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DESCRIPTIVE GEOMETRY 
 
 BY 
 
 ADAM V. MILLAR 
 
 ASSISTANT PROFESSOR OF DRAWINO UNIVERSITY OF WISCONSIN 
 AND 
 
 EDWARD S. MACLIN 
 
 INSTRUCTOR IN DRAWING UNIVERSITY OF WISCONSIN 
 
 McGRAW-HILL BOOK COMPANY 
 
 239 WEST 39TH STREET, NEW YORK 
 
 6 BOUVERIE STREET, LONDON, E. C. 
 
 1913 
 

 Copyright, 1913 
 
 BY 
 
 A. V. MILLAR 
 
 AND 
 
 E. S. MACLIN 
 
 STATE JOURNAL PRINTING COMPANY 
 
 Printers and Sterkotypers 
 madison, wis. 
 
PREFACE 
 
 In preparing the following text in descriptive geometry, the 
 authors have endeavored (1) to make the subject easier for 
 the student, (2) to help the student to visualize magnitudes in 
 space, and (3) to present the subject more nearly in accord 
 with commercial practice. 
 
 In order to accomplish these three things the ground line is 
 omitted. When the projections of several points are given 
 without the ground line being shown, the distances of the 
 points from the horizontal or vertical planes of projection are 
 not determined. The vertical projections, however, do show 
 the relative heights of the points and the horizontal projections 
 show the relative distances of the points from the vertical plane. 
 It is the relative distances of points of an object from a plane 
 with which we are concerned, since the distance of the whole 
 object from the plane of projection does not change the or- 
 thographic projection of the object on that plane. 
 
 When it is desired to locate points which are given distances 
 from the planes of projection, the ground lioe must be used. 
 Even when the ground line is not shown, it is understood to be 
 at right angles to the line joining the two projections of the 
 same point. 
 
 By the omission of the ground line, and therefore the traces 
 of a plane, the student's attention is centered on the object or 
 magnitude in space and not on the planes of projection. This 
 teaches the student to visualize the object rather than mem- 
 orize the projections of the object. The subject is thus made 
 easier because memorizing constructions and keeping the draw- 
 ing rather than the object in mind are the greatest hinderances 
 which the student encounters in mastering the subject. Since 
 the ground line is omitted in commercial work, the subject 
 
 259669 
 
IV PREFACE 
 
 taught in the above maimer is more in accord with that prac- 
 tice. 
 
 The third quadrant is used quite generally in the drafting 
 offices of this country. It seems logical, therefore, to present 
 the subject of descriptive geometry in the third quadrant, which 
 is done in the present text. Since in this text there is no 
 particular horizontal or vertical plane, objects, such as cones 
 and cylinders, can be placed in their natural positions rather 
 than being inverted to bring their bases in a horizontal plane of 
 projection. This removes one of the chief objections to the 
 third quadrant. 
 
 Although the method used here in presenting the principles 
 of descriptive geometry is new in American texts on the sub- 
 ject, it is used to some extent by French authors such as Javary, 
 '*F. J.,'' and others. It has been used at the University of Wis- 
 consin for some time and the results have been most satisfac- 
 tory. 
 
 The authors have consulted many descriptive geometries in 
 preparing the following text but are particularly indebted for 
 suggestions and problems, to the following : Phillips and Millar, 
 Fishleigh, Ames, Bartlett, and MacCord. 
 
CONTENTS 
 
 Page 
 
 Introduction 1 
 
 CHAPTER I 
 
 FIRST PRINCIPLES 
 
 Projection 5 
 
 Orthographic Projection 6 
 
 Picturing Magnitudes in Space 8 
 
 Points 9 
 
 Lines 12 
 
 Planes 16 
 
 Revolution and Counter-Revolution 19 
 
 Line Conventions 24 
 
 CHAPTER II 
 
 THE ELEMENTARY PRINCIPLES OF THE POINT, STRAIGHT 
 LINE, AND PLANE 
 
 To Find the Length of a Line and the Angles Which it Makes 
 
 V7ITH H AND V 26 
 
 To Find the Projections of a Line Which Makes a Given Angle 
 
 vriTH H OR V 27 
 
 To Find the Angle Which a Given Plane Makes with H or V 28 
 
 To Represent a Plane Which Makes a Given Angle with H or V 30 
 
 To Find the Angle Between Two Lines 31 
 
 To Find a Line Which Contains a Given Point and Makes a 
 
 Given Angle with a Given Line 33 
 
 To Represent a Plane Which Contains a Given Point and is 
 
 Parallel to Two Given Lines 35 
 
 To Find the Point in Which a Line Pierces a Plane 36 
 
 To Find the Line of Intersection of Two Planes 37 
 
 A Line Perpendicular to a Plane 40 
 
 To Find the Distance From a Point to a Plane 40 
 
 To Find a Point Which is a Given Distance From a Plane 41 
 
 To Represent a Plane Which Contains a Point and is Perpen- 
 dicular TO A Line 42 
 
VI CONTENTS 
 
 Page 
 
 To Find the Projection of a Line on a Plane 43 
 
 To Find the Angle Which a Line Makes with a Plane 44 
 
 To Find the Angle Between Two Planes 45 
 
 To Find the Common Perpendicular to Two Lines 46 
 
 Auxiliary Planes of Projection 48 
 
 Problems Intolvinq Points, Lines, and Planes 50 
 
 CHAPTER III 
 
 APPLICATIONS OF THE ELEMENTARY PRINCIPLES OF THE 
 POINT, STRAIGHT LINE. AND PLANE 
 
 Shades and Shadows 57 
 
 Plane Sections and Developments of the Surfaces of Prisms 
 
 AND Pyramids 64 
 
 Intersections of the Surfaces of Prisms and Pyramids 70 
 
 CHAPTER IV 
 
 CURVED LINES AND SURFACES 
 
 Generation and Classification of Lines 77 
 
 Projections of Curves 78 
 
 Tangents and Normals to Lines 79 
 
 Curves of Single Curvature 80 
 
 Curves of Double Curvature 85 
 
 Generation and Classification of Surfaces 86 
 
 Surfaces of Revolution 87 
 
 Tangent Planes to Surfaces. Normal Lines and Planes 89 
 
 Single Curved Surfaces 92 
 
 Warped Surfaces 99 
 
 Double Curved Surfaces 109 
 
 Problems on Tangent Planes to Surfaces 113 
 
 CHAPTER V 
 
 PLANE SECTIONS AND DEVELOPMENTS OF CURVED SURFACES 
 
 Right Cylinder 115 
 
 Oblique Cylinder 117 
 
 Oblique Cone 118 
 
 Plane Section of a Warped Surface 120 
 
 Plane Section of a Surface of Revolution 120 
 
CONTENTS Vll 
 
 Page 
 Plane Section of Any Surface by the Use of an Auxiliary Plane 
 
 OF Projection 122 
 
 Problems on the Plane Sections of Surfaces 124 
 
 CHAPTER VI 
 
 INTERSECTIONS OF CURVED SURFACES 
 
 General Method for Finding the Line of Intersection of Two 
 
 Surfaces 125 
 
 Two Oblique Cylinders 125 
 
 Cone and Cylinder 127 
 
 To Determine in Advance the Nature of the Line of Intersec- 
 tion 127 
 
 Two Cones 129 
 
 Sphere and Cone or Sphere and Cylinder 129 
 
 Problems on the Line of Intersection of Surfaces 130 
 
INTRODUCTION 
 
 It may help some instructors who contemplate using the fol- 
 lowing text-book to know how the authors have used the book 
 in their classes. With this in view, the following general 
 method for conducting the course is suggested and an outline 
 of lessons given. Each instructor will, no doubt, need to alter 
 the outline to some extent to suit the conditions under which 
 he works. 
 
 At the University of Wisconsin, descriptive geometry is 
 given as a three credit course for one semester of eighteen 
 weeks. Each week's work consists of one general lecture for 
 all students in the course, one recitation, and two two-hour 
 drafting periods for each section. One of the two-hour draft- 
 ing periods is sometimes turned into a one-hour recitation 
 period. 
 
 At the lecture, the general principles involved in the next 
 lesson are explained, general announcements made, and prob- 
 lems assigned for a home plate which is to be handed in at tho 
 beginning of the recitation period. At the recitation, the stu- 
 dents are drilled in the analyses of the problems and then sent 
 to the black board with some particular problem to solve. 
 
 In the drafting room, each student is given a slip similar to 
 the following: 
 
 Plate 1 
 
 Article 10, Problem 4 
 
 '' 10, '' 14 
 
 '' 15, '' 5 
 
 '' 15, '' 17 
 
 As far as possible duplicate slips are avoided. The student 
 solves the problems and if he has time letters the statements of 
 
INTRODUCTION 
 
 the problems. The work is done on a ll"xl5" sheet and is left 
 in pencil. Neatness, clearness, and accuracy are demanded. 
 With the exception of a few plates, the work is completed and 
 handed in at the close of each two-hour drafting period. The 
 plates are then corrected, graded, and returned to the student 
 at the next drafting period. This method of giving a plate to 
 be completed each time the student comes to the drafting room 
 has the following advantages: the student comes better pre- 
 pared for his work, he wastes no time in the drafting room, and 
 the attendance is improved. 
 
 Unannounced written quizzes are given in the drafting room 
 about every three weeks. Neatness and clearness of the con- 
 struction counts for 15% of the grade. 
 
 The following outline gives the lesson assignments for the 
 recitation work. 
 
 Lesson 1. 
 
 Articles 1 
 
 to 
 
 12 incl 
 
 u 
 
 '' 2. 
 
 
 13 
 
 
 19 
 
 
 '' .3. 
 
 
 20 
 
 
 26 
 
 
 4. 
 
 
 27 
 
 
 30 
 
 
 '' 5. 
 
 
 31 
 
 
 34 
 
 
 '' 6. 
 
 
 35 
 
 
 41 
 
 
 '' 7. 
 
 
 42 
 
 
 45 
 
 
 " 8. 
 
 
 46 
 
 
 51 
 
 
 '' 9. 
 
 
 52 
 
 
 55 
 
 
 '' 10. 
 
 
 56 
 
 
 58 
 
 
 '' 11. 
 
 
 59 
 
 
 61 
 
 
 ** 12. 
 
 
 62 
 
 
 68 
 
 
 '' 13. 
 
 
 69 
 
 
 73 
 
 
 '' 14. 
 
 
 74 
 
 
 76 
 
 
 '' 15. 
 
 
 77 
 
 
 79 
 
 
 '' 16. 
 
 
 80 
 
 
 95 
 
 
 *' 17. 
 
 
 96 
 
 
 107 
 
 
 '' 18. 
 
 
 108 
 
 
 118 
 
 
 '' 19. 
 
 
 119 
 
 
 128 
 
 
 '' 20. 
 
 
 129 
 
 
 136 
 
 / 
 

 
 INTRODUCTION 
 
 SOI 
 
 1 21. 
 
 Articles 137 to 152 inclusive 
 
 a 
 
 22. 
 
 153'' 158 '' 
 
 ( c 
 
 23. 
 
 159'' 161 " 
 
 In case the time allowed for descriptive geometry is not 
 enough to permit the giving of the course as outlined above, it 
 is suggested that shades and shadows and the latter part of 
 warped surfaces be omitted. 
 
DESCRIPTIVE GEOMETRY 
 
 CHAPTER I 
 FIRST PRINCIPLES 
 
 1. Projection. If from a point S, Fig. 1, straight lines are 
 drawn through a series of points A, B, C . . . , the points a, h, c 
 ... in which these lines pierce the plane T, are the projections 
 of the points A, B, C on this plane. 
 
 S is called the point of sight. 
 
 A, B, C, ... are points of a magnitude or object in space. 
 
 Sa, Sh, So . . . are projecting lines of the points A, B, C, 
 
 T is the plane of projection. 
 
 
 
 /b/ / 
 
 / ^ / 
 
 A\ 
 
 /n^/j^j\ /i 
 
 Fig. 1. — Perspective. 
 
 Fig. 2. — Oblique 
 projection. 
 
 %i 
 
 c 
 
 7M\ 
 
 F^G. 3. — Orthographic 
 projection. 
 
 Leaving color out of consideration, the projections a, 6, c, 
 . . . present the same appearance to the eye, situated at the 
 point of sight, as the points A, B, C, ... in space. 
 
 In perspective, Fig. 1, the point of sight is at a finite distance 
 from the plane of projection. The projecting lines diverge. 
 
 In oblique and orthographic projection. Figs. 2 and 3, the 
 point of sight is at an infinite distance from the plane of pro- 
 jection. The projecting lines are parallel. 
 
DESCRIPTIVE GEOMETRY 
 
 The projection is oblique when the projecting lines are par- 
 allel to each other and oblique to the plane of projection. 
 
 The projection is orthographic when the projecting lines are 
 perpendicular to the plane of projection. 
 
 ORTHOGRAPHIC PROJECTION 
 
 2. A point in space is not completely determined by its or- 
 thographic projection on one plane for the distance of the point 
 from the plane is not shown by its projection. All points A, B, 
 C, Fig. 4, which lie in a vertical straight line have the same 
 projection on a horizontal plane. There are two methods of 
 representing definitely a point in space. One method is to give 
 J its projection on a plane and also 
 
 A* its distance from that plane. This 
 
 ig method is used in making contour 
 
 maps; a contour line being a line 
 joining the projections of all points 
 which are a given distance above or 
 
 Ct 
 
 ; V wnicn are a given aisiance aoove or 
 
 ^^^ \l below a base plane. The other method 
 " is to use two (or more) different 
 Fig. 4. — a, 6, c, projections of planes of projection. When two pro- 
 points A, B, C, on plane T. jections are used, one is usually con- 
 sidered the principal projection. The other is supplementary 
 showing the distance of the point from the plane upon which the 
 principal projection is made. The method which uses the two 
 projections of a point is the more common. 
 
 3. PlaJies of projection. In orthographic projection, two 
 planes are generally used, Fig. 5, one horizontal and the other 
 vertical, called respectively: 
 
 (a) The horizontal plane of projection, or H. 
 
 (b) The vertical plane of projection, or V. 
 
 Their intersection is called the ground line, or G. L. 
 
 Sometimes other planes of projection are used which are per- 
 pendicular to either H or V. If the plane of projection is per- 
 pendicular to both H and V and therefore perpendicular to the 
 ground line, it is called a profile or end plane, or P. 
 
ORTHOGRAPHIC PROJECTION 
 
 Point of sight. The point of sight for the horizontal plane is 
 an infinite distance above H and for the vertical plane it is an in- 
 finite distance in front of V. In the horizontal view, the ground 
 line represents the vertical plane seen edgewise, and in the 
 vertical view, the ground line represents the horizontal plane 
 seen edgewise. 
 
 4. The four quadrants. The right dihedral angles formed by 
 the intersection of the horizontal and vertical planes of projec- 
 tion are known as the first, second, third, and fourth quadrants, 
 Fig. 5. 
 
 The first quadrant is 
 above H and in front of V. 
 The second quadrant is 
 above H and back of V. 
 
 The third quadrant is 
 below H and back of V. 
 
 The fourth quadrant is 
 below H and in front of V. 
 5. The drawing. In or- 
 der to represent both the 
 horizontal and vertical 
 
 projections of an object on 
 Fig. 5.-Principal planes of projection. ^^^ ^^^^ ^^^^^ ^^ p^p^^.^ 
 
 the planes of projection must be brought together. This is 
 accomplished by keeping one of the planes fixed and revolving 
 the other about the ground line as an axis until the two planes 
 coincide. If the horizontal plane is kept fixed, the upper part 
 of the vertical plane is revolved backward until it coincides 
 with the back part of the horizontal plane. This wiU be found 
 convenient if the work is done on a drafting board which is in a 
 horizontal position. If the vertical plane is kept fixed, the 
 front part of the horizontal plane is revolved downward until it 
 coincides with the lower part of the vertical plane. This will 
 be found convenient when working on the blackboard. By 
 either method, the first and third quadrants are opened and the 
 second and fourth closed. 
 
8 DESCRIPTIVE GEOMETRY 
 
 When the profile or end plane is used, it is brought into the 
 vertical plane by revolving it about its intersection vrith the 
 vertical plane. 
 
 It must be remembered that the projections are made vrhile 
 the planes of projection are still at right angles, the planes be- 
 ing brought together simply for convenience in making all of 
 the projections on one sheet of paper. 
 
 PICTURING MAGNITUDES IN SPACE 
 
 6. In order to solve problems intelligently, the student should 
 learn to picture to himself points, lines, and objects in their 
 proper positions v^ith reference to the planes of projection. 
 When the dravring is in a horizontal position, it is best to con- 
 sider the plane of the dravring as the horizontal plane of pro- 
 jection. The vertical plane of projection should be pictured as 
 a plane vrhich contains the ground line and is perpendicular to 
 the plane of the drawing. The student should place one of his 
 triangles to represent the vertical plane. The other triangle 
 when placed at right angles to the horizontal and vertical planes 
 of projection will represent the profile or end plane. 
 
 The first quadrant is above the drawing and in front of the 
 ground line. The second quadrant is above the drawing and 
 behind the ground line. The third quadrant is below the draw- 
 ing and behind the ground line. The fourth quadrant is below 
 the drawing and in front of the ground line. 
 
 The magnitude, such as a point, line, or object, is then pic- 
 tured in its proper position with reference to the planes of pro- 
 jection. The top or horizontal view is obtained by looking 
 straight down from above, the front or vertical view by look- 
 ing from the front, and the end or profile view by looking 
 through the end plane. After these projections are made, the 
 planes of projection are brought together by the method ex- 
 plained in Art. 5. The magnitudes should always be pictured 
 in space before the drawing is made. 
 
POINTS 
 
 POINTS 
 
 7. Let A, Fig. 6, be any point in space and H, V, and P the 
 planes of projection. The line Aa, through the point and per- 
 pendicular to the horizontal plane, is the horizontal projecting 
 line of the point and its intersection a, with the horizontal plane, 
 is the horizontal projection of the point. Similarly the line Aa\ 
 through the point and perpendicular to the vertical plane, is the 
 vertical projecting line of the point and its intersection a' with 
 the vertical plane is the vertical projection of the point. In 
 like manner the line A.a", through the point and perpendicular 
 to the profile plane, is the profile projecting line of the point 
 and its intersection a" with the profile plane is the profile pro- 
 jection of the point. 
 
 Fig. 6. — Point A in 3rd quadrant. 
 
 Fig. 7. 
 
 Any two projections of a point determine its position in 
 space, for the projecting lines intersect in the only point which 
 can have the given projections. 
 
 Since the vertical projecting line of a point in space is par- 
 allel to the horizontal plane, the distance from the point in 
 space to the horizontal plane is equal to the distance from its 
 vertical projection to the ground line. Moreover, the horizon- 
 tal projecting line of a point in space is parallel to the vertical 
 
10 
 
 DESCRIPTIVE GEOMETRY 
 
 plane, therefore the distance from the point in space to the ver- 
 tical plane is equal to the distance from its horizontal projec- 
 tion to the ground line. 
 
