
 CHANGE OF TRIM. 
 
 Longitudinal Metacentre. — We now have to deal with 
 inclinations in a fore-and-aft or longitudinal direction. We 
 do not have the same difficulty in fixing on the fore-and-aft 
 position of the centre of gravity of a ship as we have in fixing 
 its vertical position, because we know that if a ship is floating 
 steadily at a given water-line, the centre of gravity must be in 
 the same vertical line as the centre of buoyancy, by the con- 
 ditions of equilibrium laid down on p. 89. It is simply a 
 matter of calculation to find the longitudinal position of the 
 centre of buoyancy of a ship when floating at a certain water- 
 line, if we have the form of the ship given, and thus the fore- 
 and-aft position of the centre of gravity is determined. 
 
 We have already dealt with the inclination of a ship in a 
 transverse direction, caused by shifting weights already on 
 board across the deckj and in a precisely similar manner we 
 can incline a ship in a longitudinal or fore-and-aft direction by 
 shifting weights along the deck in the line of the keel. The 
 trim of a ship is the difference between the draughts of water 
 forward and aft. Thus a ship designed to float at a draught 
 forward of 1 2 feet, and a draft aft of 1 5 feet, is said to trim 3 feet 
 by the stern. 
 
 We have, on p. 93, considered the definition of the trans- 
 verse metacentre, and the definition of the longitudinal meta- 
 centre is precisely analogous. 
 
 For a given water-line WL of a vessel, let B be the centre 
 of buoyancy (see Fig. 61), and BM the vertical through it. 
 
Longitudinal Metacentre, Longitudinal BM, etc. 133 
 
 Suppose the trim of the vessel to change slightly,' the vessel 
 retaining the same volume of displacement, B' being the new 
 centre of buoyancy, and B'M the vertical through it, meeting 
 
 iw- 
 
 in 
 
 Fig. 61. 
 
 BM in M. Then the point M is termed the longitiidinal 
 metacentre. 
 
 The distance between G, the centre of gravity of the ship, 
 and M, the longitudinal metacentre, is termed the longitudinal 
 metacentric height. 
 
 Formula for finding the Distance of the Longi- 
 tudinal Metacentre above the Centre of Buoyancy. — 
 Let Fig. 62 represent the profile of a ship floating at the water- 
 line W'L', the original water-line being WL. The original 
 trim was AW — BL ; the new trim is AW - BL'. The change 
 of trim is — 
 
 (AAV - BL) - (AW - BL') = AVW -j- LL' 
 
 i.e. the change of trim is the sum of the changes of draughts 
 forward and aft. This change, we may suppose, has been 
 caused by the shifting of weights from aft to forward. The 
 inclination being regarded as small, and the displacement 
 remaining constant, the line of intersection of the water-planes 
 WL, WL' must pass through the centre of gravity of the water- 
 plane WL, or, as we have termed it, the centre of flotation, 
 in accordance with the principle laid down on p. 94.-. This 
 centre of flotation will usually be abaft the middle of lengtli, 
 and this introduces a complication which makes the calculation 
 for the longitudinal metacentre more difficult than the corre- 
 ' Much exaggerated in the figure. 
 
134 
 
 Theoretical Naval Architecture. 
 
 spending calculation for the transverse metacentre. In this 
 latter case, it will be remembered that the centre of flotation is 
 in the middle line of the water-plane. 
 
 -« 
 
 _E » 
 
 Fig. 62. 
 
 In Fig. 62 — 
 
 Let B be the centre of buoyancy when floating at the 
 water-line WL ; 
 B', the centre of buoyancy when floating at the water- 
 line W'L' ; 
 FF, the intersection of the water-planes WL, W'L' ; 
 V, the volume of either the immersed wedge FLL' or 
 the emerged wedge FWW ; 
 g, g', the centres of gravity of the wedges WEW', LFL' 
 respectively ; 
 V, the volume of displacement in cubic feet ; 
 6, the angle between the water-lines WL, W'L', which 
 is the same as the angle between BM and B'M 
 (this angle is supposed very small). 
 We have, using the principle laid down on p. 96 — 
 
 V X gg' = V X BB' 
 
Longitudinal Metacentre, Longitudinal BM, etc. 135 
 
 orBB' = .^^-"^ 
 
 V 
 But BB' = BM X 6 (61 is in circular measure) 
 
 BM X 6 = 
 
 ,_irKgl_ 
 
 V 
 
 The part of this expression that we do not know is z" X g^, 
 or the moment of transference of the wedges. At P take a 
 small transverse slice of the wedge FLL', of breadth in a fore- 
 and-aft direction, dx; length across, 2y; and distance from 
 F, X. Then the depth of the slice is — 
 
 xxQ 
 and the volume is 2y y. xQ 'K dx 
 
 This is an elementary volume, analogous to the elementary 
 area y . dx used in finding a large area. The moment of this 
 elementary volume about the transverse line FF is — 
 
 %yx .O.dxXx 
 or 2yx^ . . dx 
 
 If we summed all such moments as this for the length FL, 
 we should get the moment v X F^', and for the length FW, 
 V X F^, or for the whole length, v X gg' ; therefore, using our 
 ordinary notation — 
 
 V X gg" = J2yx^ .O.dx 
 
 = 2$Jyx^.dx (6 being constant) 
 
 We therefore have — 
 
 BMxe = ^^^^ 
 
 arBM = '-^^ 
 
 Referring to p. 99, it will be seen that we defined the 
 moment of inertia of an area about a given axis as — 
 
 jdA xy 
 
 where dA is a small elementary area ; 
 
 y its distance from the given axis. 
 Consider, now, the expression obtained, 2fyx^ . dx. The 
 elementary area is 2y . dx, and x is its distance from a 
 
136 
 
 Theoretical Naval Architecture. 
 
 transverse axis passing through the centre of flotation. We 
 may therefore say — 
 
 where lo is the moment of inertia of the water-plane about a 
 transverse axis passing through the centre of flotation. It will 
 be seen at once that this is the same form of expression as for 
 the transverse BM. 
 
 The method usually adopted for finding the moment of 
 inertia of a water-plane about a transverse axis through the 
 centre of flotation is as follows ' : — 
 
 We first find the moment of inertia about the ordinary 
 midship ordinate. If we call this I, and y the distance of the 
 centre of flotation from the midship ordinate, we have, using 
 the principle given on p. 100 — 
 
 I = lo + Aj'^ 
 or Ij = I — Ky- 
 
 The method actually adopted in practice will be best under- 
 stood by working the following example. 
 
 Numbers 
 of 
 
 Semi- 
 
 Simpson's 
 
 Products 
 for area. 
 
 Multi- 
 
 Products 
 for 
 
 Multi- 
 pliers for 
 
 Products 
 for 
 
 ordinates. 
 
 ofL.W.P. 
 
 pliers. 
 
 moment. 
 
 moment. 
 
 of inertia. 
 
 moment 
 of inertia. 
 
 I 
 
 O'O 
 
 \ 
 
 o-o 
 
 5 
 
 0-0 
 
 5 
 
 O'O 
 
 Ij 
 
 1-37 
 
 2 
 
 274 
 
 •Ah 
 
 i2"33 
 
 4J 
 
 55 '49 
 
 2 
 
 2-67 
 
 I* 
 
 4'oi 
 
 4 
 
 16-04 
 
 4 
 
 64-16 
 
 3 
 
 4-87 
 
 4 
 
 19-48 
 
 3 
 
 58-44 
 
 3 
 
 175-32 
 
 4 
 
 6-31 
 
 2 
 
 12-62 
 
 2 
 
 25-24 
 
 2 
 
 50-48 
 
 S 
 
 6-85 
 
 4 
 
 27-40 
 
 I 
 
 27-40 
 
 I 
 
 27-40 
 
 6 
 
 7-21 
 
 2 
 
 14-42 
 
 
 
 139-45 
 
 
 
 — 
 
 7 
 
 7'iS 
 
 4 
 
 28-60 
 
 I 
 
 28-60 
 
 I 
 
 2860 
 
 8 
 
 6-8; 
 
 2 
 
 1374 
 
 2 
 
 27-48 
 
 2 
 
 54-96 
 
 9 
 
 6-33 
 
 4 
 
 25-32 
 
 3 
 
 75-96 
 
 3 
 
 227-88 
 
 10 
 
 5-08 
 
 4 
 
 7-62 
 
 4 
 
 30-48 
 
 4 
 
 121-92 
 
 io| 
 
 3-56 
 
 2 
 
 7-12 
 
 4^ 
 
 32-04 
 
 4* 
 
 144-18 
 
 II 
 
 071 
 
 y 
 
 0-35 
 
 5 
 
 175 
 
 S 
 
 S75 
 
 163-42 
 
 196-31 
 
 i39'45 
 56-86 
 
 959'i4 
 
 ' This calculation for the I^.W-P. is usually performed on the displace- 
 nt sheet. 
 
 jnent sheet. 
 
Longitudinal Metacentre, Longitudinal BM, etc. 137 
 
 In column 2 of the table are given the lengths of semi- 
 
 ordinates of a load water-plane corresponding to the numbers 
 
 of the ordinates in column i. The ordinates are yx feet 
 
 apart. It is required to find the longitudinal BM, the dis- 
 placement being 91 '6 tons in salt water. 
 
 The distance apart of the ordinates being 7*1 feet, we 
 have — 
 
 Area = 163-42 X (3 X 7*i) X 2 
 = 773'S square feet 
 Distance of centre of gravity ofl s6"86x7"i 
 water-plane abaft No. 6 ordinate J ~ 1 63-42 ~ 47 e 
 
 (the stations are numbered from forward). 
 
 The calculation up to now has been the ordinaiy one 
 for finding the area and position of the centre of gravity. 
 Column 4 is the calculation indicated by the formula — 
 
 Area = 2^y . dx 
 
 Column 6 is the calculation indicated by the formula — 
 
 Moment = 2/jw . dx 
 
 It will be remembered that in column 5 we do not put 
 down the actual distances of the ordinates from No. 6 ordinate, 
 but the number of intervals away; the distance apart of the 
 ordinates being introduced at the end. By this means the 
 result is obtained with much less labour than if column 5 
 contained the actual distances. The formula we have for the 
 moment of inertia is 2^y . x^ . dx. We follow a similar process 
 to that indicated above ; we do not multiply the ordinates by 
 the square of the actual distances, but by the square of the 
 number of intervals away, leaving to the end the multiplication 
 by the square of the interval. Thus for ordinate No. 2 the 
 actual distance from No. 6 is 4X7'i = 28-4 feet. The 
 square of this is (4)^ X (Tif- For ordinate No. 4 the square of 
 the distance is (2)^ x (7'i)^- The multiplication by (7'i)^ can 
 be done at the end. In column 7 is placed the number of 
 intervals from No. 6, as in column 5 ; and if the products in 
 column 6 are multiplied successively by the numbers in 
 
138 TJuoretical Naval Architecture. 
 
 column 7, we shall obtain in column 8 the ordinates put 
 through Simpson's rule, and also multiplied by the square of 
 the number of intervals from No. 6 ordinate. The whole of 
 column 8 is added up, giving a result 959'i4. To obtain the 
 moment of inertia about No. 6 ordinate, this has to be multi- 
 plied as follows : — 
 
 (a) By one-third the common interval to complete Simp- 
 son's rule, or J x 7'i. 
 
 {b) By the square of the common interval, for the reasons 
 fully explained above. 
 
 (c) By two for both sides. 
 We therefore have the moment of inertia of the water-plane 
 about No. 6 ordinate- — 
 
 959'i4 X (I X 7-i) X (7-1)' X 2 = 228,858 
 The moment of inertia about a transverse axis through the 
 centre of flotation will be less than this by considering the 
 formula I = !„ + h.y', where I is the value found above about 
 No. 6 ordinate, and !» is the moment of inertia we want. We 
 found above that the area A = 773*5 square feet, and y = 2"47 
 feet ; 
 
 .-. I„ = 228,858 - (773-5 X 2-47^) 
 = 224,139 
 
 The displacement up to this water-plane is 9i-6 tons, and 
 the volume of displacement is — 
 
 9i'6 X 35 = 3206 cubic feet 
 
 The longitudinal BM = :^ 
 
 224139 r . e . 
 
 = > = 6g'9 feet 
 
 3200 ^ '^ 
 
 Approximate Formula for the Height of the Longi- 
 tudinal Metacentre above the Centre of Buoyancy.— 
 
 The following formula is due to M. J. A. Normand, M.I.N.A,,' 
 and is found to give exceedingly good results in practice :— 
 
 Let L be the length on the load water-line in feet ; 
 B, the breadth amidships in feet ; 
 
 ' See "Transactions of the Institution of Naval Architect?," 18S2. 
 
Longitudinal Metacentre, Longitudinal BM, etc. 139 
 
 V, the volume of displacement in cubic feet ; 
 
 A, the area of the load water-plane in square feet. 
 Then the height of the longitudinal metacentre above the centre 
 of buoyancy — 
 
 A^X L 
 
 H = O 07^1;=; ;rf 
 
 In the example worked above, the breadth amidships was 
 i4'42 feet; and using the formula, we find — 
 
 H = 6 7 "5 feet nearly 
 
 This compares favourably with the actual result of 69-9 feet. 
 The quantities required for the use of the formula would all be 
 known at a very early stage of a design and a close approxima- 
 tion to the height H can thus very readily be obtained. A 
 formula such as this is useful as a check on the result of the 
 calculation for the longitudinal BM. 
 
 We may also obtain an approximate formula in the same 
 manner as was done for the transverse BM on p. 107. Using 
 a similar system of notation, we may say — 
 
 Moment of inertia of L.W.P. about a trans- 1 , t ?, ^ b 
 verse axis through the centre of flotation j 
 
 ;/' being a coefficient of a similar nature to ;/ used on p. 103. 
 Volume of displacement = /iXLxBxD 
 
 ■'■ ^~ kx L X B X D 
 
 where /; is a coefficient obtained from the coefficients 71' and k. 
 Sir William White, in the " Manual of Naval Architecture," says, 
 with reference to the value of b, that "the value 0-075 may be 
 used as a rough approximation in most cases ; but there are 
 many exceptions to its use." If this approximation be applied 
 to the example we have worked, the mean moulded draught 
 being 5 '8 feet — 
 
 The value of H = 65 feet 
 
I40 Theoretical Naval Architecture. 
 
 This formula shows very clearly that the length of a ship 
 is more effective than the draught in determining the value 
 of the longitudinal BM in any given case. For vessels which 
 have an unusual proportion of length to draught, the values 
 of the longitudinal BM found by using this formula will not 
 be trustworthy. 
 
 To estimate the Displacement of a Vessel when 
 floating out of the Designed Trim. — The following 
 method is found useful when it is not desired to actually 
 calculate the displacement from the drawings, and a close 
 approximation is sufficiently accurate. Take a ship floating 
 parallel to her designed L.W.L. ; we can at once determine 
 the displacement when floating at such a water-line from the 
 curve of displacement (see p. 23). If now a weight already 
 on board is shifted aft, say, the ship will change trim, and she 
 will trim more by the stern than designed. The new water- 
 plane must pass through the centre of gravity of the original 
 water-plane, or, as we have termed it, the centre of flotation, and 
 
 4V 
 LL 
 
 Fig. 63. 
 
 the displacement at this new water-line will be, if the change of 
 trim is not very considerable, the same as at the original water- 
 line. Now, when taking the draught of water a vessel is 
 actually floating at, we take the figures set up at or near the 
 forward and after perpendiculars. These draught-marks should 
 be either at the perpendiculars or equally distant from them. 
 The draughts thus obtained are added together and divided 
 by two, giving us the mean draught. Now inn a line parallel to 
 the designed water-line at this mean draught, as in Fig. 63, where 
 WL represents the actual water-line, and wl the line just drawn. 
 It will not be true that the displacement of the ship is the same 
 as that given by the water-line wl. Let F be the centre of 
 
Longitudinal Meiacentre, Longitudinal BM, etc. 141 
 
 flotation of the water-line wl, and draw W'L' through F parallel 
 to WL. Then the actual displacement will be that up to W'L', 
 which is nearly the same as that up to wl, with the displacement 
 of the layer WW'L'L added. The displacement up to wl is 
 found at once from the curve of displacement. Let T be the 
 tons per inch at wl, and therefore very nearly the tons per inch 
 at W'L' and WL. SF, the distance the centre of flotation of 
 the water-plane ivl is abaft the middle of length, is supposed 
 known, and equals d inches, say. Now, the angle between wl 
 and WL is given by — 
 
 ^ , w\N -f /L 
 tan 6 = 
 
 length of ship 
 _ amount out of trim 
 length of ship 
 
 But if X is the thickness of layer in inches between W'L' and 
 WL, we also have in the triangle SFH — 
 
 tan ^ = -j very nearly (for small angles tan = sin ^ 
 very nearly) 
 
 and accordingly x may be determined. This, multiplied by 
 the tons per inch T, will give the displacement of the layer. 
 The following example will illustrate the above ; — 
 
 Example. — A vessel floats at a draught of 16' 5J" forward, 23' ij" aft, 
 the normal trim being 2 feet by the stern. At a draught of 19' 9J", her 
 displacement, measured from the curve of displacement, is 53^° tons, the 
 tons per inch is 31 'I tons, and the centre of flotation is 1 2 "9 feet abaft 
 amidships. Estimate the ship's displacement. 
 
 The difference in draught is 23' ij" — 16' 5I" = 6' 8", or 4' 8" out of 
 trim. The distance between the draught-marks is 335 feet, and wc 
 therefore have for the thickness of the layer — 
 
 12 X I2'Q X — ^ = 2-1 S inches 
 
 ^ 335 X 12 ^ 
 
 The displacement of the layer is therefore — 
 
 2-15 X 3i'i = 67 tons 
 
 The displacement is therefore — 
 
 5380 + 67 = 5447 tons nearly 
 
 Change of Trim due to Longitudinal Shift of 
 Weights already on Board.— We have seen that change 
 
142 
 
 Theoretical Naval Architecture. 
 
 of trim is the sum of the change of draughts forward and aft, and 
 that change of trim can be caused by the shift of weights on 
 board in a fore-and-aft direction. We have here an analogous 
 case to the inclining experiment in which heeling is caused by 
 shifting weights in a transverse direction. In Fig. 64, let w be 
 
 Fig. 64. 
 
 a weight on the deck when the vessel is floating at the water- 
 line WL, G being the position of the centre of gravity. Now 
 suppose the weight w to be shifted forward a distance of d feet. 
 G will, in consequence of this, move forward parallel to the line 
 joining the original and final positions of w, and if AV be the 
 displacement of the ship in tons, G will move to G' such that— 
 
 GG' 
 
 w X d 
 
 Now, under these circumstances, the condition of equilibrium 
 is not fulfilled if the water-line remains the same, viz. that the 
 centre of gravity and the centre of buoyancy must be in the 
 same vertical line, because G has shifted to G'. The ship 
 must therefore adjust herself till the centre of gravity and the 
 centre of buoyancy are in the same vertical line, when she 
 will float at a new water-line, W'L', the new centre of buoyancy 
 being B'. The original vertical through G and B meets the 
 new vertical through G' and B' in the point M, and this point 
 will be the longitudinal metacentre, supposing the change of 
 trim to be small, and GM will be the longitudinal metacentric 
 height. Draw W'C parallel to the original water-line WL, 
 
Longitudinal Metacentre, Longitudinal BM, etc. 143 
 
 meeting the forward perpendicular in C. Tlien, since CL = 
 \V\Y, the change of trim AVW'+ LL'= CL' = .v, say. The 
 angle of inclination of AV'L' to ^VL is the same as the angle 
 between VV'L' and \VC = d, say, and— 
 
 , /I i^lj X 
 
 ~ length L 
 But we also have — 
 
 therefore, equating these two values for tan 6, we have — 
 
 X _ GG' 
 L" GM 
 
 _ w X d 
 
 ~ W X GM 
 
 using the value obtained above for GG' ; or — 
 
 X, the change of trim due to the \ , 
 
 moment of transference of the 
 weight w through the distance d, 
 
 moment of transference of the \ = -^ .-, ^ X L feet 
 
 or- 
 
 „, , ^ ^ . ... 12 X w X a' X I- 
 
 1 he change of trmi m mches = ,t? t^tt — 
 
 * W X GM 
 
 and the moment to change trim i inch is — 
 
 ^ W X GM ^ 
 
 w y, d= =- foot-tons 
 
 12 X L 
 
 To determine this expression, we must know the vertical 
 position of the centre of gravity and the position of the longi- 
 tudinal metacentre. The vertical position of the centre of 
 gravity will be estimated in a design when dealing with the 
 metacentric height necessary, and the distance between 
 the centre of buoyancy and the centre of gravity is then sub- 
 tracted from the value of the longitudinal BM found by one of 
 the methods already explained. The distance BG is, however, 
 small compared with either of the distances BM or GM that 
 any small error in estimating the position of the centre of 
 gravity cannot appreciably affect the value of the mohient to 
 change trim one inch. In many ships BM approximately 
 
144 Theoretical Naval Architecture. 
 
 equals the length of the ship, and therefore GM also ; we may 
 therefore say that in such ships the moment to change trim 
 I inch = j^ the displacement in tons. For ships that are long 
 in proportion to the draught, the moment to change trim i inch 
 is greater than would be given by this approximate rule. 
 
 In the ship for which the value of the longitudinal BM was 
 calculated on p. 136, the centre of buoyancy was 2\ feet below 
 the L.W.L., the centre of gravity was estimated at i^ feet 
 below the L.W.L. ; and the length between perpendiculars was 
 75 feet. 
 
 .•. GM = 69-9 - I 
 = 68-9 feet 
 
 , , , ... Qi"6 X 68'o 
 
 and the moment to change trmi i mch = ^ 
 
 ^ 12X75 
 
 = 7'oi foot-tons 
 
 the draughts being taken at the perpendiculars. 
 
 Example. — A vessel 300 feet long and 2200 tons displacement has a 
 longitudinal metacentric height of 490 feet. Find the change of trim 
 caused by moving a weight of 5 tons already on board through a distance 
 of 200 feet from forward to aft. 
 
 Here the moment to change trim i inch is — 
 
 2200 X 490 - 
 
 = 300 foot-tons nearly 
 
 12 X 300 -^ ' 
 
 The moment aft due to the shift of the weight is — 
 
 5 X 200 = 1000 foot-tons 
 
 and consequently the change of trim aft is — 
 
 -S = 3j inches 
 
 Approximate Formula for the Moment to change Trim i inch. 
 — Assuming Normand's approximate formula for the height 
 of the longitudinal metacentre above the centre of buoyancy 
 given on p. 139 — 
 
 A^XL 
 H = 0-0735 BlTV 
 
 •we may construct an approximate formula for the moment to 
 change trim i inch as follows. 
 
 We have seen that the moment to change trim i inch is — 
 
 W X G M 
 12 X L 
 
Longitudinal Metacentre, Longitudinal BM, etc. 145 
 
 V 
 We can write W = — and assume that, for all practical pur- 
 poses, BM = GM = 0-0735-^—^ 
 
 Substituting this in the above formula, we have — 
 
 Moment to change 1 _ V f ^ A^ X L 
 
 trim I inch | ~ 35 x 12 X L ^ V °'°735b-^^- 
 
 A^ 
 or o'oooi75^ 
 
 For further approximations, see Example 18, p. 157. 
 Applying this to the case worked out in detail on p. 136 — 
 
 Area of L.W.P. = A = 773*5 square feet 
 Breadth = B = i4'42 feet 
 
 so that the moment to change trim i inch approximately 
 should equal — 
 
 0-000I7S = 7-26 foot-tons 
 
 '^ 14-42 
 
 the exact value, as calculated on p. 144, being 7'oi foot-tons. 
 
 It is generally sufficiently accurate to assume that one-half 
 the change of trim is forward, and the other half is aft. In the 
 example on p. 144, if the ship floated at a draught of 12' 3" 
 forward and 14' 9" aft, the new draught forward would be — • 
 
 I2'3"-lf"=I2'li" 
 
 and the new draught aft would be — 
 
 14' 9" 4- -iV = 14' 10 
 
 Referring, however, to Fig. 64, it will be seen that when, 
 as is usually the case, the centre of flotation is not at the middle 
 of the length, '^'^' is not equal to LL', so that, strictly speak- 
 ing, the total change of trim should not be divided by 2, and 
 one-half taken forward and the other half aft. Consider the 
 triangles FWW', FLL'; these triangles are similar to one 
 
 L 
 
14 G Theoretical Naval Architecture. 
 
 another, and the corresponding sides are proportional, so 
 
 that— 
 
 WW' ^ LL' 
 
 WF ~ LF 
 
 and both these triangles are similar to the triangle W'CL'. 
 
 Consequently — 
 
 WW' _ LU _ CL' _ change of trim 
 
 ' WF ~ LF ~ W'C ~ length 
 
 WF 
 .". WW = j -|- X change of trim 
 
 T F 
 
 and LL' = -.. -r X change of trim 
 
 that is to say, the proportion of the change of trim either aft or 
 forward, is the proportion the length of the vessel abaft or 
 forward of the centre of flotation bears to the length of the 
 vessel. Where the change of trim is small, this makes no 
 appreciable difference in the result, but there is a difference 
 when large changes of trim are under consideration. 
 
 For example, in the case worked out on p. 144, suppose 
 a weight of 50 tons is moved through 100 feet from forward to 
 aft • the change of trim caused would be — 
 
 ^0^ = i6f inches 
 
 The centre of flotation was 12 feet abaft the middle of length. 
 The portion of the length abaft the centre of flotation is there- 
 fore 1^ of the length. The increase of draught aft is there- 
 fore — 
 
 iff X ¥- = 7f inches 
 and the decrease of draught forward is — 
 
 Mf X ¥ = 9 inches 
 instead of 8|^ inches both forward and aft. The draught 
 forward is therefore — 
 
 123—9=110 
 and the draught aft — 
 
 14 9 +73 — ^S 4-3 
 It will be noticed that the mean draught is not the same as 
 
Longitudinal Metacentre, Longitudinal BM, etc. 147 
 
 before the shifting, but two-thirds of an inch less, while the 
 displacement remains the same. This is due to the fact that, 
 as the ship increases her draught aft and decreases it forward, a 
 fuller portion of the ship goes into the water and a finer portion 
 comes out. 
 
 Effect on the Trim of a Ship due to adding a 
 Weight of Moderate Amount. — If we wish to place a 
 weight on board a ship so that the vessel will not change trim, 
 we must place it so that the upward force of the added buoyancy 
 will act in the same line as the downward force of the added 
 weight. Take a ship floating at a certain water-line, and 
 imagine her to sink down a small amount, so that the new 
 waterplane is parallel to the original water-plane. The added 
 buoyancy is formed of a layer of parallel thickness, and having 
 very nearly the shape of the original water-plane. The upward 
 force of this added buoyancy will act through the centre of 
 gravity of the layer, which will be very nearly vertically over 
 the centre of gravity of the original water-plane, or, as we have 
 termed it, the centre of flotation. We therefore see that to 
 place a weight of moderate amount on a ship so that no 
 change of trim takes place, we must place it vertically over or 
 under the centre of flotation. The ship will then sink to a new 
 water-line parallel to the original water-line, and the distance 
 she will sink is known at once, if we know the tons per inch 
 at the original water-line. Thus a ship is floating at a draught 
 of 13 feet forward and 15 feet aft, and the tons per inch immer- 
 sion is 20 tons. If a weight of 55 tons be placed over or under 
 the centre of flotation, she will sink ff inches, or 2\ inches, 
 and the new draught will be 13' 2f ' forward and 15' 2f" aft. 
 
 It will be noticed that we have made two assumptions, both 
 of which are rendered admissible by considering that the weight 
 is of moderate amount. First, that the tons per inch does not 
 change appreciably as the draught increases, and this is, for all 
 practical purposes, the case in ordinary ships. Second, that the 
 centre of gravity of the parallel layer of added buoyancy is in 
 the same section as the centre of flotation. This latter assump- 
 tion may be taken as true for small changes in draught caused 
 by the addition of weights of moderate amount ; but for large 
 
148 Theoretical Naval Architecture. 
 
 changes it will not be reasonable, because the centres of gravity 
 of the water-planes are not all in the same section, but vary for 
 each water-plane. As a rule, water-planes are fuller aft than 
 forward near the L.W.P., and this more so as the draught 
 increases ; and so, if we draw on the profile of the sheer drawing 
 a curve through the centres of gravity of water-planes parallel to 
 the L.W.P., we should obtain a curve which slopes somewhat 
 aft as the draught increases. We shall discuss further the 
 methods which have to be adopted when the weights added 
 are too large for the above assumptions to be accepted. 
 
 We see, therefore, that if we place a weight of moderate 
 amount on board a ship at any other place than over the centre 
 of flotation, she will not sink in the water to a water-line 
 parallel to the original water-line, but she will change trim as 
 well as sink bodily in the water. The change of trim will be 
 forward or aft according as the weight is placed forward or 
 aft of the centre of flotation. 
 
 In determining the new draught of water, we proceed in 
 two steps : — 
 
 1. Imagine the weight placed over the centre of flotation. 
 
 2. Then imagine the weight shifted either forward or aft to 
 the assigned position. This shift will produce a certain moment 
 forward or aft, as the case may be, equal to the weight multiplied 
 by its longitudinal distance from the centre of flotation. This 
 moment divided by the moment to change trim i inch as cal- 
 culated for the original water-plane will give the change of trim. 
 
 The steps will be best illustrated by the following example: — 
 
 A vessel is floating at a draught of 12' 3" forward and 14' 6" aft. The 
 tons per inch immersion is 20 ; length, 300 feet ; centre of flotation, 12 feet 
 abaft the middle of length ; moment to change trim I inch, 300 foot-tons. 
 A weight of 30 tons is placed 20 feet from the forward end of the ship. 
 What will be the new draught of water ? 
 
 The first step is to see the sinkage caused by placing the weight over 
 the centre of flotation. This sinkage is i J inches, and the draughts would 
 then be — 
 
 12' 4j" forward, 14' •j\" aft 
 
 Now, the shift from the centre of flotation to the given position is 142 
 feet, so that the moment forward is 30 x 142 foot-tons, and the change 
 qf trim by the bow is — 
 
 30 X 142 
 
 -— -, or 141 mches nearly 
 
Longitudinal Metacentre, Longitudinal BM, etc. 149 
 
 This has to be divided up in the ratio of 138 : 162, because the centre 
 of flotation is 12 feet abaft the middle of length. We therefore have — 
 
 Increase of draught forward Jgf x 14J" = 7f say 
 Decrease of draught aft Jll X 14^" = 6|" say 
 
 The final draughts will therefore be — 
 
 Forward, 12' 4.I" + 75" = 13' oi" 
 Aft, 14' 71" - 6r' = 14' I" 
 
 Effect on the Trim of a Ship due to adding a 
 Weight of Considerable Amount. — In this case the 
 assumptions made in the previous investigation will no longer 
 hold, and we must allow for the following : — 
 
 1. Variation of the tons per inch immersion as the ship 
 sinks deeper in the water. 
 
 2. The centre of flotation does not remain in the same 
 transverse section. 
 
 3. The addition of a large weight will alter the position 
 of G, the centre of gravity of the ship. 
 
 4. The different form of the volume of displacement will 
 alter the position of B, the centre of buoyancy of the ship, and 
 also the value of BM. 
 
 5. Items 3 and 4 will alter the value of the moment to 
 change trim i inch. 
 
 As regards i, we can obtain first an approximation to the 
 sinkage by dividing the added weight by the tons per inch 
 immersion at the original water-line. The curve of tons per 
 inch immersion will give the tons per inch at this new draught. 
 The mean between this latter value and the original tons per 
 inch, divided into the added weight, will give a very close 
 approximation to the increased draught. Thus, a vessel floats at 
 a constant draught of 22' 2", the tons per inch immersion 
 being 44*5. It is required to find the draught after adding a 
 weight of 750 tons. The first approximation to the increase of 
 
 draught is -^-^— = 17 inches nearly. At a draught of 23' 7" 
 
 44'5 ' 
 
 it is found that the tons per inch immersion is 457. The 
 mean tons per inch is therefore ■|-(44'5 + 457) = 4S'ii ^nd 
 
 the increase in draught is therefore -^-^— = 16-63, or i6| inches 
 
 45 I 
 
150 Theoretical Naval Architecture. 
 
 nearly. This assumes that the ship sinks to a water-plane 
 parallel to the first water-plane. In order that this can be the 
 case, the weight must have been placed in the same transverse 
 section as the centre of gravity of the layer of displacement 
 between the two water-planes. We know that the weight and 
 buoyancy of the ship must act in the same vertical line, and 
 therefore, for the vessel to sink down without change of trim, 
 the added weight must act in the same vertical line as the 
 added buoyancy. We can approximate very closely to the 
 centre of gravity of the layer as . follows : Find the centre of 
 flotation of the original W.P. and that of the parallel W.P. 
 to which the vessel is supposed to sink. Put these points on 
 the profile drawing at the respective water-lines. Draw a line 
 joining them, and bisect this line. Then this point will be 
 a very close approximation to the centre of gravity of the layer. 
 A weight of 750 tons placed as above, with its centre of gravity 
 in the transverse section containing 'this point, will cause the 
 ship to take up a new draught of 23' 6f" with no change of trim. 
 We can very readily find the new position of G, the centre 
 of gravity of the ship due to the addition of the weight. Thus, 
 suppose the weight of 750 tons in the above example is placed 
 with its centre of gravity 16 feet below the C.G. of the ship; 
 then, supposing the displacement before adding the weight to 
 be 9500 tons, we have — 
 
 ^ „ 750 X 16 
 
 Lowermg of G = -— 
 
 ° 10250 
 
 = i'i7 feet 
 
 We also have to take account of 4. In the case we have 
 taken, the new C.B. below the original water-line was 97 feet, 
 as against io'5 feet in the original condition, or a rise of o'8 
 foot. 
 
 For the new water-plane we have a different longitudinal 
 BM, and, knowing the new position of B and of G, we can deter- 
 mine the new longitudinal metacentric height. From this we 
 can obtain the new moment to change trim i inch, using, of 
 course, the new displacement. In the above case this works 
 out to 950 foot-tons. 
 
Longitudinal Metacentre, Longitudinal BM, etc. 151 
 
 Now we must suppose that the weight is shifted from the 
 assumed position in the same vertical line as the centre of 
 gravity of the layer to its given position, and this distance must 
 be found. The weight multiplied by the longitudinal shift will 
 give the moment changing the trim either aft or forward, as the 
 case may be. Suppose, in the above case, this distance is 50 feet 
 forward. Then the moment changing trim by the bow is — 
 
 750 X 50 = 37,500 foot-tons 
 
 and the approximate change of trim is — 
 
 37)5°° -T- 950 = 395 inches 
 This change of trim has to be divided up in the ordinary 
 way for the change of draught aft and forward. In this case 
 we have — 
 
 Increase of draught forward = fi|- x 39i = 2ii inches say 
 Decrease of draught aft = ^^ X 39^ = 18 inches say 
 
 We therefore have for our new draughts — 
 
 Draught aft, 25' 2" + i6f" - 18" = 22' of" 
 Draught forward, 22' 2" + i6f" + 2ii'' = 25' 4!" 
 
 P'or all ordinary purposes this would be sufficiently accu- 
 rate ; but it is evidently still an approximation, because we do 
 not take account of the new GM for the final water-line, and 
 the consequent new moment to change trim i inch. These can 
 be calculated if desired, and corrections made where necessary. 
 . To determine the Position of a Weight on Board 
 a Ship such that the Draught aft shall remain 
 constant whether the Weight is or is not on Board. — 
 Take a ship floating at the water-line WL, as in Fig. 65. If 
 a weight w be placed with its centre of gravity in the transverse 
 section that contains the centre of flotation, the vessel will very 
 nearly sink to a parallel water-line W'L'.' This, however, is 
 not what is required, because the draught aft is the distance 
 WW greater than it should be. The weight will have to be 
 
 ' Strictly speaking, the weight should be placed with its centre of 
 gravity in the transverse section that contains the centre of gravity of the 
 zone between the water-lines WL and W'L'. 
 
152 Theoretical Naval Architecture. 
 
 moved forward sufi&cient to cause a change of trim forward of 
 WW -J- LL', and then the draught aft will be the same as it 
 originally was, and the draught forward will increase by the 
 amount W^' + LL'. This will be more clearly seen, perhaps, 
 by working the following example : — 
 
 It is desired that the draught of water aft in a steamship 
 (particulars given below) shall be constant, whether the coals 
 
 ^"so»l. 
 
 L 
 J 
 
 Fig. 65. 
 
 are in or out of the ship. Find the approximate position of 
 the centre of gravity of the coals in order that the desired 
 condition may be fulfilled : Length of ship, 205 feet; displace- 
 ment, 522 tons (no coals on board) ; centre of flotation from 
 after perpendicular, io4'3 feet; longitudinal BM, 664 feet; 
 longitudinal GM, 66i'5 feet; tons per inch, ii'4; weight of 
 coals, 57 tons. 
 
 From the particulars given, we find that — 
 
 Moment to change 1 66i'5 x 522 
 
 ■ T f = — =140 foot-tons 
 
 trim I inch | 12 X 205 ^ 
 
 The bodily sinkage, supposing the coals placed with the centre 
 of gravity in the transverse section containing the centre of 
 
 57 
 flotation, will be — = 5 inches. Therefore the coals must 
 114 •' 
 
 be shifted forward from this position through such a distance 
 
 that a change of trim of 10 inches forward is produced. 
 
 Accordingly, a forward moment of — 
 
 140 X 10 = 1400 foot-tons 
 
 is required, and the distance forward of the centre of flotation 
 the coals require shifting is — 
 
 iffi = 24-6 feet 
 
Longitudinal Metacentre, Longitudinal BM, etc. 153 
 
 Therefore, if the coals are placed — 
 
 io4'3 + 24"6 = i28'9 feet 
 
 forward of the after perpendicular, the draught aft will remain 
 very approximately the same as before. 
 
 Change of Trim caused by a Compartment being 
 open to the Sea. — The principles involved in dealing with 
 a problem of this character will be best understood by working 
 out the following example : — 
 
 A rectangular-shaped lighter, 100 feet long, 40 feet broad, 
 10 feet deep, floating in salt water at 3 feet level draught, has 
 a collision bulkhead 6 feet from the forward end. If the side 
 is broached before this bulkhead below water, what would be 
 the trim in the damaged condition ? 
 
 Let ABCD, Fig. 66, be the elevation of the lighter, with a 
 
 Fig. 66. 
 
 collision bulkhead 6 feet from the forward end, and floating 
 at the level water-line WL. It is well to do this problem 
 in two stages — 
 
 1. Determine the amount of mean sinkage due to the loss 
 of buoyancy. 
 
 2. Determine the change of trim caused. 
 
 I. The lighter, due to the damage, loses an amount of 
 buoyancy which is represented by the shaded part GB, and if 
 we assume that she sinks down parallel, she will settle down at 
 a water-line wl such that volume wG = volume GB. This 
 will determine the distance x between wl and WL. 
 
154 Theoretical Naval Architecture. 
 
 For the volume wG = wH X 40 feet X x 
 and the volume GB = GL X 40 feet X 3 feet 
 . ^^4lX_6_X_3^^f^^^ 
 94 X 40 »* 
 
 = 2-j inches nearly 
 
 2. We now deal with the change of trim caused. 
 
 The volume of displacement = 100 x 40 X 3 cubic feet 
 
 The weight of the lighter = ^°° ^ 4° X 3 ^ 3^ ^^^^ 
 
 and this weight acts down through G, the centre of gravity, 
 which is at 50 feet from either end. 
 
 But we have lost the buoyancy due to the part forward of 
 bulkhead EF, and the centre of buoyancy has now shifted 
 back to B' such that the distance of B' from the after end is 
 47 feet. Therefore we have W, the weight of lighter, acting 
 down through G, and W, the upward force of buoyancy, acting 
 through B'. These form a couple of magnitude — 
 
 \N Y.Z feet = ^^ X 3 = ^^ foot-tons 
 
 tending to trim the ship forward. 
 
 To find the amount of this trim, we must find the moment 
 to change trim i inch — 
 
 _ W X GM 
 12 X L 
 
 using the ordinary notation. 
 
 Now, GM very nearly equals BM ; 
 
 2400 
 
 .•. moment to change trim i inch = — ^ x BM 
 
 ° 12 X 100 
 
 = f X BM 
 BM = i» 
 
 where I,, = the moment of inertia of the intact water-plane about 
 
 a transverse axis through its centre of gravity ; 
 and V = volume of displacement in cubic feet. 
 
Longitttdinal Metacentre, Longitudinal BM, etc. 155 
 
 I = 12(94 X 40) X (94)' 
 V = 12,000 
 
 ,. BM = 45_Xi94)! 
 144000 
 
 and moment to alter trim i inch = -^^ 
 
 7 X 144000 
 
 = 66 foot-tons nearly 
 .'. the change of trim = l^^o. _j_ 55 
 = T-Si inches 
 
 The new water-line W'L' will pass through the centre 
 of gravity of the water-line wl at K, and the change of trim 
 aft and forward must be in the ratio 47 : 53 ; or — 
 
 Decrease of draught aft = jVo X 15^= lk inches 
 Increase of draught forward = -^-^ x 15 ^ = 8i inches 
 
 therefore the new draught aft is given by — 
 
 3' o" + 2i" - 7i" = 2' 7" 
 and the new draught forward by — 
 
 3' o" + 2I" + 8i" = 3' loi" 
 
 The same result would be obtained by considering the 
 weight of water in the compartment GB acting downwards, 
 and taking its moment about the centre of flotation K of the 
 intact part of the water-line wl. This gives a moment forward 
 of— 
 
 ( ^ 40 X 3 \ ^ foot-tons = H— foot-tons 
 ^ 35 ' 
 
 as obtained above. 
 
