Q MECHANICAL DRAWING. PREPARED FOR THE USE OF THE STUDENTS OF THE MASSACHUSETTS INSTITUTE OF TECHNOLOGY, BOSTON, MASS. LINUS FAUNCE. BOSTON : W. J. Schofield, Printer, 105 Summer Street. 1887. Copyrighted, 1887, by Linus Faunce. CONTENTS. Page. CHAPTER I. — Instruments and their Uses 5 II. — Geometrical Problems 13 III. — Inking 55 " Tinting 61 IV. — Projections 65 " Notation 68 " Projections of Straight Lines 69 " " " Surfaces 75 " " Solids 80 V. — Shadows .' . . . 96 VI. — Isometrical Drawing 112 " Oblique Projections 120 VII. — Working Drawings 121 VIII.— Examples 127 2066025 MECHANICAL DRAWING. CHAPTER I. INSTRUMENTS AND THEIR USES. 1. To do good work, good instruments are essential. An accomplished draftsman may do fair work with poor instruments, but the beginner will find it sufficiently hard to do creditable work without being handicapped by poor instruments. It is also essential that the instruments should be kept in good order. They should be handled carefully, and wiped, before being put away, with wash-leather or chamois skin. This is especially needful if the hands perspire perceptibly. 2. Pencilling. Drawings should always be first made in pencil, and inked afterwards if desired. The idea of pencilling is to locate the lines exactly, and to make them of the required length. Accuracy in a drawing can only be obtained by accu- racy in the pencil construction. There is a great tendency among beginners to overlook this important fact, and to become careless in pencilling, thinking they will be able to correct their inaccuracies when inking. This is a ereat mistake, and one to 6 INSTRUMENTS AND THEIR USES. be especially avoided. This accuracy can be obtained in pencil only by making very Jine, light lines, and to this end hard pen- cils, 6 H, should be used, and they should be kept sharp. For drawing straight lines the pencil should be sharpened to a flat, thin edge, like a wedge. The compass pencils should be sharpened to a point. A softer pencil, 4 H, sharpened to a point, should be used in making letters, figures, etc. It should be borne in mind that a 6 H pencil sharpened to a chisel point will make a depression in the paper, which can never be erased, if much pressure is put on the pencil ; hence, press very lightly when using a hard pencil, so as to avoid this difficulty. If a drawing is not to be inked, but made for rough use in the shop, or where accuracy of construction (in the draw- ing) is not essentia], or to be traced, a soft pencil would prefer- ably be used, the lines being made somewhat thicker, or heavier. 3. Compasses. In using the compasses the lower part of the legs should be kept nearly vertical, so that the needle point will make only a small hole in revolving, and both nibs of the pen may press equally on the paper. In pencilling it is not so essential that the pencil point be kept vertical, but it is well to learn to use them in one way, whether pencilling or inking. Hold the compasses loosely between the thumb and fore- finger only, and do not press the needle point into the paper. If it is sharp, as it should be, the weight of the compass will be sufficient to keep it in place. While revolving, lean the com- pass very slightly in the direction of revolution, and put a little pressure on the pencil or pen point. In removing the pencil or pen point to change them, be very careful to pull them out straight ; do not bend them from side to side, in order to get them out more easily, as it would enlarge the socket and consequently spoil the instrument for accurate work. INSTRUMENTS AND THEIll USES. 7 Id drawing a circle of larger radius than could be drawn with the compass in its usual form a lengthening bar is used. In this case steady the needle point with one hand and describe the circle with the other. The large compasses are too heavy and clumsy to make small circles nicely, hence the bow compasses should be used in mak- ing all circles smaller than three-quarters of au inch radius, or thereabouts, depending on the stiffness of the spring. Be very careful to adjust the needle point to the same length as the pen- cil or pen point. In changing the radius of the bow compasses or spacers, press the points together, thus removing the pressure from the nuts, before turning the nuts in either direction. The screw thread will last much longer if this is done. 4. Dividers or Spacers. These are used to lay off distances from scales, or from other parts of a drawing to a line, or to divide a line into equal parts. In laying off the same distance several times on a line, keep one of the points of the dividers on the line all the time, and turn the instrument in an opposite direction each time, so that the moving point will pass~alter- nately to the right and left of the line. Do not make holes in the paper in doing this, as it is impossible to ink nicely over them ; a very slight puncture is sufficient. 5. T-Squa.re. The T-square should be used with the head against the left-hand edge of the drawing board (unless the per- son is left-handed), and horizontal lines only should be drawn with it. Lines perpendicular to these should not be drawn by using the head of the T-square against an adjoining edge of the board, as there is nO pains taken to make these edges at right angles to each other, but they should be drawn by using the triangle in connection with the T-square. 8 INSTRUMENTS AND THEIR USES. Lines should be drawn with the upper edge only of the T-square. In case you wish to use the T-square as a guide for the knife in cutting paper to size, do not use the upper edge as a guide, but turn the T-square over and use the bottom edge. For, unless you are very careful, the knife will nick the edge, which would render it unfit to draw lines with. 6. Triangles. To draw lines which shall be parallel to another by means of the triangles. Let AB be the given line. Place either edge of either triangle so as to coincide exactly with the given line. Place the other triangle (or any straight edge) against one of the other edges of the first triangle. Then, holding the second triangle, or straight edge, securely in this position with the left hand, move the first one, still keeping the two edges in contact. Any line drawn along the edge which originally coincided with the line AB will be parallel to it. To draw lines which shall be perpendicular to another by means of the triangles. Let AB be the given Fine. Place the longest side of the triangle so as to coincide exactly with the given line. INSTRUMENTS AND THEIR USES. 9 Place the other triangle (or any straight edge) against one of the other edges of the first triangle. Then, holding the second triangle securely in this position with the left hand, revolve the first one so that its third edge is against the second triangle or straight edge. Any line drawn along the edge which originally coincided with the given line AB will be perpendicular to that line. The right-hand portion of the figure shows how the two tri- angles may be used, in connection with the T-square, to draw lines making angles of 15° and 75° with a given line (in this case the line which coincides with the edge of the T-square). By turning the triangles over, these angles may be drawn in the opposite direction. Lines making angles of 30°, 45°, and 60° are drawn directly by means of one triangle and T-square or straight edge. 7. Irregular Curves. To trace an irregular curve through a series of points, use that part of the edge of the curve which coincides with the greatest possible number of points (never less than three), and draw the curve through these points, then shift the curve so as to coincide with other points in the same wav, letting the instrument run back on a part of the curve already drawn, so that a continuous smooth curved line may be formed. It requires a considerable practice to draw irregular curves by means of an instrument, the tendency being to make a series of loops, on account of some of the points being covered up. There is no better way to put in a curve in pencil than by doing it free-hand, provided the hand and eye have been properly trained. Of course the curve cannot be inked in free-hand ; the irregular curve must be used, but, being no longer confined to points, it is not difficult. 10 INSTRUMENTS AND THEIR USES. 8. Scales. As it is frequently impossible to make a draw- ing on paper the real size of the object, it is customary to reduce the actual measurements by means of an instrument called a scale, — that is, the drawing may be made •£, \, •£, T l $, etc., size, according as the relative size of the object and drawing may require. If it is desired to make a drawing £ size, then 3 inches on the drawing will represent one foot on the object. It is fre- quently necessary to represent inches and fractions of an inch, hence divide the 3 inches into 12 equal parts, and each of these parts will represent one inch on the object. If each of the 12 parts are subdivided into 2, 4, or 8 parts, each part would re- present respectively £, £, or £ of an inch on the object. This may be designated, scale, 3 inches equal one foot, or £ size. Ou the scale, one inch equal one foot, the unit, one inch, is divided into 12 parts to represent inches as before. Thus, to make a scale of any unit to one foot, it is simply necessary to divide that unit into 12 parts to represent inches, subdividing these parts, as far as possible, to represent fractions of an inch. If the smallest division on a scale represents £ of an inch on the object, the scale is said to read to ■£ of an inch. The student will find on his triangular scale ten different scales, viz., t£t, •£, T 3 ^, £, §-, ^, f , 1, l£, and 3 inches to the foot, reading to 3", 1", 1", 1", 1", £", f , £", ±", and £", respectively, (the double prime over a number or fraction means inches, the single prime indicates feet). The scale should never be used as a ruler to draw lines with. 9. Needle Point. Each student should procure a fine needle, break off the eye end, and force the broken end into a .small, round piece of soft pine wood. This is to be used in pricking off measurements from the scale, marking the exact INSTRUMENTS AND THEIR USES, H Cap, . . . . 13 x 17 inches. Demy, . . 15 x 20 " Medium, . . 17x22 " Royal, . . 19x24 " Super-Royal, . 19 x 27 " Imperial, 22x30 " intersection of two lines, etc. Here, as in the case of the needle point in the compasses, it should not be forced into the paper : the finest puncture possible is sufficient. 10. Drawing Paper. This paper comes in sheets of standard sizes, as follows: — Elephant, . . 23 x 28 inches. Columbia, . . 23 x 34 Atlas, . . . 26 x 34 Double Elephant,27 x 40 Antiquarian, . 31 x 53 Emperor, . . 48 x 68 Whatman's paper is considered the best. This paper is either hot or cold pressed, the hot pressed being smooth and the cold pressed rough. The rough paper is better for tinting work, the smooth takes ink lines better than the rough, but erasures show much more distinctly on it, hence the cold pressed is better for general work. The names of the sizes of the paper given above have no reference to quality. There is very little difference in the two sides of the paper, but that one which shows the maker's name in water lines, when held up to the light, is con- sidered the right side. 11. Thumb Tacks. These are used for fastening the paper to the drawing board when it is not necessary to stretch it. 12. The geometrical problems in the next chapter are not given with the view of teaching geometry, but to give the stu- dent practice in the accurate use of his instruments. In order that the degree of accuracy of the execution of the problems may be readily seen, these problems will not be inked. 13. The plates on which these problems are to be drawn should be laid out 10'' by 14" (and cut this size when finished) 12 INSTRUMENTS AND THEIR USES. with a border line one inch from each edge. That portion of the plate within the border line is to be divided into 6 equal squares, in each of which one problem is to be drawn, beginning with No. 1 in the upper left-hand corner square, No. 2 in the upper middle, No. 3 in the upper right, etc. The number of the plate is to be printed in the upper right- hand corner of the plate, about one-eighth of an inch above the border line, and the student's name in the lower right-hand cor- ner, about one-eighth of an inch below the border line. Prou. 1, 11, etc., as the case may be, is to be printed in the upper right-hand corner of each square. The initial letters of the name should be made T 3 ^ of an inch high, and the small letters ^ of an inch high. The letters are to be made like the samples furnished in the drawing room, and they should be made as nicely as possible. CHAPTER II. GEOMETRICAL PROBLEMS. Prob. 1, To bisect a straight line AB, or arc of a circle AFB. With A and B as centres and any radius greater than one half AB draw arcs intersect- ing in C and D. Join CD. CD is perpen- B dicular to AB, and E and F are the middle \i/ points required. Note. To draw a perpendicular to a line at any point in it, as E in AB. Lay off equal distances EA and EB on each side of E, and proceed as above. Prob. 2. From a point C outside a straight line AB to draw a perpendicular to the line. Wjth C as a centre and any convenient radius cut AB in the points A and B. With B , A and B as centres and any radius draw arcs _„-' intersecting in D. Join C and D, and CD is v i / the perpendicular required. 14 GEOMETRICAL PROBLEMS. Prob. 3. To draw a perpendicular to a line ATS from a point C nearly or quite over its end. Draw a line from C to meet AB in any point B. Bisect BC in D by Prob. 1. With 1) as a centre and radius DC draw the arc CAB meeting AB in A. Draw through A and C. . r^- '* '-----■-'' Note. If a perpendicular be required at A, take any point D as a centre and radius DA and draw an arc CAB. Through B and D draw a line to meet the arc in C. Draw through C and A. Prob. 4. To draw a perpendicular to a line AB from a point A at or near its end. With A as a centre and any radius draw the arc CD. With centre D and same radius cut CD in C. With C as a centre and same \\ radius draw an arc over A, and draw a line \\ through D and C, producing it to meet this arc in E. Draw through A and E. /T Prob. 5. A second method. With centre A and any radius draw an arc CDE ; with centre C and same radius cut this arc in D ; with centre D and same radius draw \ arc EF ; with centre E and same radius draw 1 g arc intersecting EF in F. Join AF. GEOMETRICAL PROBLEMS. 15 Prob. 6. To construct an angle equal to a given angle CAB. Draw any line as DE, and take any point in it as D. With A as a centre and any radius cut the sides of the given angle in B and C. With centre D and same radius draw arc EF. With BC as a radius and E as a centre cut arc EF in F. Join DF. Note. For accuracy of construction in drawing this prob- lem the longer the radius AB is the better. Prob. 7. Through a given point C to draw a line parallel to a given line AB. o^= c_ From C as a centre and any radius \ s' \ draw the arc AD ; and from A as a centre, \ ,'' \ with the same radius, draw the arc BC. — ■- Si. With BC as a radius, and A as a centre, draw an arc cutting the arc AD in D. Draw through D and C. Prob. 8. To draw a line parallel to a given line AB at a given distance CD from it. From any two points A and B on the line as centres and with CD as a radius draw arcs E and F. At A and B erect perpendiculars to meet the arcs in E and F. Draw a line c^ through E and F. Prob. 9. To bisect a given angle BAC. With A as a centre and with any radius draw the are BC, cutting the sides of the angle '' in B and C. With centres B and C and any radius draw arcs intersecting in D. Draw AD. 16 GEOMETRICAL PROBLEMS. Prob. 1 0. To divide a given line AB into any number of equal parts. (In this case 6.) , f From A draw an indefinite line A, 1, 2 • • 5 ? , ?'