4\^^Gc l^ B43 Cornell University Library TG 327.B43 A practical treatise on segmental and el 3 1924 004 023 085 'K. Cornell University Library The original of this book is in the Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924090409313 A PR A CTICA L TREA TISE ON SEGMENTAL AND ELLIPTICAL OBLIQUE OR SKEW ARCHES SETTING FORTH THE PRINCIPLES AND DETAILS OF CONSTRUCTION IN CLEAR AND SIMPLE TERMS BY GEORGE JOSEPH BELL COUNTY SURVEYOR AND BRIDGE MASTER OF CUMBERLAND. CARLISLE : Chas. Thurnam and Sons, English Street. LONDON : E. and F. N. Spon, 125 Strand. NEW YORK : Spon and Chamberlain, 12 Cortlandt Street. 18 96. Printed hy Chas. 7'hurnam mid Sons, 11 English Street, Carlisle. T THE ME M R Y OF THE LATE ROBEET BELL (CIVIL EXaiNEER AND SURVEYOR) OF THE NOOK, IRTHINGTON, CARLISLE, THIS ESSAY IS DEDICATED BY HIS NEPHEW, GEORGE JOSEPH BELL, AS A SLIGHT TOKEN OF SINCERE AND AFFECTIONATE GRATITUDE FOR THE PATERNAL REARING AND PRACTICAL TRAINING HE SO GENEROUSLY BESTOWED ON THE AUTHOR. PREFACE. This Essay has been written at the urgent request of a few of the writer's personal friends, with a view to simplify the Oblique Bridge Problem. The request was chiefly made on account of a recent strongly expressed opinion as to the difficulty of constructing oblique arches, and also on account of the numerous accidents (some fatal) occurring on several of the County Bridges in con- sequence of so many of them being situated at the foot of steep hills, and built at right angles to the stream, irrespective of the line of roadway leading on to and off these bridges. The writer takes this opportunity of stating that he is under a deep obligation to the late Mr. Peter Nicholson, of Carlisle, for the first light he received in this practical " stone-squaring " art. His work on the Oblique Arch is perhaps for the student the most exhaustive to this day. It is not, therefore, the writer's intention in this humble attempt to travel over ground which has been much more ably investigated by the above able Engineer, and by Mr. Buck, Engineer to the London and North Western Eailway Company, Crewe, and others whom he cannot pretend to equal ; his object being to place in the hands of engineers, architects, builders, contractors, and especially working masons, in a tabulated form, the result of his calculations for 72 oblique bridges, designed to suit almost any situation, in spans ranging from 10 feet to 50 feet advancing by 5 feet, and from 35° to 70° advancing by 5°, also decimal multipliers for finding the corresponding dimensions for any span in the same degree of obliquity, thus saving much valuable time to all those engaged in practical constructions of this kind. Every theoretic and adjusted dimension necessary for enabling the draughts- man to make his projection, and thus get his templates for drawing the plan, elevation, and section, or for building the bridge from certain specified dimensions, is given in the Tables of this Essay. A Table of Theoretic Decimal Multipliers is also given for proving the theoretic dimensions, and lettered to correspond with the theoretic diagram. A Table of Multipliers is also given for finding the radius of the cylinder, and a rule for finding the radii of the end and coursing joint arcs, also multipliers for finding the dimensions of every other part of the bridge. Preface. An example in projection is given to show the mode o£ producing the templates for making the drawings of an oblique bridge, also the mode of obtaining the templates for the intradosal and extradosal angles correctly, the twisting rules, and all other templates necessary for carrying out aud completing the work. Plans, elevations, and sections of oblique bridges are given with a specification, form of tender, agreement, and bond ; also an example of a skew gusset or V-shaped piece of masonry, built obliquely against an old bridge to illustrate an inexpensive method of doing away with all dangerous corners at the ends of existing bridges. The object of the writer in making this attempt is to place reliable data, which can be readily understood, in the hands of practical workmen, so as to render the building of oblique arches as simple as the building of an ordinary square arch. No attempt is made to instr-uct highly educated engineers, to whose hands this problem has hitherto been almost exclusively confined, but this Essay has been solely written for the purpose of making the subject intelligible to the average working mason. G. J. B. Caelisle, June 1st, 1896. CONTENTS. Page I. Projection - 1 II. How to make the Drawings - 4 III. Templates and their Construction 7 IV. How to construct and use Templates in Dress- ing the Arch Stones - 12 V. Setting Out and Erecting the Bridge - 21 VI. To Construct the Template, Fig. 8, Plate 1, without using a radius ; also Explanation of Fig. 4, Plate 1, and Diagram, Plate 7 27 Calculations, Examples, Tables, &c. The Ellipse, Square, and Oblique - - 34 Directions for finding the Thickness of Courses and the Length of Checks ; and for Increasing or Diminishing the Widths of Bridges from the widths given in the Tables 35 To find the Radius of any Circle 37 Trigonometrical Functions - 39 A-djusted Intradosal and Extradosal Angles - - 40 Theoretic Intradosal and Extradosal Templates, Angle of Obliquity 50°, and 20 feet span - - 41 Adjusted Intradosal and Extradosal Templates, Angle of Obliquity 50°, and 20 feet span - 43 The Adjusted Twisting Rules by Proportion 44 Example 1, 50° and 20 feet Span - 45 Example lA, 50° and 20 feet span 48 Example to Chord 1 in 35° - 51 Example to Chord 1 in 40° 54 Contents. Example to Chord 1 in 45° Page 57 Example to Chord 1 in 50° 60 Example to Chord 1 in 55° 64 Example to Chord 1 in 60° 67 Example to Chord 1 in 65° 70 Example to Chord 1 in 70° 73 Theoretic Angles as shewn on the Diagram 76 Natural Sines, Cosines, Tangents, &c. - 78 Tables of Theoretic Multipliers, &c. 80 Table of Adjusted Multipliers for Twisting Rules 90 Adjusted Dimensions of 72 Oblique Bridges of different Spans and Angles of Obliquity 91 Table of Multipliers for finding the Lengths of Arcs and the Radius of Circles to Chord 1. - 94 Table for reducing Decimals of a Foot to Inches and Fractions of an Inch 95 Table of Fractions of an Inch with their Equivalent Decimals 96 Table of Gradients and Numbers often Calculations, with their Logarithms used in 97 Specification 98 Tender 105 Agreement 106 Bond 108 Appendix. Elliptical Skew Arches 109 Figure of Tangent Reading Tacheometer 117 The Theodolite and How to Read it 118 Full Sized Tangent Scale 120 The Tacheometer and How to use it - 121 Oblique Bridges Made Easy. CHAPTER I. PROJECTION. Before the drawings for an oblique bridgfe can be Proportions of the _ _ , . . segment, made, it is first necessary to make a projection of the segment of the cyhnder proposed to be used as the arch for the bridge. In the example given, and in all the tables of dimensions, the versed sine or rise of the arch is given as one-fourth of the direct or square span. When the segment is projected at the various angles for shewing the bridge in plan, elevation, and section, templates are got by the several mutual intersections of the lines for producing those beautifully delicate curves which give such a pleasing effect to the eye in a neatly finished drawing of an oblique arch. DIRECTIONS FOR MAKING THE PROJECTION. Draw the lines AGF and CDE parallel to each other, ^'^"Jction "'^ at a distance apart equal to the square span of the proposed bridge. In this example the square span is 20 feet. Make AGF and CDE equal to the obliquity of the bridge (CE in Fig. 4 and in the tables) plus the axial length (EH Fig. 4 and in the tables), that is, 16.781 + 35.381 = 52.162, which is the adjusted length of the line AGF or CDE in the projection. FIG. lA.— PLATE 1. On AC, as chord or span with radius of cylinder (12.5) '^° ''°o-eo'tion *^ describe the arc ABC, Divide this arc into 20 equal parts, and draw lines from each of these divisions parallel to AGF and CDE Oblique Bridges Made Easy. To construct tiie throuffli EF. From C set off the obliquity of the bridge, projection. o i ./ o ' making CD (CE in the Tables) = 16. 781 feet, and divide CD into 20 equal parts, and set up perpendiculars from each of these 20 divisions through the line AG. From A draw a line to D, and produce it until it meets a line let fall at right angles on it in the point H from F. The angle CD A (50°) is the angle of obliquity of the bridge, and DA is the length of the impost. Templates for the Divido the axial length DE or GF (EH in the Tables) heading joints and n-, ixi ti coursing joints on also into 20 equal parts, and set up perpendiculars from ^^^°' these divisions through GF. The mutual interse(;tions along either side of the oblique span line A ^ D give points through which to draw the template for the cross or head- ing joint lines on the plan. The points of intersection in the axial length, from D to F, give points through which to draw the template for the spiral or coursing joints on the plan. Template for the For the clevatiou of the face two templates are elevation of the head- . ■• « ,, i-i •/■ n •• ing joints. required, one lor the vertical projection of the cross or heading joints Aa;D, and the other for the vertical projection of the spiral bed or coursing joints DF. On the centre of the oblique span line A.xD describe the arc ABC, and again divide this arc into 20 equal parts, and parallel to A ,2; DH draw lines through these divisions of the arc and produce them beyond H, and set up per- pendiculars through A a; D from the points of intersections, giving the curve of the cross or heading joints A.xT>, and the points of intersection between these perpendiculars and the lines drawn through the division of the arc and parallel to A « D are points through which to draw this template for the elevation of the face and the cross or heading joints. Template for the From DH draw perpendiculars to the points of coursing joints, intersection along the curved hue DF, and the points of intersection between these perpendiculars and the lines drawn parallel to A a; D through the divisions in the arc are points through which to draw the template standing on DH as a base, and which is the template for the coursing joints in the elevation of the arch. Projection. For the vertical section shewing the arch cut throusrh Templates for ,, xj_ij T •! n T • drawing the sectional the crown two templates are also required, one tor shewing eievjui.m through the cross or heading joints in this position, which is the curve standing on AD, foreshortened on the obhquity with AG as base, and got from the mutual intersections of the lines drawn perpendicular from C D through AG, and lines drawn through divisions in the arc in the centre of GF, and parallel thereto. The other GF is the elevation of the curve DF, and is got from the mutual intersections of the perpendiculars set up from the divisions in DE, and from the lines drawn through the divisions in the arc described in the centre of GF, and parallel thereto. Oblique Bridges Made Easy. CHAPTER II. HOW TO MAKE THE DRAWINGS. How to make the Having now constructed the six templates required templates fur the „ , . , ■■ ■■ . , . -, i;i drawings. tor drawing the plan, elevation, and section through the crown of the arch, trace the same and gum the tracings on to a piece of thin cardboard, and when dry cut them out. A more satisfactory way is to send the tracings to a good firm of mathematical instrument makers or dealers, and have them carefully cut out in thin vulcanite. Much clearer and neater lines can be drawn with these, and the templates are permanent, whereas the edges of the card- board templates become soft and unreliable after having been used twice or thrice.* How to make the Having fixcd the paper on the board, draw two parallel lines at a suitable distance from the lower edge of the paper at a distance apart equal to the exact outside width of the bridge, then set off on the lower line the oblique span. In this example it is 26.108 feet. If it is a left-hand skew, then from the point at the left-hand end of the oblique span measure olf the obliquity of the bridge DC (which is 1G.781 in this example), and from C erect a perpendicular to meet the upper line in A, and from this point draw a line to D. The angle CDA is the angle of obliquity of the bridge {d). In this example the square span and the extreme outside width of the bridge are each 20. feet; therefore the length of the impost and the oblique span are equal, namely 26.108 feet. To prove this, take either end of the oblique span, as the case may be, for a centre, say D in the projection, and the tabulated length of the impost as radius ; the arc described will pass throuo'h the point of intersection at A made by the perpendicular set up from the obliquity and the upper parallel line. Proceed in like manner if it is a right-hand skew, but from the other end of the oblique span. * Messrs. Elliott Brothers, of 101 St. Martin's Lane, London, cut Standard temiJatcs, in metal or other material, to tracings furnished. How to Make the Drawings. Draw the imposts and divide them into the given To draw the heading number of checks on each. Produce the hne of the impost coursbg'j"o1nUines (in pencil) so as to mark off the length of the checks from ™ "'** i^'^"" which to rule in the coursing joints. From the divisions of the checks on the imposts rule the cross or heading- joints from one impost to the other with the template A a; D. Now move the paper on the board so that the lines of the impost are in a horizontal position, and with the template EDF rule the coursing joint lines, which are portions of the curve DF, moving DE along the T square from division to division with the point F or D in contact with the divisions of the checks, till the whole of the lines are ruled in, and which will now represent the coursing joints on the plan and are a representation of the coursing joints as ruled on the lagging of the centres. Replace the drawing in its original position on the To draw the elevation 1 A ■ ^^ T 1 11 "f '''® bridge. board. At a suitable distance above the plan rule a horizontal line for the springer line, and from the checks on the plan set up the foreshortened checks in the elevation with a set square, and on the springer line, which must be produced, mark out carefully the foreshortened divisions of the checks as set up from the plan. Draw the arc of the oblique face between the two points set up from the oblique span on the plan with the template standing on the base AD, and with the template standing on DH with the point D or H, as it is a right or left-hand skew, at the fore- shortened divisions of the checks rule the coursing joint lines in the elevation. These are the coursing joint lines on the plan and are parts of the curve DF set up in foreshortened elevation. Rule in the cross or heading joints with the template AD. Find the centre of the oblique arc, and from this point transfer with the compasses the divisions shewn in the face which represent the thicknesses of the arch stones as shewn in the face of the arch and are obtained by the intersection of the coursing joint lines with the face of the oblique arc. As all the joints in the face of an oblique arch radiate to a point considerably below the natural centre, and as this distance varies in accordance with the span, the angle of obliquity, and the Oblique Bridges Made Easy. The point of anofle of extrados, add this distance to the radius, and from converirence of the ° . . ..-,. •if- face lines in the that point (P) as a Centre rule m the joint hnes ni the face, through the points of intersection and the points transferred to the other lialf of the obhque arc, and set off on the face joints the thickness of the arch (in this example it is two feet), and through these points draw the outer circle of the face of the arch. To draw the sectional Qu a horizontal line lay down a plan of the impost, elevation through n p • • i • the axis. and ffom this plan set up the abutment in elevation. Draw a horizontal springer line as before, and on it set up the natural divisions of the checks from the plan, and from each end of the abutment measure off half the length of the obliquity = 8.390, set up a perpendicular, and in this space rule in half the template AG. If it is a left-hand skew the half arc will shew as hanging over to the left ; if it is a right-hand skew it will hang over to the right. From each of the checks on the impost and the divisions of the produced springer line move the template GF along the T square and rule in the coursing joints, and with the template standing on AG rule in the cross or heading joints. Templates and their Construction. CHAPTER III. TEMPLATES AND THEIR OONSTRUOTION. FIG. lA, PLATE 1. The first template required by the workman is one for To construct the ^ -" , template f(ir the angle the angle of obliquity, to enable him to set out correctly the of obliquity. acute- and obtuse angles at the ends of the imposts or abutments (see Fig. 1, Plate 1). This template can be con- structed in the following simple manner : — Draw two lines parallel to each other and the same distance apart as the external width of the bridge. On the lower line measure off the obliquity of the bridge (CE in the tables and in Fig. 4), and set up a perpendicular through the upper line. From either end of the obliquity (D in the projection) connect this point of intersection with the other end of the obliquity at A, and the template for the angle of obliquity or d in the formula is obtained ; also the length of the impost DA. The angle can be set off by the protractor equally as well and perhaps more readily. The second template is for the sprinsfer angle, and given To construct the T , ,. 1-11 template for the the chord or span and the radius, this template can be con- springer angle, structed by any workman (see Fig. 2, Plate l). In all bridges of these proportions the springer angle is the same, namely, 36° 52' 12", and may be found in two ways. By trigonometry {a), subtract the rise of the arch from the radius, thus, 12.5 — 5. = 7.5. This is the portion of the radius below, and at right angles to the chord or span AC, Fig. 3, Plate 1. Then §^ = sine of angle GCF— thus, ^5 = .600. Find this number in a table of natural sines, and the nearest number thereto is .5999549, and stands opposite 36° 52'. Again (b), pp = tangent of angle GCF thus, ^^=.750. Find this number in a table of natural tangents, and the nearest number thereto is .7499119, and this also stands opposite 36° 52'. Oblique Bridges Made Easy. To find the sine or To find the acute ano'le of a right-angled triangle, tangent of the acute o o o >-^ ^"^^"'tri'Jn^i'e"''"^'^'^ ^^^ perpendicukr divided by the hypothenuse gives the natural sine of angle opposite the perpendicular thus, §J = sine GCF, and the base divided by the hypothenuse gives the natural sine of the angle opposite base — thus, ^ = sine GFC ; or perpendicular divided by base gives natural tangent of angle opposite perpendicular — thus, ^ = tangent FCG, and base divided by perpendicular gives the natural tangent opposite base — thus, j^ = tangent CFG. See page 37. Template for the The third template is part of the arc of the circle, fixed to a piece of wood which radiates to the centre of the cylinder, of which this arc is a segment. See Fig. 3a. Templates for the The fourtli and fifth templates are for the intradosal skewbacks or checks. ^^^ extradosal angles. The theoretic angle of the intrados is the angle CDE or HEI, Fig. 4, and is found by dividing the obliquity of the bridge, CE in Fig. 4, by the length of the arc developed in the line CD — thus, ^ = 33^824 = .7238249, or HI-^EH, which is the tangent of the angle of intrados CDE or HEI. To find the On referring to a table of tangents, it will be found ThtlnSadrafangiT ^^^^ coiucides vcry nearly with the tangent of 35° 54'. See Figs. 4 and 5, Plate 1. A.nd the extradosal angle is found from the angle of intrados, thus : — ^Add the thickness of the arch to the radius for radius of the outer rim of the arch — thus, 12.5 + 2 = 14.5, and divide this by the radius 14 5 jg-E = 1.160, which being multiplied by the tangent of the intradosal angle gives the tangent of the angle of the extrados — thus, 1.160 x .7238793 = .8396999 = the tangent of 40° 1'; or it can be found by proportion thus — As the intradosal radius is to the extradosal radius so is the sine of the intradosal angle to a point produced through which to draw the extradosal angle. As 12.5 : 14.5 :: .5863724 (DH) : .6801919 (DF). See Fig. 16, Plate 1, or Diagram at the end. Consequently the theoretic angle of difference is 4° 7'. This angle of difference performs an important part, as will be seen, in subsequent calculations for working Templates and their Construction. 9 out the several parts of the proposed structure. See Tables of Theoretic and Adjusted Angles, pages 76 and 91. These templates can be accurately constructed so as to need no further adjustment, in the following simple manner :— Draw the straight line BO, Fig. 5, and let its length be equal to the length of one check on the impost. In this example the check is 2. 1 75. Upon BC as a diameter describe the semicircle BAG, with half BC or 1.0875 as radius. With either end B or as a centre, and with the thickness of a course of stones as radius 1.1924, cut the semicircle in the point A, join AB and AC, and ABC is the adjusted angle of intrados, or the template for the front of the check, which is a right-angled triangle. Let fall a perpendicular from A to D, and DA will be found by measurement to be .997. DEmaybe found thus — As the radius is to the radius added to the thickness so is DA to DE thus— As 12.5 : 14.5 : : .997 (DA) : 1.15652 (DE); see Fig. 5, Plate 1. Then measure off on the line DA, produced through A, DE = 1.15652 and join EB and EC. The angle EBC is the adjusted angle of extrados. This forms the template for the back of the check. The sine of any intradosal angle may be found by dividing the thickness of a course of stones (which is the perpendicular) by the length of the check (which is the hypothenuse of a right- angled triangle). See pages 41, 43, and 91. The sixth and seventh templates are the Twisting Rules, and these can be got from the intradosal and extradosal templates. The length of the line B A in the last figure is the distance which the rules must be placed apart at the soffit, and the length of the line BE is the distance they must be placed apart at the extrados, on the length of a check. Both rules must be in length equal to the thickness of the cylinder or depth of the arch stones — in this example it is two feet. One rule should be about 3 inches wide with parallel edges, the other must be the same width at one end, but the other end must be as much broader as the difference in the length of the lines CA and CE, which is 1.3288— 1.1924 = .1364, consequently the broader end would be 3. + .1364 = 3,1364 inches. To construct the adjusted intradosal and extradosal angles. To construct the twistinaf rules. 10 Oblique Bridges Made Easy. The two rules are framed together at the distances given above, and so that their upper edges are in the same plane, in order to make them " wind " when the parallel rule is on the apex of the check AE, Fig. 14, while the rule with the broad end is at the bottom of the check along the natural springer angle (36° 52' 12") BF, pages 41 and 43. Other methods are given in a subsequent chapter for finding the position and dimensions of these rules on page 40. theendjoinraroand '^^^ eighth and ninth templates are for the arc of the in^'^the^ed^rndToffit ^®^ ^^ coursiug joiuts, and the arc for the cross or heading of the arch stones, joints. It is shcwn in the next chapter how these two templates can be constructed from the preceding templates without calculation. To find the radius of an arc that will fit on the coursing joint line on the centres, and which is the arc produced by cutting out the skewback along the line AB of the intradosal template ABC (Fig. 14), square the length of the oblique arc ED (Fig. 4), which is the hypothenuse of a right-angled triangle, and multiply the product by the radius of the cylinder, and divide this product by CE (obliquity of bridge). Thus— ED^ X radius of cylinder 28.618^ x 12.5 „/> or. j- 5 = „- = 36.354 = radius of CE' (16.781)' the theoretic spiral arc. Or ■^^'^^'^ ^2.5 36.354 = radlus of (sine of intradosal angle /3)' (.5863724)^ the theoretic spiral arc. End Joint Arc. ED" X radius of cylinder = (28.618)'' x 12.5 ,„^,„ 2 5 =19.049 = radius ot CD (23.182) the theoretic end joint arc. Or radius _ 12.5 ~/ '■ i • , 1 ■, 1 „\2 77T7[7!777T- =19.050 = radius of the (cosine of intradosal angle|3) (.8100416) theoretic end joint arc. (This is given in subsequent calculations on pages 38 and JfY.) Templates and their Construction. 11 It will be seen the ninth template is a section of the Ternpiate for the ■■• . ■ . Rnd section or tlie end of the arch stones, and can be got by fitting a thin arch stones. piece of wood or sheet iron on the end of the check, which gives the end section. The soffit end of this template is equal to CA, Fig. 5, and the extradosal end is equal to CE ; and also, the arc which will fit square across the soffit of the arch stones, and is a segment of a circle, the theoretic diameter of which is 19.050. There should be two of these templates (see Fig. 9, Plate 1). The reason for this is explained in Chapter V., page 25. 