 Since a'h remains perpendicular to the ground line during the 
 revolution of the vertical plane from the vertical to the horizon- 
 tal position, then when the vertical and horizontal planes coin- 
 cide as they do on the drawing, Fig. 7, a'a is perpendicular to 
 the ground line, that is, the two projections of a point in space 
 are in the same line perpendicular to the ground line. 
 
 8. Figs. 8, 9, and 10 show the projections of points in the 
 first, second, and fourth quadrants respectively. The figures 
 
 ^a' 
 
 Fig. ^.— Point A 
 in 1st Q. 
 
 I 
 
 I 
 
 at 
 
 I 
 
 I 
 I 
 __L 
 
 'Fig. 9.— Point A 
 in 2nd Q. 
 
 I 
 
 I 
 ai 
 
 Fig. l(i.— Point A 
 in 4th Q. 
 
 represent the projections of the point after the planes of pro- 
 jection have been brought into coincidence. These points in 
 space should be definitely pictured from their projections. 
 "Wlien the horizontal projection is considered the principal pro- 
 jection, pictufe directly above or below it, the vertical projec- 
 tion giving the distance from the horizontal plane. When the 
 vertical projection is considered the principal projection, pic- 
 ture directly in front of or behind it, the horizontal projection 
 giving the distance from the vertical plane. 
 
 It does not matter in which quadrant the object is placed, the 
 point of sight for the horizontal view is always above H, and 
 for the vertical view it is always in front of V. 
 
POINTS 11 
 
 In the third quadrant, the point of sight is at an infinite dis- 
 tance to the right for the right end view and at an infinite dis- 
 tance to the left for the left end view. 
 
 9. Natation. Points in space are designated by capital let- 
 ters as A, B, C, etc. Horizontal projections of points, small 
 letters as a, h, c, etc. Vertical projections of points, small let- 
 ters with prime marks as a\ &', c', etc. Profile projections of 
 points, small letters with double prime marks as a", b", d\ etc. 
 
 Ground line, full. 
 
 The line joining the projections of a point will be drawn fine, 
 full in pencil, and fine, full, red in ink. 
 
 10. Problems. Picture to yourself points in the following 
 positions and then draw their projections. 
 
 1. In H V in front of V. 
 
 2. In V \\" below H. 
 
 3. V in front of V and 2" above H. 
 
 4. W below H and \\" behind V. 
 
 5. \" above H and W" behind V. 
 
 6. \\" below H and ir in front of V. 
 
 7. V above H and \\" in front of V. 
 
 8. 1" behind V and \\" below H. 
 
 9. V in front of V and 1" below H. 
 
 10. \\" behind V and T above H. 
 
 11. In V and 2" above H. 
 
 12. In H and W behind V. 
 
 13. In the third quadrant V from H and \\" from V. 
 
 14. In the first quadrant 1^'' from V and \" from H. 
 
 15. In the fourth quadrant \\" from H and \" from V. 
 
 16. In the second quadrant \" from V and \\" from H. 
 
 17. A point has its projections coinciding above the ground line. 
 Describe the location of the point. 
 
 18. The same as problem 17 except that the projections of the point 
 coincide below the ground line. 
 
12 
 
 DESCRIPTIVE GEOMETRY 
 
 LINES 
 
 11. Let AC, Fig. 11, be any straight line in space and H, V, 
 and P the planes of projection. The plane ACc, containing the 
 line and perpendicular to the horizontal plane, is the horizontal 
 projecting plajie of the line. Its intersection ac with the hor- 
 izontal plane is the horizontal projection of the line. Similarly 
 the plane ACc', containing the line and perpendicular to the 
 vertical plane, is the vertical projecting plane of the line. Its 
 intersection a'c' with the vertical plane is the vertical projection 
 of the line. In like manner the plane ACc", containing the line 
 and perpendicular to the profile plane, is the profile projecting 
 plane of the line. Its intersection a"c" with the profile plane is 
 the profile or end projection of the line. 
 
 Fig. 11. — Projections of a line. 
 
 Any two projections of a straight line determine its position 
 in space, for the projecting planes intersect in the only line 
 which can have the given projections. 
 
 "When the two projections of a line of indefinite length are 
 in the same perpendicular to the ground line, the projecting 
 planes will coincide and the line is undetermined. If, however, 
 the projections of any two points of a straight line are given, 
 the line is always determined. 
 
 * # 
 
LINES ^ 13 
 
 12. Projections of straight lines in various positions with ref- 
 erence to the the planes of projection. 
 
 If a line is oblique to H, V, and P, the projections of the Una 
 will be inclined to the ground line. 
 
 If a line is parallel to the ground line, its H and V projections 
 will be parallel to the ground line. 
 
 If a line is perpendicular to the horizontal plane, its horizon- 
 tal projection is a point and its vertical projection is a straight 
 line perpendicular to the ground line. 
 
 If a line is perpendicular to the vertical plane, its vertical 
 projection is a point and its horizontal projection is a straight 
 line perpendicular to the ground line. 
 
 If a line is parallel to the horizontal plane and oblique to the 
 vertical plane, its vertical projection will be parallel to the 
 ground line and its horizontal projection will be inclined to the 
 ground line. 
 
 If a line is parallel to the vertical plane and oblique to the 
 horizontal plane, its horizontal projection will be parallel to the 
 ground line and its vertical projection will be inclined to the 
 ground line. 
 
 13. Point on line. If a point is on a lihe in space, the hor- 
 izontal projection of the point will be on the horizontal pro- 
 jection of the line, the vertical projection of the point will be 
 on the vertical projection of the line, and the profile projection 
 of the point will be on the profile projection of the line, Fig. 12. 
 In order to represent a point of a given line which lies in a 
 profile plane, use the profile projection of the line. 
 
 Intersecting lines. If two lines, which do not lie in a profile 
 plane, intersect at a point in space, the horizontal projections 
 of the lines will intersect in the horizontal projection of the 
 point and the vertical projections of the lines will intersect in 
 the vertical projection of the point, Fig. 13. The projections of 
 the common point D must lie in the same perpendicular to the 
 ground line. 
 
 If two lines do not intersect in space, their projections will 
 
14 
 
 DESCRIPTIVE GEOMETRY 
 
 not intersect in the same perpendicular to the ground line, 
 Fig. 14. 
 
 Parallel lines. If two lines are parallel in space, their pro- 
 jections on the same plane are parallel, Fig. 15. 
 
 Pig 12. — Point C Fig. IS.— Intersect- Fig. 14.— Lines not Fig. 15.—Par- 
 on line AB. ing lines. intersecting. allel lines. 
 
 14. Notation. Lines in space are designated by capital let- 
 ters as AB, CD, MN, etc. Horizontal projections of lines, small 
 letters as dby cd, mn, etc. Vertical projections of lines, small 
 letters with prime marks as a'&', c'd\ m'n\ etc. Profile projec- 
 tions of lines, small letters with double prime marks as a"!:)", 
 c"d"; m"n\ etc. 
 
 The planes of projection will be considered transparent. 
 
 15. Problems. Picture to yourself lines in the following posi- 
 tions and then draw their projections. Show the profile pro- 
 jections of lines in the first and third quadrants in addition to 
 their horizontal and vertical projections. 
 
 1. In the first quadrant, parallel to H and oblique to V. 
 
 2. In the second quadrant, parallel to V and oblique to H. 
 
 3. In the back of H and inclined to V. 
 
 4. In the third quadrant, perpendicular to H. 
 
 5. In the third quadrant, oblique to H and V. Draw the projections 
 of a point on this line. 
 
 6. In the first quadrant oblique to H and V and in a plane perpen- 
 dicular to G. L. 
 
LINES 15 
 
 7. In a plane bisecting the faurth quadrant. 
 
 8. An oblique line intersecting the ground line. 
 
 9. Two intersecting lines in the third quadrant. 
 
 10. Two lines which are oblique to H and V and intersect on H. 
 
 11. Given the projections of a point and the projections of an oblique 
 line. Draw the projections of a line which passes through the given 
 point and is parallel to the given line. 
 
 12. Two oblique lines which are parallel, one in the first and the 
 other in the second quadrant. 
 
 13. Given the projections of a point and the projections of an oblique 
 line. Draw the projections of a line which passes through the given 
 point and intersects the given line. 
 
 14. Two intersecting lines in the first quadrant, one parallel to H and 
 oblique to V and the other parallel to V and oblique to H. 
 
 15. Two parallel lines which lie in a profile plane and are in the 
 third quadrant. 
 
 16. Two intersecting lines which lie in a profile plane and are in the 
 first quadrant. 
 
 17. Two intersecting lines in the third quadrant, one perpendicular 
 to H and the other perpendicular to V. 
 
 18. A line is given by its projections. Find the projections of a 
 point of this line which is equally distant from H and V. 
 
 19. A line of profile is given by the projections of two of its points. 
 Choose a point of this line which is equally distant from H and V. 
 
 20. A point and a line lie in a profile plane. Determine by an end 
 view whether or not the point lies on the line. 
 
 21. Draw the projections of a parallelogram when the projections of 
 two of its adjacent sides are given. 
 
16 DESCRIPTIVE GEOMETRY 
 
 PLANES 
 
 16. In space, a plane is fixed by three points, not in the same 
 straight line, a point and a line, two intersecting lines or two 
 parallel lines. If the horizontal and vertical projections of 
 these magnitudes are given, the plane which they determine in 
 space will be definitely located. When the statement ''A given 
 plane" is made, the student is expected to draw the projections 
 of the magnitudes which represent the plane. 
 
 jir^b The points A, B, and C, Fig. 16, fix 
 
 ^ ^^'^^ \ \ a plane in space. It is evident that a 
 
 straight line joining A and B, B and 
 
 •-^1 
 
 r'"""-^*>.^c C, or A and C will lie in this plane. 
 
 /< 
 
 If a point and line are given by 
 their projections, they will fix a 
 y^ ^S^^^ ' .plane. Other straight lines of the 
 
 <7>i^ — ^""^ plane can be found by joining the 
 
 Fig. 16.— Points A, B, C given point with points on the given 
 
 fix a plane. Hj^q 
 
 If the plane is fixed by two straight lines, either parallel or 
 intersecting, any number of other straight lines of the plane can 
 be found by joining a point in one of the given lines with a 
 point in the other. 
 
 If a plane is perpendicular to the horizontal plane of projec- 
 tion, all straight lines of the plane which are of indefinite 
 length, have the same horizontal projection. The vertical pro- 
 jections of these lines will be parallel or intersecting lines. 
 
 If a plane is perpendicular to the vertical plane of projection, 
 all straight lines of the plane which are of indefinite length 
 have the same vertical projection. The horizontal projections 
 of these lines will be parallel or intersecting lines. 
 
 If a plane is perpendicular to both the horizontal and ver- 
 tical planes of projection, all straight lines of the plane which 
 are of indefinite length have the same horizontal projection and 
 also the same vertical projection. 
 
PLANES 
 
 17 
 
 If one straight line of a plane is parallel to the ground line, 
 the plane is parallel to the ground line. 
 
 17. Lines in planes. The two intersecting lines MN and OP, 
 Fig. 17, fix a plane. The straight line AB, joining the point A 
 in MN with the point B in OPj is another line of the plane, ah 
 is its horizontal and a'h' its vertical projection. 
 
 A line which lies in a given plane and is parallel to H is 
 called a horizontal of that plane. In Fig. 18, HH is such a line 
 which lies in the plane of the lines MN and OP. Its vertical 
 projection Wh' is drawn first and then its horizontal projection 
 
 Fig. 17. — AB line in plane of 
 MN and OP. 
 
 Fig. 18. — HH horizontal of given 
 plane. FF frontal of given plane. 
 
 hh is found by finding the horizontal projections of the points 
 B and A where it crosses the given lines MN and OP. A line 
 which lies in a given plane and is parallel to V, is called a fron- 
 tal of that plane. FF is such a line which lies in the plane of 
 MN and OP. Its horizontal projection // is drawn first and 
 then its vertical projection ff is found. These lines which lie 
 in a plane and are parallel to H or Y play an important part in 
 the solution of many problems. 
 
 18. Points in planes. If one projection of a point which lies 
 in a plane is given, the other projection is found by finding the 
 projections of a line of the plane passing through the point and 
 
18 DESCRIPTIVB GEOMETRY 
 
 then locating the other projection of the point on the other pro- 
 jection of the line. 
 
 All the points which lie on a given plane and are a given dis- 
 tance from the horizontal plane of projection are on a horizon- 
 tal of the given plane. Likewise, all the points which lie on a 
 given plane and are a given distance from the vertical plane of 
 projection are on a frontal of the given plane. 
 
 19. Problems. In the following problems the given planes 
 may be represented by the projections of any two of their lines 
 unless it is otherwise stated. 
 
 1. Having given the horizontal projection of a line which lies in a 
 given plane, to find its vertical projection. 
 
 2. Having given the vertical projection of a line which lies in a 
 given plane, to find its horizontal projection. 
 
 3. Having given the horizontal projection of a point which lies in a 
 given plane, to find its vertical projection. 
 
 4. Find the projections of the locus of all points which lie on a given 
 oblique plane and are 1" from H; 
 
 6. Find the projections of the locus of all points which lie on a given 
 oblique plane and are li" from V. 
 
 6. Find the projections of a point which lies on a given oblique plane 
 and is I'' from H and 1" from V. 
 
 7. A point and two parallel lines are given by their projections. Is 
 the point in the plane of the two parallel lines? 
 
 8. Having given a plane which is parallel to the ground line and the 
 horizontal projection of a point of this plane. Find the vertical pro- 
 jection of the point. 
 
 9. Having given a plane which is parallel to the ground line and the 
 horizontal projection of a line of this plane. Find the vertical projec- 
 tion of the line. 
 
 10. Having given a plane which is parallel to the ground line and the 
 vertical projection of a line of this plane which is also parallel to the 
 ground line. Find the horizontal projection of the line. 
 
 11. Two lines in the third quadrant are in such a position that neither 
 their horizontal nor vertical projections intersect within the limits of 
 the drawing. Determine graphically whether or not the lines intersect. 
 
 12. A plane is represented by two intersecting lines. The horizontal 
 projection of a line which passes through their point of intersection 
 and lies in the given plane is given. Find its vertical projection. 
 
 13. A plane is represented by two parallel lines. The horizontal pro- 
 jection of another line of this plane which is parallel to the first two 
 lines is given. Find its vertical projection. 
 
REVOLUTION AND COUNTER REVOLUTION 19 
 
 REVOLUTION AND COUNTER REVOLUTION OF OBJECTS 
 
 20. An object is said to revolve about a straight line as an 
 axis when each of its points moves in the circumference of a 
 circle whose center is in the axis and whose plane is perpen- 
 dicular to the axis. 
 
 When an object is revolved about a straight line as an axis, 
 the relative position of its points is not changed. The object 
 can thus be brought into a simpler position with reference to 
 the planes of projection. The projections of the object in this 
 position are easily found and from these projections the projec- 
 tions of the object in its original position are found by the 
 counter revolution of its points. 
 
 REVOLUTION OF A POINT ABOUT AN AXIS 
 
 21. A line perpendicular to H and a point are given by their 
 projections. Revolve the point through an angle of oc ° about 
 the line as an axis. 
 
 In Fig. 19, let MN be the axis and P the given point. 
 
 The point P will move in the circumference of a circle whose 
 center is at and whose radius is OP (Art. 20). The plane of 
 the path of this point is perpendicular to MN and therefore 
 parallel to H. Therefore the horizontal projection of the path 
 is a circle with a radius op and the vertical projection is a 
 straight line through p' parallel to the ground line. Since the 
 plane of the circle is parallel to H, the horizontal projection of 
 the angle through, which the radius sweeps is equal to the true 
 size of the angle. Therefore Pi, Fig. 19, is the required posi- 
 tion of P. 
 
 22. A line perpendicular to V and a point are given by their 
 projections. Revolve the point through an angle of oc ° about 
 the line as an axis. 
 
 In Fig. 20, let MN be the axis and P the given point. 
 The point P will move in the circumference of a circle whose 
 center is at and whose radius is OP (Art. 20). The plane of 
 
20 
 
 DESCRIPTIVE GEOMETRY 
 
 the path of this point is parallel to V, and therefore the vertical 
 projection is a circle with radius o'p' and the horizontal pro- 
 jection is a straight line through p parallel to the ground line. 
 Since the plane of the circle is parallel to V, the vertical pro- 
 jection of the angle through which the radius sweeps is equal to 
 the true size of the angle. Therefore Pj, Fig. 20, is the required 
 position of P. 
 
 n 
 
 rrP(n 
 
 I 
 
 /A I' 
 
 m 
 
 o'-l 
 
 1 
 I 
 
 I 
 I 
 I 
 I 
 
 Fig. 
 
 19. — P revolved through oc" 
 about MN as axis. 
 
 Fig. 20. — P revolved through oc' 
 about MN as axis. 
 
 23. A line parallel to H and oblique to V and a point are 
 given by their projections. Revolve the point about the line as 
 an axis until the point is on the same level as the axis. 
 
 In Fig. 21, let MN be the axis and P the given point. 
 
 The point P will move in the circumference of a circle whose 
 center is at and whose radius is OP. The plane of this circle 
 is perpendicular to the axis MN and therefore perpendicular to 
 H. The horizontal projection of the circle is the straight line 
 op which passes through the horizontal projection p of the point 
 and is perpendicular to the horizontal projection mn of the 
 axis. The vertical projection of the circle is an ellipse. To 
 find the radius of the circle, let the plane of the circle be turned 
 over until it is parallel to H about a horizontal axis which 
 
REVOLUTION AND COUNTER REVOLUTION 
 
 21 
 
 passes through the center and is perpendicular to MN. When 
 the plane is in the revolved position, the point P is at pi, the 
 distance pp^ being equal to the distance d. Then op^ is the re- 
 quired radius. The point p^ is then moved around in this cir- 
 cumference until it cuts the line op extended at Pz and p^. 
 
 
 / i /y 
 
 
 Fig. 21. — Point P revolved about 
 MN as axis. 
 
 fA. ,.-' 
 
 Fig. 22. — Point P revolved about 
 MN as axis. 
 
 When the plane of the circle is counter revolved to its former 
 position at right angles to MN, the points Pg and P3 do not move 
 and therefore remain on the same level as MN. Then the points 
 P2 and P3 are the required positions of the point P. 
 
 The perpendicular distance from any point in the circumfer- 
 ence P2P1P3 to the line op represents the distance of the point P 
 above or below MN when the plane of the circle stands perpen- 
 dicular to MN. 
 
 24. A line parallel to V and oblique to H and a point are 
 given by their projections. Revolve the point about the line 
 as an axis until the point is the same distance from V as the axis. 
 
 In Fig. 22, let MN be the axis and P the given point. 
 
