LIBRARY OF CONGRESS. Shelf^L.2__ UNITED STATES OF AMEPJCA. To-day competition in every kind of trade is exceed- ingly sharp. Business is not done in the old fashioned way, — that most anything is good enough so long as the customer don't know of something better. Now-a-days your customer is very likely to know a good deal about the best goods and about the lowest prices. The live, active, wide-awake dealer must know whpre the best goods are made and where the most favorable prices can be secured. We manufacture stoves. We are not going to say here that we make the best stoves in the world and we are not going to say that we can sell at the lowest prices. We do say that we have particular facilities for making good stoves at reasonable prices. Now what we want to say to you (YOU ARE A STOVE DEALEI?, AREN'T YOU? is that you can't afford not to compare all the leading goods made in this country before you decide on your lines to carry in stock. All we ask is a fair comparison. We ask that pretty strongly and we mean it. If our goods and prices don't compare favorably, we can't ask anything more of you. It won't cost you anything (except a postage stamp or a postal card) to get our catalogue. We are the Mt. Penn Stove Works and our office and foundry are at Reading, Pennsylvania. We can't begin to give you in this space any idea of our goods (stoves, ranges and fur- naces), but we would like to send you our catalogue. This is a very modest request, — don't you think so? Digitized by tine Internet Arciiive in 2011 witii funding from The Library of Congress http://www.arcliive.org/details/tinsmitlispatternOOIitt QQQKING mJlBE BJISY ..Success the Proof of Me rit!.. ^ Over a Quarter Million NEW PRGGCSS (Evaporating) Stoves in use in the United States. For i8q4 equipped with brass hot air pipes, improved sight feed, removable drip valves, with anti-rusting points, cleanable reservoir and handsome planished steel asbestos lined oven — the best baker made. THE STAt\/DARO LIGHTING CO. CLEVELAND, OHIO. Tte Best nw (or Tiqp. PUBLISHED WEEKLY AT $2.00 PER YEAR IN ADVANCE. THIS JOURNAL IS DEVOTED TO THE Stove, TlniHardiare Interests And has a practical Tinshop Department in which are published tables, receipts, rules and diagrams for mm ew !!5i*'*'^!*1 as well as Special Departments devoted to practical papers on Cornice Making-,. Hot Air, Steam and Hot Water Heating. Answers to questions, etc., which often make a single number worth more than the price of a year's subscription. The free want and sales columns have- often secured tinners, or tinners have obtained situations, or disposed of tools- through them. American Artisan Full Size Pattern Sheet Comprises full size patterns for over 100 articles of pieced tinware. Price of entire set, $1.00, postpaid. Address, enclosing currency, express or postoffice money order or draft, DANIEL STERN, 69 Dearborn Street, - - CHICAGO, ILI^ JEWEL GBSOLEUE STOYES... ^ WE'RE' AWARDED. HIGHEST MEDAL AND DIPLOMA World's Fair, Chicago, 1 S93. SEIND FOR GATTAIaOG. George M. Clark & Company MAKERS 149-161 Superior Street, Plumbers' and Tingiars' Furnace lo. 2 Roofing and Plumbing Strong Fire; Wind-.' sfc Li.s^ht.... Our No. 1 For Tinners' shop use, has Double Burner and High Tank... Price of either... §7.00. Discounts to tlie Trade. Both the Best in their Line Thousands in use... We refer to everyone. SOLOEBiHG FURNACES... PBICE, $5.0^. Gasoline Torches For in and out door use, etc., etc, RINGE TCVE CO Manufacturers of j^ THF. . CELEBRATED... diriieal'- GASOLINE GAS STOVES. THEl TiisMiTHS' Patteri Maiual Patterns for Tinsmiths' Work. \o> BY /) JOE K. LITTLE, [C. E. FOR Tinners, Coppersmiths, Plumbers, Zinc Workers, and Sheet Metal Workers Generally. 1894. THE AMERICAN ARTISAN PRESS, 69 Dearborn Street, CHICAGO. <^^ . \> Entered according to Act of Congress in the year 1893, by JOSEPH KANE LITTLE, In the Office of the Librarian of Congress, at Washington, D. C. 3 ^ .^ . . . Preface 'T^HE purpose of this work is an eminently practical -*■ one, as it is designed not so much to furnish a batch of isolated patterns in' common use, as to lay down general geometrical principles, each one of which, when mastered by the reader, will enable him to draw a number of different patterns whose principle ot con- struction is essentially the same. The sheet metal worker who masters the geometrical constructions here- with presented, can easily develop the surface of any article with much greater ease and rapidity than by^ following the various methods in general vogue. Concerning the greater utility of the patterns here- in shown, I can speak from experience, having served a pains-taking apprenticeship in the workshop system of setting out patterns before it was my good fortune to discover the application of geometrical principles to what had been my daily toil. It has been my constant aim to make the book a satisfactory one from the stand- point of both the mathematician and the workshop, and the many enconiums bestowed on it by mechanical en- gineers and technical professors as well as by working tinners to whom advance sheets have been shown, leads me to think that I have succeeded in my aim. Each of the problems presented is complete in it- self; but although solved independently they follow each other in due order. It is my modest hope that this work will prove a useful addition to our scanty collec- tion of volumes on geometry as applied to sheet metal work. The Author. Philadelphia, Pa., Oct., 1893. CONTENTS. Book I. CHAPTEE L TAOE Classipicatioit .. « •• •• i CHAPTER II. Intboductost Pkoblems 3 Definitions, 3-5 ; Problems on Angles, Lines, Circles, Poly- gons, Ovals, Ellipses, and Oblongs, 6-20 ; Meafmrement of Angles, 20-3. CHAPTER m. Aeticles of Equal Taper or IwcLmATioN op Slant .. .. 24 Right cone defined and described, with problems, 24-7. CEAPTER IV. Patterns fob Round Articles op Equal Taper or Inclina- tion OF Slant .. .. .. .. 28 Developmeiit of right cone, and problems on same, 28-31 ; allowance for lap, seam, and wiring, 32-3 ; frustum (round equal-tapering bodies) defined and described, with pro- blems, 33-37 ; patterns for round equal-tapering bodies (frusta) in one, two, or more pieces, for both email and large wort;, 37-45. viii CONTENTS. CHAPTER V. VJiOK Eqcal-taperinq Bodies and theie Plavs „ ., .. 46 Plans and elevations, 46-50 ; plans of equal-tapering bodies ; their characteristic features, 50-6 ; problems on the plans of round, oblong, and oval equal-tapering bodies, 56-65 CHAPTEE VI. Patterns foe Flat- paced Equal-tapeking Bodies .. ., 66 Definition of right pyramid, and development, 66-7 ; pattern for right pyramid, 68-9 ; frustum of right pyramid defined and described, 69-71 ; patterns for frusta of right pyramids (hoppers, hoods, &c.) for both small and large work, 71-7 ; baking-pan pattern in one or more pieces, 77-83. CHAPTER VII. Patterns for Equal-tapebing Bodies of Flat and Curved Surfaces combined .. .. .. .. .. .. 84 Pattern for oblong body with flat sides and semicircular ends, 84r-90; for oblong body with round comers, 90-6; for oval body, 96-104. The patterns for each of these bodies are given in one, two, and four pieces, as well as for both small and large work. Book II. CHAPTER I. Patterns for Bound Articles of Unequal Taper ob Incli- nation op Slant .. .. .. .. .. .. 105 Oblique cone, definition, description, and development, 105-111 ; oblique cone frustum (round unequal-tapering body) and oblique cylinder, 111-3; patterns for roimd unequal-tapering bodies (oblique cone frusta), also oblique cylinder — work small or large, 113-23 CONTENTS. IX CHAPTEE 11. PAGB TTNEQtrAL-TAPERING BODIES AND THEIR PLANS .. .. .. 124 Plans of uuequal-tapering bodies and their characteristic features, 124-9 ; problems on plans of oblong, oval, and other unequal-tapering bodies, 130-3 ; plans of Oxford hip-bath, Athenian hip-bath, sitz bath and oblong taper bath. 134-42 CHAPTER III. rATTERys FOR Flat-faced Unequal-tapering Bodies .. ., 143 Oblique pyramids and their frusta, 143-5 ; pattern for oblique pyramid, 145-8 ; patterns for frusta of oblique pyramids (flat-faced unequal- tapering bodies) — work small or large, 148-54; pattern for a hood which is not a frustum of oblique pyramid, 154-6. CHAPTER IV. Patterns for IJNEQUAL-TAPERiNa Bodies of Flat and Curved Surfaces combined .. ,. .. ., .. .. 157 Pattern for equal-end bath (unequal-tapering body having flat sides and semicircular ends), 157-68 ; for oval bath, (oval unequal-tapering body), 168-83; for tea-bottle top (unequal-tapering body having round top, and oblong bottom with semicircular ends), 183-91 ; for oval-canister top (unequal-tapering body having round top and oval bottom), 192-205 ; for unequal-tapering body having round top and oblong bottom with round comers, 205-16. ( The patterns for each of these bodies are given in one, two, or four pieces, as well as for both small and large work.^ Pattern for Oxford hip-bath, 216-30 (two methods) ; for oblong taper bath, 230-6 (two methods). ETAL-PLATE WORK. BOOK I. CHAPTEE I. Classification. (1.) Notwithstanding the introduction of macliinery and the division of labour in the variotss "branches of metal- plate work, there is as, great a demand for good metal-plate workers as ever, if not indeed a greater demand than formerly, while the opportunities for training such men are becoming fewer. An important part of the technical edu- cation of those connected with sheet-metal work is a know- ledge of the eetting-out of patterns. Such knowledge,, requisite always by reason of the variety of shapes that are met with in articles made of sheet-metal, is nowadays especially needful ; in that the number of articles made of sheet metal, through Ihe "revival of art metal-work, the general advance of science, and the introduction of new designs (which in many cases have been very successful), ia articles of domestic use, has considerably increased. It is with the eetting-out of patterns that this volume princi- pally deals. To practical men, the advantages in saving of time and material, of having correct patterns to work from, are obvious. Whilst, however, the method of treat- ment here of the subject will be essentially practical, an THE TINSMITHS- PATTERN MANUAL. HiiKjunt of tlieory' sufficiont f6r a thorough comprehension of the rules given will he introduced, a knowledge of rules without principles being mere ' rule O'f thumb,' and not true technical education. (2.) Starting in the following pages with some introduc- tory problems and other matter, we shall proceed from these to the articles for which patterns are required by sheet-metal Avorkers and which may be thus conveniently classed and xnbdivided : \ ia. Of round surfaces. 6. Of plane or flat surfaces, divisions. | c. Of curved and I plane surface \ combined. Class T, — Patterns for Ar- ticles of e(lual taper or inclination (pails, oval teapots, gravy strainers, &c.). Class 11. — Patterns for Ar- ticles of unequal taper or inclination (baths, hoppers, canister - tops, &c.). Sub- divisions. o. Of round surfaces. 6. Of plane or flat surfaces, c. Of . curved and plane surface combined. Class III. — Patterns for Miscellaneous Articles (elbows, and' articles of compound bent surface, as vases, aquarium stands, mouldings, &c.). All these articles will be found dealt with in their several planes. We shall conclude with a few technical details in respect of the metals that metal-plate workers mostly make use of. (3.) The setting out of patterns in sheet-metal work belongs to that department of solid geometry known as " Development of Surfaces," which may be said to be the Rpreading or laying ' out without rupture the surfaces of solids in the plane or flat, the plane now being sheet metal. llii-: TlN'SMITliS' PATTERN MANIJA1-. CHAPTEK II. Introdtjctouy Problems ; with Applications. Definitions. Straiglit Line. — A straight line is the shortest distance between two points. Note. — If not otherwise stated, lines are always supposed to be straiglit. Angle. — An angle is the inclination of two lines, which meet, one to another. The lines A B, C B in Fig. 1 which are inclined to each other, and meet in B, are said to form an angle with one another. To express an angle, the letters which denote the two lines forming the angle are employed, the letter at • the angular point being placed in the middle j thus, in Fig. 1, we speak of the angle ABC. Fig. 1. Fig. 2. A Fis. 3. Perpendicular. Biglit Angles. — If a straight line, A B (Fig. 2), meets or stands on another straight line, G D, so that the adjacent angles (or angles on either side of A B) A B I>, A B C, are equal, then the line A B is said to be perpendicular to, or at right angles with (' square with ') D G, and each of the angles is a right angle. 4 THE TINSMITHS' PATTERN MANUAL. Parallel Lines. — Parallel lines are lines wHich, if produced ever so far both ways, do not meet. Triangle. — -A figure bounded by three lines is called a triangle. A triangle of whicli one of the angles is a right angle is called a right-angled triangle (Fig. 3) ; and the side which joins the two sides containing the right angle is called the hypothenuse (or hypotenuse). If all the sides of a triangle are equal, the triangle is Equilateral, If it has two sides equal, the triangle is Isosceles. If the sides are all unequal, the triangle is Scalene. Polygon. — A figure having more than four sides is called a. polygon. Polygons are of two classes, regular and irregular. Irregular Polygons have their sides and angles unequal. Regular Polygons have all their sides and angles equal, and possess the property (an important one for us) that they can always be inscribed in circles ; in other woi ds, a circle can always be drawn through the angular points of a regular polygon (Figs. 12 and 13). Special names are given to regular polygons, according to the number of sides they tpossess ; thus, a polygon of five sides is a pentagon ; of six sides, a hexagon ; of seven, a heptagon; of eight, an octagon; and so on. Quadrilaterals. — All figures bounded by four lines are Fig. 4a. Fig. 46. called quadrilaterals. The moat important of these are tiie square and oblong or rectangle. In a square (Fig. 4rt) tl.e sides are all equal and the angles all right angles, and con- sequently equal. An oblong or rectangle has all its anglos right^ angles, but only its^oppusity sides are_e(iu£il._ (Fig. dl*.) TUE TINSMITHS' PATTERN MANUAL. f) Circle. — A circle is a figure bounded by a curved line such that all oints in the line are at an equal distance from a certain point within the figure, which point is called the centre. The bounding line of a circle is called its circumference. A part only of the circumference, no matter how large or small, is called an arc. An arc containing a quarter of the circum- ference is a quadrant. An arc containing half the circumfer- ence is a semicircle. A line drawn from the centre to any point in the circumference is a radius (plural, radii). The lino joining the extremities of any arc is a chord. A chord that passes through the centre is a diarneter. A line drawn from the centre of, and perpendicular to, any chord that is not a diameter of a circle, will pass through its centre. In practice a circle, or arc, is ' described ' from a chosen, or given, centre, and with a chosen, or given, radius. If two circles have a common centre, their circumferences are always the same distance apart. In ¥ig. 5. O is the centre. AD (the curve) is an arc. A B (or B C) is a quadrant. A C B is a semicircle. O A (or B, or B C) is a radius. A D (the straight line) is a chord, A is a diameter. 6 THE TiNSMiTllS' PATTERN M\XUAL. PROBLEM I. To draw an angle equal to a given angle. Case I. — Where the * given ' angle is given by a drawing. This problem, though simple, is often very useful in practice, especially for elbows, where the angle (technically called ' rake ' or ' bevil ') is marked on paper, and has to be copied. Fig. 6. Let ABC (Fig. 6) be the given angle. With B as centre and radius of any convenient length, describe an arc cutting B A, B C (which may be of any length, see Def.) in points A and C. Draw any line D E, and with D as centre and same radius as before, describe an arc cutting D E in E. With E as centre and the straight line distance from A to C as radius, describe an arc intersecting in F the arc just drawn. From D draw a line through F ; then the angle F D E will be equal to the given angle ABC. Case II. — Where the given angle is an angle in already existing fixed work. The angle to which an equal angle has to be drawn, may be an angle existing in already fixed work, fixed piping for instance ; or in brickwork, when, suppose, a cistern may have to be made to fit in an angle between two walls. In such cases a method often used in practice is to open a two- fold rule in the angle which is to be copied. The rule is then laid down on the working surface, whatever it may be (paper, board, &c.), on which the work of drawing an angle eqxial to the existing angle has to be carried out, and lines are drawn on that surface, along eithor the outer ox inner edges of the THE TINSMITHS' PATTERN MANUAL. 7 rule. The rule being tnen removed, tne lines are proauced ; meeting, they give the angle required. Case III. — Where the given angle is that of fixed work, and the method of Case IL is inapplicable. With existing fixed work, the method of Case II. is not always practicable. A corner may be so filled that a rule cannot be appliedi The method to be now employed is as follows. Draw lines on the fixed work, say- piping, each way from the angle ; and on each line, from the angle, set off any the sa.me distance, say 6 in., and measure the distance between the free ends of the 6-in. lengths. That is, if A C, A B (Fig. 7) Fig. 7. represent the lines drawn on the piping, measure the distance between B and 0. Now on the workin g^surface on which the drawing is to be made, draw any 1 me D E, 6 in. long; and with D as centre and radius D E, describe an arc. Next, with E as centre, and the distance just measured between B and as radius, describe an arc cutting the former arc in F. Join F D ; then the angle F D E will be equal to the angle of the piping. Note. — When poiata are 'joined,* it is always by straight lines. PROBLEM IL To divide a line into any nuinhcr of equal parts. Let A B (Fig. 8) be the given line. From one of its extre- mities, say A, draw a line A 3 at any angle to AB, and en it, from the angular, point, mark off as many parts, — of any con- 8 THK TIN.-s^riTHS' I'ATTKRN MANUAL. verrient length, but all eqnal to each other, — as A B is to he divided into. Say that A B is to he divided into three equal jjarts, and that the equal lengths marked oif on A3 are A to Fig. 8. A C D B 1, 1 to 2, and 2 to 3. Then join point 3 to the B extremity of A B, and through the other points of division, here 1 and 2, draw lines parallel to 3 B, cutting A B in C and D. Then A B is divided as required. PROBLEM III. To hiseci (divide a line into two equal parts) a given line. Let A B (Fig. 9) be the given line. With A as centre, and any radius greater than half its length, describe an indefinite arc; and with B as centre and satoe radius, describe an arc intersecting the former arc in points P and Q. Draw a line through P and Q , this will bisect A B. Note. — ^It is quits as easy to bisect A B by Problem II. ; but tbe method shown gives, in P Q, not only a line bisecting A B, but a line oerpendicular to A B, This mubt be particularly remembered, I'HE T[NSM[THS- PATTERN MANUAL. 9 PEOBLEM IV. To find the centre of a given circle. Let ABC (Fig. 10) be the given circle. Take any three points A, B, C, in its circumference. Join A B, B C ; then A B, B C, are chords (see Def.) of the circle ABC. Bisect AB, B C ; the point of intersection, 0, of the bisecting lines is the centre required. Fig. 10. PEOBLEM V. To describe a circle which shall pass through any three given points that are not in the same straight line. Let A, B, C (Fig. 10) be the three given points. Join A B, B C. Now the circle to be described vsrill not be a circle through A, B, C, unless A B, B C, are chords of it. Let us therefore assume them such, and so treating them, find (by Problem IV.) O the centre of that circle. With as centre, and the distance from O to A as radius, describe a circle; it will pass- also through B and C, as required. PEOBLEM VI. Given an arc of a circle, to complete the circle of which it is a portion. Let A C (Fig. 10) be the given arc ; take any three points in it as A, B, C ; join A B, B C. Bisect A B, B C by lines 16 TUB TINSMITHS' PATTERN MANUAL, intersecting in 0. With O as centre, and to A or to any puint in the arc, as radius the circle can be completed. PEOBLEM VII. To find icheiher a given curve ia an are of a circle. Choose any three points on the given curve, and by Problem V describe a circle passing through them. If the circle coincides with the given curve, the curve is an arc. PEOBLEM VIII. To bisect a given angle. Let ABO (Fig. 11) be the given angle. With B as centre and ^ny convenient radius describe an arc cutting A B, B in D and E. With D and E as centres and any co£V3uieut distance, greater than half the length of the aro I) E as radius describe arcs intersecting in F. Join F to B j then F B bisects the given angle. PEOBLEM IX. In a given circle, to inscribe a regular polygon of any giasin number of sides. Divide (Problem II.) the diameter A C of the given circle (Fig. 12) into as many equal parts as the figure is to bava TIIK TINSMITHS' PATTERN MANUAL. ll sides, here say five. With A and C as centres, and C A as radius, describe arcs intersecting in P. Through P and the second point of division of the diameter draw a line P B cutting the circumference in B ; join B A, then B A will be one side of the required figure. Mark off the length B A from A round the circumference until a marking off reaches B. Then, beginning at point A, join each point in the circumference to the next following ; this will complete the polygon. Note. — By this problem a circumference, and therefore also one-half of it (semicircle), one-third of it, one-fourth of it (quadiant), aud so on, can be divided into any number of equal parts. PEOBLEM X. To describe any regular polygon, the length of one side being given. Let A B (Fig. 13) be the given side of, say, a hexagon. With either end, here B, as centre and the length of the given side as radius, describe an arc. Produce A B to cut the arc in X. Divide the semicircle thus formed into as many equal parts (Problem IX,, Note) as the figure is to have sides (six), and join B fo the second division point of the semicircle couniing from X. This line will be another side of the required polygon. Having now three points, A, B, 12 TIIK TINSMITHS' PATTERN MANUAL. and the second division point from X, draw a circle tKrough them (Problem V.), and, as a regular polygon oan always be inscribed in a cii'olo (cee Def.), mark off the length B A round Fig. 13. the ciroumferenoe from A until at the last marking-off, the free extremity of the second side (the side foxind) of the polygon is reached, then, beginning at A, join each point in the oirotimferenoe to the next following ; this will oomploto the polygon (hexagon). PROBLEM XL To find (he length of the cirmmference of a circle, the diameter being given. Divide the given diameter A B (Fig. li) into seven equal parts (Problem II.). Then three times A B, with C B, on© of the seven parts of A B, added, that is with one-seventh of FiQ. U. B A B added, will be the required length of the ciroumferenoe. The semicircle of the figure is superfluous, but may help to make the problem more clearly understood. THE TINSMITH.'?' PATTERN MANUAh 13 PROBLEM XII. To draw an oval, its length and width being given. Draw two lines A B, C D (tho axes of the oval), perpendi- cular to one another (Fig. 15), and intersecting in 0. Fig. 15. Make A and B each equal to half the length, and C and D each equal to, half the width of the oval. From A mark off A E equal to C D the width of the ovalj'and divide E B into three equal parts. With 0* as centre and radius equal to two of the parts, as E 2, describe arcs cutting A B in points Q and Q'. With Q and Q' as centres and QQ' as radius describe arcs intersccHng C D in points P and P'. Join. P Q, P Q', P' Q and- P' Q' ; in these lines produced the end and side curves must meet. With Q and Q' as centres and Q A as radius, describe the end curves, and with P and P' as centres and radius P D, describe the side curves ; this will complete the oval. Note. — Unless care is taken, it may be fmind that the end and siue curves will not meet accurately, and even with care tlxis may sometimes occur. It is best if great accuracy be required in tlie length, to draw the end curves first, and tlieu draw side curves to meet them; or, if the width is most important, to draw the side curves first. The centres (P and P') for the side curves come inside or outside the curves, according as the oval is broad or narrow. This figure is sometimes erroneously called an ellipse. It is, however, a good approximation to one, and for most purposes where aa elliptical article bus to be made, is very convenient. 