4 ( 
 
 1^
 
 DESCRIPTIVE GEOMETRY. 
 
 PREPARED FOR THE USE OF THE STUDENTS 
 OF THE 
 
 MASSACHUSETTS INSTITUTE OF TECHNOLOGY, 
 
 BOSTON, MASS. 
 
 BOSTON: 
 
 W. J. SciiOFiELD, Printer, 105 Summer Street. 
 
 1886.
 
 Copyrighted, 1886, by Linus Faunce.
 
 DESCRIPTIVE GEOMETRY. 
 
 CHAPTER I. 
 
 Elementaby PmNciPLEa. 
 
 1. Descriptive Geometry is the art of representing a defi- 
 nite body in space upon two planes, at right angles with each 
 other, by lines falling perpendicularly to the planes from all the 
 points of the intersection of every two contiguous sides of the 
 body, and from all points of its contour, and of solving all graph- 
 ical problems involving three dimensions. 
 
 2. These planes are called coordinate planes, or the planes of 
 projection, one of which is horizontal and the other vertical. 
 P and Q, Fig. 1, represent two such planes, and their line of inter- 
 section G L is called the ground line. 
 
 3. The perpendicular lines from a point in space to the planes 
 of projection are called the projecting lines of that point. a a,", 
 a a''. Fig. 1, are the projecting lines of the point a. 
 
 4. Each coordinate plane is supposed to extend indefinitely in 
 both directions from the ground line; hence they will form by 
 their intersection four dihedral angles (see Fig. 1). 
 
 The first angle is above the horizontal, and in front of the ver- 
 tical plane. 
 
 The second is above the horizontal, and behind the vertical. 
 
 The third is below the horizontal, and behind the vertical. 
 
 The fourth is below the horizontal, and in front of the vertical. 
 
 The points a, 6, c, and rf, Fig. 1, are in the first, second, third, 
 and fourth angles respectively. 
 
 2065979
 
 4 DESCRIPTIVE GEOMETRY. 
 
 5. In order that the two projections of an object may be repre- 
 sented on the same sheet of paper, the upper part of the verti- 
 cal plane is revolved backward, about the ground line, as an axis, 
 until it coincides with the horizontal plane. 
 
 This being done, it is evident that all that portion of the plane 
 of the paper which is in front of or below the ground line repre- 
 sents not only that part of the horizontal plane which is in front 
 of the vertical, but also that part of the vertical plane which is 
 below the horizontal, while the portion above the ground line 
 represents that part of the vertical which is above the horizontal, 
 and also that part of the horizontal which is behind the vertical 
 plane. 
 
 6. All points in the first and second angles are vertically pro- 
 jected above the ground line ; all points in the third or fourth 
 angles below it. 
 
 Points in the first and fourth angles are horizontally projected 
 in front of the ground line ; those in the second and third 
 behind it. 
 
 Thus the points a, b, c, and t?, situated in the first, second, third, 
 and fourth angles respectively, as shown in Fig. 1, are represented 
 on the plane of the paper, as shown in Fig, 2. 
 
 7. Lines situated in either of the four angles are represented 
 by their projections in the same way as described for the point. 
 
 8. The two projections of a point are on one and the same 
 straight line perpendicular to the ground line. 
 
 9. Two projections are always necessary to definitely locate a 
 point or line in space. 
 
 10. The distance of the vertical projection of a point from the 
 ground line is equal to the distance of the point itself, in space, 
 from the horizontal plane ; and the distance of its horizontal pro- 
 jection from the ground line is equal to the distance of the point, 
 in space, from the vertical plane. 
 
 11. A point situated upon either of the coordinate planes is its 
 own ])rojection on that plane, and its other projection is in the 
 ground line. 
 
 12. A right line is determined by two points; hence, any two 
 lines drawn at pleasure, exce})t parallel to each other and perpen- 
 dicular to the ground line, will represent the projections of a line 
 in space.
 
 DESCRIPTIVE GEOMETRY. 5 
 
 13. If two lines are parallel in space, their projections upon 
 either plane will also be parallel. 
 
 14. If a right line is perpendicular to either coordinate plane, 
 its projection on that plane will be a point, and its projection on 
 the other plane will be perpendicular to the ground line, 
 
 15. If a line be parallel to either plane, its projection on that 
 plane will be parallel to the line itself, and its projection on the 
 other plane will be parallel to the ground line. 
 
 16. If a line is parallel to both planes, or to the ground line, 
 both projections will be parallel to the ground line. 
 
 17. When a line is parallel to a plane, its projection on that 
 plane will be equal to the true length of the line. 
 
 18. If a point be on a line, its projections will be on the pro- 
 jections of the line. 
 
 19. The planes which contain the projecting lines of all points 
 in a straight line,— or, in other words, which contain the straight 
 line and are perpendicular to the planes of projection, — are called 
 the projecting flanea of that line. 
 
 20. If two lines intersect in space, their projections must also 
 intersect, and the right line joining the points in which the pro- 
 jections 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 vertical 
 projections of each of the lines, hence at their intersections 
 respectively. 
 
 21. Since two points determine a straight line, it may be deter- 
 mined by its intersections with the coordinate planes. These 
 intersections will be points, and are called the traces of the line, 
 horizontal or vertical, according as they are on the horizontal or 
 vertical plane. A line can have but two traces; it may have only 
 one. The points m and (/, Fig. 3, are the traces of the line g vi, 
 likewise e and/ of the line e f. 
 
 22. In the same way, a plane being determined by two lines, it 
 may be determined by its intersections with the coordinate planes. 
 These intersections will be lines, and they are called the traces of 
 the plane. A plane can have one or two traces. The line a 6, 
 Fig. 3, is the vertical trace, and c d the horizontal trace of the 
 plane X. 
 
 23. The traces of a line lying in a plane must be in the traces
 
 6 DESCEIPTIVE GEOMETRY. 
 
 of that plane ; for every line, as e / and g m, Fig. 3, lying in the 
 plane X, must intersect the coordinate planes somewhere on the 
 lines a h and c d in which the plane X intersects the coordinate 
 planes. 
 
 24. The two traces of the same plane must necessarily meet 
 the ground line in the same point ; for, by the definition of traces, 
 that point is at once in the given plane and in each of the coor- 
 dinate planes ; hence it is at the intersection of the three planes, 
 i.e.^ on the ground line. 
 
 If the plane be parallel to the ground line, this point is at an 
 infinite distance ; hence the two traces are parallel to each other, 
 and to the ground line. 
 
 25. If a plane be parallel to either coordinate plane, it will 
 have but one trace, which will be on the other plane and parallel 
 to the ground line. 
 
 26. If a plane be perpendicular to either coordinate plane, its 
 trace on the other plane will be perpendicular to the ground line. 
 If it be perpendicular to both coordinate planes, both of its traces 
 will be in a line perpendicular to the ground line. Such a plane 
 is called a profile plane. 
 
 27. If a plane pass through the ground line, its position is not 
 determined. 
 
 28. Lines and planes are supposed to be of indefinite length ; 
 hence the projections of lines and the traces of planes may always 
 be produced indefinitely in either direction. 
 
 29. If a right line is perpendicular to a plane, its projections 
 will be respectively perpendicular to the traces of the plane. If 
 a plane is perpendicular to two other planes, it will be perpen- 
 dicular to their line of intersection. But the horizontal project- 
 ing plane of the line is perpendicular to H (Art. 19), and it is 
 also perpendicular to the given plane, since it contains a line per- 
 pendicular to the given plane ; hence it is perpendicular to their 
 line of intersection, which is the horizontal trace of the given 
 plane. But the horizontal projection of the given line passes 
 through this trace, hence it must be perpendicular to it. 
 
 In the same way it can be proved that the vertical projection 
 of the line is perpendicular to the vertical trace of the plane. 
 The converse of this must also be true, — that is, if a plane is
 
 DESCRIPTIVE GEOMETRY. 7 
 
 perpendicular to a line, the traces of the plane must be respectively 
 perpendicular to the projections of the line. 
 
 Notation. 
 
 30. We will designate a point in space by a small letter, and 
 its projections by the same letter with an Ji or v written above : 
 thus a'' represents the horizontal, and a" the vertical projection of 
 the point a. 
 
 31. A line in space will be designated by a capital letter, and 
 its projections by the same letter with an h ot v written above, 
 the same as with the point. 
 
 In speaking of a line in space, we may say the line A, or, since 
 a line is determined by two points, the line a h. 
 
 32. The horizontal coordinate plane will be designated by the 
 capital letter H, the vertical by the capital letter V, and any 
 other plane in space by any capital letter. 
 
 The traces of a plane will be designated by the same letter as 
 the plane, with H or V prefixed : thus, H P denotes the horizon- 
 tal, and V P the vertical trace of the plane P. 
 
 A plane in space is determined by two parallel or intersecting 
 lines, by a line and a point, or by three points ; hence, we may 
 speak of a plane as the plane P, the plane (A, B), the plane 
 (A, a), or the plane (a, h, c). 
 
 33. Given and required lines, if visible, are represented by full 
 lines ; if invisible, by short dashes. Auxiliary lines are repre- 
 sented by long and short dashes. 
 
 Traces of planes, if .visible, are represented by full lines; if 
 invisible, by one long and two short dashes. Auxiliary planes, — 
 that is, planes which are used simply as a means of obtaining the 
 solution of a problem, — are represented by one long and three 
 short dashes. Construction lines are represented by short dashes. 
 
 Examples. 
 
 Ex. 1. — Draw the projections of a line in the third angle parallel to V, 
 and oblique to H. 
 
 Ex. 2. — Of a line in the second angle parallel to H, and oblique to V. 
 
 Ex. 3. — Of any line in the fourth angle. 
 
 Ex. 4. — Of any line lying in V, and below H. 
 
 Ex. 5. — Of a line lying in H, and behind V.
 
 8 DESCRIPTIVE GEOMETRY. 
 
 Ex. 6. — Of a line in the third angle, and perpendicular to V. 
 
 Ex. 7. — Of a line in the fourth angle, and perpendicular to H. 
 
 Ex. 8. — Of a line lying in a profile plane, and oblique to both V 
 ,and H. 
 
 V Ex. 9. — Of a line crossing the first, fourth, and third angles. 
 • Ex. 10. — Of a line crossing the second, first, and fourth angles. 
 
 Ex. 11. — Of a line crossings the first and third angles. 
 
 Ex. 12. — Of a line crossing the second, third, and fourth angles. 
 
 Ex. 13. — Of two intersecting lines lying in the third angle. 
 
 Ex. 14. — Of two lines lying in the first angle which do not intersect. 
 
 Ex 15. — Draw the traces of a plane which is oblique to both Vand H. 
 
 Ex. 16. — Of a profile plane. 
 
 Ex. 17. — Of a plane parallel to and behind V. 
 
 Ex. 18. — Of a plane perpendicular to V, and oblique to H. 
 
 Ex. 19. — Of a plane perpendicular to H, and oblique to V. 
 
 Ex. 20. — Of a plane pai-allel to G L, and oblique to both V and H.
 
 CHAPTER II. 
 
 Problems Relating to the Point, Line, and Plane. 
 
 34. Problem 1. — To find the traces of a given line. — Let the 
 line be given by its two projections A'' and A\ Figs. 4 and 5. 
 We will first find the H trace. This, being in H, will have its V 
 projection in the ground line (Art. 11), and it must also be on 
 the V projection of the line (Art 18), hence, when the V projec- 
 tion of the line, produced if necessary, meets the ground line, as 
 at «"', we know that the line in space is passing through H at some 
 point, which may be either in front of or behind V, and since the 
 two projections of a point must be in a line perpendicular to G L 
 (Art 8), and in the two projections of the line (Art. 18) this 
 point will be where a perpendicular to G L from a" meets the H 
 projection of the line, as a'*. Hence, a'' is the H trace of the 
 line A. 
 