 It will be noticed that we have assumed that the moment to 
 change trim for the water-plane wl remains constant as the 
 vessel changes trim. The slight alteration can be allowed for, 
 if thought desirable, by taking the mean between the moment 
 to change trim for the water-planes wl and WL', and using 
 that to determine the change of trim. 
 
156 Theoretical Naval Architecture. 
 
 Examples to Chapter IV. 
 
 I . A ship is floating at a draught of 20 feet forward and 22 feet aft, when 
 the following weights are placed on board in the positions named : — 
 
 Weight Distance front C.G. of 
 
 in tons. water-plane in feet. 
 
 20 IOO\ 
 
 45 80/ 
 
 60 501 
 
 30 10/ 
 
 before 
 abaft 
 
 What will be the new draught forward and aft, the moment to change 
 trim I inch being 800 foot-tons, and the tons per inch = 35 .'' 
 
 Alls. 20' Sf" forward, 22' 3" aft. 
 
 2. A vessel 300 feet long, designed to float with a trim of 3 feet by 
 the stern, owing to consumption of coal and stores, floats at a draught of 
 9' 3" forward, and 14' 3" aft. The load displacement at a mean draught of 
 13' 6" is 2140 tons; tons per inch, 18 J ; centre of flotation, 12.J feet abaft the 
 middle of length. Approximate as closely as you can to the displacement. 
 
 Ans. 1775 tons. 
 
 3. A vessel is 300 feet long and 36 feet beam. Approximate to the 
 moment to change trim I inch, the coefificient of fineness of the L.W.P. 
 being 075. 
 
 Ans. 319 foot-tons. 
 
 4. A light-draught stern-wheel steamer is very approximately of the form 
 of a rectangular box of 120 feet length and 20 feet breadth. When fully 
 laden, the draught is 18 inches, and the centre of gravity of vessel and 
 lading is 8 feet above the water-line. Find the transverse and longitudinal 
 metacentric heights, and also the moment to change trim one inch. 
 
 Ans. I3'47 feet, 7914 feet ; 56J foot-tons. 
 
 5. A vessel is floating at a draught of 12' 3" forward and 14' 6" aft. 
 The tons per inch immersion is 20 ; length, 300 feet ; centre of flotation, 
 12 feet abaft amidships ; moment to change trim I inch, 300 foot-tons. 
 Where should a weight of 60 tons be placed on this vessel to bring her to 
 an even keel. 
 
 Ans. 123 feet forward of amidships. 
 
 6. What weight placed 13 feet forward of amidships will have the same 
 effect on the trim of a vessel as a weight of 5 tons placed 10 feet abaft the 
 forward end, the length of the ship being 300 feet, and the centre of 
 flotation 12 feet abaft amidships. 
 
 Ans. 30'4 tons. 
 
 7. A right circular pontoon 50 feet long and 16 feet in diameter is just 
 half immersed on an even keel. The centre of gravity is 4 feet above the 
 bottom. Calculate and state in degrees the transverse heel that would be 
 produced by shifting 10 tons 3 feet across the vessel. State, in inches, the 
 change of trim produced by shifting 10 tons longitudinally through 20 feet. 
 
 Ans. 3 degrees nearly ; 25 inches nearly. 
 
 8. Show why it is that many ships floating on an even keel will increase 
 the draught forward, and decrease the draught aft, or, as it is termed, go 
 down by the head, if a weight is placed at the middle of the length. 
 
 9. Show that for vessels having the ratio of the length to the draught 
 about 13, the longitudinal B.M. is approximately equal to the length. 
 Why should a shallow draught river steamer have a longitudinal B.M. 
 much greater than the length ? What type of vessel would have a longitu- 
 dinal B.M. less than the length ? 
 
Longitudinal Metacentre, Longitjtdinal BM, etc. 157 
 
 10. Find the moment to change trim i inch of a vessel 400 feet long, 
 having given the following particulars : Longitudinal metacentre above 
 centre of buoyancy, 44.6 feet ; distance between centre of gravity and centre 
 of buoyancy, 14 feet ; displacement, 15,000 tons. 
 
 Ans. 1350 foot-tons. 
 
 11. The moment of inertia of a, water-plane of 22,500 square feet 
 about a transverse axis 20 feet forward of the centre of flotation, is found 
 to be 254,000,000 in foot-units. The displacement of the vessel being 
 14,000 tons, determine the distance between the centre of buoyancy and 
 the longitudinal metacentre. 
 
 Ans. 500 feet. 
 
 12. In the preceding question, if the length of the ship is 405 feet, and 
 the distance between the centre of buoyancy and the centre of gravity is 13 
 feet, determine the change of trim caused by the longitudinal transfer of 
 150 tons through 50 feet. 
 
 Ans. 5| inches nearly. 
 
 13. A water-plane has an area of 13,200 square feet, and its moment of 
 inertia about a transverse axis 14J feet forward of its centre of gravity 
 works out to 84,539,575 in foot-units. The vessel is 350 feet long, and 
 has a displacement to the above water-line of 5600 tons. Determine the 
 moment to change trim l inch, the distance between the centre of gravity 
 and the centre of buoyancy being estimated at 8 feet. 
 
 Ans. 546 foot-tons. 
 
 14. The semi-ordinates of a water-plane of a ship 20 feet apart are as 
 follows: 0-4, 7-5, 14-5, 2I-0, 26-6, 30-9, 34-0, 36-0, 37-0, 37-3, 37-3, 
 37'3. 37'3. 37'2, 37"i. 36'8, 3S'8. 33"4. z8-8, 21 7, 11-5 feet respectively. 
 The after appendage, whole area 214 square feet, has its centre of gravity 
 6'2 feet abaft the last ordinate. Calculate — 
 
 (i) Area of the water-plane. 
 
 (2) Position of C.G. of water-plane. 
 
 (3) Transverse B.M. 
 
 (4) Longitudinal B.M. 
 
 (Volume of displacement up to the water-plane 525,304 cubic feet.) 
 Ans. (i) 24,015 square feet ; (2) i8'2 feet abaft middle ordinate ; 
 (3) I7*l6feet; (4) 447 '6 feet. 
 
 15. The semi-ordinates of the L.W.P. of a vessel 15J feet apart are, 
 commencing from forward, o"i, 2'5, 5'3, 8'i, iO'8, 13*1, I5'0, i6'4, I7'6, 
 18-3, 18-5, 18-5, 18-4, i8-i, 17-5, i6-6, 15-3, 13-3, IO-8, 7-6, 3-8 feet 
 respectively. Abaft the last ordinate there is a portion of the water-plane, 
 the half-area being 27 square feet, having its centre of gravity 4 feet abaft 
 the last ordinate. Calculate the distance of the longitudinal metacentre 
 above the centre of buoyancy, the displacement being 2206 tons. 
 
 Ans. 534 feet. 
 
 16. State the conditions that must hold in order that a vessel shall not 
 change trim in passing from river water to salt water. 
 
 17. A log of fir, specific gravity o'5, is 12 feet long, and the section is 
 2 feet square. What is its longitudinal metacentric height when floating in 
 stable equilibrium ? 
 
 Ans. i6'5 feet nearly. 
 
 18. Using the approximate formula for the moment to change trim i 
 inch given on p. 145, show that this moment will be very nearly given by 
 
 30 . ^5", where T is the tons per inch immersion, and B is the breadth. 
 B 
 
 Show also that in ships of ordinary form, the moment to change trim 
 I inch approximately equals J5J55 . L'B. 
 
CHAPTER V. 
 
 STATICAL STABILITY, CURVES OF STABILITY, CALCU- 
 LATIONS FOR CURVES OF STABILITY, INTEGRATOR, 
 DYNAMICAL STABILITY. 
 
 Statical Stability at Large Angles of Inclination. 
 Atwood's Formula. — ^We have up to the present only dealt 
 with the stability of a ship at small angles of inclination, and 
 within these limits we can determine what the statical stability is 
 by using the metacentric method as explained on p. 94. We 
 must now, however, investigate how the statical stability of a 
 ship can be determined for large angles of inclination, because 
 in service it is certain that she will be heeled over to much 
 larger angles than 10° to 15°, which are the limits beyond which 
 we cannot employ the metacentric method. 
 
 Let Fig. 67 represent the cross-section of a ship inclined 
 to a large angle 6. WL is the position on the ship of the 
 original water-line, and B the original position of the centre of 
 buoyancy. In the inclined position she floats at the water-line 
 W'L', which intersects WL in the point S, which for large angles 
 will not usually be in the middle line of the ship. The volume 
 SWW' is termed, as before, the " emerged wedge" and the volume 
 SLL' the " immersed wedge" and g, ^ are the positions of the 
 centres of gravity of the emerged and immersed wedges respec- 
 tively. The volume of displacement remains the same, and 
 consequently these wedges are equal in volume. Let this 
 volume be denoted by v. The centre of buoyancy of the 
 vessel when floating at the water-line W'L' is at B', and the 
 upward support of the buoyancy acts through B'; the downward 
 force of the weight acts through G, the centre of gravity of the 
 ship. Draw GZ and BR perpendicular to the vertical through 
 B', and gh,^K perpendicular to the new water-line W'L'. Then 
 
Statical Stability, Curves of Stability, etc. 159 
 
 the moment of the couple tending to right the ship is W x GZ, 
 or, as we term it, the moment of statical stability. Now — 
 
 GZ = BR - BP 
 
 = BR - BG sin B 
 
 so that the moment of statical stability at the angle B is — 
 W(BR - BG . sin 6) 
 
 The length BR is the only term in this expression that we 
 do not know, and it is obtained in the following manner. The 
 
 Fig. 67. 
 
 new volume of displacement W'AL' is obtained from the old 
 volume WAL by shifting the volume WSW to the position 
 LSL', through a horizontal distance JiH. Therefore the hori- 
 zontal shift of the centre of gravity of the immersed volume 
 from its original position at B, or BR, is given by — 
 
 V X hH 
 
 BR = 
 
 V 
 
 (using the principle discussed on p. 96). Therefore the 
 moment of statical stability at the angle B is — 
 
 W 
 
 ■vxkk' 
 
 . V 
 
 — BG . sin ^ I foot-tons 
 
 This is known as " Atwood's formula," 
 
i6o Theoretical Naval Architecture. 
 
 V X hH 
 The righting arm or lever = — ^ BG . sin B 
 
 If we want to find the length of the righting arm or lever 
 at a given angle of heel B, we must therefore know — 
 
 (i) The position of the centre of buoyancy B in the up- 
 right condition. 
 
 (2) The position of the centre of gravity G of the ship. 
 
 (3) The volume of displacement V. 
 
 (4) The value of the moment of transference of the wedges 
 parallel to the new water-line, viz. v X hli . 
 
 This last expression involves a considerable amount of cal- 
 culation, as the form of a ship is an irregular one. The methods 
 adopted will be fully explained later, but for the present we 
 will suppose that it can be obtained when the form of the ship 
 is given. 
 
 Curve of Statical Stability. — The lengths of GZ thus 
 obtained from Atwood's formula will vary as the angle of 
 heel increases, and usually GZ gradually increases until an 
 angle is reached when it obtains a maximum value. On 
 further inclination, an angle will be reached when GZ becomes 
 zero, and, further than this, GZ becomes negative when the 
 couple W X GZ is no longer a couple tending to right the 
 ship, but is an upsetting couple tending to incline the ship still 
 further. Take H.M.S. Captain ' as an example. The lengths 
 of the lever GZ, as calculated for this ship, were as follows : — 
 
 At 7 degrees, GZ= i,\ inches 
 
 14 
 
 
 
 = 8i 
 
 21 
 
 
 
 = io| 
 
 28 
 
 
 
 = 10 
 
 35 
 
 
 
 = 7f 
 
 42 
 
 
 
 = Si 
 
 49 
 
 
 
 = 2 
 
 S4i 
 
 
 
 = nil 
 
 Now set along a base-line a scale of degrees on a con- 
 
 ' The Captain was a rigged turret-ship which foundered in the Bay of 
 Biscay. A discussion of her stability will be found in " Naval Science," 
 vol. i. 
 
Statical Stability, Curves of Stability, etc. i6i 
 
 venient scale (say \ inch = i degree), and erect ordinates 
 at the above angles of the respective lengths given. If now 
 we pass a curve through the tops of these ordinates, we shall 
 obtain what is termed a " curve of statical stability," from 
 which we can obtain the length of GZ for any angle by drawing 
 the ordinate to the curve at that angle. The curve A, in Fig. 
 68, is the curve so constructed for the Captain. The angle 
 
 I 
 
 i 
 
 -Z£t 
 
 XI 
 
 
 r<MONARCH."— 
 
 2 
 
 -1 
 
 -IFX- 
 
 A 
 
 -Kr- 
 
 
 PTAir 
 
 1* 
 
 
 [\ 
 
 1- 
 
 z 
 u 
 
 & 
 
 y 
 
 
 
 
 
 \ 
 
 H. 21. 28. 3& 42. 49. S4i 
 
 -Angles of Inclination. 
 
 at which GZ obtains its maximum value is termed the " angle of 
 maximum stability" and the angle at which the curve crosses 
 the base-line is termed the " angle of vanishing stability" and 
 the number of degrees at which this occurs is termed the 
 " range of stability T If a ship is forced over beyond the angle 
 of vanishing stability, she cannot right herself j GZ having a 
 negative value, the couple operating on the ship is an up- 
 setting couple. 
 
 In striking contrast to the curve of stability of the Captain 
 is the curve as constructed for H.M.S. Monarch?- The lengths 
 of the righting levers at different angles were calculated as 
 follows : — • 
 
 At 7 degrees, GZ = 4 inches 
 
 „ 14 
 
 )J 
 
 )) 
 
 = 81 
 
 „ 21 
 
 )) 
 
 )J 
 
 = I2i 
 
 „ 28 
 
 )) 
 
 JJ 
 
 = i8i 
 
 » 35 
 
 )J 
 
 )> 
 
 = 2I| 
 
 ' The Monarch was a rigged ship built about the same time as the 
 Captain, but differing from the Captain in having greater freeboard. See 
 also the volume of "Naval Science " above referred to. 
 
 M • 
 
1 62 Theoretical Naval Architecture. 
 
 At 42 degrees, GZ = 22 inches 
 ij 49 jj 11 = 20 „ 
 
 jj 542" 3) It /a' " 
 
 „ 69! „ „ = nil 
 
 The curve for this ship, using the above values for GZ, is 
 given by B, Fig. 68. The righting lever goes on lengthening 
 in the MonarcKs case up to the large angle of 40°, and then 
 shortens but slowly ; that of the Captain begins to shorten at 
 about 21° of inclination, and disappears altogether at S4|-°, an 
 angle at which the Monarch still possesses a large righting lever. 
 
 Referring to Atwood's formula for the lever of statical 
 stability at the angle 0, viz. — 
 
 GZ=?:4M_BG.sine 
 
 we see that the expression consists of two parts. The 
 first part is purely geometrical, depending solely upon the 
 form of the ship ; the second part, BG . sin 0, brings in the 
 influence of the position of the centre of gravity of the ship, 
 and this depends on the distribution of the weights forming 
 the structure and lading of the ship. We shall deal with these 
 two parts separately. 
 
 (i) Influence oifoiin on curves of stability. 
 
 (2) Influence of position of centre of gravity on curves of 
 stability. 
 
 (i) We have here to take account of the form of the ship 
 above water, as well as the form of the ship below water. The 
 three elements of form we shall consider are draught, beam, 
 and freeboard. These are, of course, relative ; for con- 
 venience we shall keep the draught constant, and see what 
 variation is caused by altering the beam and freeboard. For 
 the sake of simplicity, let us take floating bodies in the form 
 of boxes.' The position of the centre of gravity is taken as 
 constant. Take the standard form to be a box : — 
 
 Draught 21 feet. 
 
 Beam S^i » 
 
 Freeboard 6| „ 
 
 ' These illustrations are taken from a paper read at the Institution of 
 Naval Architects by Sir N. Barnaby in 1871. 
 
Statical Stability, Curves of Stability, etc. 163 
 
 The curve of statical stability is shown in Fig. 69 by the 
 curve A. The deck-edge becomes immersed at an inclination 
 of 14^°, and from this angle the curve increases less rapidly 
 than before, and, having reached a maximum value, decreases, 
 the angle of vanishing stability being reached at about 38°. 
 
 Now consider the effect of adding 4^ feet to the beam, 
 thus making the box — 
 
 Draught 
 Beam 
 Freeboard .. 
 
 21 feet. 
 
 5S .. 
 6| „ 
 
 The curve is now given by B, Fig. 69, the angle of vanish- 
 ing stability being increased to about 45°. Although the 
 
 10. 20. 30. 40. 50- 60. 70. 
 
 — Angle of Inclination. — 
 
 Fig. 69. 
 
 position of the centre of gravity has remained unaltered, the 
 increase of beam has caused an increase of GM, the meta- 
 centric height, because the transverse metacentre has gone up. 
 We know that for small angles the lever of statical stability is 
 given by GM . sin B, and consequently we should expect the 
 curve B to start as shown, steeper than the curve A, because 
 GM is greater. There is a very important connection between 
 the metacentric height and the slope of the curve of statical 
 stability at the start, to which we shall refer hereafter. 
 
 Now consider the effect of adding i,\ feet to the freeboard 
 of the original form, thus making the dimensions — 
 
 Draught 
 
 Beam 
 
 Freeboard 
 
 21 feet. 
 
164 Theoretical Naval Architecture. 
 
 The curve is now given by C, Fig. 69, which is in striking 
 contrast to both A and B. The angle of vanishing stability- 
 is now 7 2°. The curves A and C coincide up to the angle at 
 which the deck-edge of A is immersed, viz. 14^°, and then, 
 owing to the freeboard still being maintained, the curve C 
 leaves the curve A, and does not commence to decrease 
 until 40°. 
 
 These curves are very instructive in showing the influence 
 of beam and freeboard on stability at large angles. We see — ■ 
 
 (a) An increase of beam increases the initial stability, and 
 therefore the slope of the curve near the origin, but does not 
 greatly influence the area enclosed by the curve or the range. 
 
 (b) An increase of freeboard has no effect on initial 
 stability (supposing the increase of freeboard does not affect 
 the centre of gravity), but has a most important effect in 
 lengthening out the curve and increasing its area. The two 
 bodies whose curves of statical stability are given by A and C 
 have the same GM, but the curves of statical stability are very 
 different. 
 
 (2) We now have to consider the effect on the curve of 
 
 statical stability of the position of the centre of gravity. If 
 
 the centre of gravity G is above the centre of buoyancy B, as is 
 
 V X hH 
 usually the case, the righting lever is less than — ^ — by the 
 
 expression BG . sin Q. Thus the deduction becomes greater as 
 the angle of inclination increases, because sin B increases as 6 
 increases, reaching a maximum value of sin ^ = i when & = 
 90°; the deduction also increases as the C.G. rises in the 
 ship. Thus, suppose, in the case C above, the centre of gravity 
 is raised 2 feet. Then the ordinate of the curve C at any 
 angle 6 is diminished by 2 x sin 6. For 30°, sin 6 = |, and 
 the deduction is there i foot. In this way we get the curve D, 
 in which the range of stability is reduced from 72° to 53° owing 
 to the 2-feet rise of the centre of gravity. 
 
 It is usual to construct these curves as indicated, the 
 ordinates being righting levers, and not righting moments. The 
 righting moment at any angle can be at once obtained by 
 multiplying the lever by the constant displacement. The real 
 
Statical Stability, Curves of Stability, etc. 165 
 
 curve of statical stability is of course a curve, the ordinates of 
 which represent righting moments. This should not be lost 
 sight of, as the following will show. In Fig. 70 are given the 
 
 curves of righting levers for a merchant vessel in two given 
 conditions, A for the light condition at a displacement of 
 1500 tons, and B for the load condition at a displacement of 
 3500 tons. Looking simply at these curves, it would be 
 thought that the ship in the light condition had the better 
 stability; but in Fig. 71, in which A represents the curve of 
 
 righting moments in the light condition, and curve B the curve 
 of righting moments in the load condition, we see that the 
 ship in the light condition has very much less stability than in 
 the load condition. 
 
 We see that the following are the important features of a 
 curve of statical stability : — 
 
 (a) Inclination the tangent to the curve at the origin has to 
 the base-line ; 
 
 {b) The angle at which the maximum value occurs, and the 
 length of the righting lever at this angle j 
 
 {c) The range of stability. 
 
1 66 
 
 Theoretical Naval Architecture. 
 
 
 
 M. 
 
 
 
 // 
 
 i 
 
 
 ^ 
 
 -^^ 
 
 l_ 
 
 X 
 
 \ 
 
 \ 
 
 \ 
 
 n 
 
 
 57.3.U 
 
 
 Fig. 72. 
 
 The angle the tangent at the origin makes with the base- 
 line can be found in a very simple manner as follows : At the 
 
 angle whose circular measure 
 is unity, viz. 5 7 '3°, erect a 
 perpendicular to the base, 
 and make its length equal to 
 the metacentric height GM, 
 for the condition at which 
 the curve has to be drawn, 
 using the same scale as for 
 the righting levers (see Fig. 
 72). Join the end of this 
 line with the origin, and the 
 curve as it approaches the origin will tend to lie along this 
 curve. 
 
 The proof of this is given below.^ 
 
 Specimen Curves of Stability. — In Fig. 73 are given 
 some specimen curves of stability for typical classes of ships. 
 
 A is the curve for a modern British battleship of about 3^ 
 feet metacentric height. The range is about 63°. 
 
 B is the curve for the American monitor Miantonomoh. 
 This ship had a low freeboard, and to provide sufficient stability 
 a very great metacentric height was provided. This is shown 
 by the steepness of the curve at the start. 
 
 C is the curve for a merchant steamer carrying a miscel- 
 laneous cargo, with a metacentric height of about 2 feet. In 
 
 ' For a small angle of inclination fl, we know that GZ = GM x 9, 
 9 being in circular measure ; 
 
 GZ GM 
 
 or -„- = 
 
 9 I 
 
 If now we express fl in degrees, say 9 = <^°, then — 
 
 GZ GM 
 
 (^° angle whose circular measure is I 
 GZ_ GM 
 
 " <f " 57-3° 
 If a is the angle OM makes with the base, then — 
 
 GM GZ 
 tana = — „=^ 
 
 and thus the line OM lies along the curve near the origin. 
 
Statical Stability, Curves of Stability, etc. 167 
 
 this ship there is a large righting lever even at 90°. It must 
 be stated that, although this curve is typical for many ships, yet 
 the forms of the curves of stability for merchant steamers must 
 vary considerably, owing to the many different types of ships 
 and the variation in loading. The curves for a number of 
 
 10. 20. 30. 40. SO. 60. 70. 80- 90. 
 
 — - Angles of Inclination. 
 
 Fig. 73. 
 
 merchant steamers are given in the " Manual of Naval Archi- 
 tecture," by Sir W. H. White, and the work on " StabiUty," by 
 Sir E. J. Reed. 
 
 D is the curve of stability for a sailing-ship having a meta- 
 centric height of 3-|- feet. For further examples see the works 
 referred to above. 
 
 Fig. 74 gives an interesting curve of stability for a vessel 
 
 70. so 90. 
 
 Fig. 74. 
 
 which is unstable in the upright condition, but stable at a 
 moderate angle of heel. This vessel has a negative meta- 
 centric height, and would not remain in the upright position, 
 but on heeling to an angle of 25° she will resist further inclina- 
 tion, and consequently, if left to herself, the vessel will loll over 
 
1 68 
 
 Theoretical Naval Architecture. 
 
 to this angle, and there be perfectly stable. Such a condition 
 is quite likely to occur in a steamship which starts on a voyage 
 with a small metacentric height, loaded with a homogeneous 
 cargo. Towards the end of the voyage the coal is nearly all 
 burnt out abreast the boilers, and this weight, low down in the 
 ship, being removed, causes the C.G. of the ship to rise, and 
 thus possibly be above the transverse metacentre. The ship 
 is then unstable in the upright condition, but would incline 
 over to the angle at which the curve of stability crosses the 
 base-line. 
 
 The curve of stability for a floating body of circular form is 
 
 very readily obtainable, because 
 the section is such that the 
 upward force of the buoyancy 
 always acts through the centre 
 of the section, as shown in Fig. 
 75. The righting lever at any 
 angle is GM . sin Q, where G 
 is the centre of gravity, and 
 M the centre of the section. 
 Taking the GM as two feet, 
 then the ordinates of the curve 
 of stability are o, i"o, 173, 2'o, 
 i'73, i"o, o at intervals of 30^ 
 The maximum occurs at 90°, and the range is 180°. The 
 curve is shown in Fig. 76. 
 
 -2Ft- 
 
 30. 60. 90. 120. 
 
 — Decrees. — 
 
 Fig. 76. 
 
 Calculations for Curves of Stability. — We now pro- 
 ceed to investigate methods that are or have been adopted in 
 practice to determine for any given ship the curve of righting 
 levers. The use of the integrator is now very general for 
 
Statical Stability, Curves of Stability, etc. 169 
 
 doing this, and it saves an enormous amount of work ; but, in 
 order to get a proper grasp of the subject, it is advisable to 
 understand the methods that vt^ere in use previous to the intro- 
 duction of the integrator. 
 
 In constructing and using curves of stability, certain assump- 
 tions have to be made. These may be stated as follows : — 
 
 1. The sides and deck are assumed to be water-tight for the 
 range over which the curve is drawn. 
 
 2. The C.G. is taken in the same position in the ship, and 
 consequently we assume that no weights shift their position 
 throughout the inclination. 
 
 3. The trim is assumed to be unchanged, that is, the ship 
 is supposed to be constrained to move about a horizontal longi- 
 tudinal axis fixed in direction only, and to adjust herself to the 
 required displacement without change of trim. 
 
 It is not possible in this work to deal with all the systems 
 of calculation that have been employed ; a selection only will 
 be given in this chapter. For further information the student 
 is referred to the Transactions of the Institution of Naval 
 Architects, and to the work by Sir E. J. Reed on the 
 " Stability of Ships." The following are the methods that will 
 be discussed : — 
 
 1. Blom's mechanical method. 
 
 2. Barnes' method. 
 
 3. Direct method (sometimes employed as a check on 
 other methods). 
 
 4. By Amsler's Integrator and Cross-curves of stability. 
 
 I. Blom's Mechanical Method. — Take a sheet of 
 drawing-paper, and prick off from the body-plan the shape of 
 each equidistant section ^ {i.e. the ordinary sections for displace- 
 ment), and cut these sections out up to the water-line at which 
 the curve of stability is required, marking on each section the 
 middle line. Now secure all these sections together in their 
 proper relative positions by the smallest possible use of gum. 
 
 ' In settling the sections to be used for calculating stability by any of the 
 methods, regard must be had to the existence of a poop or forecastle the 
 ends of which are watertight, and the ends of these should as nearly as 
 possible be made stop points in the Simpson's rule. 
 
170 Theoretical Naval Architectttre. 
 
 The weight of these represents the displacement of the ship. 
 Next cut out sections of the ship for the angle at which the 
 stability is required, taking care to cut them rather above 
 the real water-line, and gum together in a similar manner to 
 the first set. Then balance these sections against the first 
 set, and cut the sections down parallel to the incUned water- 
 line until the weight equals that of the first set. When this 
 is the case, we can say that at the inclined water-line the 
 displacement is the same as at the original water-line in the up- 
 right condition. This must, of course, be the case as the vessel 
 heels over. On reference to Fig. 67, it will be seen that what 
 we want to find is the hne through the centre of buoyancy for 
 the inclined position, perpendicular to the inclined water-line, 
 so that if we can find B' for the inchned position, we can com- 
 pletely determine the stability. This is done graphically by 
 finding the centre of gravity of the sections we have gummed 
 together, and the point thus found will give us the position of 
 the centre of buoyancy for the inclined condition. This is 
 done by successively suspending the sections and noting where 
 the plumb-lines cross, as explained on p. 49. Having then 
 the centre of buoyancy, we can draw through it a line perpen- 
 dicular to the inclined water-line, and if we then spot off the 
 position of the centre of gravity, we can at once measure oif 
 the righting lever GZ. A similar set of sections must be made 
 for each angle about 10° apart, and thus the curve of stability 
 can be constructed. 
 
 2. Barnes's Method of calculating Statical 
 Stability. — In this method a series of tables are employed, 
 called Preliminary and Combination Tables, in which the work 
 is set out in tabulated form. Take the section in Fig. 77 to 
 represent the ship, WL being the upright water-line for the con- 
 dition at which the curve of stability is required. Now, for a 
 small transverse angle of inclination it is true that the new water- 
 plane for the same displacement will pass through the centre 
 line of the original water-plane WL, but as the angle of inclina- 
 tion increases, a plane drawn through S will cut off a volume of 
 displacement sometimes greater and sometimes less than the 
 original volume, and the actual water-line will take up some 
 
Statical Stability, Curves of Stability, etc. 171 
 
 such position as W'L', Fig. 77, supposing too great a volume to 
 be cut off by the plane through S. Now, we cannot say 
 straight off where the water-line W'L' will come. What we 
 have to do is this : Assume a water-line wl passing through 
 S ; find the volume of the assumed immersed wedge /SL, the 
 volume of the assumed emerged wedge wSW, and the area of 
 the assumed water-plane wl. Then the difference of the 
 volumes of the wedges divided by the area of the water-plane 
 will give the thickness of the layer between wl and the correct 
 water-plane, supposing the difference of the volumes is not too 
 great. If this is the case, the area of the new water-plane is 
 
 Fig. 77. 
 
 found, and a mean taken between it and the original. In 
 this way the thickness of the layer can be correctly found. 
 If the immersed wedge is in excess, the layer has to be de- 
 ducted; ifithe emerged wedge is in excess, the layer has to be 
 added. 
 
 To get the volumes of either of the wedges, we have to 
 proceed as follows : Take radial planes a convenient angular 
 interval apart, and perform for each plane the operation sym- 
 bolized by ^/y^ . dx, i.e. the half-squares of the ordinates 
 are put through Simpson's rule in a fore-and-aft direction for 
 each of the planes. Then put the results through Simpson's 
 rule, using the circular measure of the angular interval. The 
 
1/2 Theoretical Naval Architecture. 
 
 result will be the volume of the wedge at the particular angle. 
 For proof of this see below.^ 
 
 The results being obtained for the immersed and emerged 
 wedges, we can now determine the thickness of the layer. This 
 work is arranged as follows : The preliminary table, one table 
 for each angle, consists of two parts, one for the immersed 
 wedge, one for the emerged wedge. A specimen table is given 
 on p. 174 for 30°. The lengths of the ordinates of each radial 
 plane are set down in the ordinary way, and operated on by 
 Simpson's multipliers, giving us a function of the area on the 
 immersed side of 550'i, and on the emerged side of 477'4. 
 We then put down the squares of the ordinates, and put them 
 through the Simpson's multipliers, giving us a result for the 
 immersed side of 17,888, and for the emerged side 14,250. 
 The remainder of the work on the preliminary table will be 
 described later. 
 
 We now proceed to the combination table for 30° (see 
 p. 175), there being one table for each angle. The functions 
 of squares of ordinates are put down opposite their respective 
 angles, both for the immersed wedge and the emerged wedge, 
 up to and including 30°, and these are put through Simpson's 
 multipliers. In this case the immersed wedge is in excess, and 
 so we find the volume of the layer to be taken off to be 7836 
 cubic feet, obtaining this by using the proper multipliers. At 
 the bottom is placed the work necessary for finding the thickness 
 of the layer. We have the area of the whole plane 20,540 
 square feet, and this divided into the excess volume of the 
 immersed wedge, 7836 cubic feet, gives the thickness of the 
 layer to take off, viz. o'382 foot, to get the true water-line. 
 
 We now have to find the moment of transference of the 
 
 ' The area of the section S/L is given by J jy' . dB, as on p. 15, and the 
 volume of the wedge is found by integrating these areas right fore and 
 aft, or — 
 
 ijjy^ .de.dx 
 which can be written — 
 
 j//)/2 .dx.de 
 
 or /(j/>''' • <^x)de 
 i.e. ijy^ . dx is found for each radial plane, and integrated with respect to 
 the angular interval. 
 
Statical Stability, Curves of Stability, etc. 173 
 
 wedges, v X kh' in Atwood's formula, and this is done by using 
 the assumed wedges and finding their moments about the line ST, 
 and then making at the end the correction rendered necessary 
 by the layer. To find these moments we proceed as follows : 
 In the preliminary table are placed the cubes of the ordinates of 
 the radial plane, and these are put through Simpson's rule ; the 
 addition for the emerged and immersed sides are added 
 together, givirig us for the 30° radial plane 1,053,633. These 
 sums of functions of cubes are put in the combination table for 
 each radial plane up to and including 30°, and they are put 
 through Simpson's rule, and then respectively multiplied by 
 the cosine of the angle made by each radial plane with the 
 extreme radial plane at 30°. The sum of these products gives 
 us a function of the sum of the moments of the assumed im- 
 mersed and emerged wedges about ST. The multiplier for the 
 particular case given is o"3878, so that the uncorrected moment 
 of the wedges is 3,391,662,^ in foot-units, i.e. cubic feet, multi- 
 plied by feet. 
 
 ' The proof of the process is as follows : Take a section of the wedge 
 Sffv, Fig. 78, and draw ST perpendicular to S/. Then what is required is 
 the moment of the section about ST, and this 
 integrated throughout the length. Take P and 
 P' on the curved boundary, very close together, 
 and join SP, SP' ; call the angle P'Sl, fl, and 
 the angle PSP', d9. 
 
 Then the area PSP' = |/ . rfe SP = j/ 
 
 The centre of gravity of SPP' is distant 
 from ST, \y . cos fl, and the moment of SPP' 
 about ST is— 
 
 (Jy2 . M) X (Ik . cos 9) Fig. 78. 
 
 or \}fl . cos 8 . d0 
 
 We therefore have the moment of /SL about ST — 
 
 5/_j/3 . cos 9 . de 
 
 and therefore the moment of the wedge about ST is — 
 
 /(j/j/S . cos 9 . <ie)dx 
 orljfy^.cose .dx.de 
 
 i.e. find the value of J fy' . cos 9 . dx for radial planes up to and including 
 the angle, and then integrate with respect to the angulalr interval. It will 
 be seen that the process described above corresponds with this formula. 
 

 PRELIMINARY TABLE FOR 
 
 STABILITY. 
 
 
 
 WATER 
 
 SECTION INCLINED AT 30°. 
 
 
 
 
 
 
 Immersed 
 
 Wedge. 
 
 
 o 
 
 i 
 
 c 
 'a 
 O 
 
 .a 
 1 
 
 EC4 
 
 u 
 
 3-3 
 
 1 
 
 "5. 
 
 
 
 ii' 
 
 "S3 
 
 J.I 
 "1 
 
 a 
 
 "a, 
 "a 
 
 •s 
 II 
 
 r 
 
 I 
 
 4-S 
 
 i 
 
 I'l 
 
 20 
 
 i 
 
 s 
 
 91 
 
 i 
 
 23 
 
 ij 
 
 II-3 
 
 1 
 
 II-3 
 
 128 
 
 I 
 
 128 
 
 1,443 
 
 I 
 
 1,443 
 
 2 
 
 i8-4 
 
 i 
 
 9-2 
 
 339 
 
 * 
 
 169 
 
 6,230 
 
 i 
 
 3.115 
 
 2^ 
 
 24-4 
 
 I 
 
 24-4 
 
 595 
 
 I 
 
 595 
 
 14,527 
 
 I 
 
 14,527 
 
 3 
 
 28-6 
 
 i 
 
 21-2 
 
 818 
 
 f 
 
 ^^3 
 
 23.394 
 
 1 
 
 17.546 
 
 4 
 
 33 '8 
 
 2 
 
 67-6 
 
 1,142 
 
 2 
 
 2,284 
 
 38,614 
 
 2 
 
 77.228 
 
 5 
 
 35-5 
 
 I 
 
 35-5 
 
 1,260 
 
 I 
 
 1,260 
 
 44,739 
 
 I 
 
 44,739 
 
 6 
 
 35-6 
 
 2 
 
 71-2 
 
 1,267 
 
 2 
 
 2,534 
 
 45,118 
 
 2 
 
 90,236 
 
 7 
 
 35-6 
 
 I 
 
 35 -6 
 
 1,267 
 
 1 
 
 1,267 
 
 45,118 
 
 I 
 
 45,118 
 
 8 
 
 35 '6 
 
 2 
 
 71-2 
 
 1,267 
 
 2 
 
 2,534 
 
 45,118 
 
 2 
 
 90,236 
 
 9 
 
 35-6 
 
 I 
 
 35-6 
 
 1,267 
 
 I 
 
 1,267 
 
 45, "8 
 
 I 
 
 45,118 
 
 lO 
 
 35-6 
 
 2 
 
 71-2 
 
 1,267 
 
 2 
 
 2,534 
 
 45,118 
 
 2 
 
 90,236 
 
 n 
 
 33-8 
 
 1 
 
 25-3 
 
 1,142 
 
 i 
 
 857 
 
 38,614 
 
 f 
 
 28,961 
 
 "J 
 
 31-0 
 
 I 
 
 31-0 
 
 961 
 
 I 
 
 961 
 
 29,791 
 
 I 
 
 29,791 
 
 12 
 
 26-9 
 
 * 
 
 13-5 
 
 724 
 
 i 
 
 362 
 
 19.465 
 
 i 
 
 9.732 
 
 I2| 
 
 21-6 
 
 I 
 
 21-6 
 
 467 
 
 I 
 
 467 
 
 10,078 
 
 1 
 
 10,078 
 
 '3 
 
 14-3 
 
 i 
 
 3-6 
 
 20s 
 
 i 
 
 51 
 
 2,924 
 
 i 
 
 731 
 
 550- 1 
 
 17,888 
 
 598,858 
 
 
 
 
 Emer 
 
 3ED V 
 
 l^EDGE. 
 
 
 I 
 
 47 
 
 i 
 
 I '2 
 
 22 
 
 i 
 
 5 
 
 104 
 
 i 
 
 26 
 
 li 
 
 93 
 
 I 
 
 9-3 
 
 86 
 
 I 
 
 86 
 
 804 
 
 I 
 
 804 
 
 2 
 
 13-8 
 
 1 
 
 6-9 
 
 190 
 
 i 
 
 95 
 
 2,628 
 
 i 
 
 1,314 
 
 2J 
 
 17-8 
 
 I 
 
 17-8 
 
 317 
 
 I 
 
 317 
 
 5,640 
 
 1 
 
 5,640 
 
 3 
 
 21-4 
 
 i 
 
 161 
 
 458 
 
 f 
 
 343 
 
 9,800 
 
 i 
 
 7.350 
 
 4 
 
 27-3 
 
 2 
 
 S4'6 
 
 745 
 
 2 
 
 1,490 
 
 20,346 
 
 2 
 
 40,692 
 
 S 
 
 32-8 
 
 I 
 
 328 
 
 1,076 
 
 I 
 
 1,076 
 
 35,288 
 
 I 
 
 35.288 
 
 6 
 
 367 
 
 2 
 
 73 '4 
 
 1,347 
 
 2 
 
 2,694 
 
 49,431 
 
 2 
 
 98,862 
 
 7 
 
 38-2 
 
 I 
 
 38-2 
 
 1.459 
 
 I 
 
 1,459 
 
 55.743 
 
 I 
 
 55.743 
 
 8 
 
 36-S 
 
 2 
 
 73-0 
 
 1.332 
 
 2 
 
 2,664 
 
 48,627 
 
 2 
 
 97.254 
 
 9 
 
 33 '6 
 
 I 
 
 33-6 
 
 1,129 
 
 I 
 
 1,129 
 
 37,933 
 
 I 
 
 37.933 
 
 lO 
 
 29-3 
 
 2 
 
 58-6 
 
 858 
 
 2 
 
 1,716 
 
 25. '54 
 
 2 
 
 50,308 
 
 II 
 
 23-8 
 
 f 
 
 17-8 
 
 566 
 
 1 
 
 425 
 
 13,481 
 
 1 
 
 10,111 
 
 Hi 
 
 20-6 
 
 I 
 
 20-6 
 
 424 
 
 I 
 
 424 
 
 8,742 
 
 I 
 
 8,742 
 
 12 
 
 i6'9 
 
 § 
 
 8-5 
 
 286 
 
 J 
 
 143 
 
 4,827 
 
 J 
 
 2.4'3 
 
 I2j 
 
 12-9 
 
 I 
 
 I2-9 
 
 166 
 
 I 
 
 166 
 
 2,147 
 
 I 
 
 2,147 
 
 13 
 
 8-4 
 
 i 
 
 2-1 
 
 71 
 
 i 
 
 18 
 
 593 
 
 i 
 
 148 
 
 477-4 
 
 14,250 
 
 Emerged 
 
 454.775 
 
 
 
 
 
 
 
 Immersed 
 
 598,858 
 
 1 
 
 
 
 1,053,633 
 
 ' The multipliers used here are half the ordinary Simpson's multipliers ; 
 the results are multiplied at the end by two to allow for this. 
 
COMBINATION TABLE FOR STABILITY. 
 
 CALCULATION FOR GZ AT 30°. 
 
 
 Immersed Wedge. 
 
 
 functions 
 of ordi- 
 >r both 
 es. 
 
 
 Cfl 
 
 lit 
 
 HI 
 
 0. s 
 
 E 
 
 sof 
 ordi- 
 
 lumes 
 es. 
 