14 TliE TIKSMITHS' PATTERN MANUAL. PEOBLEM XIIi: To draw an egg-shaped oval, having the length and width given. Make A B (Fig. 16) equal to the length of the oval, and from A set off A ec^ual to half its width. Through draw Fig. 16. an indefinite line Q Q' perpendicular to A B, and with as centre and A as radius describe the eemicircle CAD. Join D B ; and from D draw D E perpendicular to Q Q' and e<^ual to D. Also from E draw E G- parallel to Q Q' and n\E TINSMITHS PATTERN MAN'UAli. 15 intersecting D B in Gr, and from G draw G F parallel to D E and intersecting Q Q' in F. From B set off B P equal to D F, and join P F. Bisect F P and througli the point of bisection draw a line cutting Q Q' in Q. Join Q P and produce it indefinitely, and with Q as centre .and Q D as radius descriho an arc meeting Q F produced in H, Make Q' equal to Q, and join Q' P and produce it indefinitely. With Q' as centre, and Q' (equal to Q D) as radius, describe an arc meeting Q' P produced in H'. And with P as centre and P B as radius describe an arc to meet the arcs D H and C H' in H and H' ; and to completo the egg-shaped oval. PEOBLEM XIY. To describe an ellipse. Before working this as a problem in geometry, let us draw an ellipse non-geomatrically and get at some sort of a defini- tion. This done,, we Mdll solve the problem geometrically, and follow that with a second mechanical method of de- scribing the curve. METHOD I.— Mechanical. A. Irrespective of dimensions. — On a piece of cardboard or smooth-faced wood, mark off any two points F, F' (Fig. 17) and fix pins securely in those points. Then take a piece of thin string or silk, and tie the ends together so as to form a loop ; of such size as will pass quite easily over the pins. Kes:t, placQ the point of a pencil in the string, and take up the slack so that the string, pushed close against the wood, shall form a triangle, as say, F D F', the pencil point being at D. Then, keeping the pencil upright, and always in the string, and the string taut, move the pencil along from left to right say, so that it shall make a continuous mark. Let us trace the course of the mark. Starting from D, the pencil, constrained always by the string, moves from D to F, 16 THE TINSMITHS' PATTERN MANUAT.. then on to B, T', C, P^, P^, A, P*, and D again, describing a curve- which returns into itself ;* this curve is an ellipse. Having d^a^vn the ellipse, let us remove the string and pins, draw a line from F to F', and produce it both ways to terminate in the curve, as at B and A. Then A B is the major axis of the ellipse, and F, F' are its foci The mid- point of A B is the centre of the ellipse. Any line through the centre and terminating botli ways in the ellipse is a diameter. The major axis is the longost diameter, and is commonly called" the length of the ellipse. The diameter through the centre. at right angles to the major axis is the shortest diameter, or minor axis, or width of the ellipse. Referring to the Fig. : — ADPBC is an ellipse. F, F' are its foci (singular^/ocus), A B is the major axis. C D is the minor axis. O is the centre. We notice with the string and pencil that ^vhen the pencil i)oint reaches P, the triangle formed by the string is F P F' ; when it reaches P', the triangle is F P' F' ; when it reaches P- the triangle is F P^ F' ; and when P^ is reached, it id F P^F'. Looking at these triangles, it is obvious that THE TINSMITHS' PATTERN MANUAIi. 17 F F' is ono side of eacH of them ; from which it follows, seeing ths:t the loop of string is always of one. length, that the siira of the other two sides of any of the triangles is equal to the sum of the other two sides of any other of them ; that is to say, F D. added to DF' is equal to F P added to P F', is equal to P P' added to F F', and go on. Which leads us to the following definition. Definition, EUi})se. — The ellipse is a closed curve (that is, a curve returning into itself), such that the sum of the distances of any point in the curve from certain two joints (foci), inside the curve is always the same. B. Length and loidtTi given. — ■ Knowing now what an ellipse is, we can work to dimensions. Those usually given are the length (major axis), and width (minor axis). Draw A B, C D (Fig. 17), the given axes, and with either ex- tremity, C or D, of the minor axis as centre, and half A B, the major axis as radius, describe an arc cutting AB in F and F'. Fix pins securely in F, F' and D (or C). ' Then, having tied a piece of thin string or silk firmly round the three pins, remove the pin at D (or C) ; put, in place of it, a pencil point in the string j and proceed to mark out the ellij)se as above explained. METHOD II. — Geometrical. — The Solution of the Problem. Length and Width- gi\^en. Draw A B, C D (Fig. 17), the major and minor axes. With C or D as centre, and half the ijiajor axis, B say, As radius, describe arcs cutting A B in F and F'. On A B^and between O and F', mark points — any number and anywhere, except that it is advisable to mark the points closer- to each other as they approach F'. Let the points here be 1^ 2, and 3. With F and F' as centres and. A 2, B 2 as radii respectively, describe arcs intersecting in P; with same centres and A3, B 3 as radii respcL-tively, describe arcs intersecting in P'. With F' and F as centres and A3, B 3 as radii respectively, 18 T1!K TIN.'.MITIIS' PATTKUN MANUAL. describe arcs intersecting in P^. With same centres and A 2, B 2 aa radii respectively, describe arcs intersecting in P*. .Similarly obtain P^. We have thus nine points, D, P» B, P', Cj P^, P^j A and P*, through wh"° jh an even curve may 1)6 drawn which will be the ellipse required. A greater number of points through which to draw the ellipse may of course be obtained by taking more points between and F', and proceeding as explained. METHOD III.— SIechanical. — Length and Width '^ GIVEN. As it is not always possible to proceed as described at end of Method L, for pins cannot always be fixed in the material to be drawn upon, we now give a second mecha- nical method. Having drawn (Fig. 186) AB, CD, the Fia. 18a. JTnP Fig. 18&. jfU given axes, then, on a strip of card or stiff paper X X (Fig. 18a), mark off from one end P, a distance PF equal to half the major axis (length), and a distance .P E equal to half the minor axis (width). Place the strip on the axes in such a position that the point B is on the major axis, and the TIIR TlXi^MITtlS' PATTERN ^MANUAL. 19 point F on the minor, and mark a point against the point P. Now shift X X to a position in which E is closer to B, and F closer to C, and again mark a point against P. Proceed similarly to mark other points, and finally draw an even curve through all the points that have been obtained. The following problems deal with sha,pes often required It/ the metal-plate worTier^ and will give him an idea of how to adapt to his requirements the problems that precede. The explanation of the measurement of angles that concludes the chapter will further assid him in his worTs. PROBLEM XV. To draw an oblong with round corners. Draw two indefinite lines A B, CD (Fig. 19) perpen- dicular to one another and intersecting in 0,. Make OA Fra 19. and 0. B each equal to half the given length; and OC and O D each equal to half the given width. Through and D n 2 25 tUE TlXSMITllS- PATTERN MANUAL draw lines parallel to A B, and throngh A and B draw lines jiai'allel to C D. We now liave a rectangle or oLlong, and re(|mre to roimd tlie corners, which aie quadrants. Mark oft" from E along E D and E A equal distances E (t and E F according to the size of corner required. With F and G as centres and E F or E G as radius, describe arcs intersecting in O'. With O' as centre and same radius descril>e the corner F G. The remaining corners can be drawn in similar manner. ITtOBLEM X¥L To draw a figure Tiacing straifjJit sides and semicircular ends {ohlong with semicircular ends). Draw a line AB (Fig. 20) equal to the given length, make AO and BO' each equal to half the given widLh. Fio. 20. I) f; .( "\ n € y F G Through and 0' draw indefinite lines perpendicular to A B; with O and 0' as centres and OA as radius describe arcs cutting the perpendiculars through O and O' in D F and GE. Join D E, GF; this will complete the figure required. ANGLES AND THEIR MEASUREMENT. The right angle BOG (Fig. 5) subtends the quadrant B C. If we divide that quadrant into 90 parts and call the partb THE TIK.SMITllS- PATTERN MANUAL. I'l degrees, tlaen a rigM angle subtendiS or contains 90 degrees (written 9(J^), or as usually espressed, is an angle of 90 degi-ees, the degree being the unit of measurement. If each division point of the quadrant is joined to O, the right anglo) is divided into 90 angles, each of which subtends c* is an angle of 1 degree. That is to say, an angle is measured by the number of degrees that it contains. Suppose the quadrant B A is divided as was B C, then BOA also is an angle of 90 degrees. If the division is continued round the semicircle ADC. this will contain 180 degrees, and the whole circumference has been divided into 360 degrees. As an angle of 90, which is a fourth part of 360 degrees, subtends a quadrant or fourth part of tlie cii'cumference of the circle, so an angle of 60, which is a sixth part of 360 degrees, subtends a sixth part of the circumference, and similarly an angle of 30 degrees subtends a twelfth part, an angle of 45 an eighth part, and so on. And this angular measurement is quit$ independent of the dimensions of the circle; the quadrant ahvays subtends a right angle; th.e 60 degrees angle always subtends an arc of one sizth of the circum- ference ; and the like with other angles. From our defini- tion p. 5 we have it tliat a chord is the line joining the extremities of any arc. The chord of a sixth part of the circumference of any circle, we have now to add, is equal to the radius of that circle. This being the case, and as an angle of 60 degrees subtends the sixth part of the circumfer- ence of a circle, it follows that an angle of 60° subtends a chord equal to the radius. SCALE OF CHORDS. Construction. — We have now the knowledge requisite fcr setting out a scale of chords, by wliich angles may be drawn and measured. On any line B (Fig. 21) describe a semicircle A E, and fi-om its centre C draw C A j)erpendicular to B. Divide A into nine equal parts. Then, as A, being a qxiadrantj ^ THE TINSP4IT1IS' PATTERN MANUAL. ^atains 90^, each of the nme divisioBS will contain 10''. The points of division, from O, of the quadrant, are marked 10, 20, 30, &0.J up to 90 at A. With O as centre, describe arcs from each of these division points, cutting the line O B. Note that the arc from point 60 cuts O B in C, the centre of the semi- circle ; the chord from O to 60 (not drawn in the Fig.), that Fia. 22. O 10 20 30 ne. It is nee ssary, therefore, to define the ri^ht cone and exolain some of its properties. Det-inition. (6.) Bight Coup. — A right cone Is a solid fi.^ure gcnoratsd or formed by the' revolution of a rfght-arglcd triangle about one of the sides containing the right angle. The side about which the triangle revolves is the axis of the cone ; the other side containing the right angle being its radius. The point of the cone is its apex; the circular end its hase. The hypotenuse of the triangle is the slant of the cone. From the method of .formation of the right cone, it follows that the axis is perpendicular to the base. The height of the cone is the length of its axis (7.) Referring to Fig. la, OBE represents a coae gene- THE TINSMITHS' PATTERN MANUAL. 25 rated or formed by the revolution of the right-angled tri- angle A B (Fig. 16) about one of its sides containing the right angle, hero the side A, Similarly the cone O D F, Fig. 2a, is formed by the revolution of C D (Fig. 26) about its side . O C. As will be seen from the figs., O A, O C are respectively the axes of the cones B E, O D F, as also their heights. Their bases are respectively B G E H, D K F L, F'G. 25. and the radii of the bases are A B and C D. The slants of the cones are O B and O D, the apes in either being the point O. Other lines will be seen in figs,, namely, those repre- senting the revolving triangle in its motion of generating 26 TtiK TJNS-MrTUS' PAT'I EKN MAXl'AL. the cone. The sides of tliese triangles that start from the apex and terminate in the base are all equal, it must he home in mind ; and each of them is the slant of the cone. Likewise their sides that terminate in A are all equal, and each shows a radius of the base of the cone. How these particulars of the relations to one another of the several parts of the right cone apply in the setting-out of patterns will be seen in the problems that follow. PEOBLEM I. To find the height of a cone, the slant and diamsier of the hass being given. m Draw any two lines A, B A (Figs. 3 and 4) perpen- dicular to each other and intersecting in A. On either line Fig. 3. Fig. 4. _.4 iLouoA-ue of£as& 3 mark off from A half the diameter of the base, in other words, the radius of the base, as A B. With B as centre, and radius equal to the slant, describe an arc cutting A in 0. Then A is the height of the cone. TIIK TTNSMITHS' PATTERN MANUAL. 27 rEOBT.EM II. TofindtTie slant of a cone, the height and diameter of the base being given. Draw any two lines O A. B A (Figs. 3 and 4") perpendicular to each other and intersecting in A. On either line mark off from A half the diameter of the base (radins of the base), as A Bj and make A O on the other line equal to the height of the cone ; join B. Then B is the rec[nired slant THE TINSMITHS' PATTERN MANUAL. CHAPTEK lY. Patteens FOR Round Articles of Equal Taper or Inclination of Slant. (Class I. Subdivision a.) (8.) If a cone has its inclined or slanting surface painted say, white, and be rolled while wet on a plane so that every portion of the surface in succession touches the plane, then the figure formed on the plane by the wet paint (see Fig. 5) Fig. 5. will be the pattern for the cone. As the cone rolls (the figure represents the cone as rolling), the portion of it touching the plane at any instant is a slant of the cone (see §7.). (9.) Examining the figure formed by the wet paint, we find it to be a sector of a circle, that is, the figure contained between two radii of a circle and the arc they cut off. The length of the arc here is clearly equal to the length of the circumference of the base of the cone, and the radius of the THE TINSMITHS' PATTERN MANUAL. 29 aic evidently equal to the slant of the cone, v From this it is obvious that to draw the pattern for a cone, we require to know the slant of the cone (which will be the radius for the pattern), and the circumference of the base of the cone. PEOBLEM III. To draio the -pattern for a cone, in one piece or in several pieces^ the slant and diameter of the base being given. Pattern in one Piece. — With O A (Fig. 66) equal to the slant as radius, describe a lung arc ACE. What has now Fig to be done is to mark oflf a length of this arc equal to the cir- cumference of the base of the cone. The best and quickest way for this is as follows. Draw a line F B (Fig. 6a) equal to the given diameter of the base, and bisecl; it in G ; then G B is a radius of the base. From G draw G E perpendicular 30 T1[E TIX.'^MITIIS' PATTEKN MANlAl. to F B ; and with G as centre and radius G B describe from B an arc meeting GE in E. The arc B E is a quadrant (quarter) of the circumference of the "base of the cone. Divide this quadrant into a number of equal parts, not too Fia. Ga. many, Bay four, by points 1, 2, 3. From A (Fig. 66) mark off along arc ACE four parts, each equal to one of the divisions of the quadrant, as from A to B. Take this length A B equal to the four parts, that is, equal to the quadrant, and from B set it off three times along the arc towards E as from B to C, C to D, D to E. Join E to O; then A CEO will be the pattern required. Note. It must be noted that when tnis pattern is bent round to form the cone, the edges O A and O E will simply butt up against each other, for no allowance has been made for lap or seam. Let us call the junction of O A and O E the line of butting. Nor, further, has any allowance been made for wiring of the edge ACE. These most essential matters will be referred to immediately. Pattern in mobe than one Piece.-— If B be joined to 0, then the sector O A B will be the pattern for one-qimrter of the cone. If C be joined to O, then the sector O A G is the pattern for one-half of it. Similarly A D will givg THE TENSMrXHS' PATTERN MANUAL 31 three- quarters of the cone. A cone pattern can thus be made in one, two, tkree, or four pieces. If the cone is required to be made in three pieces, then instead of dividing, as above, a quadrant of the circumference of the base, divide one-third of it into parts,. say five ; set off five of the parts along ACE from A, and join the last division point to the centre ; the sector so obtained will be the pattern for Gne-tMrdof the cone. If required to be made in five pieces, divide a fifth of the circumferenea of the base into equal parts, and proceed as before. Similarly for any number of pieces that the pattern may be required in. PEOBLEM IV. To draw the jpatiern for a cone, the height and the diameter of the base heing given. First find the slant. OB (Fig% 7a) by P.eoblem IL Then with A as centre and radius A B, describe B a quadrant of Fig. 7a.. FiG. 76; ^ Tiaduis afHasoQ O'O^ the circumference of the base, and proceed, as in Problem III. to draw the pattern Fig. 76 (the plain lines). » 32 THE TINSMITHS' PATTERN MANUAL. ALLOWANCE FOR LAP, SEAM, WIETNG, &o. (10.) It has already been stated that the geometric pattera Fig. 6& has no allowance for seam, wiring, or edging. (Foi the present it is assumed that these terms are understood i, we shall come back to them later .on.) In the pattern Fig. 76 the dotted line 0' B' parallel to the edge B shows ' lap ' for soldered searu. For a ' grooved ' seam not only must there be this allowance, but there must be a similar allowance along the edge O D. These allowances, it must be distinctly remembered, are always extras to the geometric pattern; that is to say, the junction line of D and B, or line of butting (see Note, Problem III ) is not interfered with. And here a word of warning is necessary. Suppose instead of marking off a parallel slip or lai3 for soldered seaHi, a slip D D' going off to nothing at the centre 0, is marked off, and that then, for soldering up, there is actually used not this triangular slip, but a parallel one as D D' O 0^, the result brought about will be that the work will solder up untrue; there will be, in fact, a * rise ' at the base of the work. We can understand the result in this way. If the parallel slip D D' 0^ used for soldering were cut off, there would remain a pattern which is not the geometric pattern, but a nondescript approximation, having a line of butting other than the true line. And it being thus to an untrue pattern that the parallel slip for seam is added, the article made up from the untrue pattern must of coUrse itself necessarily be untrue. In the fig. the dotted line parallel to the curve of tho pattern shows an allowance for wiring. For a grooved seam there must be on the edge O D an addition O D D' O^ similar to the addition on the edge O B, as above stated. (11.) In working from shop patterns for funnels, oil-bottle tops, and similar articles, workmen often find that if they take a good lap at the bottom, and almost nothing at the top of the seam, the pattern is ime. And so it is, for these patterns have the triangular slip D O D' added. Whereas, if a paralld piece D O^ O D' is uzod fox lap, tbo pattoni ig THE TINSMITHS PATTERN MAXIMAL. 33 untrue. Which again is the cas3, because, now, in addition to D D', an extra triangular piece D O^ is used, and this extra is taken off the geometric paitern. Consequently, the line of butting is interfered with; that is to say, the two lines OB and D, instead of meeting, overlap; OB forming a junction, with O^ IT instead of with D ; with which U B must always form a junction, for the pattern to he true. In setting out patterns, to. prevent error, the best rule to follow and adopt is, to first mark them out independent of any allowance for seams, or wiring, or edging, and to afterwards add on whatever allowances are intended or requisite. In future diagrams, allowances, where sbown, will be mostly shown by dotted lines. Definitiox. (12.) ri:usTUM. — If a right cone is cut by a plane parallel to Fig. 8a. Fifi. 86. ,'- £ M J} RauUus cf Base of un.:zci Cona its base, the part containing the apes is a. complete cone, as QCGDL (Fig. 8a), and the part CABD containing tho 34 THE TINSMITHS' P ATT I UN MA'NUAL. base A H B K is a frustum of the cone. In otliei' words a frustum of a right cone is a solid having cii-cnlar ends, and of equal taper or inclination of slant eveiywhere hetween the ends. Conversely a round equally tapering body having top and base parallel is a frustum of a right cnne. (13.) Comparing such a solid with round ai tides of equal ta])er or inclination of slant, as pails, coffee-pots, gravy I Figs. 9. straiuirs, and so on (Fig. 9), it will be seen that they are portions (frusta) of right cones. (14.) In speaking bere of metal- plate articles as portions of cones, it must be remembered that all our patterns are of surfaces, seeing that we are dealing with metals in sheet ; and that those patterns when formed up are not solids, but merely simulate solids. It is, however, a con- venience, and leads to no confusion to entirely disregard the distinction ; the method of expression referred to is therefore adopted throughout these pages. ( 15.) By Fig. 86 is shown the relations of the cone O A B of Fig. 8a with its portions O C D (complete cone cut oflp), and A B D (frustum). The portion O D is a complete cone, US it is the solid that would be formed by the revolution of the right-angled triangle OFD (both figs.) around OF. The triangles OFG and OFO (Fig. 8a) represent the THE TINSMITHS' PATTEHN MANIA K. 35 trian, le F D in progress of revolution. The triangle E B (both figs.) is the triangle of revolution of the uncut cone O AB (Fig. 8a) and O E H, O E A represent E B in pro- gress of revolution. The height of the cone OAB being OE (both figs.), the height of the cone O C D is OF (both figs.). The radius for the construction of pattern of the uncut cone OAB will be O B (both figs.) , for the pattern ef CD, the cone cut off, the radius will be O D (both figs.). In F E, or D M, we have the height of the frustum. Just as (§ 8) the portion of the rolling cone touching the plane at any instant is a slant of the cone, so the slant of a frustum is that portion of it, which, if it were set rolling on a plane, would at any instant touch the plane. D B is a slant of the frustum C A B D, The extremities of a slant of a frustum are * corresponding points.' Other details of coBa and frustum are shown in Fig. 8&. Fig. 10. (16.) It is obvious that, if the patterns for the cones A B, O G D (Fig. 8a) be drawn (Fig. 10) from a common centre O, the figure A G D B will be the pattern for the frustum D 2 36 THE TINSMITHS' PATTERN MANUAL. A C D B (Fig. 8a). From wbich we see that in order td draw the pattern for the frnstnm of a cone, we mnst know the slant of the con(9 of which the frustum is a portion, that is, we must know the radius for the construction of the pattern of that cone, and also the slant (radius for pattern) of the cone cut off. PROBLEM V. ^ Given the dimensions of the ends of a round equal-tapering body (frustum of right cone), and its upright height. To find the slant, or the height, of the cone of which it is a portion. Draw any two lines A, A B (Fig. 11) at right angles to each other and intersecting in A^ From A on either line, Fig. 11. C,^< of Larcje cruJL Bay on B A, mark off A B equal to half the diameter of the larger of the given ends, and from A on the other line make A C eq.ual to the given upright height. Draw a line C D tUE TINSMITHS' PATTERN MANUAL. 