 If we put H in the place of V, and V in the place of H, in the 
 preceding statement, we show how to find the V trace of the line, 
 — that is, produce A^ until it meets G L at 6^ ; erect a perpendicu- 
 lar at that point, and l/\ the point where it intersects A''', will be 
 the V trace of the line. 
 
 35. Problem 2. — G-iven the traces of a line to construct its pro- 
 jections. — Let a* and h'\ Figs. 4 and 5, be the traces ; since these 
 points lie in the coordinate planes, their V and H projections 
 respectivel}^ will be in G L at a" and V (Art. 11) ; joining a" with 
 i'', and a'' with 6'', we have A'' and A'\ the two projections required. 
 
 36. Problem 3. — To find the true length of a line joining two 
 given points in space. — When a line is parallel to either coordi- 
 nate plane its projection on that plane is equal to the true length
 
 10 DESCRIPTIVE GEOMETRY. 
 
 of the line (Art. 17). Hence, we have only to revolve the line 
 abont an axis through either end until it is parallel to one of the 
 planes, and find its projection on that plane. Let A"" and A'', Fig. 
 6, be the projections of the line joining the points a and h. Revolve 
 the line about a vertical axis through the point a; every point in 
 the line, excej^t the one in the axis, moves in a horizontal circle, 
 of which the H projection is a circle and the V projection a line 
 parallel to G L. When the H projection becomes parallel to 
 G L, the line is parallel to V (Art. 15) ; a" and a!" do not move ; 
 V has described the arc of a circle whose radius is a^ 6*, and is 
 found at 5/'; h" moves in a line parallel to G L, and is found per- 
 pendicularly above 5/ at h^'. Therefore, a'' b" represents the true 
 length of the line. 
 
 The line could have been revolved about an axis perpendicular 
 to V, until it was parallel to H, in the same way. 
 
 37. Second method. — The true length of a line may also be 
 obtained in another way by revolving the horizontal projecting 
 plane of the line about its horizontal trace, into the horizontal 
 plane, when every line of the projecting plane, hence the line in 
 question, will be shown in its true length on H. 
 
 Let the line be given as in Fig. 7. The horizontal projecting 
 lines of each end of the line a and b are perpendicular to the hori- 
 zontal trace a'' b'' of the horizontal projecting plane, therefore they 
 will be so after revolution. Hence, draw the lines «/' a'' and bf' b^ 
 perpendicular to «''5'', and make them equal respectively to the 
 lines a"-' r and //'s, which represent the heights of the points in 
 space above H ; the line joining these two points a/' and bj' will 
 be the true length required. 
 
 This result could have been obtained by revolving the vertical 
 projecting plane until it coincides with V, when a'' a^ is made 
 equal to a} r and 5'' 6/' equal to b'' s, the true length being a'' 5/'. 
 
 38. If the line passes through the plane into which it is 
 revolved, as in Fig. 8, the point in which it pierces the plane being 
 in the axis, does not move, and the two ends of the line revolve 
 in opposite directions; that is, a'' af" is still made equal to a"r, 
 and b''b/' equal to b''s; but they are lai-d off in opjjosite direc- 
 tions, and af' bf', the true length of the line, is found to pass 
 through c'\ the horizontal trace of the line.
 
 DESCRIPTIVE GEOMETRY. 11 
 
 89. Problem 4, — To pass a plane through two intersecting 
 lines. — The traces of a line lying in a plane must be in the traces 
 of the plane (Art. 23), hence we have only to find the traces of 
 each line, and a line drawn through the two horizontal traces will 
 be the horizontal trace of the plane, and one through the vertical 
 traces will be the vertical trace of the plane. 
 
 Let A and B, Fig. 9, be the two lines intersecting in e. The 
 traces of A are J'' and a", and of B are c* and d" (Art. 34). H P 
 is drawn through b'' and c\ and V P through a" and d". They 
 should intersect the ground line in the same point. 
 
 40. If we wished to pass a plane through three points, draw 
 two lines through the points, and proceed as described above (see 
 Fig. 10) ; a, 6, and c are the given points. 
 
 41. Only one plane can be passed through two lines, but an 
 infinite number can be passed through one line. Thus, in Fig. 11, 
 the planes P, Q, R, S, etc., each contain the line B. 
 
 42. Problem 5. — G-iven one projection of a line, or point, 
 lying in a plane, to find the other projectioji. — Let A'', Fig. 12, be 
 the vertical projection of a line lying in the plane P, given by its 
 traces V P and H P. The traces of the line must lie in the traces 
 of the plane (Art. 23) ; therefore, the vertical trace of the line A 
 must be at «" where A'' intersects V P, and its horizontal projec- 
 tion is at a'\ The horizontal trace must be in a line perpendicu- 
 lar to G L at the point h" where A" intersects it, and on H P, 
 hence at their intersection h''. A^ drawn through a* and h'' will 
 be the horizontal projection of the line. 
 
 If the horizontal projection of the line had been given, the ver- 
 tical projection would be found in the same way. 
 
 43. If one projection of a point were given, we simply draw 
 any line through the point, and find the other projection of the 
 line as just described. 
 
 44. If the line is horizontal, its vertical projection is parallel to 
 the ground line, and its horizontal projection is parallel to the hori- 
 zontal trace of the playie in tvhich it lies. — In Fig. 13, A" is the 
 vertical projection of such a line lying in the plane P, a" and a'' are 
 found as in Art. 42. Since A'' is parallel to G L, b" and 5^ are 
 at an infinite distance from a", hence A'' is parallel to H P.
 
 12 DESCKIPTIVE GEOMETKY. 
 
 45. Problem 6. — Given the projections of a pointy or line, 
 lying in a plane, to find its position ivhen the plane shall have been 
 revolved to coincide ivith either coordinate plane. — Let &' (^, Fig. 
 12, be the projections of a point in the phme P. If we revolve the 
 plane about its horizontal trace until it coincides with H, the point 
 will move in a plane perpendicular to P, and hence to H P, and 
 will be found somewhere in ^'' <?, and at the same distance from 
 H P that it was in space, which is equal to the hypothenuse of a 
 right angled triangle, of which c" e (the height of the point above 
 H) is one side, and c'' d (the distance of its horizontal projection 
 from H P) is the other ; hence the poii>t revolves to c. 
 
 In revolving a line it is only necessary to find the revolved 
 position of two points, and the position of the line is determined. 
 
 In the same figure we have found the revolved position of one 
 point c of the line A, but the point h being in H P does not 
 move in revolution ; hence 5^ e is the revolved position of the 
 line A. 
 
 46. Proble^m 7. — To find the true size of the angle made hy 
 tivo intersecting lines. — If we pass a plane through the two lines, 
 and then revolve the plane into one of the coordinate planes, the 
 angle between the lines will be shown in its true size. 
 
 Let A and B, Fig. 14, be the given lines intersecting at a. Pass 
 a plane through these lines by Art. 39. Its horizontal trace is 
 HP. It is not necessary to find VP. Revolve this plane about 
 H P into H. The point of intersection a revolves to a, ; the points 
 c^ and V" do not move, hence the two lines will be found at c^ a, 
 and V- a. and 8 will be the ano;le souorht. 
 
 47. Problem 8. — To find the true size and shape of any plane 
 surface. — Let the surface be given as in Fig. 15. Pass a plane 
 through it (to do this, pass a plane through an}'- two of its edges, 
 as a (^ and d <?), H P is its horizontal trace. Revolve this plane 
 about H P into H. The point a is found at «/', b at h\ c at c/', 
 and d at dl'. Hence aj" 5/' c,* dj' is the true size and shape of the 
 surface abed. 
 
 48. Problem 9.- — To find the shortest distance from a jjoint to 
 a line, and draw the projections of it in its position in space. — Let
 
 DESCRIPTIVE GEOMETRY. 13 
 
 w, Fig. IG, be the given point, and A the given line. The short- 
 est distance from w to A will be a perpendicnlar from n on to A. 
 The relative positions of these two lines will only be shown when 
 the plane containing them has been revolved into one of the 
 coordinate planes. Therefore, pass a plane through the line A 
 and point n (to do this, draw a line through n parallel to A). 
 H P is the horizontal trace (the vertical trace is not needed). 
 Revolve this plane about H P into H. The point n revolves to 
 ti,, the point d, in the line A, to d,, «* remains stationary; hence 
 the line A is found at A,. From n, drop a perpendicular on to 
 A,, and this, /i, <?, , is the length required.. 
 
 To tind the projections of this in its correct position, revolve 
 the plane back to its original position ; c, revolves back to c^, and 
 n'' </' is the horizontal projection. The vertical projection of e'' is 
 found at c'', hence n" c'' is the vertical projection sought. 
 
 49. Problem 10. — To find the angle tvhic/i a given line makes 
 tvifh an oblique j^lane. — The angle is that made by the line with 
 its projection on the oblique plane. This projection, together 
 with the line itself, and a perpendicular from any point on the 
 line to the plane, form a right angled triffhgle in which the angle 
 made by the line with the perpendicular is the complement of the 
 angle sought. 
 
 Let A, Fig. 17, be the given line, and P the given plane. 
 
 From any point a of the line drop a perpendicular B on to the 
 plane (Art. 29). Find the angle which A and B make with each 
 other (Art. 46). 8 is this angle, and its complement j3 ia the 
 angle sought. 
 
 50. Problem 11. — To find the line of intersection of tivo planes. 
 — Since the line of intersection lies in both planes, its horizontal 
 and vertical traces must lie in the horizontal and vertical traces 
 of both planes respectively, hence at their intersections. 
 
 Let P and Q, Fig. 18, be the given planes. V P and V Q inter- 
 sect at a", hence a" is the vertical trace of the line of intersection. 
 V' is the horizontal trace, since it is the intersection of H P and 
 H Q. The projections A'' and A'' of the line are found from its 
 traces by Art. 35.
 
 14" DESCRIPTIVE GEOMETRY. 
 
 The preceding explanation applies equally well to Fig. 19, 
 where the planes are assumed in a different position. 
 
 51. If the horizontal traces of the given planes are parallel, aa 
 in Fig. 20, a'' is the vertical trace ; but, since H P and H Q will 
 never intersect, there will be no horizontal trace, hence the line 
 must be a horizontal of each plane (Art. 44), and A"" will be 
 drawn through a" parallel to G L, and A^ parallel to H P and 
 HQ. 
 
 52. Let the plane P, Fig. 21, be intersected by the horizontal 
 plane Q. Since the line lies in Q and R, it must be a horizontal 
 of the plane P. This line is vertically projected in the vertical 
 trace of the plane Q at A% and horizontally projected in A'' par- 
 allel to H P. 
 
 63. Let the planes P and Q be situated so that the horizontal 
 traces do not intersect within the limits of the paper, as in 
 Fig. 22. 
 
 Pass a horizontal auxiliary plane X through the two planes ; 
 it will cut a horizontal line out of each plane (Art. 52), B out of 
 P, and C out of Q. Where these two lines intersect, at w, will be 
 one point of the line of intersection, a will be another ; hence A 
 is the line of intersection. 
 
 If the planes are so situated that neither the vertical nor the 
 horizontal traces intersect within the limits of the paper. Fig. 23, 
 pass two horizontal auxiliary planes, X and Y, through them. 
 The plane X gives the point n on the line of intersection, as 
 described above, and the plane Y the point m ; hence the line A, 
 drawn through these two points, is the line required. 
 
 Let the planes be so situated that they cross the ground line at 
 the same point, as in Figs. 24 and 25. We still make use of the 
 same principle of using horizontal auxiliary planes. By this means 
 we obtain the point n. The planes crossing the ground line at 
 the same point o, their line of intersection must pass through the 
 ground line at that point; hence o will be a point on both the 
 vertical and horizontal projections of the line. Therefore, A" and 
 A"" are the projections of the line of intersection. 
 