 1, 
 
 js 
 z: a. 
 
 11 
 
 OJ Cll 
 
 'ft 
 
 1 
 
 Funclion 
 
 squares of 
 
 nates for vo 
 
 of wedg 
 
 Sums of 
 
 of cubes 
 
 nates f( 
 
 sid 
 
 1 
 
 
 
 — 
 
 1S.340 
 
 I 
 
 IS.340 
 
 974,388 
 
 [ 
 
 10 
 
 — 
 
 iS.7t)0 
 
 4 
 
 63,040 
 
 990,153 
 
 4 
 
 20 
 
 — 
 
 16,840 
 
 li 
 
 25,260 
 
 1,034,251 
 
 ij 
 
 2'; 
 
 — 
 
 '^'791 
 
 2 
 
 35,402 1,066,771 
 
 2 
 
 30 
 
 55° 
 
 17,888 
 
 4 
 
 8.944 
 
 1.053,633 
 
 * 
 
 ;5>a 
 
 974.388 
 3,960,612 
 I.SSI.377 
 2,133.542 
 
 526,816 
 
 " O M 
 
 0-8660 
 0-9397 
 0-9848 
 0-9962 
 
 I -0000 
 
 -3° 
 
 III 
 
 843,820 
 
 3,721,787 
 
 1.527.796 
 
 2,125,434 
 
 526,816 
 
 Immersed wedge 147,986 
 Emerged wedge 134,522 
 
 8,745.653 
 Multiplier f 0-3878 
 
 Multiplier * 
 Volume of layer 
 
 13,464 
 0-582 
 
 Uncorrected moment 3,391,662 
 Correction for layer 13,502 
 
 7,836 cubic feet Corrected moment 3,378, 160 
 
 Volume of displacement 398,090 
 
 Longitudinal interval = 30 feet 
 Circular measure 10° = 0-1745 
 
 BR 
 
 BG X sin 30° 
 
 GZ 
 
 8485 
 
 5-950 
 
 2-535 
 
 * Multiplier = J X 2 x (J X 30) X (J X 0-1745) = o'582 
 t Multiplier = | X 2 x (J X 30) X (J X 0-1745) = 0-3878 
 BG = n-90 ; sin 30° = 0-5 ; BG sin fl = 5-95 feet 
 
 Area and Position of C.G. of Radial Plane. 
 
 
 Emerged Wedge. 
 
 
 
 
 — 
 
 15,340 
 
 I 
 
 15,340 
 
 10 
 
 — 
 
 15,157 
 
 4 
 
 60,628 
 
 20 
 
 — 
 
 14,766 
 
 I* 
 
 22,149 
 
 25 
 
 — 
 
 14,640 
 
 2 
 
 29,280 
 
 30 
 
 477 
 
 14,250 
 
 i 
 
 7,125 
 
 Immersed wedge 
 Emerged wedge 
 
 134,522 
 
 Functions 
 of area. 
 
 550 
 
 477 
 
 1,027 
 
 Functions 
 
 of moment. 
 
 17,888 
 
 14,250 
 
 3,538 
 
 Area = 1027 X 2 x (J X 30) 
 = 20,540 square feet 
 
 C.G. of radial plane on immersed side = — 
 
 •^ 1027 
 
 Xi= 1-723 feet 
 
 Thickness of layer = ' ^ = 0-382 feet 
 20540 
 
176 Theoretical Naval Architecture. 
 
 We now have to make the correction for the layer. We 
 already have the volume of the layer, and whether it has to be 
 added or subtracted, and we can readily find the position of 
 the centre of gravity of the radial plane. This is done at the 
 bottom of the combination table from information obtained on 
 the preliminary table. We assume that the centre of gravity 
 of the layer is the same distance from ST as the centre of 
 gravity of the radial plane, which may be taken as the case, 
 unless the thickness of the layer is too great. If the layer 
 is thick, a new water-line is put in at thickness found, and the 
 area and C.G. of this water-line found. The mean between 
 the result of this and of the original plane can then be used. 
 The volume of the layer, 7836 cubic feet, is multiplied by the 
 distance of its centre of gravity from ST, viz. i'723 feet, giving 
 a result of 13,502 in foot-units, i.e. cubic feet multiplied by 
 feet. The correction for the layer is added to or subtracted 
 from the uncorrected moment in accordance with the following 
 rules : — 
 
 If the immersed wedge is in excess, and the centre of gravity 
 of the layer is on the immersed side, the correction for the layer 
 has to be subtracted. 
 
 If the hmnersed wedge is in excess, and the centre of gravity 
 of the layer is on the emerged side, the correction for the layer 
 has to be added. 
 
 If the emerged wedge is in excess, and the centre of gravity 
 of the layer is on the emerged side, the correction for the layer 
 has to be subtracted. 
 
 If the emetged wedge is in excess, and the centre of gravity 
 of the layer is on the immersed side, the correction for the layer 
 has to be added. 
 
 We, in this case, subtract the correction for the layer, 
 obtaining the true moment of transference of the wedges as 
 3,378,160, or z^ X hH in Atwood's formula. The volume of 
 displacement is 398,090 cubic feet; BG is 11 '90 feet; sin 30° 
 = o'5. So we can fill in all the items in Atwood's formula — 
 
 GZ = ^ - BG sin 
 or GZ = 2-535 feet 
 
Statical Stability, Curves of Stability, etc. iff 
 
 In arranging the radial planes, it is best to arrange that the 
 deck edge comes at a stop point in Simpson's first rule, because 
 there is a sudden change of ordinate as the deck edge is passed, 
 and for the same reason additional intermediate radial planes 
 are introduced near the deck edge. In the case we have been 
 considering, the deck edge came at about 30°. The radial 
 planes that were used were accordingly at — 
 
 0°, 10°, 20^ 25°, 30°, 35°, 40°, 50°, 60°, 70°, 80°, 90° 
 
 Barnes's method of calculating stability has been very largely 
 employed. It was introduced by Mr. F. K. Barnes at the Insti- 
 tution of Naval Architects in 1861, and in 1871 a paper was 
 read at the Institution by Sir W. H. White and the late Mr. 
 John, giving an account of the extensions of the system, with 
 specimen calculations. For further information the student is 
 referred to these papers, and also to the work on " Stability," 
 by Sir E. J. Reed. At the present time it is not used to any 
 large extent, owing to the introduction of the integrator, 
 which gives the results by a mechanical process in much less 
 tim.e. It will be seen that in using this method to find the 
 stability at a given angle, we have to use all the angles up to 
 and including that angle at which the stability is required. 
 Thus a mistake made in the table at any of the smaller angles 
 is repeated right through, and affects the accuracy of the 
 calculation at the larger angles. In order to obtain an inde- 
 pendent check at any required angle, we can proceed as 
 follows : — 
 
 3. Triangular or Direct Method of calculating 
 Stability. — Take the body-plan, and draw on the trial plane 
 through the centre of the upright water-line at the required 
 angle. This may or may not cut off the required displace- 
 ment. We then, by the ordinary rules of mensuration, dis- 
 cussed in Chapter I., find the area of all such portions as S/L, 
 Fig. 77, for all the sections,^ and also the position of the centre 
 of gravity, g, for each section, thus obtaining the distance Sh. 
 
 ' The sections are made into simple figures, as triangles and trapeziums, 
 in order to obtain the area and position of C.G. of each. 
 
 N 
 
1/8 
 
 Tlieoretical Naval Architecture. 
 
 This is done for both the immersed and emerged wedges. The 
 work can then be arranged in tabular form thus : 
 
 Number of 
 section. 
 
 Areas. 
 
 Simpson's 
 multipliers. 
 
 Products for 
 volume. 
 
 Levers as 
 S/i. 
 
 Products for 
 
 moment about 
 
 ST. 
 
 I 
 2 
 
 etc. 
 
 A, 
 A. 
 etc. 
 
 1 
 
 4 
 
 etc. 
 
 A: 
 4A2 
 etc. 
 
 etc. 
 
 AlJTi 
 
 4A2^j 
 etc. 
 
 The volume of the wedge = Si X -g- common interval 
 The moment of wedge about ST = Mj X -I common interval 
 
 This being done for both wedges, and calculating the area 
 of the radial plane, we can find the volume of the layer and the 
 uncorrected moment of the wedges. The correction for the 
 layer is added or subtracted from this, exactly as in Barnes's 
 method, and the remainder of the work follows exactly the 
 methods described above for Barnes's method. 
 
 There is the disadvantage about the methods we have 
 hitherto described, that we obtain a curve of stability for one 
 particular displacement, but it is often necessary to know the 
 stability of a ship at very different displacements to the ordinary 
 load displacement, as, for example, in the light condition, or 
 the launching condition. The method we are now about to 
 investigate enables us to determine at once the curve of stability 
 at any given displacement and any assumed position of the 
 centre of gravity. 
 
 4. Amsler's Integrator. Cross-curves of Stability. 
 — The Integrator is an extension of the instrument we have 
 described on p. 77, known as the planimeter. A diagram of 
 one form of the integrator is given in Fig. 79. A bar, BB, has 
 a groove in it, and the instrument has two wheels which run in 
 this groove. W is a balance weight to make the instrument 
 run smoothly. There are also three small wheels that run on 
 the paper, and a pointer as in the planimeter. By passing the 
 pointer round an area, we can find — 
 
 (i) A number which is proportional to the area, i£. a 
 function of the area. 
 
Statical Stability, Curves of Stability, etc. 179 
 
 (2) A function of the moment of the area about the axis 
 the bar is set to. 
 
 (3) A function of the moment of inertia of the area about 
 the same axis. 
 
 The bar is set parallel to the axis about which moments 
 are required, by means of distance pieces. 
 
 (i) is given by the reading indicated by the wheel marked A. 
 
 (2) is given by the reading indicated by the wheel marked M. 
 
 (3) is given by the reading indicated by the wheel marked I. 
 The finding of the moment of inertia is not required in our 
 
 present calculation. 
 
 Now let M'LMW represent the body-plan '^ of a vessel 
 inclined to an angle of 30° ; then, as the instrument is set, the 
 
 Fig. 79. 
 
 axis of moments is the line through S perpendicular to the 
 inclined water-line, and is what we have termed ST. What 
 we want to find is a line through the centre of buoyancy in the 
 inclined position perpendicular to the inclined water-line. By 
 passing the pointer of the instrument round a section, as 
 W'L'M, we can determine its area, and also its moment about 
 the axis ST by using the multipliers \ and doing this for all the 
 sections in the body, we can determine the displacement and 
 also the moment of the displacement about ST.^ Dividing the 
 
 ' The body-plan is drawn for both sides of the ship — the fore-body in 
 black say, and the after-body in red. 
 
 ' This is the simplest method, and it is the best for beginners to employ ; but 
 ■certain modifications suggest themselves after experience with the instrument. 
 
i8o 
 
 Theoretical Naval Architecture. 
 
 moment by the displacement, we obtain at once the distance 
 of the centre of buoyancy in the inclined condition from the 
 axis ST. It is convenient in practice to arrange the work in 
 a similar manner to that described for the planimeter, p. 79, 
 and the following specimen calculation for an angle of 30° will 
 illustrate the method employed. Every instrument has multi- 
 pliers for converting the readings of the wheel A into areas, 
 and those of the wheel M into moments. The multipliers 
 must also take account of the scale used. 
 
 30° 
 
 Sections. 
 
 Areas. 
 
 Moments. 
 
 » 
 
 i 
 
 
 
 bJ9 
 
 « 
 
 y-i 
 
 s 
 
 
 
 
 
 
 
 
 
 
 
 
 &J 
 
 a* 
 
 % 
 
 ^ 
 
 SJ 
 
 K^ 
 
 
 Initial readings 
 
 « 
 
 
 
 ^1 
 
 £ 
 
 ^ 
 
 3 
 
 11 
 
 S 
 
 ft 
 
 3.146 
 
 1 
 
 — '3900 
 
 — 
 
 
 
 — 
 
 I and 17 
 
 3,210 
 
 64 
 
 I 
 
 64 
 
 3910 
 
 — 10 
 
 1 
 
 -ID 
 
 2, 4, 6, 8, 10, 12, 14, and 16 
 
 8,8.19 
 
 5649 
 
 4 
 
 22,S96 
 
 .•S124 
 
 + 7»6 
 
 4 
 
 ,SI44 
 
 3. 5. 7. 9. ". 13. and 15 
 
 14.345 
 
 548b 
 
 2 
 
 10,972 
 
 2381 
 
 + 743 
 
 2 
 
 14S6 
 
 33.632 
 
 4620 
 
 Multiplier for displacement = o'o2 
 Multiplier for moment = o'2i33 
 
 Displacement = 33,632 x 0-02 = 672-6 tons 
 
 = 4620 X o'2i33 = 985 foot-tons 
 
 ^^^ = 1-46 feet 
 
 GZ = 
 
 672-6 
 
 In this case the length of the ship was divided into sixteen 
 equal parts, and accordingly Simpson's first rule can be 
 employed. The common interval was 8-75 feet. The multi- 
 plier for the instrument was yoFo fo"^ the areas, and yffo for the 
 moments, and, the drawing being on the scale of i inch = 
 I foot, the readings for areas had to be multiplied by (4)^ =16, 
 
 and for moments by (4)' = 64. 
 ment in tons is therefore — 
 
 The multiplier for displace- 
 
 X 16 X (I X 8-75) X ^ = 0'D2 
 
Statical Stability, Curves of Stability, etc. i8i 
 and for the moment in foot-tons is — 
 
 yf^ X 64 X (i X 875) X is = 0-2133 
 We therefore have, assuming that the centre of gravity is at S — 
 
 GZ = 
 
 985 
 672-6 ' 
 
 1-46 feet 
 
 Now, this 672-6 tons is not the displacement up to the 
 original water-line WL, and we now have to consider a new 
 conception, viz. cross-curves of stability. These are the con- 
 verse of the ordinary curves of stability we have been consider- 
 ing. In these we have the righting levers at a constant 
 displacement and varying angles. In a cross-curve we have 
 the righting levers for a constant angle, but varying displace- 
 ment. Thus in Fig. 79, draw a water-line W"L" parallel to 
 
 Cross Curves of Stability. 
 
 
 
 C.G.iM L.W.L. 
 
 
 
 60 
 
 
 
 
 45 
 
 
 
 
 1 
 1 
 
 *5 
 
 ^ 
 
 
 
 
 T" ~- 
 
 - 
 
 
 
 -^ 
 
 1 
 
 3(^ 
 
 
 
 
 — h 
 
 S 
 
 9Q 
 
 
 
 
 t 
 
 
 is 
 
 
 
 
 
 •Sj 
 
 9P 
 
 
 • 
 
 
 i| 
 
 f 
 
 Tons 
 
 4000. 
 
 DlSPUACEMEWT. 
 
 Fig. 80. 
 
 W'L', and for the volume represented by W"ML" find the 
 displacement and position of the centre of buoyancy in exactly 
 the same way as we have found it for the volume WML'. 
 The distance which this centre of buoyancy is from the axis 
 gives us the value of GZ at this displacement, supposing the 
 centre of gravity is at S. The same process is gone through 
 
1 82 
 
 Theoretical Naval Architecture. 
 
 for one or two more water-lines, and we shall have values of 
 GZ at varying displacements at a constant angle. These can 
 be set off as ordinates of a curve, the abscissae being the 
 displacements in tons. Such a curve is termed the " cross- 
 curve of stability" at 30°, and for any intermediate displace- 
 ment we can find the value of GZ at 30° by drawing the 
 ordinate to the curve at this displacement. A similar process 
 is gone through for each angle, the same position for the centre 
 of gravity being assumed all through, and a series of cross- 
 curves obtained. Such a set of cross-curves is shown in 
 Fig. 80 for displacements between 3000 and 5000 tons at 
 angles of 15°, 30°, 45°, 60°, 75°, and 90°. At any intermediate 
 displacement, say at 4600 tons, we can draw the ordinate and 
 measure off the values of GZ, and so obtain the ordinates 
 necessary to construct the ordinary curve of stabiUty at that 
 displacement and assumed position of the centre of gravity. 
 The relation between the cross-curves and the ordinary curves 
 of stability is clearly shown in Fig. 81. ^^''e have four curves 
 
 15. 30. 4-5. 60. 75. 
 
 Degrees of Inclination. 
 
 Fig. 81. 
 
 of stability for a vessel at displacements of 1500, 2000, 2500, 
 and 3000 tons. These are placed as shown in perspective, 
 Now, through the tops of the ordinates at any given angle we 
 can draw a curve, and this will be the cross-curve of stability 
 at that angle. 
 
 It will have been noticed that throughout our calculation 
 
Statical Stability, Curves of Stability, etc. 183 
 
 we have assumed that the centre of gravity is always at the 
 point S, and the position of this point should be clearly stated 
 on the cross-curves. It is evident that the centre of gravity 
 cannot always remain in this position, which has only been 
 assumed for convenience. The correction necessary can 
 readily be made as follows : If G, the centre of gravity, is 
 below the assumed position S, then GZ = SZ + SG . sin 0, and 
 if G is above S, then GZ = SZ — SG . sin B for any angle 6. 
 Thus the ordinates are measured from the cross-curves at the 
 required displacement, and then, SG being known, SG sin 15°, 
 SG sin 30°, etc., can be found, and the correct values of GZ 
 determined for every angle. 
 
 Dynamical Stability. — The amount of work done by 
 a force acting through a given distance is measured by the 
 product of the force and the distance through which it acts. 
 Thus, a horse exerting a pull of 30,000 lbs. for a mile does — 
 
 30,000 X 1760 X 3 = 158,400,000 foot-lbs. of work 
 
 Similarly, if a weight is lifted, the work done is the product of 
 the weight and the distance it is lifted. In the case of a 
 ship being inclined, work has to be done on the ship by some 
 external forces, and it is not always possible to measure the 
 work done by reference to these forces, but we can do so by 
 reference to the ship herself. When the ship is at rest, we 
 have seen that the vertical forces that act upon the ship are — 
 (i) The weight of the ship acting vertically downwards 
 
 through the centre of gravity ; 
 (2) The buoyancy acting vertically upwards through the 
 centre of buoyancy ; 
 these two forces being equal in magnitude. When the ship is 
 inclined, they act throughout the whole of the inclination. 
 The centre of gravity is raised, and the centre of buoyancy is 
 lowered. The weight of the ship has been made to move 
 upwards the distance the centre of gravity has been raised, and 
 the force of the buoyancy has been made to move downwards 
 the distance the centre of buoyancy has been lowered. The 
 work done on the ship is equal to the weight multiplied by the 
 rise of the centre of gravity added to the force of the buoyancy 
 multiplied by the depression of the centre of buoyancy ; or — 
 
184 Theoretical Naval Architecture. 
 
 Work dojie on the ship = weight of the ship mttltiplied by the 
 vertical separation of the centre of gravity and the centre 
 of buoyancy. 
 
 This calculated for any given angle of inclination is termed 
 " the dynamical stability " at that angle, and is the work that 
 has to be expended on the ship in heeling her over to the 
 given angle. 
 
 Moseley's Formula for the Dynamical Stability 
 at any Given Angle of Inclination. — Let Fig. 67, p. 159, 
 represent a vessel heeled over by some external force to the 
 angle 6; g, g' being the centres of gravity of the emerged! 
 and immersed wedges ; gh, g'h! being drawn perpendicular tO' 
 the new water-line W'L'. The other points in the figure have 
 their usual meaning, BR and GZ being drawn perpendicular 
 to the vertical through B'. 
 
 The vertical distance between the centres of gravity and 
 buoyancy when inclined at the angle 6 is B'Z. 
 
 The original vertical distance when the vessel is upright 
 is BG. 
 
 Therefore the vertical separation is — 
 
 B'Z - BG 
 and according to the definition above — 
 
 Dynamical stability = W(B'Z - BG) 
 where W = the weight of the ship in tons. 
 
 Now, B'Z = B'R + RZ = B'R + BG . cos 6 
 
 Now, using V for the volumes of either the immersed wedge 
 or the emerged wedge, and V for the volume of displacement 
 of the ship, and using the principle given on p. 96, we have — 
 
 V X {gh + g'M) = V X B'R 
 
 orB'R^ ^X^y i^) 
 
 Substituting the above value for BZ, we have — 
 
 a 6 = —^ — - — ^ + BG cos 8 
 
Statical Stability, Ctirves of Stability, etc. 185 
 
 which is known as Moseley' s foiinnla. 
 
 It will be seen that this formula is very similar to Atwood's 
 formula (p. 159), and it is possible to calculate it out for varying 
 angles by using the tables in Barnes's method of calculating 
 stability. It is possible, however, to find the dynamical 
 stability of a ship at any angle much more readily if the 
 curve of statical stability has been constructed, and the 
 method adopted, if the dynamical stability is required, is as 
 follows : — 
 
 The dynamical stability of a ship at any given angle 
 is equal to the area of the curve of statical 
 stability up to that angle (the ordinates of this curve 
 being the actual righting moments). 
 
 As the demonstration of this is somewhat difficult, it is 
 given in Appendix A, p. 247. 
 
 To illustrate this principle, take the case of a floating body 
 whose section is in the form of a circle, and which floats with 
 its centre in the surface of the water. The transverse meta- 
 centre of this body must be at the centre of the circular section. 
 Let the centre of gravity of the vessel be at G, and the centre 
 of buoyancy be at B. Then for an inclination through 90° 
 G will rise till it is in the surface of the water, but the centre of 
 buoyancy will always remain at the same level, so that the 
 dynamical stability at 90° = W x GM. 
 
 Now take the curve of statical stability for such a vessel. 
 The ordinate of this curve at any angle 6 = W x GM . sin Q, 
 and consequently the ordinates at angles 15° apart will be 
 W . GM . sin 0°, W . GM . sin 15°, and so on; or, o, o"258 
 W.GM, 0-5 W.GM, 0707 W.GM, 0866 AV.GM, 0-965 
 W . GM, and W . GM. If this curve is set out, and its area 
 calculated, it will be found that its area is W x GM, which is 
 the same as the dynamical stability up to 90°, as found above. 
 It should be noticed that the angular interval should not be 
 taken as degrees, but should be measured in circular measure 
 (see p. 86). The circular measure of 15° is o'26i8. 
 
1 86 Theoretical Naval Architecture. 
 
 The dynamical stability at any angle depends, therefore, 
 on the area of the curve of statical stability up to that angle ; 
 and thus we see that the area of the curve of stability is of 
 importance as well as the angle at which the ship becomes un- 
 stable, because it is the dynamical stability that tells us the 
 work that has to be expended to force the ship over. For full 
 information on this subject, the student is referred to the 
 " Manual of Naval Architecture," by Sir W. H. White, and Sir 
 E. J. Reed's work on the " Stability of Ships." 
 
 Examples to Chapter V. 
 
 1. A two-masted cruiser of 5000 tons displacement has its centre of 
 gravity at two feet above the water-line. It is decided to add a military 
 top to each mast. Assuming the weight of each military top with its guns, 
 men, and ready-ammunitiun supply to be 12 tons, with its centre of gravity 
 70 feet above the water-line, what will be the effect of this change on — 
 
 (1) The metacentric height of the vessel? 
 
 (2) The maximum range of stability, assuming the present maximum 
 
 range is 90°, and the tangent to the curve at this point inclined 
 at 45° to the base-line ? 
 (Scale used, J inch = 1°, J inch = I'j foot GZ.) 
 
 Ans. (l) Reduce 0^325 foot, assuming metacentric curve horizontal ; 
 (2) reduce range to about 86f°, assuming no change in cross- 
 curves from 5000 to 5024 tons. 
 
 2. The curve of statical stability of a vessel has the following values of 
 GZ at angular intervals of 15° : o, 0-55, I '03, 0-99, o'66, o'24, and — 0'20 
 feet. Determine the loss in the range of stability if the C.G. of the ship 
 were raised 6 inches. 
 
 Ans. 16°. 
 
 3. Obtain, by direct application of Atwood's formula, the moment of 
 stability in foot-tons at angles of 30°, 60°, and 90°, in the case of a prismatic 
 vessel 140 feet long and 40 feet square in section, when floating with 
 sides vertical at a draught of 20 feet, the metacentric height being 2 feet. 
 
 4. A body of square section of 20 feet side and 100 feet long floats with 
 one face horizontal in salt water at a draught of 10 feet, the metacentric 
 height being 4 inches. Find the dynamical stability at 45°. 
 
 Ans. 171 foot-tons. 
 
 5. Indicate how far a vessel having high bulwarks is benefited by them 
 as regards her stability. What precautions should be taken in their 
 construction to prevent them becoming a source of danger rather than of 
 safety ? 
 
 6. Show from Atwood's formula that a ship is in stable, unstable, or 
 neutral equilibrium according as the centre of gravity is below, above, or 
 coincident with the transverse metacentre respectively. 
 
 7. A vessel in a given condition displaces 4600 tons, and has the C.G. 
 in the 19-feet water-line. The ordinates of the cross-curves at this dis- 
 placement, with the C.G. assumed in the ig-feet water-line, measure as 
 follows: 0-63, 1-38, 2-15, 2-o6, I '37, 0-56 feet at angles of 15°, 30°, 
 45°, 60°, 75° and 90° respectively. The metacentric height is 2'4 feet. 
 
Statical Stability, Curves of Stability, etc. 187 
 
 Draw out the curve of stability, and state (r) the angle of maximum 
 stability, (2) the angle of vanishing stability, and (3) find the dynamical 
 stability at 45° and 90°. 
 
 Ans. (l) So|° ; (2) loop ; (3) 3694, 9650 foot-tons. 
 
 8. A vessel has a metacentric height of 3-4 feet, and the curve of stability 
 has ordinates at 15°, 30°, 37!°, 45°, and 60° of 0-9, i'92, 2-02, 1-65, and 
 — 0*075 feet respectively. Draw out this curve, and state the angle of 
 maximum stability and the angle at which the stability vanishes. 
 
 ^ns. 3SJ°, 591°. 
 
 9. A vessel's curve of stability has the following ordinates at angles of 
 15°, 30°, 45°, 60°, and 75°, viz. 0-51, 0'97, o'QO, o"53, and O'cS feet 
 respectively. Estimate the influence on the range of stability caused by 
 lifting the centre of gravity of the ship 0'2 feet. 
 
 Ans. Reduce nearly 6°. 
 
 10. A square box of 18 feet side floats at a constant draught of 6 feet, 
 the centre of gravity being in the water-line. Obtain, by direct drawing or 
 otherwise, the value of GZ up to 90° at say 6 angles. Draw in the curve 
 of statical stability, and check it by finding its area and comparing that 
 with the dynamical stability of the box at 90°. 
 
 (Dynamical stability at 90° = 3 X weight of box.) 
 
 11. A vessel fully loaded with timber, some on the upper deck, starts 
 from the St. Lawrence River with a list. She has two cross-bunkers extend- 
 ing to the upper deck. She reaches a British port safely, with cargo undis- 
 turbed, but is now upright. State your opinion as to the cause of this. 
 
 12. Show by reference to the curves of stability of box-shaped vessels 
 on p. 163 that at the angle at which the deck edge enters the water the 
 tangent to the curve makes the maximum angle with the base-line. 
 
CHAPTER VI. 
 
 CALCULATION OF WEIGHTS AND STRENGTH OF BUTT 
 CONNECTIONS. STRAINS EXPERIENCED BY SHIPS. 
 
 Calculations of Weights. — ■ We have discussed in 
 Chapter I. the ordinary rules of mensuration employed in find- 
 ing the areas we deal with in ship calculations. For any 
 given uniform plate we can at once determine the weight 
 if the weight per square foot is given. For iron and steel 
 plates ,of varying thicknesses, the weight per square foot is 
 given on p. 36. For iron and steel angles and X bars of 
 varying sizes and thicknesses tables are calculated, giving the 
 weight per lineal foot. Such a table is given on p. 189 for 
 steel angles, etc., the thicknesses being in -^oths of an inch. It 
 is the Admiralty practice to specify angles, bars, etc., not in thick- 
 ness, but in weight per lineal foot. Thus an angle bar 3" X 3" 
 is specified to weigh 7 lbs. per lineal foot, and a Z bar 6" X 
 3o" X 3" is specified to weigh 15 lbs. per lineal foot. When the 
 bars are specified in this way, reference to tables is unnecessary. 
 The same practice is employed with regard to plates, the thick- 
 ness being specified as so many pounds to the square foot. 
 
 If we have given the size of an angle bar and its thick- 
 ness, we can determine its weight per foot as follows : Assume 
 the bar has square corners, and is square at the root, then, if 
 a and h are the breadth of the flanges in inches, and t is the 
 thickness in inches, the length of material t inches thick in the 
 
 section is ((/ + 5 — /) inches, or feet ; and if the bar 
 
 is of iron, the weight per lineal foot is — 
 
 ( j X 40 X ^ lbs. 
 
X 
 u 
 
 z 
 
 z 
 
 <; 
 
 t. 
 o 
 
 z 
 
 y 
 5 
 
 n|<:4 
 
 roro ro 
 
 Olio 
 
 w ro ro CO ro 
 
 "Is 
 
 a. N -^00 o wit^ 
 
 f^ fOOp fO ON ^'(J- OS 
 
 ^\b i^ a. b N ro 
 
 « « « d ro COfO 
 
 ':\% 
 
 »p M ir> p -i^op p f^ « 
 O W CO io^b r^ ON b M 
 
 SIS 
 
 "O MOO Tt-OvO MOO 'd-OvD 
 
 op « lo pN cD*p P p !^ ^ ;* 
 vo 00 a. b w ro iovb r^. on b 
 
 ■:|s 
 
 on\o •■d-w e»^ ■'i-i-i ON»o ^w ON 
 
 fO Vd- iri r^oo On w « rr> *-«ty3 t^oO 
 Mi-iMi-iMMr4C4NM<NMN 
 
 is 
 
 ro M NN M OnOO Ir^vO inmfOM m O 0\ 
 <M 'T-'O Op ON>-( mu-ir^ONi-i roLo r^oo 
 b w N (H^ Vf-y3 i^oo ON b N ro V ln\0 
 
 MMMMMMMMMCtMOMNM 
 
 12 13 
 20 20 
 
 ti M M fOtO^Tj-iT) irjio vo t^ t^CC 00 On 
 
 y^vp t^op pN p M w CO ;5f unvp f^oo on O 
 00 ON O ►- N 'rhirj^b i^OO On O w N ro io 
 
 S^S;8 s S-S"§ s-s ;r-£'S 8 3 it's 
 
 »b J>.oo b 1-t w CO V mvo r-^cJo on b « « co 
 
 -IS 
 
 M m on N vo On covO O co I^ O ^ t^ m ■^oO f 
 lOTfpcoN MtH op ONOp 00 t^vp vo UI ;ii- r(- 
 invO rN.00 ON b M « CO ro Tf- ii-i\b t-*00 On b m 
 
 o|o 
 
 fHJCT 
 
 u->0 mo i^O mo "^0 "j^O mo "j^O loOm 
 N M p\00 vp ii-1 CO N p ON f^vp .'^ to •-< p Op t^ m 
 
 Vj- io ln\o i>*oo On b w Ih M ro V mvb i^ Kob ON 
 
 HS 
 
 ■<:|- O t^ CO O vo CO ONVO C40nu-)M00iow00 Tj-i-rt^ 
 
 M pN»p -^^w (3^!^.^« p r^y^pp^^ r^r p^*P 
 
 CO fO Vl- irjvb yD i^OO On b b i-i « CO CO Vt ioo vO i>. 
 
 «|s 
 
 OOVO rl-« OOOvO Tl-N OOOvO ThN OOOvO Ti-i-i ooo 
 ^ 00 m CI ON u^ M p^"^ CO pNvp f^^Pf^pp f^;*^ f^ 
 M N CO Tj- Vt- ^n^b M3 r^oo 00 ONb ►"• « w cocoVj-iom 
 
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 VOvOu^iO-^'^cocoNNMwOOON OnOO 00 l>. t-*VO 
 
 i-t w CO CO Tt Tf mvb yD t>. i^oo bN'oNbbM««c*^eo 
 
 HS 
 
 CO Td- mvD r^oo on •-' n co ^ mvo i^oo on o •-• « co 
 t^ <S *>. M t^ w f^ tooo coop coop coop coop Tf ON ;d- ON 
 
 M M (M CO CO W io iovb «i> i^ i^OO 00 ON On b b w '^ 
 
 HS 
 
 ON "-1 -^vO On m -^vO 00 w rJ-\0 On m rl-vo 00 m covo OS 
 ^ ON CO t-^ w vp p ;«tO0 P f^ r '^ P .'TOO « r^ M to ON 
 Ih Ui « N cOCOT^Tt-^m iovo vb t-» i^ t^OO 00 On ON ON 
 
 HS 
 
 N vo M TtOO « vo O ■^00 W vo O •'too M vO O r}-00 N 
 
 « inON« lopNwys p*?*^ P^ P*P P* P'^ p p'O O 
 
 M M M M M M cOcOCoV-'tVhioio invo vo t^ t-» t>i(>0 
 
 "is 
 
 •-d- O "T-t \0 N 1>- COOO -«J-ONmQvO wrN.M00 COON-*** 
 pNN .Tt-f^pNOl ^-^J^pNN ^tt^p Ny^t^p N JOf^p 
 b w w w w N W « M COCOCoWWrnirjio io^O 
 
 «|s 
 
 *O00 ONM com^qp 
 b b b M W W W *M 
 
 HS 
 
 co>-i ooo r^to^M 
 CO ^ m tn'p f^op pN 
 
 b b b b b b b b 
 
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 ra CO f n f*^ rt- ^ to li-i^D vO *>. t-^00 OOOnOnOO'-''-<C^ 
 
I go Theoretical Naval Architecture. 
 
 If the bar is of steel, the weight per lineal foot is — 
 
 'a^b-t\ „ ^,, 
 1 X 4o'8 X t lbs. 
 
 12 / ^ 
 
 Thus a 3" X 3" X f" steel angle bar would weigh 7-17 lbs., 
 and a steel angle bar 3" X 3" of 7 lbs. per foot would be 
 slightly less than f inch thick. 
 
 It is frequently necessary to calculate the weight of a 
 portion of a ship's structure, having given the particulars of 
 its construction; thus, for instance, a bulkhead, a deck, 
 or the outer bottom plating. In any case, the first step must 
 be to find the area of plating and the lengths of angle bars. 
 The weight of the net area of the plating will not give us the 
 total weight .of the plating, because we have to allow for butt 
 straps, laps, rivet-heads, and in certain cases liners. The method 
 employed to find the allowance in any given case is to take a 
 sample plate and find what percentage the additions come to 
 that affect this plate, and to use this percentage as an addition 
 to the net weight found for the whole. To illustrate this, take 
 the following example : — 
 
 A deck surface of 10,335 square feet is to be covered with /5-incli 
 steel plating, worked flush, jointed with single-riveted edges and butts, 
 Find the weight of the deck, allowing 3 per cent, for rivet-heads. 
 
 ^-inch steel plates are 1275 lbs. per square foot, so that the net 
 weight is — 
 
 10.335x1275^ 8.8 to„, 
 2240 
 
 Now, assume an average size for the plates, say 16' X 4'. f-inch rivets 
 will probably be used, and the width of the edge strip and butt strap will 
 be about 5 inches. The length round half the edge of the plate is 20 feet, 
 and the area of the strap and lap belonging to this plate is — 
 
 20 X ft = 8-33 square feet 
 
 The percentage of the area of the plate is therefore — 
 
 8'33 
 
 -^ X 100 = 13 per cent. 
 04 
 
 Adding 3 per cent, for rivet-heads, the percentage to add to the net weight 
 is 16 per cent., or 9-4 tons. The total weight is therefore 68'2 tons. 
 
 It is usual to add 3 per cent, to allow for the weight of rivet- 
 heads. For lapped edges and butt straps, both double riveted, 
 
Calculation of Weights, etc. 191 
 
 the percentage ^ comes to about 10 per cent, for laps, s| per cent, 
 for butt straps, and 3 per cent, for liners as ordinarily fitted to 
 the raised strakes of plating. No definite rule can be laid 
 down, because the percentage must vary according to the 
 particular scantlings and method of working the plating, etc., 
 specified. 
 
 The length of stiffeners or beams required for a given area 
 can be very approximately determined by dividing the area in 
 square feet by the spacing of the stiffeners or beams in feet. 
 For wood decks, 3 per cent, can be added for fastenings. 
 
 Example.— The beams of a deck are 3 feet aparl, and weigh 22 lbs. 
 per foot run ; the deck plating weighs 10 lbs. per square foot, and this is 
 covered by teak planking 3 inches thick. Calculate the weight of a part 
 1:4 feet long by 10 feet wide of this structure, including fastenings. 
 
 (S.andA.Exam.\%<iT.) 
 
 Net area of deck = 54 x 10 = 540 
 Add for butts and laps 7 per cent. = 37 '8 
 
 .577-8 
 (Assume single-riveted butt straps and single-riveted laps.) 
 
 Weight of plating = 577-8 X 10 
 = 5778 lbs. 
 Running feet of beams = ^ = 180 
 
 Weight of beams = 180 X 22 
 = 3960 Ibs.^ 
 Total weight of plating and beams = 9,738 lbs. 
 Add 3 per cent, for rivet-heads = 292 ,, 
 
 10,030 ,, 
 
 Weight of teak » = 540 x f = 6750 lbs, 
 Add 3 per cent, for fastenings = 202 ,, 
 
 Weight of wood deck 6952 „ 
 
 Summary. 
 
 Plating and beams 10,030 lbs< 
 
 Wood deck 6,952 „ 
 
 Total ... 16,982 „ = 7-6 tons. 
 
 ' A number of percentages worked out for various thicknesses, etc., 
 will be found in Mr. Mackrow's " Pocket Book." 
 
 ^ No allowance made for beam arms, which should be done if a whole 
 deck is calculated. 
 
 " Teak taken as 50 lbs. per cubic foot. 
 
192 
 
 Theoretical Naval Architecture. 
 
 Use of Curves. — For determining the weight of some of 
 the portions of a ship, the use of curves is found of very great 
 assistance. Take, for instance, the transverse framing of a ship. 
 For a certain length this framing will be of the same character, 
 as, for example, in a battleship, within the double bottom, 
 where the framing is fitted intercostally between the longi- 
 tudinals. We take a convenient number of sections, say the 
 sections on the sheer drawing, and calculate the weight of the 
 complete frame at each section. Then along a base of length 
 set up ordinates at the sections, of lengths to represent the 
 calculated weights of the frames at the sections. Through the 
 spots thus obtained draw a curve, which should be a fair line. 
 The positions of the frames being placed on, the weight of each 
 frame can be obtained by a simple measurement, and so the 
 total weight of the framing determined. The curve AA in 
 Fig. 82 gives a curve as constructed in this way for the transverse 
 framing below armour in the double bottom of a battleship. 
 Before and abaft the double bottom, where the character of the 
 framing is different, curves are constructed in a similar manner. 
 
 Weight of Outer Bottom Plating. — The first step 
 necessary is to determine the area we have to deal with. We 
 
 Fig. 82. 
 
 can construct a curve of girths, as BB, Fig. 82 ; but the area giv,en 
 by this curve will not give us the area of the plating, because 
 although the surface is developed in a transverse direction 
 
Calculation of Weights, etc. 193 
 
 there is no development in a longitudinal direction. (Strictly 
 speaking, the bottom surface of a ship is an undevelopable 
 surface.) The extra area due to the slope of the level lines is 
 allowed for as follows : In plate I., between stations 3 and 4, 
 a line fg is drawn representing the mean slope of all the level 
 lines. Then the ordinate of the curve of girths midway 
 between 3 and 4 stations is increased in the ratio T^T • h- This 
 done all along the curve will give us a new modified curve of 
 girths, as B'B', Fig. 82, and the area given by this curve will give 
 a close approximation to the area of the outer bottom of the 
 ship. This is, of course, a net area without allowing for butt, 
 straps or laps. Having a modified curve of girths for the 
 whole length, we can separate it into portions over which the 
 character of the plating is the same. Thus, in a vessel built 
 under Lloyd's rules, the plating is of certain thickness for one- 
 half the length amidships, and the thickness is reduced before 
 and abaft. Also, in a battleship, the thickness of plating is the 
 same for the length of the double bottom, and is reduced 
 forward and aft. The curves AA and BB, Fig. 82, are con- 
 structed as described above for a length of 244 feet. 
 
 Weight of Hull. — 'By the use of these various methods, 
 it is possible to go right through a ship and calculate the 
 weight of each portion of the structure. These calculable 
 portions for a battleship are — 
 
 (i) Skin-plating and plating behind armour. 
 
 (2) Inner bottom plating. 
 
 (3) Framing within double bottom, below armour, behind 
 armour, and above armour. Outside double bottom, below 
 and above the protective deck. 
 
 (4) Steel and wood decks, platforms, beams. 
 
 (5) Bulkheads. 
 
 (6) Topsides. 
 
 There are, however, a large number of items that cannot be 
 directly calculated, and their weights must be estimated by 
 comparison with the weights of existing ships. Such items 
 are stem and stem posts, shaft brackets, engine and boiler 
 bearers, rudder, pumping and ventilation arrangements, pillars, 
 paint, cement, fittings, etc. 
 
 o 
 
194 Theoretical Naval Architecture. 
 
 It is, however, a very laborious calculation to determine 
 the weight of the hull of a large ship by these means; and 
 more often the weight is estimated by comparison with the 
 ascertained weight of existing ships. The following is one 
 method of obtaining the weight of steel which would be used in 
 the construction of a vessel : The size of the vessel is denoted 
 by the product of the length, breadth, and depth, and for known 
 ships the weight of steel is found to be a certain proportion of 
 this number, the proportion varying with the type of ship. 
 The coefficients thus obtained are tabulated, and for a new ship 
 the weight of steel can be estimated by using a coefficient 
 which has been obtained for a similar type of ship. The weight 
 of wood and outfit can be estimated in a similar manner. 
 