37 parallel to A B, or, which is the same thing, at right angles to A O, and make G D equal to half the diameter of the smaller endi Join B D, and produce it, meeting A in O. Then O A is the height of the cone of which the tapering body is a portion, and O B the slant. PEOBLEM VL To draw the pattern for a frustum of a cone, the diameters of the ends of the frustum and its upright height being given. The Frustum. — Draw any two lines A, B A (Fig. 12a) perpendicular to each other and meeting in A ; on one of the FiQ. 12a. jl Mall Longer Diajn£ter perpendiculars, eay B A, make A B equal to half the longer diameter (radius), and on the other make A C equal to the given upright height. Dfkw a line C D perpendicular to A O and make C D equal to half the shorter diameter. Join B D, and produce it, meeting A produced in 0. With A THE TINSMITHS' PATTERN MANUAL. as centre, and radius A B, describe quadrant B E, . which divide into any convenient number of equal parts, here fonr. To draw tho pattern (Fig. 12o) tako any point 0' as .-^ THE TINSM1T11S5' PATTERN MANUAL, 39' centre, and with radius OB (Fig. 12a) describe an arc ACE; also with same centre and radius, OD (Fig. 12a), describe an arc A' (J' E'. 'From any point in the outside curve, as A, draw a line, tlirough O', and cutting the inner arc in A'. From A mark off successively jjarts equal to those into which the qiiadrant B E (Fig. 12a) is divided, and the same number of ^them, four, to B. And- from B, along the outer curve, set off B C, CD, D E, each equal to A B. Join EO', cutting the inner curve in E'. Then A A' E' E is the pattern required. Just as O B (Fig. 12a) is the slant of the cone that would be generated by the revolution of right-angled triangle OAB around OA, so DB is the slant of the frustum of which A A' E'E (Fig. 12&) is the pattern. In the pattern the slant D B appears as, A A', B B', G C, &c. Parts' of the Frustum — If B be joined to 0', the figure A A' B' B will be one-quarter of the pattern of the frustum ; and if C be joined to 0', the figure A A' C 'C will be pattern for one-half of it, and so on. The paragraph *' Pattern in more than one Piece " in Problem III. should be re-read in connection with' this " Parts of a Frustum." (17.) The problem next following is important, in thdt, in actual practice, the slant of a round equal-tapering botly is very often given instead of its height, especially' in cases where the taper or inclination of the" slant is great ; as for instance in ceiling-shades. The only difference in the work- ing out of the problem from that of Problem VI. is that the radii required for the pattern of the bdJy are found from other data. Let us take the problem. PROBLEM VIL To draw the pattern for a round, equal-tapering body (frustum o/ right cone), (lie diameter of the ends and the slant being given. To find the required radii, draw any two lines O A, B A (Fig. 13) perpondictilar to one another, and meeting- in A 40 1:iiE TINSMMHS' PATTERN MANUAL. On either line, as A B, make A B equal to half the longer of the given diameters and A C equal to hajf the shorter. From C draw GD perjrundicular to A B. With B as centre and^ Half Shorter A^Biaineler C \B I ; W Ealf Longer Diameter > the given slant as radius, describe an arc cutting C I) in E. Join B E and produce it to meet' A O in O. Then B and O E are the required radii. By E F being drawn parallel to •A B, comparison may be made between this Fig. and Fig. 12a, and the difference between Problems VI. and VJI. clearly apprehended. -To draw the pattern, proceed as in Pbobi,km VI. (18.) For large work and for round equal-tapering bodies which approximate to round bodies without any taper at all, the method of Problem VI. i>< often not available, for want of space to use the long radii that are necessary for the curves of the patterns. The next problem shows how to deal with such cases ; by it a working-centre and long radri can be dispensed, with. The method gives_fairly. good results. THE TlNSMItHS' PATTERN MANUAL. 41 PilOBLEM VIIL To draio, Without long radii, the pattern for a round equal- tapering body [frustum of rigid cone), the diameters of the ends and the upright height being given. First draw one-quarter of the plan. (To do this, we fore- stall for convenience what is taught in the following chapter.) Draw any two lines B 0, C (Fig. 14) perpendicular to each other and meeting in O. With O as centre and radius equal to half the longer diameter, describe an arc meeting the lines B 0, C in B and C. ^With as centre and radius equal to 42 THE TIN>!MITUS' PATTERN MANUAL, "Jialf tlie shorter diameter describe an arc B' C. This completes the one-quarte*r plan. Now divide B C, the largest arc, into any nnmLer of equal parts, say four ; and join the points of division to hy lines cutting B'C in 1', 2', 3'. Join 3'G, and ihroxagh 3' draw 3'E perpendiciilar to 3' C, and equal to the given npright height. Join C E ; then C E may bo taken as the true ieiigth of 3'. Through C' draw C D perpendicular to C and equal to the upright height. Join C D ; then C D is the true length of C C. If it is inconvenient to find these iruo lengths on the plan, it may be done apart from it, as by the triangle~B P and Q. • " To set out the pattern. Draw (Fig. 15) any line C C equal to C D(rig. 14). With C and C as centres and radii respectively CE and 3 (Fig. 14) describe arcs intersecting in 3 (Fig. 15). With and C' as centres and radii respec- tively CE and C'3' (Fig. 14) dcsc-ribe arcs intersecting in 3' (Fig. 15). Then and 3 ato two points in the outer curve of tlio pattern, and C 3' two points in the inner cuive. To find points 2 and 2', proceed as just explained, and with the same radii^ but 3' and 3 as centres instead of C and 0. Similarly, to find points 1' and 1, and B' and B. A curved line drawn from O through 3, 2, and 1 to B will be the outer jQurve of_oiie.- quarter of the rc(j[uired pa^ttern, and a curved THE TINSMITHS' PATTERN MANUAL. 43 line from C'througli 3', 2', and 1' to B' its inner curve; that is C C B' B is one-quarter of tlio pattern. Four times the quarter is of course the required pattern complete. Note. — In cases where this method wiU be most useful, the pattern is generally required so tl at the article can be made in two, three, four, or more pieces. If the pattern is required in three pieces, one-third of tlie plan must be drawn (see end of Problem III., p. 31) instead of a quarter, as in Fig. 14; tlie remainder of the construction will then be as described above. (19.) It is often desirable in the case of large work to know what the slant or height, whichever is not given, of a round equal-tapering body (frustum of right cone) will be, before starting or making the article. Here the following- problems will be of service. PROBLEM IX. •To find the slant of a round equal-tapering body (^frustum of right cone), the diameters of the ends and the height being given. Mark off (Fig. 16) from a point m any line O B the lengths of half the shorter and longer diameters, as O C, O B. From C draw C D perpendicular to B. Make C D equal to the given height, and join B D. Then B D is the slant required. 44 Tlir. 'ilNSAIlTll.S' PAT'IKRN MANUAL. PKOBLEM X. To find tJie height of a round equal-tapering body (frustum of right cone), the diameters of the ends and the slant being given. Mark off (Fig. 17) from a point in any line B tlie lengths of half the shorter and longer diameters, as in Puoblem IX., and from C draw C D perpendicular to O B. With B as centre and radius equal to the given slant, describe an arc cutting C D in E. Then C E is the height required. Essentially this problem has already been given, in the working of Problem VII. ^IG 17. D ^ ^ Fig la B \ B JL PEOBLEM XI. Given the slant and the inclination of the slant of a round equal- tapering body ; to find its height. Let A B (Fig. 18) be the slant, and the angle that A B makes with C A the inclination of the slant. From B let fall B D perpendicular to A 0. Then B D is the height required. (20.) In the workshop, the inclination of the slant of a tapering body is sometimes spoken of as the body being so many inches " out of flue." This will be explained in the following chapter. If the inclination of the plant THE TINSMFTHS' PATTERN MANUAL. 45 is given in these terms the problem is worked thus. From any point D in any .lin& C A (Fig. 18) make D A equal in length to the number of inches the body is " out of flue," and draw D E perpendicular to A. With A as centre and radius equal to the given slant, describe an arc intersecting D E in B, Thou B D will be the height required. 48 THE TINSMITHS' PATTERN M VNl'AU CHAPTEK Y .Equal-tapeeing Bodies of which To? and Base aee Parallel, and their Plans. (21.) First let us understand what a plan is. Fi^;. 19 jrepresents an object Z, made of tin, say, having six faces, Fia. Id. of which the A B C D and G H K L faces are parallel, as also the B D K n and C A L G. The ABCD and CDKL faces are square. The ABCD face has, soldered flat on it centrally, a smaller square of tin ahcd with a central THE TIN.^MITHS' PATTERN MANt'U.. 47 circular holo in it. Now suppose wires (rc'presented in the fig. by dotted lines), soldered perpendicnlarly to the A B C D face, at A, B, C, D, a, 6, c, d, E, and F (the points E and -'F are points at the extremities of a diameter of the circular hole). Also suppose wires soldered at G and H parallel to the other wires, and that the free ends of all the wii-es are cut to such length that they will, each of them, butt up against a flat surface (plane), of glass say, X X X X, parallel to the A B C D face. Lastly suppose that all tlie points wheie the wfres touch the glass are joined by lines corresponding to edges of Z (see the straight lines in tho Fig. 20. ^gure on the plane) ; also that E' and F' are joined, that the line juiuing them is bisected, and a circle described pa.-sing through B' and F'. Then the complete representa- tion obtained is a projection of Z. Instead of actually pro- jecting tho points by wires, we may make tho doing of it another supjiosition luay, find, as if by wires, the required points, and diaw the projection. Tho ABCD face being, 48 THE TINSMITHS' PATTERN MANUAL. eay 2 inclies siuare, the flat piece, say 1 inch squa^'e, and the hole ^ inch diameter, and the back face GHK L, say 2|- in. by 2 in., then the projection that is tipon the glass ■would be as shown in Fig. 20. The plane X XXX is here supposed vertical, and the projection G' C D' H' is therefore an elevation ; if the plane "were horizontal, the projection would be a plan, and we might regard A B C D as the top of the body, and G H K Las its base, or vice versa. We may define a plan then as the representation of a body obtained by projecting it on to a horizontal plane, by lines perpen- dicular to the plane. (22.) The plane X X X X was supposed parallel to the A B C D face of Z ; the plan A' B' C D' of it is therefore of the same shape as A B C D, and in fact A B C D may be said to be its own plan. Similarly the G' H' D' C is the plan of the back-face G H K L and is of the same ^ape as that face. But the plan of the face A G 11 B to which the plane is not parallel is by no means the same shape as that face, for the long edges B H and A G of the face A G H B are, in plan, the short lines B' H' and A' G'. We need not, however, go farther into this, because in the case of the bodies that now concern us, the hoiizontal plane on which any plan is drawn is always supposed to be parallel to the principal faces of the body, so that the plans of those faces are always of the same shape, as the faces. In this paragraph the plane X X X X is supposed to be horizontal. (22a.) We are now in a position to explaia the getting at the true length of C C in the fig. of Problem VI IL, p. 42 ; or, putting the matter generally, to explain the finding the triia lengths of lines from their apparent lengths in their plans and elevations. Horizontal lines being esceptedj there is, manifestly, for any line, however positioned in space, 'a vertical plane in which its elevation will appear as (if not a point) a vertical line. Let B E (Fig. 17, p. 44) be any line in the plane of the paper, and let C D be the vertical plane seen edgeways on which the elevation E C of B E is a vortical line. Then if B be a horizontal plane seen edgeways'' THE TINSMITUS' PATTERN MANUAL. 49 pa.'-sing through C, the line joining the B extremity of B E to the C extremity of its elevation will be the plan of B E. We get thus the figure ECB, a figure in one plane, tlio plane of the paper, a right-angled triuntile in fact, of which the E C side is the elevation of B E, the C B side its plan, and the hypotenuse the line itself; a figure, which, as combining a line, its plan, and its elevation, we have under no other conditions than when t'le elevation in question is a vertical line. In the plane passing through E and B E, that is, in the plane in which these lines wholly lie, we have in the line that we get by joining C with B the plan, full length, of BE. In 1 espect of tliis plan of B E, we are cone erned with no other measurement, because, in a right-angled triangle representing a line and its plan and elevation, no other measurement of the plan lino can come in. Not so, however, with the elevation line of B E. Here other mea- surement of it than its length can and does come in, because th.it longth varies according to the position of the vertical plane with regard to it ; the plan length is always the same, But to have in the three side^ of E C B, the representation of B E, and its plan and elevation, it is evident that the plane which contains B E and its plan C B must also wholly contain the elevation E ; that is, the plane must be perpen- dicular to the piano of the triangle. Now, no matter on what vertical plane the line B E is projected, although the length of the projection will vary, the vertical distance between its extremities, that is, its height, never varies. Hence, if, in any right-angled triangle, we have in the hypotenuse the representation of a line, in one of its sides the plan of the line, and in the other side, not necessarily the elevation that comes out vertical, but the height of any elevation of the line, it comes to the same thing as if in the latter side we had the actual elevation that is veitical. And hence, further, if we have given the plan-length of an unknown line, and the vertical distance between its extre- mities, we can, by drawing a line, say C B, equal to the given plan-length, then drawing from one of its estremities and at 50 THE TINSMITHS' PATTERN MANUAL. right angles to it, a Ime, say CE, equal to the given vertical distance, and finally joining the free extremities, as by BE, of these two lines, construct a right-angled triangle, the hypotenuse, BE, of which must be the true length of the unknown line ; for there is no other line than B E of which C B and C E can be, at one and the same time, plan and elevation. V,'e have explained this true-length matter fully, because we have to make use of it abundantly in problems to come. (23.) Proceeuirig to the bodies wo have- to con.-ider, wo Fig. 21. take first a frustum of a cone, Fig. 21a. To draw its plan, lot us suppose the estremities of a diameter of its smaller face top (namely points A and F of the skeleton drawing Fig. 216) (neither drav/ing is to dimensions), to be projected, in the way just explained, on to a plane parallel to the face, then, also as there explained, we can dra%v the circle which is a projection of that face. Suppose the smaller circle of Fig. 22 to be that circle, and to be to dimensions. Projecting now, similarly, the extremiiies of a diameter of the larger face (base), namely the points G and J) of the skeleton drawing, on to the same plane, we can get the projection of the larger face. Ltjt the larger circle of Fig. 22 be that projection. The two circular projections will be concentriG (having tke The TINSMITHS* PATTHUN MANUAL. 51 ep.me centre) because the body Z is of equal taper, and tbey will, together, be the plan of Z, that u Fig. 22 is that plan. A C and F D each show the slant, and B A and E F thn height. B C and D E each show the distances between the plans of corresponding points. Fig. 22. (24.) Turn to the skeleton drawing of Z. Here A C shovra a slant of the frustum (§ 15), A B its height (see J) M, Fig. 8&), and A and are 'corresponding points' (§ 15). Looking at C D E B as at the plan of the firustum, we have, in the point B, the plan of the point A. Joining B C, we get a right-angled triangle ABC; the slant A C is its hypotenuse, the height A B k one of the sides containing the right angle, and the other side ccataining the right angle, B C, is the distance between the plans of the corresponding points A and C, as also between plans of corresponding points of Z any- where. I'his distance is that of how much the body ie * out of flue ' (a workshop expression that was referred to at tho end of the previous chapter), in other words, how much A C is out of parallel with A B. What points, in the plan of a frustum, are the plans of corresponding points is showu 52 THE TIMSMITHS' PATTERN MANUAL. by the fig., as the line joining the plans of correspondlug l)oints (the line joining B and C or that joining D and E, for instance) will always, if produced, pass through the centre of the circles that constitute the plan of the frustum ; the centre of the circles being the plan of the apex of the cone of which the frustum is a part. Which leads us to this ; that the distance, actually, between the plans of corresponding points in the plan of a frustum is equal to half the difference of the diameters of its two circles ; for, the difference between E B and D is the sum of D E and BC, and DE and BC are equal j in other words, either D B or B C is half the difference betv/een E B and D G. Pig. 2Rrt. (25.J Let us now consider another equal-tajiering body v/hich has top and base parallel, and we will suppose it to have Bat parallel sides, flat ends, and round (quadrarit) corners. Such a body is ropresented, except as to dimen- Bions, in Z, Figs. 23a and 6; Fig. 236 being a skeleton drawing of the body represented in Fig. 23a. Extending our definition of ' slant ' to apply to such a body, a * slant ' becomes the shortest line thut can be drawn anywhere on the slanting surface ; and ' corresponding points ' become, in accordance, the extreme points of such line. Either of the lines F A, G B, E C, H L, or MO represent a slant of the body, and F and A are corresponding points ; as also are G and B, E and 0, H and L, and M and 0, The height THE TINSMITHS' PATTERN MANUAL. 53 of the body is represented hj eitlier of the lines F A', G B', EC, HL', or MO'. The plane €or the plan being parallel to the M P Q E face (here the top) of Z, the plan of that face is of the same shape as the face« The round-cornered rect- angle A' F' a' D' B' C of Fig. 29 is the plan to dimensions. For the same reason the plan of the OATS face (here the base) is of the same shape as that face. The round-cornered rectangle AF GD B G of Fig. 29 is the plan t<) dimensions. How actually to draw these plans we shall deal with presently as a probL-m. The two circles constituting the plan of the frustum were concentric, that is, symmetrically disposed with respet;t to one another, because the frustum Fig. 2.36. was an equal-tapering body ; and the plans of top and base of the body we are now dealing with are symmetrical to each other for the same reason. The two plans (Fig. 29) together are the plan "of the body Z. (26.) Looking at A B G D A' B' C' D' of the skeleton draw- ing (Fig. 236) as at the plan of Z,- we have, just as with the cone frustum, in the point A' the plan of F, in the point B' the plan of G, in the point C' the plan of E, in the point L' the plan of H, and in the point 0' the plan of M. Further as in the ease of the frustum, if we join any point in the plan of the base, as A, to the plan of its corresponding point A', then we have a right-angled triangle, F A A', of which the 5-f THE TINSMITHS' PATTERN MANUA!,. hypotennse F A represents the slant of the 'bofly; F A', onA of tliG sides containing the right angle, its height, and A' A, the other side containing the right angle, the distance between the plans of the corresponding points F and A, which is also the distance between B and B', C and C', L and L', O and O', and between plans of corresponding points of the body anywhere, the body being of equal taper. As with Fig. 216 what points, in the plan, are the pLxris of corresponding points is clear froia the fig. Wh re the plan of the body consists of straight lines, the plans of corresponding points are always the extremities of lines joining these straight lines perpen- dicularly; the extremities of AA', B B', C C, and L L', for instance. "Where the plan of the body consists of arcs, the plans of corresponding points (compare with cone frustum) are the extremities of lines joining the arcs, and which, pro- duced, will pass through the centre from which the arcs are described ; the line 0' for instance. To mate all this quite plain, leference shotsld again be made to Fig. 29 ; also to Fig. 28, which is the plan of an eq"al-taperingobody with top and base parallel, end having flat sides, and semicircular ends. In Fig. 29, A A', B B', C C, D D', are lines. joining the plan lines of the flat sides and ends perpendicularly, and the extremities of each of these lines are plans of correspond- ing points, that is to say, A and A' are plans of corresponding points, as are also B and B', C and C, and D and D'. Also F and F' a'^e plans of corresponding points, being the extremities of the line F F' which is a line joining ihe ends of the arcs which are the plans of one 'of the quadrant corners of the body. Siniilarly G and Q' are plans of corre- sponding points. In I'ig. 28, F F', G G', D D', E E', are lines joining perpendicularly the plan lines of the flat sides of the body at their extremities where the semicircular ends begin ; and F and F', G and G', D and D', E and E' are plans of corresponding points. ALso A and A' are plans of corre- sponding points, and B and B', seeing that the lines joining these points, produced, pass respectively through and 0', the centres from which the semicircular ends are described. THE TINSMITHS' PATTERN MANUAL? 55 In the cone frustum, the actual distance between the plans of corresponding points was, we saw, ecjual to half the difference of the diameters of the' two circles constituting its plan. Similarly with the body Z of Fig. 23a, ani indeed with any equal- tapering body of whith the top and base are parallel, if we have the lengths of the top and baso given, or their widths, the distance between the plans of oorresponding points (number of inches ' out of flue ') is always equal to half the difference between. the given lengths or widths. Thus, tiie distance between the plans of corresponding points of Z is equal to half the difference between A B and A' B' (Fig. 29) or between CD and CD'; and the distance-be- tween the plans of corresponding points of the body of which Fig. 28 is tlie plan, is equal to half the difference between A B and A' B' of that fig., or between F D and F' D'. Summarising we have a. In the plans of equal-tapering bodies which have their tops and bases parallel, there' is, all round, an equal distance bstween the plans of corresponding points of the tops and bases. 6. Conversely. — If, in the plan of a tapering boly with top and base parallel, there is an equal distance all lound between the plans of corresponding points of the top and base, then the tapering body is an eqiial-taporing body, that is, has an eqnal inclination of slant all round. c. The plan of a round equal-tapering body having top and base parallel, consists of two concentric circles. The plan of a portion of a round equal-tapering body having top and base parallel, consists of tv/o ares having the same centre. T'ao corncra of tha body Z (Fig. 23a) are portions (quarters) of a round equal-iapering body ; tlitir jilans are arcs (quadranta) of circles having tho sanie cuntre. d. Conversely. — If the plan of a tapering body having top and base parallel, consists of two concentric circles, then the body is a frustum of a right cone. Also if the plan of a lapsring body having top and base parallel, consists of two 56 THE TINSMITHS' PATTERN MANUAL. Fig. 21. arcs halving the same centre, then tlie "body is a portion of a frustum of a right cone. The plan of each enti of tho tapering body represented in plan in Fig. 28 consists of two ai-cs (semicircles) having the same centre; the ends are portions (halves) of a frustum of a right cone. The plan of each corner of the tapering body Z (Fig. 23a) consists of two arcs (quadrants. Fig. 29) having the same centre ; the coiners are portions (quarter.