«C-flRLF .tjV>. n^ 131 GIFT OF J.B. Rogers ft*s»^#' CASSELL'S TECHNICAL MANUALS Applied Mechanics. Bricklayers, Drawing for. Building Construction. Cabinet Makers, Drawing for. Carpenters and Joiners, Drawing for. Gothic Stonework. Handrailing and Staircasing. Linear Drawing. Linear Drawing and Pro- jection. The Two Volumes in One. Machinists, Drawing for. Model Drawing. Orthographic and Isometrical Projection. Practical Perspective. Stonemasons, Draw^ing for. Systematic Drawing and Shading. MANUALS OF TECHNOLOGY. Edited by Professor Ayrton, F.R.S., and Richard Wormell, D.Sc, M.A. Illustrated throughout. The Dyeing of TextileFabrics. By Prof. Hummel. Watch and Clock Making. By D. Glasgow, Vice-Presi- dent of the British Horo- logical Institute. Steel and Iron. By Prof. W. H. Greenwoou, F.C.S., M.I.C.E., &c. Spinning Woollen and Worsted. By W. S. B. McLaren. Design in Textile Fabrics. By T. R. AsHENHURST. With Coloured Plates. Practical Mechanics. By Prof. Perry, M.E. Cutting Tools Worked by Hand and Machine. By Prof. Smith. Applied Mechanics. By John Perry, M.E. Illustrated. 7s. 6d. Practical Electricity. By Prof. W. E. Ayrton. Illustrated. Numerical Examples in Practical Mechanics and Machine JJesign. By R. G. Blaine, M.E. With Diagrams. Cassell & Company, Limited, Ludgate Hill, London, E.C. I— 12.99 CASSELL'S TECHNICAL MANUALSa^ ^ — k-"-4c — -P^^\ , LINEAR DRAWING SHOWING THE APPLICATION OP PRACTICAL GEOMETRY ,',^' TO TRADE AND MA1^UFA5^L^RE:5 , '^ \' \ ',> ,' ,' BY ELLIS A. DAVIDSON AUTHOK OF "projection," " BUILDING CONSTRUCTION," "DRAWING FOR CARPENTERS AND JOINERS," " DRAWING FOR MACHINISTS," "practical PERSPECTIVE," ETC. ETC THIR TY-SE VEN TH THO USA ND CASSELL AND COMPANY, Limited LONDON, PARIS, NEW YORK b' MELBOURNE vV^ ^y o • tC/>-*po^ First Edition September 1868. Re/>yinted Februxry 1869, December 1869, 1870, 1872, 1873, 1875, 1877, 1878, 1881, 1883. Januxry 1884, August 1884, 1886, 1887, i888, 188;, 1890, 1891, 1893, 1895, 1897, 1899. CLASSIFIED CONTENTS. RIGHT LINES. To bisect a Line To erect a Perpendicular at end of a Line The same by another method ... ... ... To draw a Line Perpendicular to another from an outlying point To draw a Line parallel to another, at a given distance ... To draw a Line parallel to another and passing through given point ... To divide a Line into any number of equal parts To divide a Line proportionally to another To draw a "Tangent " to a Circle at a given point To find a Mean Proportional between two given Lines ... To divide the space contained between two Lines into equal parts by Parallel Lines FIG. PAGB I I 2 2 3 3 6 7 7 8 a 8 8 9 9 lO 10 • 49 36 . 104 93 ANGLES. Various kinds of Angles ... ... ... To bisect an Angle ... ... ... To construct an Angle similar to another To trisect a Right Angle How to measure or construct Angles of various degrees ... To measure or construct Angles by means of the " Protractor TRIANGLES. Various kinds of Triangles To construct an Equilateral Triangle To construct a Triangle of given dimensions To inscribe a Circle in a Triangle To draw a Circle through any points, and thus touching th three Angles of a Triangle To construct a Triangle similar to another To inscribe a Square in a given Triangle ... ... ... To construct a Triangle, when Base and Angles are given To construct an Isosceles Triangle on a given Base, and with given Vertical Angle Ditto ditto To in.scribe an E^quilateral Triangle in a Square ... To construct an Equilateral Triangle of a given altitude Xo construct an Equilateral Triangle about a Circle — 13 .. i5« 14 .. 18 17 • 39 26 40 27 " 44 30 12 13 12 14 13 . i5i> 14 e . 16 ^5 20 17 ■ 36 25 • 43 29 45 32 . 46 34 • 47 35 • 48 35 • 50 37 8690S2 FIG. PAGR 51 38 52 39 67 55 69 57 70 57 7^ 58 73 60 108 96 109 9^ no 98 Iv CLASSIhlED CONTENTS, To inscribe a Triangle similar to another /« a given Circle To rtVjcr//'^ a Triangle ci(^^«^ a given Circle To inscribe an Equilateral Triangle in a Regular Pentagon ... To inscribe an Equilateral Triangle in a Hexagon to touch the side5 • To inscribe in a Regular Hexagon the largest Equilateral Triangle it will contain ... Within an Equilateral Triangle to inscribe six equal Circles ... In an Equilateral Triangle to inscribe the three largest Circles it will contain ... ... ... ... ... ... To construct an Equilateral Triangle, equal to a given Triangle To construct a Triangle equal in area to a given Polygon To construct a Triangle equal to a given Rectilineal Figure ... Ditto ditto ditto To divide a Triangle into two equal parts by a Line parallel to or perpendicular to one of its sides SQUARES. Definition of Squares To construct a Square on a given Line To construct a Square on a given Diagonal 'J'o describe a Square about a Circle To describe a Circle about a Square To inscribe a Square in a Circle To inscribe a Square in any Triangle ' To inscribe a Square in a given Trapezium To inscribe an Equilateral Triangle in a Square To inscribe a regular Octagon in a Squire To inscribe a Square in a Regular Pentagon To construct a Square equal in area to a given Rectangle To construct a Square equal in area to a given Quadrilateral Figure 112 100 To construct a Square which shall be equal to two other Squares added together 113 loi To construct a Square; outer line or ctr- cuMiference. t Perpendicular means "square" with another line : to say that a line is " perpendicular, does not necessarily mean that it is upright, for the sides LINEAR DRAWING. The same, by another method. (Fig. 3.) Let it be required to erect a perpendicular* from the point C, which may be at the end or at any other part of a hne, A B. Fig. 3- From C, with any radius, draw an arc,t D d. From D, with radius C D, draw an arc cutting the arc D d in E. From E, with the same radius, describe the arc C G, which will cut arc D <^ in F, From F, with the same radius, describe the arc E H. cutting F G in I. Draw a line through the point I to meet the point C, and this line will be the perpendicular required. or edges of a carpenter's or mason's square are perpendicular to each other in whatever position the square may be placed — The line formed by a cord having a weight at its end is, when still, a really upright line, and this is called vertical. * The teacher is advise! to draw perpendiculars to the line A B when [)laced vertically or obliquely, with the view of shoitmig the pupils that a in:. m.-iy be perpendicular to another though not upright. Explain, when two lines are perpendicular to each other, they form a " right angle." t Arc. Part of a circle. LINEAR DRAWING. Definitions concerning Parallelograms. A parallelogram is a four-sided figure whose opposite sides are parallel to each other. When the four sides are equal, and the four angles are right angles, the figure is called a SQUARE, as A When one pair of sides is of a different length to the other, but the sides remain parallel to each other in opposite pairs, the angles being right angles, the figure is called a RECTANGLE, or Parallelogram, as B. When the four sides are equal and the opposite sides parallel to each other, but the angles not right angles, the figure is called a RHOMBUS, or Lozenge, as C. In this figure the opposite angles will be equal to each other, as « ^, <^ b. EZO When the figure is formed of two pairs of parallel sides, each pair being of a different length, but neither of the angles being right angles, it is called a RHOMBOID, as D. LINEAR DRAWING. To construct a Square on the given line A B. (Fig. 4.) Erect a perpendicular at A (prob. 2). Fig. Make the perpendicular A C equal in length to A B. This is best done by using A as a centre, and A B as radius, then describing the arc (in this case a quadrant, or quarter of a circle) A C, which wil) cut off the perpen- dicular at the required length. From B, with radius A B, describe an arc ; and from C, with same radius, describe another arc, cutting the former one in D. Draw the Hnes C D and D B. Then A B C D will be a square.* • In a square, all the four sides must be coual, and all the angles must be ridkt angles. If both these conditions be fulfilled, both the diagonals will be equal. Diagonals are lines crossing to opposite angles, as A D and B C LINEAR DRAWING. To construct an Oblong of a given size (in this case 2|in. long, and i^in. wide).* (Fig. 5.) Draw the hne A B 2^ in. long. At B erect the perpendicular B C I7 in. high. From C, with radhis A B, describe an arc {c). From A, with radius B C, describe an arc ( cutting the line A B in the point D. * Parallel means running in the same direction, and keeping the same distance apart from each other. The metal rails on a railway are parallel to each other, so are the "ruts," or marks on a road made by the wheels of a cart which has passed over it. LINEAR DRAWING. 9 From D, with the same radius, describe the corre- sponding arc (F), cutting the line A B in the point G. Measure (with compasses) the length of the arc F, viz., from G to C, and mark off this length on the arc E, viz., from D to H. Draw a line through C and H, which will be parallel to A B. To divide the line A B into any number ol equal parts (in this case ten). (Fig. 9.) Draw a line (C D), parallel to A B. (The line C D may be any length — that is, it may be drawn indefinitely for the present.) From C set off along this line the number of parts into which the line A B is to be divided — viz., i to 10. These parts may be miy convenient size, but must be all equal. Draw C A and 10 B, and produce* both lines until they meet in E. From each of the points, i, 2, 3, &c., draw lines to the point E, which passing through A B will divide it into 10 equal parts. * To "produce" a line means to carry it on further, or to make it longer in the same direction. B 2 K) LINEAR DRAWING. Application No. 1 of the foregoing figure. (Fig. TO.) This problem may also be used for dividing a line proportionally to another — that is, to find divisions on a line, which shall be in the same proportion to it, that certain divisions are to another line either larger or smaller. Thus, let it be required to cut off from A B a part which shall have the same proportion to it that^ the division E D has to the line C D. Place A B parallel to C D. Join C A and D B, and produce the lines until they meet in F. Fig. lo. From E draw E F, which passing through A B will cut off G B, which will have the same proportion to A B that E D has to C D. This process is constantly used in finding the propor- tions of architectural mouldings, windows, mechanical details, &c., in making reduced or enlarged drawings. LINEAR DRAWING. II This problem is also most useful in findir.g a particular point in a line which may be so small as to render accurate division very difficult. Example : The length from A to B in a spur wheel (Fig. ir), including a tooth and a space^ is called the pitchy and the circle on which such distances are set off is called the pitch circle. Fig. 12. Now, although in many drawings the space and tooth are made equal, they are not so in a real spur wheel, the space being a very little larger than the tooth. This small difference is most important, for if the tooth and space were equal, the tooth of a wheel when in gear with another would not clear itself. The difference of one- eleventh is found in practice to be sufficient for all pur- poses. Thus, if the "pitch" is divided into ii equal parts, the tooth will be five-elevenths, and the space six- elevenths. But dividing the space A B (which in many cases is by far smaller than given above) will be found liable to 12 LINEAR DRAWING. some inaccuracy ; by this problem, however, the required point of division may be found with ease and exactness. Let A B (Fig. 12) be the length of the pitch, measured from A B in Fig. 11. Draw any line, C D, parallel to A B, and seL off on it 1 1 equal divisions {any length). Draw C B and D A, and produce the lines to meet in E. From point 5 draw a line to E, which will divide A B as required, the one part being ^ and the other ^ Set off these lengths on the pitch circle.* To construct an Equilateral Triangle on the given line A B. (Fig. 13.) From A, with radius A B, describe an arc. From B, with the same radius, describe a corresponding arc, cutting the former one in C. Lines joining A C and B C will complete the triangle, which will be equilateral — that is, all its sides will be equal. A triangle having only two of its sides equal, is called an Isosceles Triangle (A). When all three sides are of unequal length, the figure is called a Scalene Triangle, as B. * For full instruction concerning the modes of drawing the various forms of teeth of wheels, the student is referred X.p the volume on Mechanical Drawing. LINEAR DRAWING. 13 In a right-angled triangle, one of the angles, as C, is a right angle.* A right-angled triangle may be either isosceles, as D, or scalene, as E. The longest side of a right-angled triangle, viz., the side opposite to the right angle, viz., F, is called the Hypothenuse. To construct a Triangle of given dimensions. (Fig. 14.) Fig. 14. Let it be required that the sides of the triangle should be if', i", and i?". Make A B i^ in. long. From B, with a radius of \\ in., describe an arc. • When a line, C D, stands perpendicularly on another line, A B, it divides the space into two right angles; if produced beyond C, four right angles will be formed : but if the line C E be drawn, dividing the space unequally, the angle A C E is an obtuse for wide) angle, being more than a right angle, and the remaining portion, B C E. is an acute (or sharp) angle, being less than a right angle. '4 LINEAR DRAWING. From A, with a radius of i in., describe an arc cutting the former one in C. Draw A C and B C, which will complete the triangle of the required dimensions. To bisect an Angle, ABC. (Fig. 15^.) A Fig. \yi. From B, with any radius, describe an arc, cutting the lines B A and B C in D and E. From D and E, with any radius, describe arcs cutting each other in F. Draw B F, which will bisect the angle. To inecribe a Circle in the Triangle ABC. (Fig. 15A) c Bisect any two of the angles (by Fig. 15^). Produce the bisecting lines until they meet in D. LINEAR DRAWING. 1|; From D, with the radius D E, which is a perpendicular from D on A B, a circle may be described which will touch all three sides of the triangle. This is called the inscribed circle. To draw a Circle through three points, however they may be placed (provided they are not in an abso- lutely straight line). (Fig. 16.) Let A B and C Idc the three given points. Join A B and B C. Bisect A B and B C, and produce the bisecting lines until they cut each other in the point D. Fig. 16. Then D will be equally distant from each of the three points. Therefore, from D with radius D A, D B, or D C, a circle may be drawn which will pass through the three given points. It will be evident that if A and C were joined, the figure would be a triangle ; and thus this problem serves also for describing a circle which shall touch the three aigles of a triangle. This is called the circicjnscribingcircXe. The Gothic Trefoil.* (Fig. 17.) This figure will serve as an application of the con- struction of the equilateral triangle and the bisecting of • Trefoil. A figure much used in Gothic architecture. It is formed of three leaves, or lobes (hence its name', meeting at a centre, as in the three-leaved clover. It is sometimes enclosed in a circle, as in window tracery, but not always, as in many wall piercings. I6 LINEAR DRAWING. angles. It is here introduced with the view of showing students the importance of absolute accuracy in the early problems, as well as in the subsequent operations. Construct an equilateral triangle, a b c. Bisect the angles, and produce the bisecting lines, d e f. Observe, that in an equilateral triangle, the lines which bisect the angles will, if produced, bisect the sides opposite to the angles as well, and thus the points g h i are obtained. From a b and c, with radius a g, equal to half the side of the triangle, describe the arcs j k /, and the others, which it will be plain are concentric* with them. * Concentric. Drawn from the same centre. LINEAR DRAWING. tj The arcs m and n, and those corresponding to them, are also drawn from the same centres. The outer circles and the arcs p q, &c., are drawn from the centre of the triangle o. To construct on the given line D E an Angle similar to the angle ABC. (Fig. i8.) From B, with any radius, describe an arc cutting the sides of the angle in c d. From E, v/ith the same radius, describe an arc cutting E D in F. Measure the length from point c io d. Mark off the same on the arc from F — viz., to point G- Draw a line from E through G. The angle F E G will be equal to A B C. On the given line A B, to construct a Triangle similar* to C D E. (Figs. 19 and 20.) Fig. 19. Fig. 20. At A construct an angle similar to the angle H C G — viz., J A I. • When a figure is said to be similar to another, it means that it is of the same skafit. When it is said to be equal, it means that it is of the same area — that is, it contains precisely the same space. A figure may be equal to another without being similar in shape, as in i8 LINEAR DRAWING. At B construct an angle similar to the anj^le K D L -viz., M B N. Produce the lines A J and B M until they meei in O ; which will complete the triangle required. Definitions concerning Four -sided Figures which are not parallelograms. Fig. Fig. 23. A figure having four sides, which are neither equal nor parallel to each other, is called a TRAPEZIUM, as A. But any two of its adjacent (or adjoining) sides may be equal to each other, so long as they are not parallel to the opposite sides, as in the figure B. If any two of the sides are parallel to each other, the figure is called a TRAPEZOID, as C. To construct a Trapezium similar and eaual to another (C D E F). (Fij^s. 24 and 25.) Fig. 25. Draw A B equal to C D. At A construct an angle similar to that at C. Fig. 105, where the square is equal to the rectangle; and a figure may be similar without oeing equal, as in Figs. 26 and 27. *' Similar and equal" means being of both the same shape and stzc Zi another figure, as in Figs. 24 and 25. LINEAR DRAWING. 19 Make A G equal to C E. At B construct an angle similar to that at D. Make B H equal to D F. Join H G, and the trapezium on A B will be similar and equal to C U E F. It is 'dvisable that the students shonld be repeatedly exercised in con- struct ng figures similar and equal to each other ; and as the correct result of the higher figures depends on the refinement of their construction, the most intetise accuracy should, as the pupils advance, be insisted upon. To construct, on the given diagonal A B, a Trapezium similar to another (C E D F). (Figs. Draw the diagonal C D in the given trapezium. From C and D, with any radius, draw arcs cutting the diagonal C D in G and H, C E and C F in I and J, and D E and D F in L and K. From A and B, with the same radius, describe arcs cutting diagonal A B in M and N. From the point M, cut off on the arc the length G J, viz., to O, and also the length G 1, viz., to P. From the point N, cut off on the arc the length H K, viz., to Q, and also the length H L, viz., to R. Draw B Q and B R. Also A O and A P. Produce these lines until they meet in S and T. A T B S will be the trapezium required. This result would be the same whatever might be the length of ihe diagonal or the relative sizes of the figures, as an angle is not altered by the length of the lines of which it may be formed. 20 LINEAR DRAWING. To construct a Trapezium from the following given dimensions (Fig. 28)— Fig. 28. Sides C A and C B are to be adjacent to each other, forming an angle similar to A C B. C A is to be if inches long. C B „ i^ AD „ I B D „ li „ Before commencing to work out any questions, the student is recommended to think over the given con- ditions, and to consider most carefully which would be the best starting-point. Now, in the figure here required, the fiist fixed condition is, that the sides C A and C B are to make an angle similar to the given angle A B C. — Therefore at any point, construct this angle {a C b) and produce the lines until C A is if inches, and C P r^- inches long, viz., to A and B. From A, with i inch radius, describe an arc. From B, with i^ inch radius, describe another arc cutting the former in D. Draw A D and B D, which will complete the figure from the given dimensions. LINEAR DRAV/ING. 21 To construct a Square on a given diagonal A B. (Fi?. 29.) ^— y c~"~~ X -v. y ^r ^ X^ ^ V X ^ / / / \ x \ f / \ X \ ' X / \ \ ^ [/ 1 1 \ \ \ K. \ /\ ' \ \ / / 1 ^ X \ / / / ^ x \ / / / ^ \ \ ' y ^ '^ \ ^ ' / / X • /. — "^ E Fig. 29, Bisect the diagonal A B in the point C. From C, with radius C A, describe a circle cutting the bisecting line in D and E. Draw AD, D B, BE, E A, which will complete the square on the given diagonal A B. To construct a Parallelogram when the diagonal A B and the length of one pair of sides C are given. (Fig. 30.) Fig. 30. Bisect A B in the point O. From O, with radius O A, describe a circle. From A and B set off the length of the line C on the circle— viz., A D and B E. Join these points, and the required figure will be completed 22 LINEAR DRAWING. To describe a Square about a Circle. (Fig. 31.) E < F ^ 1 i. \~^ n kV U' G c > H Fig. 31. Draw two diameters* A B and C D at right angles to each other. From A and C, with radius equal to the radius of the circle (O A), describe arcs cutting each other in E. From C'and B with same radius, describe arcs cutting each other in F. From A and D with same radius, describe arcs cutting each other in G. From D and B describe arcs cutting each other in H. Draw E F, F H, H G, and G E, which will complete the square about the circle. To describe a Circle about the Square A B C D. (Fig. 32.) Fig. 32. Draw the diagonals A D and B C. From their intersectionf (O), with radius O A (O B^ • Diameter. A line drawn across a circle, and passing through the centre, it is thus equal to two of the radii fplural of rad»/>r}, and is the longest straight line that can be drawn in a circle. t Intersect. To cut through. LINEAR DRAWING. 23 O C, or O D), describe the circle touching the four angles of the square. To find the Centre of a Circle. (Fig. 33.) Fig. 33- Draw a chord,* as A B, and bisect it by a line cutting the ciicle in C and D. Bisect C D by the line E F. The intersection O is the centre of the circle. To inscribe a Square in a Circle. (Fig. 34.) A c F'g- 34- Find the centre of the circle, and draw two diameters at right angles to each other. • Chord. A line cutting off any part of a circle. The pajts into which the circle is thus divided arj called segvtents. A part of a circle contained between two radii, as D O E, in Fig. 34, is called a sector. C 24 LINEAR DRAWING. JF'iom their extremities draw lines A B C D, which will form the square in the circle. To construct a Gothic Quatrefoil.* (Fig. 35.) Construct a square on the diagonal AB(see Fi.:;. 29). Bisect the sides by the lines E G, F H, cutting the lines A C, C B, B D, and D A, in ij, k, L From A, C, B, and D, with radius A z — that is, half the side of the square— draw the arcs /', ;«, ;;, o, and those concentric with them. The outer circles are drawn from the centre O. * Quatrefoil. A figure based on four leaves or lobes. See remarks on the Trefoil, Fig. 17. LINEAR DRAWING. 25 To inscribe a Square in any Triangle, ABC. (Fig. 36.) E Fig. 36. From C drop a perpendicular, C D. From C draw a line parallel to A B — viz., C E. From C, with radius C D, describe a quadrant cutting C E in F. Draw F A, cutting C B in G. From G draw G H parallel to A B. And from G and H draw lines G I and H J parallel to C D, which will complete the square in the triangle. To inscribe a Square in a given Trapezium, A B C D. (Fig. 37; Fig. 37- Draw the diagonals A C and B D. Draw D E at right angles and equal to D B. Draw E A, cutting C D in F. c 26 LINEAR DRAWING. Draw F G parallel to A C. Draw G H and F I parallel to D B. Join H I, which will complete the square in the trapezium. To inscribe a Circle in a given Trapezium, ABC D, of which the adjacent sides are equal. (Pig. 38.) Fig. 38. Draw the diagonal A B, which " will bisect the angles C B D, and CAD. Bisect the angle A D B. Produce the bisecting line until it cuts A B in O. Then O is the centre from which a circle may be described, touching all four sides of the trapezium. To trisect* a Right Angle, ABC. (Fig. 39.) Fig. 39- From B, with any radius, describe the quadrant D E, * Trisect. To cut into three equal parts. LINEAR DRAWING. 27 From D, with the radius D B, describe an arc cutting E D in F. From E, with the same radius, describe an arc cutting E D in G. Draw Hnes B F and B G, which will trisect the right angle. The Measurement of Angles. (Fig. 40.) D Fig. 40. Angles are estimated according to the position which the two lines of which they are formed occupy as radii of a circle. The circle being divided into 360 equal parts, called " degrees," it will be evident that the lines A, O, C, con- tain 90 degrees (written 90°) or a right angle. Similarly B O C is a right angle. Now, if these right angles oe trisected (as per last problem), each of the divisions will contain 30°, thus :- A O E is an angle of 30° A O F „ „ 60° A O C „ „ 90° A O G „ „ 120° A O H 150° 28 LINEAR DRAWING. A O B is in reality not any angle at all, being a per- fectly straight line ; but the slightest divergence from it would cause it to become an angle ; as 179°, &c. Each of these angles being again divided into three parts will give tens, which may again be divided into units ; and thus angles may be constructed or measured with the greatest accuracy. Example No. 1 of the foregoing. (Fig. 41.) Fig. To find the angle contained by the lines ABC. Erect a perpendicular at B. Draw the quadrant D E, and trisect it. Divide the arc G E into three equal parts by points H and I. (70° and 80°.) Bisect the arc H I, and it will be seen that the line B C falls precisely on the bisecting point. A B C is therefore an angle of 75°. Had the line B C not fallen exactly in the bisecting point, further subdivision would have been necessary. LINEAR DRAWING. Example No. 2. (Fig. 42.) 