/ 1\ at any angle with AB. At B draw Ba • • • e, A, -,' ,' / / / 1 b making the angle ABe equal to the angle ',/ / / ;- a BA5. With any distance as a unit, lav off on i, jib •' J ;\! d b the lines from A and B as many equal spaces as the number of parts required less one. Join ]r, '2d, ?>c, etc. Prob. 11. Another method. (say) five parts. ■i 9-' To divide a line AB into Draw A 1 • • • • 5 at any angle to AB, and lay off on it five equal spaces, using any con- venient unit. Join 5B. and through the points 1, 2, o, 4 draw lines parallel to Bo, meeting AB in points a, b, c, d. Prob. 12. To divide a line AC into the same proportional parts as a given divided line AB. Draw from a point A the lines AC and AB, making any angle. Join B and C. Through the points 1, 2, and 3 on AB draw lines parallel to BC, meeting AC in points a, b, and c, the required points of division. GEOMETRICAL PROBLEMS. 17 Prob. 13. C'a. \b. Second method. Let M be the line to be divided into parts proportional to the parts Al, 12, 23, and 3B of the line AB. Draw M parallel to AB at CD by Prob. 8. Draw lines through AC and N BD to meet in E. Through E draw lines to ^ 1, 2, and 3, cutting CD in a, b, and c, the re- quired points of division. Note. If the parts on AB are equal, the parts on CD will be equal. --D -r Prob. 14. To draw on a given line AB as an hypothenuse a right triangle with its sides having the proportion of 3, 4, and 5. Divide AB by Prob. 11 into 5 equal parts. With centre B and a radius equal to three of the parts draw an arc, and with centre A and a radius equal to four of the parts cut this arc in C. Join AC and BC. Prob. 15. To trisect a right angle CAB. With centre A and any radius draw the arc of the quadrant, cutting the sides in C and B. With centres C and B and the same radius cut the arc in points 1 and 2. Join Al and A2. Note. The angle 1AB is an angle of 60°, the construction of which is apparent. 18 GEOMETRICAL PROHLKMK. Prob. 16. and N. C Bi To find a mean -proportional between two lines M Upon an indefinite line AB lay off AC equal to M, and CB equal to N. Bisect AB in E by Prob. 1, and with centre E and radius EB draw a semicircle. At C draw CD perpen- dicular to AB (Prob. 5). CD is the mean pro- portional required. Prob. 17. and P. To find a fourth proportional to three lines M, N, Draw AB equal to M, and at any conve- nient angle draw AC equal to N. Upon AB produced make BD equal to P. Join BC, and draw DE parallel to BC, to meet AC produced in E. CE is the fourth pro- ,b__jL portional required. Prob. 18. At a point A in a line AC to make an angle of 30°. With any point B as a centre and radius AB draw a semicircle ADC. With centre C and same radius cut this arc in D. Draw - AD. DAC = 30°. Prob. 19. Having given the sides AB, M, and N of a tri- angle to construct the figure. ~^c/ With centre A and radius M draw an arc. With B as a centre and radius N draw an arc to cut the first arc at C. Join AC and BC. GEOMETRICAL PROBLEMS. 19 Prob. 20. On a given side AB to construct a square. 1Q Draw BD at right angles to AB and equal ) to AB (Prob. 5). With A and D as centres and radius AB draw arcs intersecting in C. Join AC and CD. Prob. and M. t 21. To construct a rectangle of given sides AB At B draw BD perpendicular to AB by Prob. 5, and equal to M. With A as a cen- tre and radius equal to M draw an arc, and from D as a centre and a radius equal to AB cut this arc in C. Join AC and CD. Prob. 22. On a given diagonal AB to' construct a rhombus of given side AC. With centres A and B and radius AC draw arcs intersecting in C and D. Join AC, AD, 6 BC, and BD. Prob. 23. On a given base AB to construct a pentagon. With centres A and B and radius AB draw circles intersecting in 1 and 2. Join 1 and 2. With centre 2 and same radius draw the circle 3A5B4, giving points 3 and 4. Produce 35 to C and 45 to E. With centres C and E and radius AB draw arcs intersecting in D. Draw BCDEA. Note. This is an approximate method. 20 GEOMETRICAL PROBLEMS. Prob. 24. To construct a regular hexagon of given side AB. With A and B as centres and radius AB draw arcs intersecting in 0. With centre O and radius AB draw a circle, and lay off BC, CD, etc., each equal to AB. Join the points B, C, D, E, F, and A. Note. The radius of any circle goes around the circumfer- ence as a chord six times exactly. Prob. 25. On a given base AB to construct a regular poly- gon of any number of sides (in this case 7). With centre A and radius AB draw a semicircle and divide it at points 1, 2, 3, 4, \| etc., into as many equal parts as there are sides in the required polygon. Draw a line from the second point of division 2 to A. 2 A is one side of the required polygon. Bisect AB and A2 hy perpendiculars, by Prob. 1, meeting in D. With D as a centre and radius DA draw the circle BA2, etc. Apply AB as a chord to the circle as many times as there are sides in the j^olygou. J-^---- rF^B Prob. 26. On a given line AB to construct a polygon of any number of sides. An approximate method. Bisect AB, and produce the bisecting line indefinitely. With centre A and radius AB draw the arc BC, cutting the bisecting line in C. Divide the arc BC into six equal parts, in points 1, 2. 3, etc. To construct a pentagon. With cen- tre C and radius CI draw an arc, cutting the bisecting line in a point below C, GEOMETRICAL PROBLEMS. 21 which is the centre of the circle circumscribing the pentagon. If the polygon is to have more sides than six, set up from C on the line Cab as many parts of the arc CB as added to six make the number of sides of the required polygon ; thus, for a seven- sided polygon set up one division as Ca ; for an eight-sided set up two divisions as Cab, and so on. a, b, c, d, etc., are the cen- tres for the circumscribing circles of the polygons, each side of which is equal to AB. PrOB. 27. On a given base AB to construct a regular octa- gon. At A and B erect perpendiculars by Prob. 5 to AB, and bisect the exterior right angles, making the bisectors AC and BD each equal to AB. Draw CD, cutting the perpendiculars in E and F. Lay off EF a b"" "' from E to G and from F to H. Draw an indefinite line through GH. Make GK, GL. HN and IIM each equal to CE or FD. Connect C. K, L, M, N, and D. Note. This may be done by Prob. 25. Check the con- struction by seeing if AN, CM, BD, and KL are parallel. CA, KB, etc. should be parallel ; so also AD, CN, and KM. Prob. 28. To construct an octagon within a square ABCD. Draw the diagonals AC and BD. With A, B, C, and D as centres and the half of the diagonal of the square as a radius draw the arcs EF, GH, KL, and MN. Join the points MG, FL, HN, and KE. D c ^~~-% L ',-' H X B.---5/' » V- - - L i 22 GEOMETRICAL PROBLEMS. Prob. 29. The altitude AB of an equilateral triangle being given, to construct the triangle. Draw CD and EF, both perpendicular to ^ AB. With A as a centre and any radius describe the semicircle CGHD. With C and D as centres and the same radius draw arcs cutting the semicircle in G and H. Draw AGE and AHF. Prob. 30. Given the base AB of an isosceles triangle, and the angle at the vertex M, to draw the triangle. With centre B and any radius draw a semicircle cutting the base produced at C. Make the angle DBC equal to M by Prob. 6. Bisect ABD by Prob. 9. Make the angle FAB equal to EBA, and produce AF and BE to meet in G. Prob. a square 31. Within a regidar hexagon ABCDEF to inscribe St- Draw a diagonal AD. Bisect AD by a perpendicular 213 (Prob. 1). Bisect by Prob. 9 the right angles 21 A and 2 IB, and produce the bisectors to meet the sides of the hexagon in points G, H, L, and K. Join G, H, L, and K. Prob. 32. On a given diagonal AB to construct a square. Bisect AB in O (Prob. 1) ; and with centre and radius OA draw a circle to cut the bisect- ing line in C and D. Draw AC, CB, BD, and DA. GEOMETRICAL PROBLEMS. 23 Prob. 33. Within a given triangle ABC to inscribe a square. Draw AD perpendicular to and equal to AB. From C draw CE perpendicular to AB (Prob. 2). Draw DE cutting AC in F. From F draw FK perpendicular to AB. Make KH equal to KF, and from H with radius HK cut BC in G. Join EH B FG and GH. Prob. 34. About a given circle FEGD to circumscribe an equilateral triangle. Draw the diameter DE. With centre E and radius of the given circle draw the cir- cle AFG. Prolong DE to A. With centres D, F, and G and radius DG draw arcs inter- secting at B and C. Join AB and AC. To circumscribe a circle about a triangle ABC. By Prob. 1 bisect two of the sides AB ^j. and BC by perpendiculars meeting in 0. With centre O and radius OA draw the circle. Note 1. If any three points are given not in the same straight line, a circle is passed through them by joining the points and proceeding as above. Note 2. If any circle is given, its centre is found by taking any three points in its circumference, and proceeding as above. ■24 GEOMETRICAL PROBLEMS. Puob. 36. To inscribe a pentagon within a circle. Draw any diameter AB, and a radius CD perpendicular to it. Bisect BC in E. With centre E and radius ED draw the arc DE. B With centre D and radius DF draw the arc FG. DG is the side of the required pen- tagon. To inscribe a hexagon within a circle. Draw the diameter AB, and with centres A and B and the radius of the given circle draw arcs COD and EOF, cutting the circum- ference in C, D, E, and F. Join AD, DF, FB, etc. Note. Joining points A, E, and F gives an inscribed equilateral triangle. Prob. 38. To construct a regular polygon of any number of sides, the circumscribing circle being given. Draw any diameter AB and divide it into as many equal parts as there are sides in the required polygon (in this case 5). With A and B as centres and radius AB draw arcs intersecting in C. Draw a line from C through the second point of division of AB to meet the circumference in D. AD is one side of the required polygon. Lay off AD as a chord as many times as there are sides in the required polygon (in this case o). Note. This is an approximate method. ;£' GEOMETRICAL PROBLEMS. 25 Prob. 39. To circumscribe a hexagon about a circle. Draw the diameters AB and CD perpen- dicular to each other. Divide each quadrant into thirds, by Prob. 15, at points E, F, etc. Join B and E, cutting CD in G. With G as a centre and radius GE draw arc ED, cutting COD in D. With centre O and radius OD cut the diameters produced in points H, K, C, L, and M. Join points H, D, M, L, C, and K. Note. Practically the 60° triangle placed on a T-square whose blade is parallel to COD will give HD, DM, etc., by making it tangent to the circle at E, F, etc. KH and LM are drawn by the T-square. Prob. 40. a circle. Prob. 41. without it. To circumscribe a square, also an octagon, about Draw the diameters AB and CD at right angles to each other. With centres A, B, C, and D and radius OA describe arcs intersect- ing in points E, F, G, and H. These points con- nected give a square about the circle. Inscribe an octagon in the square by Prob. 28. To draw a tangent to a circle B from, a point A Draw AB and bisect it in D. With centre D and radius DA draw a semicircle cutting the given circle in C and E. Join AC. By joining AE a second tangent is found, equal to AC. Notk. To draw a tangent to a circle from a point C on the circumference. Join BC, and at C draw AC perpendicular to BC. For the tangent is perpendicular to the radius at the jioiuf of tangency. 26 GEOMETRICAL PROBLEMS. Prop.. 42. To draw a tangent to the arc of a circle when the centre is not accessible. Let C be the point upon the given arc, AB, at which the tangent is to be drawn. Lay off equal distances upon the arc from C to A and B. Join A and B. Through C draw a line par- allel to AB by Prob. 7. Prob. 43. To draw a tangent at a given point A on a circle when the "preceding method is not applicable. From A draw any chord AB. Bisect AB in C, and the arc ADB in D by Prob. 1. With A as a centre and a radius AD draw the arc EF. With D as a centre and radius DF draw an arc cutting EF at E. Join AE. Prob. 44. To draw a tangent to two given circles, A and B. Through A and B draw a line. Make DH equal to the radius BF. Draw the circle A-HC, and from B draw the tangent BC by Prob. 41. Draw AC and produce it to E. Make the angle FBK equal to CATI. Join EF. Prob. 45. To draw a tangent to two given circles which shall pass between them. Join A and B, and draw AD and BC perpendicular to AB. Draw DC, cut- ting AB in E. Draw a tangent from E by Prob. 41 to the given circles. Join the tangeut points F and G. FG is the required tangent. GEOMETRICAL PROBLEMS. 27 Prob. 46. To draw a circle tangent to a given line AB at a given point B in it, which shall also pass through a fixed paint C without the line. Draw BD perpendicular to AB, at the point B. Join CB, and draw a perpen- dicular to it at its middle point, Prob. 1. The intersection of this perpendicular and BD gives D the centre required. Prob. 47. To draw a circle of a given radius AB, which shall be tangent to a given circle C, and also to a straight line DE. Draw GH parallel to DE, by Prob. 8, at the distance AB. With a radius CM, equal to the radius of circle C plus AB, draw an arc to meet GH in 0. With the centre O and radius AB draw the required circle. Note. If two circles are tangent, the straight line connect- ing the centres passes through the point of tangency. This point it is very important to locate precisely in all cases of tan- gency. Prob. 48. To draw a circle tangent to a given circle D, and also tangent to agiven line AB, at a given point B on the line. Draw BC perpendicular to AB, and make BC equal to the radius of D. Join DC, and at its middle point draw a perpendicular to meet BC in E, the required centre. Note. By laying off BC above the line AB, and proceeding as above, another circle is found helow AB. 28 < ; E( tM ETUICAL PROBLEMS. Prop.. 49. To draw a circle tangent to a given circle A, and a given line BC, at a given point E on the circle. Draw AE and produce it. At E draw a perpendicular to AE, meeting BC in B. Draw the bisector of the angle EBC, meeting AE produced in D. DE is the radius required! Note. By bisecting EBM another circle, whose centre is G, is found, enclosing the circle AE. Prob. 50. To draw a circle of given radius R tangent to two given lines. AB (aid CD. Draw lines parallel to AB, and CD, at the given distance R from them by Prob. 8, meeting in E, the required centre. Prob. 51. To draw any number of circles tangent to eacli other, and also to two given lines AB and AD. Bisect BAD by AC, Prob. 9. Let one of the circles be 45ED, drawn by _\c taking C as a centre, and a radius / equal to the perpendicular CD from C to AD. At E draw EF perpendic- ular to AC. With centre F and ra- dius FE draw the arc DEG, cutting AD in G. At G make GH perpendicular to AC. H is a centre required. Repeat the process. GEOMETRICAL PROBLEMS. 