 Another method is described by Mr. J. Johnson, M.I.N. A., 
 in the Transactions of the Institution of Naval Architects for 
 1897, in which the sizes of vessels are represented by Lloyd's 
 longitudinal number^ modified as follows : In three-decked 
 vessels, the girths and depths are measured to the upper deck 
 
 ' Lloyd's numbers — 
 
 1. The scantlings and spacing of the frames, reversed frames, and floor, 
 plates, the thickness of bulkheads and the diameter of pillars are regulated 
 by numbers, which are produced as follows : — 
 
 2. For one and two decked vessels, the number is the sum of the 
 measurements in feet arising from the addition of the half-moulded breadth 
 of the vessel at the middle of the length, the depth from the upper part of 
 the keel to the top of the upper-deck beams, with the normal round-up, 
 and the girth of the half midship frame section of the vessel, measured from 
 the centre line at the top of the keel to the upper-deck stringer plate. 
 
 3. For three-deck steam-vessels, the number is produced by the 
 deduction of 7 feet from the sum of the measurements taken to the top of 
 the upper-deck beams. 
 
 4. For spar-decked vessels and awning-decked steam-vessels, the 
 number is the sum of the measurements in feet taken to the top of the main- 
 deck beams, as described for vessels having one or two decks. 
 
 5. The scantlings of the keel, stem, stern-frame, keelson, and stringer 
 plates, the thickness of the outside plating and deck ; also the scantlings 
 of the angle bars on beam stringer plates, and keelson and stringer angles 
 in hold, are governed by the longitudinal number obtained by multiplying 
 that which regulates the size of the irames, etc., by the length of the vessel. 
 
 The measurements for regidating the above scantling numbers are taken 
 as follows : — 
 
 I. The length is measured from the after part of the stem to the fore part 
 of the stern-post on the range of the upper-deck beams in one, two, and 
 three decked and spar-decked vessels, but on the range of main-deck beams 
 in awning-decked vessels. 
 
 In vessels where the stem forms a cutwater, the length is measured from 
 
Calculation of Weights, etc. 195 
 
 without deducting 7 feet. In spar and awning-deck vessels, 
 the girths are measured to the spar or awning decks respec- 
 tively. In one, two, and well-decked vessels, the girths and 
 depths are taken in the usual way. Curves are drawn for each 
 type of vessel, ordinates being the weight of iron or steel in 
 tons for vessels built to the highest class at Lloyd's or Veritas, 
 and abscissae being Lloyd's longitudinal number modified as 
 above. These curves being constructed for ships whose weights 
 are known, it is a simple matter to determine the weight for a new 
 ship of given dimensions. For further information the student 
 is referred to the paper in volume 39 of the Transactions. 
 
 To calculate the Position of the Centre of Gravity 
 of a Ship. — We have already seen in Chapter III. how to find 
 the C.G. of a completed ship by means of the inclining experi- 
 ment, and data obtained in this way are found very valuable in 
 estimating the position of the C.G. of a ship that is being 
 designed. It is evident that the C.G. of a ship when com- 
 pleted should be in such a position as to obtain the metacentric 
 height considered necessary, and also to cause the ship to float 
 correctly at her designed trim. Suppose, in a given ship, the C.G. 
 of the naked hull has been obtained from the inclining experi- 
 ment (that is, the weights on board at the time of the experi- 
 ment that do not form part of the hull are set down and their 
 positions determined, and then the weight and position of the 
 C.G. of the hull determined by the rules we have dealt with in 
 Chapter III.). The position of the C.G. of hull thus determined 
 is placed on the midship section, and the ratio of the distance 
 of the C.G. above the top of keel to the total depth from the 
 top of keel to the top of the uppermost deck amidships will 
 
 the place where the upper-deck beam line would intersect the after edge of 
 stem if it were produced in the same direction as the part below the 
 cutwater. 
 
 . 2. The breadth in all cases is the greatest moulded breadth of the vessel. 
 3. The depth in one and two decked vessels is taken from the upper 
 part of the keel to the top of the upper-deck beam at the middle of the 
 length, assuming a normal round-up of beam of a quarter of an inch to a 
 foot of breadth. In spar-decked vessels and awning-decked vessels, the 
 depth is taken from the upper part of the keel to the top of the main-deck 
 beam at the middle of the length, with the above normal round-up of 
 beam. 
 
196 Theoretical Naval ArchitecUire. 
 
 give us a ratio that can be used in future ships of similar type 
 for determining the position of the C.G. of the hull. Thus, in 
 a certain ship the C.G. of hull was 20'3 feet above keel, the 
 total depth being 34*4 feet. The above ratio in this case is 
 therefore 0-59, and for a new ship of similar type, of depth 39's 
 feet, the C.G. of hull would be estimated at 39-5 X 0*59, or 23'3 
 feet above the keel. For the fore-and-aft position, a similar ratio 
 may be obtained between the distance of the C.G. abaft the 
 middle of length and the length between perpendiculars. In- 
 formation of this character tabulated for known ships is found 
 of great value in rapidly estimating the position of the C.G. 
 in a new design. 
 
 For a vessel of novel type, it is, however, necessary to cal- 
 culate the position of the C,G., and this is done by combining 
 together all the separate portions that go to form the hull. 
 Each item is dealt with separately, and its C.G. estimated as 
 closely as it is possible, both vertically and in a fore-and-aft 
 direction. These are put down in tabular form, and the total 
 weight and position of the C.G. determined. 
 
 In estimating the position of the C.G. of the bottom plating, 
 we proceed as follows : First determine the position of the 
 C.G. of the several curves forming the half-girth at the various 
 stations. This is not generally at the half-girth up, but is some- 
 where inside or outside the line of the curve. Fig. 83 represents 
 the section AB at a certain station. The curve is divided into 
 four equal parts by dividers, and the C.G. of each of these parts 
 is estimated as shown. The centres of the first two portions 
 are joined, and the centres of the two top portions are joined 
 as shown. The centres of these last-drawn lines, gi, g^, are 
 joined, and the centre of the line ^1^21 viz. G, is the C.G. of 
 the line forming the curve AB, and GP is the distance from the 
 L.W.L. This done for each of the sections will enable us to 
 put a curve, CC in Fig. 82, of distances of C.G. of the half- 
 girths from the L.W.L. ^ We then proceed to find the C.G. of 
 the bottom plating as indicated in the following table. The 
 area is obtained by putting the half-girths (modified as already 
 
 ' This assumes the plating of constant thickness. Plates which are 
 thicker, as at keel, bilge, and sheer, can be allowed for afterwards. 
 
Calculation of Weights, etc. 
 
 197 
 
 explained) through Simpson's rule. These products are then 
 multiplied in the ordinary way to find the fore-and-aft position 
 of the C.G. of the plating, and also by the distances of the C.G. 
 
 Fig. 83.' 
 
 of the sections below the L.W.L., which distances are measured 
 off from the curve CC and are placed in column 6. The 
 remainder of the work does not need any further explanation. 
 
 Calculation for Area and Position of C.G. of Bottom Plating 
 FOR A Length op 244 Feet. 
 
 Modified 
 half-girths. 
 
 Simpson's 
 multipliers. 
 
 Products. 
 
 Multiples 
 for leverage. 
 
 Products. 
 
 'i^^^^ ^'"^-'- 
 
 41 -s 
 
 Sii 
 
 53 -o 
 
 49-6 
 37-5 
 
 I 
 4 
 
 2 
 
 4 
 I 
 
 41 -s 
 204-4 
 
 io6-o 
 
 198-4 
 37-S 
 
 2 
 
 I 
 
 
 
 I 
 2 
 
 83-0 
 204-4 
 
 287-4 
 198-4 
 
 75 "0 
 
 18-1 751 
 21-6 4.415 
 
 22-2 1 2,354 
 
 21-3 1 4,226 
 19-0 ! 712 
 
 587-8 
 
 273 '4 
 14-0 
 
 12,458 
 
 ' The C.G. of wood sheathing, if fitted, can be obtained from this 
 figure by setting off normally to the curve from G the half-thickness of the 
 sheathing. 
 
igS 
 
 Theoretical Naval Architecture. 
 
 Common interval = 6i feet 
 Area both sides = 587-8 X j X 6i x 2 
 = 23,904 square feet 
 
 C.G. abaft middle of length of plating = 
 
 i4'o 
 
 X 61 
 
 587-8 
 = 1-45 feet 
 
 C.G. below L.W.L. = ^^^ = 21-2 feet 
 587-8 
 
 Calculation for the Position of the C.G. of a Vessel. 
 
 Items. 
 
 c 
 
 Fkom L.W.L. 
 
 From Middle of Len gth. 
 
 Below. 
 
 Above. 
 
 Before. 
 
 Abaft. 
 
 
 
 
 
 
 
 
 
 
 
 
 H 
 
 
 
 
 
 
 
 
 ^■ 
 
 
 
 ^ 
 
 
 u 
 
 C 
 
 \^ 
 
 c 
 
 i^ 
 
 fl 
 
 
 
 <u 
 
 u 
 
 u 
 
 u 
 
 u 
 
 u 
 
 v 
 
 u 
 
 
 
 > 
 
 B 
 
 
 s 
 
 ►J 
 
 s 
 
 
 
 > 
 
 
 > 
 
 1 
 
 
 
 
 ^ 
 
 
 " 
 
 
 s 
 
 
 Equipment — 
 
 
 
 
 
 Water 
 
 25 
 
 4-0 
 
 100 
 
 — 
 
 — 
 
 — 1 — 
 
 12-0 
 
 300 
 
 Provisions 
 
 30 
 
 4-5 
 
 13s 
 
 — 
 
 — 
 
 25-0 7S0 
 
 — 
 
 
 Officers' stores 
 
 15 
 
 2-0 
 
 30 
 
 — 
 
 — 
 
 
 125-0 
 
 1.87s 
 
 Officers.men.and effects 
 
 30 
 
 — 
 
 
 6-0 
 
 180 
 
 55-01650 
 
 — 
 
 
 Cables 
 
 30 
 
 40 
 
 120 
 
 — 
 
 — 
 
 85-02550 
 
 — 
 
 — 
 
 Anchors 
 
 10 
 
 — 
 
 — 
 
 15-0 
 
 «So 
 
 goo 
 
 900 
 
 — 
 
 — 
 
 Masts, yards, etc. ... 
 
 25 
 
 — 
 
 — 
 
 45-0 
 
 1125 
 
 — 
 
 
 7-0 
 
 175 
 
 Boats 
 
 ID 
 
 — 
 
 — 
 
 2I-0 
 
 210 
 
 — 
 
 — 
 
 20-0 
 
 200 
 
 Warrant officers' stores 
 
 20 
 
 '"5 
 
 30 
 
 
 
 .^ 
 
 65-0 
 
 1300 
 
 — 
 
 — 
 
 Armament 
 
 175 
 
 
 
 4-0 
 
 700 
 
 — 
 
 — 
 
 5-0 
 
 875 
 
 Machinery 
 
 450 
 
 4-0 
 
 1800 
 
 
 — 
 
 — 
 
 — 
 
 33-0 
 
 14,850 
 
 Engineers' stores 
 
 50 
 
 0-5 
 
 25 
 
 — 
 
 — 
 
 — • 
 
 — 
 
 70-0 
 
 3.500 
 
 Coals 
 
 300 
 
 0'2 
 
 60 
 
 — 
 
 — - 
 
 3-0 
 
 900 
 
 — 
 
 
 Protective deck 
 
 210 
 
 — 
 
 
 
 I'S 
 
 315 
 
 
 
 15-0 
 
 3.150 
 
 Hull 
 
 1250 
 
 — 
 
 """ 
 
 1-5 
 
 187s 
 
 "— 
 
 — 
 
 II-5 
 
 14.375 
 
 Total... 
 
 2630 2300 
 tons 
 
 4555 
 2300 
 
 8050 
 
 2630)2255 
 
 0-857 ft. 
 
 above L.W.L. 
 
 C.G. above L.W.L. = 0-857 
 Trans, met. „ „ = 2-97 
 
 Trans, met. above C.G = 2-97 — 0-857 
 = 2-113 feet. 
 
 39.300 
 8,050 
 
 2630)31,250 
 
 11-88 ft. 
 abaft mid. 
 length. 
 
Calculation of Weights, etc. 199 
 
 Calculation for C.G. of a Completed Vessel.— By the 
 use of the foregoing methods we can arrive at an estimate of 
 the weight of hull, and also of the position of its C.G. relative 
 to a horizontal plane, as the L.W.P., and to a vertical athwart- 
 ship plane, as the midship section. To complete the ship for 
 service, there has to be added the equipment, machinery, etc., 
 and the weights of these are estimated, as also the positions of 
 their centres of gravity. The whole is then combined in a 
 table, and the position of the C.G. of the ship in the completed 
 condition determined. 
 
 The preceding is such a table as would be prepared for a 
 small protected cruiser. It should be stated that the table is 
 not intended to represent any special ship, but only the type of 
 calculation. 
 
 The total weight is 2630 tons, and the C.G. is o'8s7 foot 
 above the L.W.L. and 11 '88 feet abaft the middle of length. 
 The sheer drawing enables us to determine the position of the 
 transverse metacentre, and the estimated G.M. is found to be 
 2 'IIS feet. The centre of buoyancy calculated from the sheer 
 drawing should also be, if the ship is to trim correctly, at a 
 distance of 11 "88 feet abaft the middle of length. 
 
 Strength of Butt Fastenings. — Fig. 84 represents two 
 plates connected together by an ordinary treble-riveted butt 
 strap. The spacing of the rivets in the line of holes nearest 
 the butt is such that the joint can be caulked and made water- 
 tight, and the alternate rivets are left out of the row of holes 
 farthest from the butt. Such a connection as this could con- 
 ceivably break in five distinct ways — 
 
 1. By the whole of the rivets on one side of the butt 
 shearing. 
 
 2. By the plate breaking through the line of holes, AA, 
 farthest from the butt. 
 
 3. By the hutf strap breaking through the line of holes, BB, 
 nearest the butt. 
 
 4. By the //a/^ breaking through the middle row of holes, 
 CC, and shearing the rivets in the Hne A A. 
 
 5. By the strap breaking through the middle row of holes, 
 CC, and shearing the rivets in the line BB. 
 
200 
 
 Theoretical Naval Architecture. 
 
 It is impossible to make such a connection as this equal to 
 the strength of the unpunched plate, because, although we might 
 
 ii 
 
 il 
 
 T 
 
 ±L± 
 
 % 
 
 '\\ 
 \\ 
 
 II 
 
 % 
 
 Kl 
 
 Fig. 84. 
 
 put in a larger number of rivets and thicken up the butt strap, 
 
 Fig. 85. 
 
 there would still remain the line of weakness of the plate 
 through the line of holes, AA, farthest from the butt. 
 
Calculation of Weights, etc. 201 
 
 The most efficient form of strap to connect two plates 
 together would be as shown in Fig. 85, of diamond shape. 
 Here the plate is only weakened to the extent of one rivet-hole. 
 Such an efficient connection as this is not required in ship con- 
 struction, because in all the plating we have to deal with, such 
 as stringers and outer bottom-plating, the plate is necessarily 
 weakened by the holes required for its connection to the 
 beam or frame, and it is unnecessary to make the connection 
 stronger than the plate is at a line of holes for connecting it 
 to the beam or frame. In calculating the strength of a butt 
 connection, therefore, we take as the standard strength the 
 strength through the line of holes at a beam or frame, 
 and we so arrange the butt strap that the strength by any 
 of the modes of fracture will at least equal this standard 
 strength. 
 
 Experimental Data. — Before we can proceed to calcu- 
 late the strength of these butt connections, we must have some 
 experimental data as to the tensile strength of plates and the 
 shearing strength of rivets. The results of a series of experi- 
 ments were given by Mr. J. G. Wildish at the Institution of 
 Naval Architects in 1885, and the following are some of the 
 results given : — 
 
 Shearing Strength of Rivets in Tons. 
 (Pmi heads and cmmtersunk points. ) 
 
 
 
 
 .Single shear. 
 
 Double shear. 
 
 i incli iron : 
 
 rivets 
 
 in iron plates . . . 
 
 lO'O 
 
 18 
 
 ■ . " 
 
 3) 
 
 steel „ ... 
 
 8-4 
 
 — 
 
 ; inch steel 
 
 J) 
 
 M M 
 
 ii'S 
 
 21-2 
 
 K J) 
 
 )) 
 
 >J »» 
 
 15-25 
 
 — 
 
 10 „ 
 
 1^ 
 
 !> I) 
 
 20-25 
 
 — 
 
 It will be noticed that the shearing strength of the steel 
 rivets of varying sizes is very nearly proportionate to the sec- 
 tional area of the rivets. Taking the shearing strength of a 
 |-inch steel rivet to be 11 "5 tons, the strength proportionate to 
 the area would be for a |-inch steel rivet, 15 '6 tons, and for a 
 i-inch steel rivet 20*4 tons. Also, we see that the double shear 
 of a rivet is about i'8 times the single shear. 
 
202 Theoretical Naval Architecture. 
 
 The following results were given as the results of tests of 
 mild steel plates : — 
 
 Unpunched aSJ tons per square inch. 
 
 Holes puttered 22 ,, ,, 
 
 or a depreciation of 22 per cent. 
 
 HoXei drilled 29J tons per square inch. 
 
 Holes punched small, and the hole then \ „ 
 countersunk / 
 
 The following give the strength of the material of the plates 
 after being connected together by a butt strap : — 
 
 Holes punched the full size, the ) . . , 
 
 . -^ . . ' > 24'Q tons per square mch. 
 
 rivets havmg snap pomts ) 
 
 Holes punched small and then coun- \ 
 
 tersnnk, the rivets being pan- ]- 28*9 ,, „ 
 
 head, with countersunk points I 
 
 It appears, from the above results, that if a plate has the 
 holes drilled or has them punched and countersunk in the 
 ordinary way as for flush riveting, the strength of the material 
 is fully maintained. Also that, although punching holes in a 
 plate reduces the strength from 285- to 22 tons per square inch, 
 a reduction of 22 per cent., yet when connected by a butt strap, 
 and riveted up, the strength rises to 24*9 tons per square inch, 
 which is only 1 2 per cent, weaker than the unpunched plate, the 
 process of riveting strengthening the plate. 
 
 In an ordinary butt-strap, with the holes spaced closely 
 together in order to obtain a water-tight pitch for the rivets, it 
 is found that the punching distresses the material in the neigh- 
 bourhood of the holes, and the strength is materially reduced, 
 as we have seen above. If, however, the butt strap is annealed 
 after punching, the full strength of the material is restored. It 
 is the practice, in ships built for the British Admiralty, for all 
 butt straps of important structural plating to have the holes 
 drilled or to be annealed after punching.^ In either case the 
 
 ' In ships built for the British Admiralty, for plating which forms an 
 important feature in the general structural strength, such as the inner and 
 outer bottom plating, deck plating, deck stringers, etc., the butt straps 
 must have the holes drilled, or be annealed after the holes are punched. 
 
Calculation of Weights, etc. 203 
 
 strength is restored. For ships built to the rules of Lloyd's 
 Register, butt straps above If of an inch in thickness are 
 annealed or the holes rimed after punching.^ 
 
 In our calculations of the strength of butt straps, we there- 
 fore can assume that the strength of the material between the 
 rivet-holes is the same as the strength of the material of the 
 unpunched plate. 
 
 Again, the plating, in the cases we have to deal with, has 
 the riveting flush on the outside, and the holes are made with 
 a countersink for this purpose. Here also we can assume that 
 the strength of the material is the same as the strength of the 
 material of the unpunched plate. 
 
 The specified tests for the tensile strength of steel plates 
 are as follows : — 
 
 For ships built for the British Admiralty, not less than 26 
 and not more than 30 tons per square inch of section. 
 
 For ships built to the rules of Lloyd's Register, not less 
 than 28 and not more than 32 tons per square inch of 
 section. 
 
 The plates tested above showed a tensile strength of 2?>\ 
 tons per square inch, or nearly midway between the limits laid 
 down by the British Admiralty. It seems reasonable, there- 
 fore, in calculating the ultimate strength of riveted joints, to 
 take as the strength of the material the minimum strength to 
 which it has to be tested. Thus, in a ship built for the British 
 Admiralty, we can use 26 tons as the strength per square inch 
 of section, and in a ship built under Lloyd's rules, we can use 
 28 tons per square inch of section. 
 
 The following two examples will illustrate the methods 
 adopted in calculating the strength of butt fastenings : ^ — 
 
 In such bottom plating, the countersunk holes must be punched about 
 J inch less in diameter than the rivets which are used, the enlargement of 
 the holes being made in the countersinking, which must in all cases be 
 carried through the whole thickness of the plates. 
 
 ' In ships built to the rules of Lloyd's Register, stringer plates, sheer- 
 strakes, garboard strakes, and all butt straps, when above Jg of an inch in 
 thickness, are carefully annealed, or the holes sufficiently rimed after 
 punching, to remove the injurious effect of the punching. 
 
 - Admiralty tests, etc., adopted. 
 
204 Theoretical Naval Architecttire. 
 
 I . A steel stringer plate is 48 inches broad and -j^ inch thic3c. Sketch 
 the fastenings in a beam and at a butt, and show by calculations that the 
 butt connection is a good one. 
 
 (S. and A. Exam., 1897.) 
 
 For a ^-inch plate we shall require |-inch rivets, and setting these out 
 at the beam, we require 9 rivets, as shown in Fig. 84. The effective breadth 
 of the plate through this line of holes is therefore — 
 
 48 - 9(1) = 41^ inches 
 
 and the strength is — 
 
 41 J X /g X 26 = 470 tons 
 
 and this is the standard strength that we have to aim at in designing the 
 butt strap. 
 
 ( 1 ) As regards the number of rivets. The shearing strength of a |-inch 
 rivet being 11 '5 tons, the number of rivets necessary to equal the standard 
 strength of 470 tons is — 
 
 -3^- = 40' 8, say 41 rivets 
 
 If we set out the rivets in the strap as shown in Fig. 84, leaving the 
 alternate rivets out in the line AA, it will be found that exactly 41 rivets 
 is obtained, with a four-diameter pitch. So that, as regards the number of 
 rivets the butt connection is a good one. 
 
 (2) The strength of the plate in the line A A is the same as at the beam, 
 the same number of rivet-holes being punched in each case. 
 
 (3) If the strap is ('5 inch thick, the strength of the strap in the line B13 
 is given by — 
 
 {48 - 16(f)} X 1^ X 26 = 410 tons 
 
 This is not sufficient, and the strap must be thickened up. If made \ incli 
 thick, the strength is — 
 
 {48- 16(f)} Xi X26 = 468 
 
 which is very nearly equal to the standard strength of 470 tons. 
 
 (4) The shear of the 9 rivets in the line AA is I03"5 tons, so that the 
 strength of the plate through the line of holes CC and the shear of the 
 rivets in the line AA are — 
 
 410 + 103-5 = 5>3'S tons 
 
 (5) Similarly, the strength of the strap through the line CC and the 
 shear of the rivets in the line AA are — 
 
 468 + 184 = 652 tons 
 
 The ultimate strengths of the butt connection in the five different ways it 
 might break are tlierefore 471J, 470, 468, 513J, 652 tons respectively, and 
 thus the standard strength of 470 tons is maintained for all practical 
 purposes, and consequently the butt connection is a good one. 
 
 2. If it were required to so join two plates as to make the strength at 
 the butt as nearly as possible equal to that of the unpierced plates, what 
 kind of butt strap would you adopt ? 
 
Calculation of Weights, etc. 205 
 
 Supposing the plates to be of mild steel 36 inches wide and J inch thick, 
 give the diameter, disposition, and jjitch of rivets necessary in the strap. 
 
 (S. and A. Exam., 1895.) 
 
 The first part of this question has been already dealt with on p. 201. 
 To lessen the number of rivets, it is best to use a double butt strap, as 
 Fig. 85, so as to get a double shear of the rivets. Each of the butt straps 
 should be slightly thicker than the half-thickness of the plate, say y'j inch. 
 
 The standard strength to work up to is that of the plate through the 
 single rivet-hole at the corner of the strap. |-inch rivets being used, the 
 standard strength is — 
 
 (36 - J) X i X 26 = 457 tons 
 
 The single shear of a |-inch rivet is 1 5i tons, and the double shear may be 
 taken as — 
 
 I5'25 X l'8 = 27i tons 
 
 and consequently the least number of rivets required each side of the 
 butt is — 
 
 ^^ = 1 6 '6, say 17 rivets 
 27-5 
 
 The strength of the plate along the slanting row of holes furthest from 
 the butt must be looked into. The rivets here are made with a water-tight 
 pitch, say from 4 to 4I diameters. If we set out the holes for a strap 2 feet 
 wide, it will be found that the strength is below the standard. A strap 
 3 feet wide will, however, give a strength through this line of about 465 
 tons, which is very near the required 457 tons. There are 13 rivets along 
 the edge of the strap, and the inside may be filled in as shown, giving a 
 total number of rivets, each side of the butt, of 19. 
 
 Strains experienced by Ships. — The strains to which 
 ships are subjected may be divided into two classes, viz. — 
 
 1. Structural strains, i.e. strains which aifect the structure 
 of the ship considered as a whole. 
 
 2. Local strains, i.e. strains which affect particular portions 
 of the ship. 
 
 1. Structural Strains. — These may be classified as 
 follows : — 
 
 {a) Strains tending to cause the ship to bend in a fore-and- 
 aft direction. 
 
 {b) Strains tending to change the transverse form of the ship. 
 
 {c) Strains due to the propulsion of the vessel, either by 
 steam or sails. 
 
 2. Local Strains. — These may be classified as follows i-^ 
 
 (a) Panting strains. 
 
 (b) Strains due to heavy local weights, As masts, engines, 
 armour, guns, etc. 
 
2o5 
 
 Theoretical Naval Architecture. 
 
 {c) Strains caused by the thrust of the propellers. 
 
 {d) Strains caused by the attachment of rigging. 
 
 {e) Strains due to grounding. 
 
 We will now deal with some of these various strains to 
 which a ship may be subjected in a little more detail. 
 
 Longitudinal Bending Strains. — A ship may be regarded as 
 a large beam or girder, subject to bending in a fore-and-aft 
 direction. The support of the buoyancy and the distribution 
 of weight vary considerably along the length of a ship, even 
 when floating in still water. Take a ship and imagine she is 
 cut by a number of transverse sections, as in Fig. 86. Each 
 
 ^ 
 
 © 
 
 ® 
 
 ® 
 
 ® 
 
 © 
 
 
 ~ 
 
 
 
 
 
 
 _£_ 
 
 Fig. I 
 
 of the portions has its weight, and each has an upward support 
 of buoyancy. But in some of the portions the weight exceeds 
 the buoyancy, and in others the buoyancy exceeds the weight. 
 The total buoyancy of all the sections must, of course, equal the 
 total weight. Now imagine that there is a water-tight bulkhead 
 at each end of each of these portions, and the ship is actually 
 cut at these sections. Then the end portions (i) and (5) have 
 considerable weight but small displacement, and consequently 
 they would sink deeper in the water if left to themselves.'^ In 
 the portions (2) and (4), on the other hand, the buoyancy might 
 exceed the weight (suppose these are the fore-and-aft holds, and 
 the ship is light), and if left to themselves they would rise. The 
 
 ' Strictly speaking, each portion would change trim if left to itself, but 
 We suppose that the various portions are attached, but free to move in a 
 Vertical direction. 
 
Calculation of Weights, etc. 
 
 207 
 
 midship portion (3) lias a large amount of buoyancy, but also 
 a large weight of engines and boilers, and this portion might 
 very well have to sink a small amount if left to itself. In any 
 actual ship, of course, it is a matter of calculation to find how 
 the weight and buoyancy vary throughout the length. This 
 case is somewhat analogous to the case of a beam supported 
 and loaded as shown in Fig. 87. At each point along the 
 
 w. 
 
 Fig. 87. 
 
 beam there is a tendency to bend, caused by the way the 
 beam is loaded and supported, and the beam must be made 
 sufficiently strong to withstand this bending tendency. In the 
 same way, the ship must be constructed in such a manner as to 
 effectually resist the bending strains that are brought to bear 
 upon the structure. 
 
 When a vessel passes out of still water and encounters 
 waves at sea, the strains to which she is subjected must differ 
 very much from those we have been considering above. Sup- 
 pose the ship to be end on to a series of waves having lengths 
 
 Fig. 88. 
 
 from crest to crest or from trough to trough equal to the length 
 of the ship. We will take the two extremes. 
 
 (i) The ship is supposed to have the crest of the wave 
 amidships. 
 
2o8 
 
 Theoretical Naval Architecture. 
 
 (2) The ship is supposed to have the trough of the wave 
 amidships. 
 
 (i) This is indicated in Fig. 88. At this instant there is an 
 excess of weight at the ends, and an excess of buoyancy amid- 
 ships. The ship may be roughly compared to a beam supported 
 at the middle, with weights at the end, as in Fig. 89. The con- 
 
 B. 
 
 V777f777777Trrr 
 
 Fig. 89. 
 
 sequence is that there is a tendency for the ends to droop 
 relatively to the middle. This is termed hogging. 
 
 (2) This is indicated in Fig. 90. At this instant there is an 
 
 ^ 
 
 '"'"''''^^rrr^nrrrmrrrrr^' 
 
 m 
 
 77?p 
 
 y 
 
 Fig. go. 
 
 excess of weight amidships, and an excess of buoyancy at the 
 ends, and the ship may be roughly compared to a beam sup- 
 ported at the ends and loaded in the middle, as Fig. 91. The 
 
 Fig. gi. 
 
 •consequence is, there is a tendency for the middle to droop 
 relatively to the ends. This is termed sagging. 
 
 We have seen above how the ship may be compaTed to a 
 beam, and in order to understand how the material should be 
 
Calculation of Weights^ etc. 209 
 
 disposed in order best to withstand the bending strains, we will 
 consider briefly some points in connection with ordinary 
 beams.^ 
 
 Take a beam supported at the ends and loaded at the 
 middle. It will bend as shown exaggerated in Fig. 92. The 
 
 Fig. 92. 
 
 resistance the beam will offer to bending will depend on 
 the form of the section of the beam. Take a beam having 
 a sectional area of 16 square inches. We can dispose the 
 material in many different ways. Take the following : — 
 
 (a) 8 inches wide, 2 inches deep (a. Fig. 93). 
 
 {V) 4 inches wide, 4 inches deep {b. Fig. 93). 
 
 (c) 2 inches wide, 8 inches deep {c. Fig. 93). 
 
 {il) 8 inches deep, with top and bottom flanges 5 inches wide 
 and I inch thick {d. Fig. 93). 
 
 Co-) 
 
 (l^) 
 
 Fig. 93. 
 
 Then the resistances of these various sections to bending 
 compare as follows : — 
 
 If (a) is taken as i, then {b) is 2, {c) is 4, and (d) is 6-|. 
 
 We thus see that we can make the beam stronger to resist 
 bending by disposing the material far away from the centre. 
 The beam {i) is 6-| times the strength of {a) against bending, 
 although it has precisely the same sectional area. A line 
 
 ' The subject of beams will be founcj fully discussed in works on 
 Applied Mecbwjcs, 
 
210 Theoretical Naval Architecture. 
 
 drawn transversely through the centre of gravity of the section 
 of a beam is termed the neutral axis. 
 
 These principles apply equally to the case of a ship, and 
 we thus see that to resist bending strains the material of the 
 structure should be disposed far away from the neutral axis.^ 
 In large vessels, and those of large proportion of length to 
 breadth or length to depth, Lloyd's rules require that partial 
 or complete steel decks shall be fitted on the upper decks, the 
 upper-deck stringer made wider and thickened up, the sheer 
 strake doubled or made thicker, the plating at the bilge 
 thickened up or doubled, and the keelsons increased in 
 strength. These are all portions of the structure farthest away 
 frortt the neutral axis. 
 
 For hogging strains, the upper portions of the vessel are 
 in tension and the lower portions are in compression. For 
 sagging strains, the upper portions are in compression and the 
 lower portions are in tension. Thus the portions of the struc- 
 ture that are useful in resisting these hogging and sagging strains 
 are the upper and main decks and stringers, sheer-strake and 
 plating below, plating at and below the bilge, both of the inner 
 and outer bottom, keel, keelsons, and longitudinal framing. 
 
 Strains tending to change the Transverse Form of th-e Ship.— 
 Strains of this character are set up in a ship rolling heavily. 
 Take a square framework jointed at the corners, and imagine 
 it to be rapidly moved backwards and forwards as a ship does 
 when she rolls. The framework will not break, but will distort, 
 as shown in Fig. 94. There is a tendency to distort in a similar 
 way in a ship roUing heavily, and the connections of the beams 
 to the sides, and the transverse structure of the ship, must be 
 made sufficiently strong to prevent any of this racking taking 
 place. Transverse bulkheads are valuable in resisting the 
 tendency to change the transverse form. In ships built to 
 Lloyd's Register, the ordinary depth of beam arms is 2 J times the 
 depth of the beam ; but in sailing-ships, which only have one 
 transverse bulkhead, the collision bulkhead, when the length of 
 
 ' There are other strains, viz. shearing strains, which are of importance 
 (see "Applied Mechanics," by Professor Cotterill, and a paper read at the 
 Institution of Naval Architects in 1890, by the late Professor Jenkins). 
 
Calailation of Weights, etc. 
 
 211 
 
 the midship upper-deck beam exceeds 36 feet, the bracket 
 knees to each tier of beams must not be less than three times 
 the depth of the beam, and the depth at the throat not less 
 than one and three-quarters the depth of the beam.^ 
 
 A ship, when docked, especially if she has on board heavy 
 weights, as armour or coals, is subjected to severe strains tend- 
 ing to change the transverse form. If the ship is supported 
 wholly at the keel, no shores being supposed placed in position, 
 the weight either side the middle line tends to make the sides 
 drop, and bring the beams into tension. A ship when docked, 
 
 • • 
 
 Fig. 94- 
 
 however, is partially supported by shores as well as at the keel 
 as the water leaves, so that this case is an extreme one. 
 
 Panting. — This term is used to describe the working in 
 and out of the plating, and it is usually found at the fore and 
 after ends of the ship, where the surface is comparatively flat. 
 The forward end especially is subject to severe blows from the 
 sea, and special attention is paid to this part by working special 
 beams and stringers to succour the plating. In vessels built to 
 the rules of Lloyd's Register, the following rules have to be 
 carried out to provide suflScient local strength against panting : — 
 
 All stringers, where practicable, to extend fore and aft, and 
 to be efificienriy connected at their ends with plates forming 
 hooks and crutches of the same thickness as the floor-plates 
 amidships, and those below the hold beams should be spaced 
 about 4 feet apart. In vessels whose plating number ^ is 24,000, 
 
 ' There is also in sailing ships a couple due to the sails, tending to 
 distort the sections. 
 « See p. 194. 
 
212 Theoretical Naval Architechire. 
 
 or above, an additional hook or crutch should be fitted at the 
 ends of the vessel, between each tier of beams, to the satisfac- 
 tion of the Surveyors. 
 
 The depth for regulating the number of tiers of beams to 
 alternate frames in the fore peak to be taken at the collision 
 bulkhead. All vessels to have, in addition, provision made to 
 prevent panting by extra beams, bracket knees and stringer 
 plates being fitted before and abaft the collision bulkheads. 
 Panting beams and stringers to be fitted at the after end 
 where considered necessary by the Surveyors. 
 
 The stringer plates on the panting beams to be attached to 
 the outside plating when fitted in continuation of intercostal 
 stringers. These plates are to extend abaft the collision bulk- 
 head for a length of not less than one-fourth the midship 
 breadth of the vessel, and be efficiently supported by brackets 
 at alternate frames. Panting beams and stringers to be fitted 
 at the after end where considered necessary by the Surveyors. 
 
 The other local strains mentioned on pp. 205, 206 have to 
 be provided for by special local strengthening. 
 
 For a full discussion of the whole subject of the strains 
 experienced by ships, and the stresses on the material compos- 
 ing the structure, the student is referred to the "Manual of 
 Naval Architecture," by Sir W. H, White. 
 
 Examples to Chapter VI. 
 
 1. The area of the outer bottom plating of a ship, over which the 
 plating is worked 25 lbs. per square foot, is 23,904 square feet, lapped 
 edges and butt straps, both double-riveted. Estimate the difference in 
 weight due to working the plating with average-sized plates 20' X 4J', or 
 with the average size 12' X 3'. 
 
 Ans. About 20 tons. 
 
 2. Steel angle bars 3I" X 3" are specified to be 8J lbs. per lineal foot 
 instead of ^ inch thick. Determine the saving of weight per 100 lineal 
 feet. 
 
 Ans. 52 lbs. 
 
 3. Determine the weight per lineal foot of a steel T-t" 5" X 4" ""• §"• 
 
 Ans. I4'4S lbs. 
 
 4. For a given purpose, angle bars of iron 5" x 3" X tV °^ °^ ^'^^' 
 S" X 3" X ijV' can be used. Find the saving of weight per 100 feet if steel 
 is adopted. 
 
 Ans. 95 lbs. 
 
 5. A mast 96 feet in length, if made of iron, is at its greatest diameter, 
 
Calculation of Weights, etc. 213 
 
 viz. 32 inches, y'g inch thick, and has three angle stiffeners 5" x 3" X fV'- 
 For the same diameter, if made of steel, the thiclcness is Jg inch, with three 
 angle stiffeners 5" x 3" X I5". Estimate the difference in weight. 
 
 Ans. About I ton, 
 
 6. At a given section of a ship the following is the form : The lengths of 
 ordinates 3 feet apart are 19'6, 18-85, '7'8, •6-4, I4'S, ii'S, 7-35, and 
 I'O feet respectively. Estimate the vertical position of the centre of 
 gravity of the curve forming the section, supposing it is required to find the 
 vertical position of the centre of gravity of the bottom plating of uniform 
 thickness. 
 
 Alls. About 12^ feet from the top. 
 
 7. The half-girths of the inner bottom of a vessel at intervals of 51 feet 
 are 26'6, 29'8, 32-0, 32'8, and 3I'2 feet respectively, and the centres of 
 gravity of these half-girths are l8'6, 20"6, 21 '2, 20'0, 17 '4 feet respectively 
 below the L.W.L. Determine the area of the inner bottom and the 
 position of its centre of gravity both longitudinally and vertically. If the 
 plating is of 15 lbs. to the square foot, what would be the weight, allowing 
 14J per cent, for butts, laps, and rivet-heads. 
 
 Ans. 12,655 square feet ; 105 feet from finer end, 20 feet below the 
 L.W.L. ; 97 tons. 
 
 8. The whole ordinates of the boundary of a ship's deck are 6'5, 24, 
 29. 32. 33'S. 33'S> 33'S' 32. 3°> 27, and 6-5 feel respectively, and the 
 common interval between them is 21 feet. 
 
 The deck, with the exception of 350 square feet, is covered with % inch 
 steel plating worked flush jointed, with single riveted edges and butts. 
 Find the weight of the plating, including straps and fastenings. 
 
 Ans. 45 tons. 
 
 9. A teak deck, 2j inches thick, is supported on beams spaced 4 feet 
 apart, and weighing 15 pounds per foot run. Calculate the weight of a 
 middle-line portion of this deck (including fastenings and beams) 24 feet 
 long and 10 feet wide, 
 
 Ans. I '55 tons. 
 
CHAPTER VII. 
 
 HORSE-POWER, EFFECTIVE AND INDICATED— RESIST- 
 ANCE OF SHIPS— COEFFICIENTS OF SPEED— LA W OF 
 CORRESPONDING SPEEDS. 
 
 Horse-power. — We have in Chapter V. defined the " work '' 
 done by a force as being the product of the force and the 
 distance through which the force acts. Into the conception 
 of work the question of time does not enter at all, whereas 
 " power " involves not only work, but also the time in which 
 the work is done. The unit of power is a "horse-power" 
 which is taken as " 33,000 foot-lbs. of work performed in i 
 minute^' or "550 foot-lbs. of work performed in i second^ 
 Thus, if during i minute a force of i lb. acts through 33,000 
 feet, the same power will be exerted as if a force of 33 lbs. acts 
 through 1000 feet during i minute, or if 50 lbs. acts through 
 1 1 feet during i second. Each of these will be equivalent to 
 I horse-power. The power of a locomotive is a familiar in- 
 stance. In this case the work performed by the locomotive 
 — if the train is moving at a uniform speed— is employed in 
 overcoming the various resistances, such as the friction of the 
 wheels on the track, the resistance of the air, etc. If we 
 know the amount of this resistance, and also the speed of the 
 train, we can determine the horse-power exerted by the loco- 
 motive. The following example will illustrate this point : — 
 
 If the mass of a train is 150 tons, and the resistance to its motion 
 
 arising from the air, friction, etc., amount to 16 lbs. weight per ton when 
 
 the train is going at the rate of 60 miles pef hour on a level plain, find the 
 
 horse-power of the engine which can just keep it going at that rate. 
 
 Resistance to onward motion = 150 x 16 
 
 = 2400 lbs. 
 
 Speed in feet per minute = 5280 
 
 Work done per minute = 2400 X 5280 foot-lbs. 
 