29 l-'S- 42. To construct at a given point B an angle of a required number of degrees, say 100° . At B erect a perpendicular, B C. Trisect the right angle, carrying on the arc beyond the perpendicular, C. Divide any one of the three divisions into three equal parts representing tens. Set off one of these tens beyond C, viz., to D. Draw B D. Then A B D will be an angle of 100°. To construct a Triangle, when the length of the base and the angles at the base are given. (Fig. 43-) 1 r t^'K 43- Let it be required that the base should be 2*5 (2 deci- 30 LINEAR DRAWING. nial 5, or 2 and 5 tenths^ which is 2\) inches long, that the angle at A should be 50°, and that at B 45°. Draw the base 2'5 inches long. At A erect a perpendicular ; draw a quadrant and trisect it in E D. Divide the middle portion, D E, into three equal parts, and the second division from E will be 50°. Draw a line from A through point 50 and produce it. At B erect a perpendicular, and bisect the right angle thus formed (as 45° is one-half of 90°). Produce the bisecting line until it meets the line of the opposite angle in F. Then A B F will be the required triangle. Note.— All the three angles of a triangle are always equal to two right angles, that is i8o°, and therefore, as one of the above angles is 50°, and the other 45° — total 95° — the vertical angle, that is, that opposite the base, will be 85°. The Protractor. (Fig. 44.) For measuring and constructing angles, there is, in most cases of mathematical instruments, a brass semi- circle called a Protractor. This has a short line marked at C, and two rows of figures round the rim— the one reading from right to left, and the other the reverse way. In order to measure an angle by means of the pro- tractor, place the edge A B on the straight line which is to form one of the sides of the angle, with the point C exactly against the point of the angle to be measured. Then the line C D will be seen to correspond with the point 60°, and B C D is therefore an angle of 60° ; or, reading from the left side, A C D is an angle of 1 20°. In constructing an angle, place C against the point at which it is desired to construct an angle ; mark a point on your paper exactly against the figure corresponding to the number of degrees required ; remove the protractor, and draw a line through the point thus obtained, to C, which will give the desired angle. Protractors are sometimes made of wood or ivory, and of a rectangular form, as E F. These are used in a manner precisely similar to the semicircular instruments, but are not generally thought as useful or exact in practice. LINEAR DRAWING. 31 32 LINEAR DRAWING. To construct an Isosceles Triangle on a given base, and having a given vertical angle (say 30°). (Fig. 45-) Before commencing to work this figure, it is desirable that attention should be called to the principle upon which the construction is based. It has been shown (page 30) that all the angles of a tri- angle, of whatever shape it may be, will always be equal to two right angles (viz., 180°). Every straight line then is equal to the bases of two right angles ; for a perpendicular drawn at any point will at once form two right angles, equal to 180°, upon it. LINEAR DRAWING. 33 Now let it be supposed that ;^t8o are to be divided between three persons — that one of them is to receive ^30, and the remainder to be equally divided by the other two. It will be seen at once that, when the first condition has been fulfilled, and ^30 deducted from ^180, the remainder will be ^150, or ^^75 for each of the remaining claimants. It is on a similar principle that this operation is based ; and this mode of procedure is rendered necessary be- cause we cannot commence by constructing the vertical angle — for, as the base A B is fixed, we should not know where to commence the vertical angle, so that the sides Fig. 2. might not cut through A B (Fig. i), or pass beyond i1 (Fig, 2), and thus we are compelled to construct the angles at the base firstly, and of such a number of de- grees, that they should meet in the required angle. Now it has been shown that 180° stand on every line. Produce A B (Fig. 45), and at A construct an angle of 30''— viz., CAD. So that out of the whole sum of 180° we have set aside 30°, the fixed number. Bisect the remaining angle D A B in E. Draw A E. At B construct an angle A B F, similar to the angle B AE. Produce lines A E and B F, which will meet in G, and will form the required angle of 30°. 34 LINEAR DRAWING. To construct an Isosceles Triangle when the vertical angle is given in lines and not in de- grees. (Fig. 46.) Let A B be the given base, and C the given vertical angle. Produce A B towards D. At A construct the angle DAE, similar to the angle C (the given vertical angle). Bisect the angle E A B in F. At B construct an angle similar to the angle F A B — viz., angle G B A. Produce A F and B G until they meet in H. Then the vertical angle at H will be similar to the given angle C. Within the given Square, A B C D, to inscribe the largest Equilateral Triangle it will contain. (Fig. 47.) Trisect the right angle D A B. Bisect the angles E A F and G A H by the lines A I and A J. Join I J. Then — A I J is the largest equilateral triangle that can be con- tained in the square A B C D. The principle on which this construction is based, is, that as the angle of the square is 90°, and that of the LINEAR DRAWING, 35 equilateral triangle is 60°, there is an overplus of 30°. If, then, the two outer angles (E A F and G A H) which are each 30°, are bisected, and half of each added to the c I angle F' A G (30°) an angle of 60° is obtained centrailv placed, leaving 15° on each side. It will be seen that the sides of the equilateral triangle are larger than those of the containing square. To construct an Equilateral Triangle of the given altitude (or height) A B. (Fig. 48.) At A and B draw lines C D and E F at right angles to A B. 36 LINEAR DRAWING. From A, with any radius, describe the semicircle G H From G and H, with radius A G, cut the semicircle in I and J. From A draw lines through I and J, cutting E F in K and L. A K L will be the equilateral triangle of the required altitude. To draw a Tangent* to a circle at a given point, C. (Fig. 49.) Fig. 49. (i) Draw a radius from the centre O to the point C At C construct a right angle, O C D. Then D C is the required tangent. * A tangent is a straight line which touches a circle at one point, but does not cut off any portion of the circumference. A tangent is always at nght angles to the radius drawn from the point at which it touches. LINEAR DRAWING yi Or (2) let E be the given point. Draw radius O E and produce until E F equals E O. Bisect F O by the line G H, which will be the tangent required. To construct an Eqiiilateral Triangle about a given circle. (Fig. 50.). From any point in the circle, as A, with a radius equal to the radius of the circle, describe an arc cutting the periphery * in B and C. From B and C, with radius B C, cut the periphery in D. (It will be seen that if B C, B D, and D C are joined, an equilateral triangle will be inscribed in the circle). From B and C, with radius B C, describe arcs cutting each other in E. From B and D, with the same radius, describe arcs cutting each other in F. From D and C, with same radius, describe arcs cutting each other in G. • Periphery. The circumference, or boundary line, of a circle, ellipse, or any other regulai curvilineal figure. 3^ LINEAR DRAWING. Join F G, F E, and G E, which will complete the triangle about the circle. It will be seen that by this problem an equilateral triangle may be constructed about another, the whole consisting of four equilateral triangles. WITHIN a given circle, to inscribe a triangle similar to a given triangle, ABC. (Fig. 51.) Fig. 51. Draw a tangent to the circle. From the tangent point D, with any radius, describe a semicircle cutting the tangent in E and F. At A and B, with radius D E, describe arcs cutting the sides of the triangle in I J and G H. From the point F, mark on the semicircle the length of the arc G H— viz., F K. From E, mark on the semicircle the length of the arc I J — viz., E L. LINEAR DRAWING. l^ From D, draw a line through L cutting the circle in M. From D, draw a line through K cutting the circle in N. Draw M N, which will complete the triangle in the circle. (The various steps have here been given in detail in order to adapt the process for self-instruction, or for junior pupils ; otherwise it will be seen that the most brief mode of explaining the figure would have been : " Construct on one side of D an angle similar to A B C, and on the other an angle similar to B A C.) ABOUT a given Circle, to construct a Triangle similar to A B C, (Fig. 52.) Fig. 52. Produce the base of the original triangle ABC towards F and I. D 40 LINEAR DRAWING. Draw any radius in the circle, as O D. At O draw a line which shall make with the line D O an angle similar to that which B C makes with F B— viz., the angle DOG. In a similar manner construct at O the angle D O H, similar to the angle I A C. There will thus be three radii in the circle — viz., O D, OG, andOH. Draw tangents (viz., lines at right angles) to each of these radii, and these tangents meeting in J, K, and L, will form a triangle about the circle similar to A B C. Definitions concerning Polygons. All figures having more than four sides are called Polygons, and are distinguished by names denoting the number of their sides and angles— thus : A Polygon of 5 sides is called a Pentagon. „ 6 „ „ Hexagon. Heptagon. „ „ Octagon. „ „ Nonagon. ,, „ Decagon. „ „ Undecagon. Duodecagon. When all the sides of a polygon are equal, and all its angles equal, it is called " regular." When they are not equal, the polygon is said to be irregular. By drawing lines from the angles of a regular polygon to the centre, the figure may be divided into as many triangles as the polygon has sides. In the regular hexagon these triangles will be equilateral^ but in all other regular polygons they will be isosceles. r? 7 « 8 jj 9 « 10 » II ») 1? LINEAR DRAWING. 41 To construct a regular Polygon— in this case a pentagon— on the given line A B. (Fig. 53.) f'ig- 53- Polygons may be constructed by ^\\\\qx general methods — that is, by rules, which by simple variation as to division of parts will apply equally to all polygons — or by special rules which apply to particular figures only. The above mode of constructing a pentagon on a given line is a general one.* Produce A B on each side. From A, with radius A B, describe a semicircle cutting A B produced in C. Divide the semicircle into 5 equal parts. From A, draw A D to the second division. From B, with radius B A, describe a semicircle cutti"^ A B produced in E. From E, mark on this semicircle the length of the arc C D— viz., to F. From D and F, with radius A B, describe arcs cutting each other in G. Draw D G and F G, which will complete the pentagon on AB. * That is, any other polygon may be thus constructed. To constnict a heptagon by this method, divide the semicircle into 7 equal parts : for an ortagon, into 8, and so on ; but it must be remembered that, whatever may be the number of parts, the line A D must always be drawn to the second division. D 2 42 LINEAR DRAWING. To construct a regular Pentagon on the given line A B, by a special method. (Fig. 54.) Fig- 54- From A and B, with radius A B, describe arcs cutting each other in C and D. Draw a line through C and D perpendicular to A B. From C, with radius C A (equal to A B), describe an arc cutting the perpendicular in E, and also cutting the two previously drawn circles in F and G. Draw lines from F and G, passing through the point E, and cutting the two circles in H and I. Draw A I and B H. From H and I, with radius equal to the side of the pentagon (viz., A B, A I, or B H), describe arcs cutting each other in J. Draw H J and I J, which will complete the pentagon. LINEAR DRAWING. 43 To inscribe a regular Polygon— in this ease a pentagon— in a given circle. (Fig. 55.) Fig. 55- Draw the diameter A B, and divide it into as many equal parts as the polygon is to have sides (in this case five). From A and B, with radius A B, describe arcs cutting each other in C. From C, draw a line passing through the second division and cutting the circle in D. Draw D B, which will be one side of the polygon. Set off the length D B around the circle — viz., E F G. Join these points, and thus complete the figure. Any polygon may be thus formed, by dividing the diameter into the number of parts corresponding with the sides of the required polygon j but the line C D must, in every case, be drawn through the secoitd division. 44 LINEAR DRAWING. To inscrit)e a regular Pentagon in a circle, by a special method. ^ (Fig. 56.) Fig. 56. Draw the diameter A B. At O erect a perpendicular, O C. Bisect O A in the point D. From D, with radius D C, describe an arc cutting A B in E. From C, with radius C E, describe an arc cutting the circle in F. Draw C F, which will be one side of the pentagon. Set off the length C F around the circle — viz., GH I. Draw lines F G, G H, H I, and I C, which will complete the figure. LINEAK DRAWING. 45 Application of the foregoing principle in the construction of Grothic tracery. (Fig. 57.) Fig- 57- Draw a circle, divide it into five equal parts, and draw the radii, A B C D E. Bisect one of the radii, and set off the half on each of them-viz., F G H I J. Join these points, and a regular pentagon will be formed. Bisect the sides of this pentagon, by the lines K L M N P. Draw a small circle in the centre, and another, Q, con- centric with it. From Q to the sides of the pentagon draw lines parallel 46 LINEAR DRAWING. to O K, O L, &c., at a small dif,tance on each side ol them — viz., R S, T U, &c. Produce the sides of the pentagon indefinitely from F G H I J, and with radius H U, describe circles cutting the produced sides of the pentagon in V W and the corresponding points. Draw V C, W C, and similar lines from the other circles, and the remaining lines will be parallel to, and concentric with, those already drawn. About the given Pentagon A B C D E, to de- scribe a pentagon whose sides shall be parallel to it, and equal to the line F G. (Fig. 58.) K Fig. 58. Find the centre of the pentagon by bisecting two adja- cent angles. Draw the five radii, and produce them indefinitely. Produce one of the sides, as B C, until ic is equal to F G— viz., B H. LINEAR DRAWING. 47 From H draw a line parallel to O B, cutting the radius O C in I. From O, with radius O I, describe a circle cutting the produced radii in K L M N. Draw K L, L M, M N, N I, I K, which will complete the pentagon of the required size, described about the given pentagon. To construct a regular Hexagon on the given line A B. (Fig. 59.) Fig. 59. From A and B describe arcs cutting each other in O. From O, with radius O A, or O B, describe a circle. The radius with which a circle is struck will divide it into six equal parts, therefore set off the length O A, ivhich is equal to A B, around the circle — viz., A C E F D. Join A C, C E, E F, F D, and D B, and a regular hexagon will be formed. To inscribe a regular Hexagon in a Circle. Find the centre of the circle (figure 33), set off the radius around it, and join the points. 48 LINEAR DRAWING. Example 1 of inscribing a Hexagon in a Circle. —To draw a simple Fly-wheel. (Fig. 60.) Fig. 60. Draw the circles A and B, representing the outer and inner edge of the rim. Divide the circle B into six equal parts, and draw the radius C D E F G H. Next draw the circles I and J, representing the end of the shaft and the boss, or central part of the wheel ; the small parallelogram at the side of the inner circle represents the " key,' by which the wheel is held on the shaft. On the edge of ihe boss set off equal distances, K L. Draw the circle M, and on it, on each side of the radii, set off distances rather less than K and L — viz., N and O. Draw the sides of the arms, K N and L O, &c. With any convenient radius describe the small arcs connecting the arms with the rim at N and O. The length P O set off from P and Q on the radius, LINEAR DRAWING. 49 will give the point R, which is the centre for striking the arc, caused by the elHptical arm meeting (called pene- trating) the elliptical rim. Example 2 of the application of the Hexagon in Mechanical drawing. (Fig. 6i.) In the above drawing of a nut and bolt, the plan, that is, the appearance it would have if your eye were directly over it, and you looked down upon it, is to be drawn firstly. The two largest circles being described, the inner one is to be divided into six equal parts, and a hexagon inscribed in it. Perpendiculars drawn from each of the angles of the hexagon will give the projection of the widths of the sides of the nut. 50 LINEAR DRAWING. The rest of the figure may be copi-^d without further instructions, and the whole subject of projection, of eleva- tions, &c., from given plans, &c., will be fully treated of in the subsequent volume. To corstiruct a regular Heptagon on the given line A B. (Fig. 62.) Fig. 62. Erect a perpendicular at B, and draw the quadrant A C. Divide the quadrant into seven equal parts. Continue the arc A C beyond C, and set off on it from C, three of the divisions— viz., to D. LINEAR DRAWING. 51 Draw B D. Bisect A B and B D, and from the intersection O of the bisecting lines, with radius O A or O B, describe a circle. From A and B set off the length A B around the circle -viz., E F G H. Draw D E, E F, F G, G H, and H A, which will complete the heptagon. To construct a pentagon on this principle, divide the quadrant intoy??/^ parts, and set off one beyond C. For a hexagon, divide the quadrant into six, and set off ttuo beyond C. For an octagon, divide into eight, and set off four beyond E, &c. To inscribe a regular Heptagon in a given circle. (Fig. 63.) Fig. 63. From any point, as A, with the radius of the circle; describe arcs cutting the circumference in B and C. 53 LINEAR DRAWING. Draw the line B C and the radius A O, which will bisect B C in D. From A set off the length D B around the circle, join the points A E F G H 1 J, and the heptagon will be completed. It will be seen that B C is one side of the equilateral triangle, which could be inscribed in the circle, and thus as D B is half of B C, half of the side of the inscribed equilateral triangle gives the side of a regular heptagon. which can be inscribed in the same circle. To construct a regular Octagon on the given line A B. (Fig. 64.) Fig. 64. Produce A B on each side. Erect perpendiculars at A and B. From A and B, with radius A B, describe the quadrants C D and E F. LINEAR DRAWING. 53 Bisect these quadrants, then A G and B H will be two more sides of the octagon. At H and G draw perpendiculars, G I and H K, equal to AB. Draw G H and I K. Make the perpendiculars A and B equal to G H or I K— viz., A L and B M. Draw I L, L M, and M K, which will complete the octaofon. To inscribe an Octagon in the square A B C D (Fig. 65.) Fig. 65. Draw diagonals, A D and C B, intersecting each other in O. From ABC and D, with radius equal to A O, describe 54 LINEAR DRAWING. quadrants cutting the sides of the square in E F G H I J KL. Join these points, and an octagon will be inscribed in the square. To inscribe an Octagon in a given Circle. (Fig. 66.) Fig. 66. Draw the diameter A B, and bisect it by C D. Bisect the quadrants A C, C B, A D, and B D, in the points E 1 G H. Draw lines connecting all the eight points, which will complete the required octagon. As all other polygons may be constructed on the prin- LINEAR DRAIVING. 55^ ciples already shown, it will be unnecessary to give further examples of them. To inscribe an Equilateral Triangle in a re- gular Pentagon, A B C E D. (Fig. 67.) Fig. 67. From E, with any radius, describe a semicircle, F G. From F and G, with the same radius, describe arcs cutting the semicircle in H and 1. (The radius with which a semicircle is struck, divides it into three equal parts). From E draw a line through H and I, cutting the sides of the pentagon in J K. Draw J K, which wUl complete the equilateral triangle in the pentagon. 56 LINEAR DRAWING. To inscribe a Square in a Regular Pentagon. A B D E C. (Fig. 680 Fig. 68. Draw C D, and C F at right angles, and equal to it. Draw F E, cutting the side of the pentagon in G. Draw G H parallel to C F. And H I parallel to C D. Draw I J parallel to H G. And G J parallel to H I. Then G H I J will be the required square contained in the pentagon. LINEAR DRAWING. 57 To inscribe an Equilateral Triangle in a re- gular Hexagon, A B C D E F, so that its sides shall be parallel to three sides of the hexagon. (Fig. 69.) Fig. 69. Bisect the alternate sides, as E D, F A, and C B of the hexagon in the points G H and I. Join these points, and the hnes will form an equilateral triangle. To inscribe in a Regular Hexagon the largest Equilateral Triangle it will contain. (Fig. 70.) Fig. 70. Draw lines joining the three alternate angles of the hexagon, as A B C, which will form the required triangle. E 2 5^ LINEAR DRAWING. Within the Equilateral Triangie, A B C, to inscribe six equal Circles. (Fig. 71.) Fig. 71. Draw the lines B D, A F, and C E, bisecting the sides and angles of the triangle, and intersecting each other in O. Bisect the angle O A E, and the point (G) where the bisecting line cuts C E, will be the centre of one of the three isosceles triangles, into which the equilateral triangle has been divided. Through G draw H I, parallel to A B, and from H and I draw H J and I J, cutting B D and A F in K and L. From G H I J K and L, with radius G E, draw the six circles. LINEAR DRAWING. 59 To inscribe three equal Circles in a Circle- (Fig. 72.) Fig. 72. At any point, as A, draw a tangent, and A G at right angles to it. From A, with radius O A, cut the circle in B and C. From B and C draw lines through O, cutting the circle in D and E, and the tangent in the point F (and in another not given here, not being required). Bisect the angle at F, and produce the bisecting line until it cuts A G in H. From O, with radius O H, cut the lines D C and E B in I and J. From H 1 and J, with radius H A, draw the three required circles, each of which should touch the other two, and the outer circle. LINEAR DRAWING. To inscribe in an Equilateral Triangle, A B C, the three largest circles it will contain. (Fig. 73.) Fig- 73. Draw A G, B F, and C E, bisecting the angles and sides of the triangle, and intersecting in O. Bisect the right angle A E O. Produce the bisecting line until it cuts A G in H. Draw H I parallel to A B, H J parallel to A C, and I J parallel to B C. From H I ?jid J, with radius H K, draw the three circles, each of which should touch the other two, and two sides of the triangle. LINEAR DRAIVING. 6i To inscribe four equal Circles ia a Circle, each touching two others and the containing Circle (Fig. 74,) Fig- 74- Draw the diameters A B and C D at right angles to each other. From A B C D, with radius of the circle, describe arcs cutting each other in E F G H. Join these points, and a square will be described about the circle. Draw the diagonals E H and G F Bisect the angle C F O, and produce the bisecting line until it cuts C D in I. From O, with radius O I, describe a circle cutting the lines A B and C D in J K and L. From these centres, with radius I C, describe the four required circles. 62 LINEAR DRAWING, To inscribe seven equal Circles in a Circle. (Fig. 75-) Fig. 75. Around the circumference of the circle set off the radius, thus dividing it into six equal parts, A B C D E F, and draw the radii. Divide one of the radii, as O A, into three equal parts —viz., O G, G H, H A. From O, with radius O G, describe the central circle. From O, with radius O H, describe a circle which, cutting the radii, will give the points I J K L M. From these points, with radius O G, describe the six circles, each of which will touch the central circle, two others, and the containing circle. Similarly, a circle O G being given, to draw six equal circles to touch it and each other, divide the given circle LINEAR DRAiriNG. 63 into six equal parts. Draw radii and produce them. From G set off G H, equal to G O. From O, with radius O H, describe a circle which, cutting the produced radii, will give the centres I J K L M of the six circles. Within a Circle to inscribe any number of equal Circles, each touching two others and the containing Circle. (Fig. 76.) Divide the circle into equal sectors, corresponding to the required number of circles — viz., A B C D, &c., and bisect the sectors by the lines E F G, &c. Produce any two of the radii, as A and B, and draw the tangent H I parallel to A B. Bisect one of the angles at the base of the isosceles 64 LINEAR DRAWING. triangle thus formed, and produce the bisecting line until it cuts O F in J. From O, with radius O J, describe a circle cutting each of the lines which bisect the sectors in K L M N, &c. From these points, with radius J F, describe the re- quired circles. By drawing P O parallel to A B, and bisecting the angle at the base of the triangle, the centre for another circle may be found ; and by continuing the process as before, another series of circles may be drawn. Application of the division of a Circle in draw- ing a Back and Trundle. (Fig. 77.) v_y Fig. 77. The circle A, on which the centres of the circles repre- senting sections of the bars (or teeth of the trundle) are placed, is called the ^itch circle; and the line on which are the points of contact between the teeth of the rack and those of the wheel, is called the pitch line. The pitch circle must be divided into the given number LINEAR DRAIVIXG. 65 of teeth, and spaces BCD, and the same lengths E F G must be set off on the pitch line H I. The rest of the construction will be readily understood on reference to the figure. Numerous studies in this branch of the subject will be given in the special " Manual of Mechanical Drawing for Engineers." The above (Fig. 78) is an example of the division cf circles m drawing the Plan and Elevation of a Column, 66 LINEAR DRAWING. and is introduced here in order to impress on students the necessity of acquiring the utmost accuracy in division of spaces. The circle forming the boundary of the plan is to be divided into a number of parts, corresponding to the required number of flutes— viz., i, 2, 3, &c. ; half the width of the fillets is then to be set off on each side of these divisions, as a^ b^ &c., and semicircles drawn from the centres of the remaining spaces. The elevation of the column is projected by drawing perpendiculars from the various points in the plan. For full instruction in "projection" of elevations from plans, &c., see the second portion of this work, " Solid Geometry and Projection," and the various orders of columns will be found in the volume devoted to Architectural Drawing for Masons. To divide a Circle into any number of equal parts, having the same area. (Fig. 79.) Fig. 79. Divide the diameter A B into the required number of equal parts, A C, C D, D E, E F, F B. LINEAR DRAWING. 