 54. Let both of the planes be parallel to the ground line, but 
 oblique to both V and H, as in Figs. 26 and 27. 
 
 In this case we have to make use of a profile plane (Art. 26). 
 This profile plane cuts out of each oblique plane a line which is
 
 DESCRIPTIVE GEOMETRY. 15 
 
 projected in the traces of the profile plane. The intersection of 
 these lines will be one point in the line of intersection sought. 
 In order to get this point, it is necessary to revolve the profile 
 plane X about its vertical trace into V. The points a" and J^ being 
 in the axis, do not move. Every other point in the plane, as c^ and 
 d'\ move in horizontal circles, and are found, after revolution, at 
 (7, and d, . The revolved position of the lines will, therefore, be 
 B and C. They intersect at e% In counter revolution this point 
 is found at e" e^. A'' and A\ drawn through g" and «'' respectively, 
 parallel to the ground line, are the projections of the line of inter- 
 section of the planes P and Q. Note that, in Fig. 27, the plane 
 P crosses the second angle ; also, that when the profile plane is 
 revolved into V, all that part which is behind V rotates in the 
 same direction as that which is in front ; hence if d^ rotates to c?,, 
 c^ must rotate to c,, and d, and c, are still on opposite sides of the 
 axis. In counter revolution e\ which is on the same side of the 
 axis as <?, , will revolve to e\ and the line of intersection is in 
 the second angle. 
 
 55. Let one of the planes pass through the ground line. 
 Both of its traces will be in the ground line, and it will be 
 
 necessary, in order to determine the plane, to give the angle it 
 makes with V or H. 
 
 Let P be given, as shown in Figs. 28 and 29, and Q passing 
 through the ground line and making angle 5 with H. Pass the pro- 
 file plane X through them. It will cut from these planes lines which 
 are shown in their revolved positions at B and C respectively, 
 intersecting at e% In counter revolution this point goes to e" e^. 
 Joining €" and e'' with o, we get A"" and A*" as the two projections 
 of the line of intersection. Note particularly the construction in 
 Fig. 29. 
 
 56. Problem 12. — To find tvhere a line pierces a plane. — Pass 
 any plane through the line, and find its intersection with the 
 given plane. Since this line and the given line are in the same 
 plane they will intersect, unless they are parallel, and the point 
 where they intersect must be in the given plane, since it is on a 
 line of that plane ; hence it is the point where the line pierces 
 the plane. 
 
 Let A, Fig. 30, be the given line, and P the given plane. We
 
 16 DESCRIPTIVE GEOMETRY. 
 
 will take the plane which projects A upon H as the auxiliary 
 plane. Its traces are V X and H X. a 6 is the line of intersec- 
 tion of the planes X and P. The line A intersects the line a 5 at 
 the point c, which is, therefore, the point where A pierces P. 
 
 Fig. 31 shows the construction, if the vertical projecting plane 
 were used as the auxiliary plane. 
 
 57. In Fig. 32 the general method is shown. Pass any plane 
 X through the line A (Art. 41). Its intersection with P is the 
 line d e. c, the intersection of the lines A and d e, is the point 
 required. 
 
 58. Where the plane is determined by two intersecting lines 
 the principle remains the same, but the details of the construction 
 are different. 
 
 Let A, Fig. 33, be the given line, and let the plane be deter- 
 mined by the two lines B and C intersecting in the point a. First 
 find the line of intersection of the plane projecting A upon H 
 with the plane (B, C). The line C pierces the plane X at a point 
 whose horizontal projection is c'' (where C^ intersects H X). The 
 vertical projection of this point must be in the vertical projection 
 of the line, and perpendicularly above or below it, hence at c''. In 
 the same way we find that B pierces X at the point b. Tliere- 
 fore, b" c", b'' c'' are the projections of this line of intersection. A 
 intersects this line 5 c at the point d, which is, therefore, the point 
 where it pierces the plane (B, C). 
 
 59. Problem 13. — To find the shortest distance from a point to 
 a plane. — The shortest distance must evidently be measured 
 along a line from the point perpendicular to the plane. 
 
 Let a, Fig. 34, be the given point, and P the given plane. 
 From the point a drop the perpendicular B upon P (Art. 29). 
 B pierces P at d (Art. 56^. B, , the true length of ad, is the dis- 
 tance required. 
 
 60. Since the definition of the projection of a point on a plane 
 is where a perpendicular through the point pierces the plane, the 
 point d is the projection of the point a upon the plane P. 
 
 Hence, to find the projection of a line upon an' oblique plane, 
 it is only necessary to find where a perpendicular from each end 
 of the line pierces the plane, and join the points so found.
 
 DESCRIPTIVE GEOMETRY. 17 
 
 Let A be the given line, Fig. 35, and P the given plane. The 
 l)rojection of a upon P is g, and of h upon P is o, hence B is the 
 projection of A upon P. 
 
 61. Problem 14. — To pass a plane through a given point par- 
 allel to a given plane. — Since the required plane is to be parallel to 
 the given plane, their traces must be parallel ; therefore, the hori- 
 zontal projection of a horizontal line of the required plane will be 
 parallel to the horizontal trace of the given plane. 
 
 Let a. Fig. 36, be the given point, and P the given plane. 
 Draw the horizontal line X of the required plane through a. X'' 
 and X'' are its projections, X'' being parallel to H P. If is the 
 vertical trace of this line. Hence V Q, drawn through 5" parallel 
 to V P, and H Q parallel to H P, are the traces of the required 
 plane. 
 
 62. Problem 15. — To pass a plane through a given point per- 
 pendicular to a given line. — Let a, Fig. 37, be the given point, 
 and A the given line. The traces of the plane must be perpen- 
 dicular to the projections of the line (Art. 29). Draw the hori- 
 zontal line X of the required plane through a. X'' and X*" are its 
 projections, X'' being perpendicular to A*", b" is the vertical trace 
 of this line. Hence V P, drawn through b" perpendicular to A'', 
 and H P, perpendicular to A*", are the traces of the required plane. 
 
 68. Problem 16. — To pass a plane through a given line parallel 
 to another given line. — Let A and B, Fig. 38, b,e the given lines. 
 Through any point d of the line A draw a line parallel to B. B/ 
 and B,*^ parallel respectively to B^ and B"^ (Art. 13) will be the 
 projections of such a line. The plane P, which contains these 
 lines, A and B, (Art. 39), will contain A and be parallel to B, 
 since it contains a line which is parallel to B. 
 
 64. Problem 17. — To pass a plane through a given line p>erpen- 
 dicular to a given plane. — Let A, Fig. 39, be the given line, and 
 P the given plane. Through any point e of the line A draw a 
 line perpendicular to P. B" and B"" will be its projections (Art. 
 29). The plane Q which contains these lines, A and B (Art. 39), 
 will contain A and be perpendicular to the plane P, since it con- 
 tains a line which is perpendicular to P.
 
 18 DESCRIPTIVE GEOMETRY. 
 
 65. Problem 18. To construct the projections of the shortest line 
 which can he drawn terminating in two right lines not in the same 
 plane. — Let A and B, Fig. 40, be the given lines. The required 
 line must be perpendicular to both given lines. Pass a plane 
 through A parallel to B (Art. 63). V P and H P will be its traces. 
 Find the projection of B upon this plane (Art. 60). It being paral- 
 lel to the plane, its projection on the plane will be parallel to itself. 
 Hence, it will only be necessary to find the projection of one 
 point, as m which is at r, and D is the projection of B upon P. 
 From s, the intersection of D and A, draw a perpendicular E to 
 the plane P. It will intersect B at the point ^, and be perpen- 
 dicular to both A and B. E, will be its true length (Art. 37). 
 
 66. The line of greatest declivity of a plane is that line which 
 makes the greatest angle with H. 
 
 This is the line of intersection a b, (Fig. 41), of the plane P 
 with the plane X, which is perpendicular to both P and H. 
 
 Let B be the angle a h makes with H. Then tan. /3= ^^ But 
 
 a^b^, perpendicular to H P, is the shortest line that can be drawn 
 from a'' to H P ; hence, ^ is the greatest angle that can be made 
 by any line of the plane P with H. 
 
 67. Problem 19. — To find the angle a given plane makes with 
 either plaiie of projection. — We will first find the angle made with 
 
 . H. This angle is the one made by its line of greatest declivity 
 with H. 
 
 Let P, Fig. 42, be the given plane. Revolve the plane X, which 
 cuts from P the line of greatest declivity (Art. Q%^^ about its ver- 
 tical trace into V. d^ being in the axis does not move ; h'' moves 
 in the arc of a circle with a'' as a centre, and is found at 5, and |3 
 is the required angle. 
 
 The plane X could, just as well, have been revolved about its 
 horizontal trace into H. In that case ^'', Fig. 42, would remain sta- 
 tionary, a'' would revolve to a in a line perpendicular to a'' 5*, and 
 8 would be the angle required. Of course, § and 8 should be equal. 
 
 If the auxiliary plane X were taken perpendicular to V and P, 
 it would cut from P the line making the greatest angle with V, 
 which, being revolved into one of the planes of projection, would 
 shoAV the angle the plane makes with V,
 
 DESCRIPTIVE GEOMETRY. 19 
 
 In Fig. 43, (3 is the angle P makes with V, and d the angle it 
 makes with H. 
 
 68. If one trace of a plane and the angle it makes with the 
 other coordinate plane were given, we would determine the other 
 trace by the reverse of the last problem. That is, if H F and S, 
 Fig. 42, were given, to find VP; draw any line, as a'' b'' perpen- 
 dicular to H P ; draw a 1/ making angle 8 with a'' 6\ and a a'' jier- 
 pendicular to a* 5* ; rotate this triangle about H X as an axis 
 until it is perpendicular to H ; a is found at a", and must be one 
 point of V P ; join a" and the point where H P intersects the 
 ground line, and VP is obtained. 
 
 69. Problem 20. — Q-iven the angle a plane makes loith each 
 coordinate plane to construct its traces. — A plane is tangent to a 
 sphere at a single point. If we pass an auxiliary plane through 
 this point, the centre of the sphere, and perpendicular to H, it 
 will cut a great circle from the sphere and a line from the tangent 
 plane, which will be tangent to the circle. But this line is the 
 line of greatest declivity of the tangent plane, since it is a line 
 ci(it from it by a plane perpendicular to H and to the tangent plane 
 as it contains a normal to the tangent plane. If this auxiliary 
 plane be revolved about its vertical trace to coincide with V, the 
 line and circle will be shown in their true size and position, and 
 also the true size of the angle the plane makes with H. 
 
 An auxiliary plane passed through the tangent point, the centre 
 of the sphere, and perpendicular to V, cuts from the sphere a great 
 circle and from the tangent plane a line tangent to the circle, and 
 which makes the greatest angle with V, and this being revolved 
 into H shows the size of the angle the plane makes with V, etc. 
 
 In Fig. 44, with o as a centre, and any radius describe a circle. 
 This will represent the revolved position of both sections of the 
 sphere. Draw a b tangent to this circle so that the angle ab o is 
 equal to /3, the angle the plane makes with H. This line is the 
 revolved position of the line of greatest declivity of the plane, and 
 a will be one point on the vertical trace of this plane. Draw d c 
 tangent to the circle making the angle c d o equal to d, the angle 
 which the plane makes with V. This line is the revolved position 
 of the line of the plane which makes the greatest angle with V, 
 and c will be one point on the horizontal trace of this plane.
 
 20 DESCRIPTIVE GEOMETRY. 
 
 In the counter revolution of a h, a remains stationary and b 
 moves in the arc of a circle of which o is the centre until ab is 
 perpendicular to the horizontal trace of the plane. Therefore, 
 with as a centre, and o 5 as a radius, describe a circle, and 
 through c tangent to this circle draw H P, which will be the hori- 
 zontal trace required. The line VP, joining a and the point where 
 HP crosses the ground line, is the vertical trace of the same 
 plane. VP could be constructed independently by drawing it 
 through a tangent to the circle drawn with o as a centre, and o d, 
 as a radius. 
 
 In every case the sum of the angles made by a plane with each 
 plane of projection must be equal to, or greater than, 90 degrees. 
 