 The point P will move in the circumference of a circle whose 
 center is at and whose radius is OP. The plane of this circle 
 
22 DESCRIPTIVE GEOMETRY 
 
 is perpendicular to the axis MN and therefore perpendicular to 
 V. The vertical projection of the circle is the straight line o'y' 
 which passes through the vertical projection p' of the point and 
 is perpendicular to the vertical projection m'w' of the axis. 
 The horizontal projection of the circle is an ellipse. To find the 
 radius of the circle, let the plane of the circle be turned over 
 until it is parallel to V about an axis parallel to V which passes 
 through the center and is perpendicular to MN. When the 
 plane is in the revolved position, the point P is at p'l, the dis- 
 tance p'p'i being equal to the distance d Then o''g\ is the re- 
 quired radius. The point 'p\ is then moved around in this cir- 
 cumference until it cuts the line dp' at y'^ and p'g. "When the 
 plane of the circle is counter revolved to its former position at 
 right angles to MN, the points Pg and Pg do not move and there- 
 fore remain the same distance from Y as MN. Then the points 
 P2 and P3 are the required positions of the point P. 
 
 The perpendicular distance from any point in the circumfer- 
 ence p'zP'xP'z to the line o'p' represents the distance of the point 
 P in front of or behind MN when the plane of the circle stands 
 perpendicular to MN. 
 
 25. Problems. 
 
 Note. — Unless stated otherwise all problems are to be solved in the 
 third quadrant. 
 
 1. Revolve a point which is in the first quadrant through an angle of 
 45 degrees about an axis perpendicular to H and show its projections In 
 the new position. 
 
 2. Revolve a point which is in the fourth quadrant into H about an 
 axis in H. 
 
 3. Revolve a point which is in the third quadrant into H about the 
 ground line as an axis. 
 
 4. Having given a point in H and an axis in H 30° to V, revolve the 
 point about the axis until it is 1" above H and show its projections in 
 this position. 
 
 5. Given a point in H also a line in H oblique to V. Revolve the 
 point about the line as an axis until the point strikes V. Show its 
 projections in this position. 
 
 6. Given a point in V and a line in V oblique to H. Revolve the point 
 about the line as an axis until the point is 1" behind V. Show its 
 projections in this position. 
 
'REVOLUTION AND COUNTER REVOLUTION 23 
 
 7. Given a point in V and a line in H oblique to V. Revolve the point 
 into H about the line as an axis and show its projections in this posi- 
 tion. 
 
 8. A line lies in the front part of H and is oblique to the ground line. 
 A point lies in the back part of H. Revolve the point about the line as 
 an axis until it strikes V. Show its projections in this position. 
 
 9. Given an axis which is oblique to V and is parallel to and i" above 
 H. Revolve a point which is in the first quadrant through an angle of 
 90° about this axis. Show its projections in this position. 
 
 10. Given an axis which is oblique to V and is parallel to and i" 
 below H. Revolve a point about this axis until the point is V below 
 H. Show its projections in this position. 
 
 11. Given an axis which is oblique to V and is parallel to and f" 
 above H. Revolve a point which is in the first quadrant about the 
 axis until it is in the second quadrant. Show its projections in this 
 position. 
 
 12. Given a point in the third quadrant and an axis in H oblique to V. 
 Revolve the point about the axis and find the projections of the points 
 where it passes through H and V. 
 
 13. Given a point which is in V, and a line in H. Revolve the point 
 about the line as an axis until the point is 1" in front of V. Show its 
 projections in this position. 
 
 14. Revolve a point A which is 2" above H and is li" in front of V 
 about an axis MN which is in H and makes 30° with V until the point 
 is i" behind V and is below H. Show its projections in this position. 
 
 15. Given two points A and B on H and unequal distances from V. 
 Find a point on V which is 2'' from both A and B. 
 
 16. Two lines lie on H, oblique to V and made 60° with each other. 
 Revolve the one about the other as an axis until the H projection of 
 the angle between them is 30°. Show their projections in this position. 
 
 17. Given two intersecting lines in H. Revolve the one about the 
 other until a point of the moving line is in V. Show the projections 
 of the lines in this position. 
 
 18. Two parallel lines are in H oblique to V. Revolve one line about 
 the other as an axis until a point of the moving line strikes V. Show 
 their projections in this position. 
 
 19. Two intersecting lines are in H, oblique to V. Revolve one 
 about the other until a point of the moving line is 1" below H. Show 
 their projections in this position. 
 
 20. Given an equilateral triangle on H. Revolve the triangle about 
 one side as an axis until the H projection of the angle opposite the 
 axis is a right angle. Show the projections of the triangle in this 
 position. 
 
24 DESCRIPTIVE GEOMETRY 
 
 21. Given a right triangle in H. Revolve the triangle about the 
 hypothenuse as an axis until the H projection of the right angle is an 
 angle of 120°. Show the projections of the triangle in this position. 
 
 22. Given an isosceles triangle in H oblique to V. The angle at the 
 vertex is 30°. Revolve the triangle about the base as an axis until the 
 H projection of the 30° angle is a right angle. Shov?" the projectione 
 of the triangle in this position. 
 
 23. Given a scalene triangle on H oblique to V (all angles acute). 
 Revolve the smallest angle about the side opposite as an axis until the 
 H projection of the moving angle is a right angle. Show the projec- 
 tions of the triangle in this position. 
 
 24. Revolve a line AB which is in H about a line CD, also in H, until 
 AB makes an angle of 30° with H. Show the projections of AB in this 
 position. 
 
 25. Given an axis on H oblique to V and a point on H. Revolve the 
 point through an angle of 30° about the axis and show its projections 
 in this position. 
 
 26. Given an axis on H oblique to V and a li" equilateral triangle on 
 H. Revolve the triangle through an angle of 45° about the axis and 
 show its projections in this position. 
 
 27. Given an axis on H and a 1^" square on H with sides oblique to 
 the axis. Revolve the square through an angle of 60° about the axis 
 and show its projections in this position. 
 
 LINE CONVENTIONS 
 
 26. Pencil. The planes of projection are considered trans- 
 parent. 
 
 Retrace required lines so that they stand out from the con- 
 struction making hidden lines dashed. The dashes should be 
 about 1/8" long and 1/32" apart. 
 /AH other construction is to be in very fine full lines. 
 
 Ink. Given lines when visible, fine, full, black lines; when 
 invisible, fine, dashed, black lines. 
 
 Required lines when visible, heavy, full, black lines; when in- 
 visible, heavy, dashed, black lines. 
 
 All other construction is to be in fine, full, red lines. 
 
CHAPTER II 
 
 THE ELEMENTARY PRINCIPLES OF THE POINT, 
 STRAIGHT LINE, AND PLANE. 
 
 27. The horizontal and vertical projections of a point will fix 
 a definite point in space if the ground line is shown on the 
 drawing (Art. 7). The distances from the projections of the 
 point to the ground line are equal respectively to the distances 
 of the point from the planes of projection. If the ground line 
 is not shown, the projections of a point will not fix a definite 
 point in space, since there is nothing to show its distances from 
 the planes of projection. If, however, the projections of two 
 or more points are given without the ground line being shown, 
 the vertical projections will show the relative heights of the 
 points and the horizontal projections will show their relative 
 distances from the vertical plane. 
 
 The projection of an object on a plane does not depend upon 
 its distance from the plane, but upon the relative distances of 
 its points from that plane. For example, the horizontal pro- 
 jection of a cube does not depend upon the distance of the cube 
 from the horizontal plane, for the horizontal projection will not 
 be changed if all the corners of the cube be moved the same 
 distance up or down along vertical lines. The horizontal pro- 
 jection of the cube will be changed, however, if one or two 
 comers remain fixed while the other corners are moved. Two 
 projections of an object, when the ground line is omitted, will 
 in general definitely determine the form of the object, but will 
 not show its distances from the planes of projection. 
 
 Hereafter the ground line will be omitted from many of the 
 drawings, but it is always understood to be at right angles to 
 the line joining two projections of the same point. 
 
26 
 
 DESCRIPTIVE GEOMETRY 
 
 28. A straight line is given by its projections. Find the true 
 length of the line and the angles which it makes with H and V. 
 
 Let AB, Fig. 23, be the given line. 
 
 Analysis. Revolve the horizontal projecting plane of the line 
 about an axis which cuts the line, and is parallel to its horizon- 
 
 FiG. 23. — a&^ and a'b'^ true length of AB 
 och and ocv angles with H and V. 
 
 tal projection, until the plane is parallel to H. The horizontal 
 projection of the line in this position is the true length of the 
 line. 
 
 The angle between the line in its revolved position and the 
 axis is equal to the angle which the line makes with H. The 
 horizontal projection of the angle in this position is the true 
 size of the angle. 
 
 Construction. Let ac and a'& be the projections of the axis" 
 which passes through A and is parallel to ah. After AB is re- 
 
PROBLEMS IN POINT, STRAIGHT LINE, AND PLANE 27 
 
 volved about AC as an axis, the horizontal projection of AB is 
 a&i. &&1 is equal to h'c'. The point A, being on the axis, is sta- 
 tionary. Then h^a is the true length of the line and the angle 
 h^ab is the true size of the angle which AB makes with H. 
 
 To find the angle which the line AB makes with V, revolve 
 the vertical projecting plane of the line about an axis AE which 
 passes through A and is parallel to the vertical projection of 
 AB. When this projecting plane is parallel with Y, the ver- 
 tical projection h'^a'!}' is the true size of the angle which the line 
 AB makes with V. a'h'2 is the true length of AB and is there- 
 fore equal to ab-^^. 
 
 29. Conversely, given the horizontal (or vertical) projection 
 of a line and the angle which it makes with H (or V) to find the 
 other projections of the line. 
 
 The student should make the construction for this problem. 
 
 Note. In Fig. 23, prove that the angle h-^ac is or is not equal 
 to the angle h'a'c'. Also prove that the angle h^a'e' is or is not 
 equal to the 'angle hae. 
 
 30. Problems. 
 
 1. Find the length of a straight line joining any two points in a 
 profile plane. Find the angles which this line makes with H and V. 
 
 2. From a point on a line, given by its projections, lay off along the 
 line a length of two units. 
 
 3. A square pyramid has its base in a horizontal plane. Find the 
 true length of the slant height and a lateral edge of the pyramid. 
 
 4. The horizontal projection of a line is two units long and makes 
 45° with V. If the line makes 30° with H, find its vertical projection. 
 
 5. The horizontal projection of a line is three units long and makes 
 30° with V. If the line is four units long, find its vertical projection. 
 
 6. A line is four units long and its horizontal projection makes 15° 
 with V. Find its vertical projection when one end of the line is two 
 units higher than the other. 
 
 7. Draw the projections of a line which makes 60° with H and is 
 oblique to V. 
 
 8. Draw the projections of a line which makes 45° with V and is 
 oblique to H. 
 
 9. Draw the projections of a line which is four units long, makes 45° 
 with H and is oblique to V. 
 
28 DESCRIPTIVE GEOMETRY 
 
 10. Draw the projections of a line which is four units long, makes 
 30° with V and is oblique to H. 
 
 11. Draw the projections of a line which makes 30" with H, is oblique 
 to V and has one end three units higher than the other. 
 
 12. Draw the projections of a line which makes 60° with V, is oblique 
 to H and has one end four units in front of the other. 
 
 13. Draw the projections of a line which lies in a profile plane, makes^ 
 60° with H and is four units long. What angle does this line make 
 with V? 
 
 14. A li" cube has its base parallel to H and a side face 30° to V. 
 Find the true length of a diagonal of the cube and the angle which it 
 makes with an edge of the cube. 
 
 15. Find the true length of the hip rafter MN in the roof shown in 
 Fig. 37, page 49. Also find the true length of the jack rafter OQ. 
 
 16. Find the true length of the edge MN in the ventilator pipe shown 
 in Fig. 36, page 49. 
 
 17. Find the true length of the edge MN in the chute shown in 
 Fig. 38, page 49. 
 
 31. Two parallel lines are given by their projections. Find 
 the angles which the plane of these lines makes with the planes 
 of projection. 
 
 Let AB and CE, Fig. 24, be the given lines. 
 
 Analysis. A line of the given plane vrhich is perpendicular 
 to a horizontal will make the same angle with H as the given 
 plane. A line of the given plane which is perpendicular to a 
 frontal will make the same angle with V as the given plane. 
 Therefore, draw these lines and find the angles which they 
 make with H and V respectively. These are the required angles. 
 
 Construction, h'h' and Kh are the projections of a horizontal 
 on the given plane (Art. 17). A line which is perpendicular to 
 a horizontal has its horizontal projection perpendicular to the 
 horizontal projection of the horizontal. In like manner a line 
 which is perpendicular to a frontal has its vertical projection 
 perpendicular to the vertical projection of the frontal. There- 
 fore draw ge perpendicular to hh. Then ge and g'e' are the pro- 
 jections of a line which is perpendicular to HH and lies on the 
 given plane, oc ^ is the angle which the line GE makes with H 
 
PROBLEMS IN POINT, STRAIGHT LINE, AND PLANE 
 
 29 
 
 (Art. 28), and is therefore the angle which the plane of the 
 given lines makes with H. 
 
 By taking a line of the plane which is perpendicular to the 
 frontal FF, the angle oc ^ which the plane makes with V is found. 
 
 Fig. 24. — och angle plane makes with H. 
 ocy angle plane makes with V. 
 
 32. Conversely, to draw the projections of two lines, the plane 
 of which makes a given angle with H or V. 
 
 Analysis. Draw the projections of a horizontal which is ob- 
 lique to V. Draw the projections of a line which is at right 
 angles to this horizontal and makes the required angle with H 
 (Art. 29). These are the required lines, the plane of which 
 makes the required angle with H. 
 
 Fig. 25 shows the construction for a plane which makes the 
 angle oc with H ; HH and AB being the required lines. 
 
 By using a frontal and a line at right angles to it making the 
 given angle with V, a plane is fixed which makes the required 
 angle with Y, Fig. 26. 
 
30 
 
 DESCRIPTIVE GEOMETRY 
 
 Other lines on these planes can be drawn by joining points in 
 one of the lines with points in the other line. 
 
 Fig. 2^.— Plane oz° with H. 
 
 Fig. 2%.— Plane oc° with y. 
 
 33. Problems. 
 
 1. Find the angles which the plane of two intersecting lines makes 
 with H and V. 
 
 2. Given two intersecting lines, one parallel to G. L. and the other 
 oblique to H and V, find the angles which the plane of the lines makes 
 with H and V. What is the relation between the angles? 
 
 3. Through a given point draw two lines the plane which makes 30° 
 with H. Does this plane make 60° with V? 
 
 4. Given two intersecting lines, one oblique to G. L. and the other a 
 line of profile, find the angles which the plane of the lines makes with 
 H and V. 
 
 5. Draw the projections of (a) an equilateral triangle, (b) square, 
 (c) regular hexagon, when the plane of the triangle makes (a) 30°, 
 (b) 45°, (c) 60° with H and is oblique to V. 
 
 6. Two intersecting lines are given by their projections. Draw the 
 projections of an equilateral triangle lying in the plane of the two lines 
 so that one side of the triangle makes 30° with a horizontal of the plane. 
 
 7. Two parallel lines of profile lie in separate profile planes. The 
 plane of the lines is to be oblique to the ground line. Find the angles 
 which this plane makes with H and V. 
 
PROBLEMS IN POINT, STRAIGHT LINE, AND PLANE 31 
 
 8. A line AB makes 45" with H and is oblique to V. Represent a 
 plane which contains the line and mak:es 60° with H. (Use a right 
 circular cone with vertex on AB, base parallel to H and elements 60" 
 with H). 
 
 34. If an oblique plane is given by two intersecting straight 
 lines, the angles between the projections of these lines are not 
 usually the true size of the angles between the lines. If the 
 plane of the lines is revolved until it is parallel to one of the 
 planes of projection, then the angles between the projections of 
 the lines on that plane are the true size of the angles between 
 the lines. 
 
 If the projections of a plane figure such as a square or a 
 triangle are given, it is necessary to revolve the plane of the 
 figure until it is parallel to one of the planes of projection be- 
 fore its projection on that plane is the true size of the figure. 
 Conversely, to find the projections of a figure which lies on a 
 given plane, it is necessary to revolve the plane until it is par- 
 allel to one of the planes of projection. While the plane is in 
 this position, construct the true size of the figure on it. The 
 plane and the figure must then be counter revolved to the origi- 
 nal position of the plane and the projections of the figure in this 
 position found. 
 
 35. Two intersecting lines axe given by their projections. 
 Find the true size of the angle between them. 
 
 Let AO and BO, Fig. 27, be the given lines. 
 
 Analysis. Draw a horizontal in the plane of the given lines. 
 Revolve the given lines about the horizontal as an axis until the 
 plane of the lines is parallel to H. The horizontal projection 
 of the angle in this position is the true size of the angle between 
 the lines since they do not change their relative position dur- 
 ing the revolution. 
 
 Construction, h'h' is the vertical and lih the horizontal pro- 
 jection of a horizontal in the plane of AO and BO (Art. 17). 
 When the given lines are revolved into a horizontal position 
 about HH as an axis, their point of intersection moves to Og 
 (Art. 23). The points X and Y on the axis do not move. Then 
 
32 
 
 DESCRIPTIVE GEOMETRY 
 
 xOzV is the true size of the angle between the lines AO and BO. 
 
 The problem can be solved by revolving the plane of the 
 angle until it is parallel to V about a frontal as an axis. 
 
 36. To bisect an angle. When the angle is shown in its true 
 size, the bisector can be drawn. The bisector can then be re- 
 volved back until the plane of the angle is in its original posi- 
 
 FiG. 27. — xo^y angle tetween AO and BO. 
 
 tion and the projections of the bisector found. In general, the 
 projections of the bisector will not bisect the projections of the 
 angle. 
 
 The projection of an angle can be larger than, equal to, or 
 smaller than the angle itself. The projection of a right angle 
 is a right angle when one side of the angle is parallel to the 
 plane upon which the projection is made. 
 
 37. Problems. 
 
 1. Find the angle between two intersecting lines by using a frontal 
 as an axis. 
 
 2. Find the angle between two intersecting lines when one is oblique 
 to H and V and the other is parallel to H and makes 30° to V. 
 
 3. Find the angle between two intersecting lines when one is oblique 
 to H and V and the other is parallel to the ground line. 
 
PROBLEMS IN POINT, STRAIGHT LINE, AND PLANE 33 
 
 4. Draw the projections of the bisectors of the angles of a given tri- 
 angle. 
 
 5. Draw the projections of a triangular pyramid with base parallel 
 to H. Find the true size of the angle between any two lateral edges. 
 
 6. Given two intersecting lines AB and CD, AB parallel to H and 45'' 
 to V and CD parallel to V and 30° to H, draw the projections of the 
 bisector of the angle between them. 
 
 7. A triangle lies in a plane which is perpendicular to V and 30° 
 to H. Find the true size of the triangle by using a frontal as an axis. 
 
 8. The plane of two intersecting lines is perpendicular to H and is 
 oblique to V. Find the true size of the angle between the lines by using 
 a frontal as an axis. 
 
 9. Find the angle between two intersecting lines when one is oblique 
 to H and V and one lies in a profile plane. 
 
 10. Given both projections of three corners of a quadrilateral and the 
 horizontal projection of the fourth comer, draw the projections of the. 
 quadrilateral and find its true size and shape. 
 
 11. Letter two points on the projections of the bisector of the angle 
 between two lines which lie in a profile plane. 
 
 12. Find the true size of the angle NMP in the chute shown in 
 Fig. 38, page 49. 
 
 38. A point and a line are given by their projections. Draw 
 the projections of a line which passes through the point and 
 makes a given angle with the given line. 
 