67 From points a «;, midway between A C and F B, describe semicircles, F B and A C. From point E, describe the semicircle C B. From D, describe the semicircle F A. From b and b^ midway between C D and E F, draw the semicircles E A and D B. From C and F, draw the semicircles D A and E B, which will complete the figure. To divide a Circle into a given number of Concentric Circles, having the same area. (Fig. 80.) Fig. 80. Draw a radius A B, and on it describe a semicircle. Divide the radius A B into the number of equal parts corresponding with the number of circles required. From the points of division, i 2 3, raise perpendiculars cutting the semicircle in i -2 3. 68 LINEAR DRAWING, Through these points (from B as a centre) draw circles: then the belts C D E and the circle F wiU have the same area. The Cone and its Sections. (Fig. 8i.) Fig. 8i. A Cone is a solid, the base of which is a circle, but which tapers to a point from the base upward. If a cone be cut horizontally — that is, parallel to the base — all such sections will be circles ; but if it be cut obliquely across, as at A B, the section or cutting is called an Ellipse — as A B and C D. The line drawn from C to D is called the long diameter (or transverse axis); that across from E is called the short diameter (or conjugate axis). Both ends of an ellipse are equal, and in this its form differs from that of the Oval (see Fig. 95) which is egg- like (from ova^ an ^z^^ tl at is, broader at one end than at the other. LINEAR DRAWING. 09 A perfect Ellipse may be drawn by means of a piece of string and pins— a method which is of great service to masons, joiners, gardeners, &c. (Fig. Z2 ) Fig. Place the given diameters A B and C D at right angles to each other at their centres E. From D, with radius E A, cut the long diameter in F F. These two points, F F, are called the foci''^ of the ellipse. Place a pin in each of these, and another in D. Pass a string round the three pins, and tie it securely, thus forming a triangle of string, F F D. Take out the pin at D, and substitute a pencil, which may be drawn along, moving within the loop, and the point will thus trace a perfect ellipse. * F> bx) c :3 rt o j^ o X" "S CJ c _a "o s_< rt 'u rt U o 1^ Ur X u ^ bflpQ ;^ ■=< (U rt ,^ (U u o "T^ a, *-> c ^ \> -5 'T3 o rt c -5 Oh r3 «« vd^ <=>. •— ' d >ca £S o C O c^ ■J a; p: 'li •Z. o s^ ^ J a (U r^ c -^ '-' "r^I^ ^ p I :. > I ^% s|- r^. - -s ^•^ ^ --2 :^ o -s-^ f.-bb^^ -^ ^ ^ ^ ^^ ^ e '^ -^ "^ ^ %^^ ^% ^ ^ ^ ^1 ^. ^ tJ :^ '^ s ^ g cr .g J a ^ -r rt-;;;oS.> o^-5^^^ r-P^iStli c.S g £ S-cJ-t: So ^oo V' H " ^ o G 2 90 LINEAR DRAWING. t^C^J^/ o ^ « o LINEAR DRAWING. 9I O "^^ n> 2-13 •£ ^" S ID c ■'d o 43 > ° 1^ "^113 ^> ^ •£oa ^ u c, ^ lilt I •-'! "^fePi rl:-ll "°§ 5-2 „ :=°- .^:o s.s^ ^1 111 )X ti Q.> n-( -r* u P 4) X! - •- bi - "^ > 92 LINEAR DRAWING. To trace a Cycloid by mechanical means, (Fig. 103.) Fig. 103. Fasten a rail of wood, or any straight edge to the board. Take a circular piece of wood, and cut a small notch at any point in the edge (A), and fix a small knob or button in the centre (B). The point of a pencil held in the notch whilst rolling the disc along the straight edge by means of the knob, will describe the Cycloid.* In order to prevent the disc slipping as it rolls along, it is advisable to glue a narrow strip of sand-paper round- the edge of the disc and on that of the rail. To find a mean proportional between two given lines at A and B. (Fig. 104.) Join the lines A and B, and at their junction C, erect a perpendicular of indefinite height. Bisect the entire line which has been formed, by uniting the two lines A and B, and from the bisecting point D, with radius D A, describe a semicircle, which will cut the perpendicular C in E. * The Cycloid was invented by Galileo, an eminent mathematician and natural philosopher. He was bom in Pisa in 1564, and died in 1642. LINEAR DRAWING. 93 Then C E will be a mean proportional between A and Fig. 104. B, the application of which will be found in the following figures : — To construct a Square, equal in. area to the rectangle A B C D. (Fig. 105.) ^ — --1 G / '' ^N / / / 1 1 1 1 1 1 1 / / \ \ t « H 1, ■- 1 i / \ / i _^^ Fig. 105. Find a mean proportional between the long and. the 94 LINEAR DRAWING. short side of the rectangle, and this mean proportional will be the required length of the side of the square. The process shown in the last figure is here repeated. Produce C D, until D E equals D B. From D, with radius D B, describe a quadrant cutting C D produced, in E. Erect a perpendicular at D. Bisect C E in F. From F, with radius F C, describe a semicircle cutting the perpendicular D in G. D G will be the length of the side of the required square. From D set off D H. From H and G with radius D H, describe arcs cutting each other in I. Draw H I and G I, which will complete the square. To construct a Rectangle, eqiial in area to the Triangle ABC. (Fig. 106.) E A \ V P r 1 /■^ KX^ \ A D i"^ i \ 1 \ 1 / \ / \ / \ / \ / \ / \ / \ H / out 300 B c. His birthplace is by some said to have been Alexandria We studied geometry in Athens, then went from Greece to Alexandria, where he settled during the time of the first Ptolemy. 98 LIXEAR DRAn-IX.-r. Further, an equilateral triangle, constructed as in Fig. io8, equal to A G F, will also be equal to the pentagon. To construct a Triangle that shall be equal in area to a given Kectilineal figure, as A B C D E. (Fig. no. Fig. no. Draw B D, thus cutting off the angle C. Draw C F parallel to D B. Draw D F. The original irregular pentagon has now been reduced to the /^//;'-sided figure A F D E. Draw D A and E G parallel to it. Join D G, and the triangle G D F will be equal in area to A B C D E. By problem io6, the triangle may be converted into a rectangle, and into a square; and by problem 108, into an equilateral triangle, each of which will be equal to the original figure. LINEAR DRAWING. 99 To construct a T^'iangle equal in area to a given Rectilineal figure (continued). (Fig. in.) Fig. III. Let A B C D E F be the given figure, and let it be required that one angle of the triangle shall coincide with A. r^rom A, draw lines to the other angles of the polygon. Produce D E. Draw F G parallel to A E. Produce C D. From G, draw a line parallel to A D, cutting C D pro- duced in H. Produce B C. From H, draw a line parallel to A C, cutting B C pro- duced in I. Draw A I, then the triangle B A I is equal in arc-a to the original polygon, and the side A B, and the angle B, are common to both. As by this figure, and those referred to in the last, irregular forms may be rectified and converted into equilateral triangles, rectangles, and squares, they are of ^reat importance to land-surveyors, architects, (Sec. lOO LINEAR DRAWING. To construct a Square equal in area to a given quadrilateral flgiire, A B C D. (Fig. 112.) B E y^i — i~i" -~ _. / ^•^ ^^ / jr^ vV y X ^ k V X X ^^\ \ / / ^^X^ N / j/ ^X^^\ \ / / ^V ^\^ \ / / ^X ^^^\ "^ " A / / >y ^\}c y Nr^ y\ > ^\ J ) / \iH X \ ° / /' \ ^v \ / X 1 \ X. \ / X ' \ \^ \ / X / '\ \ \ / / v. ^v \ jf ^ ^r / >■ - XOI \-J Draw the diagonal A C, and bisect it by a per- pendicular. From B and D, draw lines parallel to A C, and cutting the perpendicular in E and F. Draw a line bisecting E F in O, and from A draw a line parallel to E F, and cutting this bisecting line in G. Find a mean proportional between O E and O G — viz., O H. Set off the length O H (the semi-diagonal) from O on E F and O G— viz., I J K. Join. H I J K, and the square will be equal to the quadrilateral figure ABC D. LINEAR DRAl^I^Q. lOI To construct a Square, which shall be equal in area to two other squarjea adde.a, to^etnsr. (Fig. 113.) ' ' ' ' '^ ' ■ ^ ' ""• E F B G Fix- "3- Place the two squares so that a side of the one, as A B, shall be at right angles to one side of the other, as B E. Draw the line A E. Now, according to Euclid (Book I., prop, xlvii.), " In any right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle." And it will be seen that A B E is a right-angled triangle, and that the squares A B C D and B E F G This proposition is said to ha\'e been discovered by Pythagoras, a disciple ofTTialcs. who, after travelling in India and Egypt in pursuit of knowledge, settled in Tarentum. in luly, where he founded the celebrated Pythagorean fcchool, 550 years B.c I02 ^«r^^4/? DRAWING. are described upon the sid/ss-of it which contain the right angles land therrforc th^ square A E H I, which is de- scribed on (the hypothenuse) A E, which subtends the right angle, is equal to the sum oi the two other squares. To construct a Square equal in area to any number of squares added together, (Fig. 114.) This is done by merely carrying on the process shown in the last figure. Let it be required to construct a square, which shall be equal to the areas of the three squares of which A, B and C are the respective sides. Place B at right angles to A, then the hypothenuse D would be the side of the square equal in area to the squares constructed on A and B. Place C at right angles to D, draw E F, and construct a square upon it ; then E F H G is equal to the squares con- structed on C and D, i-nd therefore equal to the squares constructed on all three lines. Any number of squares may be thus added together. LINEAR DRAWING. ion To divide it given Triangle, ABC, into two equal parts by a line parallel to ore of its sides. (Fig. 115.) Bisect one of the sides, as C B, in the point D, and erect the perpendicular D E equal to D C. From C, with radius C E, describe an arc cutting C D in F. From F, draw F G parallel to A B, which will divide the triangle into two parts of equal area. To divide a Triangle into two equal parts by a line perpendicular to one side. From C, draw C H perpendicular to A B. Bisect A B in I. Find a mean pr«^portional between B H and B I — viz., BJ. From B, set off B K equal to B J, and the perpendicular K L will divide the triangle as required. H I04 LINEAR DRAWING. To divide the space contained between the lines A B and C D, into equal parts, by means of lines parallel to A B. (Fig. 1 16.) /» — ^Ncx" 1, ■"^^^T^^ /Nos*- '2- -^^ /^p X y/ / k rr.^ - 3- -^ / r^y ^ . / n Vi" c*- L— y / ^^^ 5_ 6- 7_ 8 _ / y / Fig. ii6. Draw the line E F perpendicular to A B, and set off on it equal lengths corresponding to the number of spaces into which A B C D is to be divided — viz., i to 8. These spaces may be atiy size, but must be equal. From E, with radius E 8, describe an arc cutting C D in G. Draw E G. From E, with radius E 8, E 7, E 6, &c., describe arcs cutting EGinHIJKLMN. Draw lines parallel to A B through these points, and the space will be divided as required. LINEAR DRAWING. 105 To draw a Circle of a given radius, which shall touch another given circle and a straight line. (Fig. 117.) H Fig. 117. Let A be the given Circle, B C the straight line, and D E the radius of the required circle. The question here is, to find a point which shall be the centre of a circle of a given radius, which shall touch the given circle and straight line. From O, the centre of the given circle, draw a radius and produce it. From the periphery of the circle, and on this radius, set off F G, equal to D E. From O, with radius O G, describe an arc. At any point, as H, in B C, draw a perpendicular, H I, equal to D E. From I draw a line parallel to B C, cutting the arc drawn from O in J. H 2 lo6 LINEAR DRAWING. From J, with the required radius, describe a circle, which (if the work has been accurately done) will touch the given circle and straight line. To draw a Circle of a given radius D. E, which shall touch both lines of an angle, A B C. (Fig. 1 1 8,) Fig. iiS Bisect the angle by the line B F. On either of the lines of the angle erect a perpendicular equal to the given radiuF D E — viz., d e. From ^, draw a line parallel to B C, cutting the bisecting line in O. From O, with the given radius, draw the circle, which «vill touch both the lines of the angle. LINEAR DRAWING. \cri To draw a Circle which shall touch both lines of an angle, and shall pass through a given point P. (Fig. 119.) Fig. 119. Let A B C be the given angle, and P the given point, through which the required circle is to pass. Bisect the angle A B C by the line B D. From any point in B D, as E, draw a circle, touching both lines forming the angle. From B, draw a line through P, cutting this circle in F. Join F to E, the centre of the circle. From P, draw a line parallel to F E, cutting B D in G- From G, draw a line perpendicular to A B (by Fig. 6) — viz., G H. Then, with radius G H, which will be found to be equal to G P, describe a circle which will touch both lines forming the angle. io8 LINEAR DRAWING, u o o Q W >> ,0 u < c Pu « .5? 'O LINEAR LRAlVIdrC JO? U c5 X .s « Q 3 M ^ c X Z w d ^^ .s Q J ^- ^ ^ 6? p5 c t=^'~ .S .S g S •- 1 w 2 « I WW i2 .1 W cT -S g^ I 2 .S .S bJ i s 1 1 J 1 5 1 J Vi I ^ § t? d^ p *< -G < ^ < Q *H g >< « ^ s « s « . « « « " I o -I ^ I ^- e- ^ i d I a I OJ tn Si 3 rt c/) 1 s I R I I I -2 I ^ I J -I I :< I -I s I - -s - '^ - -^ -S 'M ■fiW d X rt -^ y" J S Z O" oT C OS I GS i I § i I 1 I 1 2 2 2 2 I ho LINEAR DRA WING. To drssv a Circle of a gi-vren radius, to touch two given circles, 1 and 2. (Fig. 121.) Fig. Draw any radius in each circle Ci D, and C2 D, and produce them. On these radii, beyond the circles, add to each the radius of the required circle — viz., D Ei and D E2. From Ci, with radius Ci Ei, and from C2 with radius C2 E2, describe arcs cutting each other in F. From F, with radius E D, describe the required circle, which will touch both circles. If the required circle is to include both circles, draw any radius in each, as Ci L, C2 M. Produce both these radii. On the radius Ci, set off from L the radius of the required circle — viz., to X. Diminish this by the radius of circle 2 — viz., to G. On radius M C2 set off L G— viz., M H. LINEAR DRAWING. Ill From centres Ci and C2, with radius Ci G, and C2 H, describe arcs cutting each other in J. From J, draw a line through Ci to K. With radius J K, describe the enclosing circle, which will touch circles i and 2. The Conchoid. The Conchoid is a curve which always approaches a straight line, but never reaches it, however far the curve and straight line may be produced. The straight line A B is called the asymptote, C D the diameter, and P the pole. The asymptote A B, pole P, and diameter C being given, draw C P at right angles ♦^^o A B. On each side of D, set off any number of equal parts, 1234567. From P, draw lines passing through these points. From I 2 3, &c., with radius D C, describe arcs cutting these lines \n a b c d, &c., and through these intersections trace the curve. The curve above the asymptote is called the superior Conchoid. By setting off the same lengths under the line the inferior Conchoid is obtained. The Conchoid has been used in architecture in drawing the slightly curved line which forms the profile or side of columns, called the Entasis. The Conchoid was invented by Nicomedes about A.D. 4sa .112 LINEAR DRAWING. fcJO ' (U o O w Q Q -a re (/) ^ •*: o p ^ "^ tc 1=^ c w - fe 2Hc/2c/3Pu LINEAR DRAWING. "3 114 LINEAR DRAWING. tmm LINEAR DRAWING. "5 2l6 LINEAR DRAWING. LINEAR DRAWING, ii8 LINEAR DRAWING. Printed by Cassell & COMPANY, Limited, La Belle Sauvage, London, E.C ic VJi CASSELrS TECHNICAL MANUALS ORTHOGRAPHIC AND ISOMETRICAL PROJECTION DEVELOPMENT OF SURFACES AND PENETRATION OF SOLIDS WITH ADDITIONAL CHAPTER AND PLATES ILLUSTRATIVE OF TRACES, NORMALS, TANGENT-POINTS, AND TANGENT-PLANES bV ELLIS A. DAVIDSON AUTHOR OF " LINEAR DRAWING AND PRACTICAL GEOMETRY," '* MODEL DRAWING," " DRAWING FOR CARPENTERS AND JOINERS," ETC. TWENTY-EIGHTH THOUSAND^ REVISED AND ENLARGED CASSELL AND COMPANY, Limited LONDON, PARIS, NEW YORK b' MELBOURNE First Edition December 1868. Reprinted February 1869, 1870, 1871, 1873, 1875, 1878, 1881, 1882, 1883, February 1884, September 1884, 1886, 1887, 1888, 1889, 1890, 1892, 1893, 1894, 1897, 1899. CLASSIFIED CONTENTS. Projection, the difference between — INTRODUCTION. Orthographic ~ Isometric Perspective ^ Plain Hints on Linear Drawing. The T-square, how to use Parallel Rules and Set-squares . Inking Drawings PLATB FIG. MATHEMATICAL INSTRUMENTS, &c. The Compass, description of its parts Horn Centres How to work with Compasses ... The Lengthening-bar The Bow-pencil and Bow-pen ... Spring-bows, description and use of The Drawing-pen Rules, their bevelled edges French Curves Freehand Drawing Pencils ELEMENTARY PRINCIPLES. The Planes of Projection To draw the plan and elevation of a vertical line To draw plan and elevation of a line at right angles \ to the vertical plane S To draw plan and elevation of a line parallel to | both planes ... , ... } The Intersecting Line, Its meaning ... ... ... Position of plan and elevation in drawings explained To draw a line at a simple angle— viz., inclined to the horizontal and parallel to the vertical plane... To draw a line at a "compound" angle— viz., in- clined to both planes To find the real length of a line from its projection... To draw a line parallel to the horizontal and in- dined to the vertical plane ... ^ .M M. I I 21 ' (No. IX. ' (No. 2) as ' fNa 3) I 2 z 2 23 23 2 2 '^ 2 3 2 3 26 26 «? IT CLASSIFIED CONTENTS. PLATE FIG. PAGE THE PROJECTION OF PLANES OR SURFACES. To project a plane when standing on its edge, its surface being parallel to the vertical plane ... 3 i 27 To project the same plane when at right angles, or any other angle, to the vertical plane 3 2 27 To project the plane when inclined to the hori- zontal plane 3 3 27 To project the same plane wh^n inclined to both planes of projection 3 4 29 Open door, as example of foregoing lesson 4 . — 29 Open trap-door as another example •••. 4 ~~ ^9 To project a plane square when resting on one angle, its surface being inclined to the hori- zontal plane ;• , ... 6 I 29 ,, 1, »» when inclined to both planes of projection 6 2 33 „ „ „ „ front view ... 6 3 33 SOLIDS. Definition of Cubes and Prisms .. ._. -^ — 3^ To project a cube when its axis is vertical 7 i 34 „ „ at an angle to the horizontal plane 7 ? 34 ,, „ at an angle to both planes 7 3 36 DEVELOPMENT. Definition of tefm — — 3^ To develop a Cube ... 7 4 3^ To project a square prism in three positions 8 a, 2, & 3 SECTIONS. 37 Section of a square prism from diagonal to opposite corner of other end 9 ' 39 „ ,. ,, from any angle of one end to the opposite angle of the other ... ... 9 2 41 „ ,, ,, from a line connecting middle points ot two adjacent sides, to similar points in the bottom of the two opposite sides... 9 3 41 To project r. square prism when cut so that the two parts may form a right-angled elbow 10 2 41 To develop the covering of same ... ... ... 10 3 41 To project a similar prism or pipe when two of its faces are parallel to the vertical plane 11 2 43 To develop one portion of same so as to form coal- scuttle II 3 & 4 43 PROJECTION ON THE INCLINED PLANE. To project a Cube on the inclined plane 12 "^ 43 To project a Table on the inclined plane ... ... I3 — 4' CLASSIFIED CONTENTS. v PLATE FIG. PAGE SIDE OR END ELEVATIONS. Principles explained by diagram... 14 1&2 47 To project a Triangular Prism at right angles to the vertical plane ... 15 i 50 „ ,, ,, ,, „ when its axis is inclined to the vertical plane ... ... 15 2 50 ,, „ ,, „ when its axis is parallel to the vertical plane, but inclined to the horizontal ... 15 3 50 ,, „ ,, „ when its axis is inclined to both planes ... ... To make a projection of four such prisms, as in a roof ,, ,, ,, when placed angularly ... Development of one portion of same THE PROJECTION OF POLYGONS, 15 4 so 16 I 50 16 2 50 16 3 50 NS. 17 1&2 53 17 3&4 55 18 1&2 55 18 3 57 18 4 57 To project a plane Pentagon „ ,, Hexagon To project a Pentagonal Prism placed vertically „ ,, ,, when its axis is inclined to the horizontal plane ,, „ when inclined to both planes ... To project a Hexagonal Prism in three diflFerent positions, as per last plate 19 i, 2, & 3 57 PYRAMIDS. Definitions — — 57 To project a Tetrahedron— that is, a pyramid formed of four equilateral triangles ... 20 i 59 To project a Square Pyramid when the edges of the base are parallel to the vertical plane, and when its axis is vertical 20 2 59 The same when resting on one angle of the base, its axis being oblique to the horizontal plane To project a section of this object on a given line ... To develop this object, and mark on the development the line of section To project a Hexagonal Pyramid when standing on Us base ,, ,, ,, when lying on one of its faces, its axis being parallel to the ver- tical plane ... ,, ,, ,, when its axis is at a given angle to the vertical plane To draw a section on a given line To draw the development of this Pyramid, and mark on it the section-line 21 4 63 CIRCLES AND CYLINDERS. To draw the plan of a circle when the circumference is parallel to the vertical plane ... 22 i 64 To draw the elevation of a Circle when its circum- ference is vertical, and at right angles, or any other angle, to the vertical plane ... ^ ... 2» 2 64 20 20 3 3 1* 61 20 4 61 21 I 61 21 2 63 21 21 3 5 ' 63 63 22 4 66 23 I 67 23 2 67 23 3 69 24 I 69 24 2 69 24 3 69 24 4 71 25 I 71 25 2 73 25 3 73 vi CLASSIFIED CONTENTS. PLATE FIG. To draw the plan and elevation of a circular disc, when its plane is at any angle to the horizontal and at right angles to the vertical plane 22 3 * ,» M » >' when the plane is obliquely placed in regard to both planes of projection ... ... To project a Cylinder when its axis is vertical „ „ when its axis is parallel to the vertical and at an angle to the horizontal plane .. „ „ when its axis is incluied to both planes To draw a Cylinder cut to form an elbow-joint To draw the true section on a given line To develop the Cylinder and section-line To project one of the parts To project a Cylinder cut so as to form a Double Elbow ... To develop same with section-line To draw one of the patts when lying on its section ... To project a series of Circular Blocks, as in a Speed- Pulley 26 — 74 THE PROJECTION OF CONES. Definitions — — 74 To draw a Cone of a given altitude and diameter ... 27 i 76 To project a Cone when lying on the horizontal, its axis being parallel to the vertical plane 27 2 76 To project a Cone when resting on the extremity of one diameter of the base, its axis being still parallel to the vertical plane 27 3 76 ,, ,, when resting on the extremity of a diagonal of the base, its axis being in- clined to both planes 27 4 78 SECTIONS OF CONES. Difference between an Ellipse and an Oval — — 79 To draw the Ellipse (viz., a section of a cone which cuts both slanting edges of the elevation and is oblique to the base and axis) 28 i 79 To develop the Cone, and to mark on it the section- line 28 2 81 To draw the Parabola — that is, a section of a cone parallel to the slanting side of the elevation ... 29 n 82 To draw the Hyperbola - that is, a vertical section of . a cone 29 3 84 PENETRATIONS OF SOLIDS. To draw plan and elevation of a Square Prism, penetrated at right angles by a smaller one, the axis of which is parallel to the horizontal and vertical plane ... 30 i 85 ,, ,, of the same object when the axis of the smaller prism is at an angle to the vertical plane 30 s 85 CLASSIFIED CONTENTS. To develop the surface of the larger prism, and show the shape of the apertures through which the smaller one is to pass 30 3 8.5 To draw the elevation and plan of a square prism penetrated by a smaller one, the axis of which is parallel to the vertical plane, and at an angle to the horizontal 31 r 8s To project the same object when rotated, so that the axis of the smaller prism is at an angle to both planes ... ... ... ... 31 2 8^ To draw the development of the larger prism, show- ing the apertures of penetration ... 31 3 85 To develop one of the ends of the smaller prism ... 31 4 88 To project a Square Prism penetrated by a smaller one to meet each other at their edges 32 i & 2 88 To develop the surface of the larger and one end of the smaller Prism 32 3&4 88 To draw the plan, elevation, and curves o*" penetra- tion of a Cylinder penetrated by a s nailer one ... 33 — 90 To project the same object when rotated on the axis of the larger cylinder, so that the axis of the smaller one is at an angle with the vertical plane 34 — gr To develop the surface of the larger Cylinder, show- ing the apertures of penetration, also the one end of the smaller Cylinder 35 — 96 To draw the plan and elevation of a Cone pene- trated by a cylinder, their axes being at right angles to each other 36 — 98 To develop the surface of this Cone, .showing the apertures and one end of the Cylinder 37 i loi To develop one end of the penetrating cylinder ... 37 a lor THE HELIX. Definition — — 103 To construct a Helix, and to continue the study, so as to produce a helical plane 38 — 103 To project a V-threaded screw from given data ... 39 — 105 To project a simple form of church from a given plan 40 — 107 ISOMETRICAL PROJECTION. The Elementary Principles of Isometrical Projection — — 112 To project a Cube Isometrically 41 i, 2, & 3 113 Mode of placing the T-square and set-squares for Isometrical Projection 41 4 113 Difference between a Geometrical and an Isome- trical square 41 5 113 To constnict an Isometrical scale 41 6 115 To draw the Isometrical projection of a box from given data and to a given scale ... ... ... 41 7 116 To draw the Isometrical projection of a four-armed croM. ^^ 4* — ^^^ viii CLASSIFIED CONTENTS. PLATE FIG. FAGE To draw a Circle and Cylinder by isometrical projec- tion 43 — ii8 OF TRACES, NORMALS AND TANGENT-PLANES. General information as to Traces 44 1&2 J20 Traces of a plane inclined to horizontal, but at right angles to vertical planes 45 i & 2 121 To find the Traces of lines inclined to both planes ... 46 "i 121 The Traces of a plane, and a point with its projections being given, to construct the traces of a second plane from the given point, parallel to the first... 46 2 124 The Traces of a plane, and the projections of a point being given, to construct (i) the projection of the right line falling perpendicularly from the point upon the plane, and (2) the projections of the point of coincidence of the right line with the plane .. 47 1 126 The projections of a right line being given, and also the two projections of a point, to construct the Traces of a plane, drawn from the point perpen- dicular to the right line To find the Tangent-line of a Plane and Cylinder ... To find the Tangent-point of a Plane with a Sphere To project a pile of Spheres consisting of three placed adjacent to each other, supporting a fourth, and to find the points of tangent with each other ... 