 70. Problem 21. — To find the angle made by two oblique planes. 
 — From any point in space let fall a perpendicular to each plane. 
 The supplement of the angle made by these two lines will be the 
 angle required. 
 
 71. Problem 22. — To draw the projections of any solids of defi' 
 nite size, occupying a fixed position in space, resting unth its base on 
 an oblique plane which makes known angles with both coordinate 
 planes, and construct the projections of its shadow upon this oblique 
 plane. — This problem is merely an application of several of the 
 preceding problems. 
 
 In Fig. 45, let p' be the angle the oblique plane makes with H, 
 and 8 the angle it makes with V. Find V P and H P by Art. 69. 
 Let the solid be a rectangular prism, the lowest point of which is 
 at a distance equal to a'' r above H, and a'' r in front of V. The 
 intersections a'' and a'' of the two projections respectively of the 
 horizontal line X of the plane P, which is at a distance above H 
 equal to a" r, and the line R of the plane P, which is at a distance 
 in front of V equal to t/r, are the projections of this lowest point. 
 Having found the projections of one point of the base, it is neces- 
 sary to revolve the plane P into one of the coordinate planes in 
 order to show the base of the prism in its true size and position. 
 The point a revolves to a, (Art. 45). Make a, b, c, d, equal to 
 the true size of the base of the prism, and in the desired position. 
 In counter revolution these points will be found in horizontal 
 projection at a*, b'', c'', and d'' respectively. The vertical projections
 
 DESCRIPTIVE GEOMETKY. 21 
 
 of these points are found at rt", 5", c", and d" (Art. 42). This being 
 a right prism, the elements are perpendicular to the plane, hence 
 their projections are perpendicular respectively to H P and V P 
 (Art. 29). To find n^^ the horizontal projection of the top end of 
 the element an, revolve the plane projecting this element upon 
 H, with its line of intersection with P, into H. The line of inter- 
 section revolves to V't, the point a to s, the element an to .s-?«, , 
 perpendicular to V" t, sn, being equal to the real height of the 
 prism. In counter revolution n, goes to n''. Making the other ele- 
 ments of the same length as a''n'\ and joining the tops, the hori- 
 zontal projection of the prism is completed ; e", m", n", and o" are on 
 the vertical projections of the elements vertically above e'\ ??t\ ?i\ 
 and o''. 
 
 72. Since the definition of the shadow of a point upon a plane 
 is where a ray of light through that point pierces the plane, to 
 find the shadow of this prism upon P, it is only necessary to find 
 where the ray of light, as L, through the point e pierces P (Art. 
 56) at e,; and do this for the other points which cast shadows.
 
 CHAPTER III. 
 
 Principles akd Problems Relating to the Cylinder, 
 Cone, and Double Curved Surface of Revolution. 
 
 73. A cylinder may be generated by a straight line, called the 
 generatrix^ moving along a curved line, called the directrix., with 
 all its positions parallel. The different positions of the generatrix 
 are called elements. If the curve has a centre, a line drawn through 
 it, and parallel to the elements, is called the axis of the cylinder. 
 
 Any curved section of a cylinder made by a cutting plane, and 
 taken as the limit of the elements, may be called the base of the 
 cylinder. If the plane is perpendicular to the axis, the section is 
 a right section^ and, if this section be taken as the base, the cylin- 
 der is called a right cylinder. If the right section is a circle, the 
 cylinder is a cylinder of revolution. 
 
 A cylinder may be called circular or elliptic, according as its 
 right section is a circle or an ellipse. 
 
 74. The cone differs from the cylinder only in the fact that its 
 elements, instead of being parallel, pass through a common point 
 called the vertex. 
 
 The statements and definitions relating to the cylinder, in the 
 preceding article, are equally applicable to the cone. 
 
 75. The cylinder and cone must be represented in projection 
 by their lines of apparent contour, whatever other lines may be 
 drawn on their surfaces. 
 
 These surfaces are called single curved surfaces. 
 
 76. A double curved surface of revolution, such as a sphere, 
 ellipsoid, torus, etc., is generated by a plane curve revolving about 
 a right line contained in its plane. This line is called the axis of 
 the surface, and the generating curve is called the meridian line., 
 and its plane the meridian plane. When a meridian plane is 
 parallel to V, it is called the principal meridian plane.
 
 DESCRIPTIVE GEOMETRY. ' 23 
 
 Every point in the meridian line describes the arc of a circle, 
 the centre of which is in the axis of the surface. This circle is 
 called a parallel. Hence any plane perpendicular to the axis of 
 a surface of revolution will cut a parallel from the surface. 
 
 77. The simplest way of representing a double curved surface 
 of revolution is to assume the axis perpendicular to H, when the 
 greatest parallel will be its horizontal projection, and the princi- 
 pal meridian will be its vertical projection. 
 
 78. A plane which contains two lines that are tangent to a 
 surface at a common point will be tangent to the surface, and 
 will, moreover, contain every line which is tangent to the surface 
 at that point. 
 
 In the case of a cylinder and cone, the tangent plane must con- 
 tain an element of the surface. 
 
 79. If a single curved surface and its tangent plane be inter- 
 sected by any secant plane, the line cut from the tangent plane 
 will be tangent to the curve cut from the surface. Hence if this 
 secant plane happens to be the horizontal coordinate plane, the 
 horizontal trace of the tangent plane must be tangent to the 
 base of the surface at the point where the element of contact 
 pierces H. 
 
 80. A normal to a surface at any point is a right line perpen- 
 dicular to the tangent plane at that point. 
 
 A normal plane is a plane which contains the normal line, hence 
 it will be perpendicular to the tangent plane. 
 
 81. A plane which contains one line which is tangent to a sur- 
 face, and is perpendicular to the normal at that point, must be 
 tangent to the surface at that point. 
 
 82. Problem 23. — Given one projection of a point on the sur- 
 face of a cylinder^ to find its other projection, and, second, to pass a 
 plane tangent to the cylinder through this point. — Let the cylinder be 
 given as in Fig. 46, and let a'' be the horizontal projection of a 
 point on its surface. 
 
 Through a'' draw E*" parallel to A^. This will be the horizon- 
 tal projection of two elements, one- on the upper and one on the 
 lower side of the cylinder. The base of the cylinder, since it rests 
 on H, is the locus of the horizontal traces of all the elements ; 
 hence d'' and e'', where E'' intersects the circle of the base, are the
 
 24 DESCRIPTIVE GEOMETRY. 
 
 horizontal traces of the two elements, and d"' and e" are their 
 vertical projections. E'' and E/ drawn through <:7" and e" respec- 
 tively parallel to C% are the vertical projections of the two ele- 
 ments horizontally projected in E*". a'' and a,", where a line 
 through a^ perpendicular to GL intersects E^ and E/, are the 
 vertical projections of the two points horizontally projected in a!". 
 
 Second. To drmv a plane tangent to the cylinder at the point 
 a" a!". — HP, drawn tangent to the base of the cylinder at t?'', will 
 be the horizontal trace of the tangent plane (Art. 79). The line 
 X will be a horizontal of the plane P ; its vertical trace is m" ; 
 V P, drawn through nf and the point where H P intersects the 
 ground line, is the vertical trace of the tangent plane. The verti- 
 cal trace of the plane must, of course, pass through the vertical 
 trace of the element of contact, and the vertical trace of the tan- 
 gent plane could have been found in that way. In finding the 
 plane Q tangent to the cylinder at the point a," a!\ since H Q does 
 not intersect G L within the limits of the paper, it was necessary 
 to use both methods to find V Q. 
 
 83. Fig. 47 shows the construction when the axis of the cylin- 
 der is parallel to both V and H. 
 
 Let a!' be the horizontal projection of a point on its surface. It 
 is here necessary to make use of a profile plane. To avoid con- 
 fusing the figure, assume the profile plane X, containing one end 
 of the C3dinder, and revolve it away from the figure about either 
 trace into one of the coordinate planes ; in this case about V X 
 into V. It cuts out of the cylinder a right section — a circle in 
 this case — of which c is the centre ; c rotates to c,% and the cir- 
 cular section with it. The horizontal projection of the element 
 E'' througli tlie point a meets the profile plane at a point whose 
 horizontal projection is V\ In revolution V' moves to i,/'; erect- 
 ing a perpendicular at 5,/', it cuts the circle at ^>„'' and i,,/'. These 
 are the revolved positions of the points in which the two possible 
 elements through a* meet the plane X. In counter revolution 
 they are found at ?>'' and J/' respectively. Through these points 
 draw the elements E" and E/. These will be the vertical projec- 
 tions of two elements horizontally projected at E''; aP and a/, 
 where a line through a*, perpendicular to G L, intersects E'' and 
 E/, are the vertical projections of the two points horizontally pro- 
 jected at a''.
 
 DESCRIPTIVE GEOMETRY. 25 
 
 To draw a plane tangent to the eylinder at the point a" a^. — The 
 profile plane cuts out of the plane, tangent to the cylinder along 
 the element E'' E'', a line which is tangent to the curve cut from 
 the cylinder at the point If h'^ where the element E pierces X 
 (Art. 79). Hence T, tangent to the circle at 6„", is the revolved 
 position of this line of intersection. In counter revolution d" does 
 not move, and ej^ revolves to e'' ; d" and e^ are the traces of this 
 line which lies in the tangent plane ; and, since the tangent plane 
 contains the element E which is parallel to both V and H, V P 
 and H F, drawn through d" and e* parallel to G L, are the traces of 
 the tangent plane. 
 
 In the same manner the tangent plane through the point a/' a* 
 may be drawn. Note here that, in counter revolution, m!" re- 
 volves to w\ and V Q and H Q are both found above the ground 
 line, which indicates that the plane crosses from the first, through 
 the second, and into the third angle. 
 
 84. Problem 24. — Given 07ie projection of a point on the sur- 
 face of a cone to find its other projection; and.^ second^ to pass a 
 plane tangent to the cone through this point. — Let the cone be given 
 as in Fig. 48, and let a" be the vertical projection of a point on its 
 surface. B" will be the vertical projection of two elements pass- 
 ing through a; B"" and B,*" will be the horizontal projections of 
 these elements, and a* and a,* will the horizontal projections of 
 the two points vertically projected at a". 
 
 Second. To draio a plane tangent to the cone at the point cC a^. 
 — Here, as in the case of the cylinder, H P, drawn tangent to the 
 base of the cone at c\ will be the horizontal trace of the tangent 
 plane (Art. 79), and the trace of the horizontal Y determines one 
 point s'' in V P. V P could have been determined by finding the 
 vertical trace (/" of the element of contact B. 
 
 V Q and H Q, found in the same way, are the traces of a plane 
 tangent to the cone at the point «"«,*. 
 
 85. Fiff. 49 shows the construction when the axis of the cone 
 is parallel to both V and H. Let n" be the vertical projection of 
 a point on its surface. Pass the profile plane X through the base 
 of the cone, and revolve it about V X into V. G" is the vertical 
 projection of the elements through n. In revolution d"-' moves to 
 d^ and d,% and in counter rovolution they are found in plan at
 
 26 DESCRIPTIVE GEOMETRY. 
 
 d^ and cZ,^ ; G^ and G,^ drawn through df" and c?,* and the vertex 
 o\ are the horizontal projections of the two elements vertically 
 projected in G^ and n^ and n^ are the horizontal projections 
 required. 
 
 In finding the traces of the tangent plane through the point 
 71" w'', we find one point in each trace, — that is, c" and tw*, just the 
 same as in Fig. 47 ; another point in each trace is obtained, of 
 course, by finding the two traces of the element of contact G ; 
 joining the horizontal and vertical traces, respectively, we have 
 V P and H P. The plane Q is found in the same way. 
 
 If the cone had been so situated that the traces of the element 
 of contact did not fall within easy reach, we would have made 
 use of another profile plane, getting two more points, as we did 
 t'" and yn''. 
 