 Let P be the given point, oc the given angle, and AB the given 
 line, Fig. 28. 
 
 Analysis. Eevolve the plane of the point and line about a 
 horizontal until it is parallel to H. From the revolved position 
 of the point, draw a line which makes the angle oc with the 
 revolved position of the given line. When this line is revolved 
 back with the plane to the original position of the plane, it will 
 be the required line. 
 
 Construction, h'h' is the vertical projection and hh is the hor- 
 izontal projection of a horizontal which lies on the plane of P 
 and AB and cuts AB at the point C (Art. 17). When the plane 
 is revolved about HH as an axis, B moves to Bg and P and 
 remain fixed, being on the axis. ftgC is the horizontal projection 
 of the revolved position of AB. Through p draw pe^, making 
 the required angle oc with h^c e and e' are the projections of 
 
34 
 
 DESCRIPTIVE GEOMETRY 
 
 the point E after the counter revolution of the plane. Then pe 
 is the horizontal and p'e' is the vertical projection of the re- 
 quired line. 
 
 Fig. 2S—PE oc° loith AB. 
 
 39. Problems. 
 
 1. Solve the above problem when oc is (a) 30°, (b) 45**, (c) 60°, 
 (d) 75^ 
 
 2. Draw the projections of a line which is the shortest distance from 
 a given point to a given oblique line. 
 
 3. Draw the projections of a line which is the shortest distance from 
 a given point to a given horizontal line. 
 
 4. Draw the projections of a line which is the shortest distance from 
 a given point to a given line parallel to V. 
 
 5. Draw the projections of a line which is the shortest distance from 
 a given point to a given line of profile. 
 
 6. Find the projections of a point which is three units from a given 
 point P and lies on a given line AB. 
 
 7. Find the distance between two parallel lines which are oblique to 
 H and V. 
 
PROBLEMS IN POINT, STRAIGHT LINE, AND PLANE 35 
 
 8. A point and an oblique line are given by their projections. Draw 
 the projections of an (a) equilateral triangle, (b) square, (c) regular 
 hexagon with center at the given point and side along the given line. 
 
 9. Draw the projections of a line which passes through a given point 
 P and makes (a) 30°, (b) 45°, (c) 60°, (d) 75° with a given line of 
 profile. 
 
 10. A line AB parallel to the ground line and a point P are given by 
 their projections. A 2" square lies on the plane of the point and line. 
 Draw the projections of the square when one corner is on AB and one 
 side passes through P and makes 30° with AB. 
 
 11. Find the distance from the point P to the line MN in the chute 
 shown in Fig. 38, page 49. 
 
 40. A point and two lines are given by their projections. 
 Represent a plane which contains the given point and is parallel 
 to the given lines. 
 
 Analysis. Through the given point draw a line parallel to 
 each of the given lines. The plane of these two lines is the re- 
 quired plane. 
 
 Let the construction be made in accordance with the above 
 analysis. 
 
 41. Problems. 
 
 1. Represent a plane which contains a given line and is parallel to 
 another given line. 
 
 2. Represent a plane which contains a given point and is parallel 
 to G. L. 
 
 3. Two lines parallel to G. L. are given by their projections. Rep- 
 resent a plane which contains a given point and is parallel to the 
 given lines. 
 
 4. Two lines are given by their projections, one is oblique to G. L. 
 and the other is a line of profile. Represent a plane which contains a 
 given point and is parallel to the given lines. 
 
 5. Represent a plane which contains a given point and passes at equal 
 distances from two other given points. 
 
 6. Represent a plane which contains a given line and passes at equal 
 distances from two given points. 
 
 7. Represent a plane which contains a given point and passes at 
 equal distances from three other given points. 
 
36 
 
 DESCRIPTIVE GEOMETRY 
 
 42. A plane is represented by two of its lines. Find the point 
 in which a given oblique line pierces this plane. 
 
 This problem should be thoroughly mastered. It is used in' 
 finding the plane sections of all ruled surfaces, the intersections 
 of surfaces with plane faces and indirectly to find the intersec- 
 tions of such curved surfaces as cylinders and cones. 
 
 Let AB and AC, Fig. 29, be the lines which represent the 
 plane and MN the given oblique line. 
 
 jrr ^b 
 
 Fig. 29. — MN pierces plane of AB and AC at P. 
 
 Analysis. Find the points where the lines which represent 
 the plane pierce the horizontal or vertical projecting plane of 
 the given oblique line. The line joining these two points is the 
 line of intersection of the given plane with the projecting plane 
 of the oblique line. This line of intersection cuts the given 
 line in the required point. 
 
 Construction. AB pierces the horizontal projecting plane of 
 MN at Y and AC pierces it at X. x'y\ the vertical projection 
 of the line joining these two points, intersects mV at p\ the 
 vertical projection of the required point; p is its horizontal 
 projection. 
 
 43. Problems. 
 
 1. Find where a line pierces the plane of two parallel lines. 
 
 2. Find where a line pierces the plane of a point and a line. 
 
PROBLEMS IN POINT, STRAIGHT LINE, AND PLANE 37 
 
 3. Find where a line pierces the plane of three points. 
 
 4. Find the point where a line which is parallel to G. L. pierces a 
 .plane which is oblique to H and V. 
 
 5. Find the point where a line which is parallel to H and oblique to 
 V pierces a plane which is parallel to G. L, 
 
 6. rfnd the point where a line which lies in a profile plane pierces a 
 plane which is parallel to G. L. 
 
 7. Two planes are each represented by two of their lines. Find the 
 length of the part of an oblique line included between the two planes. 
 
 8. An oblique triangular pyramid with its base parallel to H is given 
 by its projections. Find the intersection of this pyramid with a given 
 oblique plane. 
 
 9. Find the intersection of a given oblique line with a plane which 
 makes 60° with H and is oblique to V. 
 
 10. A line makes 45° with H and is oblique to V. Find the intersec- 
 tion of this line with a given oblique plane. 
 
 11. Draw the projections of a line which contains a given point and 
 touches any other two non-intersecting lines. 
 
 12. Draw the projections of a line whick contains a given point, is 
 parallel to a given plane, and touches another given line. 
 
 13. The same as problem 12 when the given line is a line of profile. 
 
 14. Given three lines, no two of which lie in the same plane, draw the 
 projections of a line which touches two of them and is parallel to the 
 third. 
 
 15. Given three lines, no two of which lie in the same plane, find the 
 projections of a line which touches all three' of them. 
 
 16. Given the projections of an oblique line and a line of profile, draw 
 the projections of a line which touches the given lines and is parallel 
 to the ground line. 
 
 44. Two planes are each represented by two of their lines. 
 Find the line of intersection of the two planes. 
 
 First Method. Let AB and BC, Fig. 30, be the lines of one 
 plane and MN and OP the lines of the other plane. 
 
 Analysis. Find the points where the two lines of one plane 
 pierce the other plane. The line joining these two points is the 
 required line of intersection of the two planes. 
 
 Construction. MN pierces the plane of AB and BC at the 
 point K (Art. 42). OP pierces the plane of AB and BC at the 
 point G. Then ¥g' and kg are the vertical and horizontal pro- 
 jections of the required line of intersection of the two planes. 
 
38 
 
 DESCRIPTIVE GEOMETRY 
 
 Second Method. Let OA and OB, Fig. 31, be the lines of one 
 plane and CD and CE the lines of the other plane. 
 
 .0 
 
 Fig. 30. — KG line of intersection of two planes. 
 Analysis. Any auxiliary plane will cut lines from the given 
 planes and these lines will intersect in a point which lies on the 
 
 Fig. 31. — KG line of intersection of two planes. 
 
 required line of intersection. A second auxiliary plane will 
 determine another point on the line of intersection. The straight 
 
PROBLEMS IN POINT, STRAIGHT LINE, AND PLANE 39 
 
 line joining the two points thus found is the required line of 
 intersection of the given planes. 
 
 The construction is much clearer when the auxiliary planes 
 are taken parallel to each other and perpendicular to one of the 
 planes of projection. 
 
 Construction. Draw ss through o and c and draw s^s^ par- 
 allel to 55 and cutting the lines of the given planes. Since the 
 plane S^ is perpendicular to H it cuts the line PQ from the 
 plane of OA and OB and the line MN from the plane of CD and 
 CE. These lines intersect at K, a point of the required inter- 
 section. Since the plane S is parallel to the plane Si, the line 
 which it cuts from the plane of- OA and OB will be parallel to 
 the line PQ. o'g\ parallel to p'q', is its vertical projection. 
 Likewise, the line which the plane S cuts from the plane of CD 
 and CE will be parallel to MN. c'g' is its vertical projection. 
 g and g' are the projections of the point of intersection of these 
 two lines. Then kg and k'g' are the horizontal and vertical pro- 
 jections of the required line of intersection of the given planes. 
 
 45. Problems. 
 
 1. Find the line of intersection of two planes when one is oblique to 
 H and V and the other parallel to H. 
 
 2. Find the line of intersection of two planes when one is oblique to 
 H and V and the other parallel to V. 
 
 3. Find the line of intersection of two planes when one is oblique to 
 the ground line and one parallel to the ground line. 
 
 4. Find the line of intersection of two planes when one is oblique to 
 H and V and the other perpendicular to H and oblique to V. 
 
 5. Find the line of intersection of two planes when each is given by 
 two parallel lines. 
 
 6. Find the line of intersection of two planes, each given by two, lines, 
 when all the lines pass through a given point P. 
 
 7. Find the line of intersection of two planes when they are both 
 parallel to the ground line. 
 
 8. Find the line of intersection of two oblique planes when a line of 
 one of the planes is a line of profile. 
 
 9. Find the line of intersection of an oblique plane and a profile plane. 
 (Use the second method.) 
 
 10. A plane is given by two parallel lines which make 30° with H and 
 are oblique to V and another plane is given by two parallel lines which 
 
40 DESCRIPTIVE GEOMETRY 
 
 make 45° with V and are oblique to H. Find the line of intersection of 
 the planes. 
 
 11. A plane makes 45° with H and is oblique to V and another plane 
 makes 60° with V and is oblique to H. Find the line of intersection of 
 the planes. 
 
 12. Two oblique lines are given by their projections. These are lines 
 of greatest slope of two planes. Find the line of intersection of the 
 planes. 
 
 46. A line perpendicular to a plane. 
 
 A line which is perpendicular to a plane is perpendicular to 
 every line of the plane and is, therefore, perpendicular to all 
 horizontals and frontals of the plane. The horizontal projec- 
 tion of a perpendicular to a plane is at right angles to the hor- 
 izontal projection of any of its horizontals (Art. 36). Like- 
 wise, the vertical projection of a perpendicular to a plane is at 
 right angles to the vertical projection of any of its frontals. 
 Therefore, to draw the projections of a perpendicular to a 
 plane, draw the projections of a horizontal and a frontal of the 
 plane. Then the horizontal projection of the perpendicular to 
 the plane is perpendicular to the horizontal projection of this 
 horizontal and its vertical projection is perpendicular to the 
 vertical projection of this frontal. 
 
 47. Given the projections of two lines of a plane and the pro- 
 jections of a point in space, find the distance from the point to 
 the plane. 
 
 Let AB and CG, Fig. 32, be the lines which represent the 
 plane and P the given point. 
 
 Analysis. Draw a perpendicular from the given point to the 
 given plane (Art. 46), and find where it pierces the plane (Art. 
 42). The length of the perpendicular from the given point to 
 the piercing point is the required distance. 
 
 Construction. HH is a horizontal and FF a frontal of the 
 given plane (Art. 17). pe, perpendicular to hhf is the horizon- 
 tal projection of the perpendicular and p'e\ perpendicular to 
 f'f\ is its vertical projection (Art. 46). The perpendicular PE 
 pierces the plane of AB and CG at E (Art. 42). P^E, the true 
 length of PE, is the required distance. 
 
PROBLEMS IN POINT, STRAIGHT LINE, AND PLANE 
 
 41 
 
 48. Conversely, to find the projections of a point which is a 
 given distance from a given plane. 
 
 Analysis. Take any point on the plane and at this point 
 erect a perpendicular of indefinite length to the plane. Select 
 any other point of the perpendicular and find the true length 
 
 FiQ. 32. — p e distance from P to plane of ABCO. 
 
 of the perpendicular from this point to the plane. Locate a 
 point on the true length of the perpendicular which is the re- 
 quired distance from the point on the plane. Counter revolve 
 the point thus found to its position on the perpendicular. This 
 is the required point. 
 
 Let the construction be made in accordance with the above 
 analysis. 
 
 This problem is used to locate the vertex of a right cone or 
 pyramid or the upper comers of a rectangular object when the 
 base of the object is on an oblique plane. 
 
42 DESCRIPTIVE GEOMETRY 
 
 49. Problems. 
 
 1. Find the distance from a point to the plane of three points. 
 
 2. Find the distance from a point to the plane of two parallel lines. 
 
 3. Find the distance from a point to a plane which is parallel to G. L. 
 
 4. A regular triangular pyramid has its base parallel to H, side of 
 base 3'', altitude i'\ Find the distance from one corner of the base to 
 the plane of the opposite face. 
 
 5. Find the projections of a point which is two units from a given 
 plane. 
 
 6. Find a plane which is parallel to and two units from a given oblique 
 plane. 
 
 7. Given the projections of any four points A, B, C, D. These points 
 are the corners of a tetrahedron. Find the true length of the altitude 
 of the tetrahedron when (a) ABC, (b) ABD, (c) AOD, (d) BOD, is the 
 
 8. Find the distance between two given parallel planes. 
 
 9. Represent a plane which contains a given line and is perpendicular 
 to a given plane. Find the line of intersection of the two planes. 
 
 10. Represent a plane which makes 30'* with H and is oblique to V. 
 Draw the projections of a perpendicular to this plane and find the angle 
 which it makes with H. 
 
 11. Find the locus of all points which are equidistant from three given 
 points. 
 
 12. A plane is represented by two of its lines, and three points are 
 given by their projections. Find a point on the plane which is equidis- 
 tant from the three given points. 
 
 50. Given the projections of a point and of an oblique line, 
 represent a plane which contains the point and is perpendicular 
 to the line. 
 
 Analysis. Since the plane is to be perpendicular to the line, 
 a horizontal of the plane will have its horizontal projection per- 
 pendicular to the horizontal projection of the line. Likewise, a 
 frontal of the plane will have its vertical projection perpendic- 
 ular to the vertical projection of the line. Therefore to rep- 
 resent the plane, draw the projections of a horizontal and of a 
 frontal which pass through the given point and are perpendic- 
 ular to the given line. These two lines will represent the re- 
 quired plane. 
 
43 
 
 Let the construction be made in accordance with the above 
 analysis. 
 
 61. Problems. 
 
 1. A given oblique line is the edge of a cube. Represent by two lines 
 the plane of the base of the cube. 
 
 2. A given oblique line is the axis of a right pyramid. Represent by 
 two lines the plane of the base of the pyramid. 
 
 3. A point and an oblique line are given by their projections. Repre- 
 sent the plane of the circle in which the point moves when it is re- 
 volved about the line as an axis. 
 
 4. A given line AB is the base of an isosceles triangle which has its 
 vertex in another given line OE. Draw the projections of the triangle. 
 
 5. Represent a plane which is twice as far from a given point as it is 
 from a given plane. 
 
 6. Given the two projections of one side of a square and the direction 
 of the horizontal projection of an adjacent side. Draw the projections 
 of the square. 
 
 7. Given the two projections of one side of a square and the horizontal 
 projection of a line which contains the opposite side. Draw the pro- 
 jections of the square. 
 
 62. Given the projections of two lines of a plane and the pro- 
 jections of a line in space, find the projection of the line on the 
 given plane. 
 
 Analysis. From any two points of the line,. erect perpendic- 
 ulars to the given plane. A line joining the points in which 
 these perpendiculars pierce the plane will be the required pro- 
 jection. The point in which the given line pierces the given 
 plane is also a point on the projection of the line on the plane. 
 
 Let the constructon be made in accordance with the above 
 analysis. 
 
 63. Problems. 
 
 1. Find the projection of a line which is parallel to H and oblique to 
 V on a plane which is oblique to the ground line. 
 
 2. Find the projection of a line which is parallel to V and oblique to 
 H on a plane which is oblique to the ground line. 
 
 3. The horizontal and vertical projections of a line are parallel re- 
 spectively to a horizontal and frontal of a given plane. Find the pro- 
 jection of the line on the plane. 
 
44 DESCRIPTIVE GEOMETRY 
 
 4. Find the projection of a line which is parallel to the ground line on 
 a given oblique plane. 
 
 5. Find the projection of a line which is oblique to H and V on a plane 
 which is parallel to the ground line. 
 
 6. Find the projection of a line which lies in a profile plane on a 
 plane which is oblique to the ground line. 
 
 7. A line 3'' long is oblique to H and V. Find the length of its pro- 
 jection on a plane which is oblique to H and V. 
 
 8. A regular triangular pyramid has its base parallel to H. Find the 
 length of the projection of one of its lateral edges on the plane of the 
 opposite face'. 
 
 9. Project a given triangle ABC upon a given plane MNO, and find 
 the true size of this new triangle. 
 
 10. Given a plane MNO and two non-intersecting lines AB and CE 
 which are oblique to the plane. Draw the projections of a line which 
 is perpendicular to the plane and touches AB and CE. 
 
 11. Given a plane MNO and a line AB which is oblique to the plane. 
 Draw the projections of a line CE which lies on the plane MNO and is 
 perpendicular to AB. 
 
 54. Given the projections of two lines of a plane and the pro- 
 jections of a line in space, find the angle which the line makes 
 with the given plane. 
 
 Analysis. The angle which a line makes with a given plane 
 is understood to be the angle which the line makes with its 
 projection on that plane. If a perpendicular be dropped to the 
 plane from any point in the given line, the angle between this 
 perpendicular and the given line is the complement of the angle 
 which the line makes with the plane. Therefore, find the angle 
 between the given line and a perpendicular to the plane from 
 any point of the line and construct its complement. This com- 
 plement is the required angle. 
 
 Let the construction be made in accordance with the above 
 analysis. 
 
 55. Problems. 
 
 1. Find the angle which a line parallel to the ground line makes with 
 a plane which is oblique to the ground line. 
 
 2. Find the angle which a line parallel to H and oblique to V makes 
 with a plane which is oblique to the ground line. 
 
PROBLEMS IN POINT, STRAIGHT LINE, AND PLANE 45 
 
 3. Find the angle which a line parallel to V and oblique to H makes 
 with a plane which is oblique to the ground line. 
 
 4. The projections of a line are parallel respectively to a horizontal 
 and a frontal of a given plane. Find the angle which the line makes 
 with the plane. 
 
 5. Find the angle which a line oblique to H and V makes with a 
 plane parallel to the ground line. 
 
 6. Find the angle which an oblique line makes with a profile plane. 
 
 7. Find the angle which a line of profile makes with a plane parallel 
 to the ground line. 
 
 8. Find the angle which a line in a profile plane makes with a given 
 oblique plane. 
 
 9. Draw the projections of a line which passes through a given point 
 and makes 30° with a given oblique plane. 
 
 10. Represent a plane which contains a given point on a given line 
 and makes a given angle with the line. , 
 
 11. Represent a plane which contains a given point P, is perpendic- 
 ular to H, and makes a given angle with a given oblique line. 
 