29 Prob. 52. To draw a circle through a given point D and tangent to two given lines AB and AC. Draw AD. Bisect BAG by AE. From any point K, on AE as a centre, draw a circle tangent to AB and AC, and cutting AD in H. Draw HK. At D draw DE, mak- ing the angle ADE equal to AHK. E is the required centre. Prob. 53. To draw a circle of a given radius R tangent to two given circles A and B. Through A and B draw indefinite Hues, and make DE and EG each equal to R. With A and B as centres, and radii AE and BG, draw arcs cut- ting each other at C, the required cen- tre. R ^ Notk. These arcs will intersect in a second centre. Prob. 54. To draw a circle through a point C, and tangent to a given circle A, at a point B in its circumference. Draw AB and produce it. Join BC and bisect it by a perpendicular meeting -?Si^ AB produced in D, the required centre. 30 GEOMETRICAL PROBLEMS. Prob. 55. Given two parallel straight lives AB and CD, to draw ares of circles tangent to them at B and C, and pasting through E, which, is anywhere on the line BC. jc e_ At B and C erect perpendiculars. Bisect BE and CE by perpendiculars (Prob. 1), meeting the perpendiculars from B and C in F and H, the re- quired centres. Draw the arcs BE This is called a reversed curve. Prob. 56. To draw a circle tangent to two given circles A and B at a given point C in one of them. Draw a line indefinitely through A and C. Make CD and CE each equal to the radius of B. Join BE and BD. Bisect BE and BD by perpendiculars meeting AC pro- duced in F and G, the centres of the required circles. Note. The tangent points should be determined accurately. Prob. 57. Same as Prob. 56. A second method. Draw a line through A and C indefinitely. Through B draw DE parallel to CA, cutting B in D and E. Draw CD to F and CE. cut- ting B in H. Through B and H draw BHK, cutting CA in K, one centre required. Through B and F draw FBL, cutting CA in L, an- other centre required. GEOMETRICAL PROBLEMS. 31 Prob. 58. Same as Prob. 56. Prob. 59. Same as Prob. 56. Method of Prob. 57. Probs. 60 and 61. Same as Prob. 56. Method of Prob. 57. Prob. 62. To inscribe a circle within a triangle ABC. Bisect two of the angles of the triangle, > ': say A and B, by Prob. 9. The bisectors meet in D, the centre of the required circle. A perpendicular from D to either side is the ~* required radius. 39 <; KO METRICAL PROBLEMS. Prob. 63. Within an equilateral triangle ABC to inscribe three equal circles, each touching two others, and two sides of the triangle. Draw the bisectors of the angles A. B, and C, cutting the sides in D, E, and F. With centres D, E, and F, and radius DF, draw arcs cutting the bisectors in H, L, and K, the required centres. Pkob, 64. Within an equilateral triangle to draw three equal circles, each touching two others and one side of the triangle. Bisect the angles A, B, and C. Bi- sect the angle DAB by AG. G is the ceutre of one of the required circles. With the centre O of the triangle as a centre, and radius OG, draw a circle cutting AD in H and BE in K. With centres G, H, and K, and radius GF, \-K draw the circles G, H, and K. Prob. 65. Within an equilateral triangle ABC to draw six equal circles which shall be tangent to each other and the sides of the triangle. Inscribe the three circles K, H, and < i by Prob. 64. Draw LGM parallel to AB, MIIN parallel to BC, and LKN parallel to AC. The points L, M, and N are the centres of the other three circles. GEOMETRICAL PROBLEMS. 33 Pkob. 66. Within a square ABCD to draw four equal cir- cles each touching two others and two sides of the square. Draw the diagonals AC and BD, and the diameters EF and GH. Draw ELI, HF, FG, and GE, giving the centres M, N, O, and P. The radius OR is found by joining OP. Prob. G7. Within a given square ABCD to draw four equal circles each touching t>vo others and one side of the square. Draw the diagonals AC and BD, and the diameters EF and GIT. Bisect the angle OAB by AK, cutting EF in K. * With radius OK and centre O draw a cir- cle cutting the diameters in the points L, M, N, aud K, the required centres. Prob. 68. Within a square ABCD to draw four equal semi- circles, each tangent to two sides of the square, and their diame- ters forming a square. Draw the diagonals and diameters. Bi- sect FC aud Bli in L and K. Draw LK, cutting GH in M. Set off SM from S on the diameters to N, O, and P. Join the points M, N, O, P. The intersections of these lines with the diagonals give the required centres 1, 2, .'3, aud 4. 34 GEOMETRICAL PROBLEMS. PrOB. 69. Within a square ABCD to draw four equal semi- circles, each touching one side of the square and their diameters forming a square. Draw the diagonals and diameters. Draw EH, HF, FG, and GE, giving the points K, L, M, and N. The lines joining these L points cut the diameters in points 1, 2, 3, and 4, the required centres. Pkob. 70. To draw within a given circle ABC three equal circles tangent to each other and the given circle. Divide the circle into six equal parts by diameters AE, DC, etc. (Prob. 24). Produce any diameter, as AE to G, mak- ing EG equal to the radius of the given circle. Join CG. Bisect the angle OGC - 5 by GH, intersecting OC in II. With centre O and radius OH draw the circle HKLM. K, L. and M are the centres of the circles. Puob. 71. To dram within a given circle ACDB four equal circles which shall be tangent to each other and the given circle. Draw the diameters AB and CD at right angles, and complete the square FBED. Draw the diagonal FE and bisect the angle FED by EG; FD and EG meet in G. With centre F and radius FG draw a cir- cle cutting the diameters in G, H, K, and L, L^ the required centres. GEOMETRICAL PROBLEMS. 35 Prob. 72. Within a given circle AFD . . . . C to draw six equal circles tangent to each other and the given circles. Draw the diameters AB, CD, and EF, di- viding the circle into six equal parts (Prob. 24, Note). Divide any radius as OB into three equal parts (Prob. 11), at points F and G. With centre O and radius OF draw a cir- cle giving the centres F, 1,2, 3, 4, 5 required. A circle of the radius FB may be drawn S b T from centre O tangent to the six circles. Note. The above is a special method. The general method, to draw any number of equal circles in a given circle, tangent to each other and the given circle, is to divide the circle by diameters into twice as many equal parts as circles required. From B, the extremity of any one of these diameters, draw a tangent ST, Prob. 41, Note. Produce the diameters on each side of AB to meet the tangent in S and T. Bisect the angle T. The bisector meets OB in F, the centre of one of the required circles. Or F may be obtained by making TM equal to TB, and at M drawing a perpendicular to OT, meeting OB in F. With centre O and radius OF draw a circle cutting the diameters in points 1, 2, 3, 4, etc., the centres required. Prob. 73. About a. given circle to circumscribe any number of equal circles tangent to each other and the given circle. ^ — i — ~„ Divide the circumference of the (riven circle by diameters into twice as many parts as circles required. From the extremity B of any diameter draw a tangent SBT (Prob. 41, Note), and produce the diameters on each side of OB to meet SBT in S and T. Produce OT, making TN equal to TB. Make NR perpendicular to TN, meeting 36 GEOMETRICAL PROBLEMS. OB produced in R, the centre of one of the required circles. The other centres are at the intersection of a circumference drawn with radius OR and centre 0, and every other diameter produced. Prob. 74. Within a given circle AC . . . E to draw any num- ber of equal semicircles touching the given circle, and their diam- eters forming a regular polygon. In the given circle let OA and OB be two radii at right angles. Divide the given circumference, commencing at B, into twice as many parts as semi- circles required, and draw diameters to the points of division. Join BA. BA cuts the first diameter to one side of OB at G. G is one end of two adjacent diameters required. Lay off OG from O on every other diam- eter in points H, K, etc. Join UK, KG, etc. 1, 2, 3, etc., are the centres of the required semicircles. Prob. 75. To divide a circle into any number of parts which shall be equal in area. Let the number of parts be four. Divide a diameter into twice as many equal parts as areas required, in this case eight, by points 1, 2, 3, 4, etc. With 1 and 7 as cen- tres and radius 01 describe a semicircle on opposite sides of the diameter ; with cen- tres 2 and 6 and radius 02 do the same thing, and so continue taking each point as a centre and the dis- tance from it to the end of the diameter as a radius. GEOMETRICAL PROBLEMS. Prob. 7G. To divide a circle into concentric rings having equal areas. Divide the radius CD into as many equal parts as areas required (Prob. 11) in points 1, 2, 3, etc. On CD as a diameter draw a semicircle, and at the points 1, 2, 3, etc. draw lines perpendicular to CD, meeting the semicircle in points A, B, etc. With centre C aud radii CA, CB, etc. draw concentric circles. Prob. 77. A chord AB and a point C being given, to find other points in the arc of the eircle passing through A. B, and C without using the centre. Draw AB, AC, and BC. Suppose four more points are required. With any radius and centres A and B draw the arcs DE and FG. Make CAE equal CBA, and GBC equal CAD. Divide the arcs DE and FG into the same number of equal parts, one more than the number of points required. Number the points 1, 2, 3, 4, from D toward E, and from G toward F. Draw lines from A and B through these points, and those passing through like-numbered points meet in points on the arc ACB. To construct a point on the curve below AB lay off GO equal to Gl on the arc FG, and D9 equal to Dl. Lines through and B, and A and 9 meet in K, a point on the curve. 38 GEOMETRICAL PROBLEMS. Prob. 7 into the same number of equal parts (say five) by the points 1, 2, 3, etc. Number the points as shown. Draw lines from A through 1, 2, 3, etc. to meet lines through the same numbers drawn from B. The lines through like numbers meet in points on the required arc. Prob. 79. To construct any number of tangential arcs of circles, having a given diameter. Suppose three (four) arcs are re- quired. Upon the given diameter as a base draw an equilateral triangle ABC (square ABCD). With each vertex as a centre, and a radius of half a side, draw arcs of circles tangent to each other, as shown. Prob.' 80. At a point C on a line AB to draw two arcs of circles tangent to AB, and to tiro parallels AD and BE. form- ing an arch. Make AD equal to AC, and BE equal to BC. Ai C make CG perpendicular to AB, and at Dand E draw the perpendiculars DF and GE, meeting CG in F and G, the required centres. GEOMETRICAL PROBLEMS. 39 Prob. 81. Jb rfraw an# number of equidistant parallel straight lines between two given parallels, AB and CD. , a fj Draw any line A5 at any angle to AB, rj^f and lay off on it any convenient dis- 2>--' y __" tance Al as a unit as many times as - lines required, plus one. Thus, if four parallels are required, lay off Al five times. AVith A as a centre and A5 as radius draw an arc cutting CD in point e. Join Ae. AVith centre A and radii Al, A2, A.'>, etc., cut Ae in a, b, c, etc. Through a, b, c, otc. draw parallels to AB and CD. Prob. 82. To draw through a point C a line to meet the inaccessible intersection of two lines, AB and DE. A b From C draw any lines CA p f yf ~T and CD to the given lines. -r — |-f Join AD, AC, and CD. Make V * ' BE parallel to AD, BF par- allel to AC, and EF parallel to CD. The intersection of EF and BF gives a point F in the required line. Draw through C and F. Prob. 83. To draw a perpendicular to a line AB, which shall pass through the inaccessible intersection of two lines, AE and CD. Produce AB to cut the lines in A and C. From A draw a perpendicular to CD, and from C a perpendicular to AE ; these per- pendiculars meet in O. Draw BO perpen- dicular to AB. OB passes through the intersection of CD and AE. -^- 40 ( l E< >M KTIUCAL PK< >BLEMS. PROB. 84. To draw an involute of a square ABCD. Produce the sides as shown. With centre C and radius CD draw the arc DE. With centre B and radius BE draw the arc EF. With centre A and radius AF draw the arc FG, etc. Note. Suppose a line to be wrapped around and in the direction of the peri- meter of any plane figure. Let the line be unwound, keeping it always straight in the process of unwinding. Any point in the line describes an involute. The involute of polygons is composed of arcs of circles, as in Prob. 84. Prop.. 85 To draw the involute of a circle. Divide the circumference into any nura- / ber of equal parts, as at A, B, C, D, etc, and draw radii to these points. At A, B, C, D, etc. draw tangents. Let the curve start at A. On the tangent at B lay off a distance from B to 1 equal to one of the parts into which the circumference is di- vided. On the tangent at C lay off a distance equal to two parts to 2. On the tangent at D three parts to 3 ; from E four parts to 4, etc. The curve through these points, 1, 2, 3, 4, etc., is the involute of the circle. GEOMETRICAL PROBLEMS. 41 Prob. 8G. To draw a spiral com posed of semicircles, the radii being in arithmetical progression. Draw an indefinite line, BAG. On the line take any two points, A and B, as centres. With A as a centre and radius AB draw a semicircle. With B as a centre and radius BC draw a semicircle, and so on, using A and B as centres, and taking the radii to the end of the diameter of the last-drawn semicircle. Prob. 87. To draw a spiral composed of semicircles, whose radii shall be in geometrical progression. Let the ratio be 2. Let AB he the radius of the first circle, A its centre. Draw the semicir- cle BLC. With B as a centre and radius BC draw the semicircle CMD. With C as a cen- tre and radius CD draw the semicircle DNE. D is the next centre, the diameter DE of the last-drawn circle becoming the radius for the next circle. So proceed. Prob. 88. To draw a spiral of one turn in a given circle. Divide the circle into any num- ber of equal parts, say twelve, by the lines OA, OB, OC, etc., and a radius OA into the same number of equal parts by the points 1, 2, '° .'5, 4, etc. With O as a centre and radius 01 draw an arc cutting OBinM; with centre. () and radius 02 an arc cutting OC in N, radius 03 an arc cutting 01) in point P, etc. Through INI, N, P, etc. draw the curve. Note. This is "the spiral of Archimedes." 42 GEOMETRICAL PROBLEMS. Prop.. .SO. T<> draw in a given circle a spiral of any num- ber of turns, sag tiro. Draw radii dividing the circle into any number of equal parts. Divide any radius, as OA,.into as many equal parrs as turns in the spiral, and divide each part into as many equal portions in points 1, 2, .'