 86. Problem 25. — G-iven one projection of a point on the sur- 
 face of a double curved surface of revolution to find its other pro- 
 jection, and to p) ass a plane tangent to the surface through this point. — 
 Let the surface be a torus, represented as shown in Fig. 50 (Arts. 
 76 and 77), and let n^ be the horizontal projection of a point on 
 its surface. Pass the meridian plane Y through n ; H Y, drawn 
 through n^ and A\ is its horizontal trace ; revolve this plane into 
 the position of the principal meridian plane ; n^ revolves in the 
 arc of a circle to 6'\ which, being in the principal meridian, is ver- 
 tically projected at s"' and s," ; in counter revolution s" and s^ move 
 to 7f and 11^ vertically above w*. Hence w" and n^ are the verti- 
 cal projections of two points on the surface of the torus horizon- 
 tally projected at w\ 
 
 Second. To draw a plane tangent to the torus at the point n. — 
 This plane must contain a line tangent to the surface, and be per- 
 pendicular to a normal at that point (Art. 81). T/', drawn tan- 
 gent to the principal meridian at s", will be the revolved position 
 of the tangent line, and N/, drawn through s" and the centre of 
 the generating circle, the revolved position of the normal. In 
 counter revolution the points m and o do not move, being in the 
 axis ; N/ moves to W, and T/ to T"". Both the tangent and nor- 
 mal, since they are in the meridian plane, are horizontally pro- 
 jected in H Y. g'' is the horizontal trace of T. Since the tangent
 
 DESCRIPTIVE GEOMETRY. 27 
 
 plane is to contain T and be perpendicular to N, H P, drawn 
 through g^ perpendicular to N\ and V P, drawn through the point 
 where H P intersects G L, perpendicular to N'', are the traces of 
 the tangent plane. 
 
 A point r" in V P could have been found by means of the hori- 
 zontal X of the plane P, and still another by finding the vertical 
 trace /" of the tangent line.
 
 CHAPTER IV. 
 
 Intersectiok of Planes and Solids, and the Develop- 
 ment OF Solids. 
 
 87. To find the intersection of any solid with any secant plane, 
 pass a series of auxiliary planes through the solid and the secant 
 plane. They will cut lines (straight or curved) from the solid, 
 and straight lines from the secant plane, the intersections of which 
 are points of the required curve of intersection. 
 
 This is a general statement applicable alike to prisms, pyra- 
 mids, cylinders, cones, or double curved surfaces of revolution. 
 While the auxiliary planes may be taken in any position, yet, for 
 simplicity, they should be chosen in such a position as to cut the 
 simplest curve from the solid, — that is, straight lines or circles if 
 possible. 
 
 In case of any solid having rectilinear elements, as the cylin- 
 der, cone, prism, pyramid, etc., the above process is practically 
 the same as finding where a certain number of elements, or edges, 
 pierce the secant plane (Arts. 56, 57, and 58), since auxiliary 
 planes may be chosen so as to cut elements, or edges, from the 
 solid. 
 
 88. The true size of the section can always be found by revolv- 
 ing it about the trace of the secant plane into, or parallel to, one 
 of the coordinate planes (Arts. 45, 46, and 47). 
 
 89. The tangent line to the curve of intersection at any point 
 is the intersection of the plane tangent to the surface at that 
 point with the secant plane ; for the line, to be tangent to the 
 curve, must lie in the plane of the curve, — that is, the secant 
 plane, and also in the plane tangent to the surface at that point ; 
 hence at their intersection.
 
 DESCRIPTIVE GEOMETRY. 29 
 
 90. To develop a body is to find, on a flat surface, the true 
 size and shape of the surface, or covering, of the object. Single 
 curved surfaces and solids bounded by planes can be developed ; 
 double curved surfaces cannot be, except approximately. 
 
 Solids bounded by planes are developed by finding the true 
 size and shape of each successive face. 
 
 Single curved surfaces, as the cylinder, cone, etc., are developed 
 by placing one element in contact with a plane, and rolling the 
 solid until every element has touched this plane. That portion 
 of the plane covered by the solid in its revolution is the develop- 
 ment of the solid. When a curved surface is developed, any 
 curved line upon it will be developed into some curve, whose 
 length will be equal to that of the original curve. 
 
 CYLINDERS. 
 
 91. To find the intersection of any cylinder with an oblique 
 plane, assume the auxiliary planes so that they will cut elements 
 from the cylinder, and the general statement for the intersection 
 of any solid with a plane (Art. 87) becomes for the cylinder as 
 follows : Pass a series of auxiliary planes through the cylinder 
 parallel to its axis and perpendicular to one of the coordinate 
 planes. Each auxiliary plane cuts two elements from the cylin- 
 der (except the two tangent planes which cut but one each) and 
 a right line from the secant plane. The intersections of this line 
 and elements give two points of required curve. 
 
 92. Problem 26. — To find the intersection of a right cylinder 
 ivith a circular base with an oblique plane, to draw a tangent to this 
 curve at any point, to find the true size of the section, to develop the 
 cylinder, to trace upon the development the curve of iyiter section, and 
 to draw a tangent to the developed curve. — Let the cylinder be 
 located as shown in Fig. 51, and P be the secant plane. The auxil- 
 iary plane X is parallel to the axis and perpendicular to H 
 (Art. 91) ; it cuts the two elements C and D from the cylinder 
 and the line X from P ; X intersects C at a, and D at b, which 
 are two points of the curve. Other points could be determined 
 in the same way. 
 
 The rest of the points were found by assuming the auxiliary 
 planes as horizontal, which may be a trifle shorter, for this case,
 
 80 DESCRIPTIVE GEOMETRY. 
 
 than the first method. The horizontal auxiliary plane R cuts out 
 of the cylinder a horizontal circle, and out of the plane the hori- 
 zontal line R ; this line intersects the circle at the points c and d^ 
 which are, therefore, two points of the curve. In the same way 
 the plane S gives the points g and k, and the plane Z the points 
 on and ??, etc. 
 
 To di'atv a tangent to the point n in the curve. — The tangent line 
 to the curve at the point n is the intersection of the plane tangent 
 to the cylinder at that point and the plane P (Art. 89). V T 
 and H T are the traces of the plane tangent to the cylinder at the 
 point n (Art. 82). T'' and T^ the intersection of the planes P 
 and T, are the projections of the tangent line. Since the point n 
 must be one point in the tangent line, it is only necessary to find 
 one other point, as r, and the tangent line is determined, — that is, 
 it is not necessary to find the vertical trace of the tangent plane, 
 which shortens the construction. 
 
 To find the true size of the section. — Each point in the curve, 
 as a, c, g., m, b, etc., revolve to a,, c,, g,, m,, J,, etc. (Art. 45), and 
 the true size of the section is shown. 
 
 The tangent line T revolves with the section. The point r* 
 does not move, since it lies in the axis H P, and the point n 
 revolves to n, ; hence T revolves to T, . 
 
 To develop the cylinder. — If we place one element of the cylin- 
 der, as D, Fig. 52, on a plane, and roll the cjdinder until the same 
 element coincides with the plane again, that part of the plane 
 covered in this revolution will be the development (Art. 90). 
 The element will, of course, be seen in its true length, and a right 
 section, being perpendicular to the elements, unrolls in a straight 
 line perpendicular to the elements, and equal in length to the 
 circumference of the right section. Hence h!" /i,\ nj" k\ kj' d\ etc., 
 Fig. 52, equal to 5* >A w*/:;\ k^'d", etc.. Fig. 51, laid off on a 
 straight line will be the development of the base, and perpendicu- 
 lars erected at these points will be the position of the several ele- 
 ments; hi" s is made equal to the height of the cylinder, and st^ 
 drawn parallel to hj' bf\ completes the development of the whole 
 cylinder. 
 
 To trace upon the development the curve of intersection. — Make 
 6,* J,", w,* w,", k,'' k", etc., equal to the heights of the points 6, n, k, 
 etc., of the curve above H. The curve traced througli these sue-
 
 DESCRIPTIVE GEOMETRY. 31 
 
 cessive points will be the development of the curve of intersec- 
 tion. 
 
 To draw the tmigent line to the j^oint n on the development. — 
 The point w," will be one point of the tangent line sought. The 
 point r being in the plane of the base must be found in the line 
 on which the base unrolls, and at a distance from nj" equal to the 
 distance of the point r from the foot of the element along which 
 the plane is tangent, — that is, make nj" r^ equal to n^ r'' and join- 
 ing r/' and n'^ we have the developed position of the tangent line. 
 This line must be tangent to the developed position of the curve. 
 
 93. Problem 27. — To find the intersection of an oblique cylin- 
 der with any oblique plane, to draw a tangeiit to any point of the 
 curve, and to develop) that part of the cylinder betweerc the base and 
 the secant plane, — Let the cylinder be given as in Fig. 53, and 
 let P be the cutting plane. 
 
 Assume the auxiliary planes, R, S, U, etc., parallel to the axis 
 and perpendicular to H (Art. 91). The plane U cuts two ele- 
 ments, F and N, from the cylinder and the line U from the plane ; 
 X and w, the intersections of this line with these elements, are two 
 points of the curve. Similarly the plane R gives the point o ; S 
 the points n and p ; W gives I and r ; X gives k and s ; Y gives 
 g and t; and Z the point /, In the construction of the different 
 lines of intersection R, S, U, etc., it is only necessary to construct 
 one of the lines, the others are drawn through the vertical projec- 
 tions of the horizontal traces, parallel to the first, since the lines 
 of intersection of several parallel planes with an oblique plane 
 must be parallel to each other. 
 
 The tangent line is found as explained in the last problem. T 
 is the plane tangent to the cylinder along the element through t, 
 and T"" and T'\ the line of intersection of the planes T and P, are 
 the projections of the tangent line to the curve at the point t. 
 
 To develop that p)<^rt of the cylinder between the base and the 
 secant p)lane. — Since neither the plane of the base nor the plane 
 of the section just found is perpendicular to the axis, the sections 
 are not right sections, hence they will not unroll in straight 
 lines. Consequently the first step necessary in the solution of 
 this part of the problem is to find a right section. K"" K^, the 
 section cut by the plane Q, which is perpendicular to the axis, is
 
 32 DESCRIPTIVE GEOMETRY. 
 
 a right section ; K' is the true size of this section ; and K, , Fig. 
 54, is its development, its length being equal to the periphery of 
 the section K'. Find the true lengths of the successive elements 
 between this right section and the base, and lay them off perpen- 
 dicular to the line K, on lines which represent the developed posi- 
 tion of the several elements. The curved line A, A, drawn 
 through the points thus found is the development of the peri- 
 phery of the base. 
 
 To develop the section cut out by the plane P, find the true 
 lengths of the successive elements between this section and the 
 base, laying them off from the curved line A, A,, and the curved 
 line I, m, n, etc., is the development of this section. This curve 
 could also have been found by finding the true lengths of the ele- 
 ments between the right section and the section cut by P, and 
 laying these lengths off from the development K, of the right 
 section. Hence the surface included between the curved lines 
 A, A, and l,m,n, etc., is the development of that portion of the 
 cylinder between the plane P and the base. 
 
 To draiv the tangent line to the point t on the development. — The 
 point e is at a certain known distance from the point f, and also a 
 certain other known distance from the point iv ; hence with w, , 
 Fig. 54, as a centre and a radius equal to w^ e^ Fig. 53, describe 
 an arc ; with t^. Fig. 54, as a centre and a radius equal to the true 
 length of the line t g. Fig. 53, describe another arc ; the point e, , 
 where these two arcs intersect, will be one point of the developed 
 position of the line. t, must be another, therefore T, , drawn 
 through t, and e,, is the tangent sought. 
 
 CONES. 
 
 94. The general statement for the intersection of any cone 
 witli any plane is as follows : Pass a series of auxiliary planes 
 through the vertex of the cone, and perpendicular to one of the 
 coordinate planes. These planes cut elements from the cone and 
 lines from the secant plane, the intersections of which give points 
 on the curve. 
 