 56. Two planes are each given by the projections of two of 
 their lines, find the angle between the planes. 
 
 Analysis. From any point in space drop a perpendicular to 
 each plane (Art. 46). The angle between these perpendiculars 
 (Art. 35), is the angle between the planes. 
 
 Let the construction be made in accordance with the above 
 analysis. 
 
 57. Problems. 
 
 1. Find the angle between two planes when one is oblique to H and V 
 and one perpendicular to H and oblique to V. 
 
 2. Find the angle between two planes when one is oblique to H and V 
 and one perpendicular to V and oblique to H. 
 
 3. Find the angle between two planes when one is oblique to H and 
 V and one parallel to H. 
 
 4. Find the angle between two planes when one is oblique to the 
 ground line and one parallel to the ground line. 
 
 5. Find the angle between two planes when they are both parallel to 
 the ground line. 
 
 6. Find the angle between a profile plane and a plane which is oblique 
 to H and V. 
 
 7. A plane is represented by a line of profile and a point. Find the 
 angle between this plane and a profile plane. 
 
46 DESCRIPTIVE GEOMETRY 
 
 8. A regular triangular pyramid has its base parallel to H. Find the 
 true size of the angle between two lateral faces. 
 
 9. In the pyramid of problem 8, find the true size of the angle between 
 the base and one of the lateral faces. 
 
 10. Given an oblique line AB and two points C and E, find the angle 
 between the planes CAB and EAB. 
 
 11. Given a plane parallel to the ground line and a point; represent 
 a plane which contains the point, is parallel to the ground line, and 
 makes 60° with the given plane. 
 
 12. An oblique plane is given by two oblique lines AB and AC. Rep- 
 resent a plane which contains AB and makes 45° with the plane ABC. 
 
 13. A line AB makes 30° with H and is oblique to V. Represent a 
 plane ABC which makes 45° with H and a plane ABE which makes 60° 
 with the plane ABC. 
 
 14. Find the angle between the faces A and B of the lamp shade shown 
 in Fig. 35, page 49. 
 
 15. Find the angle between the faces A and D of the ventilator pipe 
 shown in Fig. 36, page 49. 
 
 16. Find the angle between the bottom C and the side A in the chute 
 shown in Fig. 38, page 49. 
 
 58. Two oblique lines which do not lie in the same plane are 
 given by their projections, find the projections and the true 
 length of their common perpendicular. 
 
 Let AB and CK, Fig. 33, be the given lines. 
 
 Analysis. Pass a plane through one line parallel to the other 
 and project the second line on this plane. This projection will 
 be parallel to the line itself, since the line is parallel to the 
 plane. Where this projection cuts the first line, erect a per- 
 pendicular to the plane. This is the common perpendicular to 
 the two lines. 
 
 Construction. Through any point P of the line AB, draw EG 
 parallel to CK. Project any point of CK on the plane of AB 
 and EG. M is the projection of this point and MN, parallel to 
 CG, is the projection of CG on this plane. MN intersects AB 
 at X, and XY, perpendicular to the plane, is the required line. 
 x^y is the true length of this perpendicular. 
 
 59. Second method, when one of the lines is parallel to one of 
 the planes of projection. 
 
PROBLEMS IN POINT, STRAIGHT LINE, AND PLANE 
 
 47 
 
 Let AB and HH, Fig. 34, be the given lines. 
 
 Analysis. Draw the projections of the two given lines on an 
 auxiliary plane which is perpendicular to one of the planes of 
 projection and also perpendicular to one of the given lines. In 
 
 FiQ. 33. — XY xommon perpendicularFm. 34. — XY common perpendicular 
 to AB and CK, to AB and HH. 
 
 this auxiliary view, one of the given lines appears as a point, 
 the other as a straight line. A perpendicular from the point 
 to this line is the auxiliary view of the common perpendicular 
 to the two given lines and is the true length of the perpendic- 
 ular. Draw this view of the common perpendicular first and 
 then by projection get the horizontal and vertical views. 
 
 Construction. In Fig. 34, S is the auxiliary plane which is per- 
 pendicular to H and also to the line HH. When this plane is 
 turned over on a level with HH, y^ is the auxiliary view of HH 
 and o-i&i the auxiliary view of AB. x^y^, perpendicular to 
 difei, is the auxiliary view of the common perpendicular to the 
 two given lines and is its true length, x is the horizontal view 
 of the point X and xy, perpendicular to Kk, is the horizontal 
 
48 DESCRIPTIVE GEOMETRY 
 
 view and x'y' is the vertical view of the required common per- 
 pendicular to the lines AB and HH. 
 
 60. Problems. 
 
 1. Find the common perpendicular to two lines when one is oblique 
 to H and V and the other is parallel to H and oblique to V. 
 
 2. Find the common perpendicular to two lines when one is oblique 
 to H and V and the other is parallel to V and oblique to H. 
 
 3. Find the common perpendicular to two lines when one is parallel 
 to H and oblique to V and the other is parallel to V and oblique to H. 
 
 4. Find the common perpendicular to two lines when one of them is 
 parallel to the ground line. 
 
 5. Find the common perpendicular to two lines when one of them lies 
 In a profile plane. 
 
 6. Find the common perpendicular to two oblique lines by using a 
 plane perpendicular to one of the lines and projecting the other line on 
 this plane. 
 
 7. Given the projections of four points which do not lie in one plane, 
 find the projections of a fifth point which is equidistant from the given 
 points. 
 
 8. Represent a plane which is parallel to and equidistant from two 
 given lines. 
 
 9. Given the H and V projections of a line AB, the H projection cd of 
 another line and the H projection xy of the common perpendicular to 
 the two lines; find the V projections of CD and XY. 
 
 61. Auxiliary planes of projection. The use of an auxiliary 
 plane of projection such as was used in Art. 59 is eommon in 
 commercial practice. The position of such a plane is deter- 
 mined by the position of the object in space. The plane is 
 placed so that the projection of the object, or part of the object, 
 upon it best brings out the particular features under considera- 
 tion. These auxiliary planes are usually taken at right angles 
 to one of the principal planes of projection. 
 
 Figs. 35, 36, 37, 38, also show the use of an auxiliary plane of 
 projection. Thus in showing the true size and shape of the 
 face of the lamp. Fig. 35, the plane S is taken perpendicular to 
 V and parallel to the face A of the lamp. Then, when the face 
 A has been projected upon this plane, the plane is revolved 
 until it is parallel to V and the face of the lamp is shown in its 
 true size and shape. 
 
PROBLEMS IN POINT, STRAIGHT LINE, AND PLANE 49 
 
 Lamp shade. 
 
 
 \ 
 
 
 <t^ 
 
 
 
 p 
 
 \ 
 
 1 
 
 
 y/ 
 
 
 
 
 \ 
 
 60' 
 
 ^ 
 
 
 r,% 
 
 Hip roof 
 
 F^G. 35. — Lam'^ shade. 
 Fig. 37.— Hip roo/. 
 
 Fig. 36. — Ventilator pipe. 
 Fig. ZS.— Chute, 
 
50 DESCRIPTIVE GEOMETRY 
 
 PROBLEMS 
 
 The following problems can be solved in rectangles 10"xl4". 
 
 The ground line is to be parallel to the shorter sides of the 
 rectangle. 
 
 The profile plane P is in the center of the rectangle. 
 
 Distances are given in inches. 
 
 Points are located by giving their perpendicular distances 
 from P, V, and H. 
 
 The first distance is from P. 
 
 The second distance is from V. 
 
 The third distance is from H. 
 
 Distances are measured to the right of P, in front of V and 
 above H unless preceded by a minus sign. 
 
 1. Draw the projections of the bisectors of the angles of the triangle 
 A(— If, — f, -.2i), B(— J, — 3f, -4), C(2i, —2, ~5i). 
 
 2. The horizontal projection of a parallelogram ABCD is a 2^" square. 
 Draw the vertical projection of the parallelogram when its center is 1" 
 above, and the corner B i" below the corner A. Find the true size of 
 the parallelogram. 
 
 3. Given a point and two planes. Draw the projections of a line 
 which passes through the point and is parallel to both planes. 
 
 4. Given a plane ABC and a point P. . Draw a line through P so that 
 it will be parallel to ABC and have its projections parallel to each other. 
 
 5. The horizontal projections of two points are V apart. The ver- 
 tical projections of the same points are 2" apart. What is the greatest 
 and what is the least distance apart that the points can be placed? 
 
 6. Two intersecting lines AB and AC represent a plane which makes 
 60** with H. , AB makes 45'^ and AC 30° with H. Find the true size of 
 the angle BAG. 
 
 7. Given two points A and B, draw the projections of the locus of a 
 point which is Z" from A and 4'' from B. 
 
 8. Find a point P on a given line MN which is equidistant from two 
 given planes ABC and DEF. 
 
 9. Find the point O which is equidistant from the points A, B, C, 
 and D. 
 
PROBLEMS 51 
 
 10. Represent a plane which is equidistant from a point A and a line 
 BC and is parallel to another line DE. 
 
 11. Represent a plane which is equidistant from any three points A, 
 B, C and is parallel to a given line DE. 
 
 12. Through any three points A, B, C, take three planes which are 
 parallel and equidistant. 
 
 13. Through any three points A, B, C, take three planes which are 
 parallel and equidistant and such that one of the planes passes through 
 a fourth given point P. 
 
 14. Through any four points A, B, C, D, take four planes which are 
 parallel and equidistant. 
 
 15. Through any two points, A and B, and through any given line CD, 
 take three planes which are parallel and equidistant. 
 
 16. Find the line which will touch the three lines: A(— 2i, —3, —1) 
 B(— f. —h — 5), C(i, — 1, — 4) D(2i, —1, — i) and E(— i, — U, —4 J) 
 F(l. —i-l). 
 
 17. Through a given point take a line which touches any other two 
 non-intersecting lines. 
 
 18. Through three given lines not situated two by two in planes, take 
 three planes which have a common line of intersection. 
 
 19. Given a line AB, a point P and a distance d. Represent a plane 
 which contains AB and is the distance d from the point P. 
 
 20. Given the horizontal projection of an equilateral triangle, the ver- 
 tical projection of one corner and the true length of a side, find the ver- 
 tical projection of the triangle. 
 
 21. Given two points A and B, a plane EFG and two distances d and 
 d^; construct a triangle ABC such that AC=d, BC=d and the point C 
 lies in the plane EFG. 
 
 22. Draw a line MN which intersects two given lines AB and CD and 
 is parallel to and 1" from a given plane EFG. 
 
 23. Draw the projections of the triangle A(li, 0, — 2^) B( — 1, x, — 2) 
 C(y, z, 0), the distances x, y, and z being unknown, when AB=2|'', 
 BC=3'', and AC=3r'. 
 
 24. Draw the projections of the center and diameter of a sphere which 
 is tangent to H and V and has its center on the line A( — 2, — 2, 0) 
 B(2, 0,— 3). 
 
 25. A right square pyramid having an altitude of 2" has its axis in 
 G. L. and the sides of the base parallel to H or V. The lateral edges of 
 the pyramid make 30° with H and 45" with V. Show a profile projec- 
 tion of the pyramid. 
 
 26. At a point of outcrop 0(0, — 2, — 1) the strike of a body of ore is 
 north 45° east. The "dip" (the inclination of the ore body to the hori- 
 zontal) is 60°. A tunnel is driven at P(— li, —ih —2) south 30° west 
 
52 DESCRIPTIVE GEOMETRY 
 
 on a rising 10% grade. How far must the tunnel be driven to reach the 
 ore body? 
 
 Note.— The "strike" means a horizontal line in the plane of the ore 
 body. 
 
 27. Points on the outcrop of an ore vein have elevations of A 360', 
 B 480', and C 320'. From A to B is 300' due N., from B to C, 400' due 
 N. B. Find the strike and dip of the vein. 
 
 28. Two ore veins have a strike due N. E. and a dip of SO"* S. E. The 
 strike lines are 200' apart. What is the shortest distance between the 
 veins? 
 
 29. From the point of outcrop A, run south 30° west 100' to a drill- 
 hole B. The drillhole C is 100' from both A and B. The drill strikes 
 ore at B at a depth of 25' and at C at a depth of 50'. Determine the dip 
 and strike of the vein. 
 
 30. Points A, B, and C are upon a sidehill which is a plane dipping 
 west 30°. A is a point of outcrop of the vein. FYom A, B bears south 
 45° west 350' horizontally, while C bears south 15° east 700' on the slope 
 of the hill. At B and C, drillholes driven perpendicular to the surface 
 each encounter the vein at 175'. Determine the strike and dip of the 
 vein. 
 
 31. A horizontal tunnel cuts a vein at 1000' from the portal. The 
 vein strikes north 45° west and dips 75° to the southwest. The tunnel 
 is driven north 30° east. What will be the horizontal distance at which 
 a tunnel on a 4% grade, driven from the same point and north 15° 
 west, will cut the vein? 
 
 32. From A, B bears north 30° east 1500'; from A, C bears south 60° 
 east 1200'. At A, a vertical borehole cuts a vein at 290'; at B, an inclined 
 borehole bears north 75° east, dips 75° and cuts the vein at 515'; at C, 
 a vertical hole cuts the vein at 275'. Elevation of A, 9800'; B, 9600'; 
 C, 9900^. Required the strike and dip of the vein. 
 
 33. Given three oil wells, A, B, C: A, 550' above sea level, 1450' deep; 
 B, 325' above sea level, 975' deep; C, 425' above sea level, 1100' deep. 
 Find the depth of a well to be sunk to the same oil vein at D which is 
 613' above sea level. 
 
 34. The shaft of a mine runs N. E. from a point A whose elevation is 
 150'. The slope of the shaft from the vertical is 3 to 12. It is desired 
 to run a tunnel to meet this shaft, starting at a point B 1000' from the 
 shaft entrance, N. 15° west from it, at an elevation of 100'. What is 
 the shortest tunnel that can be run? At what grade and in what di- 
 rection from B? 
 
 35. Given the points A and B as in problem 34. A tunnel is started 
 from B with a 5% drop, running S. 30° E. A vertical shaft is sunk 
 
PROBLEMS 53 
 
 75' at A. What will be tlie length, direction, and grade of the shortest 
 tunnel from the bottom of the shaft to meet the first tunnel? 
 
 36. A borehole cuts a 6-foot core from a vein whose strike is north 
 75° east and whose dip is 60° to the northwest. The borehole bears 
 north 30" east and dips 75°. Required the true thickness of the vein. 
 
 37. A borehole bearing north 15° east and dipping 75° intersects a 
 vein whose strike is north 45° east. The true thickness of the vein has 
 been found to be 3^ feet while the core from the borehole shows 6 feet 
 of vein matter. Find the dip of the vein. 
 
 38. Given the three lines A(— i, —3, —3) B(— 2f, — 4i, —2) 
 A(— i, —3, —3) C(lf, —bh —2) A(— i, —3, —3) D(i, 0, — li). As- 
 sume on the lines in the order named, the distances 1^'\ ^'\ and li" 
 from A and through each point pass a plane perpendicular to the line 
 which contains the point. Draw the projections of the triangular pyra- 
 mid formed by the intersection of the three planes with each, other and 
 with a horizontal plane through E(0, 0, — 4). 
 
 39. A plane is given by the three points, M(— 4f, — i, 0), N(0, —5, 0), 
 and 0(0, —h — 4f ). A square which has A (If, —i, 0) B(— i, — l^s, 0) 
 for one side lies in a horizontal plane. Revolve the square about MN 
 as an axis until it comes into the given plane MNO and find its pro- 
 jections in the new position. 
 
 40. Let the square in the preceding problem be the base of a cube 
 resting on the given plane. Draw the horizontal and vertical projec- 
 tions of the cube. 
 
 41. Let the plane be given as in problem 39. A cube whose edge is 
 2Y' rests on this plane in such a position that two edges AB and EF of 
 the base make angles of 15° with the horizontal MN. The corner B of 
 the base Is at the point (—1, y, — 1). Draw the horizontal and ver- 
 tical projections of the cube. 
 
 42. Keeping the edge BA as in 41, draw the projections of the same 
 cube having one of its corners in H. 
 
 43. Keeping the edge BA as In 41, draw the projections of the cube 
 having one of its corners in a plane parallel to and li'' from V. 
 
 44. Keeping the edge BA as in 41, draw the projections of the cube 
 having the edge BF making an angle of 30° with H. 
 
 45. Keeping the edge BA as in 41, draw the projections of the cube 
 having the edge BF making an angle of 45° with V. 
 
 46. Keeping the edge BA as in 41, draw the projections of the cube 
 having the plane of its face ABFE making an angle of -30° with the 
 plane MNO. 
 
 47. Keeping the edge BA as in 41, draw the projections of the cube 
 having the plane of its face ABFE making an angle of 45° with H. 
 
54 DESCRIPTIVE GEOMETRY 
 
 418. Keeping the corner B as in 41, draw the projections of the cube 
 having one corner in H and another in V. 
 
 49. One side of an equilateral triangle lies along the line A( — 2^, — 3, 
 — i2J) B(li, — 1, — 5f ) and the vertex opposite this side is at the point 
 C(i, — 4i, — 3). Draw the projections of a regular pyramid having 
 this triangle for a base and an altitude of 4i". 
 
 50. Find the projections of a regular hexagonal pyramid standing on 
 the plane through A( — If, — 3, — 2f) parallel to the lines B( — 3^, 0, 
 —^W C(2i, —4, — 3i) and D(— li, — 4i, —W E(lf, — f, —5). One 
 side of the base lies in the line of intersection of the plane of the base 
 and a plane through A perpendicular to the line BC. The vertex is at 
 the point 0(2|, 0, 0). Find the points in which the lines BC and DE 
 pierce the pyramid. 
 
 51. Substitute a regular triangular pyramid for the regular hexagonal 
 pyramid in problem 50. 
 
 52. Substitute a regular square pyramid for the regular hexagonal 
 pyramid of problem 50. 
 
 53. Substitute a regular pentagonal pyramid for the regular hexag- 
 onal pyramid of problem 50. 
 
 54. Substitute a right circular cone for the hexagonal pyramid of 
 problem 50. The base of the cone is tangent to the line of intersection 
 of the two planes. 
 
 55. The line A(— li, 0, —W B(li, —2^, 0) coincides with the edge 
 of a cube having one corner in H and another corner in V. Edge of 
 cube 3''. Draw the projections of the cube. 
 
 56. Draw the projections of the cube of problem 55, when one edge 
 coincides with AB, one corner is at A and the comer at the other end 
 of the diagonal through A also lies in V. 
 
 57. Draw the projections of the cube of problem 55, when one edge 
 coincides with AB, one corner is at A, and the corner at the other end 
 of the diagonal of a face from A also lies in V. 
 
 58. The same as problem 55 when one corner lies at B and some other 
 corner is also in H but no corner is in V. 
 
 59. The same as problem 55 when neither A nor B is a corner of the 
 
 cube. 
 