^) of a frustunr of a right cone. The fig. annexed represents a quadrant corner in plan separately. "We conclude the chapter with some problems. PROBLEM XII. Given the height and slant of an equal-tapering body iidth top and base parallel; to find the distance between the plans of cdrrei'ponding points of the top and base {number of inches ' out of fine'). Let A' (Fig. 25) te the given height. DraTv A'B per- pendicular to A' C ; with C as centre and the given slant as radius, describe an arc cutting B A' in A. Then A A' is the distance required. ^tHE TIN.SMITH^' PATTTiRN MANUAL. 57 PEOBLEM XIII. Given the heigM of an equal-tapering body wifJi top and base parallel, and the inclination of slant (number of inches ' out of flue ') ; to find the distance hetiveen the plans of corresponding poirds of the top and base. Let G A' (Fig. 26) bo the given L&ight. Thrcragli A' draw a lino A' B perpeiidionlar to A' ; from any point, D, in A' B draw a lime D E makiug Virith A' B -an angle equal to that of the given inclination. From C draw CJ A parallel to E I> and cutting A' B in A ; then A A' is the distance required. Fia. 2G. PEOBLEM XIV. To drcLvo the plan of a round equal-tapenng body with top and base larallel [frustum of rigid cone), the diameter of either end oeing given and the height and slard. Case I. — Given the height and slant and the diameter of the smaller end. On any line O B (Fig. 17) set o& O C equal to half the given diameter, and from C draw D perpendicnlar to B. 58 THfc] TlNSMiTH.S' PATTERN MANUAL. Mark off C E equal to th.Q givon heigLt, arid with E as csnir^ and radius equal to the given slant, describe &n arc inter- seoting OB in B; then CBwill he the distance in plan between corresponding points anywhere in the frustum ; that is to say (by c, p, 55) C will be the radius for the plan of the smaller end of the frustum, and O B the radius for the plan of the larger end. Case IL — Given the height and slant and the diameter of the larger end. On any line OB (Fig. 27), set off O B equal to half the given diameter, and now work fn>m B towards instead ol from towards B ; thus. From B draw B perpendicialar to O B. Mark off B D equal to the given height, and with D as centre and radius equal to the given giant, describe an arc intersecting O B in E ; then B E will be the distance in Pig. 27. plan of corresponding points anywhere in the frustum ; that is to say (by c, p. 55) O B will be the radius for the plan of the larger end of the frustum, and O E the radius for the plan of the smaller end. THE TINSMITHS' PATTERN MANUAL. 59 PROBLEM XV. To draio the plin of a round cqual-tapering hodi) with top andhase parallel {frustum of right cone), the diameter of either e id being given, and the 'number of inches ' out of flue ' (distance between plans of corresponding points). Case I. — Given the number of inches * out of 3ue,' and the diameter of the smaller end. The radius for the smaller circle of the plan wi^l ho half the given diameter; the radius for the larger circle of the plan will be this half diameter with the addition of the number of inches ' out of flue.' Case II. — -Given the number of inches ' out o*' flue/ and the diameter of the larger end. The radius for the larger circle of the plan will be half the given diameter; the radius for the smaller circle of the plan will be the half diameter less the number of inches 'out of flue.' (27.) It should be noted that v/ith the dimensions given in this problem, we can draw plan only, we could not draw a pattern. To do that we must also have height given, for a plan of small height and considerable incliiiation of slant is also the plan of an infinite number of other frustfi (plural of frustum) of all sorts of heights and inclinationB of sLint. PROBLEM XYL To ch'dvj the plan of an oblong eqnal-tavefing body with top and base parallel, C7id having flat (plans) sides and semicircular cT.ch. Gags I. — Y/here the length and width of the top are given, and the length of the bottom. Commencing with the plan of the top, we know from § 25 that it will be of the same shape as the top ; we have there- 60 THE TINSMITHS' PATTERN MANUAL. f'jre to draw tliat sliape. On any line A B (Fig. 28) mark off A B equal toifco given length of the top, Frora A set cff AO5 and from B set off B'O' eacli equal to half the given width cf the top. Through O and 0' dra^ lines perpsa- dicular to A B ; and -with and O' as centres and O A or O B as radius describa a^cs meeting the perpendiculars in BF and EG. As DF and EG pass through the centres O aad 0' respectively they are diameters, and the arcs are Eornioircles I these diameters, moreover, are each equal to the given width. Join D E, F G, and the plan of the top is complete. D Fig. 28. E / ' / \ A' ty 61 \ a" 1 V 1 F Tlje plan of the base will be of the same shape as the base, and we will suppose it smaller than the top. What we have then to do is to draw a figure cf the same shape as the base, and to so place it in position with the plan of the top that we shall have a complete plan of the body we are dealing with. By a, p. 55, v/e know that the distances between the plans of corresponding po'cts of the top and base all round the f«il plan will bs equal. We have therefore first to ascertain the distance between the plan*? of any two corresponding points. This by § 26 will in the present instance be equal to half the difference between the given THE TINSMITHS' PATTERN MANUAL. Ql lexig'c\i3 cf the top and "base. Set off this half-diSereaca, as tha tase is smaller tlian the top, from A to A'. Then with and 0' as centresj and A' as radiua, describe the semi- circles D' A' F', E'B'G', Joia D' E', r G', axid w© have the' required plan of the hody. Case II. — Where the length and width of the top are given, and the height, and slant, or the height and the inclination of the slant (number of inches 'otit of flue'). First draw the plan of the top as in Cafce L Then if the height and .^lant are given, find by Problem XII. the distance batween the plans of corresponding . points. If the height and inclination of slant are given, find the distance" hj Problem XIII. If the inclination of tlio slant is givsn in the form of ' out of flue,' the number of inches ' out of fine ' is the required distance. Set off this distance from A to A' in the fig. of Case I., and complete the plan as ia Case I. Cass Hi.-— Where the length and width of the base (bottom) are given, and the height and slant, or the height and the inclination of the slant. On any line A B (Fig, 28) mark off A' B' equal to the given Isigth of the bottom. From A' set off A' and from B' set off B' 0' each equal to half the given width of the bottom. Through O and 0' draw indefinite lines D F, E G perpendicular to A B ; and v/ith and 0' as centres, and O A' as radius describe the semicircles F' A' D', G' B' E', join D' E', F' G*5 and we have the plan of the bottom. Xow by Problera XII. or Problem XIIL, as may be required, find the distance between the plans of corresponding points, or take the number of inches ' out of flue,' if this is what is given. Set off this distance from A' to A. With and 0' as centres and O A as radius describe semicircles meeting the perpendiculars through and O' in D and F and in E and G. Join D E, F Gf and the plan of the body is completed. 62 THE TINSMITHS' PATTERN MANUAL. PKOBLEM XVII. To draw tJie plan of an oblong equal-tapering body toiih top and base parallel, and having fiat sidett, fiat ends, and round (quadrant) corners. Case I.— Wtere tlie length and widtli of top and bottom (baso_) aro given. Draw any two lines A B, C D CFig, 29) perpendicular to each other and intersecting in 0. Make O A and B < ach eu ual to half the length of the top, which we will suppose larger than the 1 ottom, and A' and B' each equal to half the length of the bottom. Also make and D each equal to half the width of the top, and C and D' each equal to half the width of the bottom. Throxigh C, D, C, and D' draw lines parallel to A B, and through A, B, A', and B' draw lines parallel to C D and intei secting the lines parallel to A B. We have now two rectangles or oblongs, and we require to drav7 th round comers, which are quarters of circles. THE TINSMITHS' PATTERN MANUAL 63 From tlie intersecting point E along ihe sides of the rectangle mark equal distances E F and E G, according to the size of quadrant corners required. With F and G as centres and E F or E G as radius, describe arcs intersecting in O' ; and with 0' as centre and same radius, desciibe the arc F G, which will be a quadrant because if ihh points P and G be joined to O' the angle F 0' G will be a right angle (p. 21). Draw F F parallel to A B and G G' parallel to C B, and with O' PS centre and radius O' F' describe the arc F' G', which ako will be a quadrant. We have now the plan of one of tlio quadrant corners; the other corners can be drav/n in like manner. (27a.) It is important to notice that the larger corner deter- mines the smaller one. In practice it is therefore often bost to draw the smaller corner first, otherwise it may sometimes be found, after having drawn the larger corner, that it is not possible to draw the smaller curve sufiSciently large, if at all. To draw the smaller corner first, mark off from the intersect- ing point E' equal lengths E' F', E' G', according to the size determined on for the corner. With F' and G' as centres and E' F' or E' G' as radius describe arcs intersecting in O'. Then 0' will be the centre for the smaller corner. It will clsa be the centre for the larger corner, which may be described in simi'ar manner to the smaller corner in tho preceding paragraph. Case IL— Where the dimenoioiis of the top are given and tho height and slant, or the height and the inclination of ; the slant. Draw the plan of the top, A D B 0. Find the diota,nca oetween the plans of corresponding points of the top and bace by Problem XII., or Problem XIII., according to what is given ; and set off this distance, as the base is smaller than the top, from A and B inwardo towards O on the line A B, and from D and C inv/ards tov/ards O on the line D C. Complete tho plan by the aid of v/hat has already bean. (Explained. 64 THE TINSMITHS' PATTERN MANUAL. Case III. — Where tlie dimensions of tlie bottom are given, and tlie length and slant, or the height and the incli- nation of the slant. Braw the plan of the hottom, A' D' B' C'jfind the distance between the plans of corresponding points of top and bottom, set this off outwards from A', D', B', and C and complete the plan by aid of what has already been stated. PEOBLEM XVIII. To draw the plan of an oval equal-tapering body with top and base parallel, the length and width of the ton and bottom being given. Draw (Fig. 30) any two lines A B, C D intersecting each other at right angles in 0. Make A and B each equal Fig. 30. 3' ■- 3 to half the given length of the larger oval (top or bottom, as> may be), and O C and D each equal to half its given width. A B and C D v/ill be the ao:es of the \ &c., by straight lines ; then A D G G' D' A' is the pattern required. The frnstum of pyramid is here hexagonal, but by this method the pattern for any regular pyramid cut parallel tr> its base can be drawn. The next problem will show methods for larger work. (370 If 0' A D G (Fig. 34c), the pattern for the cone in which the frustum of pyramid is inscribeii, be cut out of zinc or other metal, and small holes be punched at the points A, B, C, &c., and A', B', C, &c. ; and if the cone be then made up with the lines O' A, O' B', &c., marked on it inside, and wires be soldered from hole to hole successively to form the top and bottom of the truncated pyramid, then (1) the whole pyramid of which the truncated pyramid is a portion, (2) the pyramid that is cut off, as well as (3) the truncated pyramid, will be clearly seen inscribed in the cone. The making such a model will amply repay any one who desires to be thoroughly conversant ^ith the construction of articles of the kind now under consideration. PEOBLEM XXI. To draw, without using long radii, the pattern for an equal- tapering body made up of flat surfaces, the height and top and bottom being given. Again suppose the body to be hexagonal. Case I. — For ordinarily large work. Draw (Fig 35a) the plan as by last problem. Nest join A B', and draw B' G perpendicular to A B' and eqiial to the given height of the body. Also draw B' H perpendicular to 74 THE TINSMITHS' PATTERN MANUAL. B B' ; and eqnsiX to tlie given lieigTit. Join A G and B H ; then -A G and B H ate the true lengths of A B' and B B', respectively. Fig. 35a. To draw the patterar Draw (Fig. 355) B B' equal to BH (Fig. 35tt), _With B as centre and radius equal to B A TtiK TJNft'MITHo I'ATTKUN MANUAL,, 75 (Fig. 35a), ami with B' as centre and radius equal to G A (Fig. 35a)'describe arcs intersecting in C, rigtt and left of BB'. ^Vith B as centre and the same GA as radius, and with B' as centre and radius B' A', describe arcs intersecting in C, right and left of B' B. 'Join B C. C C, C B' (Fig. 35b) ; then B B' C C is the pattern of the face B B' C G in plan (Fig. 35a). The other faces C D C D', B C B' C, &g, (Fig. 356) of the frustun) are described in exactly similar manner, G A being the distance between diagonally opposite points of any face as well as of the face B B' C C The triangles A 3' G, BB'H, can be drawn apart from the plan, as shown at K and L, (38.) It should bs observed that; if the pattern is truly Irawn, the top and bottom lines of each face will be parallel, 18 B 0, B' C, of face B C B" C (Fig. 35 h) ; and tbat this gives ^n easy method of testing whether the pattern has been •accurately drawn. Case II. — For very large work ; where it is inconvenient to diaw the whole of the plan. Draw AB (Fig. 36a) equal in length to the end-line of one of the faces of the frustum at its larger end. and produce it both ways. With B as centre and radius B A describe a semicircle, which divide into as many equal parts as the frustum has faces (Problem IX.). Hero it Is hexagonal, and the points of division working from point 1, are 2, C, 4, &c. Through the second division point, here C, draw a line to B, Ihen ABO is the angle mado in plan by two faces of the frastiim one with another, and A B, B G are two adjacent ond- lines of the plan of its larger ends. Bisect the angle ABC (Problem VIII.) by B S ; and draw a line C 0' from O making the acgie COB equal to the angle C B B (Probleia I.). On B C set off B D equal to the end-line of one of the faces of the frustum at its smaller end ; and draw D C parallel to BE, cutting 0' in C. Through C draw 0' B'' parallel to B and meeting B E in B' ; and draw B' A' parallel to B A and eq_ual to B D or 0' B'. Join A' A, and we 7B THE TINSMITHS' PATTERN MANUAI have in A'ABCC tbe plan of two adjacent faces of the tapering body or truncated pyramid. Next from B' let fall B' G perpendicular to A B, and make G F equal to tlie given height of the frustum. Join F B', then F B' is the true length of B'G. Through\B' draw B' H perpendicular to B' B and equal to the given height. Join H B, then H B is the true length of B B' one of the edges of the frustum. To draw the pattern. Draw (Fig. 366) B B' equal to H B (Fig. 36a). "With B and B' as centres and radii respectively B G and F B' (Fig. 36a) describe arcs intersecting in G (Fig. 366). Join B G and produce it making B A equal to Fia. 36&. F A B A (Fig. 36a). Through B' draw B' A' parallel to B A nA equal to B' A' (Fig. 36a)/ and join A A' ; then B A A' B' Tsill be the pattern of one face of the frustum. The adjacent i'ace B C C B' is drawn in similar manner. The fig. tihows the pattern, for two faces only of the equal-tapering body, because in cases where thfs method would have to be employed, two faces are probably the utmost tliat could be ciit out in one piece. Sometimes each face would have to be cut out separately, or perhaps even one face would have to be in portions. Any point in B' A' (Fig. 36a) instead of E' can bo chosen from "which to let fall a perpendicular to BA, andj the true length of B' G fpund as explained. The choice of j position depends upon the means at hand for drawing large) THE TINSMITHS' PATTERN MANUAI.: 77 arcs, the radius of the arc B' G(rig. 3G&) increasing In length as the point G approaches nearer to A. If the pattern bo truly drawn, B A will be perpendicular to B' G. It iQust not be forgotten that these methods are, both of them, quite independent of the number of the sides of the py.-amid. Also it should be noted that BE does not of necessity pass through a division point, nor of necessity is C C parallel to A B. These are coincidences arising from the frustum being here hexagonal. PEOBLEM XXII. To drmo (he pattern fur an ohlong or square equal-taper ivg lody with top and base parallel, and having flat sides and ends. ( The bottom is here taken as part of the body, and the whole pattern is in one piece.) Note.— This problem will be solveil in the problem next following. We adopt this course because the article there treated of is so iiiiportant an example of the oblong equal-tapering bodies in question, that it is desirable to make that, the special problem, the primary one, and this, the general problem, secondary to it. Its solution will be found at the end of Case I. PEOBLEM XXIII. To draio the pattern for a baking-pan. ^ baljing-pan has not only to be water-tight, but also to stand heat ; hence when made in one piece the corners are seamless. Case L— Where the length and width of the bottom, the width of the top, and the slant are given. Draw two lines XX, Y Y, intersecting at right angles in (Fig. 37) ', make A' and B' each equal to half the length of the bottom and C and OD' each equal to half ita width. Through C and D' draw lines parallel to Y Y ; alco through A' and E' draw lines parallel to X X and inter-- 78 THE TINSMITHS' PATTERN MANUAL, Fig. ?,S. (QUAETB« 0? FlO, 37, ENLASGEB.) Tllfi TINSMITHS' I'ATTKRN MAXUAI,. ^ ( 79 eecting the lines drawn tlirougli C and D' ; we get by this a rectangular figure, which is the shape of the bottom. Make A' A, B' B, C C, and D' D each equal to the given slant ; through B and A draw E F and G H parallel to X X ; through C and D draw S T and E P parallel to Y Y ; and make B E, B F, A H, and A G each equal to half the given width of the top. Join Q P and with Q, which is one of the angular points of the bottom, as centre and radius QP describe an arc cutting P E in P. Join Q P (the working can here bo best followed in Fig. 38) ; bisect the angle F Q P by Q 1? (Problem YIII., Chap. II.) ; and draw a line Q L making with P Q an aaglo equal to the angle F QM. The readiest way of doing this is by continuing the arc P F to L, then setting off F L equal to F M, and joining L Q. Now on B B' set oif B N equal to the thickness of the wire to be used for vfiring, and through N draw N F' parallel to E F (Fig. 37) and cutting Q L and Q P in V and F' ; make Q P' equal to Q ¥', and Q M' equal to Q V ; and join P' M' and M' F'. Repeat this construction fur the other three corners and the pattern will bo completed. This is not done in Fig. 37, for a reason that will appear presently. We shall quite under- stand the corners if we follow the letters of the Q comer in Fig. 38. These are B F F' M' P' P. The dotted lines drawn outside the pattern (Fig. 37) parallel to E F, P E, G H, and S T show the allowance for fold for wiring. It is important that the ends of this allowance for fold shall be drawn, as, fur instance, P K (both figs.), perpendicular to their respec- tive edges. Now as to bending up to form the pan. When the end adjacent and the side adjacent to the corner Q come to be loesit up on B' Q and D' Q so as to bring P Q and P Q into junctioa, it is evident that asP Q and P Q are equal .we shall obtain a true comer. To bring F Q and P Q into junction it is likewise manifest that the pattern will have also to be bent on the lines F' Q, P' Q, and M' Q. This fold on each other of P'QM' and P'QM' is generally still further bent round against Q V, 80 'i'llli' TINSMITHS' rATTERN M.ANUAl.. . (39.) The trutli of the pan when completed, and the ease with which its wiring can be carried out, depend entirely on the accuracy of the pattern at the corntsrs. This must never be forgotten. In marking out a pattern, only one corner need be drawn, as the like to it can be cut out separately and used to mark the remaining corners by. The points E, S, and T, the fixing of w^hich will aid in this marking out, can be readily found, thus: — For the R corner to come up true it is clear that D R will have to be equal to D P, from which we learn that D P is half the length of the top. Having then determined the point P we have simply to set off D R, C S, and T each equal to DP. If the pattern Fig, 37 were completed with the comers as at R, S, and T, that would be the solution of Problem XXII.-, and' the ends and sides being bent up, we should get an oblong equal -tapering body with top and bottom parallel and having flat sides and ends. Case II. — Where the length and width of the bottom, the length of the top, and the slant are given. The only difference between this case and the preceding is that jy P (Fig. 37), half the length of the top, is known instead of B F the half- width of top. To find the half- width of top ; with Q as centre and QP as radius describe an arc cutting B F in F. Then B F will be half the width of the top, just as (§ 39, previous case) we saw that D P was half the iecgth of the top. The remainder of the construction is now as in Case I. (40.) It will be evident that, in this problem, choice of dimensions is not altogether arbitrary. The lines Q F and Q P the meeting of which forms the comer, must always be of the ' same length. This limits the choice ; for with the dimensions of bottom given, and the slant, and the width of the top, the length of the top cannot be fised at pleasure, but must be such as will bring Q F and QP into junction ; and vice versa if the length of the top is given. It is un- necessary to follow the limit with other data. THE TINSMITHS' PATTERN MANUAL?" 81 Case III. — "^'liere tlie length and width of the top, and the slant and the height are given. The data 'in this case and the next are the usual data when a pan has to be made to order. The difference between this case and tlie preceding is that the size of- the bottom is not given, but has to be determined from the data. Excepting as to finding the dimensions of this, the case is the same as Cases I. and II. All that we have now to do is therefore to find the dimensions of the bottom, thus : — Fig. 39. Fxa. 40. Draw OC (Fig. 39) equal to half the given length of tha top. and through C draw C B perpendicular to C and equal to the given height. With B as centre and radius equal to the given slant describe an arc cutting C in B'. Then Q B' is the required half-length of the bottom. Next draw O E (Fig. 40) equal to half the given width of the top,. and through E draw E D perpendicular to E and equal to the given height. With D as centre and radius equal to the given slant describe an arc cutting O E in D'. Then OD' is the required half-width of the bottom. Case IV. — Where the length and width of the top and the length ?)nd inclination of the slant are given. This is a modification of Case III. Let B B' {Vii<:. 39) be the length of the given slant, and the angle that B B' make with B' C be the incliv.ation of the slant. Through B draw B C perpendicular to B' 0. Then half the given length of the top less B' C will be the required half-length of the bottom; and similarly half the given width of the top less B' C will be the req^uired half-width of the bottom. 