 95. Problem 28. — To find the intersection of a right cone ivith 
 a circular hhse ivith an oblique plane, etc., as in Problem 26. — Let 
 the cone and cutting plane P be given as in Fig. 55. The auxil-
 
 DESCmrTIVE GEOMETRY. 33 
 
 iary plane R passes through the vertex and is perpendicular to H 
 (Art. 94) ; it cuts the two elements A and B from the cone and 
 the line R from P ; R intersects A at c, and B at g^ which are two 
 points of the curve. These two points are the highest and lowest 
 points of the curve, since the line R is the line of greatest decliv- 
 ity of the plane P. Other points could be determined in the 
 same way. 
 
 The rest of the points were found by assuming horizontal auxil- 
 iary planes X, Y, and Z. The plane X cuts out of the cone a 
 horizontal circle, and out of the plane the horizontal line X ; this 
 line intersects the circle at the points d and p, which are, there- 
 fore, two points of the curve. In the same way the plane Y gives 
 the points e and w, and the plane Z the points / and m. 
 
 The tangent line T, and the true size of the section, are found 
 in the same way as described for the cylinder in Problem 26. 
 
 To develop the cone and trace on the development the curve of 
 intersection. — In a right cone all the elements pass through the 
 vertex, and are of the same length, hence the vertex remains 
 stationary and the base unrolls in the arc of a circle, of which the 
 vertex is the centre, and the slant height of the cone (that is, the 
 true length of the element) is the radius. The length of this arc 
 must be equal to the circumference of the base. TherefoKe, with 
 o\ Fig. bQ., as a centre, and a radius equal to the true length of an 
 element, describe an arc D' equal in length to the circumference 
 of the base, and o' t' T>' t' is the development of the surface of the 
 cone. Particular elements may be drawn on the development by 
 laying off distances on the arc D' equal to the distance apart of 
 the elements at the base of the cone, and joining these points with 
 the vertex o'. To find the development of the curve of intersec- 
 tion, lay off from o\ Fig. 56, on its corresponding element the true 
 length of the element from the vertex to the section, and joining 
 the points thus found, g\ /', e', etc., gives the development 
 sought. 
 
 To draw the tangent to the development at the 'point ^:>. — The 
 line C, tangent to the base of the cone, is perpendicular to the 
 element s o, hence it will be so after development. Therefore 
 draw C, Fig. 56, perpendicular to s' o', and lay off on it s' r' equal 
 to ** r\ Fig. 55; r is also a point of the tangent line T, hence r' 
 must be one point of its developed position, and p' must be
 
 34 DESCRIPTIVE GEOMETRY. 
 
 another point, consequently T' drawn through r' and f' is the 
 tangent required. 
 
 96. Fig. 57 shows the construction when the axis of the cone 
 is parallel to both V and H. The method of finding the point o," 
 is explained in the next article. Fig. 58 shows the development 
 of the cone and the tangent line. 
 
 97. Problem 29. — To find the intersection of an oblique cone 
 with any oblique plane^ to draw a tangent to any point of the curve^ 
 and to develop that ptortion of the cone between the base and the 
 secant plane. — Let the cone be given as in Fig. 59, and let P be 
 the cutting plane. 
 
 Assume the auxiliary planes R, S, X, etc., through the vertex 
 and perpendicular to H (Art. 94). The plane S cuts two ele- 
 ments, A and B, from the cone, and the line S from the plane ; 
 g and r, the intersections of this line with these elements, are 
 two points of the curve. Similarly the plane R gives the point 
 /; X the points k and p ; Y the points I and n ; and Z the 
 point m. 
 
 In the construction of the different lines of intersection R., S, 
 X, etc., it is not necessary to go through the whole process of 
 finding each of them. In the case of the cylinder, these lines of 
 intersection were parallel, here they intersect in a common point 
 0,. The auxiliary planes, being all perpendicular to H, and pass- 
 ing through the vertex of the cone, will intersect each other in a 
 common line, which must be perpendicular to H, and contain the 
 vertex; D'' will be the vertical projection of this line, and o* its 
 horizontal projection ; hence the point o, must be on this line, 
 which is common to all the auxiliary planes, and on the plane P, 
 therefore at the point where they intersect. o^ is the hoiizontal 
 projection of this point, and o/, constructed as explained in Prob- 
 lem 5, is its vertical projection. This point could be found by 
 extending one line of intersection, found regularly, until it inter- 
 sects the vertical line joining the projections of the vertex. 
 
 The tangent line T at the point ?, the true size of the section, 
 etc., are found as explained in a previous problem. 
 
 To develop that part of the cone which is between the base and 
 the secant plane. — Since the vertex remains stationary we have a 
 means of locating tlie developed positions of the elements with
 
 DESCRIPTIVE GEOMETRY. 35 
 
 respect to each other without getting a right section as in the 
 cylinder. 
 
 Draw any right line o' t\ Fig. 60, and on it lay off the true 
 length of any one of the elements, as E, Fig. 59. Find the true 
 length of the next element A, with this length as a radius and o' 
 as a centre, describe an indefinite arc. The distance these two 
 elements are apart at their lower extremities must be equal to the 
 length of that part of the circumference of the base included 
 between ^- and d/". Hence with t' as a centre, and a radius equal 
 to i''c?^ describe an arc cutting the other arc at d' ; join d' with 
 o', and we have the developed position of the element A. The 
 other elements are developed in the same way. The greater the 
 number of elements developed, the greater will be the accuracy 
 of the development. 
 
 To develop the section cut out by the plane P, find the true 
 lengths of each element from the vertex to the section o/, o g, 
 etc., and lay them off from o' on the developed positions of those 
 elements respectivel}^ o'f\ o' g\ etc. That portion of the surface 
 included between the development of the base t' c' x' d' t\ and that 
 of the section /''^' ^' etc., will be the development of that part of 
 the cone between the base and the secant plane P. 
 
 The tangent line to the point I is developed as explained in 
 Problem 27, the point e being in ^* and T"", and at a known dis- 
 tance from the points x and I. 
 
 DOUBLE CURVED SURFACES OF REVOLUTION. 
 
 98. The general statement for the intersection of a double 
 curved surface of revolution with any plane is as follows : Pass a 
 series of auxiliary planes through the solid perpendicular to its 
 axis. These planes cut circles from the solid and lines from the 
 secant plane, the intersections of which give points on the 
 curve. 
 
 99. Problem 30. — To find the intersection of a double curved 
 surface of revolution with an oblique plane, and to draw a tangent 
 to the curve at any point. — Let the double curved surface of revo- 
 lution be a torus given as in Fig. 61, and let P be the cutting 
 plane. Assume the auxiliary planes R, S, X, etc., perpendicular 
 to the axis. The plane Y cuts from the torus two circles, hori-
 
 86 DESCRIPTIVE GEOMETRY. 
 
 zontallj projected in A'' and B\ and vertically projected in V Y, 
 it cuts from the plane P the horizontal line Y ; cZ, /, ?, and w, the 
 intersections of these circles and line, are four points on the 
 curve. Similarly the plane R gives the point a; the plane X 
 gives c, g, k, and o \ the plane S gives h and 'p ; and the plane Z 
 gives e and m. 
 
 The tangent line T to any point as w, and the true size of the 
 section, etc., are shown in the iigure, and need no further ex- 
 planation. 
 
 SOLIDS BOUNDED BY PLAIN SURFACES. 
 
 100. In finding the intersection of prisms, pyramids, or any 
 solid bounded by plane surfaces with an oblique plane, pass auxil- 
 iary planes through the edges of the solid, and perpendicular to 
 V or H ; or, in other words, find where each edge of the solid 
 pierces the secant plane by Art. 56. 
 
 101. Problem 31. — To find the intersection of a prism ivith an 
 oblique plane. — Let the prism be pentagonal, and given as in 
 Fig. 62, and let P be the cutting plane. The edges A, B, G, 
 etc., pierce the plane P in the points a, h, c, etc. ; joining these 
 points we get the two projections of the intersection. Its true 
 size is shown at a, 5, c, d, e, . 
 
 The development may be found directly here, without getting a 
 right section as in the cylinder, by revolving each face about its 
 edge resting on H into H (Art. 47). a 6^/ revolves to f" g'' on-, 
 h ck g to g^- k^ r p., etc. 
 
 102. We have seen how, if we have a solid given by its pro- 
 jections, we can obtain its covering or development; hence, if we 
 have the covering given, we are able to construct the projections 
 of the solid. 
 
 103. Problem 82. — To construct the projections of a solid from 
 its covering^ and find the intersection of this solid with an oblique 
 plane. — Let the covering consist of eight regular hexagons and 
 six squares, disposed as shown in Fig. 63. Let one of the hexa- 
 gons be considered as the base, and let it be placed parallel to H, so 
 that one of its sides, a 6, is perpendicular to V. Revolve the hexa- 
 gon A and the square K until the sides a''g' and a'' k' coincide, the 
 extremities g' and k' of tiiese sides will meet at a point in space,
 
 DESCRIPTIVE GEOMETRY. 87 
 
 of winch the horizontal projection is evidently a^ ; revolve 
 square M until its side /'T coincides with the side o''/'' of the 
 revolved hexagon A, the extremities meet at a point in space of 
 which the horizontal projection is o h. It is evident that the 
 horizontal projections of the two squares K and M, in their 
 revolved positions, are a'' V' s^ x^ and e;*/^ o'^p''. To complete the 
 horizontal projection of the hexagon A, we draw through jt^ the 
 line a^ w'' parallel and equal to the line/*o''; through w'' the line 
 w*w* equal and parallel to aJ'f^; and finally join n'' and o^, and 
 w^o* will be equal and parallel to a^' ji^ (Art. 13). The vertical 
 projection of the base is the line a"/" e" at any required height 
 above H. The vertical projection of the square K, in its revolved 
 position, is the line a" a;", found as follows: Since the plane of the 
 square is perpendicular to V, the point k' moves in revolution in 
 the arc of a vertical circle, with the point a as a centre ; hence, 
 when the point has gone to a/*, its vertical projection must be at 
 a;", where a perpendicular through a^ intersects this circle. Since 
 the squares K and M are inclined at equal angles to H, their top 
 points must be the same height above H in their revolved posi- 
 tions ; hence o" and p" must be found in a horizontal line through 
 x'\ and vertically above o'' and jt?^. Hence the vertical projection 
 of the square M is e^'f d" p". The vertical projection of the 
 hexagon A is a"/" o" w" w" a;^ the points a^/^ o", and x" being co- 
 incident with those of the squares already found. The point w" 
 is obtained by dropping a perpendicular from w\ and producing 
 it until it meets a line drawn through a;" parallel to f d" ; or, 
 in other words, make a;"?^" equal and parallel to/^o"; make g" n" 
 equal and parallel to a'" a;"; and n" w" eqvial and parallel to a''/". 
 
 The other hexagons and squares are revolved after the same 
 principle, and the completed projections are obtained as shown in 
 the figure. 
 
 To find its intersection with the plane P. — Find where the 
 edges of the solid pierce the plane by Art. 56 ; some of them will 
 and some will not; hence assume any one that you think will 
 pierce it, and see if it does by actual construction. If it does not, 
 try another, and so on. The edge ep pierces the plane P at the 
 point 6 ; /o at 6 ; wn at 4, etc. ; giving 12345678 as the sec- 
 tion. Its true size is found as usual. 
 
 The lines 1, 2,, 2, 3,, 3, 4,, 4, 5,, etc., are the developed posi- 
 tions of the lines cut from the several faces by the secant plane.
 
 CHAPTER V. 
 
 llSTTERSECTIOIf OF SOLIDS. 
 
 104. To find the curve of intersection of any two solids what- 
 ever, pass a series of auxiliary surfaces (usually planes) through 
 the two solids. They will cut lines (straight or curved) from 
 each solid, the intersections of Which are points of the required 
 curve. 
 