 60. Ci—lh —If, —W Edi, —3^, — 4i) is a diagonal of a cube. 
 One corner of the cube is in a plane which makes 60° with H and con- 
 tains a horizontal passing through the point P(— 3f, 0, 0) and making 
 45° with V. Draw the projections of the cube. 
 
 61. C(— 3i, —Ih —U) E(1J, —n, — li) is a diagonal of a cube which 
 has one corner in H. Draw the projections of the cube. 
 
 62. The same as problem 61, but having one corner of the cube in V 
 In the place of in H. 
 
PROBLEMS 55 
 
 63. One edge of a 2V' cube lies along the line A(— 2i, —31, —4) 
 B(li, 0, — 2f) and an adjacent edge passes through the point 
 P(i, — li, x) in such a direction that its horizontal projection makes 
 60° with the horizontal projection of AB. Draw the projections of the 
 cube. 
 
 64. The face of the cube in problem 63 which lies on the plane ABP 
 is the base of a right square pyramid. Draw the projections of the 
 pyramid when it has an altitude of 3i". 
 
 65. The shortest line between the lines A(— i, — 3^, 0) B(|, 0, --21) 
 and C(— 2i, —2, —1) D(J, — 2f, —3) is the axis of a right rectangular 
 prism. The diagonals of the upper and lower bases coincide with AB 
 and CD respectively. The shorter edge of the base is If' long. Draw 
 the projections of the prism. 
 
 66. Substitute A(3^, 0, —If) B(3i, —2, — 3f) and C(— 2, —2h —W 
 D(li, — 5, 0) in problem 65. 
 
 67. Given a line AB, two points C and E and the distance d. Find a 
 line which passes through C, intersects AB and passes at the distance d 
 from E. 
 
 68. Find OK, the shortest line which is parallel to H and terminates in 
 the lines CD and EF. 
 
 69. Project a cube of 1^' edge on a plane perpendicular to one of its 
 diagonals. (Assume the first position of the cube so that one of its 
 diagonals shall be parallel to one of the planes of projection.) 
 
 70. A V square stick has its edges maiking 30° to V and oblique to H. 
 Find the true size of the section of this stick by the V and P planes. 
 
 71. Given the base ABC of a triangular pyramid, also the length of 
 two edges DA and DB and the altitude DO. Draw the projections of 
 the pyramid. 
 
 72. Draw the projections of a tetrahedron two edges of which are 
 3", two 4f', and two 5". (No edge is to be parallel to either H or V. 
 
 73. Find the projections of three spheres, each tangent to the other 
 two and having the following radii: iy\ 2'\ 2\'\ Construct the pro- 
 jections of a fourth sphere with a radius of \" which will be tangent to 
 the other three spheres. The plane of the centers of any three spheres 
 is not to be parallel or perpendicular to either H or V. 
 
 74. Draw the projections of a tetrahedron the sides of whose base are 
 3i", Z\'\ and ^" and the planes of whose faces make angles of 45°, 60°, 
 and 75° with the plane of the base. No two of the corners are to be the 
 same distance from H. 
 
 75. Given an acute triangle on a horizontal plane. Find a point in 
 space which joined with the vertices of the triangle will form a tri- 
 rectangular tetrahedron. 
 
56 DESCRIPTIVE GEOMETRY 
 
 76. Draw the prajections of a 3" cube having given the direction of 
 the horizontal projection of three concurrent edges of the cube. 
 
 77. Draw the projections of an equilateral triangle having given 
 both projections of one side and the direction of the horizontal projec- 
 tion of one of the other sides. 
 
 78. Having given the two projections ab and a'h' of one edge of a 
 cube and the direction dc of the horizontal projection of another edge. 
 Draw the projections of the cube. 
 
 79. Having given the directions of the projections of three concurrent 
 edges of a parallelopiped, also the direction and length of the diagonal 
 through the same corner, construct the projections of the parallelopiped. 
 
 80. Draw the projections of a rectangular parallelopiped, having 
 given one corner, aa', the horizontal projection a& of one edge, and the 
 directions ac and ad of the horizontal projections of two other edges. 
 
 81. A ray of light from a point P, which is 3'' from H and 2" from V, 
 is reflected from a point in V which is 11" from H and 2" to the right 
 of P. Where will the ray meet H? 
 
 82. A ray of light from the point A is reflected by the H plane at B 
 to the point C in V, locate C. 
 
 83. Given two points O and P in the first quadrant. Find the pro- 
 jections of a ray of light which emanates from O and after being re- 
 flected from both H and V passes through P. 
 
 84. Find the projections of a ray of light which emanates from a 
 given point and is reflected from H, V, and P. 
 
 85. A ray of light from A is reflected at B to C. Find the reflecting 
 plane T. 
 
 86. Represent a plane which passes through a given point and makes 
 given angles with the planes of projection. Between what limits may 
 the sum of these angles vary? 
 
 87. Given a line AB which is oblique to H and V. Represent a plane 
 which contains the line AB and is in such a position that AB bisects 
 the angle between a horizontal and a frontal of the plane. Let the 
 horizontal and frontal pass through the point A. 
 
 88. Given two lines AB and CD in space which do not intersect. 
 Draw a third line which makes 30° with AB and 45° with OD and in- 
 tersects them both. 
 
 89. Given the projections of a line which makes 30° with H and is 
 oblique to V. Draw the projections of a horizontal which makes 45° 
 with this line. 
 
 90. Draw the projections of a regular tetrahedron having given the 
 projections a& and a'6' of one edge and the direction ac of the horizon- 
 tal projection of another edge. 
 
CHAPTER III 
 
 APPLICATIONS OF THE ELEMENTARY PRINCIPLES OF 
 THE POINT, STRAIGHT LINE, AND PLANE 
 
 SHADES AND SHADOWS 
 
 62. In order that the projections of an object shall more 
 nearly represent the object as it appears, the shadows of dif- 
 ferent parts of the object on other parts or on the planes of 
 projection are sometimes shown. Shadows are used more in 
 architecture than any other branch of engineering but every 
 engineer should know how to find the shadow of a simple ob- 
 ject. Only the elementary principles of the subject will be out- 
 lined here. 
 
 63. Definitions. If an opaque object is placed between a 
 surface and the source of light, the object will cast a shadow 
 upon the surface. This is due to the fact that the object inter- 
 cepts the rays of light and prevents their reaching the surface 
 upon which the shadow is cast. When an opaque body is sub- 
 jected to rays of light, that portion of the surface of the body 
 which is turned away from the source of light and which, there- 
 fore, does not receive any of the rays of light, is said to be in 
 shade. The line which separates the portion of the surface of 
 an object which is turned towards the light from that portion 
 which is turned away from the light is called the shade line. 
 The shadow of the shade line of an object on a given surface is 
 the outline of the shadow of the object on that surface. 
 
 The source of light is the sun and because of its distance 
 from the earth the rays of light are assumed to be parallel. 
 For the sake of uniformity, the direction of a ray of light is the 
 direction of the diagonal of a cube which joins the upper, left 
 
58 DESCRIPTIVE GEOMETRY 
 
 hand, front corner with the lower, right hand, back corner when 
 the cube has two faces parallel to V and two parallel to H. In 
 other words, a ray of light is considered as coming over the left 
 shoulder so that its projections make angles of 45° with the 
 ground line. 
 
 64. Shadows of points and lines. To find the shadow of a 
 point on a surface, pass a ray of light through the point and 
 find where this ray pierces the surface. This piercing point is 
 the required shadow point. To find the shadow of a straight 
 line on a plane, find the shadows of two points of the line on 
 the plane and join th6m by a straight line. If the shadow of 
 the straight line falls upon two or more planes, find the shadows 
 of two points of the line on each plane and join the correspond- 
 ing shadow points. The shadow of the straight line is a broken 
 line which changes direction on the line of intersection of the 
 planes upon which the shadow is cast. To find the shadow of a 
 straight line on a curved surface, find the shadows of several 
 points of the line and join these shadow points in the proper 
 order. To find the shadow of a curved line on a surface, it is 
 usually necessary to find the shadows of several points of the 
 line on the surface and then join the shadow points in the 
 proper order. 
 
 The shadow of a plane figure on a plane which is parallel to 
 the plane of the figure is the same size and form as the figure 
 itself. 
 
 65. To find the shadow of a square prism on the plane of the 
 base. 
 
 Let the prism be given as shown in Fig. 39. 
 Analysis. Since the shadow of the prism is to be cast upon 
 the plane of the base, it is evident that the base of the prism 
 coincides with its own shadow. The shadow of the top is found 
 by passing rays of light through the points A, B, C, and D and 
 finding where these rays pierce the plane of the base. The 
 straight lines joining these piercing points, in the proper order, 
 will form the outline of the shadow of the top. The shadows 
 of the comers of the upper base joined with the shadows of 
 
SHADES AND SHADOWS 
 
 59 
 
 the corresponding corners of the lower base will be the shadows 
 of the lateral edges (Art. 64). This completes the outline of 
 the shadow of the prism. 
 
 aa 
 
 1 1 "V 
 
 'dd, 
 
 1 
 1 
 
 1 
 
 I 
 
 1 
 1 
 
 
 
 ^7 JTdl ^l^s y^s Ci 
 Fig. 39. — Shadow of square prism. 
 
 Construction, hhs is the horizontal and h'h's is the vertical 
 projection of a ray of light which passes through B. This ray 
 pierces the plane of the base at Bg. Then Bg is the shadow of 
 the point B upon the plane of the base of the prism. In like 
 manner Cs, Ds, and As are the shadows cast by C, D, and A 
 upon the plane of the base of the prism. Joining these points 
 in order gives the shadow of the top on the plane of the base. 
 
60 DESCRIPTIVE GEOMETRY 
 
 Joining Bg with B^ gives the shadovr of the vertical edge BB^. 
 The shadows of the other vertical edges are found in the same 
 manner. Then IJj^c^d^d^ is the horizontal projection of the 
 outline of the shadow of the prism on the plane of the base of 
 the prism. 
 
 The face CCiDiD of the object is turned away from the light 
 and is therefore in the shade. This face is visible in the ver- 
 tical view and the fact that it is in the shade is indicated as 
 shown in the figure. 
 
 From the figure, it is seen that a line and its shadow are par- 
 allel when the line is parallel to the plane upon which the 
 shadow is cast. 
 
 66. To find the shadow of an object on the plane of the base 
 and on the object itself. 
 
 Let the object be given as shown in Fig. 40. 
 
 This problem illustrates the finding of the shadow which an 
 opaque object will cast upon itself as well as the plane of the 
 base. It also shows the method of finding the shadow on a 
 vertical plane in addition to finding the shadow on a horizontal 
 plane. 
 
 Analysis. All of the edges shown are straight lines. If all 
 of the shadow of any edge falls on one plane, it will be neces- 
 sary to find the shadow of only two points of the edge and join 
 them by a straight line. If the shadow of an edge falls on 
 more than one plane, two shadow points should be found on 
 each plane and joined by a straight line. The shadow of any 
 point of an edge falls upon the plane which is first pierced by 
 the ray of light passing through that point (Art. 64) . 
 
 Construction. The horizontal projection of the shadow 
 
 OiB^ises '^ which the object casts upon the plane of the 
 
 base is found as in Art. 65. For this shadow it is only neces- 
 sary to find where the rays of light pierce a horizontal plane. 
 The shadow which the upper part of the object casts upon the 
 overhang EGK is found in the same manner as the shadow on 
 the plane of the base, EGK being a horizontal plane. 
 
SHADES AND SHADOWS 
 
 61 
 
 Due to the overhang EiMiKi, there is a shadow cast upon the 
 vertical faces of the object. To find this shadow, pass a ray of 
 light through the point and find where it pierces the vertical 
 face AD of the object. Since the face is vertical, the horizontal 
 
 a/ d;o/sii. 
 
 Fig. 40. — Shadow of object on plane of hase 
 and object itself. 
 
 projection of this point is Os and the vertical projection is the 
 shadow o'e. In like manner the shadows of the points P, Mi, 
 and N are found. Joining these shadow points with straight 
 lines forms the outline of the shadow of the overhang EiMiK^ 
 on the vertical faces of the object. 
 
62 
 
 DESCRIPTIVE GEOMETRY 
 
 67. To find the shadow of a cylindrical column and cap on a 
 horizontal plane and also the shadow of the cap on the column. 
 
 Let the column and cap be given as shown in Fig. 41. 
 Analysis. Since the upper surface of the cap is a horizontal 
 
 Fig. 41. — ^Shadow oj column and cap. 
 
 circle, its shadow on a horizontal plane is a circle of the same 
 radius as the cap. The shadow of the center of the circle is 
 the center of the shadow of the circle. Finding the shadow of 
 the center and drawing a circle with a radius equal to the 
 radius of the cap, will give the shadow of the top of the cap on 
 a, horizontal plane. The shadow of the lower surface of the cap 
 on a horizontal plane is found in the same way. Common tan- 
 
SHADES AND SHADOWS 63 
 
 gents to these circles drawn in the direction of the horizontal 
 projection of a ray of light completes the shadow of the cap on 
 the horizontal plane. The shadow of the column on a horizon- 
 tal plane is bounded by two lines drawn in the direction of the 
 horizontal projection of a ray of light and tangent to the base 
 of the column. 
 
 To find the shadow of the cap on the column, pass rays of 
 light through points of the lower edge of the cap and find 
 where they pierce the column. A line joining these points is 
 the boundary of the shadow of the cap on the column. 
 
 Let the construction be made in accordance with the above 
 analysis. 
 
 The side of the column and cap turned away from the source 
 of light is in the shade. GK is the shaded portion of the col- 
 umn which is visible in the vertical view and EM is the shaded 
 portion of the cap which is visible in the vertical view. 
 
 68. Problems. 
 
 1. One edge of a 1" cube lies on a horizontal plane and is oblique to 
 V. The plane of one face of the cube makes 30° with H. Draw the pro- 
 jections of the cube and find its shadow on a horizontal plane. 
 
 2. Draw the projections of a 2" cube when one diagonal is perpendic- 
 ular to H. Find the shadow of the cube on a horizontal plane which 
 passes through the lower end of the diagonal. 
 
 3. A right square P3^aniid has its vertex in a horizontal plane. The 
 side of the base is 1" and the altitude is Z" long, makes 60° with H and 
 is oblique to V. Draw the projections of the pyramid and find Its 
 shadow on a horizontal plane. 
 
 4. Draw the projections of a sphere which has a radius of \\'\ Find 
 at least twelve points In Its shadow on a horizontal plane. 
 
64 
 
 DESCRIPTIVE GEOMETRY 
 
 PLANE SECTIONS AND DEVELOPMENTS OF THE SURFACES OF 
 PRISMS AND PYRAMIDS 
 
 69. To find the line of intersection of a right prism with a 
 given oblique plane and to develop the surface of the prism. 
 
 Let the prism be given as in Fig. 42 and let MN and OP rep- 
 resent the oblique plane. 
 
 Analysis. Find the points in which the lateral edges of the 
 prism intersect the plane of MN and OP. Straight lines join- 
 ing these points of intersection is the line of intersection of the 
 prism with the oblique plane. 
 
 h, c; aj fj -J 
 
 Pig. 42. — Plane section and development of right prism. 
 
 Construction. AAj and CCi have a common horizontal pro- 
 jecting plane which cuts the line MN at E and the line OP at K. 
 e'liff the vertical projection of the line joining these points, cuts 
 the vertical projections of the edges of the prism at x' and z^ 
 which are the vertical projections of two of the required points. 
 The line OP cuts another face of the prism at R. x'r' extended 
 cuts h'h'^ at t/ which is the vertical projection of the point in 
 
PLANE SECTIONS AND DEVELOPMENTS OP SURFACES 65 
 
 which BBi cuts the plane of MN and OP. Then x'y'z' is the 
 vertical and xyz the horizontal projection of the required line 
 of intersection. 
 
 To develop the surface of the prism showing its line of inter- 
 section with the oblique plane. If one of the lateral faces of 
 the prism be placed against a plane and then the prism be re- 
 volved about an edge of this face until the next face comes into 
 this plane, and the process be repeated until each face in turn is 
 brought into the plane, the part of the plane covered by the 
 different faces of the prism will be the development of the 
 prism. In Fig. 42, (J3&3=ai&i, &3C3=&iCi, etc. To get the line 
 of intersection on the development, lay off a^x^=a\x\ hzy2==b\y'y 
 etc., and join these points by straight lines. 
 
 Note. All the problems in plane sections of prisms and pyra- 
 mids can be multiplied by changing the number of lateral faces 
 of the surface. 
 
 70. To find the line of intersection of an oblique prism by a 
 plane perpendicular to the lateral edges, and to develop the 
 lateral surface of the prism. 
 
 Let the prism and plane be given as in Fig. 43. 
 
 By finding the points in which the lateral edges of the prism 
 pierce the plane of HH and FF and joining these points in the 
 proper order, the intersection XYZ is found. 
 
 To develop the lateral surface of the prism. Since the sec- 
 tion plane is perpendicular to the edges of the prism, it is evi- 
 dent that the line of intersection will be a straight line in the 
 development. The true size of this section is x"y''z'\ This is 
 found by revolving it until it is parallel to the horizontal plane 
 about HH as an axis. In the development, lay off x^y^=^x"y[\ 
 yzZ^=y"z'\ etc., along the straight line x^x^. On a line perpen- 
 dicular to x^Xz at the point x^, lay off below the line the true 
 length x^a-i_ of the edge from X to the lower base of the prism 
 and along the same line lay off above x^x^ the true length x^a^ 
 of the edge from X to the upper base of the prism. In a similar 
 
66 
 
 DESCRIPTIVE GEOMETRY 
 
 manner, lay off at the points 2/3, Z3 and x^ the true lengths of the 
 edges through the points Y, Z, and X. In this way all points 
 in both bases of the prism are located in the development. 
 
 / .7 
 
 N*^ 
 
 ^ 
 
 ^ 
 
 Fig. 43. — Plane section and development of oblique prisrn^ 
 
 Straight lines joining in order the points in the development of 
 the lower base and also in the same way those of the upper base 
 complete the development of the prism. 
 
 71. To find the line of intersection of an oblique pyramid by a 
 plane and to develop the lateral surface of the pyramid. 
 
 Let the plane and pyramid be given as in Fig. 44. 
 
 By finding the points in which the lateral edges of the pyra- 
 mid pierce the plane of MN and OP and joining these points in 
 the proper order, the intersection XYZ is found. 
 
 To develop the lateral surface of the pyramid. If the face 
 VAB of the pyramid be placed against a plane and the pyramid 
 be rolled over about VB as an axis until the face YBC comes 
 
PLANE SECTIONS AND DEVELOPMENTS OF SUEPACES 
 
 67 
 
 into the plane and then the pyramid rolled again about VC as 
 an axis until the next face comes into the plane, the part of the 
 plane thus covered is the development of the pyramid. The 
 development is found by constructing a series of triangles hav- 
 ing given the length of the sides of each triangle. The first 
 triangle is formed of the sides VA, VB, and AB ; the second of 
 the sides YB, VC, and BC, etc. 
 
 a' y 0' 
 
 Fis. 44. — Flane section and development of otlique pyramid. 
 