82 THE TINSMITHS' PAITKRN .MANUAL. PKOBLEM XXIV. To draw (he pattern for an equal-tapering body with top and base parallel, and having flat sides and ends (same as Problem XXIL), hut loith bottom, sides, and ends in separate pieces ; the length and width of the bottom, the icidth of the top, and the slant being given. To draw tlie pattern for tlie end. Draw B B' (Fig. 41) e(|ual to the given slant, and throtigh B and B' draw lines C D and C D' perpendicular to B B'. Make B C and B D eacli equal to lialf the given width of the top ; also make B' C and B' D' each equal to half the given width of the bottom. Join C C and D D' ; thea C'GBW will be the end pattern required. Fig. 41. Fig. 42. End Patters Side PATTERy. To draw the pattern for the side. Draw E E' (Fig. 42) equal to the given slant, and through E and E' draw F G and F' G' perpendicular to E E'. Make E' F' and E' G' each equal to half the given length of the bottom, and with F' and G' as centres and radius D'D (Fig. 41) describe arcs cutting F G in points F and G. Join F'F and G'G; then F'FGG' will be the side pattern required. It should be noted that, as in the preceding problem, the width of the top determines the length of the top and mce uersa, also that lines such as D D', G G' must be equal. THE TINSMITirS' PATTEKN MANUAL. 8.'^ (41.) If the ends of tlie body are seamed (• knocked up ') on to the sides, as is usual, twice the allowance for lap shown in Fig. 41, must be added to the figure C C D' D. Similarly if the sides are seamed on to the ends, a like double allowance for lap must be added to F F' G' G, instead of to the end pattern. 84 tllE TINSMITHS' PATTERN MANUAL. CHAPTER VII. Patterns for Equal-tapeking Aktioles op Flat and Curved Surfaces combined. (Class I. Subdivision c.) From what has been stated about the plans of equal- tapering bodies, and fronid, p, 55, it will be evident that the curved surfaces of the articles now to be dealt with, are portions of frusta of cones. PROBLEM XXV» To draw the pattern for an eqnal-inperiiig body with top and bottom parallel, and having flat sides and equal semicircular ends {an '■equal-end' pan. for iiisfance), the dimensions of the top and bottom of the body and its height being given. Four cases will be treated of ; three in this problem and one in the problem following. Case I. — Patterns when the body is to be made up of four pieces, f We may suppose the article to be a pan. Having drawn (Fig. 43) the plan AECSA'D'C'B' by the method of Problem XYI. (as well as the plan, the lines of its construction are shown in the fig.), let us suppose the seams are to be at A, B, C, and D, where indeed they are usually placed. Then we shall require one pattern for the flat sides and another for the curved ends. To draw the pattern for the sides. Anywhere in AD (Fig. 43) and perpendicular to it draw THE TINSMITHS' PATTERN MANUAL. S5 FT, an3 make FH equal to the given height ;• join F'H; then F'H is the length of the slant of the article, and therefore the^ividth of the pattern for the sides. Draw A A' (Fig. 44) equal to F'H (Fig. 43). Through A and A' draW lines perpendicular to A A'. Make A D (Fig. 44) equal to A D (Fig. 43), and draw D IK parallel to Fig. 44. Side Pattekn. A A'. The rectangle ADD' A' will be the side pattern required. The fig. also shows extras for lap. To draw the pattern for the ends. Draw (Fig. 46a) two lines D A, .0 A, perpendicular to ©ne another and meeting in A ; make D A equal to D (Fig. 43), the larger of the radii of the semicircular ends ; and on A O set off A G- equal to the given height. Draw G D' perpen- dicular to A G and equal to D' (Figi 43), the smaller of the radii of the semicircular ends ; join D D' and produce it, meeting A in 0. Then with any point 0' (Fig. 456), as 86 -THE TINSMITfTS' PATTERN MANUAL. centre, and D (Fig. 45a) as radius, describe an arc D C, and with the same centre and radius D' (Fig. 45a) descrihe an arc D' 0'. Draw any line D O', cntting the arcs in points D and D'. Divide D E (Fig. 43) into aoy number of equal parts, say three. From D (Fig. 45&) mark off these three Fig. 45w. Fig. 456. dimensions to E, make E C equal to D E, and join C 0' ; then D C C D' will be the end pattern required. If R he joined to O', then D D' E' R will be half the pattern. The centre line E E' is very useful, because, in making up the article, the point E' must meet the lino SE (asfs) of Fig. 43, otherwise the body will be twisted in consequence of fche bottom not being true with the ends. Case II. — Pattern when the body is to be made up of twrt pieces. Secondly, suppose the article can be made in two pieces (halves), with the seams at E and G (Fig. 43) , the line EG (part only of it shown) being the bisecting line of the plan. It will be seen hj inspection of the fig. that we require one pattern only, namely, a pattern that takes in one entire end of the article with two half-aides attached. ■IIIK TINSMITHS' PATTKHN MAJfUAL 87 Draw tlia end pattern DD'C'C (Fig. 46) in precisely the same manner that it is drawn in Fig. 45&. Throngh C and C draw indefinite lines G C G', perpendicular to C C, and through. D and D' draw indefinite -lines D E, D' E' perpen- dicular to D D'. Make D E, D' E' each equal to D E (Fig. 43) Fig. 46. V and join E E' ; also make C Gr, C G', each equal to ( (Fig. 43), and join G G'. Then E E' G' G will be the patt( required. CG tern Gas3 III. — Pattern when the body can be made of one piece. Thirdly, suppose the article can be made in one piece, with the seam at S (Fig. 43). Then evidently the pattern will be made up of the pattern of one entire end, side patterns attached to this, and half an end pattern attached to each side pattern. Praw DD'C'C (Fig. 47) the end pattern. Through C and C draw lines perpendicular to C C, and each equal to C B (Fig. 43) ; and join B B'. Produce B B', and with B as centre a,nd D (Fig. 45a), the larger of the radii for the end pattern, as radius;, describe an arc cutting the pruduced lin© SB 'HIE TlX.SMrinS fAffEKN M\Xt"AI.. in 0« With as centre and B and O B' as radii respec- tively, draw arcs B S, B' S'. Make B S eqvial to D -E (Fig. 47) ■-.69 and join SO. Repeating tliis coustrnction for D D' A' A the other side pattern, and AA'S'S the remaining half-end pattern, will complete the pattern req^uired. nii; TlN.^MJJliS' I'ATI.KRN MANIAL 89 PEOBLEM XXVI. To draw, without long radii, the pattern for an equal-lnxiering hody with top ond bottom parallel, and having jUit slides and equal semicircular ends ; the dimensions of the top and hutiohi of the body and its height being given. TLis proLlem is a fourth case of the preceding, and exceedingly useful where the work is so large that it is inconvenient to draw the whole of the plan, and to use long radii. Draw half the plan (Fig. 48a). Divide D C into six or more equal parts, and join 1, 2, &c., to 0, Ly lines cutting D' C in 1', 2', &c., and join D 1'. Draw- D E perpendicular to Pig. 48a. Fio. 486, ^^--..^ V"~->v^ 2» A" \ y i \ S ^/ A ]>' > D 1' and equal to the given height, and join E 1'. (The line 1 to 1' appears to, but does not, coincide with E 1'.) Then E 1' may be taken as the true length of D 1'. Nest, producing as necessary, make D A equal to the given height. Joining D' to A gives the true length of D D'. To draw the 'end pattern. Draw Fig. (486) D D' equal to D'A (Fig. 48a). With D and D' as centres, and radii resjpectiyely D 1 and E 1' (Fig. 48a) describe arcs intersectingf 95 THE TJNSMitllS' PATTERN MANUAL. in point 1 (Fig. 48&). With D' and D as centres and radii respectively D' 1' and E 1' (Fig. 48a) describo arcs inter- secting in point' 1' (Fig. 486). By using points 1 and 1' (Fig. 486) as centres, instead of D and D', and repeating tlie constructiou, tlie points 2 and 2' can be found. Next, using points 2 and 2' as centres and repeating tlie construction, find points 3 and 3', and so on for the remaining points necessary to complete the end pattern, which is completed by joining the various points, as 3 to 3' by a straight line ; D, 1, 2, and 3, by a line of regular curve ; and D', 1', 2' and 3', also by a line of regular curve. Only half the end pattern is shown in Fig. 486. The side pattern can bo drawn as shown in Fig. 44. PEOBLEM XXVII. To draw Hie pattern for an equal-tapering hody with top and bottom parallel, and having flat sides and ends, and round corners {an oblong pan with round corners, for instance') ; the height and the dimensions of the tt>p and bottom being given. Again four cases will be treated of; three in this problem, tod one in the problem following. Case I. — Pattern when the body is to be made un of four pieces. The plan of the article, with lines of construction, drawn by the method of Problem XVII., is shown in Fig. 49. We will suppose the seams are to be at P, S, Q, and K, that is, at the middle of the sides and ends. The pattern required is therefore one containing a round comer, with a half-end and a half-side pattern attached to it. The best course to take is to draw the pattern for the round comer first, which, as will be eeeu from the plan, is a quarter of a frustum of a .cone. TIIL; TINSMlTIIci' PATTERN MANUAL. 91 Draw two lines CFi^. 50a) OA, C A, meeting perpendi- cularly to one another in A ; make A C equal to O A (Fig. 49) the radius of the larger arc of the plan of one of the corners, fiTid make A D equal to the given height of the hody. Draw 1)E p(!rpendii'nlar to A D and equal to the radius of the smaller arc of the plan of a corner, and join C E and produce Fiu. bO't, Fig. 506. -o ■S B J" - - it to peet AO in 0. With any point 0' (Fig. S05) as centre and radius equal to O C (Fig. 50rt), describe an arc A B, and with the sajue cunire and raJins equal to O E (Fig. 5Qa) dcsoiibe an arc A'B'. Dr^jW any line AO' cutting iho arcs ^2 THE TlN.SMITllS' PATTERN MANUAL. in the points A and A', and make A B equal to A B (Fig. 49) "by marking off the same numher of equal parts along A B (Fig. 50b) that we divide (arLitrarily) AB (Fig. 49) into. Join B O', cutting in B' the arc A' B'. Through B and B' draw B P, B' P' perpendicular to B B', make B P equal to B P (Fig. 49) and draw P P' perpendicular to B P. Through A and A' draw A B and A' R' perpendicular to A A', make A R equal to A R (Fig. 49) and draw R R' perpendicular to A R. Then R R' P' P will he the pattern of one round comer, with a half-end and a half-side pattern attached left and right. Case II.— Pattern when the body is to bo mado up of .two] pieces. I Now suppose the seams are to be at the middle ^f each end, at R and S (Fig. 49). The pattern" reqiiired will then be of twice the amount sh'^wn in Fig. 505. It will be found best to commence with the side pattern. Draw in the plan (Fig. 49), any line X X perpendicular to the Q side-line ; then X and X' will be plans of corre- sponding points (§ 26). Make XF equal to the given Fig. 51. height of the body and join X' F. Then X' F is the length of a slant of the body. Draw B B' (Fig. 51) equal to X' F (Fig. 49), and through B and B' draw B C and B' C perpen- dieiilar to B B'. Make B C equal to B (Fig. 49) and draw THE TINSMITHS' PATTERN MANUAI-. 93 C C perpendicular to B C ; then B C C B' will be the pattern for the side. We have now to join on to this, at B B' and C C, the patterns for the round corners, which can be done. thus. Produce B B' and C C',.and make B O' and C O" each equal to O C (Pig. 50a) the radius of the larger arc of the corner pattern. With O' and O" as centres and O'B as radius, describe arcs B A and C D, and with the same centres and O'B'' as radius, describe arcs B'A' and CD'. Make BA and C D equal each to B A in the plan (Fig. 49), and join A C cutting the arc B' A' in A', and D O" cutting the arc C D' in D'. Throtigh A and A' draw A R and A' E' perpendicular to A A' ; make A E equal to A E (Fig. 49), and draw E E' per- pendicular to A E. Next through D and D' draw D S and D' S' perpendicular to D D' ; make D S equal to D S (Fig. 49), and draw S S' perpendicular to D S. Then E W S' S will be the pattern lequired. It should be noted that O'B' and 0"C' should be each equal to E (Fig. 60a), or the pattern will not bo true. This gives a means of testing its accuracy. Case III.— Pattern when the body is to be made up of one piece. When the body is made in one piece it is usual to have the seam at S (Fig. 49). .It is now best to commence with the end pattern. Draw A A' (Fig. 52) equal to X'F (Fig. 49), which, the body being equal-tapering, is the length of its slant any- where, and draw A H and A' H' perpendicular to A A'. Make A H equal to AH (Fig. 49) and draw H H' perpendi- cular to A H ; then A A' H' H will be the end pattern. Next produce A A', and make A 0' equal to O (Fig. 50a). Then O' will be the centre for the arcs of the corner pattern, which, drawn by Case II. of this problem, can now be attached at A A'. Through B and B' draw B C and B' C perpendicular to B B' ; make B C equal to B C (Fig. 49), and draw Q C. perpendicular to B 0. That completes the addition of the side pattern 94 THE TINSMITHS' PATTERN MANUAL. B B' C C to the corner pattern. Nest produce C C, mark tho necessary centre, and attach to C C the comer pattern CDD'C, just as ABB' A' was attached to A A'. Then through D and D' draw D S and D' S' perpendicular to D D' ; make D S equal to D S (Fig. 49) and draw S S' perpendicular Fio. 52. to D S ; this adds half an. end-pattern to D D'. By a repeti- tion of the foregoing working on the H H' side of the end- pattern H H' A' A, the portion H H' S' S on the H H' side may be drawn, and the one-piece pattern S H A S S' A' H'" S' of tho body we are treating of completed. PBOBLEM XXYin, To dremy w?,tli©nt long radii, the pattern fcv an equal- tapesring hcd'g with top and hoitom parallel, and having flat eides and ends, and round corners ; the Imght and the dimen- sions of the top and bottom being given. This is a fourth case 6f the preceding problem, and we will apply it to the first. case of that problem", that is, when THE T1X.SM1TI1.S- l-AfTERN .MANUAL. 95 the body is to be made up of four pieces ; now, however, assuming that the pieces are so large that their patterns cannot be conveniently drawn by the method there given. Draw (Fifj. 53a) V S S' F, one-quarter of the plan of the body. Divide C D the larger arc of the plan of the round comer, into three or more equal, parts, and join each division- point to O the centre of the arcs of the plan. Draw D F perpendicular to D D' and equal to the given height of the body, and join D'F; then D' F will be the true length of DD'o Next join D2'; draw DE peroendicular to it and Fig. 586. equal to the given height, and join 2' E ; then 2' E may be taken as the true length of D 2', (In the fig. a portion of the line 2' E coincides with the line from 2 to 2'. This is, however, merely a coincidence cf the particular case.) Now draw (Fig. 536) D D' equal to D' F (Fig. 53a), and •with D and D' as centres and radii respectively D 2 and 2' E {Fig, 53a) describe arcs intersecting in point 2. Also with D' and D as centres and radii respectively D'2' and 2'B describe arcs intersecting in point 2', Kepeating this con- struction, but using points 2 and. 2' just found as' centres instead of D and D', the points 1 and 1' can be found. Simi- larly using points 1 and 1' as centres and repeating the construction we can find the points G and C. Join C C ; draw a regular curve from C through 2 and 1 to D, and another regular: curve from D' through 2' and 1' io C ; then in DCC^D' we have the Datlern of the corner. To this 96 THE TIN.SMITllS- PATTERN MANUAL. corner pattenl, to get at the pattern we require, we have to attach a half-end and a half-side pattern. Draw D S perpendicular to D D' and equal to D S (Fig. 53a). Also, draw S S' perpendicular to D S, and through D' draw D' S' parallel to D S. The half-end pattern is now attached. Through C and C draw P and C P' perpendicular to CC. Make CP equal to CP (Fig. 53a), and draw PF perpendicular to P. This attaches the half-side pattern, and completes the pattern required. If the hody is to h^ made in more than four pieces it will still be generally possible to have a complete corner in one of the pieces. The corner should always be marked out first, and whatever has to be attached should be added as described. PKOBLEM XXIX. To draw the pattern for an ovah equal-tapering "body with top and bottom parallel, the height and the dimensions of the top and bottom being given. Again we deal wit!)., four cases ; three in this problem and one in the following. Case L— Patterns when the body is to be mad© up of four pieces. The plan of the body with lines of construction, drawn by Problem XVIII., is shown in Fig. 54. It will be evident from the plan and from d, p. 55, that the body is made up of two round equal-tapering bodies (frusta of cones) ; the ends C D D' C, E E E' ¥', being equal portions of a frustum of one cone, and the sides, E G C E', D F F' D', equal portions of a frustum of another cone. Such being the case, it is clear that we require two patterns, one for the two ends, and the other for the two sides ; also that the seems should be at E, C, D, and F, where the portions meet. To draw (Fig. 555) the ends' pattern. From any point A draw two linGS A, B A (Fig. 55a) TEE TiNSMITHS' PATTERN MANUAL. Fjr,. r>l. S A 1)8 THE TINSMITHS' PATTERN MANUAL. Fig. 56a. THE TINSMITHS' PATTERN MANUAL. 99 perpendicular to each otKer ; make A B equal to Q B (Fig. 54) tho radius of the larger cnd-cUrve in the plan, and A D equal to the height of the body. From D draw D E perpendicular to- A O and equal to Q B' (Fig. 54) the radius of the smaller end- curve of the plan ; join B E and produce it, nieeting A in 0. Then with any point O' (Fig. 555) as centre, and B (Fig. 55a) as radius, describe an arc C D ; and with same centre and E (Fig. 55a) as radius describe an arc C D'. Join any point C in the arc C D to 0' by a line cutting the arc C D' in C ; divide the arc C D (Fig. 54) into any number of equal parts, and set off from C (Fig. 55i>) along C D parts equal to and as many as C D (Fig. 54) is divided into. Join D 0' cutting the arc C D' in D' ; then D D' C C will be the pattern required. To draw the pattern for the sides. From any point A draw two lines A C, A perpendicular to each other, and make A G equal to P C (Fig. 54), the radius of the larger side-curve and A B equal to the given height of the body. From B draw B E perpendicular to A B and equal to P C, (Fig. 54) the radius of the smaller side-curve. Join C E and produce it, meeting A in O. With any point 0' (Fig, 56fe) as centre and O C (Fig. 56a) as radius describe an arc C E ; and with same centre and O E (Fig. 56a) as radius describe an arc C E'. Now join C O', and, proceeding exactly f^imilarly as in drawing \the ends' pattern, complete E E' C C the pattern for the sides. Case II.— -Patterns when the body is to be made up of two pieces. In this case the seams are usually put at the ends A and B (Fig. 54). It is evident that one pattern only is now retiuired, and that this is made up of a side pattern, with, right and luft attached to it, a half-end pattern. Draw (Fig. 57) a side pattern ECC'E' in tho way described in Case I. Then with C as centre and the radius of the ends' pattern, that is B (Fig. 65a) as radius, set off the distance C 0. With as centre and radii 00 and OC describe arcs C B and C B'. Make C B equal to C B (Fig. 54) 100 TilE TINSMITHS' PAttEUN JfAKL'At. and join B 0, cutting C B' in B'. Then C B B' C is a lialf-enrl pattern attached to C C. The half-end pattern E A A' E' is Fra. 57. added at E E' by repeating the construction just described. This completes A B B' A' the pattern required. C.SJSE III. — Pattern when the body is to be made up of One piece. We will put the seam at A (Fig. 54), the middle of one end. Draw C D D' C (Fig. 58), an end pattern in the way described in Case I. Produce C C and make C equal to I'liE TINSMITHS' PATTERN MVXUAU loi 6 (Fig. 56a), the radins for a side pattern. With as centre and radii C and C descriha arcs C E, C i)'; Make V C E equal to CE (Fig. 54) and join E cutting C'E' in E'. Make E 0' ecj[Tial to B (Fig. 55a) and witii O' B as radius 10^ THE TINSMITHS' PATTERN MANUAL. describe an arc E A. With same centre and O' 1? as radiua describe an arc E' A'. Make E A equal to E A (Fig. 54) and join A 0', cutting E' A' in A', The remainder DAA'D' of the pattern can be drav/n by repeating the foregoing con- Btraction. The figure A C D A A' D' C A' wiU be the pattern required. PSOBLEM XXX» To draic, without long radii, ilie paiiern for an \ c c', and d d\ The true lengths of h h\ c c\ &c., are found as follows : From E' draw a line perpendicular to X X' and cutting A" F in F, and join EF. Then E F is the true length of EE'. Now make E' I) equal tp d d' and join D F ; then D F is the true length of dd'. Next set off E' C equal to c c', and E' B equal to 5 &' ; and join C F and B P. Then C F and B F are the true lengths respectively of cc' and 6 6'. The true length of A A' we already have in A A", and as this is the longest generating line of the frustum, E F will be the shortest. We proceed now to find the distance the points A and 6', 6 and c', c and d', &c., are apart, which we do by finding the true lengths of the lines A 6', 6 c', c d', and d E', joining the points. Through 6' draw 6' 6" perpendicular to A 6', and eyual to tho given height,- Join A 6" ; then A b" may be 120 THE TINSMITHS' PATTERN MANUAL. taten as the true lengtli of A h\ Similarly, Hirongli c', (t, and E' draw lines equal to the given height, and perpen- dicular to h c', c d', and d E' respectively. Join h c", c d", and de" ; then h c", cd", and de" may be taken as th'' true lengths required. ■ To diav? the pattern (Fig. 12) the seam to correspond with the shortest generating line. Draw A A' equal to A A" (Fig. 11) and with A and A' as centres and radii respectively^ Ah" and A' 6' (Fig. 11) describe arcs intersecting in &'. Next with 6' and A as centres and radii respectively B F and A h (Fig. 11) describe arcs intersecting in 6. Then A, 6, A', 6', aro points in the curves of the pattex"n. With h and 6' as centres and radii respectively he" and V c' (Fig. 11) describe arcs intersecting in c\ With c' and 6 as centres and radii respectively C F and b c (Fig. 11) describe arcs intersecting in c; and with c and c' as centres and radii respectively c d" andc'd'(Fig. 11) describe arcs intersecting in d'. With d' and c as centres and radii respectively D F and cd (Fig. 11) describe arcs intersecting in d. Similarly find E' and E. Draw unbroken curved lines through A & c d £ and A* V d d' E THE TINSMITHS' PATTERN MANUAL. 121 and join EE'; iliat will give us Lalf tlie pattern. By like procedure we find the otlier half of the pattern, that to the left of A A'. (65.) The lines 6 h\ c c\ &c., and the dotted lines' A h\ h c', &c., are drawn in Fig. 12 simply to show the position that the lines which correspond to them in Fig. 11 {bh\ A 6', &c.) take upon the developed surface of the tapering body. It is evident that it is not a necessity to make distinct operations of the two halves of the pattern ; for as the points 6', 6, c\ c, &c., are successively found, the points on th'^e left of A A' corresponding to them can be set off. PROBLEM V. To draw Oie pattern of an oblique cylinder {inclined circular pipe for example), the length and inclination of the axis and the diameter being given. Draw (Fig. 13) any line A' G', and at the point A' in it FiQ. 13, A' B" C" D' £' F make the angle G' A' A" equal to the given inclination of the axis. Make A' G' equal to the given diameter, and draw a line G' G" parallel to A' A". Make A' A" and G' G" each equal 122 THE TINSMITHS' PATTERN MANUAL. to the lengtli of the cylinder (the length of a cylinder is the length of its axis) and join A" G". Then A' A" G' G'! is the elevation of the cylinder. Now on A' G' describe* a semi- circle, and divide it into any number of equal parts, in the points 1, 2, 3, &c. : through each point draw lines perpendi- cular to A' G', meeting it" in points B', C',D',&c., and through B', C, D', &c., draw lines parallel to A' A". Draw any line A G perpendicular to A' A" and G' G", cutting the lines B' B", C C", &c., in points B, G, &c. Next makg B b equal to B'l, Gc equal to C'2, Bd equal to D'3, Ee equal to E' 4, and F/ equal to F' 6, and draw a curve from A through the points h, c, d, &c., to G. It is necessary to remark that this curve is not a semicircle, but a semi-ellipso (half an ellipse). To dra^ the pattern (Fig. 14). Draw any line A A, aud at about its centre draw any line G" G' perpendicular tp THE TINSMITHS' PATTEUN MAXUAL. 