49 • 1 132 To draw a Triangular Pyramid to exactly cover this pile of spheres, and to find the Tangent-Points in plan and elevation 49 2 135 Questions for Examination - — 137 47 2 128 48 I 130 48 2 132 INTRODUCTION, In another volume* the construction of figures which possess length and breadth was taught, all such figures being considered as traced upon one flat surface called a plane^ thus showing their exact forms as they are really known to be. It now becomes necessary to treat of the delineation of Solids — that is, bodies which possess not only length and breadth, but thickness as well ; and the science by which lines are so disposed that the representation of the object may seem to stand out, or project from the flat surface of the paper, is called Projection^ which is a branch of Solid or Descriptive Geometry. The subject may be divided into — Orthographic Projection^ by means of which objects are projected by parallel lines from given plans, eleva- tions, or other data, the object being placed in any given position. Isometrical-\ Projection^ by means ol which a view of an object is projected at one definite angle, a uniform • " Linear Drawing." t From two Greek words, meaning equal measure*. X INTR ODUC TION. scale, proportionate to the real measurement, being re- tained throughout. Perspective, by which objects are drawn as they appear to the eye of the spectator from any point of view that may be selected. The present volume is devoted to the study of the first and second of these divisions ; combining also the mode of obtaining required sections, the methods of describing the peculiar curves generated by one solid body inter- secting or penetrating another, and the development of surfaces — that is, the construction of the exact shape which a metal plate or other material is to be cut, so as to form or cover the required object in the most ready and accurate manner, and with the least waste — a branch which will be further considered in a subsequent volume, devoted to the technical drawing adapted to the require- ments of the metal plate-worker, boiler-maker, and tin- man. The lessons are given in as simple a manner as possible, so that the student may be able to follow them v/ith interest, and may be led to desire still further in- struction than is here afforded ; and it is hoped that the pleasure and benefit he receives from knowledge may awaken in him that spirit of enthusiasm which is the mainspring of all progress. It has been from the want of enthusiasm that our workmen have been content with the small amount of knowledge which they have obtained from their " mates " in the shop. It has been this apathy which has caused so many to be satisfied with the " rule INTRODUCTION. xi of thumb " instead of the rule of science. It is not the province of a work of this kind to dilate on the natural history of enthusiasm ; but our object is to warm up the spirit of our fellow-countrymen — to convince them, that if they will but study the principles of the sciences on which their trades are based, they will, with their acknow- ledged manual superiority, hold their own against the men of every country in the world. Let us, therefore, interpret interest in their occupation to imply enthusiasms and let us translate enthusiasm to mean that spirit which urges every man to do his work as well as it can possibly be done, and to develop the mental powers with which his Creator has endowed him to their fullest extent, so that when he leaves the workshop of life he may, in the words of Longfellow, leave "footprints in the sands of time." With the view of increasing the range of information given in the book, a special chapter has been added in this edition, explanatory of Traces, Normals, Tangent- points, Tangent-lines, and Tangent-planes, together with the method of projecting groups of Spheres and finding their tangent-points with each other, and with a pyra- midical covering, and the book will now, it is hoped, be found more useful to students preparing for examination than any other extant ELLIS A. DAVIDSON. A FEW PLAIN HINTS ON LINEAR DRAWING. Let your paper be rather smaller than your drawing- board, so that the edges may not project. To fasten the paper down, wet the back, and then paste the edges to the board ; let it lie flat whilst drying. This is only necessary when the drawing is likely to be some time in hand ; for exercises such as are contained in this volume, it will be sufficient to fasten the paper down by means of drawing-pins^ which may be bought at one half-penny each. The best T-squares are those where the blade is screwed over the butt-end, as in the illustration, as this allows of the " set-square " (or triangle) passing freely along ; whilst when the blade is mortised into the butt-end, the set square is stopped when it comes against the projecting The T-square is to be worked against the left-hand edge of the drawing-board, and should be used for horizontal lines only— perpendiculars are best drawn by working the set-square as above, against the T-square; 14 PROJECTION. for if the T-square be used for perpendicular as well as horizontal lines, the slightest inaccuracy in the truth of the edges of the board would prevent the lines being at right angles to each other. There are in some cases of mathematical instruments an implement called a " parallel rule," made of two flat pieces of ebony or ivory, connected by two bars of brass. The student is not advised to use these in obtaining parallel lines, as unless the instrument be in very good order, and very carefully used, the lines drawn will not be parallel. The best way to draw lines parallel to each other is by means of two set-squares.* Thus, let it be required to draw several lines parallel to A B. Place the edge of one of your set-squares, C, against the line, and place the other set-square, U, against the first ; hold D firmly down, and move C along the edge of D, and thus any number of parallel lines may be drawn ; and if lines at right angles to the parallels are required, it is only necessary to hold C, and place D on it, as shown in the dotted portion of the figure. In inking the drawings, use Indian ink, not writing ink, which rusts the steel of the instruments, and so destroys their refinement. Indian ink may be obtained from two- pence the stick. If you intend inking the drawings, you must work the original pencilling very lightly. * Get two set-squares (about sixpence each), the one having angles of 45', i5°j and 90°, and the other 30°, eo**, 90'. PROJECTION. 15 From the very onset aim at refinement, neatness, and absolute accuracy. Do not be satisfied if your work is nearly right. Try again, and, if necessary, again; and, with increased care and perseverance, success will be the certain result. A PLAIN DESCRIPTION OF THE MATHEMATICAL INSTRUMENTS MOSTLY USED. The most important instrument is the compass. A com- plete pair of compasses consists of the body of the in- strument, and three movable parts — viz., the steel, the pencil, and the inking-legs which are fixed in their places either by a screw, or by the end of the leg fitting ac- curately into a socket in the end of the shorter leg of the compass, and kept in its place by a projecting ledge, which runs in a slit in the upper side of the socket. This is by far the better method, and is used in nearly all modern instruments; its advantages over the screw form are, firstly, that the movable leg only remains firm in its place as long as the thread of the screw is in good order, but the very force used to tighten the pressure, wears the thread away, and then the leg shakes. The con- sequence of this can be very well imagined, when we re- member that one of the leading purposes of compasses is to draw circles, for unless the leg be absolutely firm, the circle will not be true, and the point of the pencil or ink- ing-leg will not meet the starting-point, and so an ugly break will be caused ; and secondly, that the screw being but small, is very liable to be lost. Be careful that in drawing the movable legs out, you do not wrench or bend them from side to side with the view of getting them out more easily, for by that means you will widen the socket, and cause the instrument to l6 PROJECTION. work inaccurately : the proper way is to draw the leg straight out. The steel point is used when distances are to be accurately measured or divided, and therefore com- passes which have both points of steel are called " dividers." A pair of these is found in most cases of instruments. The pencil-leg is used for drawing arcs, circles, &c. Be careful that you keep it exactly the same length as the steel one — this is accomplished by drawing the pencil out a little after each sharpening. In very old- fashioned instruments, the pencil is held in a split tube, which is tightened around it by means of a sliding- ring ; but in those of modern make, a short split tube is placed at the end of a solid leg, and the cheeks of this " cannon-leg " are tightened by a screw. This is by far the better construction, as by its means the pencil is not only more firmly held, but the points of the com- pass may be brought more closely together than in the older form. The use of the inking-leg (as its name implies) is to repeat the pencil work in ink ; the ink must be Indian ink, as already mentioned, and it is advisable to mix a small quantity of indigo with it, as otherwise it has a tendency to turn brown. When you mix the Indian ink, do not rub it very hard, as by that means you roughen the edges, and break off small pieces —they may be small indeed (and do we not frequently find failures caused by very trifling obstacles ?) — but they work between the nibs of the pen, and cause roughnesses and irregularities of thickness which materially damage a drawing. On examining the inking-leg, you will find a joint in it, the purpose of this is to enable you to bend the leg at that point, so that the part which contains the ink may be kept perpendicular to the surface of the paper whilst describing a circle, for, if the inking-leg were kept straight as the steel one, when the compass is opened to any extent, only one of the nibs (the inner one) will touch the paper, and thus the outer edge of the circle drawn will be ragged and rough. In drawing circles, be careful to lean as lightly as possible on the steel point, so that your centre may not be pricked through the paper, for then, as each concentric circle is drawn, the hole will become larger, PRoyECTioy. 17 until all chance of following the exact curve will be lost, and when you come to ink the drawing you will find the difficulty still further increased. " Horn centres" are sometimes used. These are small circular pieces of horn with three needle-points fixed in them ; one of these may be placed o\ er the centre on the paper, and pressed down ; the horn being transparent, the centre-point will be visible through the small plate, and the steel point of the compass may be placed exactly over it. This is all very well in large drawings, and where the circles to be drawn are at some distance from the centre, but where numerous small circles, immediately surrounding the centre, are required, as in the projections of the sections of cones, the horn plate is useless, as it will cover some of the space on which circles are to be drawn ; and further, the point resting on it is raised above the surface on which the other is working, and in small circles this will be a disadvantage. The student is therefore reminded of the old adage, " prevention is better than cure," and he is assured that if from the outset he endeavours to lean lightly on the instrument, practice will soon pice him beyond the necessity for the aid of the horn centre. The following hints will be found useful : — 1. See that the steel point of your compass is round and not triangular, which latter form opens the little hole made by far more than if it were round. 2. See that this point is not too thin ; it should be rather a blunt point than otherwise, only just sharp enough to prevent it slipping away from the centre. Should either of these two faults exist, they may be easily remedied by drawing the point a few times over an oil-stone, remembering to keep turning it round whilst moving it along. 3. Hold the compass loosely between the thinnb and forefinger only, allowing the instrument to rest with equal weight on both points, and merely using the finger and thumb to support and guide it. When a circle is required of a larger radius than could be reached with the compass in its usual form, a "lengthen- ing-bar" is used ; this is an extra brass rod, which fits into the socket in the leg of the compass, and has at its other end a socket into which the end of the pencil or inking-leg fits. This forms a pair of compasses with one B 1 8 PROJECTION. leg very much longer than the other, and which is, there- fore, rather awkward to manage. Here again the student is reminded that the pencil-leg and inking-pen must be bent at the joint, so that they may be perpendicular to the surface of the paper. The full-sized compass is, however, not well adapted for drawing small circles, and, therefore, a complete case of instruments contains the bow-pencil and the bow-pen. These are simply small pairs of compasses, the first of which has a pencil and the other an inking-leg. These will be found very useful, and may be purchased separately if not in the case. For still smaller purposes, " spring-bows " are used ; these constitute in themselves a small set consisting of dividers, pencil, and inking-bows. The legs instead of being united by a hinge-joint, are made in one piece, so as to form a spring, which by its action tends to force the points apart ; they are then acted upon by a nut, which, screwing upon a bar fixed in one leg and passing through the other, closes the legs in the most minute degree pos- sible. These will be found of immense service in the higher branches of mechanical and architectural drawing where very small arcs and circles are required, as in the delineation of the teeth of wheels, mouldings, and other architectural details. Another important instrument is the drawing-pen, which is something like the inking-leg of the compass already described ; it is, however, generally smaller in its nibs, and is fitted on to an ivory or ebony handle. The ink should be placed between the nibs by means of a camel's hair brush. The pen should be held nearly up- right^ with its flatter side next to the rule, the end of the middle finger resting on the head of the screw. Before you ink any line of your drawing, be careful to try your pen on another piece of paper, in order that you may ascertain whether the line drawn by the pen would be of the proper thickness, and if not, the pen may be ad- justed by means of the screw, which acts in a way similar to the screw on the spring -bows already described. Before putting your inking-leg or drawing-pen away, be sure to wipe it well, and finally to pass a piece of paper between the nibs, so as to remove any ink that may have dried, or any grit which may have been deposited. PROJECTICR, 19 The rule, or straight-edge, which you use when inking your Hnes, must have a bevelled edge ; and further, the bevel must be XyirYv^di downwards towards the paper; this will avoid any smearing which might occur if the edge of the rule were to touch the paper whilst the line is wet. Scales of different sorts are used in mechanical and architectural drawing ; but as the subject of the present volume does not necessarily involve working " to scale," the uses and construction of these will be found appended to the volumes on the above-named sections of scientific drawing. The protractor (the brass semicircle used in measuring and constructing angles) has been described, and its uses fully explained in connection with Fig. 44, page 31, of the volume on "Linear Drawing;" repetition here is therefore unnecessary, and we proceed to mention what are called " French curves." These are rules cut into an ahnost endless variety of shapes, one of which is here shown : they are used in inking curves. To do this, you must turn your French curve about, until some part of it corresponds with the form already drawn in pencil, which may then be repeated in ink, the pen being guided by the French curve. If you cannot find any portion of your rule which will correspond with the whole of your pencilled curve, draw as much of it as you can, and then find the remainder at some other part of your French curve, or on another one. As these useful im- plements may be had in innumerable patterns, from five shillings per dozen, the student is advised to provide himself with two or three of them ; but the author wishes it to be plainly understood that he does not imply that by means of French curves freehand drawing may be dis- pensed with. On the contrary, he urges this practice on B 2 23 PROyECTIOS\ all studen's : for there is such variety of form in drawing that no mechanical means can possibly supersede the. necessity for the accurate and refined education of the eye which is obtained by that study ; and further, a little practice will enable students to draw many curves bv hand in less time than it would take them to find their places on the French curve. It is not within the province of this work to recommend the instruments of any particular maker, nor to suggest the prices which should be given, as this last depends on the means at the command of the purchaser. The cheapest sets of instruments are the French or Swiss, which for the price (from is. 6d., upwards) are quite as much as could be expected ; but smallness of price is not always real cheapness, and a good article, manu- factured by, and bearing the name of, a respectable English house, will be found by far the most economical in the end. As to pencils, the degrees most generally useful are those marked HB and H,the latter of which being harder than the former, is more adapted for very minute work ; but, as a rule, hard pencils are not the best for mechanical drawings which are to be inked, as they are liable to make grooves in the paper, the bottom of which the nib of the drawing-pen does not touch, and hence the edges of the line will be ragged ; and further, lines which are drawn with very hard pencils are difficult to rub out. For mechanical drawing, it is best to cut ?ijiat point to the pencil ; this is done by cutting away the wood, and leaving about an eighth of an inch of lead projecting, which is then to be cut Mntil it is thinned to a flat, broad point like a chisel ; the broad side of this point is moved along against the rule, and the line thus drawn will be found to be much finer than one drawn with a round point. The chisel-point is economical in various ways, for it will not break so often, and the point once cut can be rubbed from time to time on a piece of fine glass-paper or file, or even on the edge of the drawing-paper. Much of the time which would be otherwise spent in sharpening the points is thus saved, and the expense of numerous pencils is at the same time diminished. Once again the student is urged to remember that the mere possession of a case of instruments, however good. PROJECTION. 21 will not constitute a draughtsman. The instruments arc merely the tools — the mechanical agents through which the mind acts ; and it cannot be denied that the more the mind comprehends of the subject to be drawn, the more willing and intelligent servants will the hands become, and the more accurately will they guide the compass or the drawing-pen. Geometrical drawing, then, should be looked upon as a mental exercise more than a merely manual occupation or employment, giving us not only subject for thought and earnest reflection, but enabling us to communicate our plans to others in such a manner that they can understand us and work out our designs better than they could have done from the most eloquent description. The student will, no doubt, find it difficult at first to draw very fine lines, or to get them to intersect each other exactly as required, especially if he has been engaged in some hard manual occupation during the day ; but he will find a little practice will soon overcome this, if he but starts with patience, energy, and the earnest desire to excel. ELEMENTARY PRINCIPLES, ILLUSTRATED BY Plate I. Fig. I. — If we place two planes or surfaces at right angles to each other, so as to form a floor and a wall, the floor, A B, is called the horizontal, and the wall, C D, the vertical, planes of projection. The Projection of Lines. No. I. — Let us take a piece of wire, and fix it in an upright position, a d, then the point on which the wire rests is called the horizontal projection, ox plan; and if we carry lines directly back from its extremities until Fig. I. Fig. 2. c / 1 ( ) 7i n 1 1 5 m ; s 1 i I n 2 k: 3 Plate I. PROJECTION. 23 they cut the vertical plane in c and d, the line c d \s the vertical projection, or elevation., of the wire. No. 2. — If a wire, I PROJECTION. 33 making angles of 45° with the vertical plane. As this plane is supposed to possess little or no thickness, its elevation, when lying tlat, is merely the line a' d ; the angle r, and d which lies directly beyond it, being marked J, If we now raise the square, allowing it to rest on the angle a^ the extremities of the diagonals, rt', c and b^ will travel through parts of circles. Thus, let it be re- quired that the diagonal a d shall be parallel to the vertical plane, and inclined to the horizontal at 45°. Draw a perpendicular from a to the intersecting line, and thus obtain d . From d draw a line at 45°, and with radius d c and d d describe arcs cutting the inclined line in J' and d'; the extremities of the diagonals are thus trans- ferred from the horizontal to the inclined elevation. Now, the points \ and ^/, in rising higher, will also have moved towards a in the plan, in the track indicated by dotted lines ; their present position is determined by dropping perpendiculars from j! and d' to cut the dotted lines, and the points being united by lines, the plan of the square in the required position will be obtained. Let it now be required to obtain the projection of this square, when, in addition to the diagonal a d being in- clined at 45° to the horizontal, it is inclined at 60° to the vertical plane ; in other words, keeping the square resting on the point «', inclined at its present angle, and rotating it. The plan then will be the same as in Fig. i, but turned round until d d' is at 60° to the intersecting line ; then perpendiculars raised from each of the angles, intersected by horizontals from the corresponding points in the pre- vious elevation, will give the projection in Fig. 2. The same plan turned so that a d is 2X right angles to the intersecting line, and worked out as in the last figure, will give the projection of the square when resting on one of Its angles, its plane being at 45° to both the planes of projection. It will be seen that the diagonal, c b^ has, in all three figures, remained parallel to the horizontal plane ; but in Fig. 3 it will be observed to be parallel to both planes. The student who has thoroughly mastered the fore- going lessons will have seen that, when he understood the projection of single lines, he soon comprehended the delineation of planes, since planes arc but forms C 34 PROJECTION. bounded by lines. It is hoped that the next step, the pro- jection of solids,* may be divested of some of its appa- rent dii^culties, by the reflection that soHds are made up of planes, t and that thus, when planes can be projected separately, it will be easy to work out several combined in one object. Thus a cube, or solid square, is formed of six equal squares ; and as each of these sides is parallel to the opposite one, the trouble will not be much more than projecting three planes. Cubes and Prisms, When three or more planes meet at one point, as at the corners of a cube, they form a solid angle. A prism is a solid whose opposite ends are equal and similar plane figures, and whose sides, uniting the ends, are parallelograms. The ends of prisms may be either triangles, squares, or polygons. A line drawn from the centre of one end of a prism to the centre of the other is called the axis. Plate VII.— To Project a Cube. First position. Fig. i. — When standing on the horizontal plane, its axis being vertical, and its sides at 45° to the vertical plane. Let a h c d he the plan of the cube, and e the plan of the axis. Draw perpendiculars from each of the angles of the plan, and make the height above the intersecting line equal to the side of the plan. Draw the top line, a act. t Excepting the sphere and its allied forms, no portions of which atc absolute planes. 36 PROJECTION. at 25° to the horizontal plane. Perpendiculars dropped from the angles of this elevation, intersected by horizontals drawn from the corresponding points in the plan of Fig. i, will give the plan of Fig. 2, or the view obtained by looking down on the elevation, in the direction of the arrow. The axis, e f., will now be seen. The student is urged to letter with the utmost care until he has become accustomed to follow each point through its various changes of position. In this plate, and all subsequent projections of prisms, the points of the base, or lower end, will be marked with the same letters as those of the opposite, or upper end, but in smaller characters. * Fig 3. — When the axis of the cube is at d^'^ to the horizontal and 30° to the vertical plane. Place the plan so that the line of the axis, e f, is at 30° to the intersecting line. Draw perpendiculars from the solid angles of the plan, and horizontals from the corre- sponding points in the elevation of Fig. 2. The intersect- ing of these two sets of lines will give the points for the Drojections. Shade Lines. — The light has been supposed to come in the direction of the parallel lines on the left of the point a. Thus, the sides c d and d b are in shade. This is indicated by the lines on the plan being darker than the others, and all perpendiculars rising from them will be dark also. Development. — The development is formed by the shapes of all the sides of an object being laid down on a flat surface, so that when folded, or connected, a given solid may be either constructed or covered. By " solid" is here meant an object that has the external appear- ance of solidity. Whether the body be really solid or hollow will be subsequently determined by sections or cuttings. To Develop a Cube (Fig. 4.)