 The line of intersection of the pyramid with the plane of MN 
 and OP can be layed off on the development by laying off on 
 each edge in succession the true lengths of VX, VY, and VZ 
 and joining these points by straight lines. 
 
 72. To find the line of intersection of any surface with a 
 given oblique plane by using an auxiliary plane of projection. 
 
 Fig. 45 represents an oblique prism cut by the plane of HH 
 and FF. The auxiliary plane of projection is taken at right 
 angles to the horizontal HH of the given plane. In the auxil- 
 iary view, the given plane is represented by the line x^Zz and 
 
68 
 
 DESCRIPTIVE GEOMETRY 
 
 the points in which the lateral edges of the prism pierce this 
 plane by x^, 2/2, ^z- By projection, the horizontal view xyz and 
 the vertical view x'yfz' of the intersection are found. If any 
 
 "VPz 
 
 a,' cT 
 
 Fig. 45. — Plane section of oblique prism by 
 using auxiliary plane of projection. 
 
 element of the surface is parallel to the auxiliary plane of pro- 
 jection, its view on that plane is the true length of the element. 
 The auxiliary plane of projection should always be taken at 
 right angles to a horizontal or a frontal of the plane of the 
 section. 
 
PLANE SECTIONS AND DEVELOPMENTS OP SURFACES 69 
 
 73. Problems. 
 
 1. Assume any four points, which are not in the same plane, and 
 draw through them four parallel lines in such a way that the plane 
 section of the prism, which has the four lines for edges, will be a par- 
 allelogram. 
 
 2. Given a point on the edge of any tetrahedron. Find the shortest 
 path to be followed on the faces in order to go around the tetrahedron 
 and return to the same point. The path is to cross the three faces but 
 not the base. 
 
 3. Two intersecting planes cut the ground line at the points T and S. 
 Find the projections of the shortest path on the planes from T to S. 
 
 4. Pass a plane so as to cut a regular hexagon from a 3" cube. 
 
 5. Cut a parallelogram having one side 2" long from the pyramid 
 O— ABCD. 0(U, — 5i, 0), A(— 3, —1, — 5i), B(i, 0, — 5i), C(2i, 
 —2, — 5i), D(— If, — 4i —51). 
 
 6. Given a square prism with its axis perpendicular to H. Side of 
 base 2i". Cut a parallelogram from the prism having one side 3" long 
 and one angle 60°. 
 
 7. Given a square prism with its axis perpendicular to H. Side of 
 base 2i". Cut a parallelogram from the prism having one diagonal 6i" 
 long and one angle 60°. 
 
70 
 
 DESCRIPTIVE GEOMETRY 
 
 INTERSECTIONS OF THE SURFACES OF PRISMS AND PYRAMIDS 
 
 74. To find the line of intersection of the surfaces of two 
 prisms when the lateral edges of one of them is perpendicular 
 to the horizontal plane. 
 
 Let the prisms be given as in Fig. 46. 
 
 Analysis, Since the prisms have plane faces, it is only nec- 
 essary to find the points in which the edges of each prism pierce 
 
 p' i^/ q' xizl 
 
 Fig. 46. — Intersection of prisms. 
 
 the faces of the other prism. A series of straight lines joining 
 in order these piercing points is the required line of intersec- 
 tion. 
 
 Consfruction. It is evident from the drawing that A, B, C, 
 D, E, and M are the points in which the edges of the triangular 
 prism pierce the faces of the square prism. It is also evident 
 that three of the edges of the square prism do not cut the faces 
 of the triangular prism. To find the points in which the fourth 
 
INTERSECTIONS OF SURFACES 
 
 71 
 
 edge of the square prism intersects the triangular prism, pass a 
 plane T through this edge and parallel to the lateral edges of 
 the triangular prism. This plane cuts two lines from the tri- 
 angular prism which intersect the edge of the square prism in 
 the points K and G. These are the points in which the edge of 
 the square prism intersects the faces of the triangular prism. 
 Joining all of the piercing points in the proper order gives the 
 intersection as shown in Fig. 46. 
 
 75. To find the line of intersection af the surfaces of two ob- 
 lique prisms. 
 
 Let the prisms be given as in Fig. 47. 
 
 w X' z' y' n' q^ 
 
 Fig. 47. — Intersection of bdlique prisms. 
 
72 DESCRIPTIVE GEOMETRY 
 
 Analysis. Pass a system of auxiliary planes parallel to the 
 lateral edges of both prisms. These planes will cut straight 
 lines from each prism which will intersect in points of the re- 
 quired line of intersection. 
 
 Construction. Through any point P, draw lines parallel to 
 the lateral edges of each prism, tt is the horizontal projection 
 of the line of intersection of the plane of these lines with the 
 plane of the bases of the prisms. Each of the auxiliary planes 
 cuts a line from the plane of the bases of the prisms which is par- 
 allel to tt. The planes also cut lines from the prisms which are 
 parallel to the respective edges of the prisms and which in- 
 tersect in points of the required line of intersection. The line 
 of intersection is a broken line which only changes direction at 
 the points where the edges of one prism pierce the faces of the 
 other. By using only the planes which contain edges of one or 
 the other of the prisms and joining in order the points thus 
 found, the broken line of intersection as shown in Fig. 47 is 
 found. 
 
 76. To find the line of intersection of the surfaces of a prism 
 and a pyramid. 
 
 Let the prism and the pyramid be given as in Fig. 48. 
 
 Analysis. Pass a system of auxiliary planes through the ver- 
 tex of the pyramid and parallel to the lateral edges of the prism. 
 These planes will cut straight lines from each surface which 
 will intersect in points of the required line of intersection. A 
 line through the vertex of the pyramid and parallel to the lat- 
 eral edges of the prism will lie in all of the auxiliary planes and 
 will therefore pierce the plane of the bases in a point common 
 to all the lines cut from the plane of the bases. 
 
 Construction. Through V^ draw YiS parallel to the lateral 
 edges of the prism. It pierces the plane of the bases at S. 
 Through 5, draw st, st^, stz, lines which the auxiliary planes cut 
 from the plane of the bases. The intersection of the lines which 
 these planes cut from the prism and pyramid are points on the 
 required line of intersection. By using only the planes which 
 
INTERSECTIONS OF SURFACES 
 
 73 
 
 contain edges of the prism or the pyramid and joining in order, 
 with straight lines, the points thus found the line of intersection 
 as shown in Fig. 48 is found. 
 
 Ha 
 
 F^G. 48. — Intersection of prism and pyramid. 
 
 77. To find the line of intersection of the surfaces of two 
 pjrramids. 
 
 Let the pyramids be given as in Fig. 49. 
 
 Analysis. Pass a system of auxiliary planes through the ver- 
 tices of the pyramids. These planes will cut straight lines from 
 each pyramid which will intersect in points of the required line of 
 intersection. A line joining the vertices of the pyramids will lie 
 
74 
 
 DESCRIPTIVE GEOMETRY 
 
 in all the auxiliary planes and will pierce the plane of the bases 
 in a point common to all the lines which the auxiliary planes 
 will cut from the plane of the bases. 
 
 Fig. 49. — Intersection of pyramids. 
 
 Construction. The line through the vertices Y^ and Vg pierces 
 the plane of the bases at S. Through s draw st, 5^1, st 2y lines 
 which the auxiliary planes cut from the plane of the bases. The 
 intersection of the lines which the auxiliary planes cut from the 
 pyramids are points on the required line of intersection. By 
 using only the planes which contain edges of the pyramids and 
 joining in order the points thus found by straight lines, the line 
 of intersection as shown in Fig. 49 is found. 
 
 78. To find the line of intersection of the surfaces of a prism 
 and a pyramid when the edges of the prism are parallel to the 
 horizontal plane, by the use of an auxiliary plane of projection. 
 
INTERSECTIONS OF SURFACES 
 
 75 
 
 Let the prism and pyramid be given as in Fig. 50. 
 Analysis. The general method for finding the line of inter- 
 section of a prism and a pyramid will solve this problem but a 
 
 Fig. 50. — Intersection of prism and pyramid hy using an auxiliary 
 plane of projection. 
 
 much simpler method is to use an auxiliary view taken on a 
 plane perpendicular to the axis of the prism. In this view the 
 prism is represented by a square and for this reason the points 
 in which the edges of the pyramid pierce the prism are found 
 by inspection. To find the points in which an edge of the prism 
 pierces the pyramid, use an auxiliary plane containing this edge 
 and parallel to the base of the pyramid. This plane cuts lines 
 from the faces of the pyramid which intersect the edge of the 
 prism in the required points. 
 
 Construction. In the auxiliary view, the points in which the 
 
76 DESCRIPTIVE GEOMETRY 
 
 lateral edges of the pyramid pierce the faces of the prism are 
 seen to be iCg, Wz, Pz, etc. The horizontal views x, w, p, etc., of 
 these points are found from the auxiliary view by projection. 
 The vertical views are then found from the horizontal views. 
 n^Uz is the auxiliary view of a plane which is parallel to the 
 base of the pyramid and contains the edge NN^ of the prism. 
 ut and uy are the horizontal views of the lines which this plane 
 cuts from two faces of the pyramid and these lines cut the edge 
 of the prism in the required points T and Y. Points on the 
 other edges of the prism can be found in the same way. Join- 
 ing the points thus found in the proper order gives the required 
 line of intersection. 
 79. Problems. 
 
 1. The lower base of a polyhedron is A( — 2f, — 2S, -^5), B( — li, —J, 
 —5), C(f, —21, —5), D(— 2i, —3 J, —5). The upper base of the poly- 
 hedron is in H, one corner being at A (i, — 4, 0). The other lateral 
 edges are parallel to AA . Find the line of intersection of this poly- 
 hedron with the pyramid O— BFG. 0(— 2i, — |, — i), E(— -li, — 4|, 
 —5), F(2i, — li, —5), G(3i, — 4i, —5). 
 
 2. Find the line of intersection of a cube with a square prism when 
 the axis of the prism coincides with the diagonal of the cube. Edge of 
 cube 2Y'', side of base of prism 2i"; altitude 5". Axis of prism is per- 
 pendicular to H, the plane of one face makes 15° with V. The H pro- 
 jection of one top edge of the cube is perpendicular to a face of the 
 prism. 
 
 Develop the surface of the cube showing the line of intersection. 
 
 3. The same as problem 2 except that the prism has an equilateral 
 triangle with side 2i" for a base. 
 
 4. The same as problem 2 except that the diagonal of the cube is i" 
 from the axis of the prism. 
 
 5. Find the line of intersection of a square prism and a right square 
 pyramid. Prism 2''x2''xiY\ with edges parallel to H and 30"* with V. 
 The plane of one face makes 30° with H and the lowest edge is J" 
 above the base of the pyramid. The base of the pyramid is a 2i" square 
 resting on a horizontal plane, side of base 60° with V. The axis of the 
 pyramid is perpendicular to H and is 5'' long. The axes of the prism 
 and the pyramid are i" apart. 
 
 Note. — This problem can be multiplied by changing the number of 
 faces of the prism or pyramid, by changing the distance between the 
 axes, by changing the angle which the plane of the face of the prism 
 makes with H, or by raising or lowering the prism. 
 
CHAPTER IV 
 CURVED LINES AND SURFACES 
 
 GENERATION AND CLASSIFICATION OF LINES 
 
 80. A line is the path of a point moving according to some 
 law. 
 
 Lines are of two general classes : 
 
 I. Straight Lines or Right Lines, in which a point moves 
 always in the same direction. 
 
 n. Curved Lines, in which a point moves so as to change its 
 direction continually. 
 
 Curved lines are of two kinds : 
 
 I. Curves of Single Curvature or Plane Curves, in which all 
 positions of the moving point lie in the same plane. Examples, 
 circle, ellipse. 
 
 II. Curves of Double Curvature, in which four positions of 
 the moving point do not lie in the same plane where these four 
 points are taken one right after the other along the curve. Ex- 
 ample, the edge of an ordinary screw thread. The line of in- 
 tersection of two curved surfaces is usually of this form. 
 
 Lines 
 
 OUTLINE 
 
 I Straight or Right Lines 
 
 II Curved Lines -< 
 
 Circle 
 I Curves of Single Cur- jj^^pge 
 
 vature or Plane Curves^ Parabola 
 
 II Curves of Double 
 Curvature 
 
 I Hyperbola 
 \ Helix 
 
78 
 
 DESCRIPTIVE GEOMETRY 
 
 PROJECTIONS OF CURVES 
 
 81. The cylinder formed by the horizontal projecting lines 
 of the points of a curve is the horizontal projecting cylinder of 
 the curve and its intersection with the horizontal plane is the 
 horizontal projection of the curve. Likewise, the cylinder 
 formed by the vertical projecting lines of the points of a curve 
 is the vertical projecting cylinder of the curve and its inter- 
 section with the vertical plane is the vertical projection of the 
 curve, Fig. 51. 
 
 The two projections of a curve 
 will, in general, determine its posi- 
 tion in space, for the projecting cyl- 
 inders intersect in the only curve 
 which can have the given projec- 
 tions. 
 
 If the plane of a curve of single 
 curvature is perpendicular to the 
 horizontal plane, the horizontal pro- 
 jection of the curve will be a straight 
 line. 
 
 Likewise, if the plane of the curve is perpendicular to the 
 vertical plane, the vertical projection will be a straight line. 
 
 If the plane of the curve is perpendicular to the ground line, 
 both projections will be straight lines perpendicular to the 
 ground line and the curve will be undetermined. 
 
 If the plane of the curve is parallel to the horizontal plane, 
 the horizontal projection will be equal to the curve. 
 
 Likewise, if the plane of the curve is parallel to the vertical 
 plane, the vertical projection will be equal to the curve. 
 
 The projections of a curve of double curvature are always 
 curved lines. 
 
 Fig. 51. — Projections of 
 a curve. 
 
TANGENTS AND NORMALS TO LINES 79 
 
 TANGENTS AND NORMALS TO LINES 
 
 82. If in a secant line AB, Fig. 52, the point A be kept fixed 
 and the point B moved along the curve until it coincides with A, 
 the secant AB becomes a tangent to the curve at the point A. 
 
 Two curves are said to be tangent to each other at a point 
 when they have a common tangent at that point. 
 
 If a straight line is tangent to a plane 
 
 A/ curve, the tangent will lie in the plane 
 
 Bi^,^^:::^^ of the curve. This is evident since the 
 
 Z l C^ I secant through the point of tangency is 
 
 ^^'^^^^^l.^*- \^ i^ t^® plane of the curve, and it remains 
 ^ ^^ \ in this plane as it moves to the position 
 Fig. 52. — Tangent to a of the tangent. 
 
 curve. Two straight lines tangent to each 
 
 other will coincide. In this case the secant which joins two 
 points of the given line coincides with the line and therefore 
 the tangent must coincide with the given line. 
 
 83. If two lines are tangent to each other in space, their pro- 
 jections on the same plane will be tangent. Let Fig. 52 repre- 
 sent a curve in space with its secant AB. Let these lines be 
 projected upon any plane. Then the projections of the points 
 A and B will approach each other as the points A and B in 
 space approach each other. When the secant AB becomes a 
 tangent in space, the points A and B coincide and their pro- 
 jections will also coincide in the only point common to the pro- 
 jections of the straight line and curve. The projections of the 
 lines are therefore tangent to each other at this common point. 
 
 84. Normals. If a line be drawn perpendicular to the tan- 
 gent at the point of tangency, it is called a normal to the curve. 
 There are an infinite number of normals to a curve at a point 
 on the curve. However, the normal to a plane curve is under- 
 stood to be the one lying in the plane of the curve unless other- 
 wise stated. 
 
80 DESCRIPTIVE GEOMETRY 
 
 85. Rectification of curves. To rectify a curve means to find 
 a straight line equal in length to the curve. This is accom- 
 plished approximately by dividing the curve into a number of 
 small arcs, so small that for all practical purposes the chords 
 of these arcs may be taken as equal in length to the arcs them- 
 selves. These small chords are then laid off one after another 
 along a straight line. The part of the line thus covered is the 
 rectified curve. 
 
 CURVES OF SINGLE CURVATURE 
 
 86. The most common plane curves are obtained by cutting a 
 right circular cone by a plane. These curves are the circle, 
 ellipse, parabola, and hyperbola. Fig. 53 shows the position 
 
 Fig. 53. — Plane sections of a cone of revolution. 
 
 which the cutting plane must take with reference to the plane 
 of the base of the cone in order to cut the particular curve de- 
 sired. must be less than © and oc greater than ®. 
 
 A circle is the path of a point moving in a plane so that its 
 distance from a given point is constant. 
 
 87. An ellipse is the path of a point moving in a plane so that 
 the sum of its distances from two fixed points is constant. 
 
 Construction. Let F and F^, Fig. 54, be the two fixed points 
 and let the sum of the distances from these points to any point 
 on the curve be equal to the line AB. Select any point, as 0, 
 on the line AB. With F as center and AO as radius, strike an 
 
CURVES OP SINGLE CURVATURE 
 
 81 
 
 arc; also with Fi as center and BO as radius, strike an arc 
 cutting the first arc in the points P and Q. These are points on 
 the curve since FP+FiP=AB and FQ4-FiQ=AB. Selecting 
 any other point on AB, as Oi, and going through the construc- 
 tion as before, two other points Pi and Qi are located. By con- 
 tinuing this process enough points can be located so that the 
 curve can be drawn. 
 
 --JM 
 
 Fig, 54. — Ellipse and tangents. 
 
 The points F and Fj are foci of the ellipse. AB is the major 
 axis. The minor axis CD is perpendicular to the major axis at 
 its middle point. The intersection of the axes is the center of 
 the ellipse. Lines drawn from any point on the curve, as P, to 
 the foci F and F^ are called focal radii. 
 
 If the axes of an ellipse are given, the foci can be found by 
 taking one-half the major axis as a radius and with the end of 
 the minor axis as a center, cut the major axis in two points 
 which are the foci. 
 
 88. To draw a tangent to the ellipse at a point on the curve. 
 Let P, Fig. 54, be the point on the curve. Draw the focal radii 
 PF and PF^. The line PT which bisects the angle F^PE is the 
 
82 
 
 DESCRIPTIVE GEOMETRY 
 
 required tangent. (Wood^s Co-ordinate Geometry, Art. 108A.) 
 This method applies to all conies. 
 
 To draw a tangent to the ellipse from a point without the 
 curve. Let S, Fig. 54, be the point without the curve. With 
 Fi as center and AB as radius, strike an arc. With S as center 
 and SF as radius, strike an arc cutting the first arc in the points 
 M and N. F^M and F^N intersect the curve in the points of 
 tangency Ti and Tg. ST^ and ST2 are the required tangents. 
 (Wood's Co-ordinate Geometry, Art. 108B.) This method ap- 
 plies to all conies. 
 
 89. A parabola is the path of a point moving in a plane so 
 that its distance from a given point is always equal to its dis- 
 tance from a given straight line. 
 
 Fig. 55. — Parabola and tangents. 
 