123 it and cutting it in G. From G, right and left of it, on the line A A mark distances G F, F E, E D, D C, C B, and BA equal respectively to the distances Gf,fe, cd, d c, &c. (Fig. 13). Through the points F, E, D, &c., right and left of G, draw lines parallel to G" G'. Make G G', g'^G" equal to G G', G G" (Fig. 13) respectively. Similarly make F F', F F", E E', E E", D D', D D", &c., right and left of G' G" equal respectively to F F', F F", E E', E E", D D', D D" &c. (Fig. 13). Draw an unhroken curved lino from G" through F", E", D", &c., right and left of G" and an unbroken curved line through F', E', D', &c., right and left of G\ The €gure A"G"A"A'G'A' will be the pattern required. The two parts G"A"A'G' of the pattern are alike in every tespect. 124 THE TINyMITIIS' PATTERN MAISUAL. CHAPTER II. IJNEQUAL-TAPERINa BoDIES, OF WHICH ToP AND BaSE ARE Parallel, and their Plans. (66.) Before going into problems showing how to draw tho patterns of unequal-tapering bodies with parallel ends, bodies which are (^as the student will realise as he proceeds) partly or wholly portions of oblique cones, it will be neces- sary to enter into considerations in respect of the plans of frusta of such, cones (see § 58, p. Ill), similar to those appertainiEg to the plans of frusta of right cones treated of in Chap. V., Book I. ; but to us of greater importance, because the constructions in problems for the setting out of patterns of bodies having unequal taper or inclination of slant are a little more difficult than those in problems for patterns of equal-tapering bodies. The chapter referred to may be now again read with advantage. As much use will be hereafter made of the terms Pro- poriionale Arcs and Similar Area we now define them. We also extend our explanation of Corresponding Points, Definitioks. (67.) Phoportional Arcs: Similar Arcs.— Arcs are pro- portional when they are equal portions of the circumferences of the circles of which they are respectively parts; they are similar when they are contained between the same generating lines. Similar arcs are necessarily proportional. In Fig. 15 the arcs A D and A' D' are proportional because each is a quarter of the circumference of the circle to which it belongs. They are similar because the generating lines V A and V D contain them both. (68.) CoRRESPONDiNa Points.— Points on the same gene- rating line are corresponding points (compare § 24) ; thus, r THE TINf^MlTUS' PATTERN MANUAL. 125 '-he points A and A are corresj.onding points, because they are on the same generating line V A ; also the points C and C OQ the generating line V C The point A' on V A is the Fig. 15, plan of A on V A ; the point C on V C is the plan of C on V ; and so on. (69.) From the figure, which shows a frustum of an oblique cone standing on a horizontal plane, it will be seen that the plan of a round unequal-tapering body (frustum of oblique cone) consists mainly of two circles G A D B, C A' D' B', the plans of the ends of the body. In Fig. 16 is shown a complete plan of an oblique-cone frustum. With the con- necting lines of the two sides we are not concerned, but may simply mention that they are tangents (lines which touch but do not cut) to the circles. Further from Fig. 15 it will bo seen that, completing the cone of which the tapering body is a portipn. a. The plan of the axis (line joining the centres of the ends) of a frustum is a line joining the centres of the circles which are the plans of its ends ; thus, the line O 0' is the plan of the axis 0" (s§e alsp 0', Fig. 16). 126 THE TINSMITHS' PATTERN MANUAL. Similarly, the plan of tlie axis of the complete cone, is plan, produced, of tlia axis of its frastum; thus, O Y is the plan of O v. 6, The plan, produced both ways, of the axis of a frustum contains the plans of the lines of greatest and least inclina- tion on the frustum (see § 52) ; that is to say, of the longest and shortest lines on it. Thus, O O', produced both ways, contains the plans of A A' and B B'. It is convenient to regard lines joining corresponding points of a frustum (corresponding points of a frustum are points on one and the same generating line of the complete cone) as generating lines of the frustum. (See in connection with this, § 46 j. Then lines A A', B B', fur instance, may bo spoken of as generating lines of the frustum represented. Similarly, the plans of the longest and shortest generating lines of a cone are contained in the plan, produced, of the axis of its frustum; thus, the plans of V A and Y' B are contained in O 0' produced, both ways. c. The line, produced, which joins the centres of the plans of the ends of a .frustum, contains the plan of the apex of the cone ; thus, O O', produced, contains Y the plan of Y'. d. The line, produced, which joins the plans of corre- sponding points of a frustum (see definition, §68) contains the plan of the apex of the complete cone, and, produced only as far as the plan of the apex is the plan of a generating line of the cone; thus, C and C being corresponding -points on the cone, the plan, C C, produced, of the joining C and C, contains Y ; and C Y is the plan of the generating lino CY'. e. The plans produced of ali generating lines of a frustum intersect the plan of its axis produced, in one point, and that point is the plan of the apex of the ,completo cono; for example, the plans produced, of the generating linos C C and D D' of the frustum intersect 0' produced in Y. (70.) It follows from e that the plan of the apex of the complete cone of which a given frustum is a portion can easily be found if we have given the plans of the ends of the THE TINSMITHS' PATTERN MANUAL. 127 frustum and tlie plans of two corresponding points not in the line passing through the centres of the plans of its ends. This is a valuable fact for us, as it spares us elevation: drawing which in many cases is very troublesome, and •indeed, sometimes practically impossible, as, for instance, where an unequal- tapering body is frustum of an exceed- ingly high cone the axis of which is biit little out of the perpendicular. This is a case in which although the apex caniiot be found in elevation because of the great length of the necessary lines, it can readily be found in plan, because, in plan, the requisite lines are short. An oblique cone may of course not only be exceedingly long, but also very greatly out of the perpendicular. In this case it is impracticable anyhow to find the plan of the apex. Problem IV., just solved, meets both casus. It was by e that we there foun'd the plan of the apex wlien accessible, that is, where the lines of the plan are not unduly long (see Fig. 11) by joining the plans of corresponding points c and c' (c and c' are corre- sponding points in that they are mid-points on the balf- plans of the ends of the frustum, and therefore necessarily on one and the same generating line), and producing c c' to intersect 0', the line joining the centres of the plans of tbe ends, that is to intersect the plan produced of the axis of the frustum. (71.) Passing the foregoing tinder review, it will be seen that if we have two circles which are the plans of the ends of a round unequal-tapering body (frustum of an oblique cone) standing on a horizontal plane, and the circles are in their proper relative positions as part plan of the frustum, then the line produced, one or both ways as may be necessary, which joins the centres of the circles, contains : — The Plan of the Axis of the frustum (see a, p. 125). The PiJLN of the Axis of the cone of which the frustum is a part (see a, p. 125). Tlie Plan of the Apicx of the cone (see c, p. 126). The Plans of the Longest and Shortest Gknerating lines of the frustuia (see h, p. 126). 128 THE TINSMITHS' PATTERN MANUAL. The .Plans of tlie Longest and Shortest Generating Lines of the cone, of which the fiustum is a part ("see &, p. 126). The Plans of the Lines of Greatest and Least Inclination of the frustum (see 5, p. 126). The Plans of the Lines of Greatest and Least Inclina- tion of the cone, of which the frustum is a part (see § 52, p. 107). And this is a matter of very great practical importance, as will be seen later on. (72.) As with circles "under the conditions stated, so exactly with arcs which form the plans either of the ends or of portions of the ends, of an unequal -tapering body. (73.) Further, referring to c and d of p. 55 as to round equal- tapering bodies we are now in a position to deduce (see Fig. 15) the following as to round unequal-tapering bodies. /. The plan of a round unequal-tapering body with top and base parallel (frustum of oblique cone) consists essentially of two circles, not concentric, definitely situate relatively to onQ another. See Fig. 16. Fig. 16. Similarly the plan of a portion of such round unequal- tapering body (frustum of oblique cone) consists essentially of two ai'cis definitely situate xelatively to one another, and THE TINSMITHS' PATTERN MANUAL. 129 not concentric. See Fig. 17. 0' is the axis of the complete frustum. g. Conversely. — If two circles, not having the same centre, definitely situate relatively to one another, form essentially the plan of a tapering hody having parallel ends, that body is a round unequal-tapering body (frustum of oblique cone). In Fig. 16, if the two circles represent essentially the plan of a tapering body having parallel ends, then the body, of which the circles are the essential plan, is a round uneoual- body (frustum of oblique cone). Similarly if two arcs definitely situate relatively to one another, and not having a common centre, form the essential part of the plan either of a tapering body or of a portion of Fig. 17. a tapering body having parallel ends, then that body or por tion is a portion of a round unequal-tapering body (frustum of oblique cone). \ In Fig. 17 if the arcs form the essential part of the plan, either of a tapering body or of a portion of a tapering body having parallel ends ; then the body or portion of body, of which that fig. is the plan, is a portion of a round unequal- tapering body (frustum of oblique cone). In the particular plan represented the arcs are similar; the points B and B', and A and A' are therefore corresponding points. 13a THE TINSMITHS' PATTERN MANUAL. We will now proceed to draw the plans of some imequal- tapering bodies, of which patterns will be presently set out. PEOBLEM VI. To draw the plan of an unequal-tapering body with top and hase parallel and having straight sides and semicircular ends (an " equal-end " hath with semicircular ends), from given dimensions of top and bottom. Draw A'B'C'D' (Fig; 18) the plan of the bottom by Problem XVI., p. 20. Bisect A'B' in 0, and through O draw C D perpendicular to A' B'. Make A and B each Fig. 18. equal to half the length, and and D each equal to half the width of the top. The plan of the top can now be drav/n in the same manner as that Qf the bottom, completing the plan required. - TOm xx^N^Mliiift vkli 'E-RT^ 'HK^TVAL. 131 PROBLEM VII. To draw tlie plan of an oval unequal-tapering body with, top and base parallel (aw oval bath), from given dimensions of top and bottom. Draw (Fig. 19) A'B'C'D' the plan of tlie bottom, llie given length and width, by Problem XII., p. 13 ; and make Fio. 19. O A and B eacli equal to half the length, and C and D. each equal to half the width of the top. The plan of the top can now be drawn in the same manner as that of the bottom ; this completes the plan required. 132 \THE TINSMITHS' PATTE:5N MANUAL? PEOBLEM VIII. To draw tne plan of a tapering body with top and base parallel and having oblong bottom icith se^nicircular ends and circular top [tea-bottle top), from given dimensions of top and bottom. Draw (Fig. 20) the plan of the oblong bottom by Problem XVI., p. 20, and with O tbe intersection of the axes of the oblong as centre and half the diameter of the top as radius, describe a circle. This completes the plan. Fig. 20. We here, for the first time, extend the use of the word 'axes' (see Problems XII. and XIV., pp. 13 and 15). It is I convenient to do so, and the_meaning is obvious. PEOBLEM IX. To draw the ^lan of a tapering hody with top and base parallel, the top being circular and the bottom oval (^oval canister-top), from given dimensions of top and bottom. Draw (Fig. 21) the plan of the oval bottom by Problem KII., p. 13, and with the intersection of the axes as centre. *rHE TINSMITHS' PATTERN MANUAL.^ 133 and half tlie given diameter of the top as radius, describe a circle. This completes the plan. Fig. 21, I PEOBLEM X. To draw the plan of a tapering body with top and base parallel and having oblong base with round corners and circular tov^ from given dimensions of the top and bottom. Draw C^ig. 22) %Ue plan of the oblong bottom by Problem XV., p; 19, and with the intersection of the axes of the Fig. 22. oblorg bottom as centre and half the diameter of thie top as radius, describe a circle. - This completes the pl^m. - 134' THE TINSM-ITllS' PATTERN MAXUAL. FEOBLEM XI. To draw the jplan of an (hiford hip-bath. Fig. 23 is a side elevation of the bath, drawn here only to make the problera clearer, not because it is necessary for the working. No method that involves the drawing of a full- size side elevation is pr-actical, on acco.unt of the amount of space that would be required. Via. 23. The "bottom of an Oxford hip-bath is an egg-shaped oval. The portion O X' of the top is parallel to the bottom A' B', and the whole XX' top, the portion O X E of the bath being removed, is also an egg-shaped oval. In speaking of the plan of the ' bath,' we mean the plan of the ~ X X' B' A' portion of it, as the plan of this portion is all that is necessary to enable us to get at the pattern of the bath. We will first suppose the following dimensions given : — The length and width of the bottom, and the length of the X X' top, the height of the bath in front, and the inclination- of the slant at back. First draw (Fig. 25) the plan of the bottom A'D' B' C by Froblem XITI., p. 14. To draw the horizontal projection of . THE TiNSMItllS' PATTERN MANUAL. 135 the X X* top, make (Fig. 24a) the angle A A' E equal to the given inclination of the slant at the hack. Through A' draw A' H perpendicular to A A', and equal to the height of the bath in front ; through H draw H X parallel to A A' and cutting A'E in X; and draw XA perpendicular to XH; then A A' will be the distance, in plan, at the back, between the curve of the bottom and the curve of the X X' top (Fig. 23). Make A' A (Fig. 25) equal to A A' (Fig. 24a), and make A B equal to the length of the X X' top. With as centre and A as radius describe a semicircle ; the remainder of the oval of the X X' top can now be drawn as was that of the bottom. This completes, as stated above, all that is necessary of the plan of the bath to enable its pattern to be drawn. Fig. 24a, Fig. 246. If the width of the X X' top is given, and not fno inclina- tion of the bath at back, make O A (Fig. 25) equal to half that width, and proceed as before. The seam in an Oxford hip-bath, at the sides, is on the lines of which C C and D D' are the plans. If tha length of the X X' top (Fig. 23) is not given, it can be determined ii. the follov/iug manner : — Let ^he angle X B' B (Fig. 246) represent the inclination of tl\» t;lant of the front, and B' X its length. Through X 1.1^5 THE TIXSMITIIS- PATTEKN M\NUAL draw XB perpendicular to S'B; tlien BE' will be the dis- tance in plan, at tlie front, betweca. the ciirve of the bottom and the curve of the XX' top (Fig. 23), and this distance, marked from B' to B (Fig. 25), together with the distance A' A at the back end of the plan of the bottom, fixes the length required. Fig, 25. If the lengths only of the slants of the bath at back and front are given and not their inclinations, the plan of the X X' top can be drawn as follows : — Draw two lines X B, B' B (Fig. 246) perpendicular to one another and meeting in B ; make B X equal to the given height of the bath in front, and with X as centre and radius equal to the given length of the slant at the front,-* describe an arc cutting B' B in B'. Make B'B (Fig. 25) equal to BB' (Fig. 246), and B A equal to the given length of the X X' top ; this will give the distance A' A. Now make A A' (Fig. 24a) equal to A A' (Fig. 25), dra'c^ AX and A'H per- pendicular to A A' and equal to the given height of the bath in front ; join H X and draw A' E, through X, equal to the length of the slant~at the back ; the remainder of the plan of the X X' top can then be drawn as already described. , Jt will be useful to show here in this problem how to THE TINSMITHS' PATTERN MANUAl.. 137 complete the back portion already comtnenced in Fig 2 la of the side elevation of the bath. Make A' E equal to the slant (§ 4, p. 24) at back, which must of course be given, and make XO equal to half the given width of the XX' top ; join E, and draw O O perpendicular to A A' produced ; then A',E is the elevation reouired. PEOBLEM XII. To draw the plan of an Athenian hip-hath or of a sitz-hath. Fig. 26 is a side elevation of the bath, drawn for the reason mentioned in the preceding problem. Fig. 26. The bottom of an Athenian hip-bath or a sitz-bath is an| ordinary oval. The portion X' P of the top is parallel to the bottom A' B', and the whole X X' top, the portion F X E of the bath being removed, is also an ordinary oval. Simi- larly as with the bath of the last problem ; we mean by plan of the ' bath,' the plan of the XX B' A' portion of it ; nt more being required for the drawing of the pattern of the bath. Vy© will first suppose the given dimensions to be those of 138 THE TiNSMItiiS' iPATTERN MANUAL. the bottom and tlie X X' top of the bath^also height of the bath in front. First draw A' D' B' C (Fig, 27) the plan of the bottom by Problem XII., p. 13. To draw the plan of the XX' top (Fig. 26) set off O A and B each equal to half the given length of that top, and and OD each equal to half its given width. The plan of the X X' top can now be drawt^ Fig. 27. as was that of the bottom. This completes, as stated aV^ve, all that is necessary of the plan of the bath to enabU its pattern to be drawn. If the length of the X X' top (Fig. 26) is not given but the inclination of the slant at front and back, these inclinations being the same, the required length can be determined iii the following manner : — Make the angle AA'E (Fig. 28a) equal to the given inclination. Through A' draw A' H perpendicular to A A' and equal to the given height of the bath in front ; through H draw H X parallel to A A' and cutting A' E in X, and draw X A perpendicular to A A' ; then A A' will be the distance in plan, at back and front, between the curve .of the Till!; TINSMITHS' PATTERN MANUAL. 139 bofetom and the curve of the X X' top. Make A A' (Fig. 27) and B E' each equal to A A (Fig. 28a) ; then A B will be the length required. Fig. 28a Fig. 28&. If the length of the X X' top of the bath (Fig, 26) is not given, nor the inclination of the slant at front and back, but only the length of the slant at front, the required length can be thus ascertained. Draw two lines XB, B' B (Fig. 286) perpendicular to one another and meeting in B ; make B X equal to the given height of the bath in front, and with X as centre, and radius equal to the length of the slant at the front, describe an arc cutting B B' in B'. Make A' A and B' B (Fig. 27) each equal to B B' (Fig. 286), then A B is the length wanted. The remainder of the plan can be drawn as described above. By a little addition to Fig. 28a we get at the back portion of the side elevation of the b'.th. It will be usrful to do this. Produce A'X and make A'E equal to the slant at back, which must of course be given. Then, on the plan (Fig. 27), E being the meeting point of the end and side curves of the oval A D B C, draw E F perpendicular to A B. Make X F (Fig. 28a) equal to AF (Fig. 27) ; join F E ; this completes the elevation required. 140 THE TINSMITIJS' PATTERN MANUAL, PEOBLEM XIII. To draid the plan of an oblong taper oath, the she of (he top and boilom, the height, and the slant at the head being given. To draw D E F C (Fig. 30) tlie plan of the top. Draw A B equal to tlie given length of the top, and through A and B draw lines perpendicular to A B. Make A E and A D each eqnal ' respectively eqtzrt] to a a', h h', cc', d 3; (Fig, 86), aBd join A B, B 0, C D, D A, A'B', B'C C'F, &c. ^Thon. A BCD AA; D'C'B A' is tb© pattern require. A ^ (78.) TBe dotted circles (Fig. 35) tbrotigli the angular points o£ the plans of the ends show the plans of the ends of the frijstuni of the oblique cone which -u'ould envelop the frustum of the pyramid. From the simiiarit}'- of the con- struction above to that for the pattern of a frustum of an oblique cone, it will bo evident that we have treated the etlges of the frustum as generating lines (see h, p. 126) of the friistum of the oblique cone in which the frustum of the pyramid could be inscribed. Should it be inconvenient to draw X X in conjunction with the plan of the pyramid dr^w X X quit© apart, and from any THli' TINSMITHS PATIEKN MANUAL 151 point V in it draw v V perpendicular to X X : mate v x equal to the height of the frustum and draw xx parallel to X X. Make va,vb,vc,vd equal to V A, V B, V C. Y D (Fig. 35 i respectively ; and make x a', x h', x c', x d' equal to V A', V B', V C, V D' (Fig. 35) respecti vely. Join a a\ h b\ c c', and d d' by lines produced to meet u V in V, and proceed ae stated above. Case IT. — Given the dimensions of the two ends of the frustum, the slant of one face and its inclinatioQ (the slant of the face of a frustum of a pyramid is a line meeting its end lines and perpendicular to them^. Draw (Fig. 37) a line E E" equal to the given slant, make the angle E" E E' equal to the given inclination, and let fall E"E' perpendicular to E E'. Draw A B C D (Fig. 35), the plan of the large end of the frustum, and let B C be the plan of the bottom edge of the face whose slant is given. Bisect B C in E and draw E E' perpendicular to B C and equal to Fig. 37. E E' (Fig. 87). Through E' draw B' C parallel to B C ; make E' C and E' B' each equal to half the length of the top edge of the B face, through C and B' draw C D' and B' A' parallel to D and B A ; make C D' and B' A' each equal to B'C; join D' A', also A A', BB', C C, and D D' ; this will complete the plan of the frustum. E' E" (Fig. 37) is the height of the frustum. The remainder of the construction is now the same as that of Case 1. 152 THE TINSMITHS' PATTERN MANUAL. For large "work and whore the ends of a frustum are of nearly the same size, it would be inconvenient to use long radii. For unequal-tapering bodies which are not portions of oblique pyramids, as in Problem XVII., the method now given, or & modification of it^ must be used. PEOBLEM XYI. To draw, without ioag radii, the pattern for a frustum of an oblique pyramid. The plan of the fnisium and its heujht being given. Let A B C D D' A' B' C (Fig. 38) be the plan of the frustum. From any point S in B C draw E E' perpendicular to B C and B' C, the plans of the bottom and top edges of the face B B' C of the frustum. Draw E'.E" perpendicular to E E' aad Fig. 33. eoual to tiie height (which either is given or can bo found as in Case IL of last problem), and join E E", then E E" is the true length of a elant of the face B C B' C, of the frustum. Join D €!',, and find its true length (D C") by drawing C G" THE TINSMITHS' PATTERN MANUAL. m perpendicTjlar to D C and equal to the height of frustum and joining D C". Kest join D' A and B' A ; through D' and B' draw lines D' A", B' B" perpendicular to D' A and B' A respectively, and naako D' A" and B' B" each equal to the given height of the frustum ; join A A" and A B", then A A" and A B" are the true lengths of D' A and B' A respectively. To draw the pattern of tlie face B"CB'C',. draw E E' (Fig. 89) equal to E E" (Fig. 38), and through E and E' draw B C and B' O' perpendicular to E E'. Make E C, E B, E' C, and E' B' equal to E C, E B, E' C, and F' B' (Fig. 38) respec- tively; join C and BB'; this completes the pattern of tbe face. The patterns of the other faces ai-e found in the following manner:— "With C (Fig, 39) and as centres and D C" and C D (Fig, 38) as radii renpectivelyj deecribe arcs intersecting in D; join CD, draw CD' parallel to CD and equal to CD' (Fig. 38); and join D D', With D' and D (Fig. 39) as centres and A A" and DA (Fig, 38) as radii respectively describe arcs intersecting in A ; join D A, draw B' A' parallel to D A and equal to D' A' (Fig. 38), and join A 154 THE TINSMITHS' PATTERN MANUAL. to A'. Next, •witli B' and B as centres and A B" and B A (Fig. 38) respectively as radii/describe arcs intersecting in A ; ^'oin B A- and draw B' A' parallel to B A and equal to B' A' (Fig. 88). Join A A', and this will complete the pattern required. PEOBLEM XVII. To draio (he pattern for a "hood. The plan of the hood is necessarily given, or else dimensions from which to drav/ it. Also the height of the hood, or the slant of one of its faces. The hood is here supposed to he a hody of unequal taper with top and base parallel, but not a frustum of an oblique pyramid. Let A B C D A' B' C D' (Fig. 40) be the given plan of the hood (a hood of three faces), A D being the * wall line,' A B and D C perpendicular to A D and B C parallel to it, also let the length of F C", a slant of face BB'O'C, be given. Draw C F perpendicular to B C and through C draw C C" perpendicular to C'F, and with F as centre and radius THE TINSMITHS' PATTERN MANUAL. 