— A cube consists of six square sides. l^t\. abcd\iQ four of these, which, uniting at g g, will form the walls, then e and /will be the top and bottom. A very useful model may be thus formed. The strips left at the edges will be found useful in fastening the sides together. If the model is made of cardboard the lines should be cut half through, and half the thick- ness of the strips peeled off. PROJECTION. 37 Plate VIII. To Project a Square Prism.— Width of side \ inch, length li (or 1-5) inch. The prism is in this lesson placed so that its axis is vertical, and its long faces are at 45° to the vertical plane. Draw the square, Fig. r, which is the plan of the prism, its sides being at 45° to the intersecting line ; and perpen- diculars drawn from the angles will give the edges of the elevation, which are to be terminated by a horizontal at i^ inch from the base. Fig. 2. — It is now required to draw the elevation and plan when the axis, although remaining parallel to the vertical, is at 35° to the horizontal plane. Now it will be evident, that as far as the elevation is con- cerned, it will merely be altered in position^ not in forin^ which change is effected by allowing the object to rest on one angle of the base, and continuing the motion until one edge of the elevation (the edges being parallel to the axis) is at 35° to the horizontal plane. It will therefore only be necessary to copy the previous elevation, inclining it at the required angle. This motion, however, whilst causing so slight an alteration in the elevation, causes an entire change in^the plan ; for whilst in the first position the plans of the edges were mere pdiits (see Fig. i, Plate I.), which united form the base, the square of the top being immediately over this, as in a line placed verti- cally, the upper extremity is directly over the lower ; but the moment the line is inclined, the plan, which was pre- viously a point, becomes a line, the length of which increases as the object approaches the horizontal. (See Fig. 2, Plate II.) But, although the position of the lines is altered, as far as their relation to the horizontal plane is con- cerned, they still remain parallel to the vertical plane ; and if the eye were placed immediately over the object, the widths across from the front to the back would be seen to be the same throughout the motion. Therefore, from the angles of the plan of Fig. i draw horizontal lines, which will give the widths of the two upper sides ; the two under them being the same, will be hidden by them. The length of the diagonal of the top and PROJECTION. 39 bottom, which is at right angles to the vertical plane, thus remains unaltered, but the diagonal which is inclined will necessarily become shortened. This will be seen in continuing the projection of the plan. Draw perpendicu- lars from the two extremities of the line which is the edge elevation of the end, to cut the middle line of the three horizontals previously drawn in the lower plane. From the middle point of the edge elevation then draw a per- pendicular which will cut the tivo outer horizontal lines, and thus four points will be obtained, and these united will give the lozenge^ which is the plan of the square end when inclined. (Refer to Fig. i, Plate VI.) The lower end of the prism will be obtained in a similar manner. Fig. 3. — It is now required that the object shall be rotated on its solid angle, so that the axis shall be at a compound angle — that is, it shall not only be obliquely placed in relation to the horizontal, but to the vertical plane. This operation has been shown in Fig. 3, Plate II., and it is therefore only necessary to remind the student that, so long as the inclination of a line in relation to the horizontal plane is not altered, no change but that of position will occur in the plan ; for, however much the object may be rotated horizontally^ the length of the space it overhangs will not be extended, nor will the heights of any of the points be altered ; and this know- ledge is the key to the projection of Fig. 3. Place the plan (of Fig. 2) so that its axis and the edges parallel to it are at 45° to the inter-secting line, then from each point in the plan raise perpendiculars, and intersect them by horizontals drawn from the corresponding angles in the elevation of Fig. 2. Join the points so obtained, and the result will be the projection shown in the upper figure of No. 3. Plate IX.— Sections. Fig-. I is the plan and elevation of the square prism formmg the subject of the last exercise. It is required to find the true shape of a section or cutting, caused by a plane passing through the prism in the direction of the line a d. This plane of section would cut through the diagonal a c oi the top and the angle d of the bottom. Draw the dotted lines a c and d' d' at right angles to the PROJECTION. 41 line of section, and at any part draw d' e parallel to a d. Now, it will be evident that this will be the greatest length of the section, and that the width will be somewhere on each side of e ; but where? How wide will the section be ? These are questions which the student will do well to ask himself. Now it is clear that in passing through a c, the section- line cuts the object in the widest part ; therefore, if the eye be carried down from a in the elevation to ^ ^ in the plan, it will be seen that the real width on each side of the centre e \s e a and e c; therefore, if these lengths be set off on each side of e in the section-line, and the points joined to d\ then a c d' will be the true section. In Fig. 2, the section-plane passes from one angle of the top to the opposite angle of the bottom, cutting through the middle of the two edges. The length will of course be equal to that of the section-line and the width, a c, as in the last figure. In Fig. 3, the section-plane passes from a line connecting the middle points of two adjacent edges of the top, to a similar line on the two opposite edges of the base. The width of the section at its middle will be equal to the diagonal a c, and at the top and bottom it will be equal togh. It is usual to cover sections with lines at 45° to their central line. Plate X. Fig. I is the plan and elevation of a square prism, similar to that which formed the subject of the last lesson. Now, if this be made of wood, and cut so that the section passes through the axis at 45°, and a pin, c, be fixed in the centre of the section, at right angles to its surface, the upper portion may be rotated on the pin, so that the short line \a) will move to ^, and be at right angles to it, and the object will be represented by the elevation and plan in Fig. 2. Fig. 3 is the development, which will show how a metal plate may be cut without waste, so as to make a square pipe to turn a corner, or form an elbow. The true section is shown in Fig. 4, its length being equal to d k 1i i \ ! j a e Plate XIV. 50 PROJECTION. it will then be seen that the height of /' 7n is the same as /^y, and that in its motion the point / will have travelled through a quarter of a circle. If now the vertical and horizontal plane be converted into one flat surface, by withdrawing the pin r, the plan, and the front and end elevation will be found to be those represented in Fig. 2. Plate XV. Fig. I is the end elevation of a triangular prism when lying on one of its long sides, its edges being at right angles and its end parallel to the vertical plane ; thus the exact shape of the end — that of an equilateral triangle — is presented to view. But when the prism is turned so that the plan is at an angle to the vertical plane (Fig. 2), the elevation becomes materially altered ; for as the object has rotated on the point <^, a has receded, whilst / has advanced, and the apex, <■/, of the opposite end, which in Fig I. was hidden beyond r, now becomes visible, the height, however, remains the same as in the original figure. Fig. 3. — Here the plan a b c f is further rotated until its edges are parallel to the vertical plane: the elevation is then shown in the dotted parallelogram, b c d e. Let us now raise the prism at one end, so that its under side is at 20"^ to the horizontal plane. In this case it will be seen that the prism rests upon the line e f^ and the points of the end elevation will now become visible in the plan, and if this plan be turned (as at Fig. 4) at an angle (say 45°) to I L, perpendicular lines from the points of the plan intersected by horizontal lines from the elevation, will give the projection of the object at a compound angle. Plate XVI. Fig. I shows the plan and elevation of four such prisms meeting at a point, a figure which often occurs in design- ing or drawing roofs of houses, churches, &c. The plan is formed of two figures similar to the plan of the last prism, crossing each other at right angles, and from this the elevation is easily projected. Fig. 2 shows the projection of the object when placed at an angle to the vertical plane. And Fig. 3 is the de- velopment of one of the four parts of which the model is PROJECTION. 53 composed. To construct this development on a straight line, set off three spaces equal in width to the sides of the prism, a b / a, and erect perpendiculars from the points. Make these perpendiculars equal to the lines similarly lettered in the plan. Now it will be clear that when two parallelograms, like those forming the plan of the prism, cross each other, they will form four right angles at the centre. Therefore, at d' and c construct angles of 45° which will meet in c, and form the required right angle, and this will complete the under side. Draw c h and c i at right angles to ^ ^r and d c, and equal to the altitude of the triangle ^ ^ in Fig. i. Join d' i and h e. Then the right-angled triangles, d' c i and e c h, turned up at right angles io a b c d e, will form the upright sides of the mitre ; / and h will then come together. The triangular end, which is represented as bent down, being now turned upward at right angles to the under side, the two upper sides are to be bent over. Then c will meet h /,/will meet c", and da will meet d' a'. Plate XVII.— The Projection of Polygons. In this plate the mode of projecting a plane pentagon is shown. Fig. I. — Let A B C D E be the geometrical figure when lying flat on the horizontal plane with one edge, A and B, at right angles to the vertical plane ; as in all the planes in a similar position, the elevation would be merely a line marking the greatest width, as A C E. Now let it be required to construct the plan of this figure, when the plane resting on A B is raised to an angle of 30° to the horizontal plane. Then as each of the points C D and E will travel through portions of circles, draw a line at 30° at A on the intersecting line, and from A, with radius A C and A E, describe arcs cutting the line in points correspondingly lettered. This, then, will be the elevation. From e draw a perpendicular cutting E/in e'. From C and D draw lines parallel to Ey! Then perpen- diculars drawn from ^ in the elevation will cut these last- mentioned lines in c d'. Join B d', d' /, e c', (f A, which will complete the plan of the figure. Fig. 2. — Here the plan is turned so that ci b' is at 45° to the vertical plane, and it will be seen that by drawing \o ! CM PROyECTION. 55 perpendiculars from the angles of the plan, and intersect- ing these by horizontals drawn from the corresponding points in the elevation, the projection of the plane will be obtained. It must be remembered that the pentagon being " regular,"* a line joining C and D will be parallel to A B, and will remain so, however much the plane may be raised. Thus, this line c d' is represented in the elevation by the point ^ — the line itself being horizontal and at right angles to the vertical plane. (See Plate I., No. 2.) Now when the object is turned round (Fig. 2), this horizontal line becomes visible, and the perpendiculars from c' d' in- tersecting it, give the points c" and d" ^ and it will then be seen that as the line joining these points was horizontal, and parallel to A B in the previous tigure, it will re- niam so in the projection ; and this will explain the cause of two points in the plan coming on one line in the pro- jection — a case which will frequently occur in the projec- tion of polygons. Figs, 3 and 4 show the same process adapted to a regular hexagon, when resting on one of its angles, which it is expected the student will be able to work out without further instructions. Plate XVIII.— To project a Pentagonal Prism. Let A B C D E (Fig. i) be the plan, and e%a,e'% a the elevation when one of the long faces is at right angles to the vertical plane. Fig. 2 is the eleva- tion, looking directly at the point E. The mode of obtaining this elevation has been shown in Plate XIV. The upper end of the axis is shown at g in the centre of the plan,t and its position in the elevation is 2i\. f g. Now it will be remembered that the ends of a right prism are equal and similar planes, parallel to each other,;|: these ends being united by lines at right * A "regular" pentagon is one which has all its sides and angles equal. (See volume on "Lineai Drawing.") t To find the centre of n regular polygon. — Bisect two of the angles or sides which adjoin each other, and the point where the bisecting lines meet will be the centre. (See " Linear Drawing.") t When the planes forming the ends of the prism are at right angles to the long sides (that is, so that if the prism stands on one of the ends, the long sides may be verticalj, it is called " a right prism." When the planes of the PROJECTION. 57 angles to their surfaces, and it will therefore be evident, that projecting a prism is only repeating the process of projecting a plane. Thus (Fig. 3), let it be required to draw the plan of the prism when resting on A B, its axis at 45° to the horizontal, and parallel to the vertical plane. It has already been shown that the axis is parallel to the edges of a prism ; consequently, as the axis is at 60°, so will be the edges. Therefore, place the line a a SiX. 45°, and on this line construct the elevation of Fig. i ; project the ends by dropping perpendiculars from the points in the elevation, Fig. 3, and intersecting these by hori- zontals from the corresponding points in the plan of Fig. I. Unite the points of these two plans by lines re- presenting the long edges of the prism, which will then be seen to be parallel to the vertical plane. Fig. 4 shows the prism when the axis is at 45° to the horizontal and 30° to the vertical plane. In this figure it will only be necessary to place the plan of Fig. 3 at the required angle with the intersecting line ; then perpen- diculars drawn from the angles, intersected by horizontals drawn from the corresponding points in the elevation, will give the projection. Plate XIX. Shows a similar application of the projection of a hexago- nal prism when resting on one angle ; and it is hoped that the student will be able to accomplish this without instructions. Of Pyramids. A pyramid is a solid which stands on a triangle, square, or polygon, and terminates in a point, all its sides being, therefore, triangles. The axis of a pyramid is the line joining the centre of the base to the point (called the apex). When the axis rises from the centre of the base, and is perpendicular to it, the sides will be all equal triangles, and the solid is called a right pyramid. When the axis is not at right angles to the base, the pyramid is called " oblique." ends are slanting to the length of the prism, it is called " oblique " In this work all prisms are assumed to be " right." unless otherwise expressed. PROJECTION. 59 Plate XX.— The Projection of Pyramids. Let it be required (Plate XX., Fig. i) to project a pyramid whose base and sides are equilateral triangles. This solid is called a " Tetrahedron." A B C is the plan, and c a the width of the elevation. The point requiring consideration is: How high will this pyramid be? Now, although the real shape and size of the side lettered A D B in the plan is a' d' b' (the equi- lateral triangle shown in the upper corner), still the line / d' would only be the height if the three iriangiilaj- sides stood upright on their edges on the lines A B, B C, C A. But they ito not. They slant inw^ard until they all meet in a point ; which, as they are all equal, will be directly over the centre of the plan (point D). This knowledge enables us to find the exact position of the apex. Draw a per- pendicular from D, and another at a. Mark on perpen- dicular a the real height of e d (one side of the pyramid standing upright), viz., a d'. This perpendicular will be the edge elevation of the side. Then from a, with radius a d\ describe an arc, cutting the perpendicular drawn from D m d", which will be the apex. Join a d" and c d", which will complete the elevation. It will, perhaps, at first seem strange that although we k/ioiu the apex to be over the middle of the solid, yet it does not appear to be so in this elevation. The reason of this is, that the length of a d" is the projection of the height (or altitude) e d of the triangle, whilst the line c d" is the projection of the edge c d, which it will be seen is longer and slants more than e d. This appears plainly in the plan, where C D is longer than D a. Yet D is in the centre. The student is now advised to turn the plan, so that the edge A B may be parallel to I L : and in the elevation projected from this figure, the apex will be over the middle of the base. Fig. 2 is the plan and elevation of a square pyramid when the two edges, A B and C D, of the base are parallel to the vertical plane. It will be seen that the position of the apex is found in the same manner as in the last figure, by marking on a perpendicular the real altitude of the side, and then inclining this elevation of the side until it cuts the axis in e. Fig. 3 shows (in fine lines) the plan and elevation of PROJECTION. 6l the same pyramid when one of the diagonals of the plan, A D, is parallel to the vertical plane. Let it be required to draw the plan of the pyramid when resting on one angle, D ; the plane of the base being at an angle of 30^* with the horizontal plane. Place the elevation at the required angle, and project the square which forms the base of the pyramid in the manner shown in Plate VI. Produce the diagonal A D, and drop a perpendicular from the apex {e) to cut this line in e' . This will give the plan of the apex. Join this point to the points of the plan of the base, and the plan of the solid will be completed. To find the true shape of the section on the line g h. Draw lines from g h i at right angles to g h, and draw / k parallel to that line ; / k will be the length of the section. From g i h draw perpendiculars cuttmg the plan in I m ft o. Join these points, and 2ipian of the section will be obtained. On each side of/', in the upper figure, set off the length p ?n or / / of the plan —viz., p q and p r. Join j q, J r, q k, and r k^ and the true section will be completed. To draw the development of this pyramid (Fig. 4). With radius equal to one of the edges of the pyramid, describe an arc. Draw the radius e a. From a^ mark on the arc the lengths a c^ c d, and a l?,l? <{, equal to the sides of the base. Join these points, and from each of them draw lines to e. On either of the lines, such as c d, construct a square for the base of the pyramid, and this will complete the development. To find the section-line on this development, mark on a e the length a g of the elevation (Fig. 3). From / (Fig. 3), draw i i' parallel to a d. From c and b (Fig. 4) set ofT on ^ ^ and b e the length d i' from Fig. 3— viz., b r and c i'. Ox\ d e and d' e (Fig. 4) set off the length d h from Fig. 3— viz., d h and d' h. Join //, /", g, i\ h, which will give the line of section required. When the upper part of a pyramid or cone is cut off, the solid is said to be " truncated.'' Plate XXI. Fig. I is the plan and elevat'on of a hexagonal pyramid when two of the sides of the plan, B C and E F, are at right angles to the vertical plane, and its axis PROJECTION. dl vertical. Now let it be required to draw the plan of the pyramid when lying on the side B C G. The elevation (Fig. 2) will be precisely the same as in Fig. i, altered only in position. It will be self-evident that if a pyramid stood on a plan, and, whilst resting on the line B C, it were gradually turned over until it should lie on one of its triangular faces, the widths F E, B C, and A D would remain the same, notwithstanding the change of position ; for, supposing pieces of board were placed upright on the lines H H, the angles A D would touch these "wooden walls " throughout the movement ; but this is not so with regard to the widths from E to C and from F to B, which are altered according to the position of the plane of the base in relation to the horizontal plane. The points for the plan in Fig. 2 will therefore be found by producing the lines from E C, F B in the plan, and in- tersecting them by perpendiculars from the corresponding points in the elevation, A line drawn from G in the plan parallel to the intersecting line, intersected by a perpen- dicular from G' in the elevation, will give G", which will be the plan of the apex. Fig. 3 is the projection of the pyramid when lying on one of its faces, with its axis at 45° to the vertical plane. In order to test the student's comprehension of the fore- going lessons, this figure is left unlettered. It is now required to find the true shape of the section ^ ^/, H I. It will be evident that, as « ^ in the elevation represents A D in the plan, A D will be the width of the section at its base. Therefore, draw A D (Fig. 5), and erect a perpendicular at its centre. Make this perpen- dicular equal to « h, and draw a line through / parallel to A U. From A (Fig. i) draw a perpendicular cutting the radii F and E of the plan in \. Join h /, /A, h D. Then A D / // will be the plan of the section, or the view of it looking downward in the direction of the arrow. On each side of /, in the true section (Fig. 5), set off half the length of the line / h in the plan — viz., i h' . Join K A' and i' D, which will complete the form of the section. The next step is to develop the covering of such a solid. It is hoped that, after the instructions already given, this will prove an easy task. From G in the plan (Fig. i) draw 9. line, G J, perpendicular 64 PROJECTION. to F C, and equal to the height of the pyramid (« G). Draw F J, C J, which represents the section which would be bounded by a diagonal of the base and tvv^o of the edges {net sides) of the pyramid. With J F as radius, describe an arc (Fig. 4), and set off on it the lengths equal to the sides of the base. Join all these points to each other and to J'. On C B, or any other of, the sides, construct a regular hexagon, which will complete the de- velopment of the pyramid and base. Bisect E F by the line "J, and on this line set off the height, ^ I, on the eleva- tion (Fig. i), and through I draw / k. Join these points to each other and to A D ; this will give the section-line marked in the development. Of the Projection of Circles and Cylinders. We now approach a branch of our subject which is of especial importance to engineers and metal plate-workers — namely, the projection of Circles and Cylinders, and their development. As, however, the previous lessons have gradually led up to this point, it is hoped that the student will have been so prepared for the subsequent studies that he v/ill find but little difficulty in them. Plate XXII. Fig. I is the front elevation of a circular plane ; and it will be seen that the plan of this is a mere line, A B, equal to the diameter of the circle. (The aperture in a child's money-box is the plan of the penny which drops through it). To prepare this disc for projection, divide its circum- ference into any number of equal parts, 2.% ab c^ and from these points drop perpendiculars to cut the plan A B in the points similarly lettered. If, now, we rotate the disc so that its plan is at right angles to the intersecting line (Fig. 2), the elevation, too, will be a line, c c, equal to the diameter. To project this circle, transfer the points C, D D, and E E to plan A B, Fig 2. Let it then be required to find the forms of elevations when the plane of the disc is at 60° and 30° to the vertical plane. Place the plan at each of these angles, B' B". Taking A as a centre, de- scribe arcs from the points in the plan to cut the plans B' and B"* in C D' E'. From each of these points draw 66 PROJECTION. perpendiculars ; and from the points similarly lettered in the elevation, draw horizontals. The intersections of these two sets of lines will give the points c d e, &c., through which the curve is to be drawn by hand in the first instance, but it may subsequently be inked by means of the French curve, or centres may be found from which parts of the ellipse may be struck. The principle on which the projection of a circle is founded having thus been shown, Fig. 3 gives a simpli- fied method. Let it be required to draw the plan of a circle when resting on one end of a diameter which is parallel to the vertical plane, the surface being at 30° to the horizontal plane. The line A B, placed at 30° to the intersecting line, will then represent the elevation of the disc. From the centre of this line, with the radius of the circle it is intended to project, describe a semicircle, and divide it into a number of equal parts, C, D D, E E. From each of the points draw lines meeting A B at right angles in the ^oinis^ c d d e e. Draw any line parallel to the intersecting line, and draw perpendiculars to it from A and B ; then this line A' B' will be the plan of the diameter which is parallel to the vertical plane. The semicircle drawn on A B represents one-half of the disc lifted up until it is parallel to the vertical plane. The lines C D and E thus show the distance which each of these points in the circumference is from the diameter A B. Therefore, from c e, c, d d m the elevation draw perpendiculars passing through the plan of the diameter A' B' in e' d c d' e. From these points set off on the lines drawn through them, and on each side of A' B the lengths ^ E, ^ D, r C, &c., and through the points thus obtained the plan is to Idc drawn. Fig. 4 shows the mode of projecting a circle when its surface is at 30^ to the horizontal and one of its diameters at 45° to the vertical plane. Place A B at 45° to the intersecting line, and on it construct the plan by measure- ment from Fig. 3. This is best done by drawing a line, C C, at right angles to the diameter, A B, and on each side of the intersection marking off the distances, e e^d d. By drawing lines through these points at right angles to A B, and making them the same length as in the plan No. 3, the points for the present figure will be obtained. From these points in the plan draw perpendiculars, and PROJECTION. 