 „ 2400 X 5280 
 
 Horse-power = 
 
 33000 
 
 = 384 
 
Horse-power, Effective and Indicated, etc. 215 
 
 In any general case, if — 
 
 R = resistance to motion in pounds ; 
 
 V = velocity in feet per minute ; 
 
 V = velocity in knots (a velocity of i knot is 6080 feet 
 
 per hour) ; 
 then — 
 
 Horse-power = 
 
 f 33000 
 
 R X V X loi 
 
 = nearly 
 
 33000 ^ 
 
 The case of the propulsion of a vessel by her own engines 
 is much more complicated than the question considered above 
 of a train being drawn along a level plain by a locomotive. 
 We must first take the case of a vessel being towed through 
 the water by another vessel. Here we have the resistances 
 offered by the water to the towed vessel overcome by the strain 
 in the tow-rope. In some experiments on H.M.S. Greyhound 
 by the late Mr. Froude, which will be described later, the tow- 
 rope strain was actually measured, the speed being recorded 
 at the same time. Knowing these, the horse-power necessary 
 to overcome the resistance can be at once determined. For 
 example — 
 
 At a speed of 1017 feet per minute, the tow-rope strain was 10,770 lbs. 
 Find the horse-power necessary to overcome the resistance. 
 
 Work done per minute = 10,770 X 1017 foot-lbs. 
 
 -- 10770 X 1017 
 
 Horse-power = — — - 
 
 i 33000 
 
 = 332 
 
 Effective Horse-power. — The effective horse-power of 
 a vessel at a given speed is the horse-power required to over- 
 come the various resistances to the vessel's progress at that 
 speed. It may be described as the horse-power usefully 
 employed, and is sometimes termed the " tow-rope " or " tug " 
 horse-power, because this is the power that would have to be 
 transmitted through the tow-rope if the vessel were towed 
 through the water at the given speed. Effective hprse-power 
 is often written E.H.P. We shall see later that the E.H.P. is 
 eritirely different to the Indicated Horse-power (written I.H.P.), 
 
2i6 Theoretical Naval Architecture. 
 
 which is the horse-power actually measured at the vessel's 
 .engines. 
 
 Example. — Find the horse-power which must be transmitted through 
 a tow-rope in order to tow a vessel at the rate of l6 knots, the resistance to 
 the ship's motion at that speed being equal to a weight of 50 tons. 
 
 Ans. 5503 H.P. 
 
 Experiments with H.M.S. "Greyhound," by the 
 late Mr. William Proude, F.R.S.— These experiments 
 took place at Portsmouth as long ago as 187 1, and they settled 
 a number of points in connection with the resistance and pro- 
 pulsion of ships, about which, up to that time, little was known. 
 The thoroughness with which the experiments were carried 
 out, and the complete analysis of the results that was given, 
 make them very valuable; and students of the subject would 
 do well to consult the original paper in the Transactions of the 
 Institution of Naval Architects for 1874. A summary of the 
 experiments, including a comparison with Rankine's "Aug- 
 mented Surface Theory of Resistance," will be found in vol. iii. 
 of Naval Science. Mr. Froude's report to the Admiralty was 
 published in Engineering, May i, 1874. 
 
 The Greyhound was a ship 172' 6" in length between per- 
 pendiculars, and 33' 2" extreme breadth, the deepest draught 
 during the experiments being 13' 9" mean. The displacement 
 
 u J, 
 
 Active 
 
 Fig. 9s- 
 
 corresponding to this mean draught being 1161 tons; area of 
 midship section, 339 square feet; area of immersed surface, 
 7540 square feet. The Greyhound was towed by H.M.S. 
 Active. It was essential to the accuracy of the experiments 
 that the Greyhound should proceed through undisturbed water, 
 and to avoid using an exceedingly long tow-rope a boom was 
 rigged out from the side of the Active to take the tow-rope (see 
 Fig. 95). By this means the Greyhound proceeded through 
 
Horse-power, Effective and Indicated, etc. 217 
 
 water that had not been influenced by the wake of the Active. 
 The length of the boom on the Active was 45 feet, and the length 
 of the tow-rope was such that the Greyhound! s bow was 190 
 feet clear of the Active's stern. The actual stress on the tow- 
 rope at its extremity was not required, but the " horizontal 
 component." This would be the stress that was overcoming 
 the resistance, the "vertical component" being due to the 
 weight of the tow-rope. The horizontal stress on the tow-rope 
 and the speed were automatically recorded on a sheet of paper 
 carried on a revolving cylinder. For details of the methods 
 employed and the apparatus used, the student is referred to 
 
 — Speed in Kmots. — 
 
 Fig. 96. 
 
 the sources mentioned above. The horizontal stress on the 
 tow-rope was equal to the nett resistance of the Greyhound. 
 The results can be represented graphically by a curve, abscissae 
 representing speed, and ordinates representing the resistance 
 in pounds. Such a curve is given by A in Fig. 96. 
 
 It will be seen that the resistance increases much more 
 rapidly at the higher than at the lower speeds; thus, on 
 increasing the speed from 7 to 8 knots, an extra resistance 
 of 1500 lbs. has to be overcome, while to increase the speed 
 
2i8 Theoretical Naval Architecture. 
 
 from II to 12 knots, an extra resistance of 6000 lbs. must 
 be overcome. Beyond 12 knots the shape of the curve 
 indicates that the resistance increases very rapidly indeed. 
 Now, the rate at which the resistance increases as the speed 
 increases is a very important matter. (We are only concerned 
 now with the total resistance.) Up to 8 knots it was found 
 that the resistance was proportional to the square of the speed ; 
 that is to say, if Ri, Rj represent the resistances at speeds 
 Vi, V2 respectively, then, if the resistance is proportional to 
 the square of the speed — 
 
 Ri : R, : : V^^ : V,^ 
 Ri Vi= 
 "■^R^^V? 
 
 By measuring ordinates of the curve in Fig. 96, say at 5 and 6 
 knots, this will be found to be very nearly the case. As the 
 speed increases above 8 knots, the resistance increases much 
 more rapidly than would be given by the above ; and between 
 1 1 and 1 2 knots, the resistance is very nearly proportional to 
 the fourth power of the speed. 
 
 The experiments were also conducted at two displacements 
 less than 1161 tons, viz. at 1050 tons and 938 tons. It was 
 found that differences in resistance, due to differences of 
 immersion, depended, not on changes of area of midship 
 section or on changes of displacement, but rather on changes 
 in the area of wetted surface. Thus for a reduction of 19^ 
 per cent, in the displacement, corresponding to a reduction of 
 area of midship section of 16-5 per cent., and area of immersed 
 surface of 8 per cent., the reduction in resistance was about 
 \o\ per cent., this being for speeds between 8 and 12 knots. 
 
 Ratio between Effective Horse-power and Indi- 
 cated Horse-power. — We have already seen that, the 
 resistance of the Greyhound at certain speeds being deter- 
 mined, it is possible to determine at once the E.H.P. at 
 those speeds. Now, the horse-power actually developed by the 
 Greyhound's own engines, or the "indicated horse-power" 
 (I.H.P.), when proceeding on the measured mile, was observed 
 on a separate series of trials, and tabulated. The ratio of the 
 
Horse-poiiOer, Effective and Indicated, etc. 219 
 
 E.H.P. to the I.H.P. was then calculated for different speeds, 
 and it was found that E.H.P. -H I.H.P. in the best case was 
 only o'42 ; that is to say, as much as 58 per cent, of the power 
 was employed in doing work other than overcoming the actual 
 resistance of the ship. This was a very important result, and 
 led Mr. Froude to make further investigations in order to 
 determine the cause of this waste of power, and to see whether 
 it was possible to lessen it. 
 
 . E.H.P. 
 
 1 he ratio , ^t T^ at any given speed is termed the 
 
 '■pro- 
 
 I.H.P. 
 
 pulsive coefficient" at that speed. As we saw above, in the 
 most efficient case, in the trials of the " Greyhound" this co- 
 efficient was 42 per cent. For modern vessels with fine lines a 
 propulsive coefficient of 50 per cent, may be expected, if the 
 engines are working efficiently and the propeller is suitable. 
 In special cases, with extremely fine forms and fast-running 
 engines, the coefficient rises higher than this. These values only 
 hold good for the maximum speed for which the vessel is 
 designed ; for lower speeds the coefficient becomes smaller. 
 The following table gives some results as given by Mr. Froude. 
 The Mutine was a sister-ship to the Greyhound., and she had 
 also been run upon the measured mile at the same draught and 
 trim as the Greyhound. 
 
 
 
 |H 
 
 1 K 
 
 
 •s 
 
 
 v 
 
 ^^■Sbi 
 
 1 u ■ z, 
 
 i 1 c 
 
 1 
 
 
 
 fill 
 
 1 ' I 
 
 .S 
 ,1. 
 
 Ship. 
 
 
 speed d 
 eriments 
 g an esti 
 sts and : 
 
 II 
 II 
 
 I 
 
 
 
 i 
 
 J3 
 
 
 
 i:"?i 
 
 a. C re 
 
 aU 
 
 
 
 u % 
 
 
 i 
 
 e due t 
 ing ex] 
 indiidl 
 ce of m 
 
 1 ' 1 
 
 II 
 
 
 
 
 
 .1 -l 
 
 > 
 
 •3 
 
 Si 
 
 
 ■o 
 
 
 Ik I u 
 
 £ 
 
 H 
 
 
 « 
 
 10,770 
 6,200 
 
 
 < 
 
 
 Greyhound 
 
 /IOI7 
 \ 845 
 
 332-1 
 1587 
 
 7F6 
 
 453 
 
 0-422 
 0-350 
 
 Mutine 
 
 / 977 
 I 757 
 
 9,440 
 
 279-5 
 
 770 
 
 0-363 
 
 
 4.770 
 
 109-4 
 
 328 
 
 0-334 
 
220 
 
 Theoretical Naval Architecture. 
 
 Resistance. — We now have to inquire into the various 
 resistances which go to make up the total resistance which a 
 ship experiences in being towed through the water. These 
 resistances are of three kinds — 
 
 1. Resistance due to friction of the water upon the surface 
 of the ship. 
 
 2. Resistance due to the formation of eddies. 
 
 3. Resistance due to the formation of waves. 
 
 1. " Frictional resistance," or the resistance due to the 
 friction of the water upon the surface of the ship. This is 
 similar to the resistance offered to the motion of a train on a 
 level line owing to the friction of the rails, although it follows 
 different laws. It is evident that this resistance must depend 
 largely upon the state of the bottom. A vessel, on becoming 
 covered with barnacles, etc., while lying in a port, loses speed 
 very considerably, owing to the greatly increased resistance 
 caused. This frictional resistance forms a large proportion of 
 the total at low speeds, and forms a good proportion at higher 
 speeds. 
 
 2. Resistance due to eddy-making. — Take a block of wood, 
 and imagine it placed a good distance below the surface of 
 a current of water moving at a uniform speed V. Then 
 the particles of water will run as approximately indicated 
 in Fig. 97. At A we shall have a mass of water in a state of 
 
 Fig. 97. 
 
 violent agitation, and a much larger mass of water at the rear 
 of the block. Such masses of confused water are termed 
 "eddies," and sometimes "dead water." If now we imagine 
 that the water is at rest, and the block of wood is being towed 
 
Horse-power, Effective and Indicated, etc. 221 
 
 through the water at a uniform speed V, the same eddies will 
 be produced, and the eddying water causes a very considerable 
 resistance to the onward motion. Abrupt terminations which 
 are likely to cause such eddies should always be avoided in 
 vessels where practicable, in order to keep the resistance as 
 low as possible. This kind of resistance forms a very small 
 proportion of the total in well-formed vessels, but in the older 
 vessels with full forms aft and thick stern-posts, it amounted to 
 a very considerable item. 
 
 3. Resistance due to the formation of waves. — For low 
 speeds this form of resistance is not experienced to any 
 sensible extent, but for every ship there is a certain speed 
 above which the resistance increases more rapidly than would 
 be the case if surface friction and eddy-making alone caused 
 the resistance. This extra resistance is caused by the forma- 
 tion of waves upon the surface of the water. 
 
 We must now deal with these three forms of resistance in 
 detail, and indicate as far as possible the laws which govern 
 them. 
 
 I. Frictional Resistance. — The data we have to work upon 
 when considering this form of resistance were obtained by the 
 late Mr. Froude. He conducted an extensive series of experi- 
 ments on boards of different lengths and various conditions 
 of surface towed through water contained in a tank, the speed 
 and resistance being simultaneously recorded. The follow- 
 ing table represents the resistances in pounds per square 
 foot due to various lengths of surface of various qualities when 
 moving at a uniform speed of 600 feet per minute, or very 
 nearly 6 knots. There is also given the powers of the speed 
 to which the resistances are approximately proportional. 
 
 We can sum up the results of these experiments as follows : 
 The resistance due to the frictioniiof the water upon the surface 
 depends upon — 
 
 (i) The area of the surface. 
 
 (2) The nature of the surface. 
 
 (3) The length of the surface. 
 
 and (4) The resistance varies approximately as the square of 
 the speed. 
 
Theoretical Naval Architecture. 
 
 
 
 Length of Su 
 
 RFACE IN Feet. 
 
 
 2 
 
 3 
 
 20 
 
 so 
 
 Ti 
 
 
 n 
 
 h 
 S, 
 
 ■g-a 
 
 1 
 
 .•§"1 
 
 • & 
 
 Nature of surface. 
 
 %-l 
 
 T3 
 
 %% 
 
 •a 
 
 l-i 
 
 •a 
 
 l-s 
 
 T3 
 
 
 o o 
 
 §-• 
 
 oS 
 
 a <- 
 
 
 
 ig- 
 
 2 ^ 
 
 C . 
 
 
 
 n O 
 
 *' ft 
 
 
 
 
 
 
 
 
 
 "is 
 
 P.i2 
 
 ■S2 
 
 0.^ 
 
 ■S2 
 
 a.<£ 
 
 •a 2 
 
 s.^ 
 
 
 S a 
 
 c u 
 
 S a 
 
 S OJ 
 
 « 0. 
 
 B V 
 
 
 c w 
 
 
 a tfi 
 
 
 fttf] 
 
 
 p4(n 
 
 
 
 
 
 
 SS- 
 
 
 "s| 
 
 u> 
 
 "s| 
 
 in>M 
 
 S 3 
 
 
 O o 
 
 c g^ 
 
 
 
 c g^ 
 
 u 
 
 c 5^ 
 
 u 
 
 
 
 ^1 
 
 
 
 % 
 
 s| 
 
 c5 ^ 
 
 tef 
 
 (3 "^ 
 
 
 *.2 
 
 a> 
 
 |-^ 
 
 •5! 
 41 
 
 l-a 
 
 ■3 
 
 S.2 
 w 
 
 ■s 
 
 
 (2 K 
 
 « 
 
 fri g 
 
 « 
 
 &,E 
 
 M 
 
 »,2 
 
 « 
 
 Varnish 
 
 2-00 
 
 0-41 
 
 I -85 
 
 0-325 
 
 1-85 
 
 0-278 
 
 1-83 
 
 0-250 
 
 Tinfoil 
 
 2-i6 
 
 0-30 
 
 I "99 
 
 0-278 
 
 1-90 
 
 0-262 
 
 1-83 
 
 0-246 
 
 Calico 
 
 I '93 
 
 0-87 
 
 192 
 
 0-626 
 
 1-89 
 
 0-531 
 
 1-87 
 
 0-474 
 
 Fine sand 
 
 2-0O 
 
 o-8i 
 
 2-CX) 
 
 0-583 
 
 2-00 
 
 0-480 
 
 2 -06 
 
 0-405 
 
 Medium sand ... 
 
 2'00 
 
 0-90 
 
 2-00 
 
 0-625 
 
 2-00 
 
 0-S34 
 
 2-00 
 
 0-488 
 
 And thus we can write- 
 
 R=/.S(I)^ 
 
 where R = resistance in pounds ; 
 
 S = area of surface in square feet ; 
 V = speed in knots relative to still water ; 
 / = a coefficient depending upon the nature and length 
 of the surface. 
 
 This coefficient /will be the resistance per square foot given 
 in the above table, as is at once seen by making S = i square 
 foot and V = 6 knots. It is very noticeable how the resistance 
 per square foot decreases as the length increases. Mr. Froude 
 explained this by pointing out that the leading portion of the 
 plane must communicate an onward motion to the water which 
 rubs against it, and " consequently the portion of the surface 
 which succeeds the first will be rubbing, not against stationary 
 water, but against water partially moving in its own direction, 
 and cannot therefore experience as much resistance from it." 
 
 Experiments were not made on boards over 50 feet in 
 length. Mr. Froude remarked, in his report, " It is highly 
 
Horse-power, Effective and Indicated, etc. 223 
 
 desirable to extend these experiments, and the law they eluci- 
 date, to greater lengths of surface than 50 feet ; but this is the 
 greatest length which the experiment-tank and its apparatus 
 admit, and I shall endeavour to organize some arrangement by 
 which greater lengths may be successfully tried in open water." 
 
 Mr. Froude was never able to complete these experiments 
 as he anticipated. It has long been felt that experiments with 
 longer boards would be very valuable, so that the results could 
 be applied to the case of actual ships. It is probable that in 
 the new American experimental tank now under construction, 
 which is to be of much greater length than any at present in 
 existence, experiments with planes some hundreds of feet in 
 length may be carried out. 
 
 These experiments show very clearly how important the 
 condition of the surface is as affecting resistance. The 
 varnished surface may be taken as typical of a surface coated 
 with smooth paint, or the surface of a ship sheathed with 
 bright copper, the medium sand surface being typical of the 
 surface of a vessel sheathed with copper which has become 
 foul. If the surface has become fouled with large barnacles, 
 the resistance must rise very high. 
 
 In applying the results of these experiments to the case of 
 actual ships, it is usual to estimate the wetted surface, and to 
 take the length of the ship in the direction of motion to deter- 
 mine what the coefficient/ shall be. For greater lengths than 
 50 feet, it is assumed that the resistance per square foot is the 
 same as for the plate 50 feet long. 
 
 Take the following as an example : — 
 
 The wetted surface of a vessel is estimated at 7540 square feet, the 
 length being 172 feet. Find the resistance due to surface friction at a speed 
 of 12 knots, assuming a coefficient of 0-25, and that the resistance varies 
 (a) as the square of the speed, and (i5) as the I '83 power of the speed, 
 (a) Resistance = 0-25 X 7540 X ('g*)' 
 
 = 7540 lbs. 
 ib) Resistance = 0-25 x 7540 X (^'l'"' 
 = 6702 lbs.' 
 
 ' This has to be obtained by the aid of logarithms, 
 
 log (2IM) = 1-83 log 2 
 = 0-5508849 
 .-. 2' «^ = 3-5554 
 
224 Tlteoretical Naval Architecture. 
 
 It is worth remembering that for a smooth painted surface 
 the frictional resistance per square foot of surface is about ^Ib. 
 at a speed of 6 knots. 
 
 It is useful, in estimating the wetted surface for use in the 
 above formula, to have some method of readily approximating 
 to its value. Several methods of doing this have been already 
 given in Chapter II., the one known as "Kirk's Analysis" 
 having been largely employed. There are also several approxi- 
 mate formulse which are reproduced— 
 
 (i) Based on Kirk's analysis — 
 
 V 
 Surface = zLD + ^r 
 
 (2) Given by Mr. Denny — 
 
 V 
 Surface = i'7LD + y^ 
 
 (3) Given by Mr. Taylor — 
 
 Surface = i5"6 VW.L. 
 
 L being the length of the ship in feet ; 
 D being the mean moulded draught ; 
 V being the displacement in cubic feet ; 
 W being the displacement in tons. 
 
 There is also given, in Chapter II., a formula for the mean 
 wetted girth, and this multiplied by the length will give the 
 wetted surface. The formula is as follows : — 
 
 Mean wetted girth = o'ps cM + 2(1 — c)D 
 
 where c = prismatic coefficient of fineness ; 
 
 M = wetted girth on the midship section ; 
 
 D = mean moulded draught. 
 2. Eddy-7naking Resistance. — We have already seen the 
 general character of this form of resistance. It may be 
 assumed to vary as the square of the speed, but it will vary 
 in amount according to the shape of the ship and the appen- 
 dages. Thus a ship with a full stern and thick stern-posts 
 will experience this form of resistance to a much greater 
 
Horse-power, Effective and Indicated, etc. 225 
 
 extent than a vessel with a fine stern and with stern-post and 
 rudder of moderate thickness. Eddy-making resistance can 
 be allowed for by putting on a percentage to the frictional 
 resistance. Mr. Froude estimated that in well-formed ships 
 this form of resistance 
 
 Aft. 
 
 ForE 
 
 Fig. 98. 
 
 usually amounted to about 
 
 8 per cent, of the frictional 
 
 resistance. It is possible 
 
 to reduce eddy-making to 
 
 a minimum by paying 
 
 careful attention to the 
 
 appendages and endings 
 
 of a vessel. Thus shaft brackets in twin-screw ships are often 
 
 made of pear-shaped section, as shown in Fig. 98. 
 
 3. Resistance due to the Formation of Waves. — A completely 
 submerged body moving at any given speed will only experi- 
 ence resistance due to surface friction and eddy-making provided 
 it is immersed sufficiently ; but with a body moving at the 
 
 Direction of Flow. 
 
 Fig. 99. 
 
 surface, such as we have to deal with, the resistance due to 
 the formation of waves becomes very important, especially at 
 high speeds. This subject is of considerable difficulty, and 
 it is not possible to give in this work more than a general 
 outline of the principles involved. 
 
 Consider a body shaped as in Fig. 99 placed a long way 
 below the surface in water (regarded as frictionless), and 
 
 Q 
 
226 Theoretical Naval Architecture. 
 
 suppose the water is made to move past the body with a uniform 
 speed V. The particles of water must move past the body in 
 certain lines, which are termed stream-lines. These stream- 
 lines are straight and parallel before they reach the body, but 
 owing to the obstruction caused, the particles of water are 
 locally diverted, and follow curved paths instead of straight 
 ones. The straight paths are again resumed some distance at 
 the rear of the body. We can imagine these stream-lines 
 making up the boundaries of a series of stream-tubes, in each 
 of which the same particles of water will flow throughout the 
 operation. Now, as these streams approach the body they 
 broaden, and consequently the particles of water slacken in 
 speed. Abreast the body the streams are constricted in area, 
 and there is a consequent increase in speed ; and at the rear of 
 the body the streams again broaden, with a slackening in speed. 
 Now, in water flowing in the way described, any increase in 
 speed is accompanied by a decrease i?i pressure, and conversely 
 any decrease in speed is accompanied by an increase in pressure. 
 We may therefore say — 
 
 (i) There is a broadening of all the streams, and attendant 
 decrease of speed and consequent excess of pressure, near both 
 ends of the body ; and — 
 
 (2) There is a narrowing of the streams, with attendant 
 excess of speed and consequent decrease of pressure, along the 
 middle of the body. 
 
 This relation between the velocity and pressure is seen in 
 the draught of a fire under a chimney when there is a strong wind 
 blowing. The excess of the speed of the wind is accompanied 
 by a decrease of pressure at the top of the chimney. It 
 should be noticed that the variations of velocity and pressure 
 must necessarily become less as we go further away from the 
 side of the body. A long way off the stream-lines would be 
 parallel. The body situated as shown, with the frictionless 
 water moving past it, does not experience any resultant force 
 tending to move it in the direction of motion.^ 
 
 ' This principle can be demonstrated by the use of advanced mathematics. 
 " We may say it is quite evident if the body is symmetrical, that is to say, 
 has both ends alike, for in that case all the fluid action about the after 
 
Horse-power, Effective and Indicated, etc. 227 
 
 Now we have to pass from this hypothetical case to the case 
 of a vessel on the surface of the water. In this case the water 
 surface is free, and the excess of pressure at the bow and stern 
 shows itself by an elevation of the water at the bow and stern, 
 and the decrease of pressure along the sides shows itself by a 
 depression of the water along the sides. This system is shown 
 by the dotted profile of the water surface in Fig. 100. The 
 
 Fig. too. 
 
 foregoing gives us the reason for the wave-crest at the stern of 
 the ship. The crest at the bow appears quite a reasonable 
 thing to expect, but the crest at the stern is due to the same set 
 of causes. In actual practice the waves that are formed ob- 
 scure the simple system we have described above, which has 
 been termed the " statical wave." 
 
 Observation shows that there are two separate and distinct 
 series of waves caused by the motion of a ship through the 
 water — 
 
 (i) Waves caused by the bow; 
 
 (2) Waves caused by the excess pressure at the stern due to 
 the expansion of the streams. 
 
 Each of these series of waves consists of (i) a series of 
 diverging waves, the crests of which slope aft, and (2) a series 
 of transverse waves, whose crests are nearly perpendicular to 
 the middle line of the ship. 
 
 First, as to the diverging waves at the bow. " The inevi- 
 tably widening form of the ship at her entrance throws off on 
 each side a local oblique wave of greater or less size accord- 
 ing to the speed and obtuseness of the wedge, and these waves 
 form themselves into a series of diverging crests. These waves 
 
 body must be the precise counterpart of that about the fore body ; all the 
 stream-lines, directions, speed of flow, and pressures at every point must be 
 symmetrical, as is the body itself, and all the forces must be equal and 
 opposite " (see a paper by Mr. R. E. Froude, on "Ship Resistance," read 
 before the Greenock Philosophical Society in 1894). 
 
228 Theoretical Naval Architecture. 
 
 have peculiar properties. They retain their identical size for a 
 very great distance, with but little reduction in magnitude. 
 But the main point is, that they become at once disassociated 
 with the vessel, and after becoming fully formed at the bow, 
 they pass clear away into the distant water, and produce no 
 further eifect on the vessel's resistance." These oblique waves 
 are not long in the line of the crest BZ, Fig. loi, and the 
 
 waves travel perpendicular to the crest-line with a speed of 
 V cos 6, where V is the speed of the ship. As the speed of 
 the ship increases the diverging waves become larger, and con- 
 sequently represent a greater amount of resistance. 
 
 Besides these diverging waves, however, " there is produced 
 by the motion of the vessel another notable series of waves, 
 which carry their crests transversely to her line of motion." It 
 is this transverse series of waves that becomes of the greatest 
 importance in producing resistance as the speed is pushed to 
 values which are high for the ship. These transverse waves 
 show themselves along the sides of the ship by the crests and 
 troughs, as indicated roughly in Fig. loo. The lengths of these 
 waves ii.e. the distance from one crest to the other) bears a 
 definite relation to the speed of the ship. This relation is that 
 the length of the wave varies as the square of the speed at which 
 the ship is travelling, and thus as the speed of the ship increases 
 the length from crest to crest of the accompanying series of 
 transverse waves increases very rapidly. 
 
 The waves produced by the stern of the ship are not of 
 such great importance as those formed by the bow, which we 
 have been considering. They are, however, similar in character, 
 there being an oblique series and a transverse series. 
 
Horse-power, Effective and Indicated, etc. 229 
 
 Interference between the Bow and Stern Transverse Series of 
 Waves. — In a paper read by the late Mr. Froude at the Insti- 
 tution of Naval Architects in 1877, some very important 
 experiments were described, showing how the residuary resist- 
 ance ^ varied in a ship which always had the same fore and 
 after bodies, but had varying lengths of parallel middle body 
 inserted, thus varying the total length. A strange variation in 
 the resistance at the same speed, due to the varying lengths of 
 parallel middle body was observed. The results were set out 
 as roughly shown in Fig. 102, the resistance being set up on a 
 
 24-0 140. 
 
 40. 
 
 - Length of Parallel Middle 
 
 Body — 
 
 Fig. 102. 
 
 base of length of ship for certain constant speeds. At the low 
 speed of 9 knots very little variation was found, and this was 
 taken to show that at this speed the residuary resistance was 
 caused by the diverging waves only. 
 
 The curves show the following characteristics :— 
 (i) The spacing or length of undulation appears uniform 
 throughout each curve, and this is explained by the fact that 
 waves of a given speed have always the same length. 
 
 (2) The spacing is more open in the curves of higher speed, 
 the length apparently varying as the square of the speed. This 
 is so because the length of the waves are proportionate to the 
 square of the speed. 
 
 ' Residuary resistance is the resistance other than frictional. 
 
230 Theoretical Naval Architecture. 
 
 (3) The amplitude or heights of the undulations are 
 greater in the curves of higher speeds, and this is so, because 
 the waves made by the ship are larger for higher speeds. 
 
 (4) The amplitude in each curve diminishes as the length of 
 parallel middle body increases, because the wave system, by 
 diffusing transversely, loses its height. 
 
 These variations in residuary resistance for varying lengths 
 are attributed to the interference of the bow and stern trans- 
 verse series of waves. When the crests of the bow- wave series 
 coincide with the crests of the stern wave series, the residuary 
 resistance is at a maximum. When the crests of the bow-wave 
 series coincides with the trough of the stern-wave series, the 
 residuary resistance is at a minimum. 
 
 These experiments show very clearly that it is not possible 
 to construct a formula which shall give the resistance of a ship 
 at speeds when the wave-making resistance forms the important 
 feature. We must either compare with the known performances 
 of similar ships or models by using Froude's " law of compari- 
 son" (see p. 237). 
 
 The following extracts from a lecture ^ by Lord Kelvin (Sir 
 William Thomson) are of interest as giving the relative in- 
 fluence of frictional and wave-making resistance -.-^ 
 
 " For a ship A, 300 feet long, 31J feet beam, and 2634 tons 
 displacement, a ship of the ocean mail-steamer type, going at 
 13 knots, the skin resistance is 5 "8 tons, and the wave resistance 
 is 3 "2 tons, making a total of 9 tons. At 14 knots the skin 
 resistance is but little increased, viz. 6'6 tons, while the wave 
 resistance is 6'i5 tons. 
 
 " For a vessel B, 300 feet long, 46-3 feet beam, and 
 3626 tons, no parallel middle body, with fine lines swelling out 
 gradually, the wave resistance is much more favourable. At 
 13 knots the skin resistance is rather more than A, being 
 6'95 tons as against 5'8 tons, while the wave resistance is 
 only 2 '45 tons as against 3*2 tons. At 14 knots there is a 
 very remarkable result in the broader ship with its fine lines, 
 all entrance and run, and no parallel middle body. At 14 
 knots the skin resistance is 8 tons as against 6' 6 tons in A, 
 ' Third volume " Popular Lectures and Addresses," 1887. 
 
Horse-power, Effective and Indicated, etc. 231 
 
 while the wave resistance is only 3- 15 tons as against 6 '15 
 tons in A. 
 
 " For a torpedo boat, 125 feet long and 5 1 tons displacement, 
 at 20 knots the skin resistance was i'2 tons, and the wave resist- 
 ance I'l tons." 
 
 Resistance of a Completely Stibmerged Body. — The condi- 
 tions in this case are completely different from those which 
 have to be considered in the case of a vessel moving on the 
 surface. In this latter case waves are produced on the surface, 
 as we have seen, but with a completely submerged body this is 
 not so, provided the vessel is immersed sufficiently. We get the 
 clue to the form of least resistance in the shape of fishes, in which 
 the head or forward end is comparatively blunt, while the rear 
 portion tapers off very fine. The reason for the small resistances 
 of forms of this sort is seen when we consider the paths the particles 
 of water follow when flowing past. These paths are termed the 
 stream-lines for the particular form. It will be seen that no eddies 
 are produced for a fish-shaped form, and, as we saw on p. 225, 
 it is the rear end which must be fined off in order to reduce eddy- 
 making to a minimum. This was always insisted on very strongly 
 by the late Mr. Froude, who said, " It is blunt tails rather than 
 blunt noses that cause eddies." A veiy good illustration of the 
 above is seen in the form that is given to the section of shaft 
 brackets in twin-screw vessels. Such a section is given in Fig. 
 98. It will be noticed that the forward end is comparatively 
 blunt, while the after end is fined off to a small radius. 
 
 Speed Coefficients. — The method which is most largely 
 employed for determining the I.H.P. required to drive a vessel 
 at a certain speed is by using coefficients obtained from the 
 results of trials of existing vessels. They are based upon 
 assumptions which should always be carefully borne in mind 
 when applying them in actual practice. 
 
 I. Displacement Coefficient. — We have seen that for speeds 
 at which wave-making resistance is not experienced, the resist- 
 ance may be taken as varying — 
 
 (a) With the area of wetted surface ; 
 
 {b) As the square of the speed ; 
 so that we may write for the resistance in pounds — 
 
232 TJieoretical Naval ArchitecUire. 
 
 R = KiSV 
 
 V being the speed in knots, S the area of wetted surface in 
 square feet, and Ki being a coefficient depending on a number 
 of conditions which we liave already discussed in deahng with 
 resistance. 
 
 Now, E.H.P = , as we have already seen 
 
 33000 
 
 (p. 215). Therefore we may say — 
 
 E.H.P. = K^SV^' 
 
 where K2 is another coefficient, which may be readily obtained 
 from the previous one. If now we assume that the total I.H.P. 
 bears a constant ratio to the E.H.P., or, in other words, the 
 propulsive coefficient remains the same, we may write — 
 
 I.H.P. = K3SV^ 
 
 K3 being another new coefficient. S, the area of the wetted 
 surface, is proportional to the product of the length and girth to 
 the water-line ; W, the displacement, is proportional to the pro- 
 duct of the length, breadth, and draught. Thus W may be said 
 to be proportional to the cube of the linear dimensions, while S 
 is proportional to the square of the linear dimensions. Take a 
 vessel A, of twice the length, breadth, and draught, of another 
 vessel B, with every linear dimension twice that of the corre- 
 sponding measurement in B. Then the forms of the two vessels 
 are precisely similar, and the area of the wetted surface of 
 A will be 2^ = 4 times the area of the wetted surface of B, and 
 the displacement of A will be 2^ = 8 times the displacement of 
 B. The ratio of the linear dimensions will be the cube root 
 of the ratio of the displacements, in the above case ^8 = 2. 
 The ratio of corresponding areas will be the square of the cube 
 root of the ratio of the displacements, in the above case 
 (^8)^ = 4. This may also be written S''. We may accord- 
 ingly say that for similar ships the area of the wetted surface 
 will be proportional to the two-thirds power of the displace- 
 
 a 
 
 ment, or W°. We can now write our formula for the indicated 
 horse-power — 
 
Horse-power, Effective and Indicated, etc. 233 
 
 I.H.P. = ^^" 
 
 where W = the displacement in tons ; 
 V = the speed in knots ; 
 
 C = a coefficient termed the displacement coefficient?- 
 If a ship is tried on the measured mile at a known displace- 
 ment, and the I.H.P. and speed are measured, the value of the 
 
 W* X V 
 coefficient C can be determined, for C = t u t> • It is usual 
 
 l.rl.Jr. 
 
 to calculate this coefficient for every ship that goes on trial, and 
 
 to record it for future reference, together with all the particulars 
 
 of the ship and the conditions under which she was tried. It 
 
 is a very tedious calculation to work out the term W, which 
 
 means that the square of the displacement in tons is calculated, 
 
 and the cube root of the result found. It is usual to perform 
 
 the work by the aid of logarithms. A specimen calculation is 
 
 given here : — 
 
 The Himalaya on trial displaced 4375 tons, and an I.H.P. 
 of 2338 was recorded, giving a speed of i2'93 knots. Find the 
 " displacement coefficient " of speed. 
 
 Here we have— W=437S 
 
 V= 12-93 
 
 I.H.P. = 2338 
 
 By reference to a table of logarithms, we find — 
 
 log 4375 = 3'64io 
 log I2'93 = i*iii6 
 log 2338 = 3-3689 
 so that log (4375)* = f log 4375 = 2-4273 
 log (12-93)^ = 3 log 12-93 = 3'3348 
 
 = 2-3932 
 The number of which this is the logarithm is 247-3, and 
 accordingly this is the value of the coefficient required. 
 
 ' The coefficients are often termed "Admiralty constants," but it will 
 be seen on p. 235 that they are not at all constant for different speeds of the 
 same vessel. 
 
234 Theoretical Naval Architecture. 
 
 2. The other coefficient employed is the '■'■midship-section 
 coefficient." ^ If M is the area of the immersed midship section 
 in square feet, the value of this coefficient is — 
 
 M X V 
 I.H.P. 
 
 This was originally based on the assumption that the 
 resistance of the ship might be regarded as due to the forcing 
 away of a volume of water whose section is that of the im- 
 mersed midship section of the ship. This assumption is not 
 compatible with the modern theories of resistance of ships, and 
 the formula can only be true in so far as the immersed midship 
 section is proportional to the wetted surface. 
 
 In obtaining the W° coefficient, we have assumed that the 
 wetted surface of the ships we are comparing will vary as the 
 two-thirds power of the displacement ; but this will not be true 
 if the ships are not similar in all respects. However, it is 
 found that the proportion to the area of the wetted surface is 
 much more nearly obtained by using W" than by using the 
 area of the immersed midship section. We can easily imagine 
 two ships of the same breadth and mean draught and similar 
 form of midship section whose displacement and area of 
 wetted surface are very different, owing to different lengths. 
 We therefore see that, in applying these formulae, we must take 
 care that the forms and proportions of the ships are at any rate 
 somewhat similar. There is one other point about these 
 formulae, and that is, that the performances of two ships can 
 only be fairly compared at " corresponding speeds." ^ 
 
 Summing up the conditions under which these two formulae 
 should be employed, we have — 
 
 (i) The resistance is proportional to the square of the speed. 
 
 (2) The resistance is proportional to the area of wetted 
 surface, and this area is assumed to vary as the two-thirds power 
 of the displacement, or as the area of the immersed midship 
 section. Consequently, the ships we compare should be of 
 somewhat similar type and form. 
 
 (3) The coefficient of performance of the machinery is 
 
 ' See note on p. 233. ^ See p. 236. 
 
Horse-power, Effective and Indicated, etc. 235 
 
 assumed to be the same. The ships we compare are supposed 
 to be fitted with the same type of engine, working with the 
 same efficiency. Accordingly we cannot fairly compare a 
 screw steamer with a paddle steamer, since the efficiency of 
 working may be very different. 
 
 (4) The conditions of the surfaces must be the same in 
 the two ships. It is evident that a greater I.H.P. would be 
 required for a given speed if the ship's bottom were foul than 
 if it had been newly painted, and consequently the coefficient 
 would have smaller values. 
 
 (5) Strictly speaking, the coefficients should only be com- 
 pared for " corresponding speeds." ' 
 
 With proper care these formulae may be made to give 
 valuable assistance in determining power or speed for a new 
 design, but they must be carefully used, and their limitations 
 thoroughly appreciated. 
 
 We have seen that it is only for moderate speeds that the 
 resistance can be said to be proportional to the square of the 
 speed, the resistance varying at a higher power as the speed 
 increases. Also that the propulsive coefficient is higher at the 
 maximum speed than at the lower speeds. So if we try a 
 vessel at various speeds, we cannot expect the speed coefficients 
 to remain constant, because the suppositions on which they are 
 based are not fulfilled at all speeds. This is found to be the 
 case, as is seen by the following particulars of the trials of the 
 Iris. The displacement being 3290 tons, the measured-mile 
 trials gave the following results : — 
 
 I.H.P. 
 
 Speed in knots. 
 
 7556 ... 
 
 i8-6 
 
 3958 
 
 157s 
 
 1765 
 
 12-5 
 
 596 
 
 8-3 
 
 The values of the 
 
 speed coefficients calculated from the 
 
 above are — 
 
 
 
 Displacement Mid. sec. 
 
 
 coefficients. coefficients. 
 
 1 8 -6 knots 
 
 188 ... S95 
 
 IS7S » 
 
 218 ... 690 
 
 12-5 „ 
 
 243 ... 770 
 
 8-3 „ ... 
 
 214 ... 677 
 
 See p. 236. 
 
236 Theoretical Naval Architecture. 
 
 It will be noticed that both these coefficients attain their 
 maximum values at about 12 knots for this ship, their value 
 being less for higher and lower speeds. We may explain this 
 by pointing out — 
 
 (i) At high speeds, although the ''propulsive coefficient" 
 is high, yet the resistance varies at a greater rate than the 
 square of the speed, and — 
 
 (2) At low speeds, although the resistance varies nearly as 
 the square of the speed, yet the efficiency of the mechanism is 
 not at its highest value. 
 
 Corresponding Speeds. — We have frequently had to use 
 the terms " low speeds" and "high speeds'' as applied to certain 
 ships, but these terms are strictly relative. What would be a 
 high speed for one vessel might very well be a low speed for 
 another. The first general idea that we have is that the speed 
 depends in some way on the length. Fifteen knots would be 
 a high speed for a ship 150 feet long, but it would be quite a 
 moderate speed for a ship 500 feet long. In trying a model 
 of a ship in order to determine its resistance, it is obvious that 
 we cannot run the model at the same speed as the ship ; but 
 there must be a speed of the model '■'■corresponding" to the 
 speed of the ship. The law that we must employ is as follows : 
 " In comparing similar ships with one another, or ships with 
 models, the speeds must be proportional to the square root of their 
 linear dimensions" Thus, suppose a ship is 300 feet long, and 
 has to be driven at a speed of 20 knots ; we make a model of 
 this ship which is 6' 3" long. Then the ratio of their linear 
 dimensions is — 
 
 6-25 ^ 
 
 and the speed of the model corresponding to 20 knots of the 
 ship is — 
 
 20 -^ ^^48 = 2'88 knots 
 
 Speeds obtained in this way are termed " C07-responding speeds." 
 
 Example. — A model of a ship of 2000 tons displacement is constructed 
 on the J inch = i foot scale, and is towed at a speed of 3 knots. What 
 speed of the shiis does this correspond to ? 
 