1^ Oblique Bridges Made Easy. CHAPTER IV. springers, HOW TO CONSTRUCT AND USE TEMPLATES IN DRESSING THE ARCH STONES. How to work the Having now constructed all the requisite templates, the workman must proceed as follows (except in bridges built at an angle of 70°, where the checks are over three feet six inches long) : — The springer stones should be long enough to allow of two checks being cut out of each stone, and these should be laid down in a convenient place for working. The bed of the springers must be worked first ; then square both ends off the bed and from the front, then apply the template for the springer angle (Fig. 2, Plate 1) on the squared ends of the stone, and mark the angle ABC of the springer template on the squared ends, keeping the template a sufficient distance from the front of the stone so as to allow of the full intradosal or soffit curve of the checks to be cut out on the stone as shewn in Fig. 10, Plate 1. Apply the template (Fig. 3a, Plate l), whose arc is drawn by the radius of the cylinder, on the ends of the stone, with the inner edge of its radiating blade truly along the angle of the springer on the line BC, and mark both ends of the stone by the arc of this template. This gives the natural curve to the checks as if they were to fit on the centres as for a square arch. Up to this point the springer stones represent the springers for a square bridge, with the springers and first course of arch stones all in one. See Fig. 10, Plate 1. Dress this curve most carefully the full length of the stone, so that it may fit the lagging accurately when placed in position on the abutment. The practical workman will see at once that this soffit curve can be worked with the square off either or both squared ends by following the curved arc, and keeping the blade of the square parallel with the springer line. Or he can work the curve with a Templates, their Construction and Use, 13 straight-edge between the arcs marked on both ends, keeping the straight-edge parallel with the springer line. On this dressed curve apply the template for the intradosal angle (Fig. 5, Plate 1), placing the hypothenuse or longest side (BC) along the springer line, and mark out the lines of the soffit checks on this curved surface. See Figs. 11 and 14. Work a true draft at the back of the checks the full ^ow to work the springers. length of the stone, vertically above 1.60 feet from the front of the springer, and exactly two feet up along the springer angle. See Fig. 2, Plate 1. Then gauge two feet from the front apex of the check, across the check to the apex at the back, and with the extradosal template resting on the draft (see dotted line behind checks. Fig. 14) mark out the lines for the back of the check. Now work off all superfluous stone outside of these lines marked by the intradosal and extradosal templates, using a straight- edge held parallel with the soffit, and in contact at both ends with a line along the apex AE (Fig. 14) and the springer lines at the bottom of the checks BF. Test this check by reversing a similar template to the one which gave the curved arc at the ends of the springer, but with a shortened arc DA (Fig. 12), so that its arc will fit into the soffit curve, while the blade, or radiating arm AE, rests along the apex AE (Fig. 14). If the stone has been truly worked, this template should be in perfect contact with the stone while in this position. See Figs. 12 & 14, Plate 1. When the twisting rules are framed apart the exact Hmv to work the " ^ checks by twisting length of the check, the bed can also be accurately worked rules. by marking the springer angle on the squared ends of the springer stone, and the angle of the intradosal template on the soffit curve (see Fig. 11), and applying the broader twisting rule along the line of the springer angle BF (Fig. 14) at the end, and the parallel rule on the apex of the bed line AE marked on the soffit curve (Fig. 11, Plate 1 ). Work off the stone till the upper surfaces of the rules " wind " or are in the same plane, with the broader rule resting on the springer angle at the bottom of the check at BF, and the parallel rule resting on the line 14 Oblique Bridges Made Easy. marked on the soffit curve at AE by the intradosal angle along the Ime BA Fig. 11, or BA Fig. 14. While the framed twisting rules are in this position, mark a line across the bed of the check, along the outside of the parallel rule, and from this line down to the springer line BC (Fig. 2) at the bottom of the other end of the check. Work off the stone with a straight-edge held parallel to the soffit. This line marked along the outside of the parallel rule is the same as if a line had been drawn across the bed from the apex of the intradosal to the apex of the extradosal angle ; or this line can be got, by gauging across the bed off the squared end of the springer, and the end of the check can be further tested by applying the template of the end section of the arch stones (Fig. 9, Plate 1), which ought to fit the end of the check accurately. Huw the workman Having workcd the first check with the utmost care, can make his own _ ° ' twisting n,ies. from tliis Worked check the workman can proceed to make the twisting rules for himself, by which he can work all the remaining checks and the beds of all the ai'ch stones. Place on the apex of the dressed skewback or check a parallel strip of wood exactly two feet long, and say three inches broad, with its outer face exactly flush with the apex of both the intradosal and extradosal angles AE (Fig. 1 4). Place another strip of wood at the bottom of the check BF (Fig. 14), with its outer face exactly flush with the end of the stone, along the springer angle BC (Fig. 10), or BF (Fig. 14). Let this strip of wood be exactly two feet long also, and three inches broad at the soffit end or B, and say five inches broad at the extradosal end C (Fig. 10), or F (Fig. 14). The exact extra width of this end of the rule is the difference between the length of CA and CE (Fig. 5) in the intradosal and extradosal angles. "Wind" these two strips of wood in the above positions, and plane the broader end of this twisting rule down till it winds, or is in the same plane with the parallel rule on the apex of the check along AE (Fig. 14), taking care not to reduce the three-inch end of this rule at B. Now frame these two rules together, as they have been placed, and correct their Templates, their Construction and Use. 15 distances apart, both at their intradosal and extradosal extremities, by the lengths of BA and BE (Fig. 5) of the intradosal and extradosal templates. Also test the width of the broad end of the twisting rule by the difference between the lengths of CA and CE (Fig. 5). The width of this rule at the broad end will be three inches added to the diflFerence between CA and CE. See pages 41 and 43. From this check the workman can get other three templates. The radiating blade of one template must fit on the end joint along the apex AE (Fig. 1 4), and the arc must fit the soffit arris of the bed or coursing joint arc AB ; and in the other, the radiating blade must fit the bed along the apex AE, and the arc must fit the end joint arc AC. (See Fig. 14, Plate 1.) The third is for a section of the end of the arch stone, and must be cut to tit the end joint arc at the soffit, and the thickness and length of the check or arch stone, less xV of an inch to allow for a bed of mortar. (See Fig. 9.) With these templates so simply produced, and at the same time so accurate, all the stones for any oblique bridge can be dressed, and when they are placed in position they will be found to fit with the greatest accuracy. In getting the arch stones, except for 70°, they ought as a rule, if possible, to be twice the length of the back bed of the check, that is twice the length of BE, or the base of the extradosal template, so as to break the joints in the soffit of the arch properly. As the bed of the voussoir or arch stone is its greatest surface, the best way of dressing these stones is to begin by working the bed first. This is done by the use of the twisting rules, which should be framed together as above described. Before commencing to work the twisted bed, take the template, Fig. 8, Plate 1, and if the stone is for a bridge with a left-hand skew, place this template at the extreme right-hand end of the stone, but so that the end or radiating blade of the template can be fully marked along the entire depth of the bed, because when the end of the stone is worked for the cross or heading joint it will be, what is called by workmen, " in square " off the twisted bed ; or How the workman can make the other necessary templates for himself. Size of arch stones. How to work the arch stones. 16 Cbliqiie Bridges Made Easy. always place this template close to the end of the stone next to the parallel rule in the framed twisting rules, whether the parallel rule be at the right hand or the left, and mark the full size of this template accurately on the bed of this rough stone. Dress off the stone along this curved line, and work a narrow draft on the soffit, along this arc, and at the end next the parallel rule work another draft up the radiating line. Leave the other end simply well marked till the other bed is worked. How to use the Jf the bridge is a right-hand skew the twisting: rule twisting rules on the . . rough arch stones, must be uscd with the broader rule to the right hand, and if the bridge is skewed to the left frame the broader rule at the left hand. Now place the framed twisting rules on the centre of the bed of the stone which is to be worked, with their intradosal ends exactly fair on the curved draft line marked out by the other template, and along the undressed soffit rule a straight line from the end of the stone next the broader of the twisting rules, beginning the line at nothing at this end of the stone, and let the line fall towards the other end of the stone, till it is as much below the end of the parallel rule as the difference in the breadth of the ends of the rule added to the distance this line is below the soffit end of the broader rule. Reverse the twisting rules, and in like manner mark the back of the stone. The workman will see at once that this saves both time and labour, and prevents any unneces- sary stone from being dressed off the bed, and also allows the stones to be used to their utmost capacity, which is important. Now sink drafts under each rule down to these lines until the upper edges of the twisting rules " wind," or are in one plane, when it will be found that the rules are resting on the surface of the intended winding bed, and if the rules are reversed they will be found to fit equally well from either side of the stone. The workman must next proceed to dress off the superfluous stone on the other parts of the bed, until a straight-edge applied parallel to the soffit, from one draft to the other, bears equally on every part of the twisted bed, when it will be found that Templates, tkeir Construction and Use. 17 the framed twisting rules will be in contact with every part of the dressed surface, and the upper edges of the rules will be in perfect " winding," or in the same plane on every part of this twisted bed. Having dressed the first twisted bed, again apply the worto.rtrtJSt o template (Fig. 8, Plate 1— or the arc with radiating the end joints. blade in this group, which fits on to the end and on the soffit curve) on this worked twisted bed, and accurately mark thereon its full size, that is, mark by the arc, and both ends which radiate from the centre of the spiral curve, so that when the cross or heading joints are worked they may fit accurately from the soffit bed to the back of the arch. Again, carefully test the narrow clean draft along the curved soffit arris by one or both of these templates, and also test the draft along the radiating line, at the end of the arch stone next the parallel rule, by the template which fits the end and spiral arc. The stone must now be placed with its soffit upper- To work the soffit most for the more convenient working of the soffit bed, and along the soffit surface, with the dressed twisted bed next the workman, apply the template for the intradosal angle of the check with the hypothenuse or longest side BC, again next the workman and along the soffit arris, if it is a right-hand skew with the perpendicular AC to the right hand, and if it is a left-hand skew then place this template with the perpendicular AC to the left hand (see Fig. 11, Plate 1), and along the perpendicular AC and the base AB of this right-angled triangle score lines on the soffit, and with the template (Fig. 3a, Plate 1), with its arc on the lines scored by AC and the radiating blade in perfect contact with the worked twisted bed, sink curved drafts on the soffit, so that the arc and its radiating blade are both in perfect contact on the twisted bed, and the soffit at the same time along the curved or spiral arris. See Fig. 11, Plate 1. Work similar curved drafts on all the dotted vertical lines scored by the perpendicular AC, the more the better, and along the base AB of this triangular template work drafts with a straight-edge till the straight- edge is in contact with the bottoms of the curved drafts Oblique Bridges Made Easy. AC and the arris of the worked twisted bed. Now work off the superfluous stone between the straight and curved drafts, so that a straight-edge moved all over the face of the soffit and parallel to the straight drafts along AB, Fig. 11, will be in contact on every portion of the surface of the soffit. These curved lines which are worked diagonally across the soffit AC being at right angles to the springer line and the axis of the cylinder, fit on the lagging as if the arch had been built square. The practical work- man will see this at once by referring to Fig. 11, Plate 1. To set out the thick- Haviug workcd one twisted bed and the soffit bed, ness of the arch ° _ _ , ' stone. gauge the soffit, and make its width tk of an inch less than the perpendicular AC of the intradosal angle, and the same at the extrados, or tV of an inch less than the thickness of the courses given in the table of adjusted dimensions (see Fig. 5, CA and CE), so as to allow of a bed of rs of an inch of mortar between the joints. To work the twist Ncxt scorc a line square across both ends of the arch on the end joints on i rv i n m ^ ^ -tit the arch stone, stouo on the soffit bed off the dressed twisted bed, at the point exactly corresponding with the radiating lines marked on the bed by the template (Fig. 8, Plate 1). Place the stone with the unworked bed uppermost, and apply this template to the unworked bed with its spiral curve exactly coinciding with the soffit arris, and both its ends in contact with the lines scored across the soffit bed from the worked bed. Mark the stone by the radiating ends of this template up from the lines scored square across the soffit bed, and set the stone on its end, and work the cross or heading joint by a straight-edge applied from the radiating line on the worked bed to the radiating line just marked on the unworked bed, beginning at the square line across the soffit, and always keeping the straight-edge parallel to the soffit bed, as was done in working the twisted bed. Having worked both ends of the stone in this manner up the radiating lines on both ends, it will be found that the requisite wind or twist is given to the cross or heading joints to make the arch stones fit accurately into the skew- backs, and with each arch stone when placed in position on the centre. Templates, their Construction and Use. 19 Having: worked the ends of the arch stone and pre- To work the second o _ i twisttd bed. viously made two templates to represent the end section of the arch stones (see Fig. 9, Plate 1, previously referred to), mark off on the dressed ends of the arch stones the size of this template, and dress off the superfluous stone above the line so marked. The other twisted bed can now be got by using a straight-edge held parallel to the soffit arris, and removing the superfluous stone till the straight- edge is in contact with the lines marked on the ends of the arch stone, as above described. Test this bed with the twisting rules, and it will be found that the rules are in perfect contact on this bed also, and that their upper surfaces " wind," or are in the same plane. Take off a suitable chamfer, say an inch up the bed and half an inch on the sofiit, and the arch stone is finished. Before an experienced workman has finished this first eJ'TrirnJe'arthls arch stone he will have found out a much simpler and point. readier way of working the other stones. He will have seen that, if he has a curved rule made to fit in along the spiral arc of the soffit arris, and a template made with its blade to fit on the twisted bed and its arc made to fit square across the soffit arc of this dressed arch stone, and a third template made with its blade to fit the radiating end of the dressed stone and its arc to fit the spiral arc along the soffit arris, after he has worked the first twisted bed, all the other parts of the stone can be as easily worked as if they were for a square arch. The mode of getting these same templates from the checks has been previously described on page 9. See Figs. 8 and 9. By using these templates the time and labour is saved ^^^IZ^'^^^'^'^.e. of rulins the soffit of all the other arch stones by the Miy worked stones, o • T-i- J and from calculations. template for the intradosal angle, as in lig. 11, and working the same as described above. This first stone should be most carefully and accurately dressed by an experienced workman, and from this carefully dressed stone get the above three templates made to fit exactly one with its radiating blade in contact with the twisted bed, AE, Fig. 14, and its arc in contact with the soffit arc AC, parallel with the cross or heading joints square 20 Oblique Bridges Made Easy. off the twisted bed. The second should be made similar to the above, only the radiating blade must fit the draft up the radiating hne at the end of the stone, AC, Fig. 14, and its arc must fit the spiral arris of the bed or coursing joint AB. The third should be a straight-edge with one of its sides made to fit into the curve of this spiral arris, and if this rule is moved across the soffit parallel to the arris it will be in contact with every part of the soffit surface. See long rule, Fig. 8. (The foregoing explanation is for those who cannot make the necessary calculations.) Iimpi^?tempiTtes.^ ^"^ Working the other arch stones, as soon as the twisted bed is dressed, apply the first of these templates square across the soffit, square oif the twisted bed, and sink three or four curved drafts down to the spiral arc marked on the twisted bed, as previously described ; then apply the long curved rule parallel to the arris, and sink two or three longitudinal drafts by this long curved rule till the rule is in contact with the bottoms of the curved drafts worked square across the soffit by the first template ; remove the superfluous stone between the drafts, and test this dressed soffit in like manner as the first stone was worked, when it will be found to be accurately dressed. By this simple method it will be found, after the workman has worked two or three stones, that he can dress as many cubic feet per day as if he had been working square arch stones by the square and straight-edge. Radii are given in the tables of dimensions for constructing these templates on pages 91, 92, and 93. Setting Out and Erecting the Bridge. 21 CHAPTER V. SETTING OUT AND ERECTING THE BRIDGE. In the erection of all bridges it is necessary first to Plan and section prepare a plan and section of the site of the proposed bridge, to enable the engineer to determine the requisite square span, the height of the abutments and piers, and the angle of obliquity most suitable for securing a direct line of communication between the roads on both banks of the river. The plan having been completed and put into the How to set out the hands of the builder, the work should proceed as follows : — Stretch a line along the margin of the river or road which is to be spanned by the bridge, do likewise on the opposite side, making the two lines exactly parallel, and at a distance apart equal to the square span of the proposed bridge. Then apply the bevel, or template for the acute and obtuse angles (Fig. 1, Plate 1), with the stock or handle along the line, and stretch another line across the river or road in a direct line with the blade or head of the template, and stick pins in at the points where this line crosses the two parallel lines ; then measure from each pin along the lines the exact length of the impost or abut- ment, and again fix pins at these points. Should these lines, owing to the nature of the ground. To Hrd the same 1 • 1 1 1 l^ c 1 i." j_i points at the bottom have to be fixed some height above the toundations, tne of the trenches. corresponding points down in the foundations are found by using a centre plumb, held in contact with the lines at the pins, and along the lines at convenient distances. When the imposts or abutments have been built up To test the angles to near the water level, these lines, measurements, and *^*'°" angles should all be again accurately tested before pro- ceeding any further with the work. 22 Oblique Bridges Made Easy. Importance of having Haviiig prepared the impost as previously described, material ready before ii- inn ii i ii i fixing the centres, and haviDg nearly all the arch stones dressed and ready so ''"naHed dow-n.'^^'^ ^^ to cause no delay in setting and closing the aj'ch, — thus lessening the possible risk of having the centres washed away by floods while the arch is unfinished, — the centres should be placed in position and lagged with good sound battens six inches broad and three inches thick of suitable lengths for the work. The lagging should be nailed down to the centres in a few places, so as to prevent its moving after the coursing and cross joint lines have been ruled on. How to make the The Centres must be made the same as for a square arch, to the square span ; and this means that more centres are required for an oblique arch than for a square one, as the centres must necessarily reach from one obtuse angle to the other obtuse angle ; and that portion of the centre reaching beyond the face of the arch at the acute angles should be well loaded at the haunches with heavy rubble, which can be used iu the wingwalls, &c., after the arch is keyed. ""'' tZer' *' To make a really good and strong arch it is important that the centres should be made of good and perfectly sound timber, well and strongly put together, and when in position should not be more than five feet apart. The footing-beams of these centres should be supported at their ends on beams placed on stout stone corbels projecting from the face of the abutments, and these should be further supported by stout wooden props set on a sound hard footing, and wedged between the top of the j3rops and the underside of the corbels. In the river bed, between the abutments, either stone pillars or wooden piles, as the circumstances of the site demand, should be provided to bear a supporting beam for about every ten feet of the square span for the footing-beams of the centres to rest upon. This precaution prevents the possibility of the centres "sagging" and spoiling the form of the arch. t^etust'TndTeading ^^^^ ^^/^^^^ ^^i"g ^^^^^ P^^^^d in positiou, the next joint lines on the " proceeding is to mark the lagging for the guidance of the workman in the following manner : — To mark the face lines which form the extreme outside of the bridge is the first duty of the builder, and this can be Setting Out and Erecting the Bridge. 23 done easily and with the greatest accuracy, in two ways : — How to rule the face (a) By fixing a tightly-strung line over two cross bars in a ■'°'" ^ "^ ^ agg'^g- vertical line up from the ends of the imposts or abutments, and high enough for the line to be a few inches above the crown of the centre. From this line and at short distances use a centre plumb, allowing the centre point of the plumb to touch the lagging, being at the same time careful not to deflect the horizontal line while using the plumb. An assistant should carefully mark the centre points on the lagging with a pencil, and these points can be joined with a thin short flexible rule. In this way a true line is obtained on the lagging for the face of the arch, {b) By fixing a stout plank with a smooth straight side, in the same line as above described, from a point near the crown of the centre to a point vertically above the face of the end of the abutment, and proceed as above directed. With the plank there is less liability to deflection, and as a rule a truer face line can be got by its use. Now take a tape-line, or a reliable piece of properly How to find the stretched wire or cord, and measure the length of these arc. face lines over the centre, which ought to be the length of the oblique arc as found by calculation and given in the Tables. Half this length will give the crown of the arch, or centre of the arch, or key-stone. From these points, representing the centres of the two faces, mark a line on the lagging along the crown between centres of the face lines, and divide this line on the lagging into the same number of equal parts as there are checks on the impost ; from these divisions the cross or heading joints may be marked out on the lagging. To mark the bed or spiral coursing ioints, and the How to rule the ^ -Til spiral coursing joint cross or heading joints, take a clean half-inch board the and the heading exact width of the courses, less ^ of an inch to allow for the ''°'" ^' pencil line, with both its edges perfectly straight and parallel to each other, and of such length as will reach from the upper bed of the check at the acute angle to the corresponding division in the line drawn on the crown of the centre. This straight-edge should be divided into the same number of arch stones as are in the half arch, with a 24 Oblique Bridges Made Easy. half-stone to represent the key-stone if there is an odd course ; — that is if all the arch stones are strictly specified to be of a certain length, but it is not necessary except to please the eye ; it is sufficient if the joints are properly broken ; the arch is no stronger for it. Take a pencil, and with this straight-edge rule the coursing joints on the lagging, and as the straight-edge is parallel and of the exact thickness of the courses with the beds of mortar, the coursing joint lines can be kept perfectly true all over the surface of the centre from impost to impost. Importance of Jt Will \)Q Seen from the foresfoine;- directions how accurate setting out. , . . . important it is to have the bridge accurately set out from the foundations to the springers on the impost, so that all the lines shall meet and fit with mathematical accuracy. Soffit and face joint It now ouly remains to point out a very simple mode of getting the soffit and face joint angles in the headers of an oblique arch. It is not necessary to explain here at any great length to the practical workman, that the ends of the arch stones forming the quoins of the soffit of the arch with the face lines on the lagging do not form a right angle, or in other words are not square, as are the ends of all the other arch stones. Soffit face angle. To get this soffit facc angle, formed by the coursing joint line on the lagging, and the outside face line, take an ordinary bevel or shifting stock, lay the stock or handle along the coursing joint line on the lagging, and set the blade fair with the face line. Mark this angle got by the bevel on the soffit bed of the arch stone. As the angle formed between the coursing joint line and the face line on each course of stones marked on the lagging is different, the blade of the bevel must be adjusted to this angle for the soffit face of each course of arch stones ; and when the bevel is set it will fit the corresponding course of arch stones on the other face of the arch. Face joint angle. To mark accurately the face joints in the arch, first set up at the ends of each impost or abutment a straight-edge, so that if it was long enough it would be in contact with the line previously stretched across the bridge to get the face line on the lagging. On the crown of the centre fix Selling Out and Erecling Ihe Bridge. 