 While the auxiliary planes may be taken in any position, j'et, 
 for convenience, they should be chosen so as to cut the simplest 
 curves from the solid, — i.e., straight lines or circles. 
 
 105. The following is a list of special cases showing what 
 auxiliary planes should be taken to cut the simplest curves : — 
 
 First. Two cylinders with a?tes oblique to each other ; choose 
 auxiliary planes parallel to both axes. Each plane cuts elements 
 from each solid. 
 
 Second. Two cones with axes perpendicular or oblique to each 
 other; choose auxiliary planes passing through a line containing 
 the vertex of each cone. Each plane cuts elements from each 
 solid. 
 
 Third. Cylinder and cone with axes oblique to each other; 
 choose auxiliary planes passing through vertex of cone and par- 
 allel to axis of cylinder. Each plane cuts elements from each 
 solid. 
 
 Fourth. In case of the intersection of cylinders, cones, double 
 curved surfaces of revolution, prisms, and pyramids (any one 
 with one of the same or different kind) with their axes parallel ; 
 clioose auxiliary planes perpendicular to the axes. The planes 
 will cut from these solids simple figures (from the cylinder, cone, 
 and double curved surface of revolution, circles, and from the
 
 DESCRIPTIVE GEOMETRY. 39 
 
 prisms and pyramids, polygons similar to their bases). One of 
 the coordinate planes should be taken perpendicular to the axes. 
 
 Fifth. In case of two cylinders, cylinder and double curved 
 surface of revolution, cylinder and cone, whose axes are perpen- 
 dicular to each other ; choose auxiliary planes perpendicular to 
 one cylinder in first case, to axis of double curved surface of revo- 
 lution in second case, to axis of cone in third case. Coordinate 
 planes should be taken perpendicular to the axes respectively. 
 
 Sixth. In case of two double curved surfaces of revolution 
 with axes oblique and intersecting, pass a series of auxiliar}^ 
 spheres, whose centres are at the intersection of the axes, through 
 the two solids. 
 
 A prism can be substituted in place of the cylinder, and pyra- 
 mid in place of the cone, in any of the above cases. 
 
 106. The tangent line to the curve of intersection at any point 
 is the line of intersection of the planes tangent to each solid at 
 that point. 
 
 197. Problem 33. — To find the curve of intersection of tivo 
 cylinders whose axes are oblique to each other, and to draw a line 
 tangent to any point in the curve. — Let the cylinders be given as in 
 Fig. 64. This comes under the first special case (Art. 105), and 
 the auxiliary planes are taken parallel to the axis of each cylin- 
 der. To find the horizontal trace of such a plane (the vertical 
 trace is not needed) we have simply to pass a plane through one 
 axis and parallel to the other (Art. 63). H S is the horizontal 
 trace of such a plane. H R, H X, etc., drawn parallel to H S will 
 be the horizontal traces of planes which are parallel to both axes. 
 Each of these planes cuts elements from each cylinder, the points 
 of intersection of which are points of the required curve. 
 
 The plane S cuts the two elements A and B from the left cylin- 
 der, and C and D from the right cylinder ; these elements inter- 
 sect at the points 2, 8, 10, and 16, which are four of the required 
 points. The plane R being tangent to one of the cylinders only 
 gives two points, 1 and 9. The plane X gives the four points, 
 3, 7, 11, and 15. By repeating the process any number of points 
 can be found. 
 
 After finding a number of points on the curve, crossing and 
 recrossing as it is liable to do, unless some system of numbering
 
 40 DESCRIPTIVE GEOMETRY. 
 
 is followed, it is very difficult to connect the points so as to form 
 the correct curve. An examination of the system used in figures 
 64, 65, and 68 will suffice without explanation ; the same method 
 is used whether the solids are two cylinders, two cones, or cylin- 
 der and cone. 
 
 To determine whether there tvill he one or two curves. — Draw the 
 two extreme planes II and Z, each of them will be tangent to one 
 (but not the same) cylinder, and secant to the other, hence there 
 will be but one curve. 
 
 If the two extreme planes were tangent to the same body, and 
 secant to the other, it is evident that the first body would pass 
 through the second, and there would be two separate curves. 
 
 To determine what points on the curve are visible. — A point on 
 the curve is visible in plan or elevation, if it is the intersection 
 of two elements, both of which are visible in plan or elevation. 
 Thus the point 9, being at the intersection of two elements which 
 are visible in plan, must be on a visible portion of the curve in 
 plan ; these elements being invisible in elevation, the curve at 
 that point will be invisible in elevation. The two elements whose 
 intersection gives the point 5 are both visible in elevation, but 
 only one is visible in plan, therefore the point is on a portion of 
 the curve which is visible in elevation, but not in plan. 
 
 To draio a tangent line to any point of the curve. — The tangent 
 line to any point of the curve is the line of intersection of two 
 planes tangent to the solids at that point. H T is the horizontal 
 trace of the plane tangent to the left cylinder at the point 14, and 
 H Q that of the plane tangent to the right cylinder at the same 
 point ; T, the line of intersection of these two planes, is the tan- 
 gent line required. 
 
 108. Problem 34. — To find the curve of intersection., etc., of 
 tivo cones with axes oblique to each other. — Let the cones be given 
 as in Fig. 65. This comes under the second special case (Art. 
 105), and the auxiliary planes are passed through a line contain- 
 ing the vertex of each cone. 
 
 The line A passes through each vertex, and a^ is its horizontal 
 trace ; hence any line, as H X, H Y, etc., drawn through a\ will 
 be the horizontal trace of a plane which will pass through the 
 vertex of each cone, and, consequently, will cut elements from
 
 DESCRIPTIVE GEOMETRY. 41 
 
 them. H Y cuts the elements B and C from the left cone, and 
 D and E from the right cone ; these elements intersect at the 
 points 2, 4, 6, and 8, which will be points on the required curve. 
 Any number of points may be found in the same way. 
 
 Here, as in the case of the cylinder in the last problem, the two 
 extreme auxiliary planes X and Z are each tangent to one (but 
 not the same) body and secant to the other ; hence there will be 
 but one curve. 
 
 In this case one of the cones is supposed to have been removed, 
 consequently those portions of the curve will be visible if they 
 are on visible portions of the remaining cone. 
 
 T is the tangent to the point w, and is found exactly as in the 
 last problem. 
 
 109. Problem 35. — To find the curve of intersection^ etc., of 
 two cones when their axes are perpendicular to each other. — Let 
 the cones be given as in Fig. 66, the axis of one being perpendicular 
 to H, and of the other parallel to both V and H. In this case 
 we pass the auxiliary planes through a line containing each ver- 
 tex, as in the last problem, but here it is necessary to use a pro- 
 file plane and make use of the traces of the auxiliary planes on. 
 this plane, as well as their traces on H. 
 
 The horizontal trace of the line A, containing the two vertices 
 of the cones, is a*; its trace on the profile plane is b. H S, H X, 
 etc., are the horizontal traces of planes which will cut ele- 
 ments from the two cones; 6^Z*, ^''c^, etc., are their traces on 
 the profile plane. Revolving the profile plane about its vertical 
 trace into V, the point b revolves to J,", and the traces of the 
 auxiliary planes on the profile plane are shown at V S', V X', etc. 
 The plane S cuts the element D from the cone whose axis is hori- 
 zontal, and B and C from the upright cone ; the intersections of 
 these elements d and e are two points of the curve. The plane Z 
 gives the points m, w, r, and t. Any number of points can be 
 found in the same way. 
 
 Since the axes of the two cones intersect, the curve will be the 
 same on each side of the central plane Z ; hence, since the plane 
 S is tangent to one cone and secant to the other, there will be two 
 separate curves. 
 
 T is the tangent line to a point on the curve, being the inter-
 
 42 DESCRIPTIVE GEOMETRY. 
 
 section of the plane Q, which is tangent to the cone whose axis is 
 horizontal, and the plane T, tangent to the other cone at the com- 
 mon point. 
 
 Fig. 67 shows the development of the upright cone, together 
 with the holes in which the horizontal cone penetrated it, and the 
 tangent line. 
 
 110. Problem 36. — To find the curve of intersection, etc., of a 
 cylinder and co7ie with axes oblique. — Let the bodies be given as 
 in Fig. 68. This comes under the third case, and the auxiliary- 
 planes are passed through the vertex of the cone parallel to the 
 axis of the cylinder. To do this, draw the line A through the 
 vertex of the cone, and parallel to the axis of the cylinder ; a'' is 
 its horizontal trace ; any line, as H S, H X, etc., drawn through 
 this point will be the horizontal trace of such a plane. 
 
 The plane S cuts the element B from the cone, and the two ele- 
 ments, C and D, from the cylinder ; the intersections of these 
 elements at 1 and 7 are tw^o points of the required curve. Any 
 number of other points can be found in the same way. 
 
 111. Problem 37. — To find the curve of intersection of a 
 sphere and hexagonal prism. — Let them be given as in Fig. 69. 
 This is an illustration of the fourth special case, of which there 
 can be a large number. 
 
 The horizontal plane X cuts the circle X from the sphere, and 
 a hexagon from the prism ; these intersect at the point n. Other 
 points are found in the same manner. 
 
 112. Problem 38. — To find the curve of intersection of a cylin- 
 der and torus with axes parallel. — Let them be given as in Fig. 
 70. This is an illustration of the fifth special case. Here planes 
 should be taken perpendicular to the axis of the torus. 
 
 The plane Y cuts from the torus the two circles A and B, and from 
 the cylinder the two elements C and D. These circles and elements 
 intersect in the points a, a,, a„, a,„, J, 5,, all of which are points ou 
 the required curve. Other points are found in tlie same manner. 
 
 113. In cases where the bodies are bounded by planes, and do 
 not rest with their bases on either of the coordinate planes, it is 
 easier to make use of the method described in Art. 58 than to find 
 the traces of the planes containing the separate faces of the solid.
 
 DESCRIPTIVE geo:metry. 48 
 
 114. Problem 39. — To find the intersection of a triangular 
 prism with a triangular pi/ramid.-"— The points, e,f, and^, Fig. 71, 
 where the elements A, B, and C of the prism penetrate the face abc 
 of the pyramid, are found by Art. 58 ; joining these points, we have 
 efg as the intersection of the prism with the face abc of the 
 pyramid. The three elements of the prism do not leave the pyra- 
 mid in the same face ; hence find the points w aud n where the 
 elements A and C leave the face ac d; then find the point r where 
 the element B would leave the same face if that face were con- 
 sidered of indefinite extent ; joining m, w, and r, we have the inter- 
 section of the prism with the face acd when extended sufficiently. 
 That portion of the triangle mnts which actually falls on the lim- 
 ited face acd, will be a part of the intersection required; the ele- 
 ment B actually leaves the pyramid at the point o in the face bed; 
 joining this point with s and t, the intersection is completed. 
 
 Fig. 72 shows the development of the surface of the pyramid 
 with the intersections traced on it. 
 
 115. Problem 40. — To construct the projections of a right hex- 
 ugonal prism and a right hexagonal pyramid^ and find their inter- 
 section tvhen they occupy a certain fixed piosition relative to each other 
 and to the coordinate planes. — Let the lowest corner of the prism 
 rest on H ; the long diameter of the base of prism is 2^ inches ; 
 its height is 3 inches ; two of its faces are in a plane perpendicu- 
 lar to H ; its axis makes an angle of 60° with H and 25° with V. 
 The long diameter of the base of the pyramid is 2 inches ; its height 
 is 3 inches ; its axis slopes downward, forward, and to the left, mak- 
 ing angle of 45° with H, and whose horizontal projection makes an- 
 gle of 15° with G L. A point in the axis of the prism 2 inches from 
 the base is ^ inch, perpendicular distance, behind a point in the 
 axis of the pyramid which is 1 inch from the base of the pyramid. 
 