Horse-power, Effective and Indicated, etc. 237 
 
 Although here the actual dimensions are not given, yet the ratio of the 
 linear dimensions is given, viz. i 148. Therefore the speed of the ship 
 corresponding to 3 knots of the model is — 
 
 3^4^= 20| knots 
 
 Expressing this law in a formula, we may say — 
 
 where V = speed in knots ; 
 
 L = the length in feet j 
 
 f = a coefficient expressing the ratio V : ^L, and 
 consequently giving a measure of the speed. 
 
 We may take the following as average values of the co- 
 efficient " c" in full-sized ships : — 
 
 When c = o'5 to o'65, the ship is being driven at a 
 moderate economical speed ; 
 c = 07 to i"o, gives the speed of mail steamers and 
 
 modern battleships ; 
 c ■= i"o to I '3, gives the speed of cruisers. 
 Beyond this we cannot go in full-sized vessels, since it is 
 not possible to get in enough engine-power. This can, how- 
 ever, be done in torpedo-boats and torpedo-boat destroyers, 
 and here we have r = 1*9 to 2*3. These may be termed 
 excessive speeds. 
 
 We have already seen that the W' coefficient of perform- 
 ance has a maximum value at a certain speed for a given ship. 
 In the case of the Iris, we saw that this was at a speed of 
 12 knots. This maximum value of the coefficient is usually 
 found to be obtained in full-sized ships at a speed correspond- 
 ing to the value c = 07. The Iris was 300 feet long, and the 
 
 12 
 
 value of ^ at 12 knots would be , — °'°9- 
 
 V300 
 
 Proude's Law of Comparison. — This law enables us 
 to compare the resistance of a ship with that of her model, or 
 the resistances of two ships of different size but of the same 
 form. It is as follows : — 
 
 If the linear dimensions of a vessel be I times the dimensions 
 of the model, and the resistance of the latter at speeds Vi, V2, V;,, 
 
238 Theoretical Naval Architecture. 
 
 etc., are Ri, Rj, R3, etc., then at tlie ^'^ corresponding speeds" of 
 t/te ship, Vi^/, Y^tjl, Y^^/l, etc., the resistance of the ship will 
 be Rif , R/, R/, etc. 
 
 In passing from a model to a full-sized ship there is a 
 correction to be made, because of the different effect of the 
 friction of the water on the longer surface. The law of com- 
 parison strictly applies to the resistances other than frictional. 
 The law can be used in comparing ■ the resistance of two 
 ships of similar form, and is found of great value when model 
 experiments are not available. 
 
 In the earlier portion of this chapter we referred to the 
 experiments of the Greyhound by the late Mr. Froude. A 
 curve of resistance of the ship in pounds on a base of speed 
 is given by A, in Fig. 96. In connection with these experi- 
 ments, a model of the Greyhound was made and tried in the 
 experimental tank imder similar conditions of draught as the 
 ship, and between speeds corresponding to those at which the 
 ship herself had been towed. The resistance of the model having 
 been found at a number of speeds, it was possible to construct 
 a curve of resistance on a base of speed as shown by C in 
 Fig. 103. The scale of the model was -j^ full size, and 
 therefore the corresponding speeds of the ship were /<J 16, or 
 four times the speed of the model. If the law of comparison 
 held good for the total resistance, the resistance of the ship 
 should have been 16^ = 4096 times the resistance of the model 
 at corresponding speeds; but this was not the case, owing to 
 the different effect of surface friction on the long and short 
 surfaces. The necessary correction was made as follows : 
 The wetted surface of the model was calculated, and by 
 employing a coefficient suitable to the length of the model and 
 the condition of its surface, the resistance due to surface 
 friction was calculated for various speeds as explained (p. 223), 
 and a curve drawn through all the spots thus obtained. This 
 is shown by the dotted curve DD in Fig. 103. Thus at 
 250 feet per minute the total resistance of the model is given 
 by ac, and the resistance due to surface friction by ad. The 
 portion of the ordinate between the curves CC and DD will 
 give at any speed the resistance due to other causes than that 
 
Horse-power, Effective and Indicated, etc. 239 
 
 of surface friction. Thus at 250 feet per minute, these other 
 resistances are given by cd. This figure shows very clearly 
 how the resistance at low speeds is almost wholly due to 
 surface friction, and this forms at high speeds a large propor- 
 tion of the total. The wave-making resistance, as we have 
 already seen, is the chief cause of the difference between the 
 
 Speed 
 
 200. 250. 
 
 I Feet Per Minute . 
 
 Fig. 103. 
 
 curves CC and DD, which difference becomes greater as the 
 speed increases. It is the resistance, other than frictional, to 
 which the law of comparison is intended to apply. 
 
 We have in Fig. 96 the curve of resistance, A A, of the 
 Greyhound on a base of speed, and in precisely the same way 
 as for the model a curve of frictional resistance was drawn in for 
 the ship, taking the coefEcient proper for the state of the surface 
 of the ship and its length. Such a curve is given by BB, Fig.96. 
 Then it was found that the ordinates between the curves AA 
 and BB, Fig. 96, giving the resistance for the ship other than 
 frictional, were in practical agreement with the ordinates 
 between the curves CC and DD, Fig. 103, giving the resistance 
 of the model other than frictional, allowing for the "/aa/ of 
 
240 Theoretical Naval Architecture. 
 
 comparison" above stated. That is, at speeds of the ship Vifi, 
 or four times the speeds of the model, the resistance of the ship 
 other than frictional was practically 16", or 4096 times the 
 resistance of the model. 
 
 These experiments of the Greyhound and her model form 
 the first experimental verification of the law of comparison. 
 In 1883 some towing trials were made on a torpedo-boat 
 by Mr. Yarrow, and a model of the boat was tried at the 
 experimental tank belonging to the British Admiralty. In this 
 case also there was virtual agreement between the boat and 
 the model according to the law of comparison. It is now the 
 practice of the British Admiralty and others to have models 
 made and run in a tank. The data obtained are of great 
 value in determining the power and speed of new designs. 
 For further particulars the student is referred to the sources 
 of information mentioned at the end of the book. 
 
 Having the resistance of a ship at any given speed, we can 
 at once determine the E.H.P. at that speed (see p. 215), and 
 then by using a suitable propulsive coefficient, we may deter- 
 mine the I.H.P. at that speed. Thus, if at 10 knots the resist- 
 ance of a ship is 10,700 lbs., we can obtain the E.H.P. as 
 follows : — 
 
 Speed in feet per minute = 10 x -fl^ 
 
 Work done per minute = 10,700 X (10 X ■^^-) foot-lbs. 
 
 10700 X -S^-^ 
 E.H.P. = — '^^^1^^ — 6_ 
 33000 
 
 = 328 
 and if we assume a propulsive coefficient of 45 per cent. — 
 
 IHP ^ 328 X 100 
 45 
 = 729 
 
 By the use of the law of comparison, we can pass from one 
 ship whose trials have been recorded to another ship of the 
 same form, whose I.H.P at a certain speed is required. It is 
 found very useful when data as to I.H.P. and speed of existing 
 
Horse-Power, Effective and Indicated, etc. 241 
 
 ships are available. In using the law we make the following 
 assumptions, which are all reasonable ones to make. 
 
 (i) The correction for surface friction in passing from one 
 ship to another of different length is unnecessary. 
 
 (2) The condition of the surfaces of the two vessels are 
 assumed to be the same. 
 
 (3) The efficiency of the machinery, propellers, etc., is 
 assumed the same in both cases, so that we can use I.H.P. 
 instead of E.H.P. 
 
 The method of using the law will be best illustrated by the 
 following example : — 
 
 A vessel of 3290 tons has an I.H.P. of 2<pa on trial at 14 knots. What 
 would be the probable I.H.P. of a vessel of the same form, but of three 
 times the displacement, at the corresponding speed ? 
 
 The ratio of the displacement = 3 
 .'. the ratio of the linear dimensions / = ^ 3 
 
 = l'44 
 
 .'. the corresponding speed = 14 X VI '44 
 = i6'8 knots 
 
 The resistance of the new ship will be P times that of the original, and 
 accordingly the E.H.P., and therefore the I.H.P., will be that of the 
 
 original ship multiplied by /' = (l'44r = 3'6, and — 
 
 I.H.P. for new ship = 2500 X 3"6 
 = 9000 
 
 When ships have been run on the measured mile at pro- 
 gressive speeds, the information obtained is found to be ex- 
 tremely valuable, since we can draw for the ship thus tried a 
 curve of I.H.P. on a base of speed, and thus at intermediate 
 speeds we can determine the I.H.P. necessary. The following 
 example will show how such a curve is found useful in 
 estimating I.H.P. for a new design. 
 
 A vessel of 9000 tons is being designed, and it is desired to obtain a 
 speed of 21 knots. A ship of 7390 tons of similar form has been tried, and 
 a curve of I.H.P. to a base of speed drawn. At speeds of 10, 14, 18, and 
 20 knots the I.H.P. is 1000, 3000, 7500, 11,000 respectively. 
 
 Now, the corresponding speeds of the ships will vary as the square root 
 of the ratio of linear dimension /. 
 
 We have — 
 
 p = 
 
 %% 
 
 and / = 
 
 \Qr] 
 
 Vr= 
 
 1-03S 
 
242 Theoretical Naval Architecture. 
 
 therefore the corresponding speed of the 7390-ton ship is — 
 
 21 -7- 1-035 = 20-3 
 
 By drawing in the curve of I.H.P. and continuing it beyond the 20 
 knots, we find that the I.H.P. corresponding to a speed of 20'3 knots is 
 about ir,7oo. Tlie I.H.P. for the 9000-ton ship at 21 knots is accordingly — 
 
 11,700 X /^ = 11,700 X I -26 
 
 = 14,750 I.H.P. about 
 
 Examples to Chapter VII. 
 
 1. The Greyhound via^ towed at the rate of 845 feet per minute, and 
 the horizontal strain on the tow-rope, including an estimate of the air 
 resistance of masts and rigging, was 6200 lbs. Find the effective horse- 
 power at that speed. 
 
 Ans. 159 E.H.P. nearly. 
 
 2. A vessel of 5500 tons displacement is being towed at a speed of 8 
 knots, and her resistance at that speed is estimated at 18,740 lbs. What 
 horse-power is being transmitted through the tow-rope ? 
 
 A71S. 460. 
 
 3. A steam-yacht has the following particulars given : — 
 
 Displacement on trial ... ... ... ... 1 76*5 tons 
 
 I.H.P on trial 364 
 
 Speed „ I3'3 knots 
 
 Find the "displacement coefficient of speed." 
 
 A71S. 203. 
 
 4. A steam-yacht has a displacement of I43'S tons, and 250 I.H.P. is 
 expected on trial. What should the speed in knots be, assuming a displace- 
 ment coefficient of speed of 196 ? 
 
 A?is. I2'2 knots. 
 
 5. The Warrior developed 5267 indicated horse-power, with a speed of 
 i4'o8 knots on a displacement of 9231 tons. Find the displacement co- 
 efficient of speed. 
 
 Ans. 233. 
 
 6. In a set of progressive speed trials, very different values of the 
 " displacement coefficient " are obtained at different speeds. Explain the 
 reason of this. A ship is 225 feet long, at what speed would you expect 
 the coefficient to have its maximum value ? 
 
 Ans. About loj knots. 
 
 7. Suppose we took a torpedo boat destroyer of 250 tons displacement 
 and 27 knots speed as a model, and designed a vessel of 10,000 tons dis- 
 placement of similar form. At what speed of this vessel could we compare 
 her resistance with that of the model at 27 knots ? 
 
 Ans. 50 knots. 
 
 8. A ship of 5000 tons displacement has to be driven at 21 knots. A 
 model of the ship displaces loi lbs. At what speed should it be tried ? 
 
 Ans, 3 knots. 
 
 9. A ship of 5000 tons displacement is driven at a speed of 12 knots. 
 A ship of 6500 tons of similar form is being designed. At what speed of 
 the larger ship can we compare its performance with the 5000-ton ship ? 
 
 Ans. I2'53 knots. 
 
 10. A vessel 300 feet long is driven at a speed of 15 knots. At what 
 
Horse-power, Effective and Indicated, etc. 243 
 
 speed must a similar vessel 350 feet long be driven in order that their 
 performances may be compared ? 
 
 A lis. i6'2 knots. 
 
 11. A vessel 300 feet long has a displacement on the measured-mile 
 trial of 3330 tons, and steams at 14, 18, and 20 knots with 2400, 6000, and 
 9000 I.H.P. respectively. What would be the I.H.P. required to drive a 
 vessel of similar type, but of double the displacement, at 20 knots ? 
 
 Ans. 13,000 I.H.P. about. 
 
 12. A vessel of 3100 tons displacement is 270 feet long, 42 feet beam, 
 and 17 feet draught. Her I.H.P. at speeds of 6, 9, 12, and 15 knots are 
 270, 600, 1 350, and 3060 respectively. What will be the dimensions of a 
 similar vessel of 7000 tons displacement, and what I.H.P. will be required 
 to drive this vessel at 18 knots ? 
 
 Ans. 354 X SS X 22"3 , about 9600 I.H.P. 
 
 13. A vessel of 4470 tons displacement is tried on the measured mile at 
 progressive speeds, with the following results : — 
 
 Speed. I.H.P. 
 
 8-47 48s 
 
 io'43 881 
 
 i2'23 1573 
 
 i2'93 2117 
 
 A vessel of similar form of 5600 tons displacement is being designed. 
 Estimate the I.H.P. required for a speed of 13 knots. 
 
 Ans. 2290 I.H.P. 
 
APPENDIX A 
 
 Proof of Simpson's First Rule. — Let the equation of the curve 
 referred to the axes Ojt, Oy, as Fig. 31, p. 51, be — 
 
 a^, «!, flj being constants ; then the area of a narrow strip length y 
 and breadth ^x is — 
 
 and the area required between x = o and ;ir = 2^ is the sum of all 
 such strips between these limits. Considering the strips as being a 
 small breadth Ax, we still do not take account of the small triangular 
 pieces as BDE (see Fig. 12), but on proceeding to the limit, i.e. 
 making the strips indefinitely narrow, these triangular areas dis- 
 appear, and the expression for the area becomes, using the formulse 
 of the calculus — 
 
 '2// 
 
 y . dx 
 o 
 
 or, putting in the value forj/ given by the equation to the curve — 
 
 k 
 (Aq + a^x + a^x'')dx 
 
 /: 
 
 which equals — 
 
 \a^ + \a^x^ + 4«3«^]q 
 
 which has to be evaluated between the limits x =%h and x = 
 The expression then becomes — 
 
 Now, from the equation to the curve, when — 
 
 x = o,y =■ a^i 
 
 X = /t, y = ao + a-Jt + aji'' 
 
 x = 2k,y = aa + 'laji + i^aji^ 
 
246 Appendix. 
 
 But calling the ordinates in the ordinary way, j/j, jj, andj/g — 
 when X = o,y = j/^ 
 
 x = 2h,y =y^ 
 therefore we have— 
 
 ao + a^h + aji^ = y^ 
 «„ + ^a-Ji + 4^2^^ = y^ 
 
 from which may be obtained the following values for the constants 
 ^0, «!, a-i ■ — 
 
 or substituting above— 
 
 Area = y,2h + ~ ■ ^(4j/, - sy^ - y,) + — ■ -^(^3 - 2/2 +71) 
 
 which is the expression known as Simpson's First Rule. 
 
 Proof of Simpson's Second Bule. — Let the equation to the 
 curve be — 
 
 y = aQ + a^x + a^x"^ + a^x 
 
 then the area will be given by — 
 
 («(, + a^x + a,^x^ + a^x^dx (i) 
 
 o 
 
 The area given by the rule is — 
 
 P(j/l + 3J'2 + 3/3+;'4) (2) 
 
 From (i) we find that the area is — 
 
 /: 
 
 2,a^k + 9^1 ^ + ^ar,h? + ^a^k^ (3) 
 
 Now- 
 
 y^ = ai, + a-Ji + aji'' 4- ajfi 
 /s = ^0 + 2^1^ 4- i^ajt''' 4- %aji^ 
 ^4 = «o 4- -ia-Ji 4- ^aji'' + ^Taji'' 
 
 These arc four equations for determining a^, a^, a.^, a-^. 
 
 If wc substitute the values given above forj^j'^, etc., in the 
 
Appendix. 247 
 
 expression for the area as given by Simpson's second rule in (2), we 
 find the result is the same as the expression (3). This shows that 
 Simpson's second rule will correctly integrate a curve whose 
 equation is that given above. 
 
 The truth of the five-eight rule given on p. 12 can be proved in a 
 similar manner. 
 
 The dynamical stability of a ship at any given angle of 
 heel is equal to the area of the curve of statical stability up 
 to that angle. 
 
 Referring to Fig. 67 showing a ship heeled over to a certain 
 angle 9, imagine the vessel still further heeled through a very small 
 additional angle, which we may call di. The centre of buoyancy 
 will move to B" (the student should here draw his own figure to 
 follow the argument). B'B" will be parallel to the water-line W'L', 
 and consequently the centre of buoyancy will not change level 
 during the small inclination. Drawing a vertical B"Z through 
 B", we draw GZ', the new righting arm, perpendicular to it. Now, 
 the angle ZGZ' = d9, and the vertical separation of Z and Z' = 
 GZ X di. Therefore the work done in inclining the ship from the 
 angle 9 to the angle + d6 is — 
 
 W X (GZ . de) 
 
 Take now the curve of statical stability for this vessel. At the 
 angle the ordinate is W x GZ. Take a consecutive ordinate at the 
 angle 9 + de. Then the area of such a strip = W x GZ x ^9 ; but 
 this is the same as the above expression for the work done in 
 inclining the vessel through the angle de, and this, being true for 
 any small angle de, is true for all the small angles up to the angle 9. 
 But the addition of the work done for each successive increment of 
 inclination up to a given angle is the dynamical stability at that 
 angle, and the sum of the areas of such strips of the curve of 
 statical stability as we have dealt with above is the area of that 
 curve up to the angle 9. Therefore we have the dynamical stability 
 of a ship at any given angle of heel is equal to the area of the 
 curve of statical stability up to that angle, where the ordinates of 
 this curve represent the righting moments. 
 
 The area of a curve of displacement divided by the load 
 displacement gives the distance of the centre of buoyancy 
 below the L.W.L. 
 
 Let OBL, Fig. 104, be the curve of displacement of a vessel 
 constructed in the ordinary way, OW being the load mean 
 draught, and WL being the load displacement. 
 
 Take two level lines, AB, A'B', a small distance ^s apart. 
 
248 
 
 Appendix. 
 
 Call the area of the water-plane at the level of AB, A square feet, 
 and the distance of this water-plane below the L.W.L., z. The 
 volume between the water-lines AB, A'B' will be A x ^2, or, sup- 
 
 L c'c. 
 
 Fig. 104. 
 
 posing they are indefinitely close together, A x dz. The moment 
 
 of this layer about the L.W.L. will be A. s. dz. 
 
 The difference between the lengths of A'B' and AB will evidently 
 
 be the weight of the volume of water between those two level lines, 
 
 A X dz 
 or . Draw B'C, BC vertically as shown. Then the breadth 
 
 c^\. ^ ■ -orr- . Ax dz , ^, j.^,. ^ . . A X z X dz. 
 
 of the strip B C is , and the area of this strip is 
 
 ^ 35 ' ^35 
 
 The area of the curve will be the area of all such strips, or — 
 
 fA.z .dz 
 
 l'~ 
 
 35 
 
 The moment of the volume of displacement about the L.W.L. 
 is given by — 
 
 ^ JA.z.dz 
 
 and the distance of the C.B. below the L.W.L. is found by dividing 
 this moment by the*load displacement in cubic feet, or — 
 
 ' A. z .ds 
 
 li 
 
 WLx 35 
 
 The area of the curve divided by the load displacement in tons is- 
 
 fA.z.dz 
 
 £ 
 
 JL 
 
 which is the same thing. 
 
 WL 
 
Appendix. 
 
 249 
 
 Normand's Approximate Formula for the Position of the 
 Centre of Buoyancy. — This formula was given on p. 63. 
 
 V 
 
 -^ is known as " Rankine's mean depth," and we may for con- 
 venience say — 
 
 The formula then becomes — 
 
 Distance of CB below) 
 the LWL in feet 
 
 \<'.--) 
 
 In Fig. 105, let ABC be the curve of water-plane areas, DC 
 being the mean draught d. Draw the rectangle AFCD. Make DE 
 
 Fig. 105. 
 
 = D. Draw EG parallel to DA, cutting the diagonal FD in H. 
 Finish the figure as indicated. The assumption made is that the 
 C.G. of the area DABC, which will give the vertical position of 
 the centre of buoyancy, is in the same position as the C.G. of the 
 area DAHC. This is seen, on inspection, to be very nearly the 
 case. These two figures have the same area, as we now proceed 
 to show. 
 
 By using the principle of similar triangles, we have — 
 
 AF _ GF _ ED 
 AD ~ GH HE 
 and therefore — 
 
 GF X HE =^ED X GH 
 or EC X HE = AG X GH 
 
2 so Appendix. 
 
 or the triangles AGH, HEC are equal in area ; therefore — 
 Area AH CD = area AGED 
 
 and this latter area equals the volume of displacement, being a 
 
 V 
 rectangle having its adjacent sides equal to A and -^ respectively. 
 
 The area of the curved figure DABC also evidently represents 
 the volume of displacement. Therefore the figures DABC and 
 DAHC are equal in area. 
 
 We now have to determine the position of the C.G. of DAHC 
 in relation to the L.W.L. 
 
 Area AGH ^ | x AG x GH ^ i GH 
 area AGED ~ AG x AD 2 ' AD 
 
 _ I GF ^ ^( d- D \ 
 ~ 2 ' AF ~ n d ) 
 or the triangle AGH = li ~, j x rectangle AGED 
 
 We may regard the figure DAHC as made by taking AGH away 
 from the rectangle AE, and putting it in the position HEC. The 
 
 shift of its C.G. downwards during this operation is — , there- 
 fore the C.G. of the whole figure will shift downwards, using the 
 principle explained in p. 96, an amount equal to x say, x being given 
 by- 
 
 AGED xx= AGH x - 
 3 
 
 or putting in the value found above, for the triangle AGH — 
 
 The C.G. of AGED is at present a distance — below the L.W.L. ; 
 therefore the C.G. of DAHC is below the L.W.L. a distance— 
 
 and taking the assumption given above, this is the distance of the 
 C.B. below the L.W.L. very nearly. 
 
 Strength of Rudder-heads. — For vessels built to Lloyd's rules, 
 the diameter of the rudder-head is determined by the longitudinal 
 number (see p. 194) and is given in the tables. For steam- vessels 
 
Appendix. 251 
 
 the diameters of rudder-heads are calculated by the following 
 formula — 
 
 where D = feet draught, B = breadth of rudder in inches, and 
 S = speed in knots ; but in no case is the diameter to be less than 
 that given in the table. 
 
 The following is the method adopted for determining the size of 
 rudder-head, having given particulars of size, speed of ship, and 
 maximum angle of rudder. To obtain the twistiiig motnent acting 
 on the rudder, we must know (i) the pressure of the water on the 
 rudder ; (2) the position of the centre of effort of this pressure. 
 
 As regards (l), the pressure in pounds on a plane moving uni- 
 formly broadside on (see Fig. 106), by the formula — 
 
 P = viik.v^ 
 
 where A is the area of the plane in square feet ; 
 V is the velocity in feet per second. 
 
 -«« 
 
 Fig. 106. Fig. 107. 
 
 This supposes the rudder to be round to an angle of 90°. For 
 an angle fl, the normal pressure is given by — 
 
 P' = P X sin 9 (see Fig. 107) 
 
 With regard to the position on the rudder at which this normal 
 pressure may be regarded as acting, it is found that for values of 
 9 between 30° and 40°, the centre of effort is about four-tenths the 
 breadth of the rudder from the leading edge, the rudder being 
 hinged at the fore side. 
 
 By this means we are able to determine the twisting moment on 
 the rudder-head. If a screw steamship is being propelled through 
 the water at a certain speed the sternward velocity of the water 
 from the propeller past the rudder, is greater than the speed of the 
 ship. The velocity of water past the rudder can be taken as 10 
 per cent, greater than the velocity of the ship. Take the following as 
 an example. A rectangular rudder 13 feet broad, hinged on the 
 
252 Appendix. 
 
 fore side, 180 square feet in area ; maximum angle, 35° ; speed of 
 
 ship, i6 knots. 
 
 16 X lOI , ^ J 
 
 16 knots = ;: = 27 feet per second 
 
 60 
 
 Velocity of water past rudder = 27 + 27 
 
 = 297, say 30 feet per second 
 
 'T.i , , , 180 X 30 X 30 . „ 
 The pressure on the rudder = 112 x ^- x sm 35° 
 
 = 46'5 tons 
 
 Taking the centre of effort as i^ x 13 = 5'2 feet, or 62-4 inches 
 from the leading edge — 
 
 The twisting moment = 46*5 x 62'4 
 
 = 2901 inch-tons 
 
 To determine the size of rudder-head for a twisting moment of 
 T inch-tons, we use the following formula : — 
 T = tV/^s 
 
 where d = diameter in inches, 
 
 /= strength of material per square inch, allowing a factor of 
 safety. 
 
 For wrought iron, / is taken as 4 tons 
 cast steel, / „ ,,5 „ 
 
 phosphor bronze,/ „ „ 3 „ 
 
 For a twisting moment of 2901 inch-tons, the diameter of a 
 rudder-head, if of wrought iron, will be given by — 
 
 </3 = ?9^I 
 
 ox d = 15^ inches 
 
 Example. — A rudder is 243J square feet in area, and the centre of 
 pressure is estimated to be 6 "12 feet abaft the centre of rudder-head, at 35°. 
 If the speed of the ship is 19 knots, estimate the diameter of the rudder- 
 head, taking the stress at 4 tons and 5 tons. 
 
 Alts. 20-1", 187." 
 
 Launehing Calculations. — Before starting on these calcula- 
 tions, it is necessary to estimate as closely as possible the launching 
 weight of the ship, and also the position of the centre of gravity, 
 both vertically and longitudinally. The case of the Daphne, 
 which capsized on the Clyde ^ on being launched, drew special 
 
 ' See Engineering, 1883, for a report on the Daphne, by Sir E. J, 
 Reed. 
 
Appendix. 
 
 253 
 
 attention to the necessity of providing sufficient stability in the 
 launching condition. A ship in the launching condition has a light 
 draught, great freeboai-d, and high position of the C.G. It is 
 possible, by the use of the principles we have discussed at length, 
 to approximate to the metacentric height, and if this is not con- 
 sidered sufficient, the ship should be ballasted to lower the centre 
 of gravity. It has been suggested that a minimum G.M. of i foot 
 should be provided in the launching condition. If the cross-curves 
 of stability of the vessel have been made, it is possible very quickly 
 to draw in the curve of stability in the launching condition, and in 
 case of any doubt as to the stability, this should be done. 
 
 iW-ufc 
 
 if'- 
 
 AW-Wl. 
 
 W. 
 
 X^ 
 
 fe 
 
 Water Linf — slir — 
 
 W^ 
 
 -TTTTTTrTrT^ 
 
 y y-^^-^ 
 
 Fig. io8. 
 
 We now have to consider the longitudinal stabiUty of a vessel 
 as she goes down the ways, and describe the calculations that are 
 made to see if the vessel will " tip" * (/.«. the stern drops over the 
 end of the ways, and the forward end lifts up off the ways), and to 
 ascertain the pressure that comes on the fore poppets when the 
 vessel lifts. After the ship has run a certain distance (see Fig. 108) 
 * Tipping is shown by the second diagram in Fig. 108. 
 
254 Appendix. 
 
 the C.G. will pass beyond the end of the ways, and will give rise to 
 a " tipping moment ; " but by this time a portion of the ship is in 
 the water, and the upward support of the buoyancy will give rise to 
 a moment in the opposite direction. Call W the weight of the 
 ship, and at the position shown in Fig. io8, let d be the distance of 
 the C.G. beyond the end of the ways, w the support of the 
 buoyancy, and d the distance of the centre of buoyancy beyond 
 the end of the ways. Then the support of the ways at this par- 
 ticular position of the ship's travel is W — le^, and if the vessel tipped, 
 i.e. if the moment of the weight W x (^ were greater than the 
 moment of the buoyancy w x d, this upward support would all be 
 concentrated at the end of the ways, and very possibly the bottom 
 of the vessel would be forced in, or the end of the ways give way, 
 and so cause a stoppage of the ship, or at any rate make the ship 
 slide down the rest of the distance on her keel. If the vessel 
 stopped on the ways, she would be in a very critical condition as the 
 tide fell. We therefore see that at all points the moment of the 
 buoyancy should exceed the moment of the weight about the end 
 of the ways. In order to see if this is so, we proceed as follows : 
 Assume a height of tide that may safely be expected for launching, 
 and take a series of positions as the ship goes down the ways, and 
 determine the buoyancy and the distances of the C.G. and C.B. 
 from the end of the ways, viz. w, d, and d' as above. 
 
 Then on a base-line (Fig. 109) representing the distance the ship 
 has travelled, we draw two curves : first, a curve AA, giving the 
 moment of buoyancy about the end of the ways ; and second, BB, 
 giving the moment of weight about the end of the ways (this latter 
 being a straight line, since the weight is constant). The distance 
 between these curves will give us the " moment against tipping " at 
 any particular place, and the minimum distance between them gives 
 the margin of safety. If a curve of moment of buoyancy is obtained, 
 as the dotted curve A'A' crossing BB, then there would be a 
 " tipping moment.'" This might be the case if the ways were not 
 long enough, or the tide might not rise so high as expected, or the 
 position of the centre of gravity might be wrongly estimated. The 
 remedy would be either to lengthen the ways or to place some 
 ballast right forward. Either of these would push the point G, at 
 which the centre of gravity of the ship comes over the end of the 
 ways, further along. Ships have been launched successfully which 
 had an adverse tipping moment ; but the velocity was very great, 
 and owing to this the ships were safely launched. 
 
 There is another point that has to be considered. As the ship 
 goes further down the ways, there comes a certain position in which 
 
Appendix. 
 
 2SS 
 
 the moment of the buoyancy about the fore poppet equals the 
 moment of the weight about the fore poppet. At this point the 
 stern of the ship will begin to lift ; and if W is the weight of 
 the ship, and W the buoyancy, the ship will be partially supported 
 by the fore poppets, and the amount of this support is W — W. 
 Thus we have a great strain, coming both on the slip and on the 
 vessel, concentrated over a small distance. The amount of this 
 can be determined as follows : In a similar manner in which we 
 
 
 
 ^ 
 
 P 
 
 
 E 
 
 
 
 E. 
 
 U) 
 
 z 
 
 
 - 
 
 
 
 
 1 
 
 
 »2 
 
 
 _ 
 
 
 ^^^"""^"----^ 
 
 
 t 
 
 
 f- 
 
 
 
 
 ^'^'■^•*»^ 
 
 
 s 
 
 
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 a. C 
 
 
 '""^^-.,^ 
 
 5 
 
 
 u. 
 
 
 
 
 
 "*«'.....,^^ 
 
 ir 
 
 
 
 -^...^..^^^ 
 
 
 
 ^^'^^.^^^^ 
 
 LI 
 
 
 500.000 
 
 S;0DO. 
 
 0L~- . 
 
 B^^^ 
 
 ^^ 
 
 --I 
 
 ■-C 
 
 
 r^ 
 
 ^ 
 
 ~^^-^ 
 
 ^-__^ 
 
 -r 
 
 
 *■ 
 
 
 ■A 
 
 
 ff- 
 
 ^»~ 
 
 V,^^^ 
 
 ^^'"— "■*»,..,^^ 
 
 (/> 
 
 
 z 
 
 
 Z 
 
 
 b. 
 
 8 • 
 
 l^i^^S,,,^^ 
 
 ^^'*"''^. 
 
 
 
 u 
 
 o 
 S 
 
 
 -|2 
 
 — I 
 
 
 3 
 ? 
 
 
 ^^^^ 
 
 
 u. 
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 O 
 
 
 i 
 
 
 ''■^"'*^5?5>-^ 
 
 U 
 
 
 
 -i 
 
 
 i 
 
 u 
 
 
 
 ^~^\ 
 
 
 ______ 
 
 
 J 
 
 
 & 
 
 
 1 ! 
 
 P 1 *^ 
 
 
 J_ 
 
 250. 
 
 Distance Ship Travels. 
 
 200. 
 
 Fig. log. 
 
 constructed the curves A and B we can construct E and F ; E gives 
 the moment of the weight of the ship about the fore poppet, which 
 moment is a constant quantity, and F gives the moment of the 
 buoyancy about the fore poppet. E and F intersect at the point P ; 
 and this is the place where the moments of weight and buoyancy 
 are equal, and therefore the place where the stern will begin to lift. 
 Now construct also C and D, C being the weight of the ship, in 
 this case 6ooo tons, and D being the buoyancy as the ship moves 
 down the ways. Then the intercept between these, at the distance 
 given by the point P, viz. aa, will give the weight borne by the fore 
 poppets when the stern begins to lift. In the case for which the 
 curves are given in Fig. 109, this weight was 675 tons. The launch- 
 ing curves for H.M.S. Sanspareil are given in Mr. Mackrow's 
 
2S6 Appendix. 
 
 " Pocket Book,'' and in that case the weight of the ship was 5746 
 tons, and the weight on the fore poppets 870 tons. 
 
 The internal shoring of the ship must be specially arranged 
 for in the neighbourhood of the fore poppets, and the portion 
 of the slip under them at the time the stern lifts must be made of 
 sufficient strength to bear the concentrated weight 
 
APPENDIX B 
 
 SYLLABUS OF THE EXAMINATIONS IN NAVAL 
 ARCHITECTURE CONDUCTED BY THE DE- 
 PARTMENT OF SCIENCE AND ART. 
 
 {From the Directory of the Department of Science and Art, by 
 permission of the Controller of Her Majesty's Stationery Office.) 
 
 First Stage or Elementary Course. 
 
 No candidate will be permitted to pass who fails to obtain marks 
 in any one of the three sections of this stage. 
 
 I. Practical Shipbuilding. — Students should be able to 
 describe the methods usually adopted by the workmen in forming 
 and combining the several parts of a steel or iron ship's hull, includ- 
 ing the transverse and longitudinal framing, stems and sternposts, 
 inner and outer bottom plating, beams, pillars (both fixed and 
 portable), deck plating and planking, and the wood and copper 
 sheathing of sheathed ships : also the methods adopted in forming 
 and combining the framing and bottom planking of wood and com- 
 posite ships. Also description of tools used in plating, planking, 
 and caulking ships. 
 
 II. Ship Calculations. — Fundamental conditions which must 
 be fulfilled by bodies when floating freely and at rest in still water. 
 Calculations relating to the computation of the areas of plane sur- 
 faces and the displacement of ships by applications of Simpson's 
 and the five-eight-minus-one (or one-twelfth) rules : " tons per inch 
 immersion ; " and the sinkage of a vessel in passing from sea into 
 river water. Also a knowledge of the specific gravities of materials 
 used in shipbuilding, and simple calculations based thereon. 
 
 III. Drawing. — Students will be required to make sketches to 
 scale. 
 
 It is intended that the examination in Practical Shipbuilding 
 shall be one principally relating to steel and iron ships, but one or 
 
 s 
 
2S8 Appendix. 
 
 two questions may be set on the important parts of wood and 
 composite vessels mentioned above. 
 
 Second Stage or Advanced Course. 
 
 In addition to the subjects for the Elementary Stage, students 
 presenting themselves for examination in the Advanced Course will 
 be expected to have received instruction in the following — 
 
 Practical Shipbuilding. — The structural details of water- 
 tight bulkheads ; methods of testing water-tight work ; the longi- 
 tudinal and transverse stresses to which ships are liable in still 
 water and amongst waves, and the structural arrangements which 
 give the necessary strength to resist those stresses : also the various 
 local stresses to which a ship is liable, and the special arrangements 
 worked to meet them ; the qualities of the various materials used in 
 shipbuilding, and the tests to which these materials are subjected ; 
 precautions to be observed when working steel plates and angles 
 hot ; effect of annealing. 
 
 Laying Off. — Comprising a knowledge of the work carried on 
 in the Mould Loft and at the Scrive Board in connection with order- 
 ing materials and laying off the several parts of an iron or steel 
 mercantile vessel ; also the similar work relating to warships, both 
 sheathed with wood and unsheathed. 
 
 Ship Calculations.— Curves of " displacement " and " tons 
 per inch immersion ; '' definitions of centre of flotation and centre 
 of buoyancy ; use of Simpson's and other rules for finding the 
 position of the centre of gravity of plane areas, and for calculating 
 the position of the centre of buoyancy ; graphic or geometrical 
 method of calculating displacement and position of centre of buoy- 
 ancy ; definitions of the terms " metacentre " and " metacentric 
 height ; " rules for calculating positions of transverse and longitu- 
 dinal metacentres ; metacentric diagrams, their construction and 
 use ; tensile strength of material between widely and closely spaced 
 holes punched and drilled in steel plates ; shearing strength of iron 
 and steel rivets, and spacing of rivets ; strength of butt fastenings ; 
 more advanced weight calculations, such as those for the weight of 
 a deck or bulkhead. 
 
 Honours. 
 
 The examination in Honours will be divided into two parts, 
 which cannot both be taken in the same year, and no candidate who 
 has not been successful in Part I. can be examined in Part II. A 
 certificate or medal will only be given when a success in Part 11. 
 has been obtained. 
 
Appendix. 259 
 
 Part I. 
 
 In addition to the subjects prescribed for the preceding stages, 
 this examination will embrace questions upon some or all of the 
 subjects specified below : — 
 
 Practical Shipbuilding. — Important fittings of ships, inclu- 
 ding ventilating and coaling arrangements, anchor and capstan 
 gear, masts and mastwork, etc. ; methods adopted for preventing 
 deterioration of hull, both when being built and whilst on service ; 
 launching arrangements ; principles of water-tight subdivision of 
 war and merchant ships. 
 
 Ship Calculations.— Definition of " change of trim ; " 
 moment to change trim one inch ; change of trim due to moving 
 weights already on board, and that due to the addition or removal 
 of weights of moderate amount ; displacement sheet, general arrange- 
 ment of calculations usually made thereon ; approximate and 
 detailed calculations relating to the weight and position of centre of 
 gravity of hull ; inclining experiment made to ascertain position 
 of centre of gravity of a vessel and precautions necessary to be 
 observed to ensure accuracy ; calculations for strength of bottom 
 plating. 
 
 Part II. 
 
 Those candidates for Honours who successfully pass the above- 
 mentioned examination may sit for examination in the following 
 subjects, relating to the higher branches of Theoretical Naval 
 Architecture, including those enumerated below ; — 
 
 Proofs of Simpson's and the five-eight-minus-one rules. 
 
 Heeling produced by the pressure of wind on sails. 
 
 Calculation of the shearing forces and bending moments set up 
 in a ship in still water, and also when floating amongst waves. 
 
 Construction of equivalent girder. 
 
 Statical and dynamical stabihty : Atwood's and Moseley's 
 formulas, and methods of calculating stability based thereon. Ex- 
 perimental methods of obtaining the stability at any angle. Use 
 of Amsler's integrator for calculating stability. Ordinary curves of 
 stability ; their construction and uses. Cross-curves of stability. 
 
 Reech's method of obtaining the co-ordinates of the centre of 
 buoyancy corresponding to any given angle of heel. 
 
 Effect of free liquid in hold on stability; stability of petroleum- 
 carrying vessels. 
 
 Rate of inflow of water through hole in bottom, and calculation 
 
26o Appendix. 
 
 of alteration of trim and heel due to admission of water through 
 damage to bottom. 
 
 Construction of curves of buoyancy, curves of flotation, surfaces 
 of buoyancy, surfaces of flotation. The " metacentric " or " locus of 
 pro-metacentres." 
 
 Radius of curvature ( R = — ) at any point of the curve of buoy- 
 ancy, and Leclert's formula [r = -ttt or = R + '^p-r ) for radius of 
 
 curvature at any point of the curve of flotation. 
 
 Loss of initial stability due to grounding. 
 
 Launching curves made to ensure that the length of ground- 
 ways proposed is sufficient, etc. 
 
 Froude's experiments on frictional resistance of water. 
 
 The resistances experienced by ships in their passage through 
 the water. Methods of calculating the indicated horse-power 
 required to drive a vessel at any given speed. Froude's law of 
 "corresponding speeds." Effective horse-power. Propulsive co- 
 efficient ; definition of, and values in typical ships. 
 
 Methods of measuring speed of ships on their trial trips ; pre- 
 cautions necessary to ensure accuracy. Progressive trials. 
 
 Calculations relating to the steering of ships. Methods of deter- 
 mining necessary size of rudder-head. 
 
 ± 
 ' m 
 
 for a vessel rolling unresistedly in still water. Effect on time of 
 oscillation of raising or " winging " weights. Curves of " extinc- 
 tion " (or " declining angles "). Causes operating to reduce the 
 amplitude of oscillation of a vessel set rolling in still water or 
 amongst waves. Usefulness of bilge keels ; formula for calcitlating 
 resistance due to bilge keels when vessel is rolhng. Effect of syn- 
 chronism between motion of ship and that of waves amongst which 
 she is rolling. Definitions of " effective wave-slope " and " virtual 
 upright." Methods of observing the rolling motions of ships. 
 
 Definitions of " stiff" and " steady " vessel ; elements of design 
 affecting these qualities. 
 
 Methods of calculating wetted surfaces. 
 