25 another straight-edge in a vertical position and sight them, when they ought to be found in a perfect line across the bridge one with another. Now take the two templates representing the end ^ace joint angle. section of the arch stones (Fig. 9, Plate 1), and set them against the upper bed of the skewback ; and against these two templates lay the template for the coursing joints (Fig. 8, Plate 1) Avith the radiating lines, which is supposed to be the size of an arch stone, so as to represent an arch stone laid in position on its bed. In the centre of this template make a nick in the coursing joint arc, always placing this template so that the nick shall be fair on the face line on the lagging. Now take a thin straight-edge about two feet six inches long, two inches broad, and a quarter of an inch thick, slope one of its sides at the end till it forms an acute angle with the other face (see Fig. 8a, Plate 1) so as to allow one face only to come in contact with the face line on the lagging at the nick in the other template, so that its whole length is in line with those fixed in a vertical position at the ends of the impost and the one on the crown of the centre. From the nick on the face line upwards mark this position of the straight- edge on the template. The angle of the face or joint line is got for the upper bed of the stone and the angle for the under bed for the next course above it. Make a nick square across the upper edge of the template, so that the angles can be accurately marked down the other side of this template ; set the bevel to this line as before, and the face joint lines for the upper and under beds of the headers can be marked on the stone. As these lines are the same on both faces of the arch, be careful to mark them permanently so that all the lines radiate to the centre nick in the template, as shewn in Fig. 8, and they will be found to correspond accurately with the lines in the Diagram for getting the face lines of an oblique arch by construction. See Fig. 6, Plate 2. At page 364 of the "Civil Engineer and Architect's Journal" for 1841, this figure is explained. Mr. Peter 20 Oblique Bridges Made Easy. Nicholson also gives a plan for getting the varying angles formed between the coursing joints and the face lines in the headers in oblique arches. IN FIG. 6, PLATE 2, To conetnict a figure r q = radius = 12.5 = r q orge ^I^S ^®^ °'^^ ro = radius x sine of adjusted extradosal angle (<^) = 12.5 x .6050624 = 7.56.32800 = ro radius x sine of adjusted extradosal angle (4>) y* g =^ . L cot. 6 X tan. <^ 12.5 x .6050624 ' ^-^^^^^^^ ^ uMOmi ^ r s .8390996 X .7599587 .6376810 Draw the horizontal line t tt, so that it is the tangent of the spiral arc t' u' at s. From s set up the perpendicular s r 7^ making 5^=11.8606011 and r o = 7.56328. From r as a centre and r ^ as a radius describe the quadrant q m n, making the divisions in the quadrant abcdefghijkl equal to the thickness of a course of arch stones. Describe the quadrant q n and connect the divisions a b c, cfr., with the centre, intersecting the inner arc in a' b' c', ^x. ; from these latter divisions draw horizontal lines, and from the points a b c, ^c, in the face of the half arch draw vertical lines, and from the points of intersection with the horizontal lines a" b" c", ^-c, draw lines to 5. (See the Fig.) The radius of the curve of the spiral arc is equal to the radius of the cylinder divided by the sine of the adjusted intradosal angle squared ; these radii are given in the Tables of Adjusted Dimensions on pages 91, 92, and 93. The angle formed between the curve t' u' and the line s a is the angle at the first face joint on the obtuse angle of the abutment, and each succeeding line converging at s is the next corresponding face joint angle up to the centre of the arch. For finding the corresponding angles formed between the face-joint lines and the spiral arc, in skew elliptical arches, from a figure similar to Fig. 6, Plate 2, and for full and clear calculations for all the angles and for all the dimensions of the several parts of a skew elliptical arch, the twisting rules, the varying points of convergence or eccentricity, and other templates, see the article, " Oblique Elliptical Arches," by Mr. James Morice, of Perth, dated August 5th, 1845, and given in the September Number of the Engineer and Architect's Journal for that year. Also, see Appendix, page 109. To Construct the Template, Fig. 8. 27 CHAPTER VI. General and Concluding Remarks. TO OONSTRUOT THE TEMPLATE, FIG. 8, PLATE 1, WITHOUT usma a radius. ALSO EXPLANATION OF FIG. 4. PLATE 1, AND DIAGRAM, PLATE 7. Draw two horizontal parallel lines, the same distance To construct Fig. s apart as the length of the twisting rules or the depth of "' radius.'"^* the arch stones (2 feet), and in the centre of the parallel lines erect a perpendicular. On each side of this perpen- dicular set off on the lower line half the soffit length of the arc of this template, multiply this soffit length by GH in the same span and angle of obliquity in the Adjusted Tables, and set off on the upper parallel line half the product of GH multiplied by the soffit length, on each side of the perpendicular ; join these points between the upper and nether parallel lines, and the necessary radiation is given to the template for the length of the arch stones. Or, as the internal spiral radius is to the external spiral radius, so is the soffit length of the template to its length at the extrados. This also gives the amount of radiation across the bed of a 2 feet arch stone. See pages 37, 42, 44, and 90. To cut the curve of this arc. Fig. 8, Plate 1, or the To find the curve of *-; -,. . . ''h^ spiral arc. arc of the long 6 feet rule without using a radius, divide half the length of the rule (that is 3 feet) by the radius of the spiral arc, and the quotient is the natural sine of the angle CFG, Fig. 3, Plate 1. Find the angle corresponding to this quotient in a Table of Natural Sines, and multiply the spiral radius by the cosine of this angle, and the product is the length of FG, Fig. 3, and the difference between this product and the spiral radius is the amount of rise in the centre of a rule 6 feet long, thus, 3 -=-41. 617 = .0720859 the sine of 4° 8' very nearly, the cosine of 4° 8' is .9973990, and the spiral radius multiplied by this cosine is 41.617 x 28 Oblique Bridges Made Easy. .9973990 = 41.508 ; therefore the rise in the centre of the long rule is equal to 41.617 — 41.508 = .109 or cV of an inch. Having found this rise, any experienced joiner can draw this curve out with the cyclograph, or trammel, as it is called by some joiners. Mathematics ViQ. 4, Plate 1, illustratcs the mathematical principles go^ erning the obluiue & ' ' 1 Jr aich problem. whicli govcm the necessary calculations in connection with the construction of oblique arches. The span AC is equal to chord 1. The rise GB is equal to one-fourth of the span, or .25; consequently CD is equal to 1.15912, as is subsequently explained, because half the arc CB is subtended by an angle of 53° 7' 48", therefore the whole arc ABC is subtended by an angle of 106° 15' 36", and by Tables of Radius Unity this arc is equal to the radius .625 multiplied by 1.8545917. Hence the length of any arc where the rise is one-fourth of the span is equal to that span multiplied by 1.15912, because 1.15912 is the length of the arc to chord or span 1. See pages 37 and 38. The obhquity of the arch CE is equal to the chord or span multiplied by the cotangent of the angle of obliquity (50 degrees in this example, that is 1 multiplied by .8390996). Hence any given span multiplied by the cotangent of any given angle of obliquity is equal to CE in that angle of obliquity. The oblique span AE is equal to the chord or span multiplied by the cosecant of the angle of obliquity, that is 1 multiphed by 1.3054073 ; hence any span in this angle of obliquity multiplied by the cosecant of the angle of obhquity is equal to the oblique span ; and ED is the development of the obhque arc or heading spiral standing on AE. Therefore ED is the hypothenuse of the right-angled triangle CDE, two sides of which are now known ; and the tangent of the angle CDE is equal to CE divided by CD, and is equal to the tangent of 35° 54' very nearly, which is the tangent of the front angle of the check or the theoretic angle of intrados, and the angle HEI is equal to the angle CDE ; therefore the hypothenuse ED is equal to CE divided by the sine of 35° 54', or CD divided by the cosine of the same angle is also equal to ED. Description of Fig. 4, Plate 1. 29 The theoretic axial lene^th EH is the lenarth it would Mathematics of the ° ° problem. require an abutment to be extended so as to allow a full spiral course of arch stones to pass over the centre and rest on the opposite abutment, and EI represents the length of such spiral course, or the plan of this spiral course is more accurately represented by the spiral line DF, Fig. 1a in the Projection. And this theoretic axial length is equal to the length of the arc divided by the tangent of CDE or HEI, and is equal to 1.15912 divided by .7238793, there- fore the theoretic axial length is 1.6012614. The adjusted axial length is equal to the length of the arc divided by the tangent of the adjusted intradosal angle. But as the heading spiral or oblique arc must be divided by a certain number of courses in proportion to the span, it is almost invariably found in practice that the theoretic thickness of a course of arch stones will not supply a divisor for the heading spiral without leaving a remainder somewhat less than the theoretic thickness of a course of arch stones, but this test division indicates clearly what the correct or adjusted number of courses ought to be, and it is usual to select a number greater than that given in the quotient, hence it will be seen that the theoretic angle of intrados or the angle of the coursing joints cannot be adhered to in practice with strict mathematical accuracy ; consequently this angle must be adjusted to fit a number and thickness of courses that will divide the heading spiral without leaving a remainder, and as the number of checks on the impost represents a similar number of courses in the face of the arch, this space- in the face is termed the divergence of the courses, and is equal to the impost multiplied by the sine of the intradosal angle. Having determined the number of courses by this test division, divide the number of checks on the impost by this number of courses, and the length of the heading spiral or oblique arc multiplied by this quotient is equal to this new or adjusted divergence, and the sine of this new or adjusted angle is equal to the new divergence divided by the length of the impost, and the adjusted thickness of courses is equal to the length of the check multiplied by the sine of the new intradosal 30 Oblique Bridges Made Easy. angle, or the divergence divided by the number of checks on the impost. ^^'^'prb""'' The length of the impost is 1.3054073 = cosecant e, and as there are 12 checks, the length of check is .1087839, and the theoretic thickness of the courses is this length of check multiplied by the sine of the theoretic intradosal angle. 1087839 x .5863724 = .0637878, and the heading spiral divided by this thickness is nearly 23, therefore the adjusted number of courses ought to be 24. And 12^-24 = .5, hence the heading spiral 1.430938 x .5 = .715469 the adjusted divergence ; and .715469 ^ 12 = .0596224 the adjusted thickness of course, or 1.087839 x .5480499 (sine 33° 14') = .0596190, or practically the same as the divergence divided by the number of checks on the impost ; and this adjusted thickness of courses, multiplied by any additional number of feet of increased span, gives the ratio of pro- gression in the number of courses for each additional foot of span, thus, 1.192448 x 5 feet = 5.96224, hence in this angle of obliquity for every additional 5 feet of span we have 6 additional courses, or the ratio of increase in the number of courses is 6 for every 5 feet increase of span, and the thickness of courses multiplied by the number of courses is equal to the length of the heading spiral. See Tables of Dimensions of Bridges on pages 91, 92, and 93. Both the number and thickness of courses are governed by the adjusted intradosal angle. But some writers on this subject recommend selecting an odd number of courses for a key-stone in the centre of the arch, but this arbitrary selection of an odd number of courses disturbs the beautiful harmony traceable in every part of the problem, therefore the natural number of courses as found above should be taken whether it is an odd or an even number. See page 35. The angle GFK is a right-angle, and CFG is 53° T 48", therefore the springer angle KFO at the bottom of the checks is 36° 52' 12". From the relative position to each other of these similar right-angled triangles CDE and HEI, Fig. 4, it is clear that EI, the theoretic coursing joint line, must be at right-angles to the face of the oblique arc ED, therefore Description of Diagram, Plate 7. 31 the lateral pressure orthrust of the arch,if built theoretically, Mathematics of the must be absolutely parallel to the face, or exactly in the same line as if the arch had been built on the square with parallel courses. But owing to the necessity of adjusting the intradosal angle, it Avill be seen that the new coursing joint line which causes the axial length to be increased, EI', is not at right-angles with the face line of the oblique arc ED, therefore the thrust is not exactly parallel Avith the face of the arc ED, but as this variation is so very slight it does not in the least affect the stability of the structure, the outward pressure being equal to the angle of difference between the theoretic and adjusted intradosal angles, or only 2° 40' from a right-angle in this example. With respect to the direction and force of the lateral pressure or thrust of segmental arches, it may be stated here that it has been found by a severe actual test that if the square or direct thickness of the abutments is made equal to one-fifth of the square or direct span, and if they are built with strong rubble and cement mortar, they will resist of themselves, without any further backing up or support, the lateral pressure of any segmental arch of these propor- tions. An arch of 40 feet direct span at an angle of 50°, which was closed on Wednesday, had the entire centering removed on the Saturday following ; this was done owing to a threatening flood, and the centres were carefully removed by degrees, so as to avoid the risk of having them violently washed out by the flood. Careful observation was made while the centres were being removed. The subsidence on the crown of the arch did not exceed one-fourth of an inch, not a single stone in any part of the structure was moved or nipped, and the abutments were at the time only walled up level with the back of the springers. This test of itself, apart from the calculated power of resistance of abutments of the above dimensions, is sufficient to demonstrate the ability of abutments of these proportions to resist the force of the lateral pressure of the arch and its superin- cumbent load. 32 Ghliqite Bridges Made Easy, To find multipliers from the diagram, by construction, for giving all the necessary dimensions of the bridge. The natural sine, cosine, and tangent of the angle. Tangent of the theoretic intradosal angle. Tangent of the theoretic extradosal angle. Radiation of the twisting rules, and the increased breadth of one rule at one end. The Diagram on Plate 7 gives a practical illustration of the way to find, by construction, all the Theoretic Multipliers given in the Theoretic Tables. AM2 is a right- angled triangle, and its base AM is equal to the length of the arc to chord 1. M2 is equal to the obliquity of the arch or cotangent of the angle of obliquity .8390996, as is shewn on the tangent scale JK. Join A2 and it will be found by scale to measure 1.430938. See Table of Theoretic Multipliers, pages 80, 81, and 82, &c. This right-angled triangle is the same as CDE, Fig. 4, Plate 1. The angle MA2 or /3 is 35° 54'. Erect a perpendicular from the base AM through the point where A2 cuts the arc at h, this perpendicular is the natural sine of the angle MA2, and will be found by scale to measure .5863724, A.d is the cosine and may be read from the base scale and is equal to .8100416, the tangent may be read from the tangent scale at L where A2 cuts the tangent line at .7239. It will be seen on the Diagram that the cotangent of the angle of obliquity divided by the length of the arc to Chord 1 . in any angle of obliquity and any ratio of rise to span is equal to the tangent of the intradosal angle. See the horizontal line 1, 2, 3, 4, 5, 6, 7, and similarly in the other angles of obliquity. The extradosal angle f A e is found by construction from the following stating : — As the radius of the cylinder 12.5 : the external radius of the cylinder 14.5 :: .5863724 {dh) : .680189 {df), produced A or the sine of the intradosal angle to/, and draw a line from A through/ to the tangent line at J, on which tangent scale the angle can be read off and will be found to be .8395955. Erect another perpen- dicular from the line AM to the point e where the line A fi/ J cuts the arc at e, and the perpendicular c e is the sine of the extradosal angle and A c is the cosine. The angle of difference between the intradosal and extradosal angles is ^ A /. Erect a perpendicular on the line A2 at g, to cut the point/, and the scaled distance g his equal to the amount of radiation of the twisting rules at the back of the arch stone, and the scaled distance Description of Diagram, Plate 7. 33 of the perpendicular gf'is, equal to the increased breadth- Value of ) = 37° 14' Sine ,ji,(CE) = .6050624 Cosine (CA) = .7961780 Tangent (JK) = .7599587 Cotangent = 1.3158610 Secant (AJ) = 1.2560005 Cosecant = 1.6527221 The Angle of Difference is = 37° 14' — 33° 14 {8) = 4°0' Sine S = .0697565 Cosine = .9975641 Tangent = .0699268 DAH is the adjusted intradosal angle {p), and DH is the sine of - = CAE is the adjusted extradosal angle (<^), and CE is the sine of = GAF is the adjusted angle of difference (S), and is equal 37° 14' — 33" 14' The distance which the twisting rules are placed apart at the soffit, multiplied by the calculated or scaled distance between the points G and H on the diagram, is equal to the divergence of the twisting rules at the extrados, that is, the increased distance which they must be placed apart at the back of the arch stone, to cause the rules to give the proper radiation to the ends of the dressed arch stones, so as to harmonise with the twist. And the soffit distance apart which the rules are placed, multiplied by the calculated or scaled distance between the points G and F, is equal to the increased breadth which must be added to the breadth of one rule at one end, so as to produce the required twist to the bed of the checks or arch stones. sec. ^ Or 14.5 ^ 12.5 (1.16) x .6552129 = 7600469 = 40° 1' 4° 7' = 28.6187748 = 28.618789 = 28.618760 L1924483 = 26.1081472 = 26.1081460 = 15.3090969 = 15.3090962 1.2757580 2.1756788 = 14.3093824 = 33° 14' 1.1923805 35° 54' 35° 54' 33° 14' 33° 14' 33° 14' 32.0252191 35.3814773 35.3814767 37° 14' 37° 14' 37° 14' Example 1. 47 The Adjusted Angle of Difference (8) is = 37° 14' - 33° 14' = 4° 0' The Cosine of 8, or the Adjusted Angle of Difference = .9975641 The Tangent of 8, or „ „ „ = .0699268 P, or the Theoretic Point of Convergence of the Face Twines below the centre of the Cylinder is = (C x Cotangent- &) -^ Arc x External Piadius (R + £) = 20. X (.8390996)'-' ^ 23.1824 x 14.5 = 8.8077817 Or Radius x Cotangent d x Tangent = 12.5 x .8390996 x .8395955 = 8.8063030 P, or the Adjusted Point of Convergence = Radius x Cotangent 6 x Tangent c/j = 12.5 X .8390996 x .7599587 - = 7.9710130 The Theoretic Tangent of the Angle formed between the first joint in the face, at the acute end of the Impost and the horizontal line shewn on the elevation at the end of the Impost, is equal to the distance of the Converging Point (P) + R — Rise -;- by half the Oblicjue Span ; thus, (8.8077817 + 12.5 - 5.) ~ 13.0540736 = 1.2492484 Tangent of - =51° 20' The Tangent of the Adjnsted Angle is = to (7.9710130 + 12.5 - 5.) ~ 13.0540736 = 1.1851483 Tangent of = 49° 50' The Theoretic Radius of the End Joint Arc is = the length of the Heading Spiral (D E) squared, and multiplied by the Radius of the Cylinder, and the product divided by the square of the circular Arc (C D) = 19.0500139 Or the Radius -^ by the Cosine of /3 squared = 12.5 -^ .6561674 = 19.0500168 The Theoretic Radius of the Bed Joint or Spiral Arc is = to D E squared x Radius ^ C E squared = 36.3517286 Or the Radius -^ by the Sine of ^ squared = 12.5 -^ .3438326 - = 36.3549006 The Adjusted Radius of the End Joint is = the Radius -^ Cosine (i squared = 12.5 ^ .6996412 ' = 17.8663006 The Adjusted Radius of the Bed or Spiral Arc is = the Rixlius -^ Sine fi squared = 12.5 H- .3003586 - ' = 41.6169205 TWISTING RULES. The Thickness of the Cylinder, or the depth of the Bed of the Arch Stones is 2 feet, hence the Rules must be exactly 24 inches long, and if placed 30 inches {x) apart at the Soffit, their distance apart at the Extrados will be .r = a; x (Secant <^ x Cosine 8 -f- Secant P) = 30. X .054951 = Divergence The Extra Breadth of one Rule at one end will be = a; x (1.054951 x Tangent 8) = 30. x .075924 = Increased Breadth The Adjusted Radiation and the Adjusted Breadth of the Rules are found by using the Secants, Cosines, and Tangents of the Adjusted Angles as above, which are more fully explained elsewhere. Adjusted Multipliers are : As Secant j8 1.1955350 : Secant , (37° 14'). That is, .625 x .8390996 x .7599587 = .3985506 = P and P x 20 = 7.97 10120 Or 12.5 X .8390996 X .7599587 ' =7.9710125 Or P may be found by proportion— See Fig. 7, Plate 2, & Figs. 2, 13, 14, & 15, Plate 1. Fig. 2, Plate 1. The springer angle y is 36° 52', sine = .5999549, cosine .8000338, tangent .7499119. Fig. 7, Plate 2B'. B D (t) is 2 feet (the depth of the arch stones). B C = B D X cosine 7 = 2. x .8000338 = 1.6000676 (>^ee square arc Fig. 7, Plate 2, and B A Fig. 2, and B E' Fig. 13, Plate 1). Fig. 13, Plate 1. B C = B E' x cosecant 6 = 1.6000676 x 1.3054073 = 2.0887399 or 2.08874 (see B C oblique arc Fig. 7). Fig. 7, Plate 2. CD = e x cotangent 6 x tangent (j^ = 2. x .8390996 x .7599587 = 1.2753620. Fig. 13, Plate 1. Or the amount cut off the back of the springer (E C Fig. 13) x tangent 4> = 1.6781992 x .7599587 = 1.2753620 C D. It will be seen from the above that 1.6781992, the length of the portion cut off from the back of the check to form the angle of obliquity at the end of the impost, is equal to the depth of the arch stone multiplied by the cotangent of the angle of obliquity, and this length multiplied by the tangent of the extradosal angle gives the height of C D Fig. 7, Plate 2, or the height of the point C in Figs. 14 and 15, Plate 1. In every case in arches of these projDortions B' G and C D in the square arc in Fig. 7 are constants, B' C is 1.6000676, and C D is 1.1999098, or 1.20. But B C and C D in the oblique arc vary with the span and angle of obliquity. Having found the values of B C and C D, both in the square and oblique arcs (Fig. 7, Plate 2). State, As 2.08874 (B C) : 13.054073 (Half the oblique span) :: 1.2753620 (0 D) Or, As 1.6000676 (B' C) : 10. (Half the square span) :: 1.2753620 (C D) Again, As 2. (Depth of arch stones B D) : 12.5 (Eadius) :: 1.2753620 (C D) 7.9706 (R P Fig. 7) 7.9706 (RPFig. 7) 7.9710 (RP Fig. 7) e 36° to Chord 1. 51 e 35° TO CHORD 1. EXAMPLE TO CHORD 1. 6 = Angle of Obliquity - P = Theoretic Angle of Intrados - - - 13 = Adjusted Angle of Intrados = Adjusted Angle of Extrados ... 8 = Theoretic Angle of Difference, equal 55° 1'— 50° 56' - - S = Adjusted Angle of Difiference, equal 53° 28' - 49° 19' •y = Natural Springer Angle, or the Angle at the bottom of the Check - 0) = Theoretic Angle produced on the end of Impost by cutting the Skewback to the Angle of 6 b) = Adjusted Angle produced on the end of Impost by cutting the Skewback to the Angle oi 6 - - - R = Radius £ = Depth of Arch Stones - ... ... W = Width of Bridge, including Parapets C = Chord or Square Span a = Length of Arc (AM) = 35° = 50° 56' = 49° 19' = 55° 1' = 53° 28' = 4° 5' = 4° 9' = 36° 52' 12" = 62° 9' 61° 7' .625 .100 1.000 1.000 L15912 Oblique Span (A E) = C x Cosecant of 61 = 1. x 1.7434468 - Or C ^ Sine 6> = 1. -4- .5735764 Or again, the Obliquity (C E) ^ Cosine 6 = 1.4281480 -=- .8191520 - Obliquity of the Arch (C E) = C x Cotangent of ^ = 1. x 1.4281480 Length of Arc (C D) is found thus : Half C -=- Radius = .50 -i- .625 = .80 or Sine of A B = half the Arc = 53° 7'.805, therefore the whole Arc A B C = 53° 7'.805 x 2 = 106° 15'.61, or - - Hence the Springer Angle at the bottom of the Check is (see Fig. 4, Plate 1) - And by Tables of Radius Unity the Arc is = .625 x 1.8545917 - Tangent of the Theoretic Angle of Intrados C D E (^) = (C x Cot. 9)^ a = (1. x 1.4281480) -r 1.15912 = 1.2321830 tan. . . - - Length of Heading Spiral or Oblique Arc = Arc x Secant of ^ = 1.15912 x 1.5867369 (A 2) 32 is the correct number of Courses, and their thickness = 1.83921847 -=- 32 Length of Impost (W) -^ Sine 9=1.^ .5735764 .... Or Length of Impost (W) x Cosecant 6' = 1. x 1.7434468 1.7434468 1.7434468 1.7434468 1.4281480 106° 15' 36" 36° 52' 12" 1.15912 = 50° 56' 1.83921847 .05747557 1.7434468 1.7434468 52 Oblique Bridges Made Easy. The Divergence of the Courses is that part of the face of the Oblique Arc which is intercepted between the Horizontal Springer Line and the Coursing Joint Line, which springs from the acute quoin on the end of Impost, and is equal to W x Cosecant x Sine /8 = L x L7434468 X .7764132 Or the Impost x Sine /3 = 1.7434468 x .7764132 The number of Courses that will spring from, or intersect the Impost is equal to : As the length of the Heading Spiral is to the length of the Theoretic Divergence, so is the full number of Courses in the Heading Spiral to the number of Courses in the length of the Divergence, or the number rising from the Impost, thus. As 1.83921847 : 1.3536351 :: 32 : Therefore 23 is the correct number of Checks on the Impost, and the correct length of the Check is equal to the Impost divided by 23 = 1.7434468 -=- 23 Hence the actual or adjusted Divergence will be 23 -^ 32 (.71875) multiplied by the length of the Heading Spiral = .71875 x 1.83921847 Or the number of Checks multiplied by the thickness of one course = 23 x .05747557 And making 1.32193827 = Sine, and the length of the Impost = Radius Thus 1.32193827 -r 1.7434468 = .7582326 = Sine of the adjusted angle or ^8 Or making the thickness of a Course of Arch Stones = Sine, and the length of Check = Radius, thus .05747557 -^ .07580203 = .7582326 Sine of j8 And to this Angle, the Extradosal Angle, or -p the Axial Length, and the Converging Point (P) for the face lines must all be adjusted :— The Tangent of /3 or the Theoretic Angle of Intrados = The Cotangent of /3 „ ,, ,^ = The Tangent of ^ or the Adjusted Angle of Intrados = The Cotangent of f3 „ „ j, = The Secant of (3 „ „ The Theoretic Axial Length is equal the Arc x Cot. /3 = .8117124 Or, As the Cotangent 0, is to the length of the Arc, so is the length of the Arc to the Axial Length, that is. As 1.4281480 : 1.15912 :: 1.15912 : .9407702 And the Adjusted Axial Length is equal to the Arc x Cot. [3 = 1.15912 x .8596297 - As similar Arcs are to one another as their radii, it follows that having found the Intradosal Angle the Extradogal Angle may be found by proportion, as the Tangents of the Intradosal and Extradosal Angles are develop- ments of their respective Arcs, therefore, As .625 : .725 :: 1.2319634 : 1.4290775 = the Tangent of Extradosal Angle or (actual Tan. 1.4290326) - - . . 