!, 4, etc. as the circle is di- vided into. With centre and 01, 02, 03, etc. as radii, draw- arcs to meet the radii OB, OC, OD, etc. in points of the required curve. PrOB. 90. Given the axes AB and CD of an ellipse to draw the curve, and at any point on, the curve to draw a tangent. Place the axes AB and CD at right angles to and bisecting < each other at 0. "With centre C and radius OA cut AB in F and F', which are the foci. Between and F', or F, take any point G, dividing AB into two parts. With centres F and F and radius AG draw arcs on either side of AB. With the same centres and radius BG draw arcs intersecting those drawn with radius AG, at points L, M, N, and P, which are points on the curve. Take any other point, T on AB, and repeat the above operation ; and so on until as many points as are neces- sary are found. Through the points draw the curve. FM plus F'M equal AB. Let M be the point at which the tangent is required. Produce FM and draw the bisector of the angle SMF'. MR is the tangent required. GEOMETRICAL PROBLEMS. 43 Note. For drawing the ellipse and similar curves through a series of points the so-called French Curves are to he used. Note. The major axis is sometimes called the transverse, and the minor the conjugate, axis. Prob. 91. To draw an ellipse by means of a trammel, the axes being given. Let the semi-axes be OA and OB. Mark off on the straight edge of a slip of paper or card MP equal to OA, and NP equal to OB. Keep the trammel with the point N always on the major axis, aud the point M on the minor axis, and P will be a point in the curve. Fmd as many points as necessary, and draw the curve. — S Prob. 92. To draw an ellipse, having given the axes. M B Let the semi-axes be OA and OB. With radii OA and OB and centre O draw circles. Draw any radii, OM, ON, etc. Make MP, NT, etc. perpendicular to OA, and HP. KT, etc. parallel to OA. P, T, etc. are points on the curve. 93. To draw an ellipse, having given the axes. Place the axes at right angles at their centres, and on them construe! a rectangle, one half being shown in BDEC. Divide OA and DA into the same number of equal parts by points 1, 2, 3, etc. Draw lines through C and 1, 2, .">, etc. to meet lines from B drawn to 1, 2, 3, etc. on A I). £ P, Q, 11, etc. are points on the curve. 44 G EOMETRI C A L PROBLE M S . Prob. 94. To draw a curve approximating to an ellij>s<>. Draw the squares ABDC and BP^FD, and their diagonals, intersecting in G and II. With centres G and H and radius GA draw the arcs AC and EF. With centres B and 1) and radius DA draw the arcs CF and AE. Prob. 95. To draw on a given line, AB as a major axis, a curve approximating an ellipse. Divide AB into three equal parts (Prob. 11) by points C and D. With centres C and D and radius CA draw two circles intersect- ing in E and F. Through C and D draw EGG, EDII, FCK, and FDL, meeting the circles in points G, H, K, and L. With centres E and F and radius EG draw the arcs GH and KL, completing the curve. Prob. 96. Given the major and minor axes of an ellipse, to draw the curve approximately. Let C A be the semi-major axis, and BC the semi-minor axis. Join A and B. Make CD equal to BC, and BE equal to AD. Bisect AE by a perpendicular, meeting BC pro- duced in F. With centre F and radius FB draw the arc BH, and with centre G and radius GH draw the arc HA. Note. One quarter of the whole curve is only shown, leav- ing to the student the construction of the full ellipse. GEOMETRICAL PROBLEMS. 45 Prob. 97. Having the axes given, to draw a curve of tan- gential arcs of circles approximating to the ellipse. -J^s^ AO is the semi-major axis, and OB the semi-minor axis. Draw a rectangle T < \ with the axes as sides. AOBC is one- * • \ r-' "i > n I \ quarter of the rectangle. Draw AB. \"V*f ~£ From C draw CMP perpendicular to \ * [ AB, and meeting BO produced in P. \'»j Make OE equal to OB. On AE as a i p diameter draw a semicircle AKE. Pro- duce OB to K. Make OL equal to BK. With centre P and radius PL draw the arc LN. Make AD equal to OK, and with centre M and radius MI) draw the arc DN, meeting LN in N. Draw NMR and PNS. With centre M and radius MA draw AR; with centre N and radius NR draw RS, and with centre P and radius PS draw an arc from S through B. Repeat in each of the quadrants. Prob. 98. To draw a parabola when the abscissa, AB and the ordinate BC are given. p / t £.__c Draw the rectangle ABCD, and divide I*--*'- -j^JU^r-^ ; ad and DC into^the same number of r^v-^''''' ' equal parts. Through the points of divis- %■?'' i ion on AD draw parallels to AB, and a Bi ',\v I from A draw lines to the points on DC. ^^~\^^ ; The first line above AB meets the line ■ _Trr^r-r^j from A to the first point of division from I) in a point P on the curve. The second parallel to A meets the second from A to DC and so on. P, Q, R, and C are points in the curve. Repeat the same below AB. 46 (;]<:< (METRICAL PROBLEMS. Prob. 99. To draw a parabola when the directrix AC and the focus D are given, and to draw a tangent at any point L on the curve. This curve is such that its apex E is always half way between A and D, and the distance from D to any point upon the curve, as F, is always equal to the horizontal distance from F to the directrix. Thus DF equals FG, and DH equals HK, etc. Through D draw BDA perpendicular to AC. This is the axis of the curve. Draw parallels to AC through any points in AB, and with centre D and radii equal to the hori- zontal distances of these parallels from AC cut the correspond- ing verticals, which will give points on the curve. To draw the, tangent at L. Draw the ordinate LN, meeting AB in N. Produce BA to the right. Make ET equal to EN. Draw LT, the tangent required. Prob. 100. To draw an hyperbola when the diameter AB, the abscissa BC, and the double ordinate DE are given. Complete the rectangle BCDF, and divide CD and DF each into the same num- ber of equal parts. Draw BL, BM, and B>\ inter- secting lines AK, AH. and AG respectively in points on the curve. Repeat below and in the other half of the curve as indicated. GEOMETRICAL PROBLEMS. 47 To draw an oval on the diameter of a given Let AB be the diameter. Draw the circle sn ACB. Make OC perpendicular to AB. Draw the lines BCD and ACE indefinitely. With t centres A and B and AB as a radius draw the arcs BE and AD. With centre C and radius CD draw the arc DE. Note. The centres A and B may he taken anywhere on the line AOB produced. Prob. 102. Upon a given line AB to draw an oval. Bisect AB at C, and draw the perpendicular CD. With B as a centre and radius AB describe an arc AD. Bisect the quadrant AE ivi F. Through F draw CFG. AG is the first part of the curve. Bisect CB in II, and draw HD. K is the second centre. Bisect EL in M, and draw KMN. GN drawn from K is the second part of the curve. Bisect CH in O, and draw DO. P is the third centre. From P through E draw PET. NT is the third part of the curve. From E with a radius ET carry the curve to the line DC, and repeat the operation for the other half of the curve. Draw a semicircle on the diameter AB for the other part of the oval. The cycloid is the path described by any point in the circum- ference of a circle which rolls along a straight line. An epicycloid is the patli described by any point in the cir- cumference of a circle which rolls along the outside of another circle. %^D 48 GEOMETRICAL PROBLEMS. A liypocycloid is the path described by any point in the cir- cumference of a circle which rolls along the inside of another circle. The rolling circle is called the generatrix, or generating cir- cle, and the line (straight or curved) on which it rolls is called the directrix. Every point in the tire of a wheel which rolls along the ground in a straight line describes a cycloid. Hence it is easily seen that in one revolution or turning around of the circle, or wheel, the circumference will roll out into a straight line. To lay oif the circumference on the straight line, either calculate its length or divide the circle into equal ares, and lav off on the straight line as many divisions as there are in the circle; each division on the straight line being equal to l lie length of one division on the arc. The arc is more than the chord, so a distance greater than the chord must lie taken, and this is obtained by judgment or by approximation. By dividing the arc into a number of smaller arcs, so that the chord practically coincides with the arc. the small chord being laid off as many times as there are small arcs, a length of straight line is obtained very nearly equal to the arc given. Pkob. 103. To construct a cycloid. "c Let AB be the directrix, and AD the generating circle. Divide the rolling circle into any number of equal parts, say 12, and lay off these lengths of arcs along AB, ~->_i <\ *% " * "IT giving points a, b, c^etc. Through 1, the centre of the generating cir- cle, draw a line parallel to AI>. This is the line of centres. On ( I E< (METRICAL PROBLEMS. 49 this lay off 12, 2.">, .'!4, etc., equal to Aa, ab, etc.. and with centres 2, 3, etc. draw the generating circle in all its posi- tions tangent to AB at points a, b, c, etc. Draw through the points of division on the rolling circle parallels to AB, to meet the different positions of the rolling circle in points P, R, S, T, and U. These parallels are drawn in the figure from points p, r, s, t, and u. Repeat the process for the other half of the curve. Another method is to take the chords of the arcs By?, Br, Us, etc., and with centres a, b, c, etc. cut the respective circles in points P, R, S, etc. The chord aP equals By?; bR equals Br; cS equals Bs, etc. Note. When the number of divisions of the rolling circle is large the curve may be drawn by arcs of circles by taking a as a centre, and radius a A, and drawing from A to P. Pro- duce Pa and Rb to meet, giving the centre for arc PR ; lib and Sc meet at the centre of arc Rs, etc. PrOB. 104. To construct an exterior epicycloid. Let R-A be the rolling circle on the outer circumference of the direct- ing circle. Divide R-A into any number of equal parts (say 12), and lay off these parts on Aab, etc., giving points a, b, c, d, etc. With the centre of the directing circle as a centre, draw an arc from R giving the line of centres R123 etc. Draw from the centre of the directing circle radial lines through a,, b, c, d, etc., meeting the line of centres in points I, 2, .'!, etc., the centres of the different positionsof the rolling circle. With 50 GEOMETRICAL PROBLEMS. centres 1, 2, 3, etc. and radius RA draw the several positions of the rolling circle. With the centre of the directing circle as a centre, draw arcs through the points of division of the cir- cle R-A to meet the several positions of the rolling circle in points C, D, E, F, etc., which are points on the curve. Draw through C, D, E, F, etc. The points of the curve may be obtained by drawing from a as a centre and radius equal to the chord of one division of R-A an arc to meet the second position of the rolling circle in C ; from b with radius equal to the chord of two divisions an arc to meet the third position in D ; from c with radius equal to the chord of three divisions to meet the fourth position in E, etc. Or from a lay off on the rolling circle tangent at a one part of R-A (in this case T \r) ; on the one tangent at b two [•arts of R-A ; on the one tangent at c three parts, etc. Prob. 105. To construct a hypocycloid. See the curve on the interior of the directing cirele in figure with Prob. 104. Aa, ab, be, etc. are equal parts of the circum- ference of the rolling circle. R123 etc. is the line of centres. The process and directions are the same as for the epicycloid. Note. When the diameter of the rolling circle is equal to the radius of the directing circle the hypocycloid becomes a straight line. GEOMETRICAL PROBLEMS. 51 Prob. 106. O lay off three spaces, points of the curve. To construct an interior epicycloid. Let the circle A-C roll on B-CMN. On the circumference of B-C lay off equal arcs CM, MN, NO, etc. Draw from M, N, O, etc. radial lines through B, and make MD, NE, OF each equal to the radius CA. With centres D, E, F, etc. draw the cir- cles tangent at M, N, 0, etc. Lay off Ml equal to CM, giving pofnt 1 ; from N lay off two divisions each equal to MC, giving point 2 ; from giving point 3, etc. 1, 2, 3, etc. are Prob. 107. To draw a scroll for a stair-railing. The circle ABCD is the eye of the scroll. Draw the diameters AC and BD at right angles. Draw the chord AD, and bisect it in 6. Draw a line 65 parallel to AC. Bisect AE in F. Bisect EF in 3. Make E4 equal to E3. Draw 32 and 45 paral- lel to BD, and 21 parallel to AC. From 6 draw an indefinite line parallel to BD and produce 65, 21, etc. From point 1 with radius IB draw the arc BH. From 2 and radius 2H draw HJ, and so on. The arc BR of the inner curve is drawn from P with radius PB; the arc RS is drawn from 6 with radius 6R. 52 CK.n.MKTIMCAL PROBLEMS. Prob. 108. A Another method. Suppose the scroll to come between the outside of the rail AB and the line CD. Draw A9 perpendicular to AB and CD, and divide it into nine equal parts. On the sixth division as a side draw the square etbcQ. With centre a and radius aA draw the quadrant AE. With centre 6 and radius 6E draw arc EF. With centre c and radius cFdraw FG ; with centre 6 and radius 0>G draw arc GH ; with centre e and radius eH draw arc HK ; and with centre/ and radius /K draw the arc KL. For the inner curve start with a as a centre and radius a\ and proceed as lor the outer. Prob. 109. being given. To construct a spired, its greatest diameter AB Divide AB into eight equal parts by points 1, 2, 3, etc. On 45 as a diameter draw a circle CDEF. This circle is the eye of the spiral. In- scribe a square CDEF, as shown in the enlarged drawing of the eye. Draw the central diameters of the square. 1-2.10 and 11,9. Divide these diam- eters into six equal parts, and number as shown. With 1 as a centre and ra- dius 1H draw an arc HK to meet a horizontal produced from 2 through 1. With 2 as a centre and radius 2K draw the arc KL to meet a vertical through 3 and 2. With centre 3 and radius 3L draw arc LM meeting a horizontal produced through -1 and 3. With 4 as a centre and radius 4M draw arc MN to meet a line drawn through 5 and 4 at. N, and so proceed. GEOMETRICAL PROBLEMS. 53 The curve may be commenced by taking 12 as a centre and radius 12 A and drawing from without toward the centre. Prob. 110. To describe an Ionic volute. Let AB be the vertical measure of the volute. Di- vide AB into seven equal parts, and from C, the lower extremity of the fourth di- vision, draw CF perpendic- ular to AB, of indefinite length. From any point on CF as a centre, with a ra- dius equal to one-half of one of the divisions of AB, draw the circle HIJK, form- ing the eye of the volute. Draw the diameter HJ perpendicu- lar to CF. Draw the square HIJK, bisect its sides, and draw the square 12 L M 11. Draw the dividing lines of the square as shown in the smaller figure, and extend them. The divis- ions corresponding to 12 N are equal. The divisions OP and RL are each equal to one-half of 12 N. From 1 as a centre and radius 1H draw the arc US ; from 2 as a centre and radius 2S the arc ST ; from centre 3 and radius 8T the arc TU ; and so proceed in the order of the numerals. In drawing the inner curves the dots on the diagonals in the small figure indicate the centres. The division of the square, of which 12 N is one side, shows how these centres are found. 