 The vibration of steamships : how caused, and methods of 
 minimizing. 
 
 Those candidates who answer the questions in the written 
 papers in a sufficiently satisfactory way may be called upon to sit for 
 a practical examination at South Kensington. This examination 
 
 Time of complete oscillation ( t = 2ir a / — or = i-ioS— r^ 
 
 \ V gjn V;, 
 
Appendix. 261 
 
 will be held on two consecutive days. The time allowed for work 
 on each day will be seven hours. 
 
 Candidates will be required to make a sheer draught of a vessel 
 from particulars to be furnished. 
 
 Candidates must themselves provide all drawing instruments, 
 moulds, battens, straight-edges, squares, etc., and all other neces- 
 saries, except drawing boards, drawing paper, batten weights, and 
 drawing battens and straight-edges over two feet in length, which 
 will be furnished by the Department. 
 
 Neatness and accuracy in drawing will be insisted on, 
 
APPENDIX C 
 
 Questions set on Calctilations in Examination Papers of the Science 
 and Art Department} 
 
 This selection of questions is reproduced by permission of the 
 Controller of Her Majesty's Stationery Office. 
 
 The complete examination set in 1898 is given, together with 
 the general instructions and rules. This was the first examination 
 set on the new syllabus given on p. 257. 
 
 Elementary. 
 
 The total area of the deck plan of a vessel is 4500 square feet. 
 What would be the surface of deck plank to be worked if there are — ■ 
 
 4 hatchways, each 4' x ^V 
 2 „ „ 10' X 6' 
 
 and two circular skylights, each 4 feet in diameter, over which no 
 plank is to be laid ? 
 
 Write down and explain Simpson's second rule for finding the 
 area of a plane surface. 
 
 The half-ordinates of the water-plane of a vessel in feet are 
 respectively, commencing from abaft, 2, 6"5, 9"3, 107, 11, 11, 10, 
 7'4, 3'6, and o'2, and the common interval between them is 15 feet. 
 Find the area of the plane in square feet. 
 
 A piece of African oak keelson is 24' x 12" x 10". What is 
 its weight ? 
 
 ' The complete examination papers set by the examiners of the Science 
 and Art Department are published yearly, price 6(/. Questions in Naval 
 Architecture set at the Science and Art Examinations, with answers, are 
 issued for Elementary, Advanced, and Honours by Mr. T. H. Watson, 
 10, Neville Street, Newcastle-on-Tyne, 
 
Appendix. 
 
 263 
 
 A steel plate is of the form and dimensions shown. What is 
 its weight ? 
 
 Fig. 
 
 Write down and explain Simpson's first rule for finding the area 
 of a plane surface. 
 
 The half-ordinates of a deck plan in feet are respectively \\, 5i, 
 loj, 13J, 142, 14I, 12I, 9, and 3j, and the length of the plan is 128 
 feet. Find the area of the deck plan in square yards. 
 
 Referring to the previous question, find the area in square feet 
 of the portion of the plan between the ordinates \\ and 5J. 
 
 The areas of the water-line sections of a vessel in square feet are 
 respectively 2000, 2000, 1600, 1250, and 300. The common interval 
 between them is \\ feet. Find the displacement of the vessel in 
 tons, neglecting the small portion below the lowest water-section. 
 
 A steel plate ^ of an inch thick, is of the form and dimensions 
 shown below. Find its weight in pounds. 
 
 kA. 
 
 1 
 
 4. 1 
 
 ! 
 
 
 
 ■^ 
 
 .1^ 
 
 ^ 
 
 IT— 
 
 
 - — 
 
 14: 
 
 
 
 
 
 
 
 Fig. 
 
 Write down — 
 
 (1) Simpson's first rule, 
 
 (2) Simpson's second rule, 
 
 for finding the areas of plane surfaces, and state under what con- 
 ditions each rule is applicable. 
 
264 
 
 Appendix. 
 
 The half-ordinates of the midship section of a vessel are 
 127, i2*8, I2'9, 12-9, 12-9, 12-8, 12-5, 11-9, 10-4, 5 '9, and 1-4 feet 
 respectively, and the common interval between them is 18 inches. 
 Find the area of the section in square feet. 
 
 A Dantzic oak deck plank is 25' 6" long and 4J" thick. It is 
 8 inches wide at one end, and tapers gradually to 6 inches at the 
 other end. What is its weight ? 
 
 What is meant by " tons per inch immersion " ? 
 
 The " tons per inch immersion " at the load water-plane of a 
 ship is 30-5 . What is the area of the load water-plane? and what 
 would be the displacement in cubic feet and in tons, of a layer 
 4 inches thick in the vicinity of this plane ? 
 
 Write down and explain Simpsort's first rule for finding the area 
 of a plane surface. 
 
 The half-ordinates of the load water-plane of a vessel are 
 o-i, 26, 5, 8-3, 10, 10-8, II, II, 10-5, 9'6, y6, 5-5, and 0-4 feet 
 respectively, and the common interval between the ordinates is 
 9 feet. Find the area of the load water-plane. 
 
 What is meant by " tons per inch immersion " ? 
 
 Referring to the previous question, what number of tons must 
 be taken out of the vessel to lighten it 3J inches ? and what weight 
 would have to be put into the vessel to increase her draught of 
 water by 2 inches ? 
 
 Define displacement. 
 
 A cylindrical pontoon is 50 feet long and 4 feet in diameter. It 
 floats in sea-water with its axis at the surface of the water. What 
 is its displacement in cubic feet and in tons ? 
 
 A steel plate, % of an inch thick, is of the form and dimensions 
 shown below. Find its weight in pounds. 
 
 Fig. 
 
Appendix. 265 
 
 Write down — 
 
 (i) Simpson's first rule, 
 (3) Simpson's second rule, 
 for finding the areas of plane surfaces, and clearly explain in what 
 cases each of these rules is applicable. 
 
 The semi-ordinates of the boundary of the deck of a vessel in 
 feet are respectively o-i, o'5, ii-6, 15-4, i6-8, 17, 16-9, 16-4, 14-5, 9-4, 
 and o*i, the semi-ordinates being 11 feet apart. Find the area of 
 the deck in square yards. 
 
 What is meant by the displacement of a vessel ? 
 
 A vessel is of a rectangular section throughout ; it is 80 feet 
 long, 15 feet broad, and draws in sea- water 6 feet forward and 
 8 feet aft. What is its displacement in cubic feet and in tons ? 
 
 A rectangular " steel " deck-plate is 14' 3" long, 3' 3J" wide, and 
 ^" thick. A circular piece 13 inches in diameter is cut out of the 
 centre of the plate. What is the weight of the plate ? 
 
 Write down and explain — 
 
 (i) Simpson's first rule, 
 (2) Simpson's second rule, 
 for finding the areas of plane surfaces, and state under what con- 
 ditions each rule is applicable. 
 
 The half-ordinates of the midship section of a vessel are i2'8, 
 12'9, 13, 13, 13, I2'9, I2'6, 12, lo'5, 6'o, and i'5 feet respectively, 
 and the distance between each of them is 18 inches. Find the 
 area of the midship section in square feet. 
 
 A Dantzic oak plank is 24 feet long and 3J inches thick. It is 
 7 inches wide at one end, and tapers gradually to 5| inches at the 
 other end. What is its weight in pounds ? 
 
 Having given a deck plan of a ship with ordinates thereon to 
 find its area, how would you know which of Simpson's rules it 
 would be necessary to use ? 
 
 The semi-ordinates of the load-water plane of a vessel are o'2, 
 3'6, 7'4, 10, II, 107, 9-3, 6'5, and 2 feet respectively, and they 
 are 1 5 feet apart. What is the area of the load water- plane ? 
 
 What is meant by " tons per inch immersion " ? 
 
 Referring to the previous question, what weight must be taken 
 out of the vessel to lighten her 3^ inches ? 
 
 What additional immersion would result by placing 5 tons on 
 board ? 
 
266 Appendix. 
 
 Give approximately the weights per cubic foot of — 
 
 (1) Teak. 
 
 (2) Dantzic oak. 
 
 (3) English elm. 
 
 (4) Iron. 
 
 (5) Steel. 
 
 A solid pillar of iron of circular section is 6' 10" long and 2\" in 
 diameter. What is its weight ? 
 
 Write down and explain Simpson's first rule for finding the 
 area of a plane surface. 
 
 The half-ordinates of a transverse section of a vessel are I2"2, 
 I2"2, I2'i, ii'8, ii'2, 100, and 7'3 feet in length respectively, 
 and their common distance apart is 16 inches. Neglecting the 
 portion below the lowest ordinate, find the total area of the section 
 in square feet. 
 
 The water-planes of a vessel are 4 feet apart, and their areas, 
 cohimencing with the load water- plane, are 12,000, 11,500, 8000, 
 3000, and o square feet respectively. Find the displacement of the 
 vessel in tons. 
 
 What weight would have to be taken out of the vessel referred 
 to in the previous question, in order to lighten her 4 inches from 
 her load water-plane ? 
 
 A wrought-iron armour plate is 15' 3" long, 3' 6" wide, and 4J" 
 thick. Calculate its weight in tons. 
 
 The half-ordinates of the midship section of a vessel are 22"3, 
 22"2, 217, 20'6, I7'2, I3"2, and 8 feet in length respectively. The 
 common interval between consecutive ordinates is 3 feet between the 
 first and fifth ordinates, and i' 6" between the fifth and seventh. 
 Calculate the total area of the section in square feet. 
 
 Write down and explain Simpson's second rule for finding the 
 areas of plane surfaces. 
 
 Obtain the total area included between the first and fourth 
 ordinates of the section given in the preceding question. 
 
 The " tons per inch immersion " of a vessel when floating at a 
 certain water-plane is 44"5. What is the area of this plane ? 
 
 A Dantzic fir deck-plank is 22 feet long and 4 inches thick, and 
 tapers in width from 9 inches at one end to 6 inches at the other. 
 What is its weight ? 
 
Appendix. 267 
 
 Advanced. 
 
 The half-ordinates of the load water-plane of a vessel in feetj 
 commencing from abaft, &re respectively 2J, 8, nj, 13^^, 13I, 12J, 
 9i, 41^, and \, the common interval between the ordinates being 
 16 feet. Find — 
 
 (i) the area of the plane ; 
 
 (2) the longitudinal position of its C.G. abaft the foretnost 
 ordinate. 
 
 Write down and explain Simpson's second rule for finding the 
 area of a plane surface. 
 
 The half-ordinates of a water-plane of a vessel in feet are 
 respectively, commencing from abaft, 2, 6"5, 9'3, 107, 11, 11, lo, 
 7*4, 3-6, and o'2, and the common interval is 14 feet. Find the 
 area of the plane in square feet. 
 
 Having given the dimensions on the load water-plane, state the 
 practical rules by which a close approximation may be made to the 
 weight which must be added, or removed, to change the mean 
 draught of water one inch, in three different types of vessels. 
 
 Define displacement. The areas of the vertical traverse 
 sections of a ship up to the load water-plane in square feet are 
 respectively 25, 100, 145, 250, 470, 290, 220, 165, and 30, and the 
 common interval between them is 20 feet. The displacement in 
 tons before the foremost section is 5, and abaft the aftermost 
 section is 6. Find the load displacement of the ship in tons and in 
 cubic feet. 
 
 Write down and explain the formula giving the height of the 
 transverse metacentre above the centre of buoyancy. 
 
 State clearly the use that can be made of this height by the 
 naval architect when he knows it for any particular vessel. 
 
 What principle should be followed in arranging the fastenings 
 in a stringer plate at the beams and at the butts ? 
 
 A stringer plate is 42" x J". Show the riveting in a beam and 
 at a butt, stating size and pitches of rivets, and show that the 
 arrangement you give is a good one. 
 
 The half-ordinates in feet of the load water-plane of a vessel are 
 respectively 0-2, 4, 8-3, 11-3, 13-4, 13-4, io'4, 7'2, and 2-2, the length 
 of the plane being 130 feet. Find — 
 
 (i) the area of the plane ; 
 
 (2) the distance of its C.G. from No. 5 ordinate ; 
 
 (3) the decrease in draught of water by removing 25 tons from 
 
 the vessel, 
 
268 Appendix. 
 
 A ship passes from sea-water to river-water. Show how an 
 estimate may be made for the change in the draught of water. 
 
 What will be the mean draught in river-wsi&r of a ship whose 
 mean draught in sea-wdA&c is 25 feet, her length on the water-line 
 being 320 feet ; breadth extreme 48 feet, and displacement 7500 
 tons ? 
 
 The areas of five equidistant water-planes of a vessel in square 
 feet are respectively — 
 
 (i) 4100 
 
 (2) 3700 
 
 (3) 3200 
 
 (4) 2500 
 
 (5) 1400 
 
 the common interval between them is 2 feet, and the displacement 
 below the lowest water-plane is 50 tons. Find — 
 
 (i) the tons per inch at each of the water-planes ; 
 
 (2) the displacement in tons up to each of the first four water- 
 planes. 
 
 Explain clearly, illustrating your remarks with rough sketches, 
 how curves of displacement and curves of tons per inch immersion 
 are constructed, and state what use is made of them. 
 
 A portion of a cylindrical steel stern shaft-tube \\ inches thick, 
 is i5|- feet long, and its external diameter is 15 inches. Find its 
 weight. 
 
 When would you use — 
 
 (i) Simpson's first rule, 
 (2) Simpson's second rule, 
 for finding the area of a plane surface ? 
 
 The half-ordinates of a water-plane of a vessel, in feet, are 
 respectively, commencing from "forward," 0-3, 3'8, 7'6, io'2, ii"5, 
 if5, II, 9'5, 67, and 2, and the common interval between them 
 is 16 feet. Find — 
 
 (i) the area of the plane in square feet ; 
 
 (2) the distance of the centre of gravity of the plane " abaft " 
 the " foremost " ordinate. 
 
 Explain briefly the method of finding the displacement of a ship 
 from her drawings. 
 
 What is the " centre of buoyancy '' ? 
 
 The load displacement of a ship is 5000 tons, and the centre of 
 buoyancy is 10 feet below the load water-line. In the light condi- 
 tion the displacement of the ship is 2000 tons, and the centre of 
 
Appendix. 269 
 
 gravity of the layer between the load and light lines is 6 feet below 
 the load-line. Find the vertical position of the centre of buoyancy 
 below the light line in the light condition. 
 
 The ordinates of the boundary of the deck of a ship are 6"S, 
 24, 29. 32, 33"5> 33"5) 33"5> 32, 3°, 27, and 6-5 feet respectively, 
 and the common interval between them is 21 feet. 
 
 The deck, with the exception of a space of 350 square feet, is 
 covered with f-inch steel plating, worked flush-jointed with single 
 riveted edges and butts. Find the weight of the plating, including 
 straps and fastenings. 
 
 In arranging the butt fastenings of the bottom plates of a ship 
 what principles would guide you in determining the rows of rivets 
 and the spacing of the rivets, in the butt straps ? 
 
 The half-ordinates of a vessel's load water-plane are o'l, 2 '6, 
 5, 8'3, 10, io'8, II, II, 10-5, 9"6, 7'6, 5-5, and 0-4 feet respectively ; 
 the common interval between the ordinates is 9 feet. Find — 
 
 (i) the area of the load water- plane — 
 
 (2) the distance of its centre of gravity " abaft " the " foremost " 
 
 ordinate ; 
 
 (3) the " tons per inch immersion " at the load water-plane. 
 Define " displacement " and " centre of buoyancy." 
 
 The transverse sections of a vessel are 25 feet apart, and their 
 "half-areas below the L.W.L. are i, 37, 81, 104, 107, 105, 88, 48, 
 and 6 square feet respectively. Find — 
 
 (1) the displacement in tons up to the L.W.L. ; 
 
 (2) the longitudinal position of the centre of buoyancy "abaft " 
 
 the " foremost " section. 
 
 How is a " curve of displacement " constructed ? What are its 
 uses? 
 
 A transverse bunker is filled with coal stowed in the ordinary 
 way. Find the weight of the coal from the following particulars :— 
 
 The tranverse section is of the same form throughout the length 
 Of the bunker, which is 12 feet long. 
 
 The semi-ordinates in feet of the transverse section are 
 respectively 6, 9, loj, ii|, 12?^, 12I and 12 feet, the common interval 
 being 2 feet. 
 
 44 cubic feet of coal as ordinarily stowed weighs i ton. 
 
 What principle should be followed in arranging the fastenings 
 in a stringer plate at the beams and at the butts ? 
 
 A stringer plate is 40" x f". Show the riveting in a beam and 
 
270 Appendix. 
 
 at a butt, staling sizes and pitches of rivets, and show that the 
 arrangement you give is a good one. 
 
 The semi-ordinates of the load water-plane of a vessel in feet 
 are respectively o-i, 5, ir6, 15-4, i6-8, 17, i6'9, i6'4, I4'5, 9'4, and 
 o'l, and the common interval is 11 feet. Find — 
 
 (i) the area of the plane in square feet ; 
 
 (2) the distance of its G.C. from the 17-feet ordinate, stating 
 
 whether the G.C. is before or abaft that ordinate ; 
 
 (3) the tons per inch. 
 
 How is a curve of "tons per inch" constructed? What use is 
 made of such a curve ? 
 
 The tons per inch at the successive water-planes of a vessel, 
 which are ij feet apart, are respectively 6'5, 6'2, S"6, 4"5, and o. 
 Construct the curve of tons per inch on a scale of i inch to 
 I foot of draught and i inch to i ton. 
 
 Define displacement. 
 
 Having given the length at L.W.L., breadth extreme, and the 
 mean draught of water, give approximate rules for finding the dis- 
 placement in tons of — 
 
 (i) gun vessels of the Royal Navy ; 
 
 (2) mercantile steam-ships having high speeds. 
 
 A rectangular pontoon 100 feet long, 50 feet wide, 20 feet deep, 
 is empty and floating in sea-water at a draught of 10 feet. What 
 alteration will take place in the floating condition of the pontoon, 
 if the centre compartment is breached and in free communication 
 with the sea, if — 
 
 (i) the pontoon were divided into five equal water-tight com- 
 partments by transverse bulkheads of the full depth of the 
 pontoon ; 
 
 (2) the water-tight bulkheads referred to in (i) ran up to and 
 stopped at a deck which is " not " water-tight, 12 feet from 
 the bottom of the pontoon .? 
 
 Referring to the fourth question back, if a deck surface of equal 
 area to that of the load water- plane therein mentioned were covered 
 with f-inch steel plating worked flush jointed with single riveted 
 edges and butts, what would be the weight of the plating, including 
 straps and fastenings ? 
 
 When would you use — 
 
 (i) Simpson's first rule, 
 (2) Simpson's second rule, 
 for finding the area of a plane surface ? 
 
Appendix. 27 1 
 
 The half-ordinates of a water- plane are 15 feet apart, and their 
 lengths, "commencing from forward," are respectively rg, 6'6, 11, 
 14-5, 17-4, 19-4, 20-5, 20-8, 20-3, i8-8, 15-8, 10-6, and 2-6. Find— 
 
 (i) the area of the plane in square feet ; 
 
 (2) the distance of the centre of gravity of the plane abaft the 
 foremost ordinate. 
 
 Define displacement and centre of buoyancy. 
 
 The areas of the vertical transverse sections of a vessel in square 
 feet up to the load water-plane, " commencing from forward," are 
 respectively 25, 100, 145, 250, 470, 290, 220, 165, and 30, and the 
 common interval between the sections is 20 feet. Neglecting the 
 appendages before and abaft the end sections, find — 
 
 (i) the displacement of the vessel in tons ; 
 
 (2) the longitudinal position of the centre of buoyancy abaft the 
 foremost section. 
 
 A portion of a cylindrical steel stern shaft-casing is I2f feet long, 
 \\ inches thick, and its external diameter is 14 inches. Find its 
 weight in pounds. 
 
 State the conditions under which a ship floats freely and at rest 
 at a given water-line in still water, and describe what calculations 
 have to be made in order to ascertain that the conditions will be 
 fulfilled. 
 
 The semi-ordinates in feet of the load water-plane of a vesSel 
 are, commencing from forward, o, 07, 3, 7, 8'S, 8, 6'5, 5, 2-5, and 
 I respectively, and the total length is 126 feet. Find — 
 
 (i) the area of the plane ; 
 
 (2) the longitudinal position of its centre of gravity " abaft " the 
 
 foremost ordinate ; 
 
 (3) the increase in draught caused by placing 20 tons on 
 
 board. 
 
 If a deck surface of equal area to the load water-plane, referred 
 to in the previous question, were covered with i|;^-inch steel plating, 
 worked flush, jointed with single riveted edges and butts, what 
 would be the weight of the plating, including straps and fastenings ? 
 
 How is a curve of tons per inch immersion constructed ? 
 What are its uses? Draw roughly the ordinary form of such 
 curves. 
 
 A stringer plate is 40 inches wide and \ inch thick. Show the 
 rivets in a beam and at a butt, and prove by calculation that the 
 arrangement is a good one. 
 
272 Appendix. 
 
 The transverse sections of a vessel are 20 feet apart, and their 
 areas up to the load water-line, commencing from forward, are 3, 
 35; 83) 136, 17s, 190, i/g, 146, 98, 50, and 11 square feet respectively. 
 Find — 
 
 (i) the displacement of the vessel in tons ; 
 
 (2) the distance of the centre of buoyancy from the foremost 
 section. 
 
 A teak deck, 2j inches thick, is supported upon beams spaced 
 4 feet apart and weighing 15 lbs. per foot run. Calculate the weight 
 of a middle-line portion of this deck (including fastenings and 
 beams), 24 feet long and 10 feet wide.' 
 
 Show by sketch and description how a curve of displacement is 
 constructed, and state its uses. 
 
 If it were required to so join two plates as to make the strength 
 at the butt as nearly as possible equal to that of the unpierced 
 plates, what kind of butt strap would you adopt .■' 
 
 Supposing the plates to be of mild steel 36 inches wide and 
 \ inch thick, give the diameter, disposition, and pitch of rivets 
 necessary in the strap. 
 
 The half-ordinates of the load water-plane of a vessel are 12 feet 
 apart, and their lengths are 0-5, 3-8, 77, 11 '5, 14-6, i6'6, I7'8, i8'3, 
 i8'5, i8'4, i8-2, 17-9, 17-2, 15-9, 13-4, 9'2, and o'j feet respectively. 
 Calculate — 
 
 (i) the total area of the plane in square feet ; 
 
 (2) the longitudinal position of its centre of gravity with refer- 
 
 ence to the middle ordinate ; and 
 
 (3) the tons per inch immersion at this water-plane. 
 
 The beams of a deck are 3 feet apart, and weigh 22 lbs. per 
 foot run ; the deck plating weighs 10 lbs. per square foot, and this 
 is covered by teak planking 3 inches thick. Calculate the weight 
 of a part 54 feet long x 10 feet wide of this structure, including 
 fastenings. 
 
 A vessel has the " tons per inch " specified below at the several 
 water-planes, viz. 17, 16, I4"6, 127, 97, 4'5, and o, and the planes 
 are 3 feet apart. Calculate the displacement of the vessel in tons. 
 
 State— 
 
 (i) the shearing stress of a f-inch steel rivet ; 
 
 (2) the ultimate tensile strength of mild steel plates. 
 
 What reduction is allowed for in calculating the strength of the 
 material left between closely spaced punched holes in mild steel 
 plates ? 
 
Appendix. 273 
 
 Honours. 
 
 The "tons per inch" of five equidistant water- planes of a ship 
 are respectively 9'8, 8'8, 7'6, 5'9, and 3*4, the water-planes being 
 2^ feet apart. Below the lowest of the planes mentioned is an 
 appendage of 60 tons. Calculate the displacement in tons up to 
 each of the water-planes. 
 
 Referring to the previous question, construct the curve of dis- 
 placement on a scale of f inch per foot of draught, and f inch per 
 100 tons of displacement, the lowest water-plane mentioned being 
 3 feet above the keel. 
 
 Obtain the expression for the height of the transverse metacentre 
 above the centre of buoyancy. 
 
 The displacement of a vessel is 400 tons, and the transverse 
 metacentre is 5f feet above the centre of buoyancy. A weight of 
 12 tons, already on board, is moved 8 feet across the deck : find the 
 inclination of the vessel to the upright, the C.G. of the vessel being 
 3 feet above the C.B. 
 
 tan 4° = 0-0699 
 
 tan 5° = 0-0875 
 
 tan 6° = 0-1051 
 
 What is meant by " moment to change trim " ? Write down 
 
 and explain the expression which gives the moment to alter trim 
 
 one inch. 
 
 Suppose a weight of moderate amount to be put on board a 
 ship, where must it be placed so that the ship shall be bodily deeper 
 in the water without change of trim? Give reasons for your 
 answer. 
 
 A transverse iron water-tight bulkhead is worked in a ship at 
 a station whose semi-ordinates are (commencing from below) 6, 9, 
 loj, 1 1 J, ii\, ll\, and 12 feet respectively, the common interval 
 being 2 feet. Find the weight of the bulkhead, the following par- 
 ticulars being given : — 
 
 „ , . , . , el inch for lower 5 feet. 
 
 Plates, lap jomted, lap butted, smgle riveted \^^-^^^^ for upper 7 feet. 
 
 Angle bar stiffeners 24 inches apart on \ ,„ ,„ . „ 
 
 one side of bulkhead / "^ 
 
 Distinguish between hogging and sagging strains. A vessel 
 has an excess of weight amidships : to what conclusion would you 
 generally arrive as to the strains produced ? 
 
 Point out, illustrating your remarks by a simple example, that 
 your general conclusion might not be correct in some cases. 
 
 T 
 
274 
 
 Appendix. 
 
 What are the causes which influence \}a& forms of curves of 
 stability ? 
 
 Give an example of such curves for— 
 
 (1) a low freeboard mastless armourclad ; 
 
 (2) a high freeboard cruiser with large sail spread. 
 Explain how lifeboats are designed to automatically right them- 
 selves when capsized, and to free themselves of water when 
 swamped. 
 
 The half-ordinates of a portion of a deck plan of a vessel, com- 
 mencing from abaft, are 2, 8, and 11^ feet respectively, and the 
 common interval is 16 feet. On the beams^between the two after- 
 most ordinates, steel plating \ inch thick is to be worked. What is 
 the weight of the plating ? 
 
 From the particulars given below, find the displacement in tons 
 of the vessel up to L.W.L. 
 
 
 Half-ordinates in feet at stations. 
 
 I 
 
 2 
 
 3 
 
 4 
 
 5 
 
 L.W.L. 
 
 2 W.L. 
 
 3 W.L. 
 
 0-2 
 
 0-2 
 0'2 
 
 7-4 
 5-3 
 
 2-0 
 
 ii-o 
 
 ID'S 
 
 7-6 
 
 9-3 
 2-6 
 
 2-0 
 
 0-3 
 0-3 
 
 Stations apart, 30 feet ; water-lines apart, 3 feet. (Appendages 
 before and abaft end ordinates and below No. 3 W.L. being 
 neglected.) 
 
 Describe clearly how you would proceed to obtain from the 
 drawings of a vessel, a close approximation to the area of " wetted 
 surface " for a given draught of water forward and aft. 
 
 A pontoon raft is formed by two cylindrical pontoons, each 
 50 feet long and 4 feet in diameter. The distance between the 
 centres of the pontoons is 8 feet throughout their lengths. In sea- 
 water the raft floats with the axes of the pontoons at the surface of 
 the water. Find the height of the transverse metacentre above the 
 centre of buoyancy of the raft. 
 
 What is meant by " change of trim " .? 
 
 It is desired that the draught of water aftm.^. steam-ship shall 
 be constant whether the coals are in or out of the ship. Show how 
 the approximate position of the C.G. of the coals may be found, 
 in order that the desired condition may be fulfilled. 
 
Appendix. 
 
 275 
 
 Briefly describe an experimental method of obtaining a curve 
 of stability for a ship. Compare the method with the calculation 
 method. 
 
 How would you proceed in arranging the fastenings in a 
 stringer plate at the beams and at the butts ? 
 
 A stringer plate is 38 inches wide and /j inch thick. Sketch 
 the riveting in a beam and at a butt, and show that the arrange- 
 ment is a good one. 
 
 Show how a comparison may be made between the turning 
 effects on a ship of — 
 
 (i) A narrow rudder held at a certain angle by a given force at 
 the end of a tiller ; and — 
 
 (2) A broader rudder of equal depth held by an equal force at 
 a smaller angle. 
 
 From the particulars given below find — 
 
 (1) the displacement in tons of the vessel up to L.W.L. ; 
 
 (2) the distance of the centre of buoyancy abaft the foremost 
 
 station ; 
 
 (3) the depth of the centre of buoyancy below L.W.L. 
 
 
 Half-ordinates in feet at stations. 
 
 I 
 
 2 
 
 8-9 
 6-4 
 2-4 
 
 3 1 4 
 
 s 
 
 L.W.L. 
 2W.L. 
 3W.L. 
 
 0-3 
 
 0-3 
 0-3 
 
 13-2 
 
 12-6 
 9-0 
 
 1 1 -2 
 
 7-8 
 31 
 
 2-4 
 0-4 
 0-4 
 
 Stations apart, 27 feet ; water-lines apart, 3 feet. (Appendages 
 before and abaft end ordinates and below No. 3 W.L. being 
 neglected.) 
 
 A pontoon raft is formed by three cylindrical pontoons, each 
 50 feet long and 4 feet in diameter. The outer pontoons are each 
 8 feet from the middle one (centre to centre) throughout their 
 lengths. Find the displacement of the raft, and the height of the 
 transverse metacentre above the centre of buoyancy, the pontoons 
 floating with their axes at the surface of the water. 
 
 Explain briefly how the vertical position of the C.G. of a ship 
 and its equipment is accurately obtained. 
 
 How is a metacentric diagram constructed, and what are its 
 uses? 
 
2/6 Appendix. 
 
 In what classes of ships would you expect the metacentre to fall 
 quickly as the draught lightens, and after reaching a minimum 
 height to gradually rise again ? 
 
 A steel ship is found, on her first voyage at sea, to be structurally 
 weak longitudinally. How would you attempt to effectually 
 strengthen the ship with the least additional weight of material, 
 giving your reasons ? 
 
 What is meant by " curves of weight " and " curves of buoyancy " 
 as applied to the longitudinal distribution of weight and buoyancy 
 in ships? 
 
 Show how these curves are obtained ; and draw, approximately, 
 on the same scale and in one diagram, such curves for any type of 
 vessel with which you are acquainted, mentioning the type you 
 have taken. 
 
 The half-ordinates of a vessel's load water-plane are o'l, 2'6, 
 5, 8'3, lo, 10-8, II, II, 10-5, 9*6, 7-6, 5-5, and o"4 feet respectively, 
 the common internal being 9 feet. The water-planes of this vessel 
 are if feet apart, and the " tons per inch " for those below the load 
 water-plane are 3'5, 3'o, 2'4, and o'g respectively. The keel appen- 
 dage is of 5 tons displacement, and at 7'5 feet below the load 
 water-plane. Find — 
 
 (i) the total displacement in tons ; 
 
 (2) the vertical position of the centre of buoyancy below the 
 load water-plane. 
 
 Explain how the height of the transverse metacentre above the 
 centre of buoyancy of a ship may be found — 
 (i) accurately ; 
 
 (2) approximately and quickly ; 
 and state clearly what use is made of the result by the naval 
 architect. 
 
 What is meant by change of trim ? 
 
 Write down the expression which gives the " moment to change 
 trim," and explain by means of a diagram how the expression is 
 obtained. 
 
 A ship is 375 feet long, has a longitudinal metacentric height 
 (G.M.) of 400 feet, and a displacement of 9200 tons. If a weight 
 of 50 tons, already on board, be shifted longitudinally through 
 90 feet, what will be the change in trim ? 
 
 Under what circumstances may it be expected that the cargoes 
 of vessels will shift ? 
 
 In a cargo-carrying vessel, the position of whose C.G. is known, 
 
Appendix. 
 
 277 
 
 show how the new position of the C.G. due to a portion of the 
 cargo shifting may be found. 
 
 A ship of 4000 tons displacement, when fully laden with coals, 
 has a metacentric height of 2| feet. Suppose 100 tons of coal to 
 be shifted so that its C.G. moves 18 feet transversely and 4J feet 
 vertically ; what would be the angle of heel of the vessel, if she were 
 upright before the coal shifted ? 
 
 tan 10° = 0-I763 
 tan 11° = o"i944 
 tan 12° = o'2i26 
 tan 13° = 0-2309 
 
 What portions of a properly constructed steel ship are most 
 effectual in resisting longitudinal bending .' Why .'' 
 
 What principle should be followed in arranging the fastenings 
 in a stringer plate at the beams and at the butts ? 
 
 A stringer plate is 40" x f ". Show the riveting in a beam and 
 at a butt, stating size and pitches of rivets, and show that the 
 arrangement you give is a good one. 
 
 Enumerate the strains to which ships are subjected which tend 
 to produce changes in their " transverse " forms. 
 
 When are the most severe transverse strains likely to be 
 experienced by a ship at " rest " ? Point out in such a case the 
 forces acting on the ship, and state what parts of the ship, if she be 
 properly constructed, will effectually assist the structure in resisting 
 change of form. 
 
 From the particulars given below, find — 
 
 (i) the displacement in tons up to the L.W.L. ; 
 
 (2) the depth of the centre of buoyancy below the L.W.L. ; 
 
 (3) the longitudinal position of the centre of buoyancy with 
 
 respect to No. 3 station. 
 
 Forward. 
 
 Half-ordinates in feet at stations. 
 
 Aft. 
 
 I 
 
 2 
 
 3 
 
 . 1 5 
 
 L.W.L. 
 
 2 W.L. 
 
 3 W.L. 
 
 4 W.L. 
 
 5 W.L. 
 
 0-3 
 03 
 0-3 
 0-3 
 03 
 
 80 
 7-0 
 6-0 
 
 4-5 
 
 2'2 
 
 12-0 
 12-0 
 
 n-5 
 
 10-3 
 
 8-3 
 
 lo-o 
 9-0 
 7-0 
 S-a 
 27 
 
 2'0 
 
 o'3 
 0-3 
 0-3 
 
 
 Stations apart, 25 feet ; water-lines apart, il feet. (Appendages 
 
278 Appendix. 
 
 before and abaft end ordinates and below No. 5 W.L. being 
 neglected.) 
 
 Define centre of gravity. Write down and explain the rule for 
 finding the " transverse " position of the C.G. of the " longitudinal " 
 half of a water-plane. 
 
 The ordinates of half a water-plane in feet are respectively— 
 o-i, 5, 11-6, I5'4, i6-8, 17, 16-9, i6'4, 14-5, 9-4, and o'l, and the 
 common interval is 1 1 feet. Find the " transverse " position of the 
 C.G. of the half water-plane. 
 
 The semi-ordinates of the boundary of a ship's deck in feet are 
 respectively, commencing from forward, o'3, g'2, 17, 22'5, 26, 28, 
 29, 29-5, 29-5, 29-5, 29-5, 29-5, 29-3, 29, 28-5, 27'5, 25-5, 21, and 11-5 ; 
 the common interval being 18 feet. 
 
 A steel stringer plate is worked on the ends of the deck beams 
 on each side of the ship, of the following dimensions : 54" x |" for 
 half the length amidships, tapering gradually to 32" x |" at the 
 fore extremity, and to 40" x ^" at the after extremity. The butts 
 are treble chain riveted. Find approximately the weight of the 
 stringer plate, including fastenings and straps. 
 
 Define the term " metacentre." Prove the rule for finding the 
 height of the transverse metacentre above the centre of buoyancy. 
 State clearly what use is made of the result by the naval architect. 
 
 What is a metacentric diagram ? How is such a diagram con- 
 structed ? For what classes of ships are such diagrams specially 
 useful 'i 
 
 Draw a typical metacentric diagram for merchant ships of 
 deep draught in proportion to their beam when fully laden, with 
 approximate vertical sides between the load and light lines. 
 
 Sketch the water-tight subdivision of an efficiently subdivided 
 steam mercantile ship. 
 
 In some ships the transverse water-tight bulkheads are so badly 
 arranged that it would be preferable, as a safeguard against rapid 
 foundering, if the vessels were seriously damaged below the water- 
 line, to dispense with these bulkheads. Explain, with sketches, 
 how this conclusion is arrived at. 
 
 What portions of a ship's structure offer resistance to cross- 
 breaking strains at any transverse section ? 
 
 Explain clearly how you would proceed with the calculations, 
 and state what assumptions you would make, in finding the strength 
 of the midship section of an iron ship against cross-breaking strains. 
 
 Define ''displacement," "centre of buoyancy," and "tons per 
 inch immersion." 
 
Appendix. 
 
 279 
 
 From the particulars given below, find : — 
 (a) the displacement of the vessel in tons ; 
 {b) the longitudinal position of the centre of buoyancy, with re- 
 spect to No. 3 station ; 
 (c) the tons per inch at the L.W.L. 
 
 Forward. 
 
 
 Half-ordinates in feet at stations. 
 
 
 Aft. 
 
 I ! 2 
 
 3 
 
 ♦ 
 
 s 
 
 L.W.L. 
 
 2 W.L. 
 
 3 W.L. 
 
 4 W.L. 
 
 5-0 
 5-0 
 4'8 
 4'i 
 
 15-4 
 15-4 
 15-0 
 
 •3-3 
 
 17-0 
 17-0 
 16-4 
 14-5 
 
 16-4 
 
 15-8 
 14-2 
 
 9'3 
 
 9'4 
 6-1 
 
 3-3 
 13 
 
 
 Stations apart, 22 feet ; water-lines apart, i \ feet. The appen- 
 dage below No. 4 W.L. is 50 tons, and has its centre of gravity 5 J 
 feet before No. 3 station. Appendages before and abaft end sections 
 may be neglected. 
 
 Explain fully the work necessary to obtain the " correct " ver- 
 tical position of the centre of gravity of a ship and her lading. 
 
 Show how the expression which gives the height of the trans- 
 verse metacentre above the centre of buoyancy is obtained, and 
 state clearly what use is made of this information by the naval 
 architect when he knows it for any particular ship. 
 
 What is a metacentric diagram ? Explain how such a diagram 
 is constructed. 
 
 A vessel of box form is 20 feet wide, 10 feet deep, and the C.G. 
 of the vessel and its lading is at the middle of its depth for all 
 variations in draught of water. Construct to scale the metacentric 
 diagram of the vessel. 
 
 How are curves of stability constructed, and what are their 
 uses? 
 
 In constructing such curves for warships which are completed 
 state — 
 
 (a) what facts must be known ; 
 
 ib) what assumptions must be made ; 
 
 {c) what justification there is for making some of the assump- 
 tions which may be open to objection. 
 
 What are the most important changes, from a designer's point 
 of view, that have taken place in recent years, in— 
 
 (fl) ships of the Mercantile Marine ; 
 
 <j)) armoured ships of war ? 
 
28o 
 
 Appendix. 
 
 From the particulars given below, find — 
 
 (i) the area of the L.W. plane, and tons per inch immersion at 
 the L.W.L. ; 
 
 (2) the displacement of the vessel in tons ; 
 
 (3) the position of the centre of buoyancy in relation to the 
 
 L.W.L. and No. 3 station. 
 
 
 
 Half-ordinates in feet 
 
 at stations. 
 
 
 
 Aft. 
 
 
 
 
 I 
 
 
 5 
 
 4 
 
 3 
 
 2 
 
 
 L.W.L. 
 
 2-0 
 
 9'3 
 
 II-O 
 
 7-6 
 
 0-2 
 
 
 2 W.L. 
 
 05 
 
 8-2 
 
 II-O 
 
 6-4 
 
 02 
 
 
 3 W.L. 
 
 03 
 
 6-5 
 
 lo-s 
 
 5-3 
 
 0'2 
 
 
 4 W.L. 
 
 03 
 
 4-6 
 
 9'5 
 
 4-0 
 
 0-2 
 
 
 5 W.L. 
 
 0-3 
 
 2-S 
 
 7-6 
 
 2'0 
 
 0-2 
 
 
 Stations apart, 25 feet; water-lines apart, li feet; all appen- 
 dages being neglected. 
 
 How would you obtain a close approximation to the area of 
 wetted surface of a ship ? 
 
 What use could be made of this information ? 
 
 Obtain the expression which gives the height of the transverse 
 metacentre above the centre of buoyancy. 
 
 The displacement of a vessel is 400 tons, the transverse meta- 
 centre is 5| feet, above the centre of buoyancy, and the centre of 
 gravity is 3 feet above the centre of buoyancy. A weight of 12 tons 
 is moved 8 feet across the deck. Find the inclination of the 
 vessel. 
 
 tan 4° = o'o699 
 tan 5° = o'o875 
 tan 6° = o'losi 
 
 Obtain the expression which gives the " moment to change 
 trim one inch." 
 
 In what position would you place a moderate weight on board 
 a ship in order that there should be no change of trim ? Give your 
 reason. 
 
 How would you rapidly determine whether any practicable 
 alteration in the distribution of weights which can be shifted would 
 enable a ship to go over a bar which would be an impossibihty in 
 her existing trim .'' 
 
 How is the information obtained for constructing curves showing 
 the longitudinal distribution of weight and buoyancy of a ship 'i 
 
Appendix. 281 
 
 How are the curves made, and what are their uses? What checks 
 would you adopt to verify the accuracy of the curves, and what 
 guarantee would you have that the conditions of the checks are 
 correct ? 
 
 What are the most severe strains likely to be experienced by 
 a ship at rest ? Point out the forces acting on the ship, and state 
 what parts of the structure operate in resisting change of form. 
 