1.2319634 .8117124 1.1632916 .8596297 1.5340297 1.15912 X 1.3536351 1.3536351 = 23.5514833 .07580203 1.32193827 1.32193811 = 49° 19' = 49° 19' .94087207 .9407702 .99641397 = 55° 1' e 35° to Chord 1. 53 = (.725 -=- .625) X 1.2319634 1.4290775 = Tan. 4. Secant = 1.7441715 Tangent <^^ = 1.16 x 1.1632916 Then (R + e -=- R) x Tangent /? 1.3494182 = Tan. ^ ' .... Secant of <^ or Adjusted Extradosal Angle = 1.6798525 Theoretic Angle of Difference = 8 = 55° 1' — 50° 56' Cosine 8 = .9974615, Tangent of 8 = .0713885 Adjusted Angle of Difference = 8 = 53° 28 - 49° 19' Cosine S = .9973780, Tangent of 8 = .0725581 Distance of the Theoretic Point of Convergence of the Face Line helow the centre of the Cylinder = (C x Cot. 2 8) -4- Arc and x (R + «) = 1. x (1.4281480)2 -H 1.15912 x .725 = P - Or the Radius multiplied by the Cotangent of 6, and by the Tangent of the unadjusted Angle of Extrados (55° 1') = .625 x 1.4281480 x 1.4290326 The Adjusted Point of Convergence (P) is equal the Radius multiplied by the Cotangent of 6 and by the Tangent of the Adjusted Angle <^ That is, .625 x 1.4281480 x 1.3497794 The Theoretic Tangent of the Angle formed between the first joint in the face at the acute end of the Impost and the horizontal line shewn on the elevation at the end of the Impost, is equal to the distance of the (Converging Point + R - Rise) -^ by half the Oblique Span. Thus (1.2757219 + .625 - .25) -h .8717234 = 1.8930534 And the Adjusted Angle is equal to (1.20479814 + .625 - .25) -^ .8717234 = 1.8122699 The Adjusted Radius of the End Joint Arc is equal to the Radius divided by the Cosine of the Intradosal Angle squared (^) = .625 -^ (.6518778)3 The Adjusted Radius of the Spiral Arc or Coursing Joint Arc is equal to the Radius divided by the Sine of the Intradosal Angle squared (^) = .625 H- (.7583240)2 = 55° 1' = 53° 28' = 4° 5' = 4° 9' = 1.2757219 1.2755437 1.20479814 62° 9' 61° 7' 1.4707799 1.0868521 TWISTING RULES. As 1.5340297 : 1.6798525 :: .9973780 : 1.0921874 = A G GH = AG-AH= 1.0921874 - 1. G F = A G X Tan. 8 = 1.0921874 x .0725581 GH G F .0921874 .07924704 54 Oblique Bridges Made Easy. e 40° TO CHORD 1. EXAMPLE TO CHORD 1, 4> 8 6 7 0) R = e = W = C = Angle of Obliquity Theoretic Angle of Intrados Adjusted Angle of Intrados Theoretic Angle of Extrados Adjusted Angle of Extrados Theoretic Angle of Difference, equal 50° 2' - 45° 8' Adjusted Angle of Difference, equal 47° 39' -43° 24' Natural Springer Angle, or the Angle at the bottom of the Check Theoretic Angle produced on the end of Impost by cutting the Skewback to the Angle oi d Adjusted Angle produced on the end of Impost by cutting the Skewback to the Angle of 6 Radius Depth of Arch Stones Width of Bridge, including Parapets Chord or Square Span Length of Arc - - - - (A M) 40° 45° 48' 43° 24' 50° 2' 47° 39' 4° 14' 4° 15' 36° 52' 12" 58° 23' 56° 53' .625 .100 1.000 1.000 1.15912 Oblique Span (A E) = C x Cosecant of ^ = 1. x 1.5557238 Or C -=- Sine 6=1.- .6127876 Or again, the Obliquity (C E) ^ Cosine d = 1.1917536 -r .7660444 Obliquity of the Arch (C E) = C x Cotangent of 6» = 1. x 1.1917536 Length of Are (C D) is found thus, half C ^ Radius = .50 ^ .625 = .80 or Sine of A B = half the Arc = 53° 7'.805, therefore the whole Arc A B C = 53° 7'.805 x 2 = 106° 15'.61, or Hence the Springer Angle at the bottom of the Check is (see Fig. 4, Plate 1) And by Tables of Radius Unity the Arc is = .625 x 1.8545917 - Tangent of the Theoretic Angle of Intrados C D E (0) = (C x Cot. 9) -h a = (1. X 1.1917536) -- 1.15912 = 1.0281537 Tan. Length of Heading Spiral, or Oblique Arc = Arc x Secant of = 1.15912 x 1.4343805 (A 2) 28 is the correct number of courses, and their thickness = 1.6626191 -i- 28 - Length of Impost (W) -=- Sine 61 = 1. h- .6427876 Or Length of Impost (W) x Cosecant = 1. x 1.5557238 1.5557238 1.5557238 1.5557239 1.1917536 106° 15' 36" 36° 52' 12" 1.15912 45° 48' 1.6626191 .0593792 1.5557238 1.5557238 9 40° to Chord 1. 55 The Divergence of the Courses is that part of the face of the Oblique Arc which is intercepted between the horizontal Springer Line and the Coursing Joint Line, which springs from the acute quoin on the end of the Impost, and is equal to W x Cosecant 6 x Sine = 1. x Lo557238 x .7169106 Or the Impost x Sine /3 = 1.5557238 x .7169106 The number of Courses that will spring from, or intersect the Impost is equal to. As the length of the Heading Spiral is to the length of the Theoretic Divergence, so is the full number of Courses in the Heading Spiral to the number of Courses in the length of the Divergence, or the number rising from the Impost, thus, As 1.6626191 : 1.11531488 :: 28. Therefore 18 is the correct number of Checks on the Impost, and the correct length of the check is equal to the impost divided by 18 = 1.5557238 + 18 Hence the actual or adjusted Divergence will be 18 ^ 28 (.6428571) multiplied by the length of the Heading Spiral = .6428571 x 1.6626191 Or the number of Checks multiplied by the thickness of one Course = 18 x .0593792 And making 1.068256 = Sine, and the length of the Impost = Eadius Thus 1.06882640 -^ 1.5557238 = .6870283 = Sine of the Adjusted Angle or ;3 Or, making the thickness of a Course of Arch Stones = Sine, and the length of the Check = Eadius, thus, .05937920 H- .0864291 = .6870278 Sine of S And to this Angle, the Extradosal Angle, or 4> the Axial Length, and the converging point (P) for the face lines must all be adjusted ; — 1.0283226 .9724575 .9456530 1.0574704 1.3763210 1.15912 X The Tangent of /3 or the Theoretic Angle of Intrados The Cotangent of /3 „ „ ,, „ The Tangent of B or the Adjusted Angle of Intrados - The Cotangent of /3 „ „ „ „ The Secant of ji „ „ „ „ The Theoretic Axial length is equal the Arc x by Cot. /3 .9724575 - Or, As the Cotangent B is to the length of the Arc, so is the length of the Arc to the Axial Length, that is. As 1.1917536 : 1.15912 :: 1.15912 : 1.1273800 - - - - - And the Adjusted Axial Length is equal to the Arc x Cot. ^ = 1.15912 x 1.0574704 .... .... As similar Arcs are to one another as their radii, it follows that having found the Intradosal Angle the Extradosal Angle may be found by proportion, as the Tangents of the Intradosal and Extradosal Angles are developments of their respective Arcs, therefore, As .625 : .725 :: 1.0283226 : 1.1928542 = the Tangent of Extradosal Angle, or ^, or the (External Radius -^ by the Internal Eadius) x Tan. /3 = Tan. 4> (actual Tan. 1.1931626) 1.11531488 1.11531488 = 18.782899 .0864291 1.0688264 1.0688256 = 43° 24' = 43° 24' 1.1271949 1.1273800 1.22573509 = 50° 2' 56 Oblique Bridges Made Easy. = (.725 -^ .625) X 1.0283226 = 1.1928542 = Tan. Secant <)!> = 1.5562634 /8 = Tangent <^ = 1.16 x .9456530 Then (R + e ^ E) x Tangent 1.09695748 = Tan. st. Secant of = Theoretic Angle of Extrados (/) = Adjusted Angle of Extrados S = Theoretic Angle of Difference, equal 45" 1' - 40° 47' S = Adjusted Angle of Difference, equal 43° 22' — 39° 9' 7 = Natural Springer Angle, or the Angle at the bottom of the Check 0) = Theoretic Angle produced on the end of Impost by cutting the Skewback to the Angle oi 6 = 0) = Adjusted Angle produced on the end of Impost by cutting the Skewback to the Angle oi 6 = R = Radius - - e = Depth of Arch Stones - - .... W = Width of Bridge, including Parapets - - - - C = Chord or Square Span - a = Length of Arc - - - - - (A M) 45° 40° 47' 39° 9' 45° 1' 43° 22' 4° 14' 4° 13' 36° 52' 12" 54° 45' 53° 47' .625 .100 1.000 1.000 1.15912 Oblique Span (A E) = C x Cosecant of 5 = 1. x 1.4142136 Or C ^ Sine 5 = 1. -r .7071068 Or again, the Obliquity (GE) - Cosine ^ = 1. -=- .7071068 Obliquity of the Arch (C E) = C x Cotangent of 5 = 1. x 1.000 Length of Arc (CD) is found thus: Half C -^ Radius = .50 -^ .625 = .80 or Sine of A B = Half the Arc = 53°7'.805, therefore the whole Arc ABC = 5.3°7'.805 x 2 = 106° 15'.61, or H:ence the Springer Angle at the bottom of the Check is (see Fig. 4, Plate 1 And by Tables of Radius Unity the Arc is = .625 x 1.8545917 Tangent of the Theoretic Angle of lutrados C DE (/?) = (C x Cot. 6) ^ a = (1. X 1.000) H- 1.15912 = .8627234 = Tan. Length of Heading Spiral or Oblique Arc = Arc x Secant of (3 = 1.15912 x 1.S206810 Secant (3 (A 2) ... 24 is. the correct number of Courses, and their thickness = 1.5308277 -h 24 Length of Impost (W) H- Siije 5 = 1. -f- .7071068 - Or Length of Impost (W) X Cosecants =1. X 1.4142136 1.4142136 1.4142135 1.4142135 1.000 106° 15' 36" 36° 52' 12" 1.15912 40° 47' L5308277 .06378448 1.4142135 1.4142136 58 Oblique Bridges Made Easy. The Divergence of the Courses is that part of the face of the Oblique Arc which is intercepted between the Horizontal Springer Line and the Coursing Joint Line which springs from the Acute Quoin on the end of the Impost, and is equal to W x Cosecant 6 x Sine/3 = 1. x 1.4142136 X .6532004 Or the Impost x Sine /3 = 1.4142136 x .6532004 The number of Courses that will spring from, or intersect the Impost is equal to : As the length of the Heading Spiral is to the length of the Theoretic Divergence, so is the full number of Courses in the Heading Spiral to the number of Courses in the length of the Divergence, or the number rising from the Impost, thus, As 1.5308277 : .92376488 :: 24 Therefore 14 is the correct number of Checks on the Impost, and the correct length of the Check is equal to the Impost divided by 14 = 1.4142136 ^ 14 Hence the Actual or Adjusted Divergence will be 14 -H 24 (.58333) multiplied by the length of the Heading Spiral = .68333 x 1.5308277 Or the number of Checks multiplied by the thickness of one Course = 14 x .06378448 And making .89297772 = Sine, and the length of the Impost = Kadins, thus, .89297772 ~ 1.4142136 = .6314235 Sine of the Adjusted Angle, or/3 Or making the thickness of a Course of Arch Stones = Sine, and the length of Check = Eadius, thus, .06378448 ^ .1010152 = .6314344 Sine of /3 And to this Angle, the Extradosal Angle, or <^ the Axial Length, and the Converging Point (P) for the Face Lines must all be adjusted : — The Tangent of /3 or the Theoretic Angle of Intrados = .8626694 The Cotangent of ^ „ „ = 1.1591927 The Tangent of ^ or the Adjusted Angle of Intrados = .8141280 The Cotangent of /3 „ „ = 1.2283081 The Secant of /3 ' „ „ = 1.2894977 The Theoretic Axial Length is equal to the Arc x Cot. (3 = 1.15912 x 1.1591927 Or, As the Cotangent 6 is to the length of the Arc, so is the length of the Arc to the Axial Length, that is. As 1.000 : 1.15912 :: 1.15912 : 1.3435591 And the Adjusted Axial Length is equal to the Arc x Cot. /3 = 1.15912 x 1.2283081 As similar Arcs are to one another as their radii, it follows that having found the Intradosal Angle the Extradosal Angle may be found by proportion, as the Tangents of the Intradosal and Extradosal Angles are develop- ments of their respective Arcs, therefore, As .625 : .725 :: .8626694 : 1. 0006965 = the Tangent of Extradosal Angle, or - - (^ Or the (External Radius H- by the Internal Eadius) x Tan. fi = Tan. (^ (actual Tan. 1.0005819) - .... = (.725 -f .625) X. 8626694 = 1.0006965 = Tan. (^ Secant./) = 1.414G251. .92376488 .92376488 14 4825946 .1010152 .89297772 .89298272 39° 9' 39' 9' 1.3436434 1.3435591 1.4237564 45' r 45° r 9 45° to Chord 1. 59 Then(R + e ^ R)xTangent^_ = T;ingent(^ = 1.16 x .8141280 = .9-143884 = Tan. 4> ' ' - - Secant of <^ or Adjusted Extradosal Angle = 1.3755645 Theoretic Angle of Difference = 5 = 45° 1' — 40° 47' ... Cosine 8 = .9972717, Tangent S = .0740203 Adjusted Angle of Difference = 8 43° 22' - 39° 9' Cosine 8 = .9972931, Tangent 8 = .0737279 Distance of the Theoretic Point of Convergence of the Face Line below the Centre of the Cylinder = (C x Cot.^ d) -^ Arc and x (R + e) = 1. x (1,000)' -=- 1.15912 X .725 = P - - Or the Eadius multiplied by the Cotangent of d and by the Tangent of the Unadjusted Angle of Extrados (45° 1') = .625 x 1.000 x 1.0005819 The Adjusted Point of Convergence (P) is equal the Radius multiplied by the Cotangent of and by the Tangent of the Adjusted Angle <^, that is, .625 X 1.000 X .9445516 . - - - The Theoretic Tangent of the Angle formed between the first joint in the face at the acute end of the Impost and horizontal line shewn on the elevation at the end of the Impost, is equal to the distance of the (Converging Point + R — Rise) -=- by half the Oblique Span, thus, (.62547446 + .625 —.25) -4- .7071068 = 1.4148845 And the Adjusted Angle is equal to (.59034475 + .625 - .25) -r .7071068 = 1.3652036 - The Adjusted Radius of the end joint Arc is equal to the Radius divided by the Cosine of the Intradosal Angle squared (^) = .625 -r (.7754957)' The Adjusted Radius of the Spiral Arc, or Coursing Joint Arc, is equal to the Radius divided by the Sine of the Intradosal Angle squared (0) = .625 ^ (.6313528)'' 43° 22' 4° 14' 4° 13' 62547446 .6253636 .59034475 54° 45' 53° 47' 1 0392530 1.5679806 TWISTING RULES. As L2894977 : 1.3755645 :: .9972931 : L0638568 = AG GH = AG - AH = 1.0638568 — 1. = GF = AG X Tan. 8 = 1.0638568 x .0737279 GH GF .0638568 .07843592 60 Oblique Bridges Made Easy. EXAMPLE. (d 50° Calculated to Chord 1.) Decimal Dimensions of a Segmental Oblique Bridge, peoposed to be eeected over the Eae Burn, at Hempland, Ford, in the Longtown Union, Cumberland, at an Angle of 50° AND 20 feet Span. Illustrating the origin and value op the Decijial Multipliers, given in THE Tables, foe finding corresponding dimensions of the full size of any Bridge in the same degree of Obliquity. 6 = Angle of Obliquity - - - - (3 = Theoretic Angle of Intrados (3 = Adjusted Angle of Intrados

And to this Angle, the Extradosal Angle, or = 1.3057261 Tangent fi Tangent 1.16 X .6552129 = Then(R + ^ ^ R) x .7600469, Tan. Secant of ^, or Adjusted Extradosal Angle = 1.2560005 Theoretic Angle of Difference = S = 40° 1' — 35° 54' Cosine 8 = .99741 Tangent S = .07197 Adjusted Angle of Difference = ,8 = 37° 14' - 33° 14' ... Cosine S = .9975641, Tangent of 8 = .0699268 Distance of the Theoretic Point of Convergence of the Face Lines below the centre of the Cylinder = (C x Cotangent 6)- -r Arc and x (R + e) = 1 X (.8.390996)2 ^ 1.15912 = .6074332 x .725 = P Or the Radius multiplied by the Cotangent of 0, and by the Tangent of the Unadjusted Angle of Extrados (40° 1') = .625 x .8390996 x .8395955 The Adjusted Point of Convergence (P) is equal to the Radius multiplied by the Cotangent of 9 and by the Tangent of the Adjusted Angle (<^) That is .625 x .8390996 x .7599587 The Theoretic Tangent of the Angle formed between the first joint in the face, at the acute end of the Impost and the horizontal line shewn on the elevation at the end of the Impost, is equal to the distance of the Converging Point = (P + R — Eise) -j- by half the Oblique Span, thus, (.440389 + .625 - .25) -^ .6527036 = 1.2492484 Tangent of And the Adjusted Angle is equal to (.3985506 + .625 - .25) ^ .6527036 = 1.1851483 Tangent The Theoretic Radius of the End Joint Arc is equal to the length of the Heading Spiral (D E) squared, multiplied by the Eadius of the Cylinder, and the product divided by the square of the circular Arc (CD) = (1.430938)2 X .625 -H (1.15912)2- Or Eadius -^ Cosine 2^ = .625 -^ .6561674 The Theoretic Eadius of the Spiral Arc or Coursing Joint Arc is equal to the length of the Spiral Arc (D E) squared, multiplied by the Eadius of the Cylinder, and the product divided by the square of the Obliquity (C E) = (1.430938)' X .625 -^ (.8390996)2 OrRadins -^ Sine 2^ = .625 ^ .3438326 The Adjusted Eadius of the End Joint Arc is = Eadius -^ Cosine 2^ = .625 -=- .6996318 The Adjusted Radius of the Spiral Arc is = Radius -r Sine 2^ = .625 -^ .3003478 ... ..-.'.- 40° 1' 40° 1' 37° 14' 4° 7' 4° 0' .4403890 .4403151 .3985506 51° 20' 49° 50' .9524997 .9525008 1.81765 1.81776 .8933270 2.080920 e 50° to Chord 1. 63 TWISTING RULES. As the Depth of the Bed of the Arch Stones is equal to .1, the Rules must be exactly .1 in length, and if placed 1. apart [x) at the SofiSt their distance apart at the Extrados will be found thus, x^ is equal to x multiplied by the fourth term in the following stating — As the Secant of /3 is to the Secant of so is the Cosine of 6 to the length of the line A G on the Diagram, thus, As 1.2345044 : 1.3057261 :: .9974199 : 1.0549635 (A. G) x.^ = 1.0549635 x 1. x.^ Therefore the increased distance apart at the Extrados is equal to 1 .0549635 — 1. And the increased breadth of one end of one of the Twisting Rules is equal to x.-^ X by the Tangent of the Angle of Difference (Sj = 1.0549635 X .0719733 - A G on the Diagram — 1. is equal to 1.0549635 — 1. and is equal to the multiplier G H And A G multiplied by the Tangent of S is equal to 1.0549635 x .0719733 and is equal to the multiplier G F Or, As the Internal Radius is to the External Radius so is the Sine of /? ( = DH) on the Diagram toDF, As.625 : .725 :: .5863724(DH) : .6801919 (DF) HF = DF — DH = .6801919 — .5863724 = .0938195 = H F GH = HFxSine of ,8 = .0938195 x .5863724 GH G F = H F X Cosine of /J = .0938195 x .8100416 G F Any distance which may be found most convenient to place the Twisting Rules apart at the Soffit, multiplied by G H, under the corresponding Span and Angle of Obliquity found in the Table of Multipliers, will give the increased distance which the Rules should be placed apart at the Extrados, that is the amount of radiation across the Bed of the Arch Stone. This radiation can also be found by proportion as previously explained. And this Soffit distance apart, of the Twisting Rules, multiplied by G F, gives the increased breadth of the end of one of the Rules, which increased breadth gives the accurate twist to the Bed of the Skewback Check, and to the Bed of the Arch Stones, when the Rules are framed together and the Stones dressed thereto. One Rule must be parallel, and one end of the broadened Rule must be exactly the same width as the parallel Rule. 1.0549635 .0549635 .0759292 .0549635 .07592920 .0550131 .0759976 For Adjusted Multipliers see Example 1 and Table nf Adjusted Multipliers. 64 Oblique Bridges Made Easy. 9 55° TO CHORD 1. EXAMPLE TO CHOED 1. 9 = Angle of Obliquity = 55° (i = Theoretic Angle of Intrados - = 31° 8' /? = Adjusted Angle of Intrados = 30° 34' 07746 x .517034 The number of Courses that will spring from or intersect the Impost is equal to ; As the length of the Heading Spiral is to the length of the Theoretic Divergence, so is the full number of Courses in the Heading Spiral to the number of Courses in the length of the Divergence, or the number rising from the Impost, thus. As 1.3541652 : .63117880 :: 24 Therefore 11 is the correct number of Checks on the Impost, and the correct length of the Check is equal to the Impost divided by 11 = 1.2207746 -> U Hence the actual or adjusted Divergence will be 11 -h 24 (.458333) multiplied by the length of the Heading Spiral = .458333 x 1.3541652 Or the number of Checks multiplied by the thickness of one Course = 11 x .05642355 And making .62065900 = Sine, and the length of the Impost = Eadius, thus, .6206590 ^ 1.2207746 =.50841408 = Sine of the Adjusted Angle or ^ Or making the thickness of a Course of Arch Stones = Sine, and the length of Check = Eadius, thus, .05642355 h- .11097950 = .50841416 = Sine of ,8 And to this Angle, the Extradosal Angle, or c^ the Axial Length, and the Con- verging Point (P) for the Face Lines must all be adjusted : — = .60403-23 = 1.6555405 = .5906134 = 1.6931550 „ - = 1.1613889 by Cot. /3 = 1.15912 x The The Tangent of /8 or the Theoretic Angle of Intrados The Cotangent of /8 „ ,, „ The Tangent of ji or the Adjusted ,, ,, The Cotangent of /3 „ ,, ,, The Secant of fi „ ,, „ Theoretic Axial Length is equal the Arc 1.6555405 Or, As the Cotangent 9 is to the length of the Arc, so is the length of the Arc to the Axial Length, that is. As .7002075 : 1.15912 :: 1.15912 ; 1.9188014 And the Adjusted Axial Length is equal to the Arc x Cot. ,6 = 1.15912 x 1.6931550 As similar Arcs are to one another as their Radii, it follows that having found the Intradosal Angle the Extradosal Angle may be found by proportion, as the Tangents of the Intradosal and Extradosal Angles are develop- ments of their respective Arcs, therefore. As .625 : .725 : : .6040323 : .7006774 = the Tangent of Extradosal Angle or <^ or the (External Eadius ^ by the Internal Eadius) x tan. (i = tan. .^(actual tan. .7006411) = (.725 -r .625) X .6040323 = .7006774 = tan. = Theoretic Angle of Extrados - - (f> = Adjusted Angle of Extrados - - - - - S = Theoretic Angle of Difference, equal 30° 2' — 26° 29' S = Adjusted Angle of Difference, equal 28° 16' - 24° 52' y = Natural Springer Anglo, or the Angle at the bottom of the Check (0 = Theoretic Angle produced on the end of the Impost by cutting the Skewback to the Angle of (1) = Adjusted Angle produced on the end of the Impost by cutting the Skewback to the Angle of 6 E = Radius - - - - e = Depth of Arch Stones W = Width of Bridge, including Parapets C = Chord or Square Span a = Length of Arc (AM) 60° 26° 29' 24° 52' 30° 2' 28° 16' 3° 33' 3° 24' 36° 52' 12" 45° 18' 44° 35' .625 ,100 1.000 1.000 1.15912 Oblique Span (A E) = C x Cosecant of ^ = 1. x 1.1547005 - Or C H- Sine ^ = 1. h- .8660254 Or again, the Obliquity (C E) -^ Cosine 9 = .5773503 -;- .5000000 Obliquity of the Arch (CE,) = C x Cotangent of (9 = 1. x .5773503- Length of Arc (C D) is found thus : Half C -^ Radius = .50 -h .625 = .80 or Sine of A B = half the Arc = 53° 7'.805, therefore the whole Arc A BC = 53° 7'.805 x 2 =106° 15'.61, or Hence the Springer Angle at the bottom of the Check is (see Fig. 4, Plate 1) - And by Tables of Radius Unity the Arc is = .625 x 1.8545917 Tangent of the Theoretic Angle of Intrados C D E (/3) = (C x Cot. 6) -i- a = (1. X .5773503) - L15912 = .4980936 Tan. Length of Heading Spiral or Oblique Arc = Arc x Secant of /3 = 1.15912 x 1.1172384 (A 2) - 24 is the correct number of Courses, and their thickness = 1.2950133 -i- 24 Length of Impost (W) ^ Sine ^ = 1. -h .8660254 Or Length of Impost (W) x Cosecant ^ = 1. x 1.1547005- 1.1547005 1.1547005 1.1547006 .577350 106° 15' 36" 36° 52' 12" 1.15912 26° 29' 1.2950133 .05395888 1.1547005 1.1547005 68 Oblique Bridges Made Easy. The Divergence of the Courses is that part of tlie face of the Oblique Arc which is intercepted between the horizontal Springer Line and the Coursing Joint Line which springs from the acute quoin on the end of the Impost, and is equal to W x Cosecant 9 x Sine y8 = 1. x 1.1547005 X .4459375 Or the Impost x Sine /3 = 1.1547005 x .4459375 The number of Courses that will spring from or intersect the Impost is equal to : As the length of the Heading Spiral is to the length of the Theoretic Divergence, so is the full number of Courses in the Heading Spiral to the number of Courses in the length of the Divergence, or the number rising from the Impost, thus, As 1.2950133 : .51492425 :: .24 : Therefore 9 is the correct number of Checks on the Impost, and the correct length of the Check is equal to the Impost divided by 9 = 1.1547005-7-9 Hence the actual or Adjusted Divergence will be 9 -h 24 (.375) multiplied by the length of the Heading Spiral = .375 x 1.2950133 Or the number of Checks multiplied by the thickness of one Course = 9 x .05395888 And making .48562996 = Sine, and the length of the Impost = Eadius, thus, .48562996^1.1547005 = .4205678 = Sine of the Adjusted Angle, Or making the thickness of a Course of Arch Stones = Sine, and the length of Check = Eadius, thus, .05395888 -^ .12830005 = .4205678 Sine of/3 And to this Angle, the Extradosal Angle, or the Axial Length, and the Converging Point (P) for the Face Lines must all be adjusted : — The Tangent of /3 or the Theoretic Angle of Intrados The Cotangent of ^ „ „ „ The Tangent of /3 or the Adjusted Angle of Intrados The Cotangent of ^ „ „ ,, The Secant of /? ,, „ „ The Theoretic Axial Length is equal the Arc x by Cot. /3 2.007156 Or, As the Cotangent 6 is to the length of the Arc, so is the length of the Arc to the Axial Length, that is, As .5773503 : 1.15912 2.32711262 And the Adjusted Axial Length is equal to the Arc x Cot. /? 2.1576015 As similar Arcs are to one another as their Eadii, it follows that having found the Intradosal Angle the Extradosal Angle may be found by proportion, as the Tangents of the Intradosal and Extradosal Angles are develop- ments of their respective Arcs, therefore, As .625 : .725 :: .4982185 : .5779334 = the Tangent of Extradosal Angle, or ^ Or the (External Eadius -^ by the Internal Eadius) x Tan. fi =Tan. 4> (actual Tan. .5781262) .... = (.726 -^ .625) X .4982185 = .57793346 = Tan. -'.-.' Secant of <^ or Adjusted Extradosal Angle = 1.1353921 Theoretic Angle of Difference = 8 = 30° 2' - 26° 29' - Cosine S = .9980811, Tangent 8 = .0620386 Adjusted Angle of Difference = 8 = 28° 16' - 24° 52' Cosine 8 = .9982398, Tangent of 8 = .0594109 Distance of the Theoretic Point of Convergence of the Face Lines below the centre of the Cylinder = (C x Cot.^ 6) -4- Arc and x (R + e)= 1. x (.5773503)2 -=- 1.15912 x .725 = P - Or the Radius multiplied by the Cotangent of 9, and by the Tangent of the Un- adjusted Angle of Extrados (30° 2'j = .625 x .5773503 x .5781262 The Adjusted Poiat of Convergence (P) is equal the Radius multiplied by the Cotangent of 6 and by the Tangent of the Adjusted Angle That is .625 x .5773503 x .5376943 - - - - The Theoretic Tangent of the Angle formed between the first joint in the face at the acute end of the Impost, and the Horizontal Line shewn on the elevation at the end of the Impost, is equal to the distance of the (Converging Point + R — Rise) h- by half the Oblique Span. Thus (.20849151 + .625 - .25) -^ .5773502 = 1.0106372- And the Adjusted Angle is equal (.19402370 + .625 - .25) h- .5773502 = .9855780 - - . - - - - The Adjusted Radius of the End Joint Arc is equal to the Radius divided by the Cosine of the Intradosal Angle .squared (ji) =.625 -4-(.9072888)2 The Adjusted Radius of the Spiral Arc, or Coursing Joint Arc, is equal to the Radius divided hy the Sine of the Intradosal Angle squared {fi) = .625 -H (.4205080)' - - .... 28° 16' 3° 33' 3° 24' .20849151 .20861331 .19402370 45° 18' 44° 35' .75925726 3.53452869 TWISTING RULES. As 1.1021849 : 1.1353921 :: .9982398 : 1.0283152 = AG GH = AG-AH= 1.0283152 - 1. GF = AG X Tangent 8 = 1.0283152 x .0594109 - G H .0283152 GF. 06109313 70 Oblique Bridges Made Easy. e 65° TO CHOED 1. EXAMPLE TO CHORD 1. e = Angle of Obliquity - - = 65° (i = Theoretic Angle of Intrados - - = 21° 55' ji = Adjusted Angle of Intrados - - - = 22° 11' <^ = Theoretic Angle of Extrados - - = 25° 1' 4> = Adjusted Angle of Extrados - = 25° 19' S = Theoretic Angle of Difference, equal 25° 1' — 21° 55' = 3° 6' S = Adjusted Angle of Difference, equal 25° 19' — 22° 11' - = 3° 8' y = Natural Springer Angle, or the Angle at the bottom of the Check - = 36° 52' 12" (o = Theoretic Angle produced on the end of Impost by cutting the Skewback to the Angle oi d . = 42° 48' (1) = Adjusted Angle produced on the end of Impost by cutting the Skewback to the Angle oi d - - = 42° 55' R = Radius - - = .625 c = Depth of Arch Stones - = .100 W = Width of Bridge, including Parapets = 1.000 C = Chord or Square. Span - = 1.000 a = Length of Arc - - - (A M) = 1.15912 Oblique Span (A E) = C x Cosecant of 6^ = 1. x 1.1033779 Or C -f- Sine ^ = 1. -f .9063078 Or again, the Obliquity of (C E) -f Cosine d = .4663077 ~ .4226183 Obliquity of the Arch (C E) = C x Cotangent = 1. x .4663077 Length of Arc (C D) is found thus : Half C -^ Radius = .50 -;- .625 = .80 or Sine of A B = half the Arc = 53° 7' .805, therefore the whole Arc ABC = 53° 7' .805 x 2 = 106° 15' .61, or Hence the Springer Angle at the bottom of the Check is (see Fig. 4, Plate 1) - And by Tables of Radius Unity the Arc is = .625 x 1.8545917 Tangent of the Theoretic Angle of Intrados C D E (/8) = (C x Cotangent 0) -^ a = (1. x .4663077) -f 1.15912 = .4022945 Tangent - Length of Heading Spiral or Oblique Arc = Arc x Secant of ,8 = 1.15912 x 1.0779025 (A 2) 24 is the correct number of Courses, and their thickness = 1.2494183 -i- 24 Length of Impost (W) -r- sine 9=1.