54 GEOMETRICAL PROBLEMS. Fillet ( Bead Torus Cyma Recta Cyma Reversa The Roman moldings are given above, the method of con- struction being evident. All the arcs are arcs of circles, and the angles are 45°, except in the flatter form of the cyma recta, where the line of centres is at 30°. CHAPTER III. 14. It is supposed that the student has now become familiar with the use of the instruments necessary for the construction of a drawing in pencil, and has acquired a certain degree of proficiency in handling them which is necessary for accurate work. The next step is to learn to ink a drawing after it has been pencilled. Starting with a good pen, in good condition, and a smooth, well-ground black ink, it only remains for the student to learn to make a clean, sharp, even line. This may seem at first like an easy thing to do ; nevertheless, the ability to make a good ink line every time comes to most students very slowly, and after a great deal of practice. Therefore, before beginning to work on the plates, which are to be finished carefully and handed in for inspection, it will save a considerable time and paper to make lines against the triangle without regard to their length, direction, or location, until the student is thoroughly familiar with his pen and can make a fair line. Several hours, if necessary, in this preliminary practice will be invaluable. 1.5. India Ink. A special ink, called India ink, is always used in making drawings. It comes either in the stick form, and has to be ground as used, or in the liquid form. In the 56 INKING. latter form it is held in solution by an acid which corrodes the pen and eats into the fibre of the paper, so that if it is desired to erase a line it is much more difficult than if made with the ground ink. This ink is also very liable to rub off like soot. The only advantage it has is the saving of time in preparation. This kind of ink cannot be recommended for anything except coarse, rough work. It should not be used for tinting. To prepare the stick ink for use place a small amount of water in the ink slab or saucer (the slab should be perfectly clean), then grasp the ink firmly and, with a rotary motion, grind until the liquid is black and a little sluggish in its motion. After it has reached the point when it is black enough the grinding should cease, as a continuation only makes the liquid thicker, thus causing it to flow less freely from the pen. The liquid will look black in the slab after a very little grinding, but the necessary consistency will not be reached for some time. In ordex to determine when the proper point is reached make a heavy line with the drawing pen on a piece of paper and wait for it to dry ; do not go over this line a second time. If the ink has not been ground sufficiently, it will look pale after it is dry, in which case more grinding is necessary. If at any time the ink becomes too thick, it can be diluted by putting in more water and mixing thoroughly. The stick should always be wiped dry after using, to prevent its crumbling. The ground ink should be kept covered as much as possible to prevent evap- oration, which would soon cause it to become too thick for use. It is not advisable to prepare a large quantity of ink at once, as the greater the amount the longer it takes, and freshly ground ink is preferable. If carefully covered, however, it may be kept two or three days. In case the ink becomes dry in the saucer it should all be washed out, as it is almost impossible to redissolve it entirely so that there will not be little scales which get into the pen and cause the ink to flow irregularly. INKING. 57 16. Drawing Pi-:n. This instrument, commonly called a right-line pen, is one of the most important of the drawing instruments, and it is very essential that it be of good quality. The screw is used to adjust the distance between the nibs, in order to make the line of the desired weight. The ink may be placed between the nibs by means of a brush or strip of paper, but it is more convenient to dip the pen into the ink, being careful to wipe the outside of the nibs before using. While inking the pen should be held so that both nibs rest on the paper evenly, and it should be inclined a little to the right, or in the direction of the line, with its flatter side against the triangle or straight edge, the end of the middle finger rest- ing on the head of the screw. A slight downward pressure is necessary (the greater the rougher the paper), but do not press against the ruler, as the lines would be uneven in thickness. The ruler is simply a guide for the pen. The lines should always be drawn from left to right (relative to the person and not to the drawing). If it is desirable to go over a line a sec- ond time for any reason, it should be drawn in the same direc- tion ; never go backward over it. In inking a curved line by means of the right-line pen and irregular curve, it is necessary to constantly change the direc- tion of the pen so that the nibs shall always be tangent to the curve. This requires considerable practice to do nicely. In case the ink does not flow freely from the pen, moisten the end of the finger and touch it to the end of the pen, and try it on a piece of waste paper. If this fails the pen should be wiped out cleau and fresh ink put in. In making fine lines the nibs of the pen are near to each other, consequently the ink dries between them quite rapidly, hence it will be found advis- able to clean out the pen thoroughly quite frequently to insure perfect lines. This is one of the secrets of being ahle to make 58 INKINC. good fine lines ; they should also be made more rapidly than heavy lines; the heavier the line the slower the pen should be moved. Do not keep the point of the pen too near the straight edge, as the ink is liable to flow against it, thus causing a blot. Especial attention should be given to the care of the pens ; they should always be carefully wiped after using, and should not be put away with any ink dried on them, nor allowed to get rusty on the inside of the nibs. Any old piece of cotton cloth will answer to wipe the pens and stick of ink on. 17. How to Sharpen the Pen. To make good lines the pen must be kept in first-class condition, — that is, not only clean, but sharp; and every draftsman should be able to sharpen bis own pen. The curve at the point of the nibs of the pen should always lie a semi-ellipse, with its long diameter coinciding with the axis of the pen ; it should not be a semicircle, nor should it be pointed. The student is advised to look carefully at the points of his new pen, so as to get a correct idea of the proper curve before it becomes changed by wear. When this curve becomes changed by wear, or if, from any other cause, one nib is longer than the other, the nibs should be screwed together, then, hold- ing the pen in a plane perpendicular to the oil-stone, draw it back and forth over the stone, changing the slope of the pen from downward and to the right to downward and to the left, or vice versa, for each forward or backward movement of the pen, so as to grind the points to the proper curve, making them also of exactly the same length. This process, of course, makes the points even duller than before, but it is a necessary step. Next separate the points a little by means of the screw, and then place either blade upon the stone, keeping the pen at an angle of about 15° with the INKING. 59 face of the stone, move it backward and forward, at the same time giving it an oscillating motion, until the points are sharp. This is quite a delicate operation, and great care should be exer- cised at rirst. The pressure upon the stone should not be very great, and it is well to examine the point verj' often so as to be sure and stop when each nib has been brought to a perfect edge, otherwise one nib is liable to be longer than the other and the pen will not work well, even if each nib is sharp of itself. Although the points want to be brought to a perfect edge, they should not be sharp like a knife, as in that case they would cut the paper. It will probably be necessary to try the pen with ink to be sure that it is in good condition. Sometimes a slight burr is formed on the inside of the blades ; this is re- moved by separating the points still farther, so as to insert the knife-edge of the oil-stone between them and draw it carefully through. One motion should be sufficient to remove the burr. The pen should never be sharpened by grinding the inside of the blades other than just indicated. 18. Inking a Drawing. In inking a drawing it is prefer- able to ink all the circles and arcs first, as it is easier to make the straight lines meet the arcs than the reverse. Of a number of concentric circles the smallest should be inked first. Here, as in the case of the pencil compasses, the pen point should be kept nearly vertical, the top of the compass being inclined a little toward the direction of revolution, and there should be a slight downward pressure on the pen point, but none on the needle point. Where a large number of lines meet at a point care should be taken to avoid a blot at their intersection. These lines should be drawn from rather than toward the point, and each line should be thoroughly dry before another is drawn. 60 INKING. In case of two lines meeting at a point neither line should stop before reaching the point, nor go beyond it. Either of these defects gives a very ragged appearance to the drawing. 19. Stretching Paper. For ordinar}' small line drawings it is usually sufficient to fasten the paper to the board by means of thumb tacks, but for large drawings, or those which are to be tinted at all, it is necessary to stretch the paper by wetting it and fasten it to the board with mucilage. To do this lay the paper on the board, fold over about one-half an inch along each edge of the sheet ; do not cut the corners ; next wet the upper surface of the paper, except that portion folded over ; do not rub the surface of the paper, simply press the sponge against it on all parts ; apply the mucilage to one of the edges and fasten that edge down, beginning at the middle and rubbing toward either end ; do the same with the opposite edge next, giving a slight pull to the paper as it is fastened down ; repeat this process for the two remaining edges. It is very important that the edges of the paper where the mucilage is to be applied should be kept dry, so that the muci- lage will be ready to act as soon as it can dry, and to facilitate this the less mucilage you can use and accomplish the result the better. If a large quantity of mucilage is used it will moisten the edges of the paper so much that it will be likely not to stick, as the body of the paper will dry as soon as the edge, and therefore pull them up. The drying of the mucilage can be hastened by rubbing the edge briskly with a piece of thick paper under the lingers until it becomes hot. The board should never be placed near the fire or radiator to hasten the drying, as it would dry the paper before the mucilage set, caus- ing the edges to be pulled up. The board should be left to dry in a horizontal position, and all the superfluous water should TINTING. 61 be removed with a sponge, so as to avoid water marks in the paper, which always show in tinted drawings. In some cases, when the mucilage sets slowly, it may be necessary to moisten the centre of the paper sometime after stretching, to prevent its pulling up the edges by drying too rapidly. 20. Correcting and Cleaning Drawings. Pencil lines are removed by meaus of a piece of rubber. When a mistake is made in inking, or it is desired to change a completed draw- ing, it becomes necessary to erase an ink line. This can be done by means of a rubber ink eraser, the same as in the case of pencil lines, except that much more rubbing is necessary. Ink lines can also be removed, and more quickly, by means of a knife ; in this case care should be taken not to use the point of the knife, as V-shaped holes are made which will always show. The flat portion of the knife should be used. After erasing an ink liue, the surface which has been made rough by scratching should be rubbed down with some hard, perfectly clean, rounded instrument before inking other lines over it. A drawing can be cleaned by means of India rubber, or stale bread crumbled on the drawing and rubbed over it. Although dirt can be removed from a drawing, it should be the aim of the draftsman to keep it as clean as possible. Therefore, the drawing should be kept covered when not being worked upon, and, if the drawing is a large one, all except that portion which is in use should be kept covered. tinting. 21. Tinting may be done in colors or India ink, as desired. The method of putting on the tint is the same in either case ; consequently, we will take up only the India-ink tint here. If a drawing is to be tinted, the paper must be stretched as 62 TINTING. explained in the first part of this chapter. Especial care should be taken to keep the paper perfectly clean. That por- tion of the drawing which is to be tinted must not be touched with the India rubber, as the surface is thereby made rough and will not take a uniform tint. Hence, in laying out the work pencil lines must not be made on the surface to be tinted. 22. Preparation of the Tint. Clean the ink slab, water glass, and the brushes thoroughly, also be sure that there are no scales on the stick of ink which could possibly come off. Fill the slab about half full of water, and grind the ink as pre- viously explained until it is black, but not thick. Fill the water glass about half full of clean water, and with a brush transfer enough of the ink in the slab to the glass to make a light tint. It is hard to get an ink which is absolutely free from specks, therefore, it is well to let the ink, after it is prepared in the slab, stand a short time to allow these specks to settle to the bottom ; then, in transferring the ink to the glass, do not plunge the brush down to the bottom of the slab, thus taking up this sediment, but let the brush fill from the surface of the liquid. The mixture in the water glass is the one to be used for tinting, and it is better not to make it as dark as you wish it upon the drawing when finished, as it is much easier to put on a light tint evenly than a dark one. The required depth of shade can be obtained by successive washes. Let each wash dry thoroughly before putting on another. A smoother effect can usually be obtained, especially on a large surface, by going over the surface to be tinted with clean water first, and then letting it dry. 23. Laying on the Tint. Having laid out the surfaces to be tinted, incline the board so as to slope like an ordinary TINTING. 63 desk ; then dip your brush into the tint you have mixed, and take up as much of the liquid as it will carry, begin in the upper left-hand corner of the surface and draw it along the upper . boundary, holding the brush nearly vertical and leaving quite a puddle as you proceed. Lead this puddle gradually downward by going across the surface from left to right, about a quarter of an inch at a time, dipping the brush frequently in the tint so as to keep the puddle about the same size all the time ; it should not be, large enough to run down at any point. This puddle should not be left standing at any place any longer than is absolutely necessary, as it is very apt to leave a streak ; therefore, having commenced on a surface, finish it as quickly as possible, — do not let anything interrupt you. When you get to the bottom dry your brush on a piece of blotting paper, and with the dry brush take up the superfluous tint, as you would with a sponge, until it looks even. Iu laying on the tint do not bear on with the brush, as the brush marks would be liable to show, but use the point only, just touching it to the paper so as to wet it, and the tint will follow along of itself (the board being properly inclined). In following the boundaries of the surface to be tinted let the brush be pointed towards the boundary from the inside of the surface. This gives what is called a, flat tint, and is used to represent surfaces which are parallel to the plane of projection. 