6j from the points correspondingly lettered in the elevation, draw horizontals, and the intersections will give the points through which the projection of the circle is to be drawn. Of Cylinders. A cylinder is a solid body of the character of a prism, but having its ends circles. The axis, or line on which a cylinder might be turned, unites the centres of the ends ; and if the ends are at right angles to the axis, the solid is called a ri^ht cylinder. If the ends are at an angle with the axis, so that if the cylinder were placed on one of them it would be slanting instead of upright, it is called an oblique cylinder. In the first case, the ends would be circles ; but in the second, although all the sections at right angles to the axis are circles, the ends being at an angle to it are ellipses. It will be readily understood that all sections passing from one end of a cylinder to the oiher^ parallel to the axis, will be parallelograms ; and by rolling up a rectangular piece of paper, it will be seen that the surface or development of a cylinder is a parallelo- gram, the height of which is equal to the length of the cylinder, and the breadth to its circumference. Plate XXIII. Fig. I is the plan and elevation of a cylinder when standing on its base, and it will be evident that then, although the cylinder might be rotated on its axis, that axis would remain at right angles to the horizontal and parallel to the vertical plane. Fig. 2 shows the elevation of the cylinder when its axis is at 45° to the horizontal, and parallel to the vertical plane. To project the plan of this, on A B describe a semicircle which will represent half of the end. Divide this semicircle into any number of equal parts, E, U, C, &c., as in Fig. 3 in the last plate. From these divisions, draw lines parallel to the axis of the cylinder, which, passing from end to end, will give the same points in both. Draw a line for the axis of the plan parallel to the intersecting line, and perpendiculars from the various points in the elevation. Mark off the lengths C C, &c., on each side of the axis, as in Fig. 3, Plate XXII., and E 2 o 4-3 r— ( PROJECTION. 69 through the points thus obtained draw the ellipses forming the plans of the ends. Unite these by lines, parallel to the axis, which will complete the plan. Fig. 3 is the projection of the cylinder when the axis is at 45° to both of the planes of projection. No descrip- tion of the working is deemed necessary, as it is simply a repetition of Fig. 4 in the last plate, and will no doubt be readily understood. Plate XXIV. On referring to Plate X., the student will be reminded that, if a solid be cut across, the parts will, when rotated on a centre, form an "elbow" — that is, they may be joined so as to turn a corner. This principle holds equally good in relation to cylinders. Fig. I is the plan and elevation of a cylinder which it is required to cut so that the parts may be joined to form an angle of 90°. The following rule must be impressed on the minds of students — viz., Whatever may be the required angle, the section must be made at half that angle with the axis. Thus, if a pipe is to follow two walls which meet at an angle of 120°, each part must be cut at 60°. And, therefore, in the present figure, draw the section-line A B at 45° (half of 90° required). If now the upper part of the cylinder be rotated on a centre (C), the point B will meet A, and the line B F will become A G. Now divide the plan into any number of equal parts, E D, &c., and carry up perpendiculars from these points to cut the section-line in d' d' e' e'. To find the true section, draw A B, Fig. 2, equal to the section-line A B in Fig. i, and set off on this line all the distances, '^\\ \\\ I i ' 1 ! Ik I\l 1 ! il l\[ 1 i ! ! i Ki 1 ! i 1 i \ 1 ! 03 11! \ii! ll! \i! li! i N! 1!1 1 Nl \^\\ '\ \\\ \ l\ il { 1 1 ji 1 1 111 1 !i 1 1 1 1 M \\ \ I !i I 1 1 ! 1 ' 1 yf 1 1 1 1 1 1 ii : i V J : — \-y 1 1 1 \/\ ^ !il !/ li \/\ - \\V \\/ \ V A -^ PROJECTION. 73 The development of the piece of metal of which this double-elbow is to be cut must now receive our attention. Produce the base-line of the cylinder, and at any part of it erect a perpendicular, from which set off on each side the same number of equal parts as that into which the half-plan of the cylinder is divided, and draw perpen- diculars from the points. From the top of the original erect cylinder draw a horizontal line, and this, uniting the two external perpendiculars, will complete the general development. Returning now to the elevation of the original cylinder, it will be seen that the perpendiculars which were drawn from the points of division in the plan, cut both of the section-lines, and from these points of in- tersection draw horizontal lines to cut the perpendiculars in the development, that drawn from the highest point (viz., the point where the section-line starts from the side of the elevation) to cut the central perpendicular., and that from each of the other points to cut the next pair of perpendiculars in succession ; through these points the curve, which is the development of the section-line, is to be drawn. The lower section-line will of course be developed in precisely the same manner from the corresponding points of intersection occurring on it. Fig. 3 is the elevation and plan of one of the ends of the above object when resting on its section. On a hori- zontal line mark off the length of the section-line in the elevation^ and at the extremities draw lines at 45° ; make these equal to the length of the longer and shorter sides of one of the end pieces of the elevation, and join them by a line which will (if their lengths be correct) be at right angles to them. Divide this line into two equal parts, and from the bisecting point draw a line parallel to the sides already drawn : this will be the axis. On each side of this draw lines parallel to it, and at distances apart corresponding with those in the elevation, to meet the intersecting line. Drop perpendiculars from these inter- sections, passing through a horizontal line drawn in the lower plane ; on these set off from the horizontal line the widths of the corresponding lines in the plan ; join the extremities by tracing an ellipse to touch each, and this will be the true section on which the object now rests. From each of the points through which the ellipse has been traced draw horizontal lines, and intersect these by 74 PROJECTION. perpendiculars drawn from the points occurring in the e7Ld of the object ; through these intersections draw the eUipse, which will be the plan of the end. Plate XXVI. —Plan and Projection of a Speed-pulley. This is a further application of the lesson illustrated by Plate XXI 1 1., and consists of the projection of three pairs of parallel circles. Having drawn the plan, describe on A B a semicircle, and divide it into any number of equal parts — c d e f ^^. From each of these points, draw lines meeting A B at right angles in c' d' e f g'. These lines will cut C D in c" d" e" J" g". Draw a line, X X, at a height above I L equal to the radius of the circle— viz., e e. From A B, and all the points between them, draw perpendiculars passing through X X, and on these per- pendiculars, set off on each side of X X, distances corre- sponding to the distance between the point similarly lettered in the semicircle, and the line A B, as e e\ d d\ SiC, and this will give the points A B' C D' E' F' G', through which the ellipse is to be drawn. From each of the points last mentioned, draw horizontal lines, and intersect them by perpendiculars from the points C, d", oints front which the elevation of the cylinder would be projected if the axis of the smaller one were parallel to the vertical plane. In that case the perpen- diculars would be raised from the points where the axis of the smaller cylinder cuts the circumference of the plan of the larger one ; but if this were done in the present position of the plan, the elevation would be narrower than the cylinder. All vertical sections of a cylinder are parallelograms, and all those which pass through the centre are equal. Still, reference to Plate III., Fig. 2, will remind the student that the real size of a plane is only obtained in the elevation when it is parallel to the vertical plane ; and it will be seen that the elevation of the plane, of which the diameter of the plan, which is at an angle with the vertical plane, is the elevation, would not therefore be the projection of the largest section, and would not represent the true width any more than the elevation of the open door in Plate IV. represents its real width, and it thus becomes necessary to draw the dotted line referred to, so that the elevation may represent the greatest width of the cylinder. Now draw another diameter in the plan at right angles to the axis of the smaller cylinder, and the extremities of this line will be the front and back lines of the larger cylinder, which, if the axis of the smaller one were parallel to the vertical plane, would be the centre of the elevation ; but as it has of course rotated with the object, it is central no longer, but its relation to the heights remains the same, however the larger cylinder may be turned on its axis. On this perpendicular, therefore, set off from the inter- PROJECTION, 95 secting line the real height, and draw the horizontal line, which represents the top of the larger cylinder. Mark on the perpendicular, too, the height at which the axis of the smaller cylinder intersects that of the larger, and draw a horizontal through the point. Returning now to the plan, the preparation for the projection of the circular end of the smaller cylinder, as shown in Plate XXII , is necessary. On the line which forms the end in the plan draw a semicircle, and divide it into any number of equal parts. Through these points of division draw lines parallel to the axis of the smaller cylinder, which will be seen to pass through the plan of the larger one, and the intersections will be the plans of points " common to both " cylinders. Now, from the points where these lines meet the straight line, which is the plan of the end of the smaller cylinder (on which the semicircle has been drawn), raise perpen- diculars passing through the horizontal line which has been drawn across the elevation, and above and below this horizontal set off on the perpendiculars the lengths of the lines drawn from the points in the plan from which they started to the semicircle. Join the points thus ob- tained, and the projection of the end will be obtained. From each of the points through which the ellipse has been drawn now draw horizontal lines, and raise perpen- diculars from the points in the opposite end of the plan ; the curve of the end of the smaller cylinder which is turned away must then be traced through these points, and it will be observed that, as only one side of the ellipse could really be seen in this position of the object, the other half is drawn in dots. It now remains to find the shape of the curve of penetration — that is, the curve generated where the smaller cylinder penetrates the larger, and this will be accomplished by finding the eleva- vations of the points which in the plan were spoken of as *' common to both " cylinders. From these points — that is, from the points where the lines drawn parallel to the axis of the smaller cylinder cut the circle, which is the plan of the larger one, erect perpendiculars cutting the horizontal lines in the elevation, which are in fact the elevations of the lines in the plan. The curves must then be traced through the intersections of these two sets of lines. The perpendiculars must ^6 PROJECTION. be drawn not on]y from the points on the front of the plan, but from those on the back part, and these cutting the horizontals will give the points through which the curve on the other side of the cylinder is to be drawn. The reason why the perpendiculars at the back are to cut the same horizontals as those in the front, is that already pointed out — viz., that a point is not altered in height when the object on which it exists rotates on its axis in the manner shown in the diagram. Plate XXXV. It is now necessary to develop the larger cylinder, and to draw accurately upon the development the form of the aperture through which the smaller one shall pass. Now it must be borne in mind that this aperture, notwith- standing that it is to contain a cylinder, will not be a circle when the surface through which it is pierced is laid out flat. This will be evident on referring to the plan in Plate XXXIII., where the length of the straight line e to e' is the 7'eal width of the penetrating cylinder. Whereas the distance between e and e\ when measured on the circum- ference of the plan, would be much more, but as the axes of the two cylinders penetrate each other at right angles, the diameter in the elevation will remain unaltered. The development of the general form of the cylinder will be accomplished by the method shown in Fig. 3, Plate XXIV.* On this development draw a centre line A° represent- ing A in the plan. The outer perpendiculars B' B" will represent B in the plan. On each side of A° set off the lengths g f e, and erect perpendiculars ; then the heights of the points correspondingly lettered in the elevation, marked off on these perpendiculars, will give points through which the development of the aperture may be traced. It now only remains to develop the form of one of the ends of the penetrating or smaller cylinder. To do this, draw a horizontal line and erect a perpendicular E, and * The difference between this distance on the curve and on a straight lint would be considerable, therefore divide it into several smaller parts, x x x, and set them off separately, by which means the difference will be lissened. T^ 9^ PROJECTION. on each side of thiu point set off the distances/;^, c g,fEj into which the end of the smaller cylinder is divided, and from these points erect perpendiculars. On these set off the lengths of the lines between E E' and the plan of the larger cylinder — viz., E e, ¥f, Gg, C B, &c. The curve uniting the extremities of these perpendiculars will give the form in which the piece of metal is to be cut, so that when rolled and joined at its outer edges, it may form a part of a cylinder of the required size which will exactly fit to the aperture in the larger cylinder already ex- plained. Plate XXXVI.— To draw a Cone penetrated by a Cylinder, their axes being at right angles to each other. Draw in the first place the mere elevation of the cone, ABC, and of the cylinder, D D' E E', intersecting each other in F F' G G' ; and from these the general plan may be projected in the horizontal plane. The next problem for solution is the curve which will be generated by the intersection of the cylinder (which is a round body of equal diameter) with the cone (which is a round body of ever decreasing diameter). At D D' draw the perpen- dicular H I equal to the altitude of the cone, and from J, the middle line of the elevation of the cylinder, describe a semicircle equal to half the end of the cylinder. From I draw a line touching this semicircle in c, and reaching the intersecting line in C. Between D and and c and D' mark off any number of divisions, 2lS b d. It must, of course, be understood that the greater the number of divi- sions marked off, the greater will be the number of points subsequently obtained, and of course, the greater the accuracy of the intersecting curve and development ; but the object of the author is to make the operations as clear as possible, and therefore, in order to avoid one set of lines passing over another, and causing difficulties and con- fusion, he has only marked one division {b) in the upper and one {d) in the lower portion of the elevation. The student, who is expected to work this figu/e to a much Note. — To prove that I c' is a true tangent, draw c J, whidi wil! be at right angles to it. (See Tangents, in " Linear Drawing.") Plate XXXVI. G 2 lOO PROJECTION. larger scale, will, however, do wisely to use many more points, all of which are worked in the same manner. From I draw a line through b cutting the intersecting line in b\ and from I draw a line through d cutting the intersecting line in d' . Through the centre of the plan draw the line X X and carry perpendiculars to it from d b' d'; and from D', with radius D <^, D <^, D c, draw arcs cutting I H produced in points similarly letteied. From these points draw lines parallel to X X, cutting the plan of the cone in points to which (in order that the same line may be followed throughout) the same lettering is given — viz , d' b' c . From these points carry perpen- diculars cutting the base line of the elevation of the cone in d' b' c\ and draw lines from these points to the apex, C, of the cone. Intersect these lines by others drawn from b c d '\Vi the original semicircle, and through the points thus obtained the curve of penetration, starting at F and G, and ending in F and g', is to be drawn. It is now necessary to show on the plan the curve formed at the junction or penetration of the two bodies. Four points in these curves may at once be found by dropping per- pendiculars from F f' and G g' in the elevation to cut X X in F F' and G G'. Now it will be remembered that every horizontal section of a right cone is a circle, and thus the lines parallel to the base on which the points bed exist, are really edge elevations of circles, the diameter of which is regulated by their position on the cone. The length from point i on the edge of the cone to the axis, is thus the radius of the circle on which the point <^, and the corresponding point beyond it, are placed. Therefore, with this radius describe a circle from the centre of the plan, and drop a perpendicular from ^, cutting it in b b. Draw a circle from the same centre of the plan with radius 2 2', and a perpendicular from ^, cutting it in c c. Draw a circle from the same centre with radius 3 3', and a perpendicular from d, cutting it in d d. Draw the curve Y d c bY' d c b, which will be the plan of the aperture required. (Of course the corre- sponding lines on the other side will give a similar result.) PROJECTION. lOI Plate XXXVII.— To Develop the Surface of this Cone. The development of the simple surface of the cone having been fully demonstrated in Plate XXVI II., Fig. 2, it is not deemed necessary to repecjt tlie process lere ; and we proceed, therefore, on the -isrumption that this simple development has been obtained, and requires the addition only of the shape of the aperture through 'vhich' the cylinder is to pass. Draw tne centre Hne', C A. On each side of A mark off the distances A' d' , A b\ A c'. from the plan, and mark the same from B and B. From all these points draw lines to C. On A and B B mark the heights A F' F from the elevation, and on the lines d C, ^ C, ^ C, in the development, mark the heights which the points d^ d, and c are in the elevation when carried out by horizontal lines to meet A C. These points joined will give the shape which the aperture must be cut on the flat material, so that when rolled, and the lines B B are brought together, the apertures shall be of the required shape and size. To Develop one end of the Cylinder. Set off on a straight line the lengths from D on the semi- circle, representing half of the cylinder (Plate XXXVI.) — viz., D b, c, X*, d D D' — and erect perpendiculars from them. Mark off on each of these perpendiculars the distance which each point correspondingly lettered is from the line D D', and unite these points by a curve drawn by hand. The limits of this volume precluding any further illus- trations of penetrations and developments, general prin- ciples only have been treated of. The subject will, however, be further enlarged upon in the future manuals, in which the studies will be worked out in a manner adapted to each special branch of industry. • The point x is introduced in order to divide the space between c and rf, which would be too lone to measure by one chord. (See note to Plate XXXV.) ■^ T K li ? d "^ p— a ^J ^^ X^V^ ^y^ n ' ' L =■ ^ 6 ' ■ ' • ^ ^ /^ PROJECTION. [03 Plate XXXVIII.— The Helix. If a piece of paper of the form of a right-angled triangle (A B C, Fig. i) be rolled round a cylinder (Fig. 2), the hypothenuse, or long side (A C), of the triangle, will gene- rate a curve winding round the cylinder like a corkscrew. This is called the Helix, and it is this which forms the thread of a screw. To describe a Helix. Let the circle A G in Plate XXXVIII. be the plan of the cylinder around which the line is winding. Let the dotted perpendiculars A and G represent the elevation, and let the distance A' to A i be the height which the curve has reached when it has travelled once round the cylinder, so as to be exactly over the point from which it started. This is called " one revolution," and is the " pitch," that is, the distance from thread to thread, in a screw. Divide the plan into any number of equal parts, as A, B, C, D, &c., and divide A' Ai into the same number, viz., «, ^, r, d^ &c. Draw horizontals from n^ d, c, ci, &c., and perpendiculars from the corresponding points of the plan ; then the intersections of B with b and C with c, &c., will give some of the required points. Now, it w'ill be seen that the points H I J K L in the plan are immediately at the back of B C D E and F, and, therefore, the same perpendiculars will pass through them, and thus the inter- ^ Plate XXXVIII, PROJECTIOX. 105 sections of these lines with the horizontals correspondingly lettered will give the remaining points required for the formation of the curve. Through all the points now obtained the curve may be traced by hand. To continue the helix, repeat the height of the pitch as A 2, A3, A 4 ; divide these spaces as before, and from the points draw horizontals to intersect the perpendiculars already drawn ; for it will be evident that the corresponding points in each revolution will be immediately over each other. Now let us suppose that, instead of a mere line being drawn round a cylinder, an inclined plane were to sur- round the smaller cylinder M N. You will understand this, perhaps, better if you cut out of paper the plan A G. Cut the smaller circle, M N, away altogether, and cut through the line A M. Place in the hole M N a cylin- drical piece of wood of the exact size. Keep the edge A M fixed, but raise the edge O P. Then a rising plane, or walk, would be formed once round the small cylinder ; and if this were constructed on a large scale, a person having travelled along this plane would have reached A i, and be immediately over the point from which he ascended. Now a staircase is ofily a?i inclined plane on which ledges, or stairs are placed to render the ascent easier. For in- stance, let it be required to reach any height by means of an inclined plane. Of course this would be more readily accomplished by means of steps on the plane. And it will thus be seen, that if steps were placed on the inclined plane which surrounds the cylinder M N, the principle of a circular staircase would be developed. This principle will be further worked out in a subsequent volume. The points for the inner curve are obtained by perpendiculars taken from inner ends of the radii, cutting the correspond' ing perpendiculars. Plate XXXIX.— To Project a Screw. In the previous plate a flat surface was supposed to encircle a cylinder. Now if a "thread" of a triangular form (of which M N O is the section) were substituted, the result would be a V-threaded screw. To draw this, project the edge N of the thread as in the previous plate. P Q represents the plan of the smaller cylmder. The radii Plato XXXIX. PROJECTION. icy from the points in the outer circle will pass through this inner one ; and from these intersections draw perpen- diculars. Now the surface of the tooth is not, as in the last figure, a flat surface, but slanting, as will be seen by the lines O N and N M in the section ; and therefore the curve for the inner helix will not start as before, from the same level as the outer one, but from the point where a perpendicular from P cuts the horizontal, ^, midway between the edges of two teeth. The curve at the back is omitted here to avoid confusion ; but the student, who is expected to work to at least twice the scale, is advised to follow the curve throughout, as the only means of ensuring correctness. The extreme points of the inner and outer curves being joined, will complete the projec- tion of the screw. The whole subject of the projection of the various forms of screws will be fully considered in the volume devoted to mechanical drawing for engineers and machinists. Plate XL— To Project a small Church from the Plan. The church, it will be seen, is made up entirely of simple solids — viz., square prisms of various lengths, triangular prisms, and a square pyramid ; and as the student has already had some practice in these, he will find, it is believed, but little (if any) difficulty in following out the instructions, although the diagram is not lettered. The building is to be considered in the first instance as formed of the square prisms only — that is, divested of the triangular prisms which form the roof, and also of the pyramid which forms the spire. These solids, then, will be represented in the plan by two rectangles crossing each other at right angles, and as they are equal in width their intersection is a sg7/are, which is the plan of the tower ; the shorter end of the longer rectangle then becomes the plan of the chancel, and the longer end the plan of the nave ; the snialler rectangles form the plans of the transepts. It is advisable now to proceed with the projection of the body of the church from the plan. This operation is very simple, requiring only that perpendiculars should be drawn from Plate XL. PROJECTION. 