^ .9063078 Or Length of Impost (W) x Cosecant 6* = 1. x 1.1033779 1.1033779 1.1033779 1.1033779 .4663077 106° 15' 36" 36° 52' 12" 1.15912 21° 55' 1.2494183 .0520590 1.1033779 1.1033779 d 66° to Chord 1. 71 The Divergence of the Courses is that part of the face of the Oblique Arc which is intercepted between tlie Horizontal Springer Line and the Coursing Joint Line which springs from the Acute Quoin on the end of the Impost, and is equal to W x Cosecant d x Sine /? = 1. x 1.1033779 X .3732577 Or the Impost x Sine /3 = L1033779 x .3732577 The number of Courses that will spring from or intersect the Impost is equal to : As the length of the Heading Spiral is to the length of the Theoretic Divergence, so is the full number of Courses in the Heading Spiral to the number of Courses in the length of the Divergence, or the number risingfrom the Impost, thus, As 1,2494183 : .4118442:: 24 Therefore 8 is the correct number of Checks on the Impost, and the correct length of the checlc is equal to the Impost divided by 8 = 1.1033779 -4- 8 - - Hence the Actual or Adjusted Divergence will be 8 -r- 24 (.3333333) multiplied by the length of the Heading Spiral = .3333333 x 1.2494183 Or the number of Checks multiplied by the thickness of one Course = 8 x .0520590 - And making .4164727 = Sine, and the length of the Impost = Eadius, thus, .4164727 - 1.1033779 = .3774524 = Sine of the Adjusted Angle, or ^ Or making the thickness of a Course of Arch Stones = Sine, and the length of Check = Eadius, thus, .0520590 -f- .1379222 = .3774519 Sine of fi And to this Angle, the Extradosal Angle or (^ the Axial Length, and the Converging Point (P) for the Face Lines must all be adjusted : — The Tangent of fi or the Theoretic Angle of Intrados = .4023354 The Cotangent of jB „ „ „ = 2.4854887 The Tangent of /3 or the Adjusted Angle of Intrados = .4077531 The Cotangent of /3 „ „ „ = 2.4524642 The Secant of ^ ' „ „ „ = 1.0799364 The Theoretic Axial Length is equal the Arc x by Cotangent ji = 1.15912 x 2.4854887 Or, As the Cotangent 6 is to the length of the Arc so is the length of the Arc to the Axial length That is, As .466,3077 : 1.15912 :: 1.15912 : 2.8812716 And the Adjusted Axial Length is equal to the Arc x Cotangent ji = 1.15912 X 2.4524642 As similar Arcs are to one another as their Kadii, it follows that having found the Intradosal Angle the Extradosal Angle may be found by proportion, as the Tangents of the Intradosal and Extradosal Angles are developments of their respective Arcs, therefore. As .625 : .725 :: .4023354 : .4667090 = Tangent of Extradosal Angle or 4>, or the (External Eadius -=- by the Internal Eadius) x Tangent fi = Tangent 4> (actual Tangent .4666618) - = (.725 -;- .625) X .4023354 = ,4667090 = Tangent - - Secant ^ = 1,1035277 .4118442 .4118442 7.911090 .1379222 .4164727 .4164720 22° 11' 22' 11' 2.8809796 2.8812716 2.8427003 25° 1' 25° 1' 72 Oblique Bridges Made Easy. Then (R + £ -^ E) x Tangent fi = Tangent 4>= 1.16 x .4077531 = .4729935 = Tangent 9^ Secant of <^ or Adjusted Extradosal Angle = 1.1062458 Theoretic Angle of Difference = S = 25' 1' — 21° 55' Cosine 8 = .9985367, Tangent of 8 = .0541581 Adjusted Angle of Diflforence = 8 = 25° 19' - 22' 11' Cosine 8 = ,9985050, Tangent of 8 = .0547416 Distance of the Theoretic Point of Convergence of the face line below the centre of the cylinder = (C x Cot.^ d) -^ Arc and x(R + e)= 1, x (.4663077)2 4- 1.15912 x .725 = P Or the Radius multiplied by the Cotangent of 6, and by the Tangent of the Unadjusted Angle of Extrados (25° 1') = .625 x .4663077 x .4666618 The Adjusted Point of Convergence (P) is equal the Radius multiplied by the Cotangent of 6 and by the Tangent of the Adjusted Angle 4' That is .625 x .4663077 x .4730538 The Theoretic Tangent of the Angle formed between the first joint in the face at the acute end of the Impost and the horizontal line shewn on the elevation at the end of the Impost, is equal to the distance of the (Converging Point + R — Rise) 4- by half the oblique span. Thus (.1360049 + .625 - .25) + .5516889 = .9262555 And the Adjusted Angle is equal to (.1378678 + .625 - .25) 4- .5516889 = .9296322 The Adjusted Radius of the End Joint Are is equal to the Radius divided by the Cosine of the Intradosal Angle squared [fi) = .625 -;- (.9259805)- The Adjusted Radius of the Spiral Arc or Coursing Joint Arc, is equal to the Radius divided by the Sine of the Intradosal Angle squared [fi) = .625 -f (.3775714)2 25° 19' 3° 6' 3° 8' .1360049 .1360049 .1378678 42° 48' 42° 55' .72891414 4.384115 As 1.0799364 : GH = AG — GF = AG X 1.1062458 :: .9985050 AH = 1.022830 — 1. Tan. 8 = 1.022830 x TWISTING RULES. 1.022830 = AG GH = .022830 0547416 GF = .0559913 e 70° to Chord 1. e 70° TO CHORD 1. EXAMPLE TO CHORD 1. 73 S s > 7 (0 R e w c a Angle of Obliquity Theoretic Angle of Intrados . . . - Adjusted Angle of Intrados Theoretic Angle of Extrados Adjusted Anglo of Extrados Theoretic Angle of Difference, equal 20° 1' - 17° 26' Adjusted Angle of Difference, equal 19° 3' — 16° 35' Natural Springer Angle, or the Angle at the bottom of the Check Theoretic Angle produced on the end of the Impost by cutting the Skewback to the Angle of 6 Adjusted Angle produced on the end of the Impost by cutting the Skewback to the Angle of 6 Radius Depth of Arch Stones Width of Bridge, including Parapets Chord or Square Span Length of Arc (AM) 70° 17° 26' 16° 35' 20° 1' 19° 3' 2° 35' 2° 28 36° 52' 12" 40° 43' 40° 30' .625 .lUO 1.000 1.000 1.15912 Oblique .Span (A E) = C x Cosecant of ^ = 1. x 1,0641778 - Or C 4- Sine 6^ = 1. -^ .9396926 - - Or again, the Obliquity (C E) -r Cosine 6 = .3639702 ^ .3420201 - Obliquity of the Arch (U E) = C x Cotangent of (9 = 1. x .3639702 Length of Arc (C D) is found thus : Half C -=- Radius = .50 -=- .625 = .80 or Sine of A B = half the Arc = 53° 7'.805, therefore the whole Arc A B C = 53° T'.SOo X 2 = 106° 15.'61, or - - '- Hence the Springer Angle at the bottom of the Check is (see Fig. 4, Plate 1) And by Tables of Radius Unity the Arc is = .625 x 1.8545917 - Tangent of Theoretic Angle of Intrados ODE (/3) =(C x Cot. 6)^ a =(1 x .3639702) -T- 1.15912 = .3140056 Tan. Length of Heading Spiral or Oblique Arc = Arc x Secant of ^ = 1.15912 x 1.0481453 (A 2) - 24 is the correct number of Courses, and their thickness = 1.2149261 -^ 24 Length of Impost (W) -r Sine 6* = 1. - .9396926 - - - Or length of Impost (W) x Cosecant 5 = 1. x 1.0641778 - 1.0641778 1.0641778 1.0641778 .3639702 106° 15' 36" 36° 52' 12" 1.15912 17° 26' 1.2149261 ,05062192 1.0641778 1.0641778 n Oblique Bridges Made Easy. The Divergence of the Courses is that part of the face line of the Oblique Arc which is intercepted between the Horizontal Springer Line and the Coursing Joint Line which springs from the acute quoin on the end of the Impost, and is equal to W x Cosecant x Sine^ = L x 1.0641778 X .2995950 Or the Impost x Sine /? = 1.0641778 x .2995950 - - - The number of Courses that will spring from or intersect the Impost is equal to, As the length of the Heading Spiral is to the length of the Theoretic Divergence, so is the full number of Courses in the Heading Spiral to the number of Courses in the length of the Divergence, or the number rising from the Impost, thus, As 1.2149261 : .3188223 :: 24 Therefore 6 is the correct number of Checks on the Impost, and the correct length of the Check is equal to the Impost divided by 6 = 1.0641778 H- 6 . - - - - Hence the actual or Adjusted Divergence will be 6 -^ 24 (.25) multiplied by the length of the Heading Spiral = .25 x 1.2149261 - Or the number of Checks multiplied by the thickness of one Course = 6 x .05062192 And making .30373152 = Sine, and the length of the Impost = Radius, thus, .30373152 H- 1.0641778 = .2854142 = Sine of the Adjusted Angle, or/3 - - - Or making the thickness of a course of stones = Sine, and the length of the Check = Eadius, thus, .05062192 -=- .17736296 = .2854143 = Sine of ;8 - - - - And to this Angle, the Extradosal Angle, or 4> the Axial Length, and the Converging Point (P) for the Face Lines must all be adjusted : — The Tangent of /3 or the Theoretic Angle of Intrados = .3140200 The Cotangent of /3 „ „ „ „ = 3.1845102 The Tangent of (3 or the Adjusted Angle of Intrados = .2977962 The Cotangent of /3 „ „ „ „ = 3.3580008 The Secant of /3 „ „ „ „ = 1.0433995 The Theoretic Axial Length is equal the Arc x by Cot. y8= 1.15912 x 3.1845102 - . - - ... Or, As the Cotangent is to the length of the Arc, so is the length of the Arc to the Axial Length, that is. As .3639702 : 1.15912 :: 1.15912 : And the Adjusted Axial Length is equal to the Arc x Cot. /3 = 1.15912 x 3.3580008 - - ' - As similar Ares are to one another as their Eadii, it follows that having found the Intradosal Angle the Extradosal Angle may be found by proportion, as the Tangents of the Intradosal and Extradosal Angles are developments of their respective Arcs, therefore. As ,625 : .725 :: .3140200 : .3642631 = Tangent of Extradosal Angle, or (actual Tan. .3642997) - - = (.725 H- .625) X .3140200 = .3642632 = Tan. <^ Secant = 1.0642905. .3188223 .3188223 6.29810 .17736296 .30373152 .30373152 16° 35' 16° 35' 3.6912291 3.6913986 3.8923258 20° 1' 20° r e 70' io Chord 1. 75 Then (R + e -^ R) x Tangent /3 = Tangent <^ = 1.16 x .2977962 = .3454435 = Tan. - ' • -' Secant of <)^, or Adjusted Extradosal Angle = 1.0579390 Theoretic Angle of Difference = 8 = 20° 0' - 17° 25' Cosine S = .9989837, Tangent S = ,0451183 Adjusted Angle of Difference = 8 = 19° 3' - 16° 35' Cosine 8 = .9990734, Tangent of 8 = .0430781 Distance of the Theoretic Point of Convergence of the face line below the centre of theCylinder = (CxCot.2 61)^ Arc and x (E + f) = 1. x (.3639702) = ^ 1.15912 X .725 = P - - Or the Radius multiplied by the Cotangent of d, and by the Tangent of the Unadjusted Angle of Extrados (20° 0') = .625 x .3639702 x .3639702 - - - - The Adjusted Point of Convergence (P) is equal the Radius multiplied by the Cotangent of and by the Tangent of the Adjusted Angle <^, that is, .625 X .3639702 x .3453040 - - - - ' - - The Theoretic Tangent of the Angle formed between the first joint in the face at the acute end of the Impost and the horizontal line shewn on the elevation at the end of the Impost, is equal to the distance of the (Converging Point + R — Rise) -r by half the Oblique Span. Thus (.08285925 + .625 - .25) -v .5320889 = .8604938 - And the Adjusted Angle is equal to (.07855022 + .625 - .25) 4- .5320889 = .8523955 .... ... The Adjusted Radius of the End Joint Arc is equal to the Radius divided by the Cosine of the Intradosal Angle squared (^) = .625 -4- (.9584056)- - The Adjusted Radius of the Spiral Arc, or Coursing Joint Arc, is equal to the Radius divided by the Sine of the Intradosal Angle squared {/3) .625 -j- (.2854096)2 . . . . ... 19° 3' 2° 35' 2° 28' .0S285923 .08279S43 .07855022 40° 43' 40° 27' .6804376 7.6726091 TWISTING RULES. As 1.0433995 ; 1.0579390 :: .9990734 : 1.0129952 = A G G-H = A G - A H = 1.0129952 - 1 - GF = A G X Tan. 8 = 1.0129952 x .0430781 GH GF .0129952 .Oi36379 7fi Oblique Bridges Made Easy. Theoretic Angles as shewn on the Diagram. To find the Extradosal Angle from the Intradosal Angle. The Tangent of the Extradosal Angle (<^) is equal to the Tangent of Introdosal Angle [P) multiplied by the External Kadius of the Cylinder (R + e) divided by its Internal Radius (R). Thus Angle of Obliquity 6 = 35° R + £ R r 8.25 -r 6.25 = = 1.32 X 11.375 ^ 9.375 = = 1.21.333 X 14.50 -^ 12.50 = = 1.16 X 17.G25 -- 15.625 = = 1.128 X 20.75 -^ 18.75 = 1.10666 X 23.875 H- 21.875 = = 1.0914 X 27.0 ^ 25.0 = = 1.08 X 30.125 - 28.125 = = 1.0711 X .33.25 -^ 31.25 = = 1.064 X l'1.32 X 1.21333 X 1.16 X 1.128 X Obliquity 6 = 40° 1.10666 X 1.0914 X 1.08 X 1.0711 X 1 1.064 X f 1.32 X 1.21333 X 1.16 X 1.128 X 6 = 45°. 1.10666 X 1.0914 X 1.08 X 1.0711 X J .064 X ri.32 X 1.21333 X 1.16 X 1.128 X e = 50°- 1.10666 X 1.0914 X 1.08 X 1.0711 X .1.064 X 1.2319634 (tan. ji) 1.0283226 (tan. (i) = .8626694 (tan. /3) .7238793 (tan. fi) = Theoretic Theoretic Intradosal Extradosal Angle = /3 Angle = 4> tan <> 50° 56' 58° 25' )) 7> )) 56° 13' 7» IT )J 55° 1' )J ); )' 54° 16' J) 51 It 53" 45' V II 11 53° 22' n 51 11 53° 4' )) 1> )1 52° 51' )j )I 11 52° 39' tail. 4> 45° 48' 53° 37' ?» T5 ;? 51° 17' '5 15 »3 50° 2' )3 ?3 3T 49° 14' Jl 7J 3? 48° 41' J) 3) )1 48° 18' 31 3T 3) 48° 0' ?J J) 3» 47° 46' 1? )3 J> 47° 35' tan. ) 5> 33 43° 40' J) J3 33 43° 17' JJ 11 TJ 42° 58' )? 31 7» 42° 44' )» 33 ') 42° 33' tan. 'k 35° 54' 43° 42' )j J) 1) 41° 18' j> 11 !) 40° 1' )) 1) 11 39° 14' )) 1) 1) 38° 42' 3? )> >7 38° 19' 53 M 11 11 ji 38° r 37° 47' 36' Theoretic Angles. 77 Theoretic Angles as shewn on the Diageam. To find the Extradosal Angle from the Intradosal Angle. The Tangent of the Extradosal Angle {) is equal to the Tangent of the Intradosal Angle (/3) multiplied by the external radius of the cylinder (U + e) divided by the internal radius (R), Angle of Obliquity 6 = 55° = 60° 61 = 65° 6 = 70° (1.32 X .6040323 (tan P) 1.21333 X )) 1.16 X H 1.128 X )J 1.10666 X )) 1.0914 X )•< 1.08 X )5 1.0711 X JJ 1 1.064 X )) ri.32 X .4982185 (tan (3) 1.21333 X J) 1.16 X •3-I 1.128 X )) 1.10666 X )1 1.0914 X ■)•> 1.08 X JJ 1.0711 X 5) 1 1.064 X )J fl.32 X 4023354 (tan. P) 1.21333 X )J 1.16 X JJ 1.128 X )5 1.10666 X )T 1.0914 X J) 1.08 X J) 1.0711 X )) ll.064 X 51 /'1.32 X .3140200 (tan P) 1.21333 X 1.16 X 1.128 X 1.10666 X 1.0914 X 1.08 X 1.0711 X U.064 X tan. (ji tan. (/> tan, (j> Theoretic Theoretic Intradosal Extradosal Angle = p Angle = (ji 31° 8' 38° 34' J) )J 36° 14' )) )1 35° 1' J) )5 34° 16' J) )5 33° 46' n 17 33° 24' JJ Tf 33° 7' )) J) 32° 54' JJ J1 32° 44' 26° 29' 33° 20' JJ JJ 31° 9' JJ JJ 30° 2' JJ JJ 29° 20' JJ JJ 38° 52' JJ JJ 28° 32' JJ JJ 28° 17' JJ II 28° 5' JJ 'J 27° 56' 21° 55' 27° 59' T) J' 26° 1' )) JJ 2.5° 1' )J JJ 24° 25' JJ JJ 24° 0' JJ JJ 23° 43' JJ )J 23° 29' JJ Jj 23° 18' JJ JJ 23° 11' ir 26 22° 31' ij ji 20° 51' )i IJ 20° 1' JI JJ 19° 30' JI II 19° 10' II II 18° 55' JJ II 18° 44' II JI 18° 36' •J 11 18° 28' 78 Oblique Bridges Made Easy. Natural Sines, Cosines, Tangents, &c. To find the natural sine, cosine, tang^ent, cotangent, secant, cosecant. Sec, of an angle containing seconds. First find the natural sine, &c., due to the given degree and minute, subtract this natural sine, Ac, from the natural sine of the angle sought, and call this difference d ; then take out the natural sine, &c., due to the next higher minute. Find the difference between these two tabulated natural sines, itc, which is the change for one minute, or 60"; call this difference 1). Then as 60 seconds are to this difference between the tabulated sines, &c., — D of the less and greater angles than the one sought, so are the seconds only of the given angle — d to a decimal quantity to be ADDED to the sine first taken out, if it is a sine, tangent, secant, &c. ; but to be SUBTRACTED from the natural sine, tangent, &c., of the angle first taken out, if it is a cosine, cotangent, cosecant, &c. (See Figs. 3 and 4, Plate 1.) Find the natural sine of the angle CFG; the sine of this angle is equal to half the chord divided by the radius CG ^ CF = .5 -=-.625 = .8000000 the natural sine of the angle CFG. Given the natural sine = .8000000 The natural sine of 53° 7' = .7998593 1 { .0001407 = d Also „ „ „ „ 53° 8' = .8000338 „ „ „ „ 53° 7' = .7998593 .0001745 = D or change for 60" Then, As 1745 (D) : 1407 (d) :: 60" : 48".378 Hence .8000000 is the natural sine of 53° 7' 48".378 Find the natural sine of the springer angle (CFK, Fig. 4) ; the natural sine of this angle is equal to CK or GF ^ FC = .375 -^ .625 = .6000000, which is the natural sine of the angle CFK. Given the natural sine = .6000000 The natural sine of 36° 52' is = .5999549 .0000451 Also „ „ „ „ 36° 53' is = .6001876 „ „ „ „ 36° 52' is = .5999549 .0002327 = D or change for 60" Then, As 2327 (D) : 451 (d) :: 60" : 11".628 Hence .6000000 is the natural sine of 36° 52' 11".628, and -(- 53° 7' 48".378 = 90° 0' 0".006 the right angle GFK. Natural Sines, Cosines, Tangents, &c. 79 Find the angle CDE, Fig. 4, whose natural tangent is equal to CE -=- CD (.8390996 1.15912) = .7239108. Given the natural tangent = .7239108 The natural tangent of 35° 54' is = .7238793 Also the natural tangent of 3:".° 55' is = .7243227 0000315 = d 7243227 „ 35° 54' is = .7238793 .0004434 = D Then, As 4434 : 315 :: 60" : 4".2625 Hence .7239108 is the natural tangent of 35° 54' 4".2625 4 Find the natural tangent of 35° 54' 4".2625. The natural tangent of 35° 54' is Also .7238793 .7243227 The change for 60" is = .0004434 = D Then, As 60" : 4".2625 :: .0004434 : .000031499 = d Hence the natural tangent of 35° 54' 4".2625 is = .7238793 .00003149 .72391079 Find the natural cotangent of 35° 54' 4".2625169. The natural cotangent of 35° 54' is = 1 3814458 „ 35° 55' is = 1.3806001 The diflference for 60" is .0008457 = D Then, As 60" : 4".2625 ::. 0008457 : .0000600 1.3814458 .0000600 Hence the natural cotangent of 35° 54' 4".2625 is = 1.3813858 6 < Find the angle whose natural cotangent is 1.3813858. Given the natnral cotangent = 1.3813858 „ „ „ „ of 35° 54' = 1.3814458 1.3814458 Therefore this difference is .0000600 = d Also the natural cotangent of 35° 55' 1.3806001 The difference for 60" is = .0008457 = D Then, As .0008457 : .0000600:: 60" : 4".2568 Hence 1.3813858 is the natural cotangent of 35° 54' 4".2568. To find secants and cosecants proceed in like manner as in the above examples. 80 Oblique Bridges Made Easy. A Table of Theoretic Multipliers and Divisors, given under each Span IN 35°, 40°, 45°, 50°, 55°, 60", 65°, and 70° of obliquity, for finding the Theoretic Dimensions of all the Bridges in the corresponding Angles and Spans given on THE Diagram. (Sec the Diagram, Plate No. 7, and Fig 16, Plate 1.) AM IK I A DH LK CE J K A2 Or A 2 AG GH GF OrGR and GF HF GH GF is the length of the Arc to Chord 1. is the Cotangent of the Angle of Obliquity or 9 is the Cosecant of the Angle of Obliquity or 6 is the Sine of the Intradosal Angle or/3 is the Tangent of the Intradosal Angle or /3 is the Sine of the Extradosal Angle or ^ is the Tangent of the Extradosal Angle or is equal to the Arc to Chord 1. multiplied by the Secant of the Intradosal orfrontAngleof the Check = 1.15912 x 1.2345044 = Heading Spiral is equal to the Arc to Chord 1. divided by the Cosine of the Intradosal Angle = 1.15912 -h .8100416 = Heading Spiral (See Diagram, lengths of arcs on the line A B and Fig 4, Plate 1). is equal, As the Secant of the Intradosal Angle is to the Secant of the Extradosal Angle, so is the Cosine of the Angle of difference (S) to the length of A G = (See Diagram, Plate 7, and Fig. 16, Plate 1) is equal to A G - A H = 1.054951 - 1. = is equal to A G multiplied by the Tangent of the Angle of difference (8) = 1.054951 X .07197 = are found, As the Internal Radius is to the External Radius, so is the Sine of the Intradosal Angle D H to D F, thus, as 12.5: 14.5::. 58637 (DH):. 680189 (D F) = is equal DF - DH = .680189 - .58637 = is equal HF multiplied by the Sine of ^8 = .093819 x .58637 = is equal HF inultiplied by the Cosine of /3 = .093819 x .81004 = (See 50° and 20 feet Span in Tables of Multipliers.) 1.15912 = AM .83909 = IK 1.30540 = IA .58637 = DH .72387 = LK .64301 = CE .83959 = JK 1.430938 = A2 1.430938 = A2 1.054951 = AG .054951 = GH .0759248 = GF .680189 = DF .093819 = HF .05501 = GH .07599 = GF Use of the Multipliers and Divisors. Use of the Multipliers and Divisors. I.ength of Arc = Chord or Span x A M Obliquity = Chord or Span x IK, in the corresponding Span and Angle of Obliquity Oblique Span = Chord or Span x I A ■ Length of Impost = the External Width of the Bridge (W) x I A Or Ijength of Impost = the External Width of the Bridge (W) -^ Sine 9 The Heading Spiral is = the Chord, or Span x A 2 Length of Check is = the Theoretic Thickness of a Course of Arch Stones -=- D H (Sine j8) Number of Checks is = Length of Impost ■- by this Length of Check = whole number for Length of Check is = Length of Impost -H by the number of Checks is the correct - Eadius of the End Joint Arc or the Template to fit the curve Square across the Soffit of the Arch Stone = Radius -f- Cosine jB squared Eadius of the Spiral or Bed Joint Arc = Radius -H Sine fi squared Point of Convergence of the Face Lines below the centre of the Cylinder = Radius X IK & by JK Axial Length = Arc x Cotangent of Intradosal Angle (^8) = Length of Arc = Obliquity = Oblique Span = Length of Impost = Length of Impost = Heading Spiral = Length of Check = Xumber of Checks = Length of Check = End Joint Eadius = Spiral Eadius = Convergence = Axial Length TWISTING RULES. The Increased Distance apart of these Eules at the Extrados is equal to their Intradosal, or Soffit Distance apart multiplied by G H The Increased Width of one Rule, at one End, is equal to the Soffit Distance apart multiplied by 6 F .... Increased Distance = Increased Width 82 Oblique Bridges Made Easy. Table of Theoretic Multipliers and Divisors in 35°. 85° 10 feet SPAN 15 feet SPAN 20 FEET SPAN 25 FEET SPAN 30 FEET SPAN 35 FEET SPAN 40 FEET SPAN 45 FEET SPAN 50 FEET SPAN / Sine .57357 .57357 .57357 .57357 .57357 .57357 .57357 .57357 .57357 Cotangent = IK S 1.42814 1.42814 1.42814 1.42814 1.42814 1.42814 1.42814 1.42814 1.42814 Cosecant = I A... 1.74344 1.74344 1.74344 1.74-344 1.74344 1.74344 1.74344 1.74344 1.74344 Sine = DH .77641 ..77641 .77641 .77641 .77641 .77641 .77641 .77641 .77641 Cosine = D A ... .63022 .63022 .63022 .63022 .63022 .63022 .63022 .63022 .63022 Tangent = LK ... y3=50°66'. Cotangent 1.23196 .81171 1.23196 .81171 1.23196 .81171 1.23196 .81171 1.23196 .8:171 1.23196 .81171 1.23196 .81171 1.23196 .81171 1.23196 .81171 Secant = AL ... 1.58673 1.58673 1.58673 1.58673 1.58673 1.58673 1.58673 1.58673 1.58673 Cosecant 1.28797 1.28797 1.28797 1.28797 1.28797 1.28797 1.28797 1.28797 1.28797 SlNE = CE .85187 .83114 .81931 .81174 .80644 .80247 .79933 .79705 .79494 Tangent = JK...,^ 1.62653 1.49472 1.42903 1.38994 1.36.382 1.34486 1.33026 1.31984 1.31031 Secant = AJ .. 1.90935 1.79838 1.74417 1.71228 1.69116 1.67590 1.66421 1.65589 1.64830 Cosine 8- Tangent .99148 .99575 .99746 .99830 .99879 .99909 .99938 .99945 .99955 .13135 .09247 .07138 .05824 .04919 .04249 .03521 .03317 .02997 A2 1.83921 1.83921 1.83921 1.83921 1.83921 1.83921 1.83921 1.83921 1.83921 AG 1.19307 1.12857 1.09643 1.07728 1.06452 1.05523 1.04817 1.04301 1.03833 GH .19307 .12857 .09643 .07728 .064.52 .05523 .04817 .04301 .03833 ^ GF .15741 .10435 .07826 .06274 .05236 .04483 .03690 .03459 .03111 Theoretic Multipliers and Divisors in 40°. 83 Table of Theoretic Multipliers and Divisors in 40°. 10 FEET SPAN 15 FEET SPAN 20 FEET SPAN 25 FEET SPAN 30 FEET SPAN 35 FEET SPAN 40 FEET SPAN 45 FEET SPAN 50 FEET SPAN 1 Sine ( .64278 .64278 .64278 .64278 .64278 .64278 .64278 .64278 .64278 COTANGKNT = I K 6- 1.19175 1.19175 1.19176 1.19175 1.19175 1.19175 1.19175 1.19175 1.19175 Cosecant = I A . . . 1.55672 1.55572 1.55572 1.55572 1.55572 1.55572 1.55572 1.55572 1.56572 Sine = D H .. r .71691 .71691 .71691 .71691 .71691 .71691 .71691 .71691 .71691 Cosine = P A ... .69716 .69716 .69716 .69716 .69716 .69716 .69716 ,69716 .69716 Tangent = L K ... /3= 45° 48'. Cotangent 1.02832 .97245 1.02832 .97245 1.02832 .97245 1.02832 .97245 1.02832 .97245 1.02832 .97245 1.02832 .97245 1.02832 .97246 1.02832 .97245 Secant = A L ... 1.43438 1.43438 1.43438 1.43438 1.43438 1.43438 1.43438 1.43438 1.43438 40° Cosecant 1.394E7 1.39487 1.39187 1.39487 1.39487 1.39487 1.39487 1.39487 1.39487 Si.\E = CE ... ' .80506 .78024 .76641 .75737 .75107 .74663 .74314 .74041 .73825 Tangent = J K ...^- 1.35719 1.24746 1.19316 1.15987 1.13760 1.12237 1.11061 1.10165 1.09450 Secant = a J ... 1.68581 1.59879 L55680 1.63144 1.51464 1.50323 1.49447 1.48775 1.48254 Cosine I Tangent \ .99070 •99542 .99727 .99820 .99873 .99904 .99926 .99941 .99951 .13727 .09599 .07472 .05999 .05036 .04366 .03841 .03433 .03113 A 2 ' 1.66261 1.66261 1.66261 J.66261 1.66261 1.66261 1.66261 1.66261 1.66261 AG 1.16435 1.10951 1.08238 1.06574 1.0.5461 1.04699 1.04112 1.03652 1.03306 l'^« .16435 .10951 .08238 .06674 .05461 .04699 .04112 .03662 .03306 ^GF [ .15983 .10650 .08011 .06393 .05311 .04571 .03998 .03558 .03215 4s ^,^,„^- 84 Oblique Bridges Made Easy. Table op Theoretic Multipliers and Divisors in 45° 45° 10 TEET 15 feet 20 FEET 25 FEET 30 FEET 35 FEET 40 FEET 45 FEET 50 FEET SPAN SPAN SPAN SPAN SPAN SPAN SPAN SPAN SPAN / Sine .70710 .70710 .70710 .70710 .70710 .70710 .70710 ,70710 .70710 Cotangent = I K 0- 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 Cosecant = I A ... ' 1.41421 1 1.41421 1.41421 1.41421 1.41421 1.41421 1.41421 1.41421 1.41421 Sine = D H .65320 .65320 .65320 .65320 .65320 .65-320 .65320 .66320 .65320 Cosine = D A .. .75718 .75718 .75718 .75718 .75718 .75718 .75718 .75718 .75718 Tangent = LK ... .86266 .86266 .86266 .86266 .86206 .86266 .86266 .86266 .86266 /3 40°47'- Cotangent 1.15919 1.15919 1.15919 1.15919 1.15919 1.15919 1.15919 1.15919 1.15919 Secant = AL 1.32068 1.32068 1.32068 1.32068 1.32068 1.32068 1.32068 1.32068 1.32068 Cosecant . 1.53092 1.53092 1.53092 1.53092 1.53092 1.53092 1.53092 1.53092 1.53092 SlNE = CE .75088 .72296 .70731 .69737 .69046 .68560 .68157 .67858 .67623 Tangent = J K ... 1.13694 1.04644 1.0n058 .97302 .95450 .94180 .93142 .92385 .91793 Secant = AJ ... 1.51414 1.44742 1.41462 1.39526 1.38242 1.37367 1.36658 1.36143 1.35742 Cosine | Tangent 1 .99054 .99536 .99727 .99820 .99873 .99904 .99927 .99942 .99952 .13846 .09658 .07402 .05999 .05036 .04395 .03812 .03404 .03084 A2 1.53082 1.53082 1.53082 1.53082 1.53082 1.53082 1.53082 1.53082 1.53082 AG 1.13563 1.09088 1.06820 L05456 1.04541 1.03912 1.03399 1.03025 1.02725 GH .13563 .09088 .06820 .05456 .04541 .03912 .03399 .03025 .02725 \cr .15723 .10535 .07906 .06326 .05264 .04536 .03941 .03506 .03168 Theoretic Multipliers and Divisors in 50" Table of Theoretic Multipliers and Divisors in 50°. 50° 10 feet SPAN 15 FEET SPAN 20 FBF.T SPAN * 25 FRET SPAN 30 FEET 35 FEET SPAN SPAN 40 FEET SP.\N 45 FEET SPAN 50 FEET SPAN / Sine " .76604 .76604 .76604 .76604 .76604 .76604 .76-04 .76604 .76604 Cotangent = I K .83909 .83909 .83909 .83909 .83909 .83909 .83909 .83909 .839119 Cosecant = I A... 1.30540 1.30540 1.30540 1.30540 1.30540 1.30540 1.30540 1.30540 1.30540 SiNE = DH ... , .58637 .58637 .58637 .58637 .58637 .58637 ..58637 .58637 .58637 Cosine = DA ... .81004 .81004 .81014 .81004 .81004 .81004 .81004 .81004 .81004 Tangent = LK ... /J = 35°54'. Cotangent .72387 1.38144 .72387 1.38144 .72387 1.38144 .72387 l.,38144 .72387 1.38144 .72387 1.38144 .72387 1.38144 .72387 1.38144 .72387 1.38144 Secant = A L 1.23450 1.23450 1.23450 1.23450 1.23450 1.23450 1.23450 1.23450 1.23450 Cosecant SisE=CE ... ■ 1.70540 1.70540 1.70540 1.7C640 1.70540 1.70540 1.70540 1.70540 1.70540 .69f88 .66000 .64301 .63248 .62524 .62000 .61589 .61267 .61014 Tangent = JK ,. ^ .95562 .87852 .83959 .81654 .80115 .79022 .78175 .77521 .77010 Secant = A J ... 