155 equal to tlie given lengtli, describe an arc cutting C C" in C". Join F C" ; then C G" is the height of the hood, which "we need. If the height of the hood is given instead of the length F C", make C C" equal to the height and join F C", which will be the true length of F C Next, through C draw C E perpendicular to CD; draw'C'C" pei^ndicular to C E, make C C" equal to the height and join EC". Now produce C B' to meet AB in G; draw B' B" perpendicular to B' G and equal to the height, and join G B". To draw Ihe pattern of the hood. Draw F C (Fig. 41) equal to F C" (Fig. 40) ; through F and C draw B C and B' C, each perpendicular to F C ; make F B equal to F B (Fig. 40) ; make F G equal to F G (Fig, 40), and C'B' equal to C B' (Fig. 40). Join B B' and C C, then B B' C' C wHl be the pattern of the face of which B B' C' G (Fig. 40) is the plan. To draw the pattern of the face C D' D C (Fig, 40). With C' and C (Fig. 41) as centres and EC" and CE Fig, 41. (Fig, 40) as radii respectively, describe arcs intersecting in E. Join G E and produce it, making C D equal to G D (Fig, 40), and through C draw C D' parallel to C D and equal to C D' (Fig. 40). , Join D D', then C C D D' is the pattern of the face of which a ODD' (Fig. 40) is the plan. With B' and B as 156 THE TINSMITHS' PATTERN MANUAL. centres and radii respectively eqnal to B" G and B G (Fig. 40), describe arcs intersecting in G. Join B G and produce it, making B A equal to B A (Fig. 40), and throngli B' draw B' A' parallel to B A and equal to B' A' (Fig, 40). Join A A' ; and the pattern for tlie hood is complete. THE TINSMITHS'. PATTERN xMANUAL.' 157 CHAPTER IV. Patteens for Unequal-tafeeing Articles of Flat and Curved Surface combined. Class II, [Subdivision c.) From what lias been stated about the plans of unequal- tapering bodies and from g, p. 129, it will bo evident that the curved surfaces of the articles now to bo dealt with are portions of f i usta of oblique cones. (79.) The advantages referred to in § 61 of Ijooking upon the oblique cylinder as frustum of an oblique con© will be evident in this chapter. For there is to each of the problems a Case where the plan arcs of the curved portions of the body treated of have equal radii. To deal with these as problems excej^tional to a general principle would be most inconvenient. . j*3 extreme cases, however, of the one principle that the curved portions of the bodies before us are portions of frusta of oblique cones, their solution presents no difficulty. It will be sufficient to take one such Case in connection with only one of the bodies. This we shall do in '^Ge IV. of the aext problem. PROBLEM XVIII. To draw (fie pattern for an unequal-tapering body with top and base paralMl and Jiaving flat sides and semicircular ends (an ' equal-end ' bath, for instance'), the dimensions of top and bottom of the body and its height being given. Five cases will bo treated of; four in this problem and One ia the problem following. 15S THE TINSMITHS' PATTERN MANUAL. Case I.™Pattenis wlien the body is to "ba mado i:p oi fciir pieces. Draw (Figo 42) the plan of the body (see Problem VI,, p. 130), preserving of its construction the centres 0, 0' and the points A, A' in which the plan lines of the sides and curves Fig. 42. A G I' A of the ends meet each Obhsr, Join A A'j as shown (four places) in the fig. The ends A D A A' D' A' and A E A A' E' A' of the body (see g, p. 129) are portions of frusta of oblique cones. Let us suppose that the seams are to be at the four A corners where they are usually placed, and to correspond with the four lines A A'. Then we shall require one pattern for the flat sides, and another for the semicircular ends. To draw the end pattern, (80.) Draw A 6 D D' A' (Fig. 43) the A 5 D D' A' portion of Fig. 42 separately, thus. Draw any line S X and with any point (to correspond with 0, Fig, 42) in it as centre and O D. (Fig. 42) as radius describe an arc (hero a quadrant) Dh A. equal to the arc D & A (Fig. 42). Make D O' eqiial to D 0' (Fig. 42), and with 0' as centre and 0' D' (Fig. 42) as radius describe an arc (here a quadrant) D' A' equal to the arc D' A' (Fig. 44). Joining A A' completes the portion of Fig. 42 required. Now divide D A into any number of equal parts, here three, in the points & and c. From D' draw D' D" per- pendicular to X X and equal to tho given height. Then P, D" are, in eieYation, the corresponding points of which THE TINSMITHS' PATTERN MANUAL. 159 D, D' are the plans. Being corresponding points, they are in one and the same generating line (§ 68). Join D D" and produce it indefinitely, then somewhere in that line will lie the elevation of the apex of the cone of a portion of which Fig. 43. A & D D' A' is the plan. Now from 0' draw 0' 0" perpen- dicular to X X and equal to the given height.. Then 0, 0" are, in elevation, the centres of the ends of the frustum in the same plane that D, D" are represented in, that is in the plane of the paper ; 0, 0' being the plans of these centres. Join 0" and produce it indefinitely ; then in this line lies the elevation of the asis of the cone of a portion of which A fe D D' A' is the plan, and necessarily therefore the elevation of the apex. That is, the intersection point P of these two lines is the elevation of the apex. Next, from P let fall P P' perpendicular to X X ; then P' will be the plan of the apex. Join D" 0". With P' as. centre and P'c, P' b, and P' A sue- 160 THE TINSMITHS' PATTERN MANUAL. cessively as radii, describe arcs cutting XX in C, B, and A". Join these points to P by lines cutting 0" D" in C, B', and A'. Next draw a line P D (Fig. 44) equal to P D (Fig. 43), and with P as centre and P C, P B, and P A" (Fig. 43} successively as radii describe arcs cc.hh, and a a. With D as centre and ladius equal Do (Fig. 43) describe arcs 'cutting arc c c in C ,and C right and left of P D. With same radius and these jii-mts C and C successively as centres describe arcs cutting arc bh in B and B ri-ht and left of P D. With B and B successively as centres and same radius describe arcs cutting arc a a in A and A right and left of P D. Join the points C, B, and A right and left of P D to P. With P as centre and P U" (Fig. 43) as radius describe an arc cutting P D iu D". With same centre, and P C (Fig. 43) as radius, describe arc c'c' cutting lines P C right and left of PD in C and C. With same centre, and P B' (Fig. 43) as radius, describe arc THE TINSMITHS' PATTERN MANUAL.^ 16I h' V cutting lines P B right and left of P D in B' and B'. Similarly find points A' and A'. Through the successive points A, B, C, D, C, B, A, draw an unbroken curved line. Also through the successive points A', B', C, D', C, B', A, draw an unbroken curved line. Then AD A A' D' A' will be the required pattern for ends of the.body. To draw the pattern for the sides. " Through A' (Fig. 42) draw A' F perpendicular to A' A' make F G equal to the given height and join A' G. Then A' G is the slant of the Jbody at the side. Next draw (Fig. 45) A A equal to A A (Fig. 42), and mate A F equal Fig. 45. to A F (Fig. 42) ; through F draw F A' perpendicular to A A and equal to.A'G (Fig. 42), and through A' draw A' A' parallel to A A. Make A' A' equal to A' A' (Fig. 42). Join A. A', A A', then A A' A' A is the pattern for the sides. Case II. — Pattern when the body is to be made up of two pieces. We will take it that the seams are to be at D D' and E E^ _(Fig. 42). It is evident that we want but one pattern, which shall include a side of the body and two half-ends. First draw as just explained A' A F A A' (Fig. 46) a f-ide- pattern of the body. Produce one of the lines A A' of this pattern, and make A P' equal to A"P (Fig. 43). With P' as centre and P B, P C, and P D (Fig. 43) successively as radii describe arcs h, c, and d, and with A as centre and A h (Fig. 43) as radius describe an arc cutting arc 6 in B. With same radius and B as centre describe an arc cutting arc c in C ; similarly with C as centre and same radius find D.( 162^ THE TINSMITHS' PATTERN MANDAL. J'oin B F, C P', D P'. Now with P B' (Fig. 43) as radius and P' as centre describe an arc. h' cutting PB in B', and with P C, P D" (Fig. 43) successively as radii describe arcs c' and d cutting P' and P' D in C and D'. Through the points Fi(J. 46. A, B, C, and D draw an unbroken curved line. Also through the points A', B', C, and D' draw an unbroken curved line. Then A D J)' A' will bo a half-end pattern attached to the right of th^ side pattern. Draw the otber half-end pattern A E E' A' in the same manner ; then E F D D' A' A' E' will be the^omplete pattern required. ^ASB ni. — Pattern when the body is to bo made up of one ^ piece. >Tn this case we will put the seam to correspond with D D' (Fig. 42).' \: First draw a E A A' E' A' (Fig. 47) an end pattern of the body in the same manner that A D A A' D' A' (Fig. 44) was arawn. ^ "With A' and A (right of E E') as centres and A' G aniAF (the small length AF) (Fig. 43) respectively aa THE TI^ii^MITHS' PATTERN MANUAL. 163 radii desirilDO arcs intersecting in F; join AF and pi'odiice it, making A A equal to A A (Fig. 42). Through A {extremity of F A') draw A' A' parallel to A A and equal to Ui THE TINSMITHS' PATTERN MANl'AL, A' A' (Fig. 42). Join A A', the extremities of the lines just drawn, and produce it indefinitely ; and make A P' equal A" P (Fig. 43). With P' as centre and P B, P C, and P D (Fig. 43) successively as radii describe arcs b, c, and d, and with A (of A P') as centre and A b (Fig, 43) as radius describe an arc cutting arc b in B. With same radius and B as centre describe an arc cutticg arc c in C, and similarly with C as centre and same radius find D. Join B P', C F, and D P'. Nsw with P' as centre and radii successively equal to P B', PC, and^PD" (Fig. 43) describe arcs h', c', and d' cutting P'B, FC, and FD in B', C, and D'. Through the points A, B, C, and D draw an unbroken curved line. Also through the points A', B', C, and D' draw an unbroken curved line. We have now in D F A A' A' D' attached to the right of the end pattern we started with, a side pattern and a half-end pattern. By a repetition of the foregoing construction we can attach A A D D' A' A' to the left of the end pattern we started with. The figure D E D D' E' D' will be the completo pattern required. Case IV.— Where the plan arcs D A, D' A' (Fig. 42) have equal radii. This is the extreme case above (§ 79, p. 157) referred to, where the cone becomes cylindrical. Problem V., p. 121, may Fig. 48. A* ,. S' CD' advantageously be compared with th& work now given. The arcs (Fig. 48) D A and D' A' (here quadrants) being equal, THE TINSMITHS' PATTERN MANUAL. 1G5 tlieir radii D, 0' D' are ako equal. Through D' and 0' draw D' D", and 0' A" perpendicular to X X and each equal to the given height of the body. Join D D", A" D" ; divide the arc D A into any number of equal parts, here three, in the points h and c; and through c, h, and A draw cC, t B, and A each perpendicular to X X and cutting it in C, B, and O respectively. The arc D A being here a quadrant the point where the line from A perpendicular to X X cuts X X is necessarily O, the centre whence the arc is drawn. Through C, B, and draw C C", B B", and O A" parallel to D D" and cutting A" D" in points C", B", and A". Also through D" draw a line D"A' perpendicular to DC and cutting the lines just drawn in C, B' and A'. Make c 2 equal to C c ; B' 1 equal to B b, and A' equal to A. From D" through 2, 1, to O draw an unbroken curved line. To draw the pattern. Draw D D" (Fig. 49) equal to D D" (Fig. 48) and through D" draw au indefinite line A' A' perpendicular to DD". Fig, 49. Mark off on A' A', right and left of D", D" C, C B', and B A' respectively equal to the distances between D" and 2, 2 and 1, and 1 and (Fig. 48) ; and through C, B', and A', right and left of D" draw indeSnite lines each parallel to D D". Make CC right and left of D D" equal to C C (Fig. 48) ; and make B' B right and left of D D" equal to B'B (Fig. 48). Also make A' right and left of D D" equal to A' (Fig. 48). 166 THE TINSMITHS' PATTERN MANUAL. Next make C'C" right and left of DD" equal to C'C' (Fig, -48). Similarly find points "B", A" right and left of D D" by making B' B", A' A" respectively equal to B' B", and A' A" (Fig. 48). Through the points 0, B, C, D, C, B, O, draw an unbroken curved line. Also through the points A", B", C", T>", C'\ B", A", draw an unbroken curved Hne. Then D A" D" A" will be the pattern required* PEOBLEM XIX. To draw, without long radii, the paUem for ai unequal' tapering hodp with top and base parallel and having fiat sides and equal semicircular ends {an ' equal-end ' bath, for instance). The dimensions of the top and hottom of the body and its height being given. This i problem is a fifth ease of the preceding, and is exceedingly useful where the work is so large that it is inconvenient to draw the whole of the plan, and to use long- radii. To draw the pattern. Fig. 50. First draw half the plan (Fig. 60). It is evident that the df awing of the side pattern presents no difficulty, as long THE TINSMITHS' PATTEHN MANUAL. 167 radii are not involved. It can be drawn as in Case I. of preceding problem. Divide the quadrants D A; D' A', each into the same number of equal pirts, here three, in the points c, h, c', b' ; join c c', h b'. Through D' draw D' E perpendicular to D' D and equal to the given height of the body. From D' along D' D mark off D' A, D' B, and D' C respectively equal to A' A, b' b, and c' c ; and join points D, C, B, and A to E, then E A, E B, E 0, and E D, will be the true leng-hs of A' A, b' b, c' c, and D' D respectively. Next join c D', and draw D' d" perpendicular to D' c and equal to the given height. Join c d", then cd" may be taken as the true length of D'c. Similarly join be' and A &', and through c' and V draw c' c" and b' b" perpendicular to c' b and 6' A respectively, and each equal to the given height. Join b c"and A b", then b c" and A b" may be taken as the true lengths of h c' and A 6' respectively. Now draw (Fig. 51) a line DD' equal to D E (Fig. 50) and with D' and D as centres and rndii respectively equal to d" c and Dc (Fig.' 50) describe arcs right and left of DD', inter- secting in c and c. With c, right of D, and D' as centres, and radii respectively equal to C E and D' c' (Fig. 50) describe arcs intersecting in c\ right of D'. With c, left of P, and D' as C(!ntres, and radii as before, describe arcs intersecting in c', left of D'. With successively c' and c right and left of D D', i6§ TllB TINSMITHS' PATTERN MANUAL. as centres and radii respectively eqttal to c" 6 and c h (Fig. 50) describe arcs intersecting in h and 6 right and left of D D'. With successively 6 and c' right and left of D D' as centres and radii respectively equal to B E and c' h' (Fig. 50) describe area intersecting in h', and 6' right and left of DD', and with suc- cessively 6' and 6 right and left of D D' as centres and radii respectively equal to A b" and 6 A (Fig. 50) describe arcs in- tersecting in A and A right and left of D D'. Similarly with A E and b' A' (Fig. 50) as radii and centres respectively A and b' describe intersecting arcs to find points A and A', right and left of D' D. Through the points A, b, c, D, c, b, A draw an unbroken curved line. Also through the points A', b', c jy, c\ b', A', draw an unbroken curved line. Join A A', A A', right and left of D D', then A D A A' D' A' will be the pattern required. The lines c c\ c D', 6 6', &c., are not needed in the working ; they are drawn here to aid the student by showing him how the pattern corresponds with the plan, line for line of same lettering (see also § 66, p. 121). PROBLEM XX. To draw the pattern for an oval unequal-tapering body with top and base parallel {an oval bath, for instance). The height and dimensions of the top and bottom of the body being given. Four cases will be treated of, -three m this problem, and one in the problem following (see also § 79, p. 157). Draw (Fig. 62) the plan of the body (see Problem VII., p. 131), preserving of its construction, the centres 0, O', P, V and the several points d and d' in which the side and end curves medt each other. Join dd', as shown (four places) in the fig. From the plan we know (see g, p. 129) that dGd d! Q' d/f d'Bdd'W d', the ends of the body are like portions qf the frustum of an oblig^ue cone; we also know that !fHE TINSMITHS' PATTERN MANUAL; 169 dAdd' A'd', dEd d''E' d', tLe Bides of the "body are like por- tions o£ the frustum of an oblique cone. In Plate I. (p. 181), is a representation of the oval unequal- Fia 52. JS / /^ " ' i" ~>^ \ ^ I ' u is \ r V o' '(? ; / N V "6^ >/ i tapering body for which patterns are required, also of two oblique cones {x and Z). The oblique cones show (except as to dimensions) to what portions of their surfaces the several portions of the surface of the oval body correspond. Thus the sides, A', of the body correspond to the portion A of cone a;, and the ends, B', B', correspond to the B portion cone Z. The correspondence will be more fully recognised as we proceed with the problem. The difference of obliquity between B' and B is seeming only, not real ; and arises simply from Z being turned round so that the whole of the dV> d(X B' ^ (Fig. 52) of the cone shall be seen. If the representation of Z showed its full obliquity, then the line on it from base to apex would be the right-hand side line of the cone, and only half of the B portion could be seen. x: ' THE TINSMITHS' PATTERN MANUAL. 171 ■ Case L — Patterns when the body is to be made up of four pieces. It is clear that we require two patterns ; one for the two (ends, and one for the two sides ; also that the seams should correspond with the four lines d d', where the portions of the reepective frusta meet each other. To draw the pattern for the ends. Draw separately (Fig. 53) the G'Gfdd' port 'on of Fig. 52, thus. Draw any line X X, Fig. 53, and with any point O (corresponding to O, Fig. 52 j in it as centre and O G (Fig. 52) as radius, describe an arc G d equal to G d of Fig. 52. Make G 0' equal to GO' (Fig. 52), and with 0' as centre and O' G' (Fig. 52) as radius describe an arc G' d' equal toG'cZ' of Fig. 52. Joining d d' completes the portion of Fig. .52 required. Now divide the arc G d into any number of equal parts, here three, in the points / and e. At G' and 0' draw G' G", O' O" perpendicular to X X, and each equal to the given height of the body. Join O O", G G" ; produce them to their intersection in S (§ 80, p. 158) ; and from S let fall S S' perpendicular to XX. Join 0"G". With S.' as centre and S'/, S'e, and S'd Buccessively as radii, describe ai;cs cutting X X in F, E, and D. Join these points to S bylines cutting 0" G""in F', E', and D'. - Next draw S G (Fig. 54) equal to S G (Fig, 53), and with S as centre, and SF, SE,and SD (Fig. 53) successively as r::dii describe arcs //, e e, ami d d. With G as centre and xadius equal to G/ (Fig. 53) describe arcs cutting arc // right and left of S G in F and F. With each of these point.s, F and F as centre and same radius describe arcs cutting arc e e right and left of S G in E and E. With same radius and each of the last-obtained_j)oint8 as centre describe arcs cutting dd jight and left of SG in D and D. Join all the points right and left of SG to S. With S as centre and SG" (Fig. 53) as radius, describe an arc cutting S G in G'. With Fame centre and S F' (Fig. 5^ as radius, describe an arc ff CTitting the lines S P right^nd left of S G in F' and F'. With same centre and S E' (^ig. 53) as radius, describe an 172 Fig. 54. THE TINSMITHS' PATTERN MANUAL. 173 arc e' e' cutting the lines S E right and left of S G in E' and E'. Similarly by arc d' d' obtain points D' and D'. Through the points D, E, F, G, F, E, D, draw an unbroken curved line. Also through points D', E', F', G', F', E', D', draw an unbroken curved line. Then D G D D' G' D' will be the required pattern for the ends of the body. It is in fact the development of the B portion of cone Z of Plate I. To draw the pattern for the sidss. Draw separately (Fig. 55) the ' Ahdd' portion of Fig, 52, thus. Draw any line X X, Fig. 55, and v/ith any point P Fig. 53. (corresponding to P, Fig. 52) in it as centre and P A (Fig. 52) ^ radius, dcGcribe an arc A d equal to A ', C, B', draw an unbroken curved line. This gives us B E B B' E' B' a complete end pattern. Now with B' on thsT light-hand side of the end pattern as centre and B' B as radius, and B as centre and h h (Fig. 62) as radius describe arcs intersecting in F. Join B F, F B' ; produce F B' indeiSnitely, and to P B' attach the half-end pattern FE E' B' in precisely the same manner that THE TINSMITHS' PATTERN MANUAL. 189 F E E' B' the half-end pattern in Fig. 65 is attached to the side pattern B F B'. By a repe|,ition of tho foregoing Fig. 66. 7\^ construction on the left of the end pattern B E B B' E' B' we can attach B B E.E' B' and complete E E E E' E' E' the pattern required. PEOBLEM XXIII. To draw, without long radii, the pattern for a tapering body with, top and base parallel, and having circular top and oblpng bottom with semicircular ends. The dimensions of the top and bottom of the body and its height being given. This problem is a fourth case of the preceding, and is exceedingly useful where the work ia so Isrge that it is 190 THE TINSMITHS' PATTERN MANUAL. incoBvenient to draw the whole of the plan, and to use long radii. To draw the pattern (with the body in four pieces, as in Case I. of preceding problem). (81.) -Draw (Fig. 67) E' E 6 A h' one quarter of the plan of the body. Divide the quadrants E 6, E' 6', each into the same nukber of equal parts, here three, in the points d, c, d', c' ; Fig. 67. join ddl c c'. Through E' draw E' F perpendicular to E' E and equal to the given height of the body. From E' along E' E mark off E' D, E C, and E' B respectively equal to d d\ c c', and h V ; and join E F, D F, C F, and B F ; then E F, D F, C F, and B F will be the true lengths of E S', d d', c c', and h V respectively. Next join d E', and draw E' e" perpen- dicular to it and equal to the given height, and join d e" ; then de" may be taken as the true length of d E'. Also join cd' and h a' ; through d' and c' draw cl' d" and c' c" perpendicular to c d' and h c' respectively, and each equal to the given hei.ght, and join c d" and h c" ; then c d" and h c" may be taken as the true lengths of d' c and c' 6 respectively. Through 6' draw THE TINSMITHS' PATTERN MANUAL. 191 V h" perpendicular to h' A and equal to the given heiglit, and join A h", then Ah" will be the true length of 6' A. Next draw (Fig, 68) E E' equal to E F (Fig. 67), and with E'- and E as centres and radii respectively equal to d e" and E d (Fig. 67) describe arcs intersecting in d, and with d and E' as centre and radii respectively equal to DF and E'd' (Fig. 67) describe arcs intersecting in d'. With d' and d as centres and radii respectively equal to c d" and d c (Fig. 67) describe arcs intersecting in c, and with c and d' as centres Fig. 68. and radii respectively equal to C F and d' c'(Fig. 67) describo arcs intersecting in c'. With c' and c as centres and radii respectively equal to h c" and c h (Fig. 67) d scribe arcs intersecting in 6. Similarly with 6 and c' as centres and radii respectively equal to B F and c' 6' (Fig. 67) describe arcs intersecting in h'. With h' and h as centres and radii respectively equal to h" A and b A (Fig. 67) describe arcs intersecting in A. Through E, d, c, b, draw an unbroken curved line. Also through E', d', c\ h\ draw an unbroken curved line. Join 6' A, b A"; then E c A 6' c' E' is the pattern required. (82.) The lines dS, cd, dW, &o., are drawn in Fig. 68 simply to shov/ the position that the lines which correspond to them in Fig. 67 {dd\ cc', dE', &c.) take upon the developed surfat-e of the tapering body. 192 THE TINSMITHS' PATTERN MANUAL. PEOBLEM XXIV. To draw the pattern for a tapering hody with top and hase parallel, and having an oval bottom and circular top (^oval canister top, for instance). The height and dimensions of the top' and bottom of the body being given. Again four cases will be treated of; three ip this problem and one in the problem following (see also § 79, p. 157). Draw (Fig. 69) the plan of the body (see Problem IX. > p. 132), preserving of its construction the centres 0, O', P, P, and the four points (d) where the end and side curves of the Fig 60. plan of the bottom meet one another.ialso the points A'E' where the axis A E cuts the circular top. Join d A' (two places) and d E' (two places). From the plan we know (see g, p. 129) that dGdA'G'E',dBdA' B' E', the ends of the body, are like portions of the frustum of an oblique cone ; we also know that dAdA'jdEd E', the sides of the body, are like portion^ of the frustum of an oblique cone. TIIP] TINSMITHS' PATTERN MANUAL. 