109 the various points. From the two front angles of the transept which faces the spectator, therefore, draw per- pendiculars, and a horizontal line cutting them off at a height above the intersecting line equal to the required height of the walls of the church. This horizontal line may be drawn of indefinite length, as it will regulate the height of the whole body of the building. A perpen- dicular drawn from the third angle of the transept (/.t'., the front left-hand corner of the square) will give the one edge of the tower of which the square is the plan, and a perpendicular drawn from the right-hand corner of the square will give, not only the side of the transept, but will, if continued, give the right-hand line of the tront of the tower : further, a perpendicular raised from the distant right-hand corner of the square will give the side, the height of which may be determined by a horizontal to form the top line of the walls of the tower. Next draw perpendiculars from the two angles of the right-hand end of the longer rectangle, and tnese carried up will give the projection of the rectangle, or wall form- ing the extreme end of the nave. We now retjrn to the plan, and draw the diagonals, which constitute the plan of the edges cf the pyramidicai spire. (See Hate XX., Fig. 2.) From their intersection draw a perpendicular, and on this mark the height of the required pyramid, this line being the axis. From the apex thus fixed draw lines to the upper angles of the projection of the tower, which will complete the spire. Again reverting to the plan, draw lines through the middle of the rectangles, which will give the plans of the ridges of the roof. (See Plate XVI., Fig. 2.) From the point where the ridge-line meets the front of the transept draw a perpendicular, and mark on this, above the top line of the walls, the perpeftdicular height shown in the dotted triangle annexed. Join this point to the upper corners of the front of the transept, and this will com- plete its gable. From the apex of this triangle draw a horizontal line, and intersect it by a perpendicular drawn from the point where the ridge-line in the plan cuts the front line of the square. This intersection will give the point where the ridge meets th_* front of the tower. From this point draw a line parallel to that side of the triangle, and this will complete the visible transept ; the opposite no PROJECTION. one is, of course, hidden by the body of the church, and could not therefore be seen in the present view. The student is, however, advised to project this object on the inchned plane, as shown in Plate XII I., when the upper portion at least of the hidden transept will be seen. The rectangular part of the wall at the end of the nave has already been projected from the plan, and it now only remains to complete it by the addition of the gable. It must be obvious that the gable-point will be imme- diately over the point where the ridge-line meets the end of the nave in the plan ; and, therefore, from this point erect a perpendicular, and carry it up between the two lines which represent the edges of the end of the nave. Draw a perpendicular, too, from the point where the ridge-line cuts the plan of the tower. A horizontal drawn from the gable-point of the transept will cut these perpen- diculars, and give the corresponding point in the end of the nave, and in the part of the roof which meets the side of the tower. Produce this horizontal until it meets a perpendicular drawn from the end of the ridge of the chancel in the plan, and this will give the distant point in the ridge, and thus complete the projection of the church.* An endeavour has been made to divest the subject of many difficulties; for, although there is no "royal road ' to learning, the road may be materially smoothened, and the traveller guided over obstacles which, unaided, might have been to him insurmountable barriers, each failure only weakening the power and the desire to make another attempt. Therefore in some cases the student has been left to complete a study, or to work out an exercise, un- assisted by instructions, and a series of examination questions are appended, with the view of enabling him to test whether he has understood the lessons. It is hoped that this system will have had the effect of giving him confidence in himself; and each success, however small, will b3 an encouragement to further eflbrt. The great difficulty in the way of studying subjects of this class, is the aptness to copy the diai^rams^ instead of working out the prwciplesj and therefore the student is * For further projections from plans and elev.'itions, see volumes on "Architectural Drawing" and " Building Ccnstruction." PROJECTION. Ill u^-ged to turn the plans and elevations in directions different from their position in the plates, and then to project them from other data. This will lead to thinking ; and whoever has learnt to think ^ and has had that thinking properly directed, has only to add energy and perseverance to accomplish that success which should be the object of the ambition of every intellectual man. ISOMETRICAL PROJECTION.* In all the previous constructions, it will have been ob- served that the projections have been obtained by the union of plans and elevatio7is. Isometrical Projection enables the draughtsman to work out views of buildings, &c., without these separate draw- ings, but still embodying both. This most useful system may be called the Perspective of the Workshop, as by its means we are enabled, not only to show in one drawing a view of the complete object, but all the lines of the pro- jection may be measured by a uniform scale ; and hence the name, Isometrical, derived from two Greek words meaning " equal measures." In this respect it differs from perspective, in which the sizes of all objects and lines diminish as they recede into the distance, according to distinct optical laws ; and it differs also from orthographic projection (which has formed the subject of our study hitherto), as m that branch of science the lengths of the lines are altered according to the angle at which the object may be placed. The whole system of isometrical projection is based on a cube resting on one of its solid angles, whilst its base is raised until the one solid diagonal — that is, the diagonal which connects the one angle of the top to the opposite angle of the bottom — is parallel to the horizontal plane. Then, if the cube be rotated on the angle on which it rests until the diagonal is at right angles to the vertical plane, the projection of the cube will be a regular hexagon. This will be clearly understood on referring to the follow- ing plate. * Invented by Professor Parish, of Cambridge, about 1820. ISOMETRICAL PROJECTION. II3 Plate XLI.— The Isometrical Projection of a Cube. Fig. I is the plan and Fig. 2 is the elevation of a cube, when raised on the solid angle ^, so that the solid diagonal, A b^ is horizontal, and thus when rotated on a^ until A B is at right angles to the vertical plane, as in Fig. 3, the point b is hidden by the point A, and the projection will be seen to be a regjilar hcxagoti. Now we know that when a regular hexagon stands on one angle, so that a line drawn from that angle to the centre may be quite upright, the two sides adjacent will be at 30° to the line on which the figure stands ; and this knowledge enables us to drau the isometrical projection of a cube without plan or elevation, but by means of the set-square of 30°, 60°, and 90°, by simply placing it with the long side of the right angle against the T-square (see Fig. 4), and having drawn one line of the hexagon, re- versing the set-square and drawing the other, then, either moving the square along until its short edge is at the point of meeting of the two previously drawn lines, or turning it so that the short edge rests on the set-square, and thus drawing the vertical line. These three lines are then to be made equal, and the upper lines of the hexagon may be drawn, by again placing the set-square in the first and second position when the T-square is moved higher up on the board. All the lines forming the piojection of the cube will thus be seen to be equal, but they will not be the real size which they would be in the plan or eleva- tion, but will all of them bear the same proportion to the original measurement, and may therefore be measured by a uniform scale throughout. To understand the construction of the isometrical scale, observe that the square, A B C D, Fig. i, is represented in the projection. Fig. 3, by the lozenge. A' b' c' , and draw p q and p r parallel to A B and A C. From h i j k draw perpendiculars to cut these lines in s t u V, and from these points draw lines parallel to the sides of the box, cutting perpendiculars drawn from / in n \n w X y 2, which will show the junction of the inner sides of the walls and the bottom, and will complete the projection. Plate XLII.— To Project a Pour-armed Cross. Plate XLII. shows the isometrical projection of a four- armed cross standing on a square pedestal. Scale, i of an inch to the foot ; side of pedestal, 8 feet ; height of ditto, 2 feet; complete height of cross, 14 feet. The pedestal having been projected in a manner pre- cisely similar to that by which the box in Plate XLI. was drawn, carry up the perpendiculars from the angles; rnake the perpendicular A B 14 feet high, and by drawing lines from B parallel to the sides of the base, complete Plate XLII. IlS ISOMETRICAL PROJECTION. the top of a block which would contain the entire object ; for, as the complete height of the cross is 14 feet, the top of the upright would be in the top of the block ; and as the arms are 8 feet long from end to end, their ex- tremities would be in the sides of the block, which may thus represent a glass case exactly containing the cross. The thickness of the central upright is 2' o" ; and as the width of the side of the pedestal is 8' o", it follows that if 3' o" be marked off from C to ^, from D to d, from C toy^ and from E to g^ the spaces d e and f g will each be 2' o". From d e and /^ draw lines parallel to the sides of the pedestal, which, crossing, will give the lozenge h i j k, which is the plan of the central upright. From d e f g draw perpendiculars to touch the edges of the top of the solid block, B F and B G in / m ?i o, and lines drawn from these points parallel to the sides will give the top of the central upright. On the front perpendicular A mark off ^ at 9' o", and P at 11' o" from the bottom, and from these points draw lines parallel to the sides C D and C E. These will give the heights of the top and bottom edges of the arms. But the arms are not so thick as the central upright, being only i' o"; therefore between ^and e, and f and g, mark off half a foot from each of the points. This will leave the spaces r s and / it each i' o" wide. From these draw perpendiculars, which, cutting the lines drawn from p and q, will give the ends of the arms ; then draw lines parallel to the sides of the pedestal, cutting hj and h k mv w and x y, and from these points draw per- pendiculars. From the angles of the ends of the arms draw lines parallel to the sides of the pedestal, cutting these perpendiculars, and these will complete the two arms which are turned towards the front. By producing these lines as shown in the diagram, the portions visible of the opposite arms may be drawn. All further detail will, it is hoped, be rendered clear by reference to the figure. The Isometric Circle. Projection does not deal with curves as such, but it becomes necessary to find points in rectilineal figures through which the curves pass, then to project the recti- lineal figure, and trace the curve through the points thus Plate XLIII. f20 ISOMETRICAL PROJECTION. obtained. Thus for isometrical purposes (as in radial perspective) the circle is enclosed in a square, as in Fig. I. Having drawn the circle, describe around it the square A B C D. Draw the diagonals, and also the two diameters, at right angles to each other, meeting the sides of the square in the tangent points E F G H. The circle not only touches at these four points, but cuts through the diagonals in the points J K L M. Draw lines through each of these points, cutting the sides of the square \Vi j k I m. Proceeding now to project the circle thus prepared, draw the diagonal C D in Fig. 2, equal to C D in Fig. i. From C and D draw lines at 30° to C D, intersecting in A and B. This will be the isometrical representation of the enclosing square The points E F G H andy k I in are obtained by mark- ing from A the distances Ay, j E, E /, and Ay, / F, and F w, and drawing lines from these points parallel to the sides of the figure. The intersections J K L M will thus be obtained through which the ellipse, which is the iso- metrical projection of the circle, is to be drawn. The study may be carried on to the projection of a cylinder by repeating the operation for the bottom, and joining the intersections by perpendiculars. The limits of this volume necessarily preclude further illustrations of this branch of projection. Various objects will, however, be delineated on this simple system in the volumes devoted to architectural and engineermg drawimr. TRACES, NORMALS, AND TANGENT-PLANES. Of Traces. Those right lines in which lines or planes, or lines or planes produced, cut the planes of projection, are called their "Traces." The point or line in which any line or plane, or any line or plane produced, intersects the horizontal plane, is called the "Horizontal Trace." The point or line in which any line or plane, or any line or plane produced, intersects the vertical plane, is called the " Vertical Trace.'' The following simple illustrations will make these defi- nitions still more obvious : — TRACES, NORMALS, AND TANGENT-PLANES. 121 Plate XLIV. Fig. I. — Let A B, CD, be the vertical and horizontal planes of projection, intersecting in I L. Let E F G H be a plane, standing on its edge perpendicular to and touching both planes. Then (the material permitting), if the plane were pushed from E F, it would pass through or penetrate the vertical plane in the line G H, which would then be the "Vertical Trace." Again, if the plane were forced downward, the line E H, through which it would pass, would be the "Horizontal Trace." A plane need not necessarily touch the planes of projec- tion ; thus, \Qt e / g hhQ 2i plane suspended in space, at right angles to both planes of projection ; then, if it were produced both ways, the line e' li would be its Horizontal Trace, and //' g' its Vertical Trace. If the co-ordinate planes of projection were then laid flat, the traces of these two planes would be those shown in Fig. 2. Plate XLV. Fig. I, A B C D.— This is a plane at right angles to the vertical, and incHned to the horizontal planes of projection. A B is then the horizontal, and B C the vertical trace. E F G H shows a plane suspended in space, which, on being produced both ways, penetrates the planes of pro- jection in J K and K M, which are therefore the vertical and horizontal traces. Fig. 2. — Here the two planes of projection are supposed to be spread out, showing the vertical an-d horizontal traces of the two planes drawn in Fig. i. Plate XLVI. Fig. I. — In this figure a b \% the plan, and a b' the ele- vation of the line A B suspended in space, and it will be clear that, since the projecting planes of A B must contain the projectors of all points in A B, the plans and eleva- tions of all points in any line, A B, must be points in the plan and elevation of A B itself ; thus, a b are in the plan a b, and a' b' in the elevation a' b' of the line A B. Now the line A B is the edge of a vertical plane, abV> A, standing on a b (its plan). And it will be clear that \{ a b Plate XLIV, Fig. I A E Fig. 2. Plate XLV. 124 TRACES, NORMALS, AND TANGENT-PLANES. is produced until it meets the intersecting line I L at e/, and if A B is produced until it meets a perpendicular raised at v in 7/', and \i a b were produced in the opposite direction until it should meet A B produced in /, a trian- gular plane would be completed, the horizontal trace of which would be / v. Again, if the elevation a' b' is produced to v' and to /' the point where the projector of / meets I L, then v' /', will be the vertical trace of the plane in which A B lies. A B will thus be seen to form, as it were, the edge caused by the meeting of two planes — the one, v t 7/, perpendicular to the horizontal and inclined to the vertical plane ; the other, / / v\ perpendicular to the vertical and inclined to the horizontal plane. Having thus explained the precise meaning of the term " Traces," we proceed to give some studies, in order that the student may, by absolute practice, at;ain proficiency in the application of the principles laid down. We take, as a basis for these studies, the admirable theories given by Monge, who, in 1794, published, in France, his " Geometric Descriptive." Fig. 2. — A plane, the horizontal trace of which is A B, and the vertical trace of which is B C, and a point the pro- jections of which are G g., being given, to construct the traces of a second plane, drawn from the given point parallel to the first. The traces of the plane sought for must be parallel to the respective traces of the given plane, because these traces, taken in pairs, are the intersections of two planes, parallel to a common plane. We have, therefore, only to find for each of them one of the points through which they respectively pass. To obtain this, from the given point conceive a hori- zontal right line in the plane sought for. This hne will be parallel to the trace A B, cutting the vertical plane in a point which will be one of those of the trace of a plane sought for on it, and we shall have its two projections by drawing the indefinite horizontal g F from the point ^, and the right line G J from the point G parallel to A B. If G be produced to meet the intersecting line, I L, of the two planes of projection, the point J will be the hori- zontal projection of the intersection of the horizoiital right line with the vertical plane. Plate XLVI. 126 TRACES, NORMALS, AND TANGENT-PLANES. This point of intersection, therefore, will be found upon the vertical line, J F, drawn from the point J. But as it must also be found upon g F, it will be dis- covered at the point (F) of intersection of the two latter right lines. Lastly, by drawing a line from F parallel to B C, we shall have upon the vertical plane the trace of the plane re- quired ; and if this trace be produced until it meets the intersecting line I L in E, and E D be drawn parallel to A B, we shall have the trace of the same plane upon the horizontal plane. If, instead of a horizontal right line in the plane sought for, a line parallel to the vertical plane were conceived, the construction would be as follows : — Draw from the point G, parallel to I L, the indefinite right line G D. From the point g draw g H parallel to C B, producing it until it cuts I L in the point H, whence draw H D per- pendicular to I L. This latter will cut G D in D, whence, if a line be drawn parallel to A B, one of the traces sought for will be obtained. And if, having produced this trace to meet I L in E, the line E F be drawn parallel to B C, we shall have the trace in the vertical plane. Plate XLVII. A plane whose two traces are A B and B C, and a point whose projections are D d, being given, to construct — 1. The projection of the right line falling perpendicu- larly from the point upon the plane, and 2. The projection of the point of coincidence of the right line with the plane. The perpendiculars D G, d g, falling from the points D and d, upon the respective traces of the plane, will be the indefinite projections of the right line required. For, if along the perpendicular a vertical plane be conceived, such plane will cut both the horizontal and the given planes in two right lines, both of them perpendicular to the common intersection A B of the two planes. Now the first of these lines, being the projection of the vertical plane, is also that of the perpendicular which is Fig. I. Plate XL VII. 128 TRACES, NORMALS, A^D TA^GENT-PLANES. included. Therefore the projection of this perpendicular must pass through the point D, and be perpendicular to AB. The same demonstration will serve for the vertical projection. As to the point of coincidence of the perpendicular with the plane, it is evident that it must be found at the intersection of this plane with the vertical plane drawn along the perpendicular, such intersection being projected indefinitely upon E F. By obtaining the vertical projection / ^ of this inter- section, we shall find it to contain that of the point re- quired; and as this point must be projected upon the right line^^, it will be found at the intersection,^, of the lines f e^ and d g. It remains, therefore, only to discover the right line f ^^ Now the intersection of the given plane with the vertical pline, which are perpendicular to each other, will meet the horizontal plane in the point E, whose vertical projec- tion, ^, will be found by dropping E e perpendicularly upon I L, and it will meet the vertical plane of projection in a point, the horizontal projection of which is the intersection of the fine I L with D G produced, if necessary, and the vertical projection of which must be at once upon the vertical line F y, and the trace C B. Of course, it will be at the point, f^ of their intersection. The vertical projection g of the foot of the perpendi- cular being found, the construction of its horizontal projection will be easy, for by dropping the indefinite per- pendicular G g upon I L, a right line will be obtained, which will contain the point required, and the line D F must also contain it ; it will be found at the point G of these two right lines. Fig. 2. — A right line whose two projections are A B. a b, and a point, the two projections of which are D d, being given, to construct the traces of a plane, drawn from the point perpendicularly to the right line. It has been shown in the preceding figure that the two traces must be perpendicular to the respective projections of the two right lines. It remains to be discovered through what points each of them ought to pass. For this purpose, let a horizontal line from the given TRACES, N-ORMALS, AND TANGENT-PLANES. 1 29 point be conceived in the plane required, produced so as to meet the vertical plane of projection, and we shall find its vertical projection by drawing the indefinite horizontal d G, through the point d, and its horizontal projection by drawing a perpendicular to A B, through the point D produced until it cuts I L in H, which Avill be the horizontal projection of the point of coincidence of the horizontal with the vertical plane of projection. This point of coincidence, which must be found in the vertical line H G, and the horizontal line d G, and consequently at the point G of the intersection of these two lines, will be one of the points of the trace on the vertical plane. We shall then find this trace by drawing the line F C from the point G, perpendicular to a b^ and if from the point C, where the first trace meets I L, C E be drawn perpendicular to A B, we shall have the second trace required. The same process would discover the point of coin- cidence of the plane with the straight line. Were it necessary to drop a perpendicular from the given point upon the right line, we should construct, as has just been described, the coincidence of the right line with the plane drawn by the given point, and which would be perpendicular to it ; and we should obtain from each of the two projections of the required perpendicular, two points through which it must pass. Of Planes — tangent to Curved Surfaces and of Normals. There is no curved surface but what may be generated in several ways, by the movement of curved lines. Therefore, if from any point of a surface two generating lines are supposed to spring in the position they would naturally have in passing each other through such point, and if the tangents be supposed in this point, to each of the two generators the plane described by such two tan- gents is the tangent-plane. The point of the surface in which the two generators cut each other, and is at the same time common to the two tangents, and to the tangent-plane, is the point of contact between the surface and the plane. The right line drawn through the point of contact perpendicularly I 130 TRACES, NORMALS, AXD TANGENT-PLANES. to the tangent-plane, is said to be normal to the surface; It is perpendicular to the ground of the surface because the direction of such ground coincides in every part with that of the tangent-plane, which may be considered as its prolongation. (Monge.) A knowledge of tangents and of normals to curved surfaces is very useful in a great number of arts ; in many it is indispensable. The several portions which compose vaults of hewn stone are called voussoirs (see " Drawing for Stone- masons," page 41, and " Building Construction," page 65), and the faces on which two contiguous voussoirs touch each other are denominated joints, whether the voussoirs form a single course, or whether they be comprised in two successive courses. The position of the joints of vaults is subject to several conditions, which must necessarily be complied with. One of the conditions required in the position of joints is, that they be all perpendicular to each other and to the surface of the vault ; any material deviation from this rule not only destroys the general symmetry of the struc- ture, but also diminishes the firmness and durability of the vault. For instance, if one of the joints be made oblique to the surface of the vault, one of the two con- tiguous voussoirs will form an obtuse angle and the other an acute one ; and in the reaction which these voussoirs would exert against each other, the two angles would present an unequal resistance, whence, in consequence of the com- parative brittleness of the material, the acute angle would bilge and spoil the shape of the vault, as well as endanger the edifice. The reduction of vaults into voussoirs, there- fore, absolutely requires a knowledge of planes tangent and normals to the curved surface of the arch. Plate XLVIII. To find the tangent-line, or the line at which a plane touches a cylinder against which it rests. Fig. I. — Let A be the end elevation of the cylinder, and B C the edge of the plane resting against it ; and let a be the plan of the cylinder and b b' c c the plan of the cylinder alluded to. a ^^ 1 Fig. I. '^ i \o ^ ^^ \ 1 V U' d a / \ L Plate XLVIII. I 2 132 TRACES, NORMALS, AND TANGENT-PLANES. Now it is required to find on the plan of the cyHnder the exact hne touched by the plane resting upon it. It will be remembered (" Linear Drawing," page 36) that a tangent is always at right angles to a radius, therefore if a radius be drawn from O, at right angles to B C, and meeting it in d, that will be the tangent-point. From d draw the perpendicular d' d", which will be the tangent- line on the plan. Fig. 2.