1.38318 1.33108 1.30572 1.29102 1.28134 1.27454 1.26930 1.26528 1.26216 Cosine ( Tangknt \ .99074 .99556 .99741 .99830 .99880 .99911 .99931 .99945 .99955 .13698 .09452 .07197 .05824 .04890 .04220 .03695 .03288 .02967 A2 ' 1.43093 1.43093 1.43093 1.43093 1.43093 1.43093 1.43093 1.43093 1.43093 AG 1.11006 1.07344 1.05495 1.04400 1.03669 1.03151 1.02748 1.02436 1.02194 GH .11006 .07344 .05495 .04400 .03669 .03151 .02748 .02436 .02194 .GF , .15205 .10146 .07592 .06080 .05069 .04352 .03796 .03368 .03032 8G Oblique Bridges Made Easy. Table of Theoretic Multipliers and Divisors in 55° 10 FEET SPAN 15 FEET SPAN 20 FEET SPAN 25 FEET SPAN 30 FEET SPAN 35 FEET SPAN 40 FEET SPAN .81915 45 FEET .SPAN 50 FEET SPAN / Sine C .81915 .81915 .81915 .81915 .81915 .81915 .81915 .81915 Cotangent = 1 K Q. .70020 .70020 .70020 .70020 .70020 .70020 .70020 .70020 .70020 Cosecant = I A... 1.22077 1.22077 1.22077 1.22027 1.22077 1.22077 1.22077 1.22077 1.22077 Sine = D H .51703 .51703 .51703 .51703 .51703 .51703 .51703 .51703 .51703 Cosine = D A ... .8.5596 .85596 .85596 .85596 .85596 .85596 .85596 .85596 .85596 Tangent = LK... ^=31°8'. Cotangent .60403 1.65554 .60403 1.65554 .60403 1.65554 .60403 1.65554 .60403 1.65554 .60403 1.65554 .60403 1.65554 .60403 1.65554 .60403 1.65554 Secant = A L . . 1.16827 1.16827 1.16827 1.16827 L16827 1.16827 1.16827 1.16827 1.16827 Cosecant 1.93411 1.93411 1.93411 1.93411 1.93411 1.93411 1.93411 1.93411 1.934U 55° SiNE = CE ... r .62342 .59107 .57381 .56304 .55581 .55048 .54634 .54317 .54072 Tangent = J K ...c/j. .79733 .73278 .70064 .68130 .06859 .65937 .65230 .64692 .64281 Secant = AJ ... [ 1.27896 1.23974 1.22102 1.21002 1.20292 1.1P7S2 1.19394 1.19101 1.18878 Cosine (\ .99159 .99604 .99770 .99850 .99894 .99921 .99940 .99952 .99961 Tangent [ .13046 .08924 .06788 .05474 .04599 .03958 .03462 .03084 .02793 A 2 L35416 1.35416 1.35416 1.35416 1.35416 1.35416 1.35416 1.35416 1.35416 AG 1.08554 1.05697 1.04275 1.03418 1.02856 1.02457 1.02107 1.01897 1.01715 GH .08554 .05697 .04275 .03418 .02856 .02457 .02107 .01897 .01715 \ GF .14161 .09432 .07078 .0566J .04730 .04055 ,03534 .03142 .02840 Theoretic Multipliers and Divisors in 60° 87 Table of Theoretic Multipliers and Divisors in 60°. 60 60' 10 FEET SPAN 15 FEET SPAN 20 FEET SPAN 25 FEET SPAN 30 FEET SPAN 35 FEET SPAN 40 FEET SPAN 45 FEET SPAN 50 FEET SPAN ' Sine [ .86602 .86602 .86602 .S6602 .86602 .86602 .86602 .86602 .86602 Cotangent = I K Qs .57735 .57735 .57735 .57735 .57735 .57735 .57735 .57735 .57735 Cosecant = IA... \ 1.15470 1.15470 1.15470 1.15470 1.15470 1.15470 1.15470 1.15470 1.15470 SiNE = DH .44593 .44593 .44593 .44593 .44593 .44593 .44593 .44593 .44593 Cosine = DA ... .89506 .89506 .89506 .89506 .89506 .89506 .89506 .89506 .89506 Tangent = LK ... /3 = 26°29'. Cotangent .49821 2.00715 .49821 2.00715 .49821 2.00715 .49821 2.00715 .49821 2.00715 .49821 2.00715 .49821 2.00715 .49S21 2.00715 .49821 2.00715 Secant = AL ... 1.11723 1.11723 1.11723 1.11723 1.11723 1.11723 1.11723 1.11723 1.11723 Cosecant 2.24246 2.24246 2.24246 2.24246 2.24246 2.24246 2.24246 2.24246 2.24246 Sine = C E .54950 .51728 .50050 .48988 .48277 .47767 .47383 .47075 .40844 Tangent = JE ... <^. .65771 .60442 .57812 .56193 .55127 .54370 .53806 .53357 .53021 Secant = A J ... 1.19fi9J 1.16847 1.15508 1.14707 1.14188 1.13825 1.13556 1.13344 1.13187 Cosine { Tangent 1. .99286 .99668 .99808 .99876 .99913 .99936 .99950 .99961 .99967 .12012 .08162 .06203 .04978 .04162 .03579 .03142 .02793 .02531 A 2 1.29500 1.29500 1.29500 1.29500 1.29500 1.29500 1.29500 1.29500 1.29500 AG 1.06366 1.04239 1.03189 1.02543 1.02117 1.01816 1.01589 1.01411 1.01276 GH .06366 .04239 .03189 .02543 .02117 .01816 .01589 .01411 .01276 \ GT .12776 .08507 .06400 .05104 .01250 .03643 .03191 .02832 .02563 Oblique Bridges Made Easy. Table of Theoretic Multipliers and Divisoks in 65° 65° 10 feet 15 FEET 20 FEET 25 FEE'I- 30 FEET 35 FEET 40 FEET 45 FEET 50 FEB! span .90630 span SPAN SPAN SPAN SPAN .90630 SPAN SPAN SPAN ; Sine .90630 .90630 .90630 .90630 .90630 .90630 .90630 Cotangent = I K .46630 .46630 .46630 .46630 .46630 .46630 .46630 .46630 .46630 Cosecant = I A... 1.10337 1.10377 1.10337 1.10337 1.10337 1.10337 1.10337 1.10337 1.10337 Sine = DH .37325 .37325 .37325 .37325 .37325 .37325 .37325 .37325 .37325 CO.SINE = D A .. .92772 .92772 .92772 .92772 .92772 .92772 .92772 .92772 .92772 Tangent = LK ... .40233 .40233 .40233 .40233 .40233 .40233 .40233 .40233 .40233 /3 = 21°55'. Cotangent 2.48548 2.48548 2.48548 2.48548 2.48548 2.48548 2.48548 2.48548 2.48548 Secant = A L ... 1.07790 1.07790 L07790 1.07790 1.07790 1.07790 1.07790 1.07790 1.07790 Cosecant 2.67911 2.67911 2.67911 2.67911 2.67911 2.67911 2.67911 2.67911 2.67911 Sine = CE .46921 .43863 .42288 .41336 .40673 .40221 .39848 .39554 .39367 Tangent = J K ... d,- .53133 .48809 .46666 .45397 .44522 .43931 .43446 .43066 .42825 Secant = A J ... 1.132.39 1.11275 1.10352 1.09822 1.09463 1.09224 1.09030 1.08879 1.08784 Cosine ( Tanoent ... ' ... [ .99439 .99744 .99853 .99917 .99933 .99950 .99962 .99970 .99975 .10628 .07168 .05415 .04074 .03637 .03142 .02735 .02414 .02211 A 2 j.24948 1.24948 1.24948 1.24948 1.24948 1.24948 1.24948 1.24948 1.24948 AG 1.04465 1.02968 1.02226 1.01800 1.01484 1.01279 1.01111 1.00979 1.00896 GH GF .04465 .02968 .02226 .01800 .01484 .01279 .01111 .00979 .00896 .11102 .07380 .05535 .04147 .03690 .03182 .02765 .02437 .02230 Theoretic Multipliers and Divisors in 70° 89 Table of Theoretic Multipliers and Divisors in 70°. 10 FEET SPAN 15 FEET SPAN 20 FEET SPAN 25 FEET SPAN 30 FEET SPAN 35 FEET SPAN 40 FEET SPAN 45 FEET SPAN 50 FEET SPAN /Sine { .939C9 .93969 .93969 .93969 .93969 .93969 .93969 .93969 .93969 Cotangent = I K .36397 .36397 .36397 .36397 .36397 .36397 .36397 .36397 .36397 Cosecant = I A . . . 1.06417 1.06417 1.06417 1.06417 1.06417 1.06417 1.06417 1.06417 1.06417 SiNE = DH .29959 .29959 .29959 .29959 .29959 .29959 .C9969 .29959 .29969 Cosine = D A ... ,95406 .95406 .95406 .95406 .95406 .95406 .95406 .95406 .95406 Tangent = LK ... ^ = 17° 26'. Cotangent .31402 3.18451 .31402 3.I8J51 .31402 3.18451 .31402 3.18451 .31402 3.18451 .31402 3.18451 .31402 3.18451 .31402 3.18451 .31402 3.18451 Seoant = AL ... 1.04814 1.04814 1.04814 1.04814 1.04814 1.04814 1.04814 1.04814 1.04814 70°r Cosecant 3.33782 3.33782 3.33782 3.33782 3.33782 3.33782 .3.33782 3.33782 3.33782 Sine = CE .38295 .35592 .34229 .33380 .32831 .32419 .32116 .31895 .31675 Tangent = JK ... <^- .41455 .38086 .36429 .35411 .34758 .34270 .33912 .33653 .33394 Secant = A J ... 1.08252 1.07007 1.06429 1.06084 1.05868 1.05709 1.05593 1.05511 1.05428 Cosine .. ... ( Tangent [ .99606 .99822 .99898 .99934 .99954 .99966 .99974 .99979 .99983 .08895 .05970 .04511 .03603 .03026 .02589 .02269 .02036 .01803 A2 1.21492 1.21492 1.21492 1.21492 1.21492 1.21492 1.21492 1.21492 1.21492 AG 1.03492 1.01910 1.01437 1.01144 1.00959 1.00819 1.00717 1.00643 1.00568 GH .03492 .01910 .01437 .01144 .00959 •00819 .00717 .00643 .00568 ^GF .09205 .06083 .04575 .03649 .03055 .02610 .02285 .02049 .01813 90 Oblique Bridges Made Easy. Table of Adjusted Multipliers for Twisting Eules. 10 FEET 15 FEET 20 FEET 25 FEET 30 FEET 35 FEET 40 FEET 45 FEET 50 FKET 1 GH { GF SPAN SPAN SPAN SPAN SPAN SPAN SPAN SPAN SPAN .18420 .12268 .09218 .07386 .06147 .05231 .04603 .04087 .03685 61=35° .15835 .10546 .07924 .06348 .05284 .04502 .03957 .03513 .03167 1 GH .15103 .10085 .07558 .06045 .05031 .04306 .03769 .03356 .03^33 = 40° 1 GF .15971 .10664 .07992 .06393 .05320 .04554 .03986 .03549 .03207 6 = 45° r GH 1 GF .12766 .15681 .08505 .10447 .06385 .07843 .05108 .06275 .04249 .05220 .03634 .04464 .03203 .03934 .02825 .03470 .02550 .03133 e = 50° ■ GH GF .09603 .14656 .06406 .09777 .04801 .07328 .03802 .05803 .03208 .04897 .02743 .04186 .02401 .03665 .02142 .03269 .01923 .02936 = 55° r GH 1 GF .08267 .13997 .05506 .09323 .04139 .07008 .03304 .05594 .02755 .04666 .02357 .03991 .02070 .03504 .01837 .03111 .01606 .02719 = 60° j' GH 1 GF .05651 .12193 .03771 .08136 .02831 .06109 .02262 .04882 .01882 .04062 .01617 .03489 .01408 .03039 .01256 .02710 .01777 .02457 = 65° r GH \ GF .04569 .11205 .03036 .07445 .02283 .05599 .01824 .04475 .01517 .03722 .01310 .03213 .01139 .02795 .01018 .02498 .00909 .02231 = 70° ( GH 1 GF .02509 .08427 .01738 .05836 .01299 .04363 .01041 .03499 .00874 .02905 .00876 .02494 .00654 .02196 .00575 .01931 .00522 .01754 Having found G H and G F in any angle of obliquity for any span, G H and G F may be found by proportion for any other span in the same angle of obliquity to the third place of decimals, from the tabulated values of G H and G F, thus — As 30 feet span : 20 feet span :: .04801 (G H) : .03200 = G H in 30 feet span. As 30 feet span : 20 feet span :: .07328 (GF) : .04885 = G Fin 30 feet span. Or, having found G H & G F in 50 feet span, thus — As 20 : 50 :: .01923 : .04807 = G H in 20 feet span. As 20 : 50 :: .02936 : .07340 = G F in 20 feet span. Other tabulated dimensions may also be similarly found by proportion, namely, the radii for the end and spiral arcs, the axial length, &c. Adjusted Dimensions of Oblique Bridges. 91 Adjusted Dimensions of 72 Oblique Bridges, of Different Spans and Angles of Obliquity. < s u o d CD 1 !1 u -a a o .3 B 'o ■■so OS Ph o ■g w ■s §^ <4-( o 1 oDh- 1! a « to a O > S3 5S Is 1° II 1 27° 9' 30.0 7.50 18.75 22.778 73.702 5.551 2.020. 23.094 9 2.566 36 1.079 34.641 38.850 17.32034.773 75,027 „ ,: 26°50' 35.0 8.75 21.875 26.574 85.990 3.388 2.020. 23.094 9 2,566 42 1.079 40,414 45.325 20.207 40.569 87.532 „ „ 26°35' 40.0 10.00 25.0 30.371 98.231 7.222 2,0 20. 23.094 9 2.566 48 1.079 46,188 51.800 23.094 46.364 100.036 „ „ 26°24' 45.0 11.2.n 28.125 34.167 110.554 8.060 2.0 20.23.094 9 2.560 54 1.079 51.961 58.275^25.980 52.160 112.541 \ 26°15' 50.0 12.50 31.250 37.963 122.830 8.897 2.0 20. 23.094 9,2.566 60 1.079 57.735 64.750 28.367,57.956 125.045 j/i all Bridr/es where the liisc is one-fuurth of the Span, the &'prinr/ir Angle = J, at the bottom of tile Checl; is 36° 52' \V' 'ind the Radius is equal to the Chord or Span multiplied bi/ .625 ; or the Versed Sine or Rise multiplied by 2 5 thus 20. X .625 ^ 12,5; and 5. x 2.5 = 12.5. The length of Ares of these proportions is equal to the Chord or Span yiidtiplied by l.],'',9r2, or the Radius multiplied by ' ~ ' 12.5 X 1.8545917 = 23,18239625. In the tabulated cakulatio Adjusted Dimensions of Oblique Bridges. 93 Adjusted Dimensions of 72 Oblique Bridges, of Different Spans and Angles of Obliquity. 3 -J ca- -6- ■ i( ll S -S3 s II ■3 1 ■s s cc M ■^ i3 ■tt S x: 13 s ^ 3.1415926 0.497150 Area of a circle to radius 1 ) Area of a circle to diameter 1 0.7853982 1.895090 Capacity of a sphere to diameter 1 0.5235988 1.718999 Capacity of a sphere to radius 1. 4.1887902 0.622089 Arc equal to radius 57°.2957795 1.758123 Length of 1. degree to radius 1. 0.01745329 2.241877 Sine of 1. second to radius 1 0.00000485 6.685575 Basis of Napierian logarithms 2.7182818 0.434294 Modulus of common logarithms 0.4342945 1.637784 Complement to the same = Nap. log. 10 2,3025851 0.362216 98 Oblique Bridges Made Easy. COPY OF SPECIFICATION. CUMBERLAND COUNTY COUNCIL. ^IJeCtftCattOn of Work to be done and the Material to be used in the erection of an Oblique Bridge at an angle of 45° over the River King at Kingsbridgeford, in the Parish of Lanercost, in the County of Cumberland, as shewn in Pink on the Plan of Site, together with the necessary approaches thereto also shewn on the Plans and Sections, and explained in this Specification. In the above Bridge the square span is 40.0 feet, oblique span 56.568 feet, rise of arch 10.0 feet, radius of cylinder 25.0 feet, external width of Bridge including Parapets 20. feet, angle of springers at the bottom of the checks 36° 52' 12", dimensions of springers 4.040 feet long, by 3.000 feet broad, by 3.333 feet thick, and the two obtuse angle springers 7.0 feet long, by 3.000 feet broad, by 3.700 feet thick, length of arch stones twice the length of the bed of the check or 3.133, depth of arch stones or breadth of bed 2.0 feet. All stone used in this Bridge to be from Leeshill Quarry, or from such quarry or quarries as may be approved of in writing by the County Surveyor and Bridge Master. Interpretation. 1. The following words when used in this Specification shall be considered as having the meanings hereby assigned to them, namely : — 1st, Board shall mean the Highways and Bridges Committee of the Cumberland County Council ; 2nd, Contractor shall mean one or more Contractor or Contractors for the works herein specified, and shall include his or their legal personal representatives ; 3rd, Surveyor shall mean George Jos. Bell of Carlisle, County Surveyor and Bridge Master, or the Surveyor for the time being of the above Authority ; 4th, the Works shall mean any Works shewn on the Drawings, or referred to in this Specification, or the Bill of Quantities, or which may be found necessary during the progress of the work. Extent of Contract. 2. This Contract is to be considered as requiring the Contractor to provide, at his own expense, the whole of the materials, and all requisite implements and machinery of every description, cartage, manual labour, centering, and coffer-dams (if required), and the forming of all temporary Bridges and Works necessary for the completion of the Works herein described, and enumerated in the Bill of Quantities, or which may be ordered from time to time by the County Surveyor and Bridge Master. The whole of the Works to be completed within the specified time and in accordance with the Plans and Specifications ; both sides of the Works to be finished in the same way as shewn on the Elevation, and at the prices set forth in the Bill of Quantities, ard in the position shewn on the Plans above-mentioned. Copy of Specification. 99 MATERIALS AND WORKMANSHIP. 3. The foundations of the abutments, the springers, arch, string courses, coping', stone, and the remaining parts of the Bridge, shall be of good, clean, hard, sound local stone, or from some other quarry approved of by the Surveyor, in this case the springers, arch stones, string courses, and coping and every part of the Bridge, shall be from Leeshill Quarry. 4. Cement shall be Portland Cement of the best quality, capable of setting Cement, perfectly hard in v^ater, and after six days' immersion to bear a tensile strain of 300 lbs to the square inch. 5. The cement mortar shall be mixed in the follov^fing proportions, namely : — Cement Mortar. Cement (by measure) one part to two parts of clean-vifashed sharp sand, mixed dry twice on a clean board, and then sufficient water added to allow of its being tempered for use at pnco. A second mixing or tempering will not be allowed. G. The lime to be used shall be the best local lime, as may be determined by the Lime. Surveyor, and shall be brought fresh from the kilns upon the Works and only in such quantities as will allow of its being used perfectly fresh and active. 7. Lime mortar shall be mixed in the following proportions, namely: — Well Lime Mortar, slacked lime (by measure) one part to three parts of clean-washed sharp sand, mixed on a clean board, and then tempered with water for use. 8. In using either class of mortar, it shall be mixed to such a consistency as will How Cement and allow of a full trowelful to be taken up the surface of the wall with ease ; stone or ^™^ be'us* df'^^ ° other work shall be free from dirt, and if dry such surface shall be well wetted with water. The mortar shall then be spread evenly on the surface, so as to allow the stone placed upon it to have a full bed of J of an inch in thickness after forcing out with a heavy wooden mallet any excess between the surfaces, and all joints must be flushed perfectly full, and all packing set in and surrounded by a full bed of mortar, and any surplus brought to the face of the work shall be neatly taken off with the trowel, and the joints tuck-pointod with cement mortar as the luorh proceeds. 9. Cement concrete for foundations and the backing of the arch, &c., shall be mixed Cement Concrete, in the following proportions, namely : — Three parts (by measure) of broken stone or clean gravel from 1 to 3 inches in diameter, to two parts of clean-washed sharp sand and one part of cement ( = 5 to 1), mixed together in a dry state, and twice wet on clean boards with sufficient water added to bring it to a proper temper, when it shall be immediately placed in site in six-inch layers, and each layer well rammed ; it must on no account be left to set, as a second mixing or watering will not be allowed. If the concrete is to be applied in water, the concrete must be lowered into position in a suitable box, the bottom of which can be released to allow the contents to be deposited on the floor of the foundations, so as to prevent washing the cement out of the concrete. 10. Eubble work consists of large stones rough from the quarry. The ordinary Rubble Work in class of stone must be self-bedded, and in size and weight shall be just within the ^°^°iJohes ^^^ capacity of an ordinary man's strength to lift. In the foundations they must be the largest and strongest stones obtainable. The stones to be placed with the broadest and truest bed downwards, and every stone shall be set solid in and surrounded by a full bed of cement mortar. All inequalities amongst the largest stones shall be levelled up with smaller stones, and all these shall also be set in and surrounded by a full bed of 100 Oblique Bridges Made Easy. cement mortar, for in no part of the structure will dry fulling or packing be allowed. Every stone throughout ihe entire structure shall he sent in and surrounded by either cement or lime mortar (as the case maybe), and a double-handled wooden mallet must be used to secure a sound and firm bed for each stone. The face work must be preserved true and fair, and the wall from face to back levelled up in the manner described above, in courses of not more than 3 feet in height. The first layer on each course to be entirely of " through stones " or " headers," which must reach not less than 3 feet into the wall should the wall be more than 3 feet in thickness, and if less than 3 feet thick, then, in such case, the headers shall reach through the full width of such wall. Springers, Coping, 11. The springers, arch stones, copings, kc, shall be of the best local stone, or such other stone as shall be approved of by the Engineer or Surveyor, and be dressed true to the dimensions shewn on the plans and sections or otherwise given in detail ; the beds and joints to be truly squared or otherwise shaped as may be required, all faces to be left broken or "rock" face; the coping shall be V-jointed at the ends, set in and jointed with cement mortar, and the V-joints shall then be run full with cement grout com- posed of one part of cement and one part of clean-washed sharp sand by measure, the whole of the coping to be tuck-pointed with cement mortar as described above; and in every class of work all stones shall he laid on their natural hed. Pointing. 12. The whole of the Bridge shall be tuck-pointed with cement mortar as the work proceeds, and in the same style and quality of workmanship as the pointing done on the County Bridges (sample of which can be seen on Campbeck and Irthing Bridges). The cost of this pointing to be included in the Schedule of Prices for rubble work, and the workmanship on both sides of the Bridge shall be of the same quality and in the same style of work as shewn on the Elevation. Dry Rubble Drain 13. This dry wall shall be built behind all abutments and wing walls, and the Plan, 12 Inches work to be executed in a similar manner to the other ordinary rubble work above described, except that no mortar is to be used in this twelve-inch drainage wall, and each tile weep hole shall be connected therewith. Clay Puddle, &c. 14^. After the whole of the work connected with the arch and backing has been cojupleted, and before any embankment is made, the whole of the upper surface of the arch and abutments shall be covered with a six-inch layer of good clay puddle (or one-inch of asphalte) to be worked and set in a manner approved of by the Surveyor. Weep Holes. 15. The Contractor shall cause to be inserted in the Bridge during the course of its construction, two-inch circular tiles as weep holes, which shall extend through each wall into the dry rubble drain wall, the ratio being for the abutment one weep hole in every lineal yard, just above the ordinary water level, and a similar course of tiles under the springers, and for the rest of the abutments and wing walls one such weep hole to every superficial yard of wall face ; also a four-inch circular tile from above the springers to the lowest point of concrete backing, as shewn on the Plan. Roadway. ^^- '^^^ roadway shall be composed of a six-inch layer of pitching set by hand, sandstone, cobbles, or any other stone may be used, and the interstices thoroughly filled up with small gravel or chippings. Upon this pitching shall be laid a four-inch bed of sound good road metal, to be approved of by the County Surveyor, and may be Whinstone or Limestone as he directs, put on at separate dates, the first two-inch coat Copy of Specification. 101 to be well consolidated before the second two-inch coat is applied, the last coat to be broken to pass through a two-inch ring in every direction. The finished surface of the road shall have a vise from each side to the centre of the roadway of one in thirty-six, that is an inch to the yard, and each coating of metal shall be covered with a sufficient binding of good clay. 17. The Contractor shall provide and build into the parapet walls in the centre Date onBuildlng, of the Bridge suitably tooled stones bearing the date of erection, the name of the Bridge, and other letters if required. 18. The whole of the workmanship and materials used in the construction of the workmanship and Bridge shall be the best of their respective kinds, and shall, at the expiration of the e. e a. s. term of maintenance, or when certified by the County Surveyor that the Works are complete, be delivered up in the most perfect order and condition, and in strict accordance with the Plans and Specifications. GENERAL STIPULATIONS AND CONDITIONS. 19. The description of the work required by the Cumberland County Council to be executed is set forth in this Specification and on the drawings, and the nearest approximation to the actual quantity of work required to be executed is set forth in the Bill of Quantities ; but the County Surveyor reserves to himself the power to vary, extend, or diminish the quantities of work without vitiating the Contract, and also to make any alteration in the quantity, quality, and description or character of the Works which he may think proper, and the Contractor shall not have any claim upon the Cumberland County Council for any such alterations other than the actual work done, calculated according to the particular price of such work as set forth in the Bill of Quantities. Should there be any discrepancies between the wording of this Specification and the dimensions written on the drawings, or between the written diniensior.s and the diagrams they refer to, in all cases the Contractor shall be regulated by the wording of the Specification in preference to the written drawings, and by the written dimensions in preference to the drawings. 20. The Contractor shall at his own cost and charges furnish all temporary works, centering, barrows, planks, carts, pumps, and other necessary implements, machinery, and materials, stDne, cement, lime, sand, fencing, &c., all horse and manual labour, lights, and watchman (if needed) ; he shall likewise erect and maintain all necessary coffer-dams, temporary bridge, embankments, and everything requisite ; and if any materials or implements shall be brouglit for use which the County Surveyor may deem of inferior description, or improper to be used in the work, the same shall be removed at once by the Contractor upon his receiving notice in writing from the Surveyor to do so, and if the directions of the Surveyor be not coiTi]3lied with within 24 hours after the giving of such written notice to the Contractor or his Agent, the Surveyor shall be at liberty to remove the defective material or implements at the Contractor's expense. 21. The Contractor shall provide at least thirteen centres for the arch, each centre to be of sufficient strength to bear one-fifth of the gross weight of the arch, and these thirteen centres shall be constructed of and lagged with approved foreign timber, the lagging shall be of battens of suitable lengths, G;^ inches broad, and 2^ inches thick. Power to vary Works. Materials, Implements, Liabour, &c. Ceaterln*?. 102 Oblique Bridges Made Easy. Centering— The centres shall be supported in the following manner, that is to say, corbels shall be left in the abutment walls of sufficient substance and strength (to be neatly broken off after the centres are removed), these corbels shall be further strengthened by having stout wooden props placed underneath each corbel, each prop shall rest on a solid footing, and wedges shall be driven between the head of each prop and the underside of each corbel, and at a suitable distance below the springer line, to support a beam on which the ends of the centre shall rest, each centre on two hard- wood wedges, as broad on the bed as the beam, and at least four inches thick at the smaller end. The beams and wedges to be well smeared with soft soap, and the beam smeared on the face where the wedges rest, and the wedges smeared on both sides so as to facilitate the slackening of the wedges, and to allow the arch to sustain itself. At least four rows of support shall be provided in the bed of the stream between the abutment walls the full width of the Bridge, either of stone or wooden piles as the case may require, and on each of these four rows of stone pillars or wooden piles the Contractor shall place a beam of approved strength and quality, on which the above thirteen centres shall rest on hard- wood wedges, as broad on the bed as the beams, and smeared with soft soap as above described. After the arch has been closed and properly keyed, each set of wedges shall, on the seventh day after the arch has been keyed, be driven back an inch daily, till the arch sustains itself and the whole of the weight is removed from the centres, when the spandrel walls may be commenced. Contractor 22. The Works will be set out by the County Surveyor and Bridge Master, flnishea accuracy who will also give the proper depths and particulars on the ground, and the Contractor will be held responsible for the Works being completed in accordance with the lines and levels so given. The Contractor shall not absent himself from the Works without leaving a fully authorised Agent to act in his absence, and he shall not re-let, sub-let, or assign the Contract, or any portion thereof, without the consent of the Surveyor in writing. Notice to 23. The Contractor must commence the Works within three days from the ""wSksf^ time a written notice to do so, signed by the Surveyor, shall have been served upon him at his usual address or place of residence, and he shall not commence without such written notice. Extra work, how 24. Upon an order in writing from the Surveyor the Contractor shall execute owe r. g^(.