24. For representing surfaces which are oblique to the plane of projection a graduated tint is necessary. There are two methods of doing this, — the French and the American. The French method consists in dividing the surface into small divisions (these divisions should be indicated only, not drawn across the surface), laying a flat tint on the first space, and when this is dry laying another flat tint on the first two spaces. 64 TINTING. Proceed iu this way until the whole surface is covererl, com- mencing at the first space each time. By this method the shading shows streaks of tint of different depths, but are almost unnoticeable if the divisions are taken quite small. The American method is most used, aud is called shading by softened tints. There are two ways of doing this : — 1st. By mixing a small amount of dark wash at first, and starting as if you were to put on a flat tirft, and then, by re- peated additions of clean water, going over a little more surface at each addition, gradually make the dark tint lighter until you are using almost pure water. 2nd. Divide the surface into divisions, as in the French method, only not so many ; put on a flat medium tint on the first space, but, instead of taking up all the tint from the bot- tom edge of the surface, leave a slight amount, touch the brush to some clean water and apply it to the lower edge of the pud- dle, thus making a lighter tint, and bring down this new tint a short distance ; repeat this a few times until the tint has prac- tically no color. Be careful to remove the most of the tint from the brush each time before touching it to the clear water. This work must be done even quicker than the ordinary flat tint. Use as little tint or water in the brush as you can and not have it dry in streaks. Let this dry, and then repeat the proc- ess, commencing at the top and going over two spaces with the flat tint and softening off the lower edge, and so on, commenc- ing at the top each time. Usually the tint should be softened out in the length of one division. If this shading is done per- fectly, there will be a gradual change in the tint from beginning to end. CHAPTER IV. PROJECTIONS. 25. Orthographic Projection, or Descriptive Geometry, is the art of representing a definite body in space upon two planes, at right angles with each other, by lines falling perpen- dicularly to the planes from all the points of the intersection of every two contiguous sides of the body, and from ail points of its contour. 26. These planes are called coordinate planes, or the planes of projection, one of which is horizontal, and the other vertical. H and V, Fig. 1, represent two such planes and their line of intersection GL is called the ground line. 27. We shall only take, in this book, just enough of the elementary principles of projections to enable the student to make working drawings of simple objects. 28. Since solids are usually made up of planes, planes of lines, and lines of points, if we thoroughly understand the prin- ciples involved in the projections of points, we ought to be able to draw the projections of lines, of planes, and of solids. The only difference being that, with a large number of points, the student is liable to get them confused. To avoid this liability it is advisable, at first, to number the points of a solid and their corresponding projections as fast as found lightly in pencil, and erase them after the problem is finished. Do not try to draw 66 PROJECTIONS. the object all at once, it is impossible ; one point at a time is all that can be drawn by anybody, and in this way the most complicated objects become simple, even though it may take a long time to complete the drawing. 29. The projection of any point in space on a plane is the point at which a perpendicular drawn from the given point In the plane pierces the plane. This perpendicular is called the projecting line of the point. Thus, in Fig. 1, a h is the projection of the point a on the plane H, and o* of the same point on the plane V. These are called respectively the horizontal and vertical projections of the point a ; aa h is called the horizontal projecting line of the point a, and aa v the vertical projecting line of the same point. The horizontal projecting line aa h is perpendicular to H, by definition, the plane V is assumed perpendicular to H, hence aa h is parallel to the plane V, and aa v is equal in length to a h b. Also the vertical projecting line, for the same reason, is parallel to H, consequently aa h is equal in length to a v b. From the definition it is readily seen that each point in a line perpendicular to a plane will have its projection on that plane in one and the same point ; hence one projection of a point does not definitely locate its position in space. 30. From the preceding article the following principles may be noted : — First, the perpendicular distance from the horizontal projec- tion of a point to the ground line is equal to the perpendicular distance of the point in space from the vertical plane ; or, briefly, the horizontal projection of a point indicates the distance of the point in space in front of V, but it conveys no idea of its distance above H. Second, the perpendicular distance from the vertical projec- tion of a point to the ground line is equal to the perpendicular PROJECTIONS. 67 distance of the point in space from the horizontal plane ; or, briefly, the vertical projection of a point indicates the height of the point in space above H, but it conveys no idea of its dis- tance in front of V. 31. If from the points a" and a h , Fig. 1, perpendiculars should be erected to each coordinate plane, they will intersect at the point a in space ; and as two straight lines can intersect at only one point, there is only one point in space which can have a h and a® for its projections. Hence two projections of a point are always necessary to definitely locate its position in space. 32. It is evident that it would be very awkward to make our drawings on planes at right angles to each other; hence the vertical coordinate plane is supposed to be revolved back- ward about its line of intersection GL with the horizontal plane until it forms one and the same surface with the hori- zontal plane, which may be considered to be the plane of the paper. In this revolution all points in the vertical plane keep the same distance from the ground line, and their relative positions remain unchanged. Thus, a v revolves to a v , a r b being equal to a v b, and as a*b was perpendicular to GL before revolution it will be so after revolution, and will form one and the same straight line with a h b. Therefore, the two projections of a point must always be on one and the same straight line, perpendicular to the ground line. Now, if we draw a line across our paper and call it GL, all that portion in front of this line will represent the horizontal plane, and that portion behind it will represent the vertical plane, and the point a located as in Fig. 1 is represented by its projections on the plane of the paper as shown in Fig. 2. 68 PROJECTIONS. 33. A point situated upon either of the coordinate planes has for its projection on that plane the point itself and its other pro- jection is in the ground line. This is readily seen by referring to Fig. 1 ; c v and c h are the projections of a point on H, and d v and d h of a point in V. These points are represented on the plane of the paper as shown by the same letters in Fig. 2. NOTATION. 34. We will designate a point in space by a small letter, and its projections by the same letter with an h or v written above ; thus a h represents the horizontal and a v the vertical projection of the point a. This point may be spoken of as the point a, or as the point whose projections are a r a h . 35. The horizontal coordinate plane will be designated by the capital letter H, the vertical by the capital letter V. 36. Construction lines are those which are made use of sim- ply to obtain required results. They are not a necessary part of the drawing, and when left on a drawing are intended to show the individual steps taken. To this end the student is expected to ink in all construction lines illustrating the special subject in hand until especially directed not to do so. That is, while on the subject of projec- tions it is not desired to have the geometrical construction lines inked in, but only those which refer to projections. When on the subject of shadows only those which show how the shadow is found are required. These lines should be inked in with a light, short dash not more than T J 5 of an inch long, and as light as the student finds he can make easily. All lines representing the projections of single lines, or edges of planes or solids, if visible, are inked in with a full, continu- PROJECTIONS. 69 cms line, a little heavier than the construction lines ; if invisi- ble, they should be marie with short dashes the same length as the construction lines, but the same thickness as the visible lines, so as to distinguish them from the construction lines. The true length of a line when found should be inked in with a long and short dash, about the same thickness as the ordinary full line. When the true length of a line is given, it is put in like an ordinary construction line. Indicate an isolated point by drawing a small cross through it. 37. In working drawings — which are practical applications of projections — horizontal projections are usually called plans, and vertical projections are called elevations. Therefore, they will be used synonymously throughout this book. 38. The student should distinguish between the terms ver- tical and perpendicular. Vertical is an absolute term, and applies to a line or surface at right angles to the plane of the horizon while perpendicular is a relative term, and applies to any line or surface which is at right angles to any other line or surface. If one point is farther from V than another, the first is said to be in front of the second point. Hence, if I say that a line slopes downward, backward, and to the left, it signifies that the line occupies such a position that the lower end is nearer V than the upper, and also that it is on the left of the upper end. Fig. 6 shows the projections of such a line. PROJECTIONS OF STRAIGHT LINES. 39. A straight line is determined by two points, therefore it is only necessary to draw the projections of each end of a straight line and join them, and we have the projections of the line. If the line is curved, it becomes necessary to draw the projections of several points and join them with a curve. 70 PROJECTIONS. 40. If we lay a fine wire, ab, Fig. 3, on a horizontal plane, and also parallel to a vertical plane, its horizontal projection will be the wire itself, that is, a^V 1 is its horizontal projection, equal in length to the wire itself, parallel to GL, and at a dis- tance from it equal to the distance of the wire from V. The vertical projection of the end a will be at aP in the ground line, and of b at b v also in the ground line, since each end is on H (Art. 9). Hence the vertical projection of the line will coincide with GL between a v and b v , and a r b v is its vertical projection. a r b v is equal in length to aW, hence is equal to the actual length of the wire. Now, suppose the wire to be revolved from right to left about a horizontal axis, through the end a, keeping the line parallel to V. If a pencil were attached to the end b at right angles to the line, so that its point touched V at b v , it would trace a circular arc b r c v d v , etc. (of which a v is the centre and a v b v , or the true length of the line, is the radius) on the vertical plane as the wire is revolved ; the end a would, of course, not move. After the wire has been revolved through an angle of 30° its vertical projection will be at a v c v , and it must be equal in length to the real length of the wire. a h will be the horizontal pro- jection of the fixed end of the wire. Since the wire is parallel to V, every point in it is at the same distance from V, hence their horizontal projections must all be the same distance from GL, that is, in a line parallel to GL ; the horizontal projec- tion of the end c must also be in a line through c* perpendicu- lar to GL, hence at c h where this parallel and perpendicular intersect ; a h c h , then, is the horizontal projection of the wire after it has been revolved through an angle of 30°. For the same reason a v d v and a h d h are the two projections of the wire after being revolved through an angle of 45°. Similarly, a v e v and a h e h are its projections after revolving through an angle of 60°. PROJECTIONS. 71 When the line has been revolved through 90° it becomes per- pendicular to H, and its vertical projection is a v f", perpendicu- lar to GL, and its horizontal projection is a point, a h , as might have been seen from the definition of the projection of a point. 41. The same reasoning applies if we take the wire lying against V and parallel to H and revolve it about a vertical axis through the end a, as in Fig. 4. The projections are given for the line lying against V, and making angles of 30°, 45°, 60°, and 90° with V, being parallel to H in each position. 42. The following principles may be noted from the pre- ceding articles : — 1st. A line situated in either plane is its own projection on that plane, and its other projection is in the ground line. 2nd. If a right line is perpendicular to either plane of pro- jection, its projection on that plane will be a point, and its pro- jection on the other plane will be perpendicular to the ground line and equal in length to the given line. 3rd. When a line is parallel to either coordinate plane its projection on that plane will be parallel to the line itself, and equal to the actual length of the line in space, and its projection on the other plane will be parallel to the ground line. 4th. If a line is parallel to both planes, or to the ground line, both projections will be parallel to the ground line. 5tk. If a line is oblique to either coordinate plane its pro- jection on that plane will be shorter than the actual length of the line itself. 6th. If a line is parallel to one coordinate plane and oblique to the other, its projection on the plane to which it is parallel is equal to the true length of the line in space, and the angle which this projection makes with the ground line is equal to the true size of the angle the line in space makes with the plane to which it is oblique. 72 PROJECTIONS. 