— To find the tangent-point of a plane with a sphere. Let A B be the elevation and a a' b b' be the plan of a plane resting against a sphere. The tangent-point in the elevation will be at C — the extremity of a radius at right angles with the tangent-hne — this being a radius of a "great circle" of the sphere — that is, a circle drawn with the radius of the sphere, the plane of the circle in the present instance passing through the vertical axis, and being in this case parallel to the vertical plane. Through C draw a horizontal line, which will be the elevation of a small circle of the sphere. From C drop a perpendicular to C in the plan; with O' C describe the circle, of which the previous line is the elevation. It will be evident that C is the tangent-point in the plan. The student who has worked through this course will find no difficulty in Fig. 3, in which is given the projection of the subject when rotated so that the plan may be at any required angle with the vertical plane, as well as being inclined to the horizontal plane. No further instructions are therefore given, as it is hoped the diagram will be self- explanatory. Plate XLIX. To project a pile of equal Spheres, consisting of three placed adjacent to each other, supporting a fourth, and to find the points of tangent with each other. Fig. I. — A, B, C are the three spheres resting on the horizontal plane, their centres being in the three angles of the equilateral triangle a b c^ the side of which is the length of a diameter of the spheres. We proceed to obtain the projection of these three spheres in the vertical plane. From b in the plan draw a perpendicular, and on it set ofT from J L the height b\ equal to the radius of the spheres, Plate XLIX. 134 TRACES, NORMALS, AND TANGENT-PLANES. and from b' describe a circle, which will be the elevation not only of the sphere B, but of C also, that sphere bein^ situated immediately behind B. From a in the plan draw a perpendicular, which being intersected by a horizontal from b' (the centres of these three spheres being on the same level), will give a', the centre of the third sphere. It will of course be seen that a portion of this last sphere is hidden by the elevation of B ; the cause of this will be seen on reference to the plan. It will easily be understood that these three spheres touch each other on the sides of the triangle which unites their centres, and the tangent-points are therefore d,e,f, in the plan, and will of course be found on a horizontal joining the centres of A' and B' in the elevation. The exact points will be obtained by raising perpendiculars from d , e,f^ to cut the horizontal last referred to ; the point d' will represent e as well, that point being situate imme- diately behind d', and the centre b' will also represent/"', for the same reason. It will of course be understood that the fourth sphere is drawn in plan from the centre of the triangle, uniting the centres of the other three, and with the same radius ; but it is necessary, in relation to the elevation, to point oul that although this sphere rests on the other three, it does not rest on the highest point of either^ but lies, as it were, in the hollow space between them. To find the exact position of the fourth sphere, draw a perpendicular from the centre of its plan and another from the centre of A {a). On the latter, draw a circle shown in dots in Fig. 2, representing the sphere as if resting on the highest point of A'. Now it will be clear that if the sphere were really so placed, and gradually rolled over in direction of the other two, it would sink into the cavity until it touched them. Therefore, from d with radius a g, describe an arc cutting the perpendicular raised from the centre of the plan in g', which will be the centre from which the eleva- tion of the fourth sphere D' is to be described. The tangent-point of the spheres A' and D' will then be seen at //, which, being projected on the plan, is at h', and as this sphere touches all the three nether ones, the corre- sponding tangent-points will be found on lines joining the centre with b and c, at distances equal to h' — viz., / and /. TRACES, NORMALS, AND TANGENT-PLANES. 1 35 From these points draw a perpendicular, and from h in the elevation draw a horizontal intersecting it in r, which will represent the two points. It is now further required to draw a triangular pyramid, which shall exactly cover this pile, and touch each sphere ; and it will be evident that the tangent-points will be in the surfaces and not in the edges of the pyramid. Therefore, draw the line E F, touching the spheres B' and D' in k and /. This line will be a true tangent — being at right angles to the radii k b' and I g". From / draw a perpendicular, cutting a line drawn from the centre of the sphere D, parallel to 1 L in /', which will be the plan of the point where the side of the pyramid touches the upper sphere ; and from k draw a perpen- dicular cutting similar lines drawn from b and c in k' and k\ which will be the corresponding tangent-points in the spheres B and C. Now from the point in D where the radius / cuts the circumference, set off the diameter on each side, thus dividing the circle into three equal parts, and from these points draw radii. On these radii set off the distance which /' is from the centre, then 7n and n will be the tangent points of the two other sides of the pyramid with the upper sphere. In the same way find the points o o',p p\ which will be the tangent-points on the lower spheres. These points, however, are in the plan, and it is required to find them in the elevation. From /, in Fig. 2, draw a horizontal line, and from m and n in the plan draw perpendiculars cutting it in ;«', which will represent in the elevation the points m and n. From k (Fig. 2) draw a horizontal, and from p and i)\ Fig. I, draw perpendiculars intersecting it in p" and i)", which will represent not only the tangent-points p, p\ in the near side of the pyramid, but o and o\ on the distant side as well. Produce the radii b k and c k', and draw the line G H at right angles to I L. Produce the radii c o\ a o, a p, and b p\ and set off on them from a b and c, the length c q, or b r — viz., j, /, 7t, v. Through these points draw lines which, meeting the line G H, will complete the triangle G H J, which is the base of the pyramid. K3G> TRACES, NORMALS, AND TANGENT-PLANES. From these three points draw lines to the centre of the upper sphere in the plan, and from this central point, which is the plan of the apex of the pyramid, draw a per- pendicular, which, cutting E F in K, will give the apex in the elevation. From J draw a perpendicular meeting I L in M. Draw M K, which will be the edge of the pyramid, and wiii complete the subject. QUESTIONS FOR EXAMINATION. The following questions are appended with the view of enabling the student to test his own knowledge, and as suggestions to teachers as to the mode of stating questions on this subject. It is hoped that the lessons given in this volume, and the appHcation of them, will have shown the constructions upon which all the questions are based. 1. Give the plan and elevation of a line 3 inches long, when parallel to the vertical and horizontal plane, and 2 inches distant from each. 2. Give the plan and elevation of this line when it is at right angles to the vertical and parallel to the hori- zontal plane, its height being 2 inches from the ground. 3. Give the plan and elevation of tne same line, when the former is a point, and the latter a vertical line 3 inches long. 4. Give the elevation* and plan of the same line when it is parallel to the vertical, but is inclined to the horizontal plane at 70°. 5. Give the plan and elevation of the line, when it is inclined at 70° to the horizontal, and 45° to the vertical plane. 6. A wire 3 inches long projects from a wall at 60° to the surface, and is parallel to the ground. Give the plan and elevation. 7. A plane 2" x 3" rests on its narrow edge in such a manner that its surface is at right angles tj both planes. Give plan and elevation. 8. Give plan and elevation of the same plane, when its sur- face is vertical, but inclined to the vertical plane at 45°. 9. Give plan and elevation of the same plane when its shorter edges are at right angles to the vertical plane, and its surface inclined to the horizontal plane at 60°. 10. Give plan and elevation when the plane rests on one • When elevation occurs before " plan," it is to suggest that the elevation should ■ be drawn first, and the plan projected from it, and vice versd, the drawings being executed in the order of their names. 138 PROJECTION. of its short edges, its surface being inclined at 60° to the horizontal plane, and its long edges being at 45° to the vertical plane. 11. A square plane of 3 inches side Hes on the horizontal plane, its one diagonal being at right angles to the vertical plane, and the other parallel to it. Give plan and elevation. 12. Give elevation and plan when the plane rests on one of its angles, its surface being inclined at 40° to the horizontal plane, but its one diagonal remaining at 90° to the vertical plane. 13. Give plan and elevation of the same plane when one of its diagonals is at 45° to the horizontal, and 60° to the vertical plane, the other diagonal being parallel to the horizontal plane. 14. A cube of 2 inches side stands on the horizontal plane, with two of its faces parallel to the vertical plane. Give its plan and elevation. 15. Draw its plan and elevation when standing on one of its sides, the opposite one being horizontal, and the others being at 45° to the vertical plane. 16. Give plan and elevation when resting on one of its solid angles, one diagonal of the base being at 50° to the horizontal, and the other at 90° to the vertical plane. 17. Draw elevation and plan of the same cube, when rest ing on one of its edges, so that two of its sides are vertical and the rest make angles of 45° with the hori- zontal, but are at right angles to the vertical plane. 18. Add the shape (the develoMnent) of the piece of metal or other substance which on being folded would form the above-named cube. 19. There is a stick of timber 2 inches square at base, and 5 inches high. Give the true shape of a section caused by a plane entering at one angle of the top, and emerging at the opposite angle of the base. 20. Give the development of one portion of this square prism. 21. Give plan and elevation of a triangular prism when resting on one of its long faces, the surface of the triangular end being at 50° to the vertical plane. — The end is an equilateral triang:le of 2 inch edge, and the length of the prism is 3I inches. 22. Give plan and elevation of the same prism when the edcf-e of the end on which it rests is at 50*^ to the QUESTION'S FOR EXAMINATION. 1 39 vertical plane, and the under side is inclined to the horizontal plane at 35°. 23. Add the development of this prism. 24. Draw the plan and elevation of a regular pentagon of I inch side when resting on one of its angles, so that its surface is at right angles to the vertical, and at 60° to the horizontal plane. 25. Give the projection of this polygon when the line joining the angle on which it rests to the middle of the opposite side is at 40° to the vertical plane, the inclination to the horizontal plane remaining the same as in the last figure. 26. There is a hexagonal prism of i inch side and 4 inches long. Draw plan and elevation when standing on its end, with two of its faces parallel to the vertical plane 27. Give the plan and elevation of the same prism, when the axis is vertical and one of its faces is at 40° to the vertical plane. 28. Give elevation and plan of the same prism when two of its faces are parallel to the vertical plane, and the prism is so inchned that the axis is at 50° to the hori- zontal plane. 29. Draw the plan and elevation when the prism rests on one of the sohd angles, and the axis is at 50° to the horizontal and 45° to the vertical plane. 30. Project the prism when lying on one of its long faces, the axis being at 40° to the vertical plane. 31. Give the tnie section caused by a plane passing from one angle of the top to the opposite angle of the bottom. 32. Draw the development of the prism, marking on it the line of section, as per last figure. 33. Give the plan and elevation of a pyramid formed of four equilateral triangles of 3 inches side, when one edge of the base is at 35° to the vertical plane. 34. Draw the plan and elevation of a pyramid, the base of which is a square of 2 inches side, and the altitude of which is 3 inches, when its axis is vertical, and two of the edges of the base are parallel to the vertical plane. 35. Draw the same pyramid when the edges of the base are at 45° to the vertical, and the axis perpendicular to the horizontal plane. 36. Give the projection when the pyramid rests on one angle of the base, the surface of which is inclined at 30° to the horizontal plane. ,,I40 PROJECTION. yj. Draw the true shape of a section caused by a plane passing from a point in one of the long edges at 2 inches, to a point in the opposite edge i inch from the bottom. Give the development of the pyramid, marking on it the line of section. 38. Give plan and elevation of a hexagonal pyramid when two of the edges of the base (i inch long) are at 20° to the vertical plane, the altitude being 2^ inches. 39. Draw elevation and plan of this pyramid when lying on one of its triangular faces, with its axis parallel to the vertical plane. 40. A circular disc (i^ radius) stands so that one dia- meter is vertical, and another at right angles to the first is at 50° to the vertical plane. Give plan and elevation. 41. Give elevation and plan of the same circular disc, when resting on the end of one diameter, which is parallel to the vertical plane, the surface being at 40° to the horizontal plane. 42. Draw the plan and elevation of the same disc, when the diameter is at 40° to the horizontal and 60° to the vertical plane. 43. A circular slab of stone, such as a mill-stone, 4 feet diameter and i foot high {J,o be represented by inches for feet), lies on the horizontal plane. Give the plan and elevation. 44. A second circular slab, 3 feet diameter and i foot high, rests on a slab, similar to the last ; their centres being coincident. Draw the plan and elevation. 45. Draw the elevation and plan of these two slabs, one placed on the other, as above, when their circular surfaces are inclined at 40° to the horizontal plane. 46. A cylinder, 4 inches long and 2 inches diameter, stands on its circular end. Give the plan and elevation. 47. Draw the plan and elevation of the same cylinder when lying on the horizontal plane, its axis being parallel to both planes of projection. 48. Give plan and elevation of the cylinder when lying on the horizontal plane, its axis being at 60° to the vertical plane. 49. Draw the plan and elevation of a cylinder 4 inches QUESTIONS FOR EXAMINATION. 14 1 long and 2 inches diameter, when the axis is inclined at 60° to the horizontal and 45° to the vertical plane. 50. Give the true section caused by a plane passing through the middle point of the axis at 45° to it. 51. Draw the development of this cylinder, marking on it the line of section. 52. A cylindrical pipe, of 2 inches diameter, is to be cut so as to turn a right angle. Give plan and elevation, showing the section-line. 53. Give the elevation and plan of one of the parts when resting on the sectional surface. 54. Give the true shape of the section, and the develop- ment, showing how both parts of the elbow may be cut out of the same piece of metal without any waste. 55. From piping of the same diameter, construct a double elbow-joint, one end of which bends one way and the other the opposite. Give development of the three parts to be cut out of one piece without waste. 56. The same piping is to be carried round three sides of a square room (size at pleasure). Give development, showing the section-line. 57. A pipe of sheet iron (2 inches diameter) is to be joined so as to turn an angle of 120°. Show on an elevation the inclination of the hne of section, and show on a development the line in which the metal must be cut to form the required parts without any waste. 58. Given a cone of 2\ inches base and 3^ inches altitude. Draw the plan and elevation of this cone when standing on its base. 59. Give elevation and plan, when the cone hes on the hori- zontal plane, its axis being parallel to thevertical plane. 60. Draw the projection of the cone, when lying on the horizontal, with its axis at 45° to the vertical plane. 61. Project the cone when resting on one end of the diameter of the base, the axis being inclined at 70° to the horizontal plane. 62. Project the cone, when the axis is inclined at 70° to the horizontal and 45° to the vertical plane. 63. Draw the true section of the same cone caused by a plane at 40'' to the surface of the base, which enters at \ inch from the bottom. Add the development, marking on it the line of section. Ia.2 PROJECTION. 64. Draw the parabola resulting from a plane entering the base of a similar cone at f inch from the centre. 65. Draw the hyperbola resulting from a section-plane entering the base of a similar cone at f inch from the axis. 66. A pipe 2 inches square is penetrated by another of I mch side. The smaller one passes through 2 sides of the larger, their axes being at right angles to each other. Give elevation and plan when tvvo faces of each of the pipes are parallel to the vertical plane. d"]. Project this olDJect when the two faces, which in the last case were parallel to the vertical plane, are at 60° to it. 68. Give the development of the larger pipe, showing the exact shape of the aperture through which the smaller one is to pass. 69. Give the elevation and plan of the object when the smaller pipe penetrates the sides of the larger at 60°. 70. Draw the development of the larger pipe, showing the apertures, and of one piece of the smaller one. 71. A square pipe of 2 inches side is penetrated by another of 1 5 inch side, their axes being at 60° to each other, and parallel to the vertical plane ; and two edges of the smaller meeting two edges of the larger pipe. Give the elevation and plan. 72. Draw the plan and elevation, when two faces of the larger pipe are parallel to the vertical plane. 73. Draw the development of the larger pipe, showing the shape of the apertures through which the smaller one is to pass, and also one of the ends of the smaller pipe. 74. A cube of 3 inches side stands on the horizontal plane, and is surmounted by a square pyramid, formed of a base and four equilateral triangles of 3J inch side. Give elevation and plan, when tvvo faces of the cube and two of the sides of the base of the pyramid are parallel to the vertical plane. 75. Draw the elevation and plan of this object, when the faces are parallel to the vertical plane, as in the last question, but when the base is inclined at 25° to the horizontal plane. 76. Draw the plan and elevation of the object, when the sides of the cube are at 50° and 40° to the vertical plane. "]"], Give plan and elevation of the object, when the faces of the cube are at 45°, and two of the sidts of the QUESTIONS FOR EXAMINATION. 143 base of the pyramid are parallel to the vertical plane, their axes being coincident. y^. Draw the shape of the piece of metal to form a gas- shade, 20 inches wide across the circular base, 6 inches across the top, and 10 inches perpendicular height. (To be worked I size.) 79. A cylindrical coal-scuttle is to be made of sheet iron ; it is to be 10 inches in diameter and 18 inches high at the highest part, the lid to be inclined at 45°. Draw the shape the metal is to be cut to form this object, and the exact shape of the lid. (To be worked h size.) 80. A cylinder, 2\ inches diameter and 6 inches long, is penetrated by another of \h inch diameter and 5 inches long, their axes being at right angles to each other, and intersecting at their centres. Show the mode of obtaining the curves of penetration. Develop the larger cylinder and one of the ends of the smaller one. 81. Draw the plan and elevation of this object when the axis of the larger is parallel and of the smaller at 60° 10 the vertical plane. 82. There is a solid cross, formed by a central cube of I inch side, on each face of which another similar cube is fixed. Give the plan and elevation when two vertical faces of the original cube are parallel to the vertical plane. 83. Draw the plan and elevation when of the two adjacent vertical sides of the original cube one is at 60° and the other at 30" to the vertical plane. 84. Project this object when resting on one angle of the base of the lowest cube, which is inclined at 30° to the horizontal plane, the diagonal being parallel to the vertical plane. 85. A cone, the base of which is 4 inches and the altitude of which is 5 inches, is penetrated by a cylinder of 2 inches diameter. The axis of the cylinder intersects that of the cone at right angles, at i inch from the ground. Draw the plan and elevation when the axis of the cylinder is parallel to the vertical plane 86 Project this object when the axis of the cylinder is at 60° to the vertical plane. ^y. Draw the development of the cone (showing the aper- tures of penetration) and of one end of the cylinder. 144 PROyECTION. 88. A thread winds round a cylinder 6 inches high, and of 2 inches diameter, and reaches the top in 6 revo- lutions. Project the helix curve thus generated. 89. Project a V-threaded screw of 3 inches diameter and \ inch pitch. 90. There is a solid formed of two equal square pyramids of 2 inches base and 3 inches altitude, which are united by their bases. Draw the elevation and plan when the object rests on one of the triangular faces of one of the pyramids, the axis of the object being parallel to the vertical plane. 91. Give the projection of the object, when resting on one of the faces of one of the pyramids. The axis is at 45° to the vertical plane. 92. Draw the elevation and plan when the object rests on an edge of one of the pyramids, the axis being at 60° to the vertical plane. 93- ■ foot. Show 20 feet. 94. Draw an isometrical projection of a plane square of 2 inches side. 95. Give an isometrical projection of a pavement consist- ing of squares of i foot side. Scale, \ inch. Show 5 squares in width and 12 in length. Tint the alternate squares in pale Indian ink. 96. Draw an isometrical projection of a cube of 2 inches edge. 97. Draw the isometrical projection of a box 3 feet square and 2 feet high, made of wood 3 inches thick. Scale, I inch to the foot. 98. There is a block of stone, 6 feet square and i foot high ; on this rests another, of the same height and 4 feet square; and on this again a third block, of the same height and 2 feet square, is placed, the centres of the three blocks being over each other. Give the isometrical view of the group. Scale, \ inch to the foot. 99. A cylinder of 2 inches diameter and 4 inches long lies so that its end is vertical. Give the isometrical pro- jection. 100. There is a stool the top of which is a square of 12 inches side, the height 18 inches, and the thick- ness of the legs 2 inches (the other measurements at pleasure). Scale, 2 inches to a foot. Draw an isometrical view of this object. Printed by Cassell & COMPANY, LIMITED, La Belle SauTage, London, E.C 10.899 Selections from Cassell ^ Company's Publications. JUustratetr, yiuc-^rt, anti otijcr Volumes. 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Ellis. Illustrated. Cloth, 2S. 6d. each. In Red Indian Trails. Uncrowning a King. Two Boj's in Wyoming Kloiioik-* INuggets. Scouts and Comrades : o Tecuniseh, Chief of th Shawanoes. Cowmen ana B-ustlers. A Strange Craft and it ■Wonderful Voyages. Pontiac.Ciiiei of the Ottawas. i :i the Days of the Pioneers, cnoa with Silence. I'he Phantom ot tlie River, i ne Great Cattle Trail. The Path in the Ravine. The Young Ranchers, 'ihe Hunters of the Ozark. The Camp in the Mountains. Ned in the Woods. A Tale of Early Days in the West. Down the Mississippi. Ned on the River. A Tale ol Indi m River Warfare. The Last War Trail. Footprints in the i'orest. The Rtibber Hunters (formerly Up the Tapajos). Ned in the BiOCK House. A Story of Pioneer Life in Kentucky. The Lost Trail. Camp-Fire and Wigrwam. Lost in the Wilds. Lost in Samoa. A Tale of Adven- ture in the Navigator Islands. Tad; or, "G-etting Even "with Him. Books by Edward S. Ellis. Illustrated. Cloth, is. 6d. each. Captured by Indians. I Wolf Ear the Indian. The Daughter of the Chieftain I Astray in the Forest The "World in Pictures." IS. 6d. each. A;l the Ri-ssias. Chats about Germany. 'i.he Eastern Wonaerlaiid (Japan). Illustrated throughout. Cheap Edition. Glimpses of South America. The Land of Temples (India). The Isles of the Pacific. Peeps into China. The Land of Pyramids (Egypt) Hall-Crown Story Books. Faii'way Island. Heroes of the Indian Empire Working to Win. ^i the South Pole. Perils Afloat and Brigands Ashore. Pictures of School Life and Boy- hood. Books lor the Little Ones. Fully Illustrated. t.assell'3 Robinson Crusoe. With 100 Illustrations. Cloth, 3s. 6d; gilt edges. 5s. The Oia j; airy Tares. With OriHii:al Uliistrations. Cloth. Is. Cassell's Swiss Family Robinson. Illustrated. Cloth, 3s. 6d. ; j^^ilt edges, 58. The buuUay Scrap Book. With Several Hundred illustrations. Pajjcr boards, 3s. 6d. ; cloth, gik edges, 5s. Albums lor Children. 3s. 6d. each. The Album lor Home, School, and Play. Containinf; Stories by Popular Authors. UlustrateJ. My Uw'i Album ol Animals. With Picture Album of All Sorts. Full-page Illustrations. The Chii-Uhac Album. Illustrated throughout Cassell & Company's Complete Catalogue luUi be sent post Free on appiicatwn to CASSELl & COMPANY, Limited, Ludgate Uuly Lon.ion. THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO 50 CENTS ON THE FOURTH DAY AND TO $1.00 ON THE SEVENTH DAY OVERDUE. MAS fVW FEB U^W8 DEC 21 1 9 42 TW^^ U.'\pr'Dili) ?n'nn t5 1 S : eoctsiLu 70A p r .' ^2|!K SApr'S'gm 26P/ia\55'W SEN T ON tU O CT \ 8 »' U.C.BEHKELEV f'-Z'kA fafw.*^ i*^3^: 869032 THE UNIVERSITY OF CALIFORNIA LIBRARY iS^ 'ecfio