|.g^ work, or day work, and shall be paid far the same according to the scheduled prices for similar work. If no price is quoted in the Schedule for such extra work or day work, then, before commencing, a price shall be agreed upon in writing for the same. No extra charges will be allowed unless the same are certified for as extras by the Surveyor. Foundations. 25. If good sound rock is not met with at a less depth than nine feet below the surface of the water (or the datum line) at the time of commencing the Works, then a concrete foundation shall be laid upon sound ground (ground to be well rammed first) or on wooden piles if found necessary, the concrete to be mixed and applied as previously described, that is, in six-inch layers, each layer to be well rammed before another layer is applied, and in the manner shewn on the Plans and Sections. If rock is found, then the same shall be quarried out so as to form a level bed or floor, and Copy of Specificalion. 103 all and every inequality shall be levelled up with rubble and cement mortar. If a rock foundation is found and approved of by the Surveyor, then the depth of such foundation shall be at any depth less than nine feet below the surface of the water (or the datum line) which the Surveyor may determine and approve, and a corresponding deduction shall be allowed by the Contractor for this less depth. 26. The Contractor, at his own expense, shall keep all the excavations free from water during the progress of the work. 27. No work shall be covered up till it has been seen by the Surveyor and directions given by him to that effect, and all filling in shall be done in six-inch level layers, and each layer shall be thoroughly rammed with a proper beater before another layer is applied. 28. The whole of the excavation for each separate foundation shall be taken out and the site inspected by the Surveyor prior to any building being done in the several trenches thus opened. 29. The Contractor to do all the filling in, in six-inch level layers, on both sides of the arch, as th3 walls advance in height ; embankments, &c., always to be kept within 3 feet of the upper surface of the walls, and he shall ram each layer solid, before applying another layer, up from the present ground level (between the wing walls) to the level of the foundation of the new road as the work proceeds. 30. The County Surveyor shall have power to order the dismissal from the Works of any incompetent workmen at his discretion. 31. The Works are to be constructed under the direction, superintendence, and to the absolute satisfaction of the County Surveyor, whose judgment as to the quality of materials and workmanship, or otherwise, shall be final and binding, as between the Highways and Bridges Committee of the Cumberland County Council and the Contractor, should any dispute arise with reference thereto. 32. Payments will be made on the certificate of the County Surveyor at the rate of £85 per cent, of the Works or on the total Contract, the remaining £15 per cent, will be retained in the hands of the Highways and Bridges Committee until the expiration of the term of maintenance, which in this case is fixed at six months, or till the County Surveyor shall certify that the works are perfect, and have been completed to his satisfaction, when the Contractor shall be entitled to the balance. 33. The Contractor, at the time of tendering for the work, shall specify the number of months in which he will bind himself to complete his Contract. In the event of the Contractor not completing the work in the specified time, or in the event of his not proceeding with the work to the satisfaction of the County Surveyor by reason of insufficient plant, the non-arrival of materials, or an insufficient number of workmen, or by any cause whatever, then the County Surveyor shall have power, after giving seven days' notice in writing, to take the work out of the Contractor's hands and to employ men and carry out the Contract and deduct all payments to the men so employed, and also to the Clerk of the Works, from the amount of the Contract, the whole of the work to be completed to the entire satisfaction of the County Surveyor in six months from the time of the written notice of commencement, and everything to be cleared off the ground and the place restored to its normal condition at the Pumping. Examination. Excavation for Abutments, Piers, &c. Earth Filling. Dismissal of Incompetent Hands. Direction of Works. Payments. Date of Completion. 104 Oblique Bridges Made Easy. expiration of the period named. Should ths specified time be exceeded, and the Surveyor not elect to carry on the work himself, then the wages of the Clerk of the Works for such extra time shall be paid by the Contractor, and failing such payment the amount of ^vages due to the Clerk of the Works for such extra time shall be deducted from the balance of the Contract due to the Contractor. Penaltyfor 34. In the event of the work not being completed in the time specified the - omp e ion. Contractor shall forfeit to the County Council the sum of £1 sterling for each and every week during which such default in the completion of the work shall continue as and for liquidated damages, and the amount of such forfeit to be deducted from the money retained in the hands of the County Council, or which may otherwise be due or become due to the Contractor^ or the County Council may recover the same by law, and the fact of the work being taken out of the hands of the Contractor by reason of such default will in no way release or relieve the Contractor from the obligation of this penalty. Extension of time 35. Should the Contractor be delayed in the execution of the work by causes ^ '°°' beyond his control, by acts or omissions of the Surveyor, or by reason of extra works in connection with the Contract, then the Surveyor shall have power, upon satisfactory proof of the justice of the claim, to grant a reasonable extension of time for the completion of the work, but such claim must be made in writing by the Contractor within 14 days from the date of the cause of such application. Contractor to keep 36. The work to be maintained in good and perfect order for six months after and maintain . ... or Works in good entire completion, or till certified by the County Surveyor that the Works are perfect months. and complete. Agreement and 37. The Contractor shall enter into an Agreement in writing, and also give a ° Bond with one or more sufficient Sureties, for the due performance of the Contract. (Signed) GEO. JOS. BELL. (Signed) THOMAS TELFER. Witness— X. Y. Z., Clerk of the County Council of Cumhcrland, May 6th, 1893. Copy of Tender. COPY OF TENDER. Tender for Woik to be done in tlie erection of an Oblique Bridge over the Kiver King, at Kingsbridgeford, in the Parish of Lanercost, in the County of Cumberland. particul:ars or work. 105 Measure. Quan- tity. Rate. £ s. n. Clearing site, setting out work, and preparing foundations for the abutments, wing wall, &c c. yds. 260 3/- 39 Cement concrete in foundations, &o 0. yds. 30 12/- 18 Of random block, in single face to level of springer c. yds. 200 14/- 140 Of stone in springer, feet by feet by feet in cement... u. ft. 400 2/9 65 Of masonry in arch , set in cement c. ft. 2700 2/3 303 15 Of rubble set in lime, tuck-pointed with cement on one face to the level of the string course .. c. yds. 260 11/- 143 Of string course, 1.5 inches by .5 inches, in lime lin. ft. 284 1/6 21 6 Of rubble in lime, tuok-poiuted with cement on both sides in parapet walls, including date stones sqr. yds, 130 10/- 65 Of coping, set and V-jointed in cement, as shewn on Plan lin. ft. 294 2/3 33 1 6 Of dry rubble drainage wall, as per Plan c. yds. 40 5/- 10 2-inch tile weep holes lin. yds. 1 Of dry rubble six-quarters fence wall roods. nil. ROADWAY. Earth embankment between wing walls and approaches... 0. yds. 360 1/3 22 10 6-inoh pitching set to curvature of road surface c. yds. 70 il- 14 2-inch broken limestone, or blue cobbles (road metal) c. yds. 50 6/- 15 6 -inches clay puddle or 1 inch of asphalte u. yds. 20 3/- 3 Centering, coffer-dams, and other incidental expenses ... ... ... 120 5 10 Total cost of completing the Works as shewn on the Plan specified for in this Specification 3 and Sec tions, a nd £ 1003 18 4 Time required to execute the Contract, six months, or till the 7th October, 1893. (Signed) THOMAS TELFEE. Langholm, N.B., March 24th, 1893, 106 Oblique Bridges Made Easy. COPY OF AGREEMENT. Hn Hgreement made the Sixth day of May, One Thousand Eight Hundred and Ninety-three, ^etweett Thomas Telfer, of Langholm, in the County of Dumfries, N.B., Builder, hereinafter called the Contractor of the one part, and George Joseph Bell, County Surveyor and Bridge Master, on behalf of the ©OttttttJ ffiownctl of Cwwberlattb', hereinafter called the Authority of the other part. ^IljevertS the said Contractor hath proposed to the said Authority, in consideration of the payment of the sum of One Thousand and Three Pounds Eighteen Shillings and Four Pence (£1,003 18s. 4d.), to execute and find materials for the whole of the Works required for the erection of a New Oblique Stone Bridge over the river King at Kingsbridgeford, in the Parish of Lanercost, in the County of Cumberland, and other Works connected therewith, as the same are set forth and described in the Specifications, Conditions, Bill of Quantities, and Schedule of Prices, and Plans hereto annexed, and which, for the purpose of identification, have been signed by the said Contractor : ^nb njljctrcrts the said Authority have accepted the said proposal, ^o\x> tjjese '^veasnia njitne«0 that the said Contractor glptlj hereby, for himself, his executors and administrators, agree with the said Authority and their successors, that he (the said Contractor) shall and will duly execute all the Works mentioned and referred to in the said Specifications, Conditions, Bill of Quantities, Schedule of Prices, and Plans, including all the additions or alterations, new or extra works that shall be ordered by the said Authority or their Surveyor, as therein provided, with such exceptions and deductions as are therein provided, in all respects in strict conformity therewith, and will maintain the same in an efiicient manner as therein provided ; g^nlr the said Authority doth hereby, for themselves and their successors, promise and agree with the said Contractor, his executors and administrators, that they will pay to the said Contractor, his executors or administrators, the said sum of One Thousand and Three Pounds Eighteen Shillings and Four Pence (£1,003 18s. 4d.) or such other sum as shall become payable to the said Contractor, his executors or administrators, under the provisions of the said Specifications, Bill of Quantities, Schedule of Prices, and Plans, at such time or times, and in such manner, and after the receipt of such Certificates, as are therein provided ; 2lnl> it is further hereby mutually agreed by and between the said Authority and the said Contractor that they respectively shall and will observe and perform all the conditions and obligations mentioned in the said Specifications, Conditions, Bill of Quantities, Schedule of Prices, and Plans, and that all the powers, liberties, rights, and privileges mentioned therein and conferred thereby shall and may be exercised by the parties hereto according to the true intent and meaning thereof : ^Cnt* it is hereby lastly declared and agreed that this Agreement shall be taken and construed to be not to the prejudice but to be in confirmation and eulargement of, and in addition to and extension of, the said Specifications and the provisions therein contained : ^vot>tt>cb Copy of Agreement. 107 always, and notwithstand ng the preceding provisions, it is here'iy agreed tliat the said Authority sliall be at liberty from time to time, and at any time, to abandon any of the Works hereby contracted for, deducting the sums mentioned in the Schedule for such abandoned Works from the amount of the Contract without any compensation to the said Contractor in case the Works so abandoned shall not have been commenced with, and in case they shall have been commenced with and in part executed, then, with such compensation as shall be settled by the said George Joseph Bell, County Surveyor and Bridge Jilaster for the time being of this Authority, whose decision shall bo final and binding and conclusive on both parties hereto, grt n>itness whereof the said Contractor and the said George Joseph Bell have hereunto set their hands the day and year first above written. Signed and Delivered by the said Thomas Telfer, in the presence of X. Y. Z.„ Clerk of the Cumber- land County Council, on the 6th day of May, 1893. THOMAS TELFER. Signed and Delivered by the said George Joseph Bell, in the presence of X. Y. Z., Clerk of the Cum- berland County Council, on the 6th day of May, 1893. GEO. JOS. BELL. 108 Oblique Bridges Made Easy. COPY OF BOND. Know all Men by these Presents, that we, Thomas Telfer, Builder, Langholm, and Walter Scott, Skinner, Langholm, in the County of Dumfries, N.B., are held and firmly bound to the Cumberland County Council in the sum of Three Hundred Pounds (^300 Os. Od.) of lawful money of the United Kingdom, for which payment, well and truly to be made, we bind ourselves and each of us, our and each of our heirs, executors, and administrators, for and in the whole jointly and severally, firmly by these Presents. Sealed with our Seals, dated the Sixth day of May, in the Year of our Lord One Thousand Eight Hundred and Ninety-three (1893). The Condition of this Obligation is such that whereas certain Articles of Agreement, bearing date the Sixth day of May, 1893, have been entered into by and between George Joseph Bell, County Surveyor and Bridge Master, acting for and in behalf of the said Cumberland County Council, and the above bounden Thomas Telfer, for the erection of a new Oblique Stone Bridge over the Eiver King at Kingsbridgeford, in the Parish of Lanercost, in the County of Cumberland. Now if the bounden Thomas Telfer shall, for and during the period during which the Baid Articles of Agreement shall, by their original terms be in force, as well as during any extension of time which, as provided in the said Articles, may be granted for their performance, duly and fully observe, comply with and perform all and singular the Terms, Conditions, and Agreements which, on the part of the said Thomas Telfer are by the said Articles of Agreement to be observed, complied with, and performed, then this Obligation shall le void, otherwise shall remain in full force and virtue. Signed and Sealed by the above bounden Thomas Telfer and Walter Scott, in the presence of X. Y. Z., Clerk of the County Council of Cumberland, on the 6th day of May, 1893. (Signed) THOMAS TELFER { L. s. (Signed) WALTER SCOTT Elliptical Skew Arches. 109 APPENDIX. ELLIPTICAL SKEW ARCHES. If it is desired to erect an elliptical arch on the skew, proceed in like manner as directed on page 35 for segmental arches ; and for finding the spiral and end joint radii of the elliptical arc, see pages 10, 38, and 47. To find the length of the elliptical arc by Tables (see Hurst's Tables, page 249) divide the rise or half the minor axis by the span or major axis ; find this quotient in the Height Column in the Tables, and opposite this quotient, in the Length Column, will be found the corresponding length of elliptical arc to chord L Multiply the span or major axis by this tabulated length of arc to chord L and the product is the length of the elliptical arc required. See Tables on page 94, for circular arcs ; and Hurst's Tables, page 245. To find the theoretic intradosal angle, divide the obliquity of the arch by the length of the elliptical arc (CE by CD, Fig. 4), or divide the cotangent of the angle of obliquity by the tabulated length of the elliptical arc to chord 1. given in the Tables, and the quotient is the natural tangent of the theoretic intradosal angle (yS). For the adjusted angles see pages 9 and 35, and for obliquity see page 28. To find the theoretic axial length, divide the length of the elliptical arc by this tangent, see pages 8, 29, and 46, axial lengths, and for the adjusted axial length divide the arc by tlie tangent of the adjusted angle. To find the points of convergence of the face-joint lines in an elliptical skew, proceed as shewn on pages 47, 49, and 50. BC on the plan is depth of arch stones x cosec 6, or BC = arch stones H- sine d. CD = arch stones x cot. 6 x tan. 4>. In the example given on Plate 8, state, As 2.828 (BC) : 7.071 (Half the oblique span of the small circle in the ellipse) :: 2.2539 (CD) : 5.635 (P). See Fig. 7, Plate 2. Or R x cot. 6 X tan. <^ = 5. x 1. X 1.126987 = 5.6349 (P). In the elevation, this point must be set off below the springer line, at a distance from the face of the springer equal to half the oblique span of the small circle in the ellipse, that is 7.07L See Plate 8. The distance of this point of convergence below the centre for the ellipse is equal to the radius of the ellipse x cot. 6» x tan. 4>^ = 28.125 x .862161 = 24.248. At the junction or overlapping of the differently dressed arch stones, between the fifth and sixth courses in this example, the arch stones will fit accurately at the soffit, but the space between the beds at the extrados, caused by the difference in the extradosal thickness of the arch stones, must be carefully pinned up with sound hard stone and the space well grouted with strong cement mortar. The point of convergence of the face lines in the sectional elevation through the axis, in the small segment, is not the same distance from the face of the springer at B, as in the elevation, because in this position the oblique arc standing on AD in the Projecti.m, is shewn fore-shortened on the obliquity AG. Therefore this distance from B is As AD : AG :: Half the oblique span = As 35.355 : 25.:: 7.071 :5. See Plate 8, and oblique span 6 45°, page 91. This point of convergence is the same distance below the sprincrer line as in the elevation, therefore proceed as directed for the elevation and rule in the face- joint lines to the same point of convergence. The distance from the face of the springer can also be found by construction. From A in the Projection, on the line AD, measure off half the oblique span (7.071) of the 110 Oblique Bridges Made Easy. small circle in the ellijase, and from the obliquity AG let fall a perpendicular on this point on AD, and the distance from A on AG to the point of departure of the perpendicular is the same distance as that found by the above stating. For finding the thickness of courses and the length of check proceed in like manner as directed for segmental skew arches. See page 35. And for using the templates on Pjate 8, see page 12. The experienced workman will see that the soffit twist of the arch stones must vary with the radii of the ellipse, and consequently, will require a separate set of twisting rules for each variation in the radii of the ellipse, and to find the radii of the sjiiral and end joint arcs, proceed as directed ou pages 9, 10, 38, and 47, by dividing the radii of the ellipse as shewn thereon. For the radiation of the twisting rules and the increased breadth of one end of one rule, see pages 41, 43, and 90, and the Eeference on page 114. The workman must mark each dressed arch stone to correspond with the twisting rules by which it is worked, as the soffit twist will only fit that part of the centre, corresponding to the radii of the twisting rules, for that particular part of the surface of the elliptical centre. It will be seen on referring to pages 9, 31, 32, 33, 40 and 80 that by making a construction on the Diagram, Plate 7, for each of the several radii of the ellipse, multipliers can be measured therefrom for an elliptical skew, as accurately as for a segmental skew, the ellipse being a succession of segments. TO CALCULATE THE RADIUS OF A FALSE ELLIPSE DRAWN FROM THREE CENTRES. Subtract the radius of one of the small circles from half the elliptical span, square this distance, and from the result subtract the square of the difference between the rise and the radius of one of the small circles, divide the remainder by twice the difference between the rise and the small radius, and add the result to the rise and the sum is the length of the third radius. Thus, Span 25 feet, Else 6.25. Eadius of small circle (which may be of any suitable diameter) 5. feet. Half the span 12.5 feet. Distance between the centre of the ellipse and the centre of the small circle is 12.5 — 5 = 7.5 feet. See the Figure on Plate 8. Hence the third radius of the ellipse is — = 7.52 =56.25 (6.25 - 5)2 = L5625 54.6875 2 (6.25 - 5) = 2.5 54.6875 —-Z — =21.875 2.5 And 21.875 + 6.25 = 28.125 the Third Eadius. If the minor axis of the ellipse be made equal to half the major axis or span, the rise or semi minor axis will be equal to one-fourth of the span, and if the radius of the small circle be made one-fifth of the span, the third radius of the ellipse is equal to the span added to half the rise, thus — 25. + (6.25 -^ 2) 3.125 = 28.125. To construct an ellipse by equilateral triangles, where the rise is a fourth of the span, rise x .634 = short radius, and rise x 3.366 = long radius. The length of an elliptical arc of these proportions, calculated to Chord 1., is 1.21136, and any desired span, multiplied by this length of arc to Chord 1., is equal to the length of such elliptical arc, 25. X 1.21136 = 30.284, the length of the arc in the above example. The Ellipse. Ill THE ELLIPSE. To find the length of any ordinate, ef or gh, drawn to either diameter, ah or cd. Knowing the absciss or space between a/ or dg, and the two diameters ah and cd, state As ah' : cd^ :: af x fb : ef ; and As cd' : ah" :: eg x gd : gh\ Square root of the fourth term equal ordinate. TO CONSTRUCT AN ELLIPSE BY ORDINATES. Divide half the span into 10 equal parts, and erect perpenliculars or ordinates at the divisions The length of these ordinates is equal to the rise of the ellipse multipliel by the respective numbers on each ordinate in the left hand division on the diagram. See the right hand division for full sized dimensions. The complement of each ordinate is shewn above it. ^ , ^ M Cj ^ Table of Ohdinates FOB OONSTKUOTING AN EllII'SE. .43589 X 6.25 = 2.72431 •6 „ „ 3.760 .71414 „ „ 4.46337 •8 „ „ 6.0 .86602 „ ,, „ 5.41262 .S1661 6.7281S .95394 „ „ 6.96212 .97979 „ „ 6.12368 .99199 „ „ 6.21868 10 „ „ „ 6.25 Given the major and minor axis to construct an ellipse. Divide half the span or major axis, and the rise or half the minor axis, each into ten or any equal number of parts or divisions, and connect the points 1 to 10 as shewn below. 112 Oblique Bridges Made Easy. TO CONSTRUCT AN ELLIPSE FROM TWO CIRCLES. Describe two semicircles, which may bear such proportion to each other as the situation demands or the taste of the engineer dictates, making the radius of the larger circle equal to half the span or major axis, and the radius of the smaller circle equal to the rise or half the minor axis. The points of inter- section of the horizontal and vertical lines drawn from any radial line will be points in the elliptical curve. See the two examples given in the figure below. The rise or radius of the small circle is one fourth of the span or 6.25, and the radius of the other is three-eights of the span or 9.375. There are various other methods of drawing an oval or false ellipse, but without known radii, there is no basis from which to calculate the radiation and twist in the arch stones for an elliptical skew arch. TO CONSTRUCT A CIRCLE BY NATURAL SINES OR ORDINATES. Divide the radius of any desired circle into ten equal parts and multiply the radius of the circle by the corresponding natural sine or ordinate, in the left hand division of the diagram below, and the product is the ordinate for the circle ; or multiply the radius by the chord of the arc or twice the natural sine of half the angle, and the product is the chord of the corresponding angle. Table op Natural Sines OR Okdinates roR Constructing Circles, the Radius being divided into 10 Equal Arcs Abscissa tor Finding the Spaces between Ordinates, the Radius being DIVIDED into 10 Equal Arcs. .4368857 X 12.6 = 6.4485713 .1000 X 12.5 = 1.2 .6000000 „ „ „ 7.50 >f ti i> 11 11 .7141401 8.9267612 ,, ,, .8000 10.00 ti )> It II II .8660254 10.8232176 .9165140 11.4564250 II ). II II II .9539382 11.9242275 ,, ,, , .9797960 12,2474376 II -,. II II II .9949869 12.4373362 „ „ „ „ „ 1.000 12.60 II II II II II If the diameter of a circle is 25. feet the radius is 12.5 feet ; the chord of an arc drawn at six tenths of the radius, at right-angles thereto, will be 20. feet, and the versed sine will be 5. feet, a fourth of the chord and one-fifth of the diameter, and this chord and its arc are subtended by an angle of 106° 15' 36", or twice half the subtending angle; because half the chord 10. feet divided by the radius 1 2.5 feet is .80 the sine of 53° 7' 48". See the figure above. The length of the chord of any arc of a circle is equal to the radius multi2)lied by twice the sine of half its subtending angle. For the lengths of arcs to chord 1. and radius 1, see pages 37, ib, 91, 92, 93, and 94. Elliptical Arches. 113 Another Way of CoNsxRucriNG a Circle by Natural Sines or Ordinates. When the quarter circle is divided into ten equal arcs, the divisions, or absoissse, on the half chord or radius become unequal in length, as will be seen in the example given below, and this method of constructing a circle is consequently less convenient for practical and ready use than the former example. The ordinates are the sines of the respective angles, and the abscissaa or spaces are the varying differences between the cosines of these subtending angles. For ordinates of this circle, multiply the radius by the corresponding sine or ordinates, as m the above example, and for the absciss* or spaces between the ordinates, multiply the radius by the abscissae or difference between the cosines. Table of Natural Sines OR Ordinates FOR Constructing Circles, the \ Circle being DIVIDED into 10 Equal Arcs. Abscissa for Finding the Spaces between Ordinates, the \ Circle being divided into 10 Equal Anns. .1669346 X 12.6 .3090170 „ ,, .4539905 ,, ,, .5877853 „ „ .707106S „ ,, .8090170 „ „ .8910065 „ „ .9510566 ,, „ .9S76833 „ „ 1.0000 „ ,, = 1.9564312 „ 3.8627126 „ 3.6748812 „ 7.3473162 , 8.83S8360 , 10.1127126 , 11.1376812 „ 11.8882062 „ 12.3461087 ,, 12.60 .0123117 X 12.6 .0366318 „ „ .0600500 „ „ .0319895 „ „ .1019102 „ „ .1193216 ,, „ .1337948,, ,, .1449735 „ „ .1626325 ,, „ ,1564345 „ ,, = 0.16S8962 , 0.4578975 , 0.7506250 , 1.0248687 , 1.2738775 , 1.4916187 , 1.6724360 , 1.8121687 „ 1.9072812 ,, 1.9554312 To Construct a Segment by Natural Sines or Ordinates. Divide the half chord or span into 10 equal parts, and if the versed sine or rise of ^^^ ^J^ i^°- fourth of the span deduct .CO, the natural sine of 36° 52' 12", from the sines of all the other angles ibov'tt ctrd li;e in the figU, and the remainders are the ordinates for constructing the segmental arc The middle ordinate of any arc is = radius x versed sine of half the subtending angle For the actual ordinates, suppose the chord or span to be 20 feet the radius of ^te cuc^e -U b 12 5 and the versed sine or rise, being one-fourth of the span, will be 5 feet._ Th- th radms multiplied by the middle ordinate in the figure, thus 12.5 x .40 = 5 feet because the sine o 90 is L and 7-60 the natural sine of 36° 52' 12" is .40, which is the ordinate above the chord line , and similarly with the' sines of all the other angles, for the other ordinates. And for the spaces or abscissae between the ordinates, multiply the abscissa m the figure by the radiufof tL drcr-d the product is the space between the actual ordinates for setting out the segmental arc. Table of Natural Sines OR Ordinates for Constructing Segments, the h Chord being divided into 10 Equal Parts. Tneu^ ofCosirif-s. ss'?