 See Fig. 73 for construction. If the axis of prism makes an angle 
 of 60° with H and 25° with V, the plane of the base makes an angle 
 of 30° with H and 65° with V. V F and H P, the traces of this 
 plane, are found by Art. 69. The projections of the prism are 
 then found by Art. 71. 
 
 Next it is necessary to locate the axis of the pyramid. To do 
 this we use the converse of the principles in Art. 65, which 
 explains how to find the perpendicular distance between two lines 
 when their projections are given ; here the projections of one line,
 
 44 DESCRIPTIVE GEOMETRY. 
 
 the direction of the projections of the other, and the perpendicu- 
 lar distance between them are given, to find the projections of the 
 second, d, the point in the axis of the prism, 2 inches from the 
 base, is found by laying off on the revolved position of the axis 
 rt,* h!" the distance a,* dl' equal to 2 inches, and in counter revohi- 
 tion dj' goes to d''. Now, through this point d, draw the line A 
 parallel to the axis of the p3'ramid by the converse of the princi- 
 ples in Art. 86. A/ is drawn through d\ making angle of 45" 
 with G L. A,'' must, of course, be parallel to G L. From d" lay 
 off on A/ a distance d'' e," equal to 2 inches, and f?" n^ equal to 1 
 inch ; in counter revolution A,"* moves through angle of 15" to 
 A\ A/ moves to A^ ef to e", ej" to g*, w," to n", and w e is a 
 line through d equal in length and parallel to the axis of the 
 pyramid. Through the line A and the axis of the prism pass a 
 plane. V Q and H Q are the traces of this plane. From the 
 point d draw the perpendicular c? o of indefinite length ; revolve 
 this indefinite line until it is parallel to V, and lay off d'' o," equal 
 to i inch on the revolved position ; in counter revolution o," moves 
 to o", and o* is found vertically below it on the horizontal projec- 
 tion of the indefinite perpendicular. Through the point o draw 
 the \mefg equal and parallel to ew, and this line will be the axis 
 of the pyramid correctly located relative to the axis of the prism. 
 
 Through the point g pass the plane R perpendicular to the axis 
 fg of the pyramid. This will contain the base of the pyramid, 
 and the projections of the pyramid can then be found the same as 
 the prism by Art. 71. 
 
 The intersection of the two bodies is obtained by finding where 
 each edge of one penetrates the faces of the other, if at all, by 
 Art. 58, the same as in Art. 114. 
 
 116. Problem 41. — To find the curve of intersection of two 
 double curved surfaces of revolution with their axes intersecting. — 
 Let the two surfaces be an ellipsoid and a sphere, given as in 
 Fig. 74, their axes intersecting at o. 
 
 Intersect these two surfaces by a series of auxiliary spheres 
 whose centres are at o; each sphere will cut circles out of the 
 surfaces, the intersections of which will be points on the required 
 curve. The sphere A cuts out of the ellipsoid the circle C, and 
 out of the sphere the circle D, their intersections e and e, are two 
 points on the curve. Other points can be found in the same way.
 
 CHAPTER VI. 
 
 Miscellaneous Problems. 
 
 117. Problem 42. — To find the right section of a cylinder ^ 
 making a certain known angle with the horizon, which will exactly 
 mitre ivlth a horizontal cylinder ivhose axis is at right ayigles to it. 
 — Let the horizontal cylinder be circular, and with its axis per- 
 pendicular to V, Fig. 75 ; the other cylinder makes an angle ^ 
 with H, and is parallel to V. The section of each cylinder on the 
 mitre plane Q is the same. 
 
 The plane P will cut a right section from the slanting cylinder, 
 whose vertical projection will be in the line V P, and whose hori- 
 zontal projection can be found as follows j The plane R is tangent 
 to both cylinders along the elements A and B, which must inter- 
 sect at the point a in the plane Q. In this Way A^ is determined, 
 and consequently 6\ a point on the horizontal projection of the 
 curve. In the same Way the plane T gives the two points d and 
 d,. Other points are found in the same way. 
 
 The true size of the section dh d, is easily found. 
 
 118. Problem 43. — »I^o find the right section of the raking 
 moidding ivhich ivill mitre with a gutter, the pitch of the roof and 
 the section of the gutter heing gi})en.-^'Y.\\\% is an application of the 
 principles of the preceding article. 
 
 Let the section of the gutter be given as in Fig. 76, and the 
 pitch of the roof equal the angle ^. 
 
 The points and planes have been lettered the same as in Fig. 75, 
 so that the explanation for that figure is equally applicable for 
 this, except that in this figure the true size of the section is shoWn 
 as revolved into a plane parallel to V instead of into H. It is 
 not necessary that the back part of the moulding be made to fol- 
 low the inside of the gutter, as it Is invisible ; hence it is made 
 out of an ordinary plank of any convenient thickness, and the 
 curve exdh {■& aW that it is necessary to find. 
 
 It is well to note that, when the plane Q makes an angle of 45°
 
 46 I)ESCRrE»TtVE GEOMETRY. 
 
 with V, as is usual in such work, the distances d'' k, x^ c, etc., are 
 equal to o r, s f, etc., respectively ; hence the true size of the sec- 
 tion can be obtained directly from the elevation without the use 
 of the plan, as follows: Draw any number of lines, as V R, V S, 
 etc., making angle ^ with the horizon ; draw any line, as V P, at 
 right angles to these lines ; from the points of intersection c?", re", 
 etc., of the line V P with V T, V X, etc., lay off along these lines 
 d" d^, x"-' x% etc., equal to or.st^ etc., the distance from the points 
 where V T, V X, etc., intersect the curve of the gutter to any ver- 
 tical line as C. A curve drawn through the points d'\ x% etc., 
 will be the curve of the moulding. 
 
 119. Problem W.~^Griven the right section of the doping part 
 qf the side wall to a flight qf steps, to find the right section of the 
 lower end of this wall at right angles to the given section. — Let 
 e d^ x^ 6, Fig. 77, be half the given section, and ^ the angle of the' 
 slope of the wall. 
 
 This problem is simply the converse of the preceding one, and 
 needs no further explanation. 
 
 The side walls of the Institute steps illustrate this problem. 
 
 120. Problem 45. — To find the covering of an eight-sided dome 
 as given in Fig. 78. ' — The right section of the portion o ah is the 
 curve a" t" s", etc. This right section unrolls in the straight line 
 e'' t, .*?,, etc., e'' ^,, t^ s,, etc., being equal to «" (5", f s", etc., respectively. 
 
 The right section of the portion o h c is shown at 5* o', and it 
 unrolls in the straight line h*" o,^^ b'^ o,/^ being equal in length to 
 the arc h'' o'. 
 
 The right section of the portion o c d is, the ellipse shown at 
 n'' o'\ which unrolls in the straight line w^ o,„*, equal in length to 
 the arc w* o". 
 
 Since most of the remaining problems are simply applications 
 of principles, already familiar, I have thought it best to give the 
 construction without the description in some cases ; a general 
 description without the construction in others ; and still others 
 without either construction or description, in order that the stu- 
 dent may accustom himself to apply these principles to the solu- 
 tion of practical problems as they may arise. 
 
 121. Problem 46. — Given the elevation of the outline of a 
 twelve-sided vase,, as in Fig^ 79, to find the elevation of the edges
 
 DESCRIPTIVE GEOMETRY. 47 
 
 A, jB, (7, and D, thus completing the elevation of the object; also 
 the development of one of the sides. 
 
 122. Problem 47. — To strike the pattern of the sheet metal 
 used in making a bath-tub. — Figs. 80, 81, and 82 show the con- 
 struction for three different styles of tubs. 
 
 123. Problem 48. — To find the iyitersection of the steam dome 
 and boiler of a locomotive ; to develop the steam dome ; and also to 
 develop the taper sheet. — Fig. 83 shows the two elevations of a 
 portion of a locomotive, and the intersection of the steam dome 
 and boiler. 
 
 P^ig. 84 is the development of the steam dome, and Fig. 85 is 
 the development of the taper sheet without the lap. 
 
 124. Problem 49. — To find the positioyi of a guide pidley, and 
 its shaft, to take a belt fi'om one pidley to another whose shafts 
 are at right angles to each other. — Let A and B, Fig. 86, be the 
 given pulleys. Assume any point c in the line a b (as the guide 
 pulley may be placed at any convenient point between the two 
 given pulleys) ; from this point draw the lines cm and c e tangent 
 to the pulleys B and A respectively. The plane P of these two 
 lines c m and c e will be the plane of the guide pulley. 
 
 The ground line may be taken in any convenient position per- 
 pendicular to the given shafts. 
 
 To find the projections of the guide pulley, its diameter being 
 known, revolve the plane P into the horizontal coordinate plane ; 
 cm revolves to cf' fnf', and c e to cf' e,'' ; C}" is then drawn equal to 
 the actual size of the guide pulley. In counter revolution its 
 centre is found at o" o\ and the circle C,** is found at C C", as 
 explained in Art. 71. 
 
 The shaft being perpendicular to the plane of the pulley, its 
 projections are respectively perpendicular to the traces V P and 
 H P, and they must pass through the centre o" o''; hence through 
 the point o draw the indefinite lines o^n" perpendicular to VP, and 
 o'' n'' perpendicular to H P, and the direction of the shaft will be 
 determined. 
 
 To show a definite length of the shaft, revolve the line rs 
 until it is parallel to V ; then, in its revolved position, lay off 
 o/ sj' and o,'' r,^ each equal to half the length of the shaft ; in
 
 48 DESCRIPTIVE GEOMETRY. 
 
 counter revolution r, goes to r, s, to s, and rs is the length 
 required. 
 
 125. Problem 50. — Given the longitudinal section and plan^ or 
 end view, of a connecting rod of varying cross section, to complete the 
 drawing of the elevation. — Let the section and plan be given as in 
 Fig. 87. The lower end from A to B is cylindrical, from C to D 
 is octagonal, from E to F is rectangular, between B and C the 
 cylinder exj^ands regularly in the arc of a circle from the cylin- 
 drical to the octagonal section, and between D and E it expands 
 in an elliptical curve from the octagonal to the rectangular 
 section. 
 
 126. Problem 51. — Given the section and plan of a portion of 
 a rocker arm, as in Fig. 88, to complete the elevation. — There is but 
 one curve to be found, and that is simply the intersection of a 
 cylinder and torus, whose axes are at right angles with each other. 
 The cylinder and torus are shown in the figure by dotted lines. 
 
 127. Problem 52. — To find the development of the conical por- 
 tion B of the double elbow shown in Fig. 89, the right sections of the 
 portio7is A and C being circular. 
 
 128. Problem 53. — To find the section on AB of the screw, as 
 given in Fig. 90 ; also, find section on CI). 
 
 129. Problem 54. — To construct the shadow of a sphere on a 
 horizontal j^lati'^ }' also, find its line of shade. — Pass a plane P, 
 Fig. 91, perpendicular to H, through the ray of light which passes 
 through the centre of the sphere. Revolve this jjlane into H. 
 The great circle which it cuts from the sphere revolves to the 
 circle of which o,^ is the centre ; the ray of light through the 
 centre revolves to of' a''; cf b'' and c?/' o\ parallel to o/' a'', are the 
 revolved positions of the two rays of light lying in the plane P, 
 and tangent to the sphere ; o'' and 6\ where these two lines inter- 
 sect H P, are the two extremities of the long diameter of the ellipse 
 which is the shadow of the sphere ; the short diameter of this 
 ellipse e/'x'' %ill, of course, be equal to the diameter of the sphere. 
 
 To get the line of shade, revolve the plane P back to its orig- 
 inal position ; the point cf' goes to c^, and cZ,* to d''; c* and d'' are 
 two points of the ellipse which constitutes the line of shade.
 
 Fig. 2. 
 
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 Fig. 13
 
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 Pig.56. 
 
 '¥ig.58.
 
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 Fig. 62.
 
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 Fig.66. 
 
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 Fig.78.
 
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 Fig.81.
 
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 Los Angeles 
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 DEC 141982 
 
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