1th. The projection of a line on a plane can never be longer than the line itself. 8th. If a point be on a line its projections will be on the projections of the line. 9th. If a line in different positions makes a constant angle with a plane its projections on that plane tvill all be of the same length, without regard to the position it mag occupy relative to the other plane. 43. If two lines intersect in space their projections must also intersect, and the straight line joining the points in which the projections intersect must be perpendicular to the ground line ; for the intersection of two lines must be a point common to both lines, whose projections must be on the horizontal and ver- tical projections of each of the lines, hence at their intersec- tions respectively. 44. If two lines are parallel in space their projections upon the vertical and horizontal planes will be parallel respectively. If one projection only of two lines are parallel, the lines in space are not parallel. 45. Any two lines drawn at pleasure, except parallel to each other and perpendicular to the ground line, will represent the projections of a line in space. 46. Prob. 1. To draw the projections of a line of a defi- nite length and occupying a fixed position in space. Let it be required to draw the projections of a line 1" long, which makes an angle of 30° with H, and whose horizontal projection makes an angle of 45° with GL, the lower end of the line being £" above H and 4/' in front of V. It is first necessary to place the line in such a position that its true length and the true size of the angle it makes with one of the coordinate planes are shown, and these are only shown PROJECTIONS. 73 when it is parallel to one of the coordinate planes. In this case it must be placed parallel to V. a v and a h , Fig. 5, are the pro- jections of one end of the line, d° being %" above GL and a h 4/' below it. Through a? draw a r b* at an angle of 30° with GL and 1" long ; through a h draw a h b j 1 parallel to GL, b' 1 be- ing found by dropping a perpendicular from b r (Art. 40). The two projections of the line, when parallel to V, are thus found to be a v b* and a h bK Now let us suppose the end a of the line to be fixed and the whole line to be revolved through an angle of 45° about a ver- tical axis through this point, the line keeping the same angle with H. The horizontal projection will not change in length (Art. 42, 9th), but will move through an angle of 45°, and will be found at a h b h . It is evident that in this revolution, so long as the angle with H does not change, every point in the line will remain at the same height above H. The point a does not move, being in the axis. We have seen that b h moves to U 1 ; b v must, therefore, be somewhere on a perpendicular through l/\ and, since the points do not change their heights, it must also be on a line through b v parallel to GL, hence at their intersection b v . Join a? and b v and we have a v b" and a ft M as the required pro- jections of the line. 47. If this line were revolved through 15° more, the point b 11 would go to c h , and b D to c v , and a?c° and a h c h would be the projections of the line making an angle of 30° with H, and whose horizontal projection made an angle of 60° with GL. If it were revolved still 30° more, the two projections would be a v d" and a h d h , each being perpendicular to the ground line. When a line is in this position, i. e., has its two projections in a Hue perpendicular to GL, it is said to be in a profile plane, a profile plane being understood to be one that is perpendicular to both V and H. 74 PROJECTIONS. When a line is oblique to only one of the coordinate planes it is said to make a simple angle ; when it is oblique to both of them it is said to make a compound angle. 48. If the angle that the line made with V had been given, it would have been necessary to have first placed the line par- allel to H, and then to have revolved it about an axis through one end perpendicular to V, in which case the length of the vertical projection would not change, and the points would not change their distances from V. In Fig. 6, a v b v and a h b h are the two projections of a line 1" long, making an angle of 45° with V, and whose vertical pro- jection makes an angle of 60° with GL. The principles and explanation for this construction are the same as for Prob. 1, if the horizontal and vertical planes are supposed to be inter- changed. 49. Prob. 2. To find the true length of a line given by its projections, and the angle it makes with either plane of projec- tion. Let a v b v and a h b h , Fig. 7, be the projections of the given line. The true length is only shown when it is parallel to one of the coordinate planes, hence this line must be revolved about an axis through either end until it is parallel to one of the planes. If it is revolved about a vertical axis through a until it is par- allel to V, the point a does not move, b h moves to b h , b v is found at 6" (where a perpendicular through b h intersects a horizontal through b v ), and a v hj is the true length. Also, the angle which a"b v makes with GL is equal to the true size of the angle the line makes with H. If it had been required to find the angle this line made with V, it would have been necessary to have revolved the line about a horizontal axis until it became parallel to H. Assuming the PROJECTIONS. 75 axis through the end b, Fig. 7, dP moves to a *, a h to a h , and a A 5 ft is the true length of the line (which of course should equal a v b"), and the angle it makes with GL is equal to the true size of the angle the line makes with V. 50. Note. The angles which the vertical and horizontal projections of a line make with GL are greater than the angles which the line in space males with H and V respectively, except when the line is parallel to one of the planes. PROJECTIONS OF SURFACES. 51. Plane surfaces are bounded by lines, therefore the prin- ciples which govern the projections of lines are equally appli- cable to these surfaces. 52. If we suppose a rectangular card abed, Fig. 8, placed with its surface parallel to V and perpendicular to H, each edge being parallel to V, it will be projected on V in a line equal and parallel to itself, hence the true size of the card itself is shown in vertical projection. Two of the edges, ab and cd, being per- pendicular to H, are projected on that plane in the points a h and d h respectively. The other two edges, ad and be, are par- allel to H as well as V, hence they will be projected in their true length on H. and, since one is vertically over the other, they will both be horizontally projected in the same line a h d h . Now, if the card is revolved about one of its vertical edges as an axis, like a door on its hinges, the vertical edge which coincides with the axis does not move ; the other vertical edge moves in the arc of a circle. The horizontal projection of the card will still be a straight line of the same length as before. Let the card be revolved through 60° ; a h , does not move ; d h moves in the arc of a circle, of which a h is the centre and a h d h the radius, to d h ; a h d h is the horizontal projection of the card in its new position ; the vertical projection of the edge cd in 7G PROJECTIONS. this jjosition is found at c t v d t v , vertically above d h , and a r b v c'd v is the vertical projection of the card after being revolved through an angle of 60°. . If the card should be revolved through 30° more, i. e., 90° in all, its surface will be at right angles with both coordinate planes, and its two projections will be found at a v b v and aV', in one and the same straight line perpendicular to GL. 53. If the card be placed on H, with one of its edges par- allel to V, a h b h c h d h , Fig. 9, will be its horizontal and a r b v its vertical projection. If this card be revolved about one of its edges which are perpendicular to V as an axis, like a trap-door on its hinges, through an angle of 30°, a r b v and a h b i h c i h d h will be its two projections. If it be revolved through 60° more, or 90° in ail, its projections will be a v e v and a h d h , which are just the same as a v b v and a h e h in Fig. 8, as they should be, since the cards are the same size in the two figures and they occupy the same relative position in each, i. e., they are in a profile plane. 54. The following principles may be noted : — 1st. When a plane surface is perpendicular to another plane its projection on that plane will be a line. 2nd. When a plane surface is parallel to either coordinate plane, its projection on that plane, will be equal to the true size of the surface and its other projection will be a line parallel to GL. 3rd. When a plane surface is perpendicular to one plane and oblique to the other, the angle which its projection on the plane to which it is perpendicular makes with the ground line is equal to the angle the surface in space makes with the plane to which it is oblique. 4th. If a plane surface, in different position*, makes a con- stant angle with a plane, its projections on that plane will all be of the same size. I PROJECTIONS. 77 55. Prob. 3. To draw the two projections of a plane sur- face, or card, of a certain size, and making a compound angle with the coordinate planes. Let the card be of the size shown in Fig. 9, and suppose it to make an angle of 30° with H, and its horizontal projection an angle of 45° with GL. Draw the horizontal projection a ll b h c !l d /l of the card equal to its true size ; a v b'' will be its vertical projection. Revolve a r b° through an angle of 30° to a v b v , and a v b v will be the vertical projection of the card when it makes an angle of 30° with H and is perpendicular to V ; al'b^c^d' 1 is its corresponding hori- zontal projection. The angle with H is still to be 30° after the card has been revolved to its desired position, hence its horizontal projection will be the same size. Therefore, make a h b] l c }l d h , Fig. 10, equal in size to a h b]'c]'d h , Fig. 9, and making the desired angle with GL. In this revolution, as long as one edge rests on H and the angle remains constant with H, every point keeps the same height above IT, therefore the vertical projections of a h and d h , Fig. 10, must be found at a v and d v ; also of b] 1 and c A at 6" and c ", whose heights above GL are equal to the height of b;\ Fig. 9, above GL. The other parallelograms in Fig. 10 represent the projections of the same card at different angles with H, the horizontal pro- jections making the same angles with GL. 56. a h b''c h d h and a v d v b v c v , Fig. 11, represent the projections of the same card when it is lying on H with one of its diago- nals parallel to V, and a''d''b'c" and a ,l b / /i c''d h are its projec- tions after being revolved about an axis perpendicular to V through the corner a through an angle of 45°. Fig. 12 repre- sents the projections of this card when, besides making an angle of 45° with H, the horizontal projection of the diagonal makes 78 PROJECTIONS. an angle of 30° with GL. The steps to obtain this are exactly the same as in Figs. 9 and 10, hence the explanation will not be repeated. 57. Prob. 4. To draw the projections of a regular pentag- onal card, the diameter of the circumscribed circle being given, in two positions. 1 st , when it is perpendicular to V and making an angle of 60° with H, one of its edges being perpendicular to V ; 2nd, when, besides making an angle of 60° with H as in 1st position, it has been revolved through an angle of 45°. The pentagon must first be drawn in its true size and posi- tion ; a^J/'c/'d^e*, Fig. 13, equal to the actual size of the card, is its horizontal projection, cj'd' 1 being perpendicular to GL, and a r b r c v is its vertical projection. Revolve a'b^'c 1 ' through an angle of 60° to a r b v c v ; each point moves in the arc of a cir- cle with a as a centre, and a r lfc v will be the vertical projection of the card when in the 1st position asked for ; a h b h c h d h e h is its corresponding horizontal projection. For the 2nd position revolve the plan just found through 45° to the position a h b h , etc., shown in Fig. 14; a r b''c'd r e v will be the corresponding vertical projection. 58. Cards of any shape and size, and occupying any position, may be drawn in the same way, care being taken to locate one point at a time. 59. If the angle the card made with V had been given, it would have been necessary to have first placed the card par- allel to V and then to have revolved it, through the angle it made with V, about an axis perpendicular to II, in which case the length of the horizontal projection, which is a straight line, would not change, and the several points would not change their respective distances from H. PROJECTIONS. 79 In the second revolution, which chauges the angle with the coordinate planes from simple to compound, the vertical projec- tion must be revolved and the corresponding plan found. The angle with V being the same the vertical projection does not change its size ; the distances of the points in front of V remain the same after revolution as before, hence are found at the same distances from GL respectively. 60. Prob. 5. To draw the projections of a circular card making a compound angle with the coordinate planes. Let the diameter of the card be given, the angle it makes with V, and the angle through which the vertical projection is to be revolved. A circle may be considered as a polygon of an infinite num- ber of sides, hence we can take as many points as we please on the circumference of the circle, and each one moves according to the principles just described. Place the card parallel to V ; a circle, a r b''cf, etc., Fig. 15, equal to the actual size of the given circle, is its vertical pro- jection, and a h b''c\ etc. is its horizontal projection. Revolve the card through the required angle about a vertical axis through a; a ,l b l 'c h , etc. is its horizontal, and a v b v c", etc. is its vertical projection. The card is to be revolved through a certain angle, still keeping the same angle with V. The size of the vertical pro- jection will, therefore, not change. Hence, revolve the vertical projection found in Fig. 15 through the required angle to the position a c b'c'\ etc., Fig. 16. None of the points change their distance from V, consequently a''b h c h , etc. is the horizontal pro- jection of the card, found as in the last problem. 61. Fig. 17 shows a somewhat shorter method of drawing the projections of a circular card making a simple angle with 80 PROJECTIONS. the coordinate planes ; in this case it makes an angle of 30° with V and is perpendicular to II. a h b h c ] \ etc., making 30° with GL, is its horizontal projection. Suppose the card re- volved about its horizontal diameter ae until it is parallel to H. It will then be shown in its true size at a h bc, etc. ; bV 1 , cc h , etc. will show the actual distances of the points b, c, etc. from the horizontal diameter ; cfe v will represent the vertical projec- tion of the horizontal diameter about which the card is revolved. Of course, b v must be found somewhere in a line through b h , perpendicular to GL, therefore, lay off tb v equal to bf/', and b* is a point of the required vertical projection ; c''s is made equal to cc\ d'r to dd h , etc. Other points may be found in the same way. It is evident that n V\ me h , etc. are respectively equal to bb h , cc b , etc., hence it is only necessary to revolve the semi- circle and the distance bb h is laid off on both sides of the diam- eter a r e^ giving the two points b 1 ' and ri". PROJECTIONS OF SOLIDS. 