~<*- t-i- ■ -o&ooooo 80 *7 36 - .ISOOOOO YS c t-s =• .:i4^oooo 71 20 '* - laoovcc SS Is IS - SI It S3'' !S se *o- so 'i io -=> *3 St ** = M SI IK ^ .+000000 .^Sooooo JGOOOOO .6^0000 .72.00000 . .soooooo .9039746 .1683766 „ .2284960 „ .2772695 „ .3166169 „ .3474183 „ .3707735 „ .3871176 „ .3967962 „ .40000 „ X 12.5 = 1.1746812 „ „ ,. 2.104707B „ „ 2.8561876 Absciss-E for finding the Sfaoes between Ordinates, the i Chord being DIVIDED into 10 Equal Parts. .080000 X 13.6 = 1.0000 3.9664487 4.3427287 .8389700 .9599400 .00 114 Oblique Bridges Made Easy. Another Way to Construct a Segment by Natural Sines or Ordinates. If half the segment is divided into 10 equal arcs, proceed as directed in the foregoing example for the ordinates ; and the spaces between the ordinates are formed by multiplying radius by each abscissaj given on the table in the figure. / T/ISLE OF Cf>SiNE.S. ,'' l\ ' 84- +f ts « .03^Sr& , ^^ f9 && Z& - ./34-3338 1 y ; ^ Yi- 3 ^f = .Z746075 1 ^ 1 68 'H- Sf-- .S€Z'>-GSZ 1 ^'^ 1 65 a,6 7- .'ff7S^8S , §: ' 1 S8 7 2ff = .SZ&IOSI ^ \ 1 sp, ts 33 - -eo^-t-jn 1 J' 1 *7 ^9 « - .67Jif*05 1 y ' fa. 10 S9 - -79/0033 1 ', 3£ SZ IP,= .7S3SS89 t' 1 Table of NATUB4L Sines OR OUDINArES foe cokstkucti Segments, the i Segment bein ihvided into 10 Equal Aecs, NO 0.8937887 AESCISSJi; FOR finding the Spaces Between Ordinates, the i Segment being divided into 10 Equal Akos. .0716015 X 12.6 = .0589956 X 12.5 = 0.7374450 .1372314 „ „ „ 1.7153925 .0663630 0.8170375 .1966268,, „ „ 2.4578325 .0711688 „ „ „ 0.8896075 .2491768,, ,, „ 3.1147087 .0783626,, „ „ 0.9545325 .2944297,, ,, „ 3.6803712 .0809006,, „ „ 1.0112575 .3319973 ,, ,, „ 4.1499662 .0847433,, „ „ 1.0592912 .3616565 „ „ „ 4.6194562 .0878579,, „ „ 1.0983237 .3828632,, „ „ 4.7S58660 .0902175,, ,, „ 1.12771S7 .3957045 , 4.9463662 .0918020 „ „ ,, 1.1475250 .40000 ,, „ „ 5.000 .0926878 „ „ „ 1.1673475 Appendix. 115 Example of Elliptical Skew Bridge, Span 25 Feet, Angle of Obliquity 45° EEFERENCE. Angle of Obliquity Span of Ellipse Span of the small Circles in tlie Ellipse Oblique Span of the Ellipse, Span x Cosec. 61 = 25x1.4142 136 Oblique Span of the small Circle, Span x Cosec. 61= 10 x 1.4142136 Theoretic Tan. /3, Obliquity -f Arc = 25 -=- 30.284 = .825518 Tan., very nearly - Width of Bridge, extreme Length of Impost, Width of Bridge H- Sine 0, or W x Cosec. 9 - Theoretic Divergence, Impost x Sine (3 = 28.284 x .636527 Theoretic Thickness of Courses, Check x Sine ji = 2.0230 x .636527 Xumber of Checks on Injjost, see pages 35 and 91 Length of Checks, Impost h- Number of Checks = 28.28427 -H 14 Adjusted Divergence of Courses, is Number of Checks ^ Number of Courses x Heading Spiral = (14 -^ 31) .4516129 x 39.2656287 Sine of the New or Adjusted Angle, is Divergence -=- Impost = 17.73281 + 28.284 = .62695, nearly Adjusted Thickness of Courses, is 39.2656287 4- 31 = 1.26663, or Check x Sine j3 = 2.0203 X .62705- Length of Elliptical Arc, is Span 25. x 1.21136, see Hurst's Tables Length of Oblique Arc or Heading Spiral, Arc x Sec. Theoretic /3 = 30.284 x 1.296589 Number of Courses, Heading Spiral -H Thickness of Courses Axial Length, Arc 4- Tan. ^ = 30.284 H- .80497 Height from Waterline to Springer, made to suit situation Rise of Arch or Half the ilinor Axis, one-fourth of Span Radius of small Cylinders in the Ellipse Radius of Ellipse, Half the Rise added to Span 3.125 + 25 See Rule, page 110 Distance of the Point of Convergence of the Face Lines below the Springer Line in the small Circle, Radius x Cot. 9 x Tan. <;i = 5. x 1. x 1.126987 Distance of P. in the Ellipse, R x Cot. d x Tan.'<^ = 28.125 x 1. x .862161 - Sine of 38° 50' is .62705, sine^ is .39319 ; Cos. is .77897, Cos.2 jg .60679 ; Tan. is .80497, Sec. Tan. Extradosal Angle 4> in small Circle, is R + e 4- R x Tan.^ = 1.4 x .80497 = 1.26968 Tan. nearly. See page 76. Actual Tan. 1.126987 ; Sec. 1.50668 Angle of Difference in small Circle - - Tan. Extradosal Angle 1.50668 :: cos. 8 .98604 : AG on Diagram. And for the ellipse. As sec. (i 1.28374 : sec. i, 1.32034 :: COS. 8 .99943 : 1.02792 AG on Diagram. GH in small circle is .15728, "and GF is .1948'; and GH in ellipse is .02792, and G F is .03469. Again by Rule of Three in the small circle. As 1.573 B A : 2.-5 desired soffit distance apart :: .2732 difference between B A and B E : .43400, the extradosal distance or radiation ; and As 1.573 BA : 2.5 desired soffit distance apart :: .3292 difference between C A and C E : .5229 the increased breadth of one rule at one end. In the ellipse. As 1.5737 B A : 2.5 the desired soffit distance apart :: .0447 difference between B A and BE : .07101 the increased extradosal distance or radiation; and As 1.5737 B A : 2.5 the desired soffit distance apart :: .0555 difference between C A and CE : .08816 the increased breadth of one rule at one end. The Combined Tacheometer and Theodolite. 117 THE PATENT TANGENT READING TACHEOMETER In the following pages will be found a description of Bell and Elliott's Patent Tangent Reading Tacheometer, fitted with Solar Attachment, of which Messrs. Elliott Brothers, Opticians to the Government, 101 and 102 St. Martin's Lane, London, W.C, are the Sole Manufacturers. 118 Oblique Bridges Made Easy. THE THEODOLITE, Directions for Readinq Degrees, Minutes, and Seconds, by the Vernier, FROM the Horizontal and Vertical Arcs. Figure 1, which is drawn larger than the actual size, so that the divisions may be read more easily, represents part of the circle or limb, and the vernier of a combined five-inch transit Theodolite and Tangent Reading Tacheometer, as supplied by Messrs. Elliott Brothers, of 101, St. Martin's Lane, London, W.C. The horizontal and vertical circles are divided into 360°, and each degree is again subdivided by shorter lines into three equal parts of 20 minutes value each. In some instruments the circles or limbs are centessimally divided, that is, into four -r^^ths, instead of 360°. The open divisions on the outer rim of the circle are the degrees, and the closer and shorter division lines, on the inside of the circle, are minutes ; therefore the spaces between the short lines on the primary scale or circle are each equal to 20 minutes ; so that if the arrow of the vernier reads past any number of degrees, and coincides exactly with the first close division, past the degree, the reading would be that number of degrees and 20 minutes, and if the arrow coincides with the second division, then, the reading is 40 minutes. But, if the vernier arrow reads past any of the 20 minutes division lines, this quantity, which cannot be estimated on the primary scale on the circle, must be read by the vernier. The vernier reads minutes and seconds, in this way : — 59 subdivisions on the circle (that is, 59 divisions each of 20 minutes value), are again subdivided into 40 equal parts, for the length of the vernier ; therefore, each of these subdivisions on the vernier is smaller by j-'jj-th of 20 minutes than the divisions on the circle or limb, consequently the value of each subdivision on the vernier is equal to (20' X 60 -r 40) = 30", so that this five-inch instrument reads angles to half minutes, by the vernier scale. The odd minutes and seconds past the 20 minutes or 40 minutes divisions on the primary scale, on the circle or limb, are read in this way : — Clamp the instrument when in correct bearing on the point or angle to be measured, and look along the vernier, from the arrow, till a line on the vernier and a line on the circle is found to coincide ; then, beginning at the vernier arrow, count the number of minutes on the vernier scale to this line of coincidence, also the seconds (if any) past this last full minute on the vernier scale, and add this quantity to the first reading taken from the primary scale, thus ; — Figure 1. The TheorMite. 119 1 * '" "T" me Mr ..LI ^\\*\Tr r Figure 2. 3° 40' is read on the primary scale, and 2' 30" is read on the vernier scale, and 3' 40' + 2' 30"== 3" 42' 30" or the actual measurement of the observed angle. The wide spaces on the vernier scale are minutes, and the spaces between the shorter and closer lines are seconds. See Figure 1, where the readings of the degrees, minutes, and seconds in this example are marked by an asterisk, and the above explanation for reading the instrument will be easily followed and understood. In Figure 2, the example shews the reading to be 4° 46' 40", equal to a gradient of 1 in 12 very nearly. To Find the Value in Seconds, of the Reading of any Vernier. Find the value of each division or subdivision, in degrees, minutes, and seconds, on the primary scale of the circle or limb ; divide this quantity by the number of divisions on the vernier, and the quotient is the required value of the vernier reading. That is, the quotient gives the number of seconds the instrument reads to. Example. — If each degree on the circle is subdivided into three equal parts, or to 20 minutes, and the vernier has 40 divisions, reduce the minutes of a subdivision on the primary scale into seconds, thus : — 20' X 60 = 1200", then 1200" -^ 40 = 30", or the reading of the vernier in seconds in Figure 1. If it is a six-inch instrument, and the vernier is divided into 60 equal parts, each of these sub- divisions on the vernier will be j'^th smaller than the divisions on the vertical circle, or the horizontal plate or limb, and for finding the value of the vernier readings of this instrument, proceed as above : — 20' X 60 = 1200", then 1 200" -h 60 = 20", therefore the six-inch instrument, so divided, will read angles to twenty seconds in Figure 2. It being understood that if one vernier is reading elevation the other will read a corresponding depression. 120 Oblique Bridges Made Easy. Zero or Departure Point. n u -gko Full-sized Tangent Scale of Bell and Elliott's Patent Tangent Reading Tacheometer, with Micrometer Drum and Vernier, The Tacheoineler. 121 How TO Bead the Angles from the Tangent Scale, and use the Instrument as a Tangent Eeading Tacheometer. The radius of the instrument is determined from the centre of its axis, and may be 6.25 inches or 100 sixteenths, from the centre of the axis to the surface of the scale, -which may be considered as radius 1. First, focus the horizontal hair-lines in the microscope, so that they come out clear and sharp ; iiext, bring the zero of the verticil circle and the arrow of the vernier into exact coincidence and adjust the instrument to a perfectly horizontal position. Bring the 500 on the micrometer drum, which is its zero, into perfect coincidence with the arrow of the vernier, and if the tangent scale is in true adjustment, the line at nothing, or 500 (that is, the zero of the tangent scale,) will appear in the centre, between the two horizontal hair-lines of the microscope. If any error is found in the departure point or zero of the tangent scale, this can be easily removed by holding the milled head tightly between the thumb and finger, and turning the drum to the right or left, the quantity required for the adjustment ; if too tight to move in this way, loosen the screw in the milled head a little, taking care to tighten it again after making the adjustment. See page 120. Having seen to the above, give the staflf-holder the accurate height of the instrument from the station point to the axis of the instrument, and let the staff-holder adjust the upper edge of the lower target on the staff, to this height of the instrument before placing the staff at the end of the line on the point to be measured. Direct the telescope, so that the centre horizontal hair-line just cuts the top of the upper target on the staff ; clamp the telescope in this position, and look through the microscope to find the line on the tangent scale, which falls exactly between the two hair-lines of the microscope. Should no line fall between the two hair-lines of the microscope, if it is an ascent or rising ground, press the tangent scale forward till either a number or an unnumbered line falls between the two hair-lines, by turning the micrometer drum to the right, and add the number read by the micrometer drum to the whole number read on the scale (remember the scale is read from the operator on rising ground), and find this number in the tabulated tangents for its equivalent angle. If it is a descent or falling ground, the scale is read to the operator, and the micrometer drum is turned to the left, and proceed as before. The numbered lines or divisions on the tangent scale are hundredths of the radius of the instrument (from the axis to the siu-face of the tangent scale), and the unnumbered lines on the scale are two- hundredths of this radius ; but as the drum and vernier divides each of these 100 parts (that is -^V of an inch) into 1,000 equal and visible parts, each numbered division on the tangent scale is equal to 1 000 so that 'if the first numbered division, or 1 from zero, falls exactly between the hair-lines of the m'icroscope, it must be read as 1,000, and its value as a tangent is 1,000 ^ 100,000 = .01000, or the tangent of 34' 22".585, but if the next unnumbered line past 1 falls exactly between the hau--lmes, then 500 must be added to 1,000 for the tangent of this increased angle, and 500 - 100,000 = .00500, or the tangent of 17' 11".323, or 1500 - 100,000 = .01500, and is the tangent of 51' 33".743. But as it generally happens in practice, that the hair-lines fall between a numbered and an unnumbered line on the tangent scale, this quantity has to be found by the drum and vernier, and added to the reading of the tangent scale. The scale on the drum is divided into 100 equal parts on the inside next the vernier, and into 50 equal parts on the outside next the operator. One complete revolution of the micrometer_ drum from zero dther to the right or left, just moves the tangent scale one of these unnumbered divisions, or 500. 122 Oblique Bridges Made Easy. on the scale. The scale on the drum is divided into 100 equal parts, and each 100th part is further divided into five parts by means of the vernier, thus making the drum and vernier read 500, which, with the 500 read by the unnumbered line on the tangent scale, divides each numbered division on the tangent scale (fV) into 1,000 equal parts. The vernier is read thus : — Suppose the drum to be standing at zero, in this position, none of the division lines of the vernier coincide, but if the drum is turned the least to the right or left, the first division on the vernier, next the arrow, will coincide with the first division on the drum scale, and this is 1 or Tinr.VffT °'' -00001 the tangent of 2".06256459, and if the second division line on the vernier and the second on the drum scale coincide, then it is 2 or T^^^fi^^iy = .00002 the tangent of 4' .1251292, and so on in succession to 5. See figures 3, 4, and 5 below, for readings of the Drum and Vernier Scales. s S f S 1 ■"IJT A 1 1 1 1 T 1 1 1 1 1 1 \ TT 1 1 TT 1 { lb 1 ] (0 » 9 1 to r« c • « 4 9 3 a lO Soo 1 so 3 t> to sa 60 ro 80 This Re.<.os 33 and i ^ AS tah .00095 — s .00001 Figure 3. s ■■ s yf s 1 II ^ 1 1 \ 1 1 1 1 1 1 1 1 ^T 1 1 1 |\ 10 t / 100 ^ 80 70 eo So 4^ 39 Xfi 'o 600 to zo 30 ^ So So TMie RCAOS \fi And $ « A3 Tah .00133 • ■ooqos i Fiffure 4- i^l • i S ii I 1,1 1 'TT .1 1 1 1 1 1 1 1 1 \ 1 1 1 1 1 1 1 1 1 1 1 1 \ r T V J \ ;ioo n so Ic • 3 » • 1 so ( 'o 40 30 20 o*< 3 M . [0 A M i f> v> THIS RKi^DS ZX% AMD |9 = A« TAM .QOZZZ * .09019 Figure S. The Tacheometer. 123 It ■will be seen when the fifth division line on the vernier coincides with the fifth line on the drum scale, the arrow of the vernier coincides with the first division on the drum ; therefore the vernier has divided the first one hundred on the drum scale into 5, and 5 -f- 100,000 = .00005 the tangent of 10".3128230 ; and in reading the drum and vernier, when the vernier arrow coincides with the first division on the drum scale it is 5, and if the arrow is past 5, then look for the line on the vernier that coincides with a line on the drum scale and add that number of the vernier to 5. Suppose the arrow to be past 5, and the third line on the vernier is in coincidence with the fourth line on the drum scale, the reading of the drum and vernier is 8. The fine divisions on the drum scale next the vernier count 5, and the open divisions count 10, so that, if the vernier arrow reads past 3 open divisions and one close division, that is 35, but if the arrow is past the 5 add the number of the vernier line that coincides, to the 35. If 1 is road on the tangent scale by the microscope, that is .01000 and 35 on the drum that is .00035, the correct reading is .01035. If it was 135 the reading would be .01135. Always make the .01000 additions of the drum readings to the tangent scale readings thus 35^. The readings of the drum .01035 will be clearly understood by referring to the examples given above. Figure 6, Let be the position of the Tacheometer, and F« a tower (standing in a valley on the opposite bank of a river, and below the position of the instrument), whose height and distance are to be measured ; wm is a ten feet staff. E is the horizontal through the axis of the instrument 0. Observe the angles m E, n E, and FOE, carefully reading their tangents from the tangent scale, with the drum and vernier. Now OT E = E tan. wi E. » E = E tan. 71 E. .-.wE— «.E = tan. 7nOE — OEtan. »0E. or OT « = O E (tan. m O E — tan. n E). hence E = 10 tan. OT E — tan. w O E And » E = E tan. n E = feet. Also E F = O E tan. F E. «r is known. = feet. 24 Oblique Bridges Made Easy. It will be seen by referring to tbe above formula and figure, that on falling ground, vertical heights are found by multiplying the horizontal distances by the tangent of the greater angle. From the foregoing formula and Figure 6, it will be seen that the horizontal distance OE, is found by dividing the 10 feet space on the staff by in n on the scale Figure 7, that is, by the difference between the tangents of the two angles, taken from the top and bottom readings of the 10 feet staff; and that the vertical height is found by multiplying the horizontal distance OE, by 6, n on the scale, the tangent of the less angle, taken from the bottom target n of the 10 feet staff. This is for rising ground. When the bottom target, n on the staff, is raised to correspond accurately with the height cf the instrument, no deduction or allowance has to be made for the actual height of the point on which the staff is placed above the point on which the instrument stands. Messrs. Elliott Brothers make such a staff. The natural tangent of any observed angle is read direct either from the drum scale, or from the drum and tangent scales combined, therefore the operator does not require the use of any book of tables in the field. For the natural tangents, from the readings of 1 to 9, perfix four ciphers thus, .00001 and .00009, so as to make five places of decimals, and from 10 to 99, perfix three ciphers, and from 100 to 500 prefix two ciphers. From 1 to 9 on the tangent scale, prefix one cipher, because 1 and 9 on the tangent scale is 1,000 and 9,000, thus : — .01000, and .09000. From 10 on the tangent scale, which is 10,000, place the decimal point m front thus : — .10000, and so on to the end of the scale (see figures 3, 4, and 5), and the operator has the natural tangents of the observed vertical angles accurately read off from the tangent scale, up to 17° 44' 41", advancing from zero, by 2".06256459, at each unit read on the scales. For finding the value of the odd seconds see the top of page 79. I n b Figure 7. Suppose in making an observation as in Fgure 7 on the upper target m, the hair-lines of the microscope fall between the numbered line 17, on the scale, page 120, which is 17,000, and the next unnumbered line, which is 16,500, the scale must be pressed forwaid or backward, as it is an ascent or descent, by turning the micrometer drum to the right or left, as the case may be, until the unnumbered line 16,500 comes between the hair-lines of the microscope, and it is found that the micrometer drum reads 235 ; this number must be added to 16,500 ; tlien the position of the point m' on the tangent scale is 16,735, which being divided by O &, the radius (100,000) becomes .16735 and is the tangent of the angle OT E. Again, if in observing the lower target n, in Figure-? the hair-lines fall between 15, and the next The Tacheometev. 125 unnumbered line on the tangent scale, press forward the scale till 15 falls between the hair-lines, and suppose 15 is road by the micrometer drum, then 15 on the tangent scale is 15,000 and + 15, read on the micrometer drum is = 15,015, and divided by 100,000 becomes .15015 the tangent of ?i OE. The tangent of m E is .16735 and the tangent of ra E is .15015, therefore mOE-TOOE = .01720 the difiference between the tangents, hence the constant 10 feet space observed on the staflf -r 01720 {mn) = 581.395 (OE), and OE (581.395) x .15015 {bn) = 87.296 (toE). This is, for horizontal distances. Divide the 10 feet space on the staflf by the diflferenoe between the tangents of the two observed angles, and for vertical heights, Multiply the horizontal distance E by the tangent of the lesser angle ?i E, this is for rising ground. Briefly, Divide 10 feet by m n for horizontal distances, and for vertical heights. Multiply the horizontal distance E O by 6 n\ and if the observations are carefully made, the results are absolutely accurate to within 2 seconds of arc. For vertical heights on falling ground, Multiply the horizontal distance by the tangent of the greater angle, m E. In other words, for vertical heights, both on rising and falling ground. Multiply the horizontal distance by the tangent of the angle read from the lower target on the staflf at n. When the Solar Attachment is applied to the Tangent Reading Tacheometer, the Meridian, Longitude, Latitude, and Time, are readily determined. See illustration of the instrument. Page 117. This useful accessory consists of a small equatorial, whose polar axis is at right-angles to the axis of the Main Telescope of the Tacheometer. Both the Polar and Horizontal axes are provided with slow motion screws. A finder is attached to the Level of the Solar Telescope, which greatly facilitates observations. The Hour Circle can be clamped in any position around the Polar Axis. Adjustment .■—B.a.ying adjusted the Tacheometer, place the Main and Solar Telescopes in a Horizontal position, and direct them to a distant object; turn the Main Telescope to different positions of the circle, and adjust the Solar Attachment (by means of the adjusting screws which are placed under the Hour Circle) until the cross wires of the Solar Telescope bisect the distant object in every position of the Main Telescope, then adjust the Level of the Solar Telescope in a similar way. Applicalion .-—Refer to the Nautical Almanac for the declination ol the sun for the day and hour of observation, and correct for refraction. Now depress the Main Telescope to this angle on the vertical circle and without diaturbing the position of the Main Telescope, bring the Solar Telescope to horizontal position in the same plane by means of its Level. The two Telescopes «ill now embrace an angle equal to the amount of declination. With the Telescopes in the above position, unolamp the Main one, and set its Vernier to the co- latitude of the position at observation. n the two Telescopes be now moved round their vertical axis the image of the sun will be broucrht into the field of the Solar Telescope, and the Main Telescope into the Meridian of the plane of obs6r°vation ; the polar axis of the equatorial being parallel to the axis of the Earth. Projection of Segment for Templates. B - Fie. 7 3I-6SS- X, Fig. e, 7 ----i.S-X Theoretic Twisting Rules. \ Fig. 6 Fig. 7 Adjusted Twisting Rules. \ ""T 1 c 0» i. i Fig. e Span = Ac 20 Oblique Span = Ai> ag.lOS Obliquity = C.j 13.78! Axial Length = Di'ir QF 36.381 Ftg. 8 A Fig, 6 B Z-I7S6788 O C Adjusted Intradosal andExtradosalAngles. E I'lysTSsr Fig. 6 B a-iyttree o c b Theoretic Intradosal and Extradosal Angles. 2-r7S67sg tr intrados. / Radius 41-817 Templates for Spiral Arc i-jasas^s Fig 8 c E END SEOnON OF AKHSTONE C A i-i9S3e9s (RADIUS l7-««« Is. c Fig. Ga / » ANOLEOr SPlllNaEll AT y^-^ T BOTTOM OF CHECK B|^..6.J. ( — i.tooo«re- Template for End Joint Arc. Fig. 2 (tAoius 12-50 Fig. 3a Template for Natural Radius. O. Bsu, Delt- ■MA- PLATE 1 COTANatNT A Fig. 3a Fig. 12 # Plan of Impost. Fig. 13. 50° 20 Ft. Span. Designed to cross the Rae Burn at Hemplands Ford In the Longtown Union. «-*■ j-i —r^ - Elevation. Fig. 2. \ "" -^ t'--t'-t'- r^- I'-n'-n- I i_.>-.i Sectional Elevation through the Axis. Fig. 4. Plan. Fig. 1. 26-108 Plan of Impost. Fig. S- r I 1 I I I I I I i_L G. Bell, Delt. Scale of Feet. Sectional Elevation through the Axis. Fig. 4. 26.108 1 Plan of Impost. Fig. 3. > "<»ir-.- ^. ^^^^^^^fp^^^^l^?^^^^^^^!?^^!^!^?^-' Sectional Elevation through the Axis. Fig. 4. zssei- Plan of Impost. Fig. 3. i.«» a-; l/> w^ i^' t 'fr.^ ^ ik^ g^ Section Parallel to the Face. Fig, 5. Scale 8 F«at to an Inch. Designed to cross the River irthing at Mumps Hall, near Gilsland Station, N.E.R. 60° 60 Ft. Span. Fisrife-' /.;s-;" ift'' PLATE 4. Section Parallel to the Face. Fig. 5. Designed to cross the River Keeir^ .■<<^'^ r SINE j3 = . 7764132 COS: .6302242 ^=35° J TAN „ = 1.2319634 COT : .8117124 I SEC, =1.6867363 C08EC: I.2B79740 .-J* J ^V'^' -vfc" .l.-^ ^^" ^W cvO" ^^'^^a^'* ,^'' .839, 'P^ :<>' o,^° <^' .-j.^- .x^" I 50" 56'=/ r SINE j3 = . 7160106 COS: .6971661 "j 5=40° J TAN „= 1.0283226 COT : .9724575 }■ 45° 48' = ^ I SEC „ = 1.4343806 C08EC : 1.3948740 J .cf 0* N^' .\'* .-^^"^ fl=45° / 0°* I 8INE/3 = .6532004 COS: .7671861 -\ TAN „ = .8626604 COT : I.IS9I927 j- 40° 47' = )8 SEC „ = 1.3206810 COSEC : 1.5300288 j .O^" ,\«> >^' .c^^"^ .tr-l" 6&* £_eL. *o^ «''> ?>^ ^e*^^ ,c* >'-" *»' <:.^;*/'*'' -^^" a9 *'* .^»^ ,-<»'^ ..^' .«.*' .*' ^ C0° .*^^ <*'' 7 K =50° I SINE j3 = .5863724 COS : .8100416 "j TAN „ = .7238703 COT : 1.3814468 j- 35° 54' = i8 SEC „ = 1.2346044 COSEC : 1.7064010 J «># .-i*- fc*^ !>^' 7 f*^ /• 8INEj8 = .58e3724 COS: .8100416 -| =50° J TAN „= .7238783 COT : 1.3814468 j- 35° 54' = /3 I 8EC „ = 1.2346044 C08EC : 1.7064010 J . f SINE )8 =.51703(4 COS: .8559884 "j fl=55° -{ TAN „= .6040323 COT : 1.6555406 j- 31° 8' = ^ i SEC „ "^ 1.1682701 COSEC : I.034II85 J r SINE j8 = . 4458375 COS: .8950641 fl=60° \ TAN „ = .4982186 COT : 2.0071516 I SEC „ == I.II72384 COSEC : 2.2424669 26° 29'=/3 t S!NE )8 = .3732577 COS : .9277277 "j 0=65° i TAN,, = .4023354 COT : 2.4854887 j- 21° 55'=/3 I SEC „ = 1.0779026 COSEC : 2.6791146 J f SINE/3 = .29959Gg COS: .0540662 5=70° i TAN „ = .3140200 COT : 3.1845102 [ SEC „ = 1.0481453 COSEC : 3.3378294 I 17° 26'=/ LEfifSTH OF /iHC l-IO l-ZO 1-30 t-^O l-SO 1-60 SA '^ I I I I _fc- I I T"=z ;:^ TP- — 5^ Tir" Plan G. Bell, DeU PLATE 8.