 What are equivalent girders for ships ? Briefly describe how 
 they are constructed. 
 
 The water-lines of a vessel are 5 feet apart, and the " tons per 
 inch" at those lines, commencing from the load water-line, are 
 33'8, 2f9, I9'5, i6'4, and 6 tons respectively. Neglecting the 
 appendage below the lowest water-line, calculate the displacement 
 of the vessel, and the vertical position of the centre of buoyancy — 
 
 (i) when floating at her load water-line ; 
 
 (2) when floating at the line next below her load water-line. 
 
 Obtain the formula giving the height of the longitudinal meta- 
 centre above the centre of buoyancy. 
 
 What is meant by " change of trim " ? 
 
 A ship 220 feet long has a longitudinal metacentric height of 
 252 feet, and a displacement of 1950 tons. Calculate the change 
 of trim due to shifting a weight of 20 tons, already on board, 
 through a longitudinal distance of 60 feet. 
 
 Describe in detail how you would proceed to obtain the statical 
 stabiHty at large angles of inclination of a vessel of known form. 
 
 Show how you would calculate the weight of the outer bottom 
 plating of a vessel, and the position of its centre of gravity. 
 
 The shearing force and bending moment operating at every 
 transverse section of a vessel floating at rest in still water being 
 required, how would you proceed to obtain this information ? 
 
 Explain clearly how the stability of a vessel at small angles of 
 inclination is affected by the presence of water on board, which is 
 free to shift transversely. 
 
 A box-shaped vessel, 105 feet long and 30 feet broad, floats at a 
 uniform draught of 10 feet. A central compartment, 20 feet long, 
 contains water which is free to shift from side to side. Calculate 
 by how much the metacentric height is reduced from what its value 
 would have been had the water \)%&a. fixed. 
 
 Write down and explain Simpson's first and second rules for 
 obtaining the areas of plane surfaces. 
 
282 Appendix. 
 
 The half-ordinates of the load water- plane of a vessel are 13 feet 
 apart, and their lengths are o'6, 4'6, 9T, 12-8, I5'5, I7'2, iS'i, i8'4, 
 i8'5, 18-5, i8'4, i8'2, 177, 16-9, 15-3, 12-5, and 7-0 feet respectively. 
 Calculate — 
 
 (i) the total area of the plane in square feet ; 
 
 (2) the area included between the third and sixth ordinates. 
 
 The water-planes of a vessel are 3 feet apart, and the displace- 
 ments up to the several planes are 2380, 1785, 1235, 740, 325, 60, o 
 tons respectively. Calculate the vertical position of the centre of 
 buoyancy, and show that the method you employ is correct. 
 
 Define the terms " metacentre " and " metacentric height." 
 
 The load displacement of a vessel is 1880 tons when floating at 
 the water-plane given in the second question above. Calculate 
 her longitudinal metacentric height, assuming the ship's centre of 
 gravity to be in the load water-line, and her centre of buoyancy 
 6 feet below that Une. 
 
 Sketch a metacentric diagram for any one type of ship, specify- 
 ing the type chosen. How is such a diagram constructed? 
 
 What is meant by the dynamical stability of a vessel at any 
 angle of inclination ? Obtain Moseley's formula for calculating its 
 value. 
 
 Explain how you would proceed to calculate the weight, and 
 position of centre of gravity, of the transverse framing of a vessel. 
 
 Show how you would calculate the " wetted surface " of a vessel 
 with considerable accuracy. Describe any method of rapidly cal- 
 culating wetted surfaces with which you may be acquainted. 
 
 1898. 
 SUBJECT IV.— NAVAL ARCHITECTURE. 
 
 Examiner : J. J. Welch, Esq., R.C.N.C. 
 
 General Instructions. 
 
 If the rules are not attended to, the paper will be cancelled. 
 
 You may take the Elementary stage, or the Advanced stage, or 
 Part I. of Honours, or (if eligible) Part II. of Honours, but you must 
 confine yourself to one of them. 
 
 Put the number of the question before your answer. 
 
 You are to confine your answers strictly to the questions 
 proposed. 
 
Appendix. 283 
 
 Your name is not given to the Examiner, and you are forbidden 
 to write to him about your answers. 
 
 The vakie attached to each question is shown in brackets after 
 the question. A full and correct answer to an easy question will 
 in all cases secure a larger number of marks than an incomplete 
 or inexact answer to a more difficult one. 
 
 The examination in this subject lasts for four hours. 
 
 FIRST STAGE OR ELEMENTARY EXAMINATION. 
 
 Instructions. 
 
 You are permitted to answer only ten questions. 
 
 You must attempt No. 11. Two of the remaining questions 
 should be selected from the Calculations ; and the rest from the 
 Practical Shipbuilding section. 
 
 Practical Shipbuilding. 
 
 I . Give a sketch of a side bar keel, and describe how the several 
 lengths are secured together, and how the work is made water- 
 tight. (8) 
 
 3. Describe the operation of getting an ordinary bar stem into 
 its correct position on the blocks. (8) 
 
 3. Sketch a transverse frame, from keel to water-tight longitu- 
 dinal (or to margin plate), of a vessel having a doulale bottom. 
 Specify usual sizes of plates and angles for a large ship. (12) 
 
 4. Describe the usual method of bending and bevelling the 
 frame angle-bars of a vessel. (8) 
 
 5. In some cases it is necessary to work the floor-plate of a 
 transversely framed vessel in two lengths. Describe, with sketches, 
 two usual methods of connecting these lengths together. (10) 
 
 6. Show how a beam of '' i '" section is secured to the framing, 
 
 • 
 giving usual disposition and pitch of rivets. (8) 
 
 7. If, after a bottom plate is in position, it is found that the rivet 
 holes do not quite correspond with those of the adjacent plates, 
 what should be done to rectify this and to ensure sound work ? 
 
 (8) 
 
 8. Name, with sketches, the finished forms of rivets employed 
 in shipwork. (8) 
 
 9. Sketch a usual shift of butts of deck plating, and show how 
 the several plates are secured together, and to the beams. Specify 
 
284 
 
 Appendix. 
 
 size and pitches of rivets which would be used, assuming the plating 
 to be T^" thick. (12) 
 
 10. Give a rough sketch of a frame of a composite vessel, 
 naming the several parts of which it is made up. (8) 
 
 Drawing. 
 
 11. What does sketch. Fig. 113, represent ? Draw it neatly in 
 pencil on a scale twice the size shown. (14) 
 
 Elementary Stage. 
 
 .N9'll. 
 
 Fig. T13. 
 
 Calculations. 
 
 1 2. State the fundamental conditions which must be fulfilled by 
 a vessel when floating freely and at rest in still water. (6) 
 
 13. The areas of successive water-planes of a vessel are, begin- 
 ning with the load water-plane, 14,850, 14,400, 13,780, 12,950, 
 11,770, 10,130, and 7680 square feet respectively, and the common 
 interval between the planes is 31 feet. Neglecting the part below 
 the lowest water-plane, calculate the vessel's load displacement in 
 tons. (12) 
 
 14. Point out why it is that a vessel sinks deeper in the water 
 when passing from the sea into a river. 
 
Appendix. 285 
 
 How much would you expect the vessel in the preceding example 
 to sink from her load-Hne under such circumstances ? (8) 
 
 15. The external diameter of a hollow steel shaft is 18 inches, 
 and its internal diameter 10 inches. Calculate the weight of a 
 20-feet length of this shafting. (8) 
 
 SECOND STAGE OR ADVANCED EXAMINATION. 
 
 Instructions. 
 
 Read the General Instructions on p. 282. 
 
 You are permitted to answer only twelve questions. 
 
 You must attempt Nos. 29 and 33. The remaining questions 
 may be selected from any part of the paper in this stage, provided 
 that one or more be taken from each section, viz. Practical Ship- 
 building, Laying Off, and Calculations. 
 
 Practical Shipbuilding. 
 
 21. Give sketches showing the sections of moulded steel in 
 general use for shipbuilding purposes, and say for what parts of the 
 structure each is used. (12) 
 
 22. For what purposes are web frames fitted in vessels ? 
 
 How is a web frame secured where it crosses a continuous 
 stringer plate ? (12) 
 
 23. At what stage of the work are the deck-beams attached to 
 the frames of a transversely framed ship, and when is the riveting 
 of beam knees performed ? 
 
 Give a sketch of a beam knee, showing the fastenings. (14) 
 
 24. Sketch a satisfactory shift of butts of bottom plating, and 
 show the riveting adopted in edges and butts, and for security to the 
 frames. Specify size and pitches of rivets, assuming the plating is 
 
 _7_" - ■ - 
 16 
 
 thick. (20) 
 
 35. Describe fully the operations of getting in place and riveting 
 up a bottom plate, and the precautions necessary to ensure sound 
 and eflScient work. (14) 
 
 26. Describe how a large transverse water-tight bulkhead is 
 constructed, and secured in position. (14) 
 
 27. What tests are applied to — 
 
 (i) mild steel plates, 
 (2) rivet steel and rivets, 
 before acceptance from the manufacturer? (16) 
 
 28. An ordinary steel deck is to be covered with planking. 
 
286 Appendix. 
 
 State the order in which this work would be proceeded with, and 
 show how the planking would be fastened. 
 
 Give, on a large scale, a sketch of one of the fastening bolts. 
 
 (20) 
 
 29. Enumerate the several causes which tend to produce the 
 transverse straining of ships, and point out the parts of the structure 
 which supply the necessary transverse strength to resist these 
 straining actions. 
 
 Laying Off. 
 
 30. How is a Scrive board constructed, and what are its uses ? 
 
 (10) 
 
 31. Describe briefly the system of fairing the body adopted — 
 (i) in the middle portion, 
 
 (2) at the extremities, of a vessel. (12) 
 
 32. How would you find the true form of the plane of flotation 
 of a vessel which has a considerable trim by the stern, and also a 
 list to starboard or port ? (12) 
 
 Drawing. 
 
 33. What does the drawing Fig. 1 14 represent ? Draw it neatly 
 in pencil on a scale twice the size shown. (24) 
 
 Advanced Stage. 
 
 Ne33. 
 
 Fig. 114. 
 
Appendix. 287 
 
 Calculations. 
 
 34. The transverse sections of a vessel are 18 feet apart, and 
 their areas up to the load water-line, commencing from forward, are 
 6'5> 55'8, 132-0, 210-9, 266-3, 289-5, 280-2, 235-7, i6i-2, 77-8, and 10-9 
 square feet respectively. Calculate the displacement of the vessel 
 in tons, and the longitudinal position of her centre of buoyancy. 
 
 (H) 
 
 35. Referring to the preceding question, calculate the volume of 
 displacement comprised between the first and sixth sections, and 
 the distance of the centre of buoyancy of that portion from the 
 first section. Ci6) 
 
 36. What is a curve of tons per inch immersion ? Show how 
 such a curve is constructed, and give a sketch indicating its usual 
 shape. (8) 
 
 37. A steel stringer plate is 48 inches wide and f j inch thick. 
 Sketch the fastenings in a beam and at a butt, and show by 
 calculations that the butt connection is a good one. (20) 
 
 HONOURS EXAMINATION.— Part 1. 
 Instructions. 
 
 Read the General Instructions on p. 282. 
 
 You are permitted to answer only fourteen questions. You 
 must attempt Nos. 70 and 74 ; the remainder you may select from 
 any part of the paper in this stage, provided that one or more be 
 taken from each section, viz. Practical Shipbuilding, Laying Off, 
 and Calculations. 
 
 Practical Shipbuilding. 
 
 60. What is the usual spacing adopted for transverse frames in — 
 (i) a first-class battleship ; 
 
 (2) a large merchant vessel ? 
 
 Give reasons for the differences noted between the two classes 
 in this respect, and also for the different spacing adopted in the 
 several parts of a battleship. (20) 
 
 61. What are the characteristic qualities of the following ship- 
 building materials : (l) Dantzic fir, (2) East India teak, (3) cast 
 steel ? 
 
 State where these materials are employed. (16) 
 
 62. Give a rough sketch of the midship section of a vessel having 
 a double bottom, and point out the order in which the work of 
 
288 Appendix. 
 
 erecting the framing of such a ship would be proceeded with in way 
 of the double bottom. (30) 
 
 63. Roughly sketch the stern-post of a screw ship, showing how 
 it is connected to the keel and bottom plating. (16) 
 
 64. What considerations govern the lengths and breadths of 
 plates used on the bottom of a ship .? Describe fully the work of 
 getting into place and riveting-up one such plate. (16) 
 
 65. Give sections of the beams commonly employed in ship- 
 building, and say where each form is employed. (12) 
 
 66. In what vessels is straining at the butts of bottom plating 
 specially liable to take place, and why ? What method of stiffening 
 butt straps has been designed to prevent the above action ? (i6) 
 
 67. Show by sketch and description how water-tight work is 
 secured — 
 
 (i) at the upper edge of a longitudinal bulkhead ; 
 
 r 
 
 form, worked above 
 
 L 
 
 1 
 
 (2) where a middle line keelson of 
 
 J 
 the floors, passes through a transverse bulkhead. (16) 
 
 68. Describe the work of laying and fastening the planking of a 
 deck, the beams of which are not covered with plating. (25) 
 
 69. Sketch, and describe the working of, a large sliding water- 
 tight door as fitted to a bulkhead between machinery compartments. 
 
 (30) 
 
 70. Enumerate the principal local stresses experienced by ships, 
 and point out what special provision is made to meet each. (25) 
 
 Laying Off. 
 
 71. What information and drawings would you require before 
 proceeding with the work of laying off a vessel on the mould loft 
 floor? 
 
 Show how the extremities of a ship are usually laid off and 
 faired. (20) 
 
 72. How is the shape of longitudinal plate frames obtained in 
 those parts of a ship where there is not much curvature ? Sketch 
 a mould for a longitudinal plate, showing the marks which would 
 be put upon it for the information of the workman. (16) 
 
 73. The lines of a vessel sheathed with wood having been given 
 to the outside of sheathing, show how you would obtain the body 
 plan to outside of framing, (i) approximately, (3) accurately. (25) 
 
Appendix. 289 
 
 Calculations. 
 
 74. The half-ordinates of the load water-plane of a vessel are 
 spaced 18 feet apart, and their lengths, commencing from forward, 
 areo-6, 3-4, 7-1, 11-4, ifro, 20-3, 24-0, 26-8, 28-8, 30-0, 30'5, 30-5, 
 30-0, 28-9, 27-0, 24-3, 21-1, 17-2, 127, 77, and 3-0 feet respectively. 
 Calculate the total area of the plane in square feet, and the 
 longitudinal position of its centre of gravity. (25) 
 
 75. Prove the formula used for calculating the distance between 
 the centre of buoyancy and the transverse metacentre of a vessel. 
 
 (12) 
 
 76. A vessel, 200 feet long between perpendiculars and of 1080 
 tons displacement, floats at a draft of 11' 3" forward and 12' 3" aft, 
 and has a longitudinal metacentric height of 235 feet. Supposing a 
 weight of 20 tons to be moved forward through a distance of 120 
 feet, what would be the new drafts of water forward and aft, 
 assuming the centre of gravity of the water-plane area is 10 feet 
 abaft the midship section .' (16) 
 
 77. The deck of a vessel is covered with j^^-inch mild steel plating, 
 and the beams, spaced 3 feet apart, weigh 20 lbs. per foot run. 
 The half-ordinates of the foremost 84 feet length of this deck are 
 o"8, 3'5, 6'5, 9'4, i2'i, i4'5, i6"6, i8'4, and 20 feet respectively. 
 Calculate the total weight of plating and beams for this portion of 
 the deck. (20) 
 
 78. Show how the work of estimating the weight and position 
 of the centre of gravity of the outer bottom plating of a vessel from 
 her drawings would be proceeded with. (16) 
 
 HONOURS EXAMINATION.— Part II. 
 Instructions. 
 
 Read the General Instructions on p. 282. 
 
 You are not permitted to answer more xhsia fourteen questions, 
 of which two at least must be taken from the Practical Shipbuilding 
 and Laying Off section. 
 
 Noi E. — No Candidate is eligible for examination in Part II. 
 of Honours who has not already obtained a first or second class in 
 Honours of the same subject in a previous year. 
 
 Those students who answer the present paper sufficiently well 
 to give them a reasonable chance of being classed in Honours, will 
 be required to take a practical examination at South Kensington. 
 Honours candidates admissible to this Examination will be so 
 informed in due course. 
 
290 Appendix. 
 
 Practical Shipbuilding and Laying Off. 
 
 84. Describe the usual method of bending and bevelling Z-bar 
 frames by hand. 
 
 What advantages are claimed to accrue from the use of bevelling 
 machines ? (30) 
 
 85. What are the reasons for working mast partners ? Sketch 
 and describe an arrangement of mast partners for a steel ship. 
 
 (20) 
 
 86. Describe, with illustrative sketches, the characteristic 
 features of the launching arrangements adopted for a large ship. 
 
 (35) 
 
 87. A raking mast, of uniform diameter at its lower end, stands 
 upon a deck which has considerable round-up and sheer. Show 
 how the true shape of the lowest plate of mast would be obtained. 
 
 (25) 
 
 Ship Calculation and Design. 
 
 88. The tons per inch immersion at the several water-planes of 
 a vessel are 29-1, 28'8, 28'2, 27'3, 26'o, 24'3, 21-9, i8'6, and I3'i 
 respectively, the common interval between the planes being 1^ feet. 
 The part of the ship below the lowest water-plane has a displace- 
 ment of 300 tons, and its centre of buoyancy is 2ii feet below the 
 load water-line. Estimate (i) the total displacement of the vessel 
 in tons ; (2) the vertical position of her centre of buoyancy. (25) 
 
 89. State and prove Simpson's second rule for the calculation of 
 plane areas, pointing out clearly the assumptions involved. (25) 
 
 90. What are curves of displacement and curves of tons fier inch 
 immersion? Give sketches showing their usual shapes. (16) 
 
 91. How would you proceed to estimate the shearing force and 
 bending moment acting at any cross-section of a given vessel, when 
 floating freely and at rest in still water ? (30) 
 
 92. Prove Atwood's formula for the statical stability of a vessel 
 at any angle of heel. Show how a curve of statical stability is 
 constructed, and explain its uses. (25) 
 
 93. A weight of moderate amount is to be placed on board a 
 given vessel in such a position that the draft of water aft will be 
 unaffected by the addition. Explain how the necessary position of 
 the weight can be calculated. (30) 
 
 94. Describe any method by which the statical stability of a 
 vessel of known form and lading can be obtained experimentally. 
 
 (25) 
 
Appendix. 291 
 
 95. What are cross-curves of statical stability ? How are these 
 related to the ordinary stability curves ? (30) 
 
 96. Point out clearly how the presence of water or other liquid 
 having a free surface in the hold of a vessel affects her stability. 
 
 (30) 
 
 97. What resistances are experienced by a vessel when being 
 towed through water at a uniform speed ? What is the relative 
 importance of these resistances (i) at low speeds, (2) at high 
 speeds? (16) 
 
 98. Write down and explain Froude's law of " corresponding 
 speeds." 
 
 A certain vessel of 1000 tons displacement can be propelled by 
 engines of 11 50 I.H.P. at 14 knots. What will be the "' correspond- 
 ing speed/*' of an exactly similar vessel of 8000 tons displacement, 
 and what indicated horse-power is likely to be required to propel 
 the larger vessel at that corresponding speed ? (35) 
 
 99. What is meant by — 
 
 (i) a j//^ vessel; 
 (2) a steady vessel ? 
 What features of the design affect these qualities ? (25) 
 
 100. A rudder hung at its forward edge and entirely below water 
 is rectangular in shape, 14 feet deep, and 10 feet broad. Calculate 
 the diameter of steel rudder-head required, the maximum speed of 
 the vessel being 14 knots, and the greatest helm angle 35°. 
 
 Note. — Sin 35° = 0-574. (30) 
 
 loi. A hole I square foot in area is pierced in a vessel 12 feet 
 below her load water-line in wake of an empty compartment. 
 Calculate the capacity in tons per hour of the pumps required to 
 just keep this leak under. (25) 
 
292 Appendix. 
 
 BOOKS ON "THEORETICAL NAVAL ARCHITECTURE." 
 
 " Transactions of the Institution of Naval Architects." 
 ■'Transactions of the North-East Coast Institution of Engineers 
 
 and Shipbuilders." 
 " Transactions of the Institution of Engineers and Shipbuilders in 
 
 Scotland." 
 " Shipbuilding, Theoretical and Practical." By Prof. Rankine and 
 
 Mr. F. K. Barnes, M.I.N.A. 
 " Naval Science." Edited by Sir E. J. Reed, K.C.B., F.R.S. 
 " Theoretical Naval Architecture." By Mr. Samuel J. P. Thearle, 
 
 M.I.N.A. 
 " Yacht Architecture." By Mr. Dixon Kemp, Assoc. I.N. A. 
 •' Manual of Naval Architecture." By Sir W. H. White, K.C.B., 
 
 F.R.S. 
 " Stability of Ships." By Sir E. J. Reed, K.C.B., F.R.S. 
 "Text Book of Naval Architecture," for the use of Officers of the 
 
 Royal Navy. By Mr. J. J. Welch, M.I.N.A. 
 " Know your own Ship," for the use of ships' officers, etc. By Mr. 
 
 Thomas Walton. 
 " Naval Architects', Shipbuilders', and Marine Engineers' Pocket 
 
 Book." By Mr. Clement Mackrow, M.I.N.A. 
 " Resistance and Propulsion of Ships.'' By Mr. D. ^^^ Taylor, 
 
 M.I.N.A. 
 " Applied Mechanics " (Appendix on " Resistance and Propulsion 
 
 of Ships "). By Professor Cotterill, F.R.S. 
 " Encyclopsedia Britannica," article on " Shipbuilding." By Sir 
 Nathaniel Barnaby, K.C.B. 
 
STEAM-TUG " . 
 
 Displacement Table. 
 
 LBNGTH BETWEEN PERPENDICULARS 
 
 IIREADTH MOULDED 
 
 DIPTH MOULDED 
 
 ft. ins. 
 
 75 o WATER-LINES APART ... 
 
 14 6 I ORDINATES APART 
 
 8 3 : 
 
 1 Forward 5 5 
 DRAUGHT OF WATER (MOULDED) ■ Aft ... 6 2 
 
 (, Mean ... 5 9^ 
 
 v- 
 
 .1 
 1 
 
 •5 
 
 1 
 
 9 
 
 1 
 
 a 
 
 i 
 
 
 Appendage b 
 
 
 
 
 
 
 
 
 
 Watbr 
 
 •LINBS 
 
 3VV 
 
 
 
 Vertical Sections. 
 
 elow 6 W.I.. 
 
 6W.L. 
 
 5* W.L. 5 W 
 
 .L. 4 W.L. 
 
 .L. 
 
 2 W.L. 
 
 L.W.L. 
 
 
 
 H 
 €« 
 
 e " 
 
 
 to 
 
 •s 
 
 1 
 
 ■IS 
 
 S 
 
 1 : 
 
 
 IS 
 
 £■3 
 
 
 1 
 
 1^ 
 
 £ 
 
 SiMrSON's MULTIPLIEKS. 
 
 
 
 
 \ 
 
 a 
 
 I* 4 
 
 a 
 
 4 
 
 
 1 
 
 1 
 
 
 o'oo 
 
 o'oo 
 
 
 0*00 
 
 - 
 
 - 
 
 0-0 
 
 o'o 
 
 0-0 
 
 o-o 
 
 Q-O 
 
 0-0 
 
 0-0 
 
 o'o 
 
 o"o o'os 
 o'ao 
 
 0-05 
 
 0-oj 
 o"io 
 
 o'os 
 
 oos 
 0-20 
 
 0-05 
 
 0-05 
 o'os 
 
 0-05 
 
 o'sso 
 
 0-550 
 
 5 
 
 2-75 
 
 a 
 
 
 0-07 
 
 o-,8 
 
 
 I'X3 
 
 o-i 
 
 o'o3 
 
 o-ias 
 
 .00 
 
 l-o6 
 
 ,64 
 3a8 
 
 a-.a 
 
 0-& 
 1-13 
 
 3-28 rjS 
 S'Sa 
 
 S-52 ' /■& 
 1 3-64 
 
 7-28 
 
 8-88 
 
 8-88 
 
 2(m 
 2-60 
 
 10-40 
 
 23-055 
 
 92-220 
 
 4 
 
 368-88 
 
 3 
 
 
 0-50 
 
 i-oo 
 
 
 3-00 
 
 o'a 
 
 o-ao 
 
 0-9S 
 0-475 
 
 1-90 
 
 3-a8 
 
 '•35 
 
 3'5a5 
 
 470 333 
 13-32 
 
 6-66 : 4'OS 
 8-10 
 
 8-10 
 
 i8-20 
 
 9' 10 
 
 4'9' 
 4-92 
 
 9-84 
 
 51-820 
 
 103-640 
 
 3 
 
 310-92 
 
 4 
 
 
 .•48 
 
 5'9» 
 
 " 
 
 11-84 
 
 o-aa 
 
 1-30 
 
 l-laS 
 
 9-00 
 
 6-8o 
 
 13-60 
 
 6-45 
 
 17-ao S3' 
 21-28 
 
 2i-a8 
 13-30 
 
 11-80 
 
 23-60 
 
 6l4 
 24-96 
 
 24-96 
 
 6-40 
 6-40 
 
 25-60 
 
 78-81S 
 
 315260 
 
 2 
 
 630-52 
 
 5 
 
 
 »-S4 
 
 S-o8 
 14-40 
 
 
 S-o8 
 
 o-a3 
 
 ■•'7 
 
 i'8i 
 
 7-a4 
 
 5-Qf ; 10-10 
 10-10 ] 
 
 8-85 
 
 11 -80 «-6j- 
 26-60 
 
 13-80 
 
 13-80 
 
 7-00 
 a8"oo 
 
 14-00 
 
 ro2 
 
 7-02 
 
 14-04 
 
 96-180 
 
 192-360 
 
 I 1 19**36 
 
 3-«o 
 
 
 ax-04 
 
 o*a6 
 
 3-74 
 
 4'JO 
 a-as 
 
 18'oa 
 
 J-«f 
 11-90 
 
 a3-8o 
 
 b-tu 
 9-93 
 
 26-48 7 '70 
 28-40 
 
 28-40' ri5 
 14-50 
 
 39 '00 
 
 29-00 
 
 29-00 
 
 Tio 
 
 7-20 
 
 28-80 
 
 103-180 
 
 412-720 
 
 
 
 1505-43 ; 
 
 7 
 
 
 3'39 
 
 6-78 
 
 
 6-78 
 
 o-a9 
 
 1-97 
 
 3-9S 
 1-975 
 
 7-90 
 
 S35 
 10-70 
 
 10-70 
 
 9-30 
 
 ia'40 6'£? 
 27-28 
 
 13-64 7-<v 
 14-08 
 
 14-08 
 
 7'// 14-22 
 
 a8'44 
 
 710 
 7-10 
 
 14-20 
 
 98-875 
 
 197-75° 
 
 I 
 
 197-75 
 
 8 
 
 4 
 
 »-6« 
 
 10-64 
 
 
 ai-a8 
 
 0-31 
 
 3-30 
 
 '■7S 
 «-375 
 
 It'OO 
 
 4:00 
 8-00 
 
 >6-oo 
 
 7'a75 
 
 19-40 S'tS 
 33-80 
 
 23-80 
 8-64 
 
 6-5' 
 13-02 
 
 26*04 
 
 27-08 
 
 27-08 
 
 6-85 
 
 27-40 
 
 87-400 
 
 349-600 
 
 ' 
 
 699*20 
 
 9 
 
 
 1-50 
 
 3-00 
 
 
 9-00 
 
 0-33 
 
 0-99 
 
 ISO 
 0-75 
 
 3-00 
 
 4-50 
 
 4-50 
 
 J-lW 
 4-53 
 
 6-04 43' 
 17-28 
 
 J-Jt5 
 10-70 
 
 10-70 
 
 6-0/ 
 24-04 
 
 12-02 
 
 6-30 
 
 6-30 
 
 12-60 
 
 68 '100 
 
 136-200 
 
 3 
 
 408*60 
 
 10 
 
 ^ 
 
 0*58 
 
 a-3« 
 
 
 9-ae 
 
 0-37 
 
 0-86 
 
 o-t7 
 o-a35 
 
 1-88 
 
 0-7* 
 1-56 
 
 3-ia 
 
 I -6a 
 
 4-32 iSs 
 
 7-40 
 
 7-40! 2-SS 
 5-70 
 
 11-40 
 
 ^■00 ! 16-00 
 16-00 j 
 
 5-03 
 
 20-12 
 
 37 545 
 
 150*180 
 
 4 
 
 600*72 
 
 ! II 
 
 , 
 
 o'S7 
 
 o-sr 
 
 
 a-8s 
 
 0-38 
 
 o'aa 
 
 o-qy 
 o-oas 
 
 0-05 
 
 o-ijs' 
 0-10 
 
 0-05 
 
 005 
 0-075 
 
 0-05 o-ig- 
 o'ao 
 
 o-os 1 0-05 
 1 o-io 
 
 0-05 
 
 O'OS 
 0-20 
 
 0-05 
 
 o-<b- 
 
 0-65 
 
 o-6s 
 
 1-350 
 
 1-350 
 
 5 
 
 6*75 
 
 
 
 
 49"99 
 
 
 49"'9 
 ai-o4 
 
 a8-i5 
 
 13-78 
 
 60-97 
 
 
 87-87 
 
 a 
 
 1 
 
 105-67 
 it 
 
 128-74 l44-'o 
 4 » 
 
 514-96 a88-ao 
 3 » 
 
 155-36 '63-70 
 
 4 ' 
 
 195183 
 = 1951-83 
 
 
 1913-02 
 1505*43 
 
 407-59 
 
 90-485 
 5 
 
 174-54^ 
 4* 
 
 58-505 
 
 4 
 
 621-44 
 
 1 
 
 163-70 
 
 
 I5«-4«S 
 
 785-43 
 
 634*oa 
 
 576-40 
 
 621-44 
 
 o = 43'4 575 
 
 Displacement of appendage = 49-99 X (} X 7-1) -i- 35 
 = 3*38 tons 
 
 Centre of gravity of appen-l 38-15X7-1 
 dage abaft No. 6 ordi->= -— i^ =4f«et 
 nale I _ ^ 
 
 Centre of gravity of appen-l _ '3'78 _ ^. j-^^, 
 daje below No. 6 W.L. / 49-99 
 
 = 5-27 feet below L.W.L. 
 
 Volume of displacement of main solid in cubic feet = 1951-83 X(} X 7-1) X (J X 1) x 2 
 
 = '951-83 X ^^— = 3079-5 cubic feet 
 
 9 
 . '4-» 
 
 Displacement of main solid in tons (salt water) = 1951 -83 X—^ -4-35 
 
 = 87-98 tons 
 Centre of buoyancy of main solid below L.W.L. = ^^^,.3^ — = »'»' f"' 
 
 Centre of buoyancy of main solid abaft No. 6 ordinate = 12Z^-^" = 1 -48 feet 
 Summary. 
 
 Item. 
 
 Tons. 
 
 Below L.W.L. 
 
 Abaft No. 6. 
 
 C.B- 
 
 Moment. 
 
 C.B. Moment. 
 
 Main solid 
 
 .\ppendage below 6 W.L. 
 
 87-98 
 3-38 
 
 2-31 
 537 
 
 194-44 
 17-81 
 
 1-48 
 4-0 
 
 130-21 
 13-52 
 
 Displacement in tons =91*36 
 
 C.B- below L.W.L. = 2*32 feet 
 C.B. abaft 6 ordinate = 1*57 feet 
 
 91-36 
 
 )2i2-as 
 2-33 feet 
 
 )i4373 
 1-57 feet 
 
 ft. ins. 
 1 o 
 
 7-1 feet. 
 
 Metacbntres. 
 
 O lA 
 
 i-s 
 
 Z o 
 
 .5-3 
 
 Transverse. 
 
 o iJ 
 
 m B 
 
 "■- Cubes 
 of 
 
 "^ ; ordinates. 
 
 O 1 
 
 Functions. 
 
 of 
 
 cubes. 
 
 Longitudinal. 
 
 Functions 
 
 of 
 ordinates. 
 
 Multi- 
 pliers. 
 
 Functions 
 forC. ofG. 
 of water- 
 plane. 
 
 Multi- 
 pliers. 
 
 Functions 
 for moment 
 of inertia of 
 water-plane. 
 
 For particulars of the calculations carried out in this portion of the table, see Chapters III. and IV. 
 
Sheer Drawing of a Tug 75 Vx 14-6x8-3.^ Moulde d Depth. 
 
 " ■ e ' /a I Fo o T . 
 
 I CAI-I 
 
 After BoDY^ 
 
 — Principal Dimensions — 
 
 rr. iNt. 
 Length. BrrwEEN.PERPENOicuLARS .75_0 
 
 Breadth.. Moulded I4_6 
 
 Depth Mouuded 3-3 
 
 DuAUCHT... Moulded... Forward 5.S 
 
 .. DX Dg. Aft .6.2 
 
 D.5 D.e Mean 5. 91!^ 
 
 Half Breadth Plan. 
 
 PLATE I. 
 
INDEX 
 
 Algebraic expression for area of a 
 curvilinear figure, 14 
 
 Amsler's integrator, 178 
 
 Angles, measurement of, 86 
 
 Area of circle, 4 
 
 figure bounded by a plane curve 
 
 and two radii, 15 
 
 portion of a figure between two 
 
 consecutive ordinates, 12 
 
 rectangle, i 
 
 square, I 
 
 triangle, 2 
 
 trapezium, 3 
 
 trapezoid, 2 
 
 wetted surface, 80, 81 
 
 Atwood's formula for statical sta- 
 bility, 158 
 
 Barnes' method of calculating sta- 
 bility, 170 
 Beams, 209 
 Bilging a central compartment, 32 
 
 an end compartment, 153 
 
 Blom's mechanical method of cal- 
 culating stability, 169 
 BM, longitudinal, 133 
 
 , , approximations, 138 
 
 , transverse, 103 
 
 , , approximations, 107 
 
 Books on theoretical naval archi- 
 tecture, 292 
 Buoyancy, centre of, 61, 62 
 , strains due to unequal distri- 
 bution of weight and, 206 
 Butt fastenings, strength of, 199 
 Butt straps, treatment of Admiralty 
 and Lloyd's, 202 
 
 Calculation of weights, 188 
 Captain, stability of, 161 
 
 Centre of buoyancy, 61, 62 
 
 , approximate position, 
 
 63. 64 
 
 flotation, 94 
 
 gravity, 45 
 
 of an area bounded by a 
 
 curve and two radii, 58 
 
 of an area with respect to 
 
 an ordinate, $1, 55 
 
 of an area with respect to 
 
 the base, 56 
 
 ot a plane area by experi- 
 ment, 49 
 
 of a ship, calculation of, 
 
 19s 
 
 of outer bottom plating, 
 
 196 
 
 of solid bounded by a 
 
 curved surface and a plane of, 60 
 
 of solids, 50 
 
 Circle, area of, 4 
 
 Circular measure of angles, 86 
 
 Coefficient of fineness, displacement, 
 
 29, 30 
 
 , midship section, 27 
 
 , water-plane, 29 
 
 speed, 231 
 
 Combination table for stability, 175 
 Comparison, law of, 237 
 Conditions of equilibrium, 88 
 
 stable equilibrium, 92 
 
 Corresponding speeds, 236 
 Crank ship, 123 
 Cross-curves of stability, 178 
 Curve of areas of midship section, 
 
 27 ■ 
 
 displacement, 22 
 
 sectional areas, 19 
 
 stability, 160, 166 
 
 , calculation of, 168 
 
294 
 
 Index. 
 
 Curve of tons per inch immersion. 26 
 Curves, use of, in calculating 
 weights, 192 
 
 Difference in draught, salt and 
 river water, 30 
 
 Direct method of calculating sta- 
 bility, 177 
 
 Displacement, 21 
 
 , curve of, 22 
 
 of vessel out of the designed 
 
 trim, 140 
 
 sheet, 64 
 
 Draught aft remaining constant, 151 
 
 Draught, change of, due to different 
 density of water, 30 
 
 Dynamical stability, 183, 247 
 
 Eddy-making resistance, 224 
 Effective horse-power, 215 
 Equilibrium, conditions of, 88 
 
 , stable, conditions of, 92 
 
 Examination of the Science and Art 
 Department, questions, 262 
 
 , syllabus, 257 
 
 Experimental data as to strength of 
 
 plates and rivets, 201 
 Experiments on Greyhound, 216 
 to determine frictional resist- 
 ance, 221 
 
 Five-eight rule, 12 
 Framing, weight of, 192 
 Free water in a ship, 124 
 Frictional resistance, 221 
 Froude, Mr., experiments of, 216, 
 221 
 
 GM by experiment, 115 
 
 GM, values of, 121 
 
 Graphic method of calculating dis- 
 placement and position of C.B., 
 72 
 
 Creylwund, H.M.S., experiments 
 on, 216 
 
 Hogging strains, 208 
 Horse-power, 214 
 
 , effective, 215 
 
 , indicated, 218 
 
 Hull, weight of, 193 
 
 Inclining experiment, 115 
 Indicated horse-power, 218 
 
 Inertia, moment of, 97 
 Integrator, Amsler's, 178 
 Interference between bow and stern 
 
 series of transverse waves, 229 
 Iron, weight of, 35, 36 
 
 Launching, calculations for, 252 
 
 Lloyd's numbers for regulating 
 scantlings, 194 
 
 Lloyd's rule for diameter of rudder- 
 head, 251 
 
 Longitudinal bending strains, zo6 
 
 Longitudinal BM, 133 
 
 metacentre, 132 
 
 metacentric height, 133 
 
 Materials for shipbuilding, weight 
 
 of. 35 
 Mechanical method of calculating 
 
 stability, 169 
 Metacentre, longitudinal, 132 
 
 , transverse, 90 
 
 Metacentric diagram, 109 
 
 height by experiment, 115 
 
 , values of, 121 
 
 Moment of an area about a line, 
 
 S° .. . 
 
 Moment of inertia, 97 
 
 of curvilinear figure, loi 
 
 , approximation to, 
 
 102 
 
 Moment to change trim one inch, 
 
 143 
 
 , approximate, 144, 
 
 IS7 
 Monarch, stability of, 161 
 Moseley's formula for dynamical 
 
 stability, 184 
 
 Normand's approximate formula 
 
 for longitudinal BM, 144 
 , for position ofC.B., 
 
 63. 249 
 
 Outer bottom plating, weight of, 
 192 
 
 Panting, 211 
 
 Planimeter, 77 
 
 Preliminary table for stability, 174 
 
 Prismatic coefficient of fineness, 30 
 
 Propulsive coefficient, 218 
 
Index. 
 
 295 
 
 Questions set in examinations of 
 the Science and Art Department, 
 262 
 
 Racking strains, 210 
 Rectangle, area of, i 
 Residuary resistance, 229 
 Resistance, 220 
 Rolling, strains due to, 210 
 Rudder-head, strength of, 250 
 
 Sagging strains, 208 
 
 Science and Art Department exami- 
 nation, questions, 262 
 
 , syllabus, 257 
 
 Shaft brackets, form of, 225 
 
 Sheer drawing, 64 
 
 Shift of C.G. of a figure due to 
 shift of a portion, 96 
 
 Simpson's first rule, 6 
 
 , approximate proof, 
 
 g 
 
 , proof, 24s 
 
 second rule, 10 
 
 , proof, 246 
 
 Sinkage due to bilging a central 
 compartment, 32 
 
 Speed, coefficients of, 231 
 
 Stability, curves of, specimen, 166 
 
 -, dynamical, 183 
 
 , , Moseley's formula, 184 
 
 , statical, 89 
 
 , , at large angles, 158 
 
 , , cross-curves of, 178 
 
 ■ , , curve of, 160 
 
 , , calculations for, 168 
 
 , , definition, 89 
 
 Steadiness, 123 
 
 Steel, weight of, 35, 36 
 
 Stiffness, 123 
 
 Strains experienced by ships, 205 
 
 Strength of butt fastenings, 199 
 
 Subdivided intervals, 13 
 
 Submerged body, resistance of, 231 
 
 Syllabus of examinations of Science 
 and Art Department, 257 
 
 Tangent to curve of centres of 
 
 buoyancy, I14 
 curve of stability at the origin, 
 
 166 
 Tensile tests for steel plates. 
 
 Admiralty, 203 
 
 , Lloyd's, 203 
 
 Timber, weight of, 35 
 Tons per inch immersion, 26 
 Transverse BM, 103 
 
 metacentre, 90 
 
 strains on ships, 210 
 
 Trapezium, area of, 3 
 
 , C.G. of, 48 
 
 Trapezoidal rule, 5 
 Trapezoid, area of, 2 
 Triangle, area of, 2 
 
 , C.G. of, 48 
 
 Trigonometry, 85 
 
 Trim, change of, 141 
 
 , moment to change, 143 
 
 Velocity of inflow of water, 35 
 Volume of pyramid, 17 
 rectangular block, 17 
 
 solid bounded by a curved 
 
 surface, 18 
 
 sphere, 17 
 
 Water, free, effect on stability, 
 
 124 
 Wave-making resistance, 225 
 Weight, effect on trim due to adding, 
 
 147. 149 
 
 of hull, 193 
 
 of materials, 35 
 
 of outer bottom plating, 192 
 
 steel angles, 189 
 
 Wetted surface, area of, 80 
 Wood, weight of, 35 
 
 PRINTED BY WILHAM CLOWES AND SONS, LIMITED, LONDON AND BECCLES,