62. A cube is a solid bounded by six equal faces, and when it is placed so that two of its faces are parallel to H. and two others parallel to V, its two projections are a'b'e'f' and a h b h c h d h , Fig. 18. The top and bottom, being parallel to H, are hori- zontally projected in one and the same square, ct!'V'c h d h , which is, of course, equal to the exact size of these faces, and their vertical projections are a v &° and e v f" respectively (Art. 30-2nd) ; the front and back faces being parallel to V are vertically pro- jected in one and the same square, a r b''c r f'\ which is also equal to the exact size of these faces, and their horizontal projections are a h b h and c h d h respectively (Art. 30-2nd) ; the left and right hand faces are perpendicular to both V and H, therefore their vertical projections are a v e v and ///'', and their horizontal pro- jections are a h d h and b h c h respectively (Art. 30-lst). PROJECTIONS. 81 The plan shows two dimensions of the cube, the length and breadth, and the elevation two dimensions, the length and thick- ness ; therefore, the three dimensions of the solid being shown in their true size in the two projections the object is com- pletely represented. In this case it does not matter which pro- jection is drawn first, as they each show two dimensions in their true size. 63. Shade Lines. In outline drawings it is customary to put in shade lines, i. e., lines heavier than the others ; they give relief to the drawing, and, when properly placed, are of assist- ance in reading it. Shade lines, or edges, are those edges which separate light, from dark surfaces. The rays of light are generally assumed to come from over the left shoulder in the direction of the diagonal of a cube, the person supposed to be facing the cube, and the cube to be in the position shown in Fig. 18. That is, the ray of light enters the cube at the upper, front, left-hand corner, whose projections are a? and a h , and leaves it at the lower, back, right-hand corner, whose projections are f 9 and c h ; the diagonal joining these points will represent the actual direction of the conventional ray of light, and its projections a v f" and a h c h are the projec- tions of this ray. The different rays of light are all supposed to be parallel to each other. It is evident from the figure that both projections of the rays of light make angles of 45° with GL. The student should distinctly understand that, although the projections of the ray of light make 45° with GL, the actual angle it makes with V or H is quite different. To find this angle apply the princi- ples of Art. 25 to Fig. 18, and we get a= 35° 15' 52" as its actual size. C>1. In the cube, Fig. 18, the top, front, and left-hand faces 82 PROJECTIONS. are light, and the bottom, back, and right-hand faces are dark, and the shade lines are, therefore, e r f'\ which separates the front from the bottom, b''f'\ which separates the front from the right-hand face, b u c'\ which separates the top from the right- hand face, and c h d'\ which separates the top from the back face. The two other shade edges which separate the left-hand face from the back and bottom faces are not seen in either projec- tion, since the top and front edges of the left-hand face are in the same plane and nearer the eye. It is for this same reason that the shade, lines mentioned above are seen only in one projection, i. e., e r f v is a shade line; it is seen in elevation, but in plan is hidden by the upper front edge of the cube a v b v -a h L h . 65. It will be noticed that the right-hand and lower edges are shaded in elevation, and the right hand and upper in plan. From this many draftsmen have adopted the arbitrary rule to shade the right-hand and lower lines in elevation, and the right- hand and upper in plan. This rule is really applicable in but few cases, except when the object is of rectangular section and so placed that its surfaces are perpendicular to one or both of the coordinate planes. Other draftsmen shade the right-hand and lower lines in both plan and elevation. This is also applicable only in the cases stated above, besides being obliged to change the direction of the ray of light in plan and elevation, or imagine the object revolved while the ray of light remains fixed. Others still^ollovv the rule given in Art. 63, except that they only put in shade lines where the dark portion of space adja- cent to the line in question is visible, and they change the direc- tion of the ray of light, or, what is practically the same thing. revolve the object. By this method the plan and elevation are shaded alike, but the right-hand line is not necessarily a shade line, while the left-hand line is necessarily not a shade line. PROJECTIONS. 83 The method taken up in this book, and the one it is expected that the student will follow in this course, is not given because it is the one most generally in use, or because it is the easiest, — quite the contrary ; but because it is, in the opinion of the writer, the only method which can be followed consistently throughout a course of projections, shadows, isometric, and working drawings. 66. Prob. 6. To draw the two projections of a right square prism with its base on II and its vertical faces oblique to V. Fig. 19. It is evident that the vertical faces will not be projected on V in their true size, the height only of the prism being shown in its real length, and also that the two ends, being parallel to H, will be projected on that plane in their true size. Hence, draw the square a h b h c h d h equal to the ends of the prism, with its edges making the same angles with GL as the vertical faces make in space with V. Through the corners a h , b h , c h , and d h , draw the perpendiculars a v e v , b'f v , etc., making each equal in length to the height of the prism. a v b r c r d" will be the vertical projection of the upper base. e v f v m v n v of the lower base, and a v c?m v e v the vertical projection of the whole prism. In this case the top and two front faces are light, while the bottom and two back faces are dark ; hence, in plan the lines a h d h and c h d h , which separate the top from the two back faces, -are shade lines ; in elevation they are behind the two front edges ab and be, consequently are not seen. The element ae separates the left front from the left back face, hence is a shade line, and a r i''\ its vertical projection, is accordingly made heavy ; its hori- zontal projection is simply a point; the element cm also sepa- rates a light from a dark surface, and its visible projection c'ni should be made heavy. The edges ef and fm separate the two 84 PROJECTIONS. front faces from the bottom, and their visible projections e r f v and /'in'' are also made heavy. 67. Prob. 7. To draw the two projections of a right regu- lar •pentagonal prism standing with -its base on H, with none of its faces parallel to V. Fig. 20. Here, as in the last problem, it is necessary to draw the hori- zontal projection first, as it shows the pentagonal end in its true size and position. a h V'c h d h e h is its horizontal projection, and a v d v ri°f v , found as in the last problem, is its vertical projection. From the shade lines shown in the figure it will be noticed that the line cd, which separates the light top from the dark right-hand face, is visible in both projections, hence it is made heavy in both. A prism of any number of sides standing on II can be drawn in the same manner. 68. Prob. 8. To draw the two pro ect ions of a right regu- lar hexagonal prism with its axis perpendicular to V. Fig. 21. Here it is necessary to draw the vertical projection first, and construct the horizontal projection from it according to the principles noted in the last two problems. 60. From an inspection of Figures 18 to 21 it is evident that the 4.V J triangle can be used to determine positiccly the light and dark fares only when these faces are perpendicular, or nearly so, to one or both of the coordinate planes. In Fig. 18 the triangle can be used iu both plan and eleva- tion, since every face is perpendicular to at least one of the coordinate planes. In Figs. 19 and 20 the faces are perpen- dicular to H only (except the top, which is, of cuurse, known to be light), hence the 45° triangle can only be used iu the plan. #. & v h y £ a * PROJECTIONS. 80 In Fig. 21 the faces are perpendicular to V only (except the front end, which must be light), hence the triangle can only be used in elevation. 70. Prob. 9. To draw the two projections of a right cylin- der standing on its base. Fig. 22. Its horizontal projection will be a circle equal to the end of the cylinder (Art. 42). Its vertical projection will be repre- sented by a v b r d v c v , a r V'' and m h e h lie in the auxiliary plane, the line m h e h also lies in the plane of the card mnop, therefore, the point r h , where these two lines intersect, is a point on the card, and must be one point of the horizontal projection of the shadow required ; r v on m v e v is its vertical projection. Assume any other auxiliary pla.ie, as d v o v , and another point, s v s h of the shadow, will be 110 SHADOWS. found in the same way. Draw an indefinite line through these points r and s in both projections, and the shadow is finished. Vertical auxiliary planes could have been taken instead of horizontal with the same result. 110. Prob. 26. To cast the shadoio of an abacus on a con- ical column. Fig. 61. Since neither projection of the surface receiving the shadow is a line, this problem must be done by the indirect method as used in the two preceding problems. Find the shadow on any horizontal plane, as a v c v . Its ver- tical projection is a v c v , and is the GL for this shadow. This plane cuts the column in a horizontal circle, of which a v c v is the vertical and aJ'V'c' 1 is the horizontal projection. The shadow of the bottom edge of the abacus on this plane is the circle d h a h e h , drawn with o*, the shadow of o the centre of the abacus, as a centre and radius equal to that of the abacus. This circle cuts the circle a ft 5*c* at the point a h , which is one point of the horizontal projection of the shadow required. a v , on the verti- cal projection of this auxiliary horizontal plane, is one point of the vertical projection of the required shadow. Any number of other points can be found in the same way. 111. Prob. 27. To cast the shadow of the prism given as in Fig. 62 on H and V, also of the pyramid on the prism and on H. The shadows of the prism on H and V and of the pyramid on H require no explanation. If the shadow of a line falls partly on two plane surfaces, A and B (no figure), which do not intersect (at that portion under discussion), A being between the line and B, that part which falls on A does not fall on B, and the shadow of the line SHADOWS. Ill on B may be said to begin at the shadow of the point where the shadow of the line leaves A, on B. Now, if B is horizontal or vertical, and A is oblique to both V and H, the shadows of the line and the plane A on B can be readily found ; and then get the shadow of the line on A by the reverse of the above process, that is, note where the shadow of the line and plane A on B intersect ; find what point on A cast this intersecting point, and that will be one point of the shadow of the line on A. Now, referring to Fig. 62, we see that the shadow of the pyramid and prism on H intersect at the four points rj, ^, x%, and z h s ; the points on the prism (this being between the pyra- mid and H) which cast these shadows are r h , t h , x h , and z h respectively (found by drawing a 45° line from the shadow to the edge casting it), and they are, therefore, points of the shadow of the pyramid on the prism. The shadow on the upper edge of the prism may be found by this same principle, that is, the shadow of this upper edge on H is bltf* ; this intersects the shadow of the pyramid on H at s'g and y h s ; s h and y h are the points which cast these shad- ows, hence they are points of the shadow required. Join r h s h , s h t h , x h y h , and y h z h , and the horizontal projection of the shadow is completed. Its vertical projection is found by projecting each one of these poiuts on to the vertical projection of the prism. The points s h and y h could also have been found by casting the shadow of the pyramid on an auxiliary horizontal plane through the top edge of the prism ; o£s h and o*y h are the shadows of the two shade elements of the pyramid on such a plane ; they intersect the upper edge of the prism at the points s h and y'\ which are the points required. A vertical auxiliary plane through this edge or the other edges of the prism might have been used with the same result. CHAPTER VL ISOMETRICAL DRAWING. 112. In all the previous constructions two projection.? have been used to represent a body in space. In isometrical projec- tions only one view is used, the body being placed in such a position that its principal lines or edges (length, breadth, and thickness) are parallel to three rectangular axes, which are so placed that equal lengths on them are projected on the plane equal to each other. Thus we have the three dimensions of a body shown on one plane in such a way that each can be meas- ured, thereby combining the exactness of ordinary projections and the intelligibleness of pictorial figures. It is used chiefly to represent small objects in which the principal lines are at right angles to each other. In large objects the drawing would look distorted. 113. If we take a cube situated as in Fig. 63, and tip it to the left, about its lower left corner e until the diagonal eg is horizontal, Fig. 64, and then turn it through an angle of 90°, still keeping eg horizontal, we obtain Fig. 65. The vertical projection in this figure is what is called an isometrical projec- tion. The edges of a cube are all of equal length, and it will be seen that they appear equal in the figure, consequently the visi- ble faces must appear equal. It will also be seen that the figure can be inscribed in a circle, and that the outline of the isomet- rical projection i» a regular hexagon, hence, that those lines V c ISOMETRICAL DRAWING. 113 which represent length and breadth make angles of 30° with a horizontal, and those which represent thickness are vertical. 114. The edges of the cube being inclined to the plane on which they are represented, appear shorter than they actually are on the object, but since they are all equally foreshortened, and since a drawing may be made to any scale, it is. customary to ignore this foreshortening, and make all the isometrical lines of the object equal to their true lengths. This will give what is called the isometrical drawing of the object, which will be somewhat larger than the isometrical projection. Fig. 65 represents the isometrical projection of the cube shown in Fig. 63, and Fig. 67 is the isometrical drawing of the same cube. 115. Definitions.