193 (83.) It is evident tliat in tliis problem the arcs dG d, E'G'A' and dBd, E' B' A' are, neither • pair, proportional (§ 67, p. 124). We have hitherto in Problems XVIII., XX., and XXII. been dealing with proportional arcs. The work- ing will therefore differ, though but slightly, from that of problems mentioned. In Plate II. (p.-203) is a representation of the tapering body ""for which patterns are required, also of two oblique cones (x and Z). The oblique cones show to what portions of their • surfaces the several portions of the tapering body correspond. Thus the sides, B', of the body correspond to the B portion of cone Z, and the ends, A', correspond to the portion A of cone x. The correspondence will be more fully recognised as we proceed with the problem. Case I. — Pattern when the body is to be made up of four pieces. We will suppose the seams are to correspond vnth. the plan lines G G', B B', A A', E E', of ends and sides, as in Problem XXII. .Then one pattern only, consisting of a half-end- pattern, with, attached, a half-side pattern will be required. To draw the pattern. Draw (Fig.'TO) separately G' Gfd A' the G' Gfd A' portion of Fig. 69, thus. Draw any line X X and with any point O in it (corresponding to 0, Fig. 69) as centre and G (Fig. 69) as radius describe an arc Gd, equal to G d oi -Fig. G9. Make G 0' equal to G 0' (Fig. 69), and with O' as centre and O'G' (Fig. 69) as radius describe an arc G'A' equal to G' A' of Fig. 69. Joining d A' completes the portion of Fig. 69 required Now divide GcZ.into any number of equal parts, here three, in the points /and e. From G' and" 0' draw G' G", 0' 0" perpendicular to X X and each equal to the given height of the body. Join G G"„ 0" ; produce them to inter.-ect in S (§ 80, p. 158) ; from S let fall S S' per- pendicular to X X, and join 0" G". Now join d to S', by a line cutting arc G'A' in d\ then (§ 68, p. 12i, and d, p. 126) 194 THE TINSMITHS' PATTERN MANUAL. d! and d are corresponding points and (§ 67, p. 124) tbe arcs Gr d, G' d' are proportional ; G G' d' £? is the plan of a portion of an oblique-cone frustnm, lying between the same generating lines, and d d' A' is plan of a portion of the same frustum outside the generating lino S' d. With S' as centre Fig, 70. and S'/, S' e, and S' d successively as radii describe arcs cutting X X in F, E, and D ; join these points to S by lines cutting 0"G" in F', E', and D', With S' as centre and radius the distance between S' and A' describe an arc A' A cutting X X in A ; from A draw A A" perpendicular to X X and cutting O" G" in A", and join A" S. Next draw any line X X (Fig. 71) and with any point P in it (corresponding to P, Fig. 69) as centre and radius PA (the radius of arc d A.d., Fig. 69) describe an arc A d equal to Ad of Fig, 69. Make A A' and A 0' respectively equal to THE TINSMITHS' PATTERN MANUAfc, 195 A A' and A 0' (Fig. 69). Divide A d into any nurater of equal parts, here three, in the points 6 and c. Prom A and 0' draw A' A", O' O" perpendicular to X X and each equal to the given height of the body. Join A A", PO"i produce Fig. 71. them to intersect in Q (§ 80, p. 158) ; from Q let fall Q Q' per- peadio'llar to X X, and Join O" A". With Q' as oentte and Q' &, Q' c, and Q' d successively as radii describe arcs cutting XX in B, C, and D, and join these points to Q. Nest draw (Fig. 72) a lino S G equal to S G (Fig. 70) and ifirith S 843 centre and S F, S E, and S D (Fig. 70) successively as radii describe arcs /, e, and d. With G as centre and radius equal to G/ (Fig. 70) describe an arc cutting arc / in F. With same radius and F'as centre describe an arc cutting arc e in E, and with S aa centre and same radius describe an arc cutting arc d in D. Joia th© points F, E, and D to S. With S as centre and S G" (Fig. 70) as radius describe an arc g' cutting S G in G'. With samo centre and S F' (Fig. 70) 196- THE TINSMITHS' PATTERN MANUAL. as radius describe arc /' cutting B P in F'. With same centra and S E' (Fig. 70) as radius describe an arc e' cutting S E in E', and with same centre and S D' (Figo 70) as radius describe arc d' cutting S D in D'. With same centre and S A" (Fig. 70) as radius describe arc a', and with D' as centre aifd radius d! A' (Fig. 70) describe an are intersecting arc a' in A'. Make D Q equal to D Q (Fig. 71) and with Q as centre and Q C, Fig. 72. ; Q B, and Q A (Fig. 71) successively as radii describe arcs c, 6, and a. With D as centre and d c (Fig. 71) as radii describe an a,rc cutting arc c in C, and with C as centre and same radius describe an arc cutting arc b in B. Similarly with same radius and B as centre find point A. Join A A'. Through the points G, F, E, D, 0, B, A draw an unbroken curved line. Also through the points G', F', E', D', A' draw an un- broken curved line. Then "GD A A'E'G' is the pattern required.' 1}HiE TINSMITHS' PATTERN MANUAL. 197 Case II. — Pattern -when tlie "body is to he made np of two pieces. We will suppose tlie seams are to correspond witli tlie lines G G' and B B'. It is evident tliat here we need but one pattern only, wHda will combine a side of the body and two half-ends, in fact will be double that of Pig. 72, First draw B D A'B' (Fig. 73) a half-end pattern exactly as the half-end pattern G D A' G' in Fig. 72 is drawn, and make DQ equal to D Q (Fig. 71). With Q as centre and Fm. 73. QC, QB, and Q A (Fig. 71) successively as radii describe arcs dd, c c, bb, and a a. With D as centre and radius equal to dc (Fig. 71) describe an arc cutting arc cc in C, and with same radius and C as centre describe are cutting arc 6 6 in B. With B as centre and same radius describe an arc cutting are a in A, and with same radius and A as .centre describe an arc cutting arc bb in B, Similarly with same radius and lOS THE TINSMITHS' PATTERN MAXtJAL. points B and C, to the right of A, successively as centres find points C and D. Join D Q and produce it indefinitely ; make D S' equal to S D (Fig. 70). With S' as centi-e and S E, S F, and S G (Fig. 70) successively as radii describe arcs e, f, and g, and with D as centre and d e (Fig. 70) as radius describe an arc cutting 'arc e in E. With same radius and E as centre describe an arc cutting arc / in F. Similarly with F as centre and same radius find point G. Join the points E, F, and G to S'. With S' as centre and S D' (Fig. 70) describe arc ff cutting S' D in D'. With game centre and S E' (Fig. 70) as radius describe arc e' cutting S' E in E'. Similarly with same centre and S F' and S G" (Fig. 70) successively as radii find points F' and G'. Through tho-points D, C, B, A, B, C, D, E, F, G, draw an unbroken curved line. ALso through points A', D', E', F', G' draw an unbroken curved line. Then B A G G' A B' is the complete pattern required. Case III. — Pattern when the body is to be made up of erne piece. In this case we will put the seam to correspond with G G' (Fig. 69). We now need an end pattern (the end dBdA B' E' in plan), with right and left a side pattern attached {dAdA', t? E d E' in plan), and joined to each of these a half- end pattern {d A' G' G, d E' G' G in plan). Fircit draw Figs. 70 and 71 ; then draw (Fig. 74) S G equal to S G (Fig. 70), and with S as centre and S F, S E, and S D (Fig. 70) STiCcessively as radii describe arcs//, ee, and dd. With G as centre and radius G/ describe arcs cutting arc// right and left of S G in F and P. With points F, F, succes- sively as centres and same radiias describe arcs enttingaro e0 right and left of S G in E and E, and with same, radius, and the last found points as centres describe arcs cutting arc d d right and left jof S G in D and B. Join all the points found to S. With S as centre and S G" (Fig, 70) as radius dt scribe an arc cutting SG in G\ With same centre and S F' (Fig. 70) as radius describe an arc /'/' catting lines SF right and -left of SG in F' and F. With same centra METAL-PLATE WORK. 199 and S E' (Fig. 70) describe an arc e' e' cutting lines S E right and left of S G in E' and E', and with same centre and S D' (Fig. 70) as radius describe arc d! d' cutting lines S D right and left of S G in D' and D'. With S as centre and D' right and left of S G as centres and radii respectively equal to S A" and d' A' (Fig. 70) describe arcs intersecting in A' and E'. Through D, E, F, G, F, E, D, draw an unbroken (Mrvedline. Also through E', D', E', F', G', F', E', D', A', draw a:a unbrolcen curved line, and join D E', D A'. This gives us D G D A' G' E' a complete end pattern. Now attach the side pattern D A D A' and the half-end pattern D G G' A' to the right and left of the complete end pattern we started with, in precisely the same manner that the side pattern D A D A' and half-end pattern D G G' A' in Fig. 73 is attached to D A', which corresponds to D A' in Fig. 74. This will complete G E G A G G' G' G' the pattern required. ,. 200 THE TINSMITHS' PATTERN MANUAL. PROBLEM XXV. To draw, without long radii, the pattern for a tapering ho^y with top and base parallel, and having an oval bottom and circular top. The height and dimensions of the top and base of the body being given. This problem is a foy.rtli case of tlio preceding, and is exceedingly useful where the work is so large that it is- inconvenient to draw the whole of the plan, and to use long radii. T)3 draw the pattern (with the body in four pieces, as in Case I, of preceding problem). (84.) Draw (Fig. 75) E (^ & A A'-d' E' one quarter of the plan of the body. Join c (the point where the end and side curves Fig. 7S. of the plan of the bottom meet) to A', the extremity of the uadrant E A'. We must now get at corresponding points in l-HE TINSMITHS' PATTERN MANUALf 20l the arcs E c, E' A'. To do thio, as the arcs are not propor- tional (§ 67), we must find the plan of the apex of the oblique cone of a portion of which E c A' E' is the plan. It is in the finding of these points that our present working differs from the working of Problems XIX., XXI., XXIlI.,and XXVII., where the corresponding points d d', cc' (Figs. 50, 58, 60, and 82) are found. With radius O E produce arc E c inde- finitely, and through draw Q perpendicular to E'E and cutting E c produced in Q. Then E Q is a quadrant, and E Q, E' A' (each a quadrant) are proportional. In Q and A' therefore we have corresponding points (§ 68), as well as in E and E', which are points on the longest generating line (6, p. 126) ; and the intersection of O 0' produced, of which E E' is part, and Q A' joined and produced will give us V, the required plan of the apex. Next divide E c into any number of equal parts, here two, in point d. Join dY, cY cutting E' A' in d' and c', respectively (the lines from d and c are not carried to V in the fig.)s thtn (^ and d', c and c' are corresponding points, Kext through E' draw E' F perpendicular to E' E and equal to the given height; from E' along E'E mark off E' D, E' C, respectively equal to d d' and c c', and join E F, DF, and CF; then EF, DF and CF will be the true lengths of E E', d d', and c c'. Join E d' and d c' ; through d' and c' draw d' d" and c' c" perpendicular to d' E and c' d re- spectively, and each equal to the given height, and join E d", t? c" ; then E(i"and dc" may be taken respectively as the true lengths of d' E and c' d. Now divide arc A c into any number of equal pans, her© two, in the point b, and join 6 A'; through A' draw three lines A' A"; one perpendicular to A' c, the second perpendicular to A' 6, and the third per- pendicular to A' A, and each equal to the given height, and join c A", h A", and A A" ; then c A", 6 A", and A A" may ha taken as the true lengths of A' c, A' b, and A' A respectively. Nest draw .(Fig. 76) E E' equal to E P (Fig. 75), and with E and E' as centres and radii respectively equal to E d" and E' d' (^Fig- 75 j de.«cribe arcs intersecting in d\ and with d' and E as centres and radii respectfully equal to DF and Ed 'm ^HE TINSMITHS* PATTERN MANUAL. (Fig. 75) describe arcs intersecting in- d. With i an^ 3! as centres and radii respectively equal to dc" and d! c' (Fig. 75) describe arcs intersecting in c', and with c' and d as centres and radii respectively equal to OF and dc (Fig. 76) de- scribe arcs intersecting in c. With c and c' as centres and radii respectively equal to cA"and c'A' (Fig. 75V describe Fis. 76. arcs intersecting in A'; and with A' and c as centres and radii h A" and c b (Fig. 75) describe arcs intersecting in &. Simi- larly with A' and b as centres and radii respectively equal to A A" and 6 A (Fig. 75) describe arcs intersecting in A. Juin A A'. Through E, d, c, b, A draw an unbroken curved line. Also through E', d', c\ A' draw an unbroken curved line. Then E c A A' E' is the pattern required. The lines d d', c c', E d', d c', &c., are not needed for the working, they are drawn for the reason stated in § 82, end of Problem XXHI. (85.) If Y is inaccessible, corresponding points e, c'. d, d' can thus be found. From the point E' along the arc E' A' set off an arc proportional to the arc E c in. the following manner. Join O c (line not ^own in fig.) and through O' draw O'c* (also not shown in fig.) parallel to Oc and cutting arc E' A' in- c' ; then arcs E' c' and E c will be pro- portional. (The student must particularly notice this THE TINSMITHS' PATTERN M\NUAL, 205 metliod of drawing proportional arcs. It is outside the scope of the book to prove the method.) Now divide arcs E' c\ E c each into the same number of equal parts, here two, in the points d and d' ; then (^, (^' and c, c' are corresponding points.. PROBLEM XXVL To draw the pattern for a tapering body witli top and base parallel, and having oblong bottom with round (quadrant) corners, and circular top. The dimensions of the top and base of the body and its height being given. Again four cases will be treated of, three in this problem, and one in the problem following so § 79, p. 157). Draw (Fig. 77) the plan of the body (see Problem X., p. 133) preserving of its construction the centres 0' and the Fig. 77. points E' B' where the flat sides and flat ends meet the circle of the top. Join B B', E E' each in four places. From the plan we know (see g, p. 129) that the round corners of the body are portions of frusCa of oblique cones, 206 THE TINSMITHS' PATTERN MANUAL. (86.) Looking at the plan, we can at once see that what we have to deal with diffors somewhat from what has as yet been before us. Hitherto a line passing through the centres of the plan arcs bisected the arcs, and the cone development was consequently identical each side of a central line. In Fig. 77, however, the line drawn through O O' does not bisect the plan arcs E B, E' B'. This affects the working but little, as will be seen. In Plate III. (p* 213) the tapering body is represented ; also an oblique cone Z, the A portion of which corresponds to the A' portion of the body, and the development of the former is the development of the latter. Cass I.— Pattern when the l)ody is to be made up of four pieces. We will suppose the seams to correspond with ihe plan lines G E', D E', F B', A B', of ends and sides, as in Problems XXII. and XXIV. just preceding. Then one pattern, com- prising a half-end, a complete corner, and a half-side, will be the pattern required. To draw the pattern. Draw separately (Fig. 78) an E E', B' B portion of Fig. 77, thus. Draw an indefinite line S' d (Fig. 78), and with any point O (corresponding to O, Fig. 77) in it as centre and B (Fig. 77) as radius describe an arc B E. Join O O' (Fig. 77) and produce it cutting arc B E in d ; make d B and d E (Fig. 78) equal respectively to length of both cc' and dd'. Next join c'd,h'c; draw c' c", b' h" perpendicular to c' d, V c respectively, and each equal to the given height ; and join d c" and c h" ; then d c", c h" may Pio. 82. (f d h V BE cZ. „^_^^ b' "^"^ ^ ~-^ntre and S C (Fig. 95 j describe an arc (not shown in the fig.) cutting S A and S C in A and C respectively ; make S B' equal to S B' (Fig. 95). 234 THE TlNSMiTHS' PATTERN MANUAL/ Through A, D, C, and B draw an unbroken curved line. Also through A', D', C, and B' draw an unbroken curved line ; this will complete the pattern for a head-corner. To attach the patterns for the flat portions of the head to A A*^ , Fig. 96. and B B' respectively ; draw through a' (Fig. 94) a line a'F^ perpendicular to a! s', and through b' draw 6' G pei-pendicular to b' D'. Now draw (Fig. 94) a' P perpendicular to a' F and equak THE TINSMITHS' PATTERN MANUAL. 235 to the given height of the bath, asud join F P, then F P is the true length of F a'. Next draw h' K perpendicular to b' G (h' K will of course coincide with the line b' B') and equal to the given height ; join G K, then G R is the true length of G 6'. Now with B' (Fig. 96) aa centre and G R (Fig. 1)4) as radius, and B as centre and radius b G (Fig. 94), describe arcs intersecting in G. Join B G, B' G. With A' (Fig. 96) as centre and F P {Fig, 9-ij as radius, and A as centre and radius a F (Fig. 94) describe arcs intersecting in F. Join A F and produce it indefinitely", and make As equal to as (Fig. 94); through A' draw A's' parallel to A s and equal to a' s' (Fig. 94); and join ss'. The pattern for the portion, seen in plan in Fig. 94, Gb ass' a' b' of the head of the bath is now completed. It is needless to work out in detail the addition of the portion (Fig, 96; sh G h'fs' of the pattern, by which we complete the head pattern G E G ^' E' B'. The extra lines ia this latter portion of the pattern appertain to the next problem. PEOE-LEM XXXI. iTo draw, without long radii, the pattern for an oblong taper] bath ; given dimensions as in Problem XXX. This problem is a second case of the preceding. Patterns when the body is to be made up of four pieces- ' with seams as in preceding problem. 1 Again, the patterns required will be three ; one for the head of the bath, one for the toe, and one for the sides. The latter pattern needs no description. The pattern for the toe can be readily drawn by Problem XXVIII,, p. 94. The pattern for the head can be drawn as follows. Draw half the plan of the bath, as the lower half of Fig. 94, and divide the arcs s h, s' h', each into any number of equal parts, here two, in respectively the points/ and/', and join//'. Draw (Fig. 94a) any two lines K S, K L perpen- dicular to each other, and make K L equal to the given 236 ' THE TINSMITHS' PATTERN MANUAL. heigttt of the batli. From E along K S mark oif K H equal to h h",, K F equal to //*, and K S equal to s s' ; and join L H, L F, L S ;■ then L H, L F, and L S are the true lengths of hH, ff, and s s' respectively. Next join (Fig. 94) /' 7*, s'/; draw //", s' s", perpendicular to /' h, s' f respectively, and each equal to the given height; and join hf",fs"; then hf", fs" may be taken respectively as the true lengths of f h and s'/. The true length of h' G may be found along /*' B' as was that of 6 G in Problem XXX. along- V B' ; it will of course be equal to G R, and we shall speak of it as G R. Next draw (see Fig. 96, left-hand portion) ss' equal to LS (Fig. 94a), and with s' and s as centres and radii respectively equal to f s" and sf (Fig. 94) describe arcs intersecting in /. V/ith / and s' as centres and radii respectively equal to LF (Fig, 94ffl) and s'/' (Fig. 94) describe arcs intersecting in/'; and with/' and / as centres and radii respectively equal to hf" SiJxdfh (Fig. 94) descrilio arcs intersecting in h ; also with h and /' as centres and radii respectively equal to LH (Fig. 94a) and/'/t' (Fig. 94) describe arcs intersecting in h'. With h' and ^ as centres and radii respectively equal to G E and h G (Fig. 94) describe arcs iutersecting in G. Through s,f,Ji, draw an unbroken curved line. Also through s',f',h' draw an un- broken curved line ; and join li G, G h'. Then shG h' f's' is the pattern for the portion of the head of the bath repre- Bt-nted in plan in Fig. 94 by the same lettering. It is unne- gsssary to pursue the pattern further. INDRX. The references are to pag-es, except those in brackets, which are to paragraphs. A. « Angle, defined, 3; to draw equal to a given angle, 6; to bisect, 10. Angles, measurement of, 20. Apex, of cone, 24, 105 (43); of pyramid, 66 (29). Arc, defined, 5; to complete circle from, g; to find if given curve is, 10. Arcs, proportionate and similar, 124. Athenian hip-bath, plan, 137. Axis, of cone, 24 (6), 105 (43); of ellipse, 16; of pyramid, 66 (29). B. Baking Pan pattern, jj. Bath, Athenian hip, plan, 137. , 'equal-end,' equal taper pattern, 84-90; four pieces, 84; two pieces, 86; one piece, 87; short radius method, 89. , — -, unequal taper, plan, 130; pattern, 157-68; four pieces. 158, two pieces, 161; one piece, 162; when ends are cylindri- cal, 164; short-radius method, 166. , oblong taper, plan, 140; representation (plate) 237; plate ex- plained, 231; pattern, 230-6; short-radius method, 235. , Oxford hip, see that heading. , oval, plan, 131; representation (plate) 18 1; plate explained, 169; pattern, 168-83; four pieces, 171; two pieces, 17^; one piece, 177; short-radius method, 177. 240 . INDEX. Bath, sitz, plan, 137. Bevel (angle). 6. Bisect a line, to, 8; an angle, 10. c. Canister-top, oval, plan, 132; representation (plate), 203; plate explained, 193; pattern, 192-205; four pieces, 193; two pieces, 197; one piece, 198; short-radius method, 200. Center of circle, 5; of ellipse, 16. Chord, defined, 5. Chords, scale of, to draw, 21; how to use, 22. Circle, defined, 5; sector of, 28 (9); to find centre, 9; to describe, that shall pass through three points, 9; to complete from arc, 10; to find if given curve is arc of, 10; to inscribe polygon in, 10; to find length of circumference geometrically, 12. , , inclined, extreme cases of oblique cone frusta, 112 (61); pattern for, 121, Circumference of circle defined, 5; to find length of, geometrically, 12. Classification of patterns, 2. Coffee-pots, 34; hexagonal, 70 (35). Cone, defined, 105 (43); axis, radius, apex, base, 24 (6), 105 (43); elevation of generatmg lines, 108 (55). , right, defined, 24 (6), 105 (44); basis of patterns for articles of equal taper, 24 (5); compared with oblique case, 106 (from 47); representation (plate), 227; development of, by paint, 28; gen- erating lines, 106(47); corresponding points of generating lines, see Corresponding points; to find height, 26; to find slant, 27; to find dimensions of, from frustum, 36. Cone, right, pattern, 29-31 ; one piece, 29; more than one piece, 30. , , frustum (round equal-rapering body), defined, 33; rep- resentation, 50; representations of round equal-tapering arti- cles, 34; relations of, with complete cone, 34 (from 13); de- velopment, 34 (15). -, , , plan, 50 (23); characteristic features, 55 {c, d)\ to draw, see equal-tapering bcdies (plans, to draw) , , , pattern, 37-43; ends and height given, 47; ends and slant given, 39; pattern for parts of, 39; short-radius method, 41. -, oblique; defined, 106 (45); basis of patterns for articles of un- equal taper, 105 (42); compared with right cone, 106 (from 47); INDEX. 241 obliquity, how measured, 107 (52); representations (plates), i8r, 203, 213, 227, 237; development of, 108 (56); generating lines, 106 (46-7); longest and shortest generating lines, 107 (from 51); lines of greatest and least inclination, 107 {c,2); height of ele- vations of generating lines, 108 (55); true lengths of elevations of generating lines, 107 (54). , , plan of axis, 125-6 (a); of generating lines, 126 (fe>iiti. m-^ . . . 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