 Construction. Let F, Fig. 55, be the given point and AB the 
 given straight line. Draw a line CD from F perpendicular to 
 AB. Through any point, as 0, on CD, draw a line parallel with 
 AB. With F as center and CO as radius strike an arc cutting 
 this line in the two points P and Q. P and Q are points on the 
 parabola, since FP and FQ are each equal to the distance of P 
 
CURVES OF SINGLE CURVATURE 83 
 
 and Q from AB. Selecting some other point on the axis and 
 going through a similar construction, two other points of the 
 curve are located. By continuing this process, enough points 
 can be located so that the curve can be drawn. 
 
 The point F is the focus, the line AB the directrix, and the 
 line CD the axis of the parabola. 
 
 90. To draw a tangent to the parabola at a point on the 
 curve. Let P^ Fig. 55, be the point on the curve. Draw the 
 focal radii FPi and FjF^, (To make this construction similar 
 to that for the tangent to the ellipse, it is necessary to consider 
 one focus Fi of the parabola at an infinite distance on the axis.) 
 The line PjT which bisects the angle FPjE is the required tan- 
 gent. 
 
 To draw a tangent to the parabola from a point without the 
 curve. Let S, Fig. 55, be the point without the curve. With S 
 as center and SF as radius, strike an arc cutting the directrix 
 in the points M and N. MF^ and NF^, parallel with the axis 
 CD of the parabola, cut the curve in the points of tangency Ti 
 and Tg. ST^ and STg are the required tangents. 
 
 91. A hyperbola is the path of a point moving in a plane so 
 that the difference of its distances from two fixed points is con- 
 stant. 
 
 Construction. Let F and Fi, Fig. 56, be the two fixed points 
 and let the difference of the distances from these points to any 
 point on the curve be equal to the line AB. Select any point, 
 as 0, on the line AB extended. With F as center and AO as 
 radius, strike an arc ; also with F^ as center and BO as radius, 
 strike an arc cutting the first arc in the points P and Q. These 
 are points on the curve since FP — ^FiP=AB and FQ — ^FiQ=AB. 
 (If Fi be taken as center and AO as radius and F as center and 
 BO as radius, the points located will be on the other branch of 
 the hyperbola.) Selecting any other point on AB, as Oi, and 
 going through the construction as before, two other points Pi 
 and Qi are located. By continuing this process enough points 
 can be located so that the curve can be drawn. 
 
84 
 
 DESCRIPTIVE GEOMETRY 
 
 The points F and Fj are the foci of the hyperbola. The point 
 C, midway between the foci F and Fi is the center of the curve. 
 The line AB is the transverse axis. The line XY, perpendicular 
 to AB at its middle point, is the indefinite conjugate axis of the 
 curve. 
 
 Fig. 56. — Hyperbola and tangents. 
 
 92. To draw a tangent to the hyperbola at a point on the 
 curve. Let P^, Fig. 56, be the point on the curve. Draw the 
 focal radii FP^ and F^P^. The line PjT which bisects the angle 
 FjPiF is the required tangent. 
 
 To draw a tangent to the hyperbola from a point without the 
 curve. Let S, Fig. 56, be the point without the curve. With 
 Fi as center and AB as radius, strike an arc. With S as center 
 and SF as radius, strike an arc cutting the first arc in the points 
 M and N. F^M and F^N intersect the curve in the points of 
 tangency. ST^ is one of the tangents. The other point of tan- 
 gency, where NFj intersects the curve, is without the limits of 
 the drawing. 
 
CURVES OF DOUBLE CURVATURE 
 
 85 
 
 CURVES OF DOUBLE CURVATURE 
 
 93. An ordinary helix is the path of a point moving on the 
 surface of a cylinder of revolution so as to intersect its elements 
 at a constant acute angle. 
 
 The axis of the cylinder is the axis of the helix. 
 Construction. Let MN, Fig. 57, be the axis of the helix and P 
 the generating point. 
 
 Suppose that for one complete turn around the axis the gen- 
 erating point moves through a vertical distance m'n\ This dis- 
 tance is the pitch of the helix. 
 
 Since the curve is on the surface of a 
 cylinder of revolution, its projection on 
 a plane perpendicular to the axis will be 
 the circle pcfg. 
 
 To draw the vertical projection of the 
 helix, divide the circle pcfg into any 
 number of equal parts, as twelve, and 
 m'7^' into the same number of equal 
 parts. Draw horizontal lines through 
 the points of division of m'n\ Since the 
 motions of the point around and along 
 the axis are both uniform, the point in 
 making one-twelfth of a complete turn 
 around the axis will rise one-twelfth of 
 the distance mV. a is the horizontal 
 and a' the vertical projection of the 
 point after making one-twelfth of a 
 turn. In the same manner, the points 
 h\ c\ d', etc., are found. The vertical projection of the curve is 
 drawn through the vertical projections of these points. 
 
 Since the curve cuts all the elements of the cylinder at the 
 same angle, it is evident that the helix will become a straight 
 line if the cylinder is opened along an element and the surface 
 rolled out into a plane. From this position it is seen, that the 
 hypothenuse of a right triangle will form a helix if the altitude 
 
 P' 
 
 Fig. 57. — Helix and 
 tangent. 
 
86 DESCRIPTIVE GEOMETRY" 
 
 of the triangle coincides with an element of a right circular 
 cylinder and the base of the triangle is wound around the base 
 of the cylinder. 
 
 94. To construct a tangent to a helix at a point on the curve. 
 Let B, Fig. 57, be the point at which the tangent is to be drawn. 
 Let the triangle referred to in the above paragraph as being 
 wound around the cylinder, be unrolled as far as the point of 
 tangency B. The hypothenuse of the triangle in this position 
 is tangent to the curve, since it touches the curve at the point B 
 and makes the same angle with the horizontal plane as the 
 curve. The base ht of. the triangle is the horizontal projection 
 of the tangent, and is equal in length to the arc hp of the circle 
 pcfg* Therefore, to get the projections of the tangent, draw its 
 horizontal projection ht tangent to the circle pcfg at the point h. 
 On this line lay off from h the true length of the arc hap of the 
 circle. This will give the point t where the tangent pierces the 
 plane of the base. The vertical projection of this piercing point 
 is at f and this joined with h' gives the vertical projection h'f 
 of the tangent. 
 
 GENERATION AND CLASSIFICATION OF SURFACES 
 
 95. A surface is the path of a line moving according to some 
 law. 
 
 The moving line is the generatrix, and its different positions 
 are the elements of the surface. 
 
 Surfaces are of two general classes : 
 
 I. Ruled surfaces, which can be generated by straight lines. 
 Examples, cylinder, cone, helicoid. 
 
 II. Double curved surfaces, which can only be generated by 
 curved lines. Example, sphere, ellipsoid. 
 
 Ruled surfaces are of three kinds : 
 
 I. Plane surfaces. 
 
 II. Single curved surfaces, which can be rolled into a plane 
 without undergoing distortion. Examples, cylinder, cone. 
 
GENERATION AND CLASSIFICATION OP SURFACES 
 
 87 
 
 III. Warped surfaces, which cannot be rolled into a plane 
 without undergoing distortion. Example, the surface of a 
 screw thread. 
 
 Single curved surfaces are of three kinds : 
 
 I. Cylinders, in which the rectilinear elements are parallel. 
 
 II. Cones, in which the rectilinear elements intersect in a 
 point. 
 
 III. Convolutes, in which the rectilinear elements are tan* 
 gent to a curve of double curvature. Example, the surface hav- 
 ing tangents to a helix for its elements. 
 
 OUTLINE 
 
 I Plane 
 
 {I Cylinder 
 II Cone 
 III Convolute 
 
 Surfaces 
 
 I Euled 
 
 III Warped < 
 
 Helicbid 
 
 Hyperbolic Paraboloid 
 Conoid 
 Cylindroid 
 
 Hyperboloid of Revo- 
 lution of one Sheet 
 
 II Double curved < 
 
 Sphere 
 
 Torus or Anchor ring 
 Ellipsoid of Revolution 
 Paraboloid of Revolution 
 Hyperboloid of Revolution of 
 two Sheets 
 
 SURFACES OF REVOLUTION 
 
 96. A surface of revolution is the path of a line revolving 
 about a straight line as an axis. 
 
 It is evident that the intersection of this surface with a plane 
 perpendicular to the axis is a circle. A surface of revolution: 
 
88 
 
 DESCRIPTIVE GEOMETRY 
 
 may be generated by a circle having its center moving along a 
 straight line, its different positions in parallel planes, and its 
 radius changing according to a given law. 
 
 The intersection of the surface with a plane containing the 
 axis is a meridian line, and the plane is a meridian plane. All 
 meridian lines of the same surface are identical. Any surface 
 of revolution may be generated by revolving its meridian line 
 about its axis. 
 
 OUTLINE 
 
 Surfaces of 
 Revolution 
 
 Ruled 
 
 ^. - -, ( Right circular cylinder 
 
 Single curved -^ -d- i. • i 
 
 * I Right circular cone 
 
 \ TKT A ^ Hyperboloid of Revolution 
 Warped | of one Sheet 
 
 Double curved ^ 
 
 ' Sphere 
 Torus 
 ElKpsoid of Revolu- ( Prolate 
 
 tion ( Oblate 
 
 Paraboloid of Revolution 
 Hyperboloid of Revolution of 
 
 two Sheets 
 
 97. If two surfaces of revolution have a common axis, what 
 will be their line of intersection? 
 
 What must be the relative position of the axis and the gen- 
 eratrix to form a cylinder of revolution? A cone of revolution? 
 A hyperboloid of revolution of one sheet ? 
 
 The cylinder and cone are the only single curved surfaces of 
 revolution. 
 
 The hyperboloid of revolution of one sheet is the only warped 
 surface of revolution. 
 
TANGENT PLANES TO SURFACES 
 
 89 
 
 TANGENT PLANES TO SURFACES. NORMAL LINES AND PLANES 
 
 98. Let ADi and ADg, Fig. 58, be any two intersecting curves 
 on a surface, and BCD a curve which, if moved along the curves 
 ADi and ADg will generate the given surface. The curve BCD 
 may vary in form as it moves. The secants AB, BC, and AC lie 
 in one plane. As the curve BCD moves toward A, the points of 
 intersection B and C will travel along the curves AD^ and ADg 
 until the secants AB and AC become the tangents AB^ and ACg 
 at the point A. The curve ADg is the position of the moving 
 curve BCD when the points B and C reach A. In this position, 
 the secant BC becomes the tangent B3C3. The tangents ABj, 
 AC2, and B3C3 all lie in one plane which is called the tangent 
 plane to the surface at the point A. The point A is the point 
 of contact. 
 
 Since AD^ and AD2 are any 
 curves of the surface, it follows 
 that, in general, the tangent at 
 A to every curve of the surface 
 through this point will lie in the 
 tangent plane. Therefore, the 
 tangent plane will contain all 
 straight lines tangent to lines 
 of the surface at the point of 
 contact. 
 
 If any plane be passed through the point of contact, it will 
 cut a straight line from the tangent plane and a line from the 
 surface, and these lines will be tangent to each other. 
 
 Since two intersecting lines will determine a plane, it follows 
 that a plane tangent to a surface at a given point is determined 
 by two straight lines tangent at this point to two lines of the 
 surface. 
 
 In any ruled surface, the rectilinear element through the 
 point of contact lies in the tangent plane; for the rectilinear 
 element and its tangent coincide (Art. 82). 
 
 Fig. 58. — Tangent plane to a sur- 
 face. 
 
90 DESCRIPTIVE GEOMETRY 
 
 99. A tangent plane to a single curved surface is tangent all 
 along a rectilinear element. In Fig. 59, let BD be the curve of 
 the base of any single curved surface as a cylinder and AB and 
 AjBi any two rectilinear elements of the surface. From any 
 point A of one element draw on the surface a curve AD^ which 
 cuts the other element at A^. The chords AA^ and BB^ Ue in 
 the same plane and will intersect at some point as Si. Now if 
 the plane ABB^Ai be revolved about AB as an axis, the point A^ 
 will approach A along the curve AD^ and B^ will approach B 
 along the curve BD. When the point A^ reaches A, the secant 
 ASi becomes the tangent AS to the curve AJD^, and at the same 
 time Bi reaches B, the secant S^B becoming the tangent SB to 
 the curve BD. The plane SAB is tangent to the cylinder at the 
 point B (Art. 98), since it contains the tangent SB to the curve 
 BD, and also the rectilinear element AB. This plane is also 
 tangent to the cylinder at A, containing the tangent SA to the 
 curve ADi and the element AB. Since A was taken as any 
 point on the element AB, the plane SAB must be tangent to the 
 surface at every point of this element. 
 
 The same demonstration ap- 
 plies to the tangent plane to a 
 cone. 
 
 In the case of the convolute, 
 the element A^B^ is a curve in- 
 stead of a straight line. The 
 curve will change its form as 
 the secant plane revolves, finally 
 becoming a straight line coin- 
 FiQ. 59.— Tangent plane to single ciding with the axis AB. Other- 
 curved surface. ^ise the demonstration given 
 
 above also applies to the convolute. 
 
 Since the plane is tangent all along the element, the intersec- 
 tion of the tangent plane with the plane of the base will be a 
 straight line tangent to the curve which represents the base 
 (Art. 98). 
 
TANGENT PLANES TO SURFACES 91 
 
 100. Tangent planes to warped surfaces. Since warped sur- 
 faces have straight line elements, the tangent plane to the sur- 
 face at any point will contain the rectilinear element through 
 that point (Art. 98). If any curve of the surface be drawn 
 through the point of contact, the straight line tangent to the 
 curve at this point and the rectilinear element through the point 
 will determine the tangent plane (Art. 98). 
 
 If the warped surface is of such a form that there are two 
 rectilinear elements through each point of the surface, then the 
 tangent plane will be determined by the two elements through 
 the point of contact. 
 
 Although the tangent plane to a warped surface contains a 
 rectilinear element of the surface, it is, in general, tangent at 
 only one point of this element. The tangent plane usually in- 
 tersects the surface. These principles will be better under- 
 stood when the tangent planes to some of the surfaces are 
 drawn. 
 
 101. Tangent planes to double curved surfaces. Through the 
 point on the surface at which the tangent plane is to be passed, 
 draw any two curves of the surface. The straight lines tan- 
 gent to these curves at their point of intersection will determine 
 the tangent plane. If care is taken to select simple curves of 
 the surface in simple positions with reference to the planes of 
 projection, the tangents to these can be easily drawn. 
 
 In the case of a double curved surface of revolution, the 
 simplest curves are usually the meridian curve and a circle 
 which lies in a plane perpendicular to the axis of the surface. 
 
 102. Normals. A straight line perpendicular to the tangent 
 plane at the point of contact is the normal to the surface at that 
 point. Any plane containing a normal line is a normal plane 
 to the surface. 
 
92 DESCRIPTIVE GEOMETRY 
 
 SINGLE CURVED SURFACES 
 
 CYLINDERS 
 
 103. A cylinder is the path of a straight line moving along a 
 curve and remaining parallel to another straight line. 
 
 The curve is the directrix and the moving line is the recti- 
 linear generatrix. 
 
 The directrix can be a closed curve as the circle or ellipse or 
 a curve which is not closed as the hyperbola or helix. 
 
 The intersection of the cylinder with a horizontal plane is 
 usually taken as the base. Any plane section may be consid- 
 ered to be the base. If this base has a center, the straight line 
 through it and parallel to the rectilinear elements is the axis of 
 the cylinder. 
 
 A right cylinder is one in which the rectilinear elements are 
 perpendicular to the plane of the base. 
 
 A section taken perpendicular to the elements of any cylinder 
 is a riifht section. If a cylinder is intersected by a plane par- 
 allel to the rectilinear generatrix it will cut rectilinear ele- 
 ments from the surface. 
 
 104. To represent the surface. A cylinder is usually repre- 
 sented by the projections of its base and its extreme elements. 
 
 In Fig. 60, AECG is the base and EF the rectilinear genera- 
 trix. The extreme elements in the horizontal view are tangent 
 to the base and parallel to the horizontal projection of the rec- 
 tilinear generatrix. The extreme elements in the vertical view 
 are drawn through the extreme points of the vertical projec- 
 tion of the base and parallel to the vertical projection of the 
 rectilinear generatrix. 
 
 Since the surface is indefinite in extent, it can be represented 
 with the upper end broken off unless some definite portion of 
 the surface is taken for a particular purpose. 
 
 105. To represent a rectilinear element of the surface. 
 Through any point M of the base draw a straight line MN par- 
 
CYLINDERS 
 
 93 
 
 allel to the rectilinear generatrix ; this will be an element of the 
 surface. 
 
 To represent a point of the surface. Locate a rectilinear 
 element, as MN, and then take any point P of this element. 
 
 106. To represent a plane which contains a given point and is 
 tangent to a cylinder. 
 
 The tangent plane to a cylinder is tangent all along a rectilinear 
 element. 
 
 Fig. 60. — Cylinder. 
 
 The intersection of the tangent plane with the plane of the hose 
 is ta/ngent to the base. 
 
 Let the cylinder be given as in Fig. 61, and let P be the given 
 point. 
 
 Since the tangent plane must contain a rectilinear element of 
 the cylinder, a straight line through P and parallel to the rec- 
 tilinear elements will lie in the tangent plane and will pierce 
 the plane of the base at M. A tangent to the base of the cylin- 
 der through the point M is a horizontal of the required tangent 
 plane. An element of the cylinder through the point of tan- 
 
94 
 
 DESCRIPTIVE GEOMETRY 
 
 gency will be the element of contact. This element, the hori- 
 zontal OM, and the line PM will determine the tangent plane. 
 
 There can be as many tangent planes to the cylinder through 
 the given point as there can be lines drawn tangent to the base 
 of the cylinder through the point where the auxiliary line 
 pierces the plane of the base. 
 
 Fig. 61. — Tangent plane to cylinder. 
 
 107. Problems. 
 
 1. A right circular -cylinder and a point are given by their projec- 
 tions. Represent a plane which contains the point and is tangent to 
 the cylinder. 
 
 2. Represent a plane which is tangent to a cylinder at a given point 
 of the surface. 
 
 3. Represent a plane which is tangent to a cylinder and parallel to a 
 given straight line. (A plane containing the given line and a line 
 parallel to the elements of the cylinder will be parallel to the required 
 tangent plane.) 
 
 4. Given the projections of a point and a right circular cylinder 
 which has its axis parallel to the ground line. Represent a plane which 
 contains the point and is tangent to the cylinder. 
 
 5. Represent a plane which makes 45° with H and is tangent to a 
 right circular cylinder. The axis of the cylinder is parallel to H and 
 makes 30° with V. 
 
CONES 
 
 95 
 
 CONES 
 
 108. A cone is the path of a straight line moving along a 
 curve, and passing through a fixed point. If the directrix is a 
 plane curve, the fixed point does not lie in the plane of the 
 curve. 
 
 The curve is the directrix, the fixed point the vertex and the 
 moving line the rectilinear generatrix of the cone. 
 
 m' g' 
 Fig. ^2.— Cone. 
 
 The directrix can be a closed curve as a circle or an ellipse 
 or a curve which is not closed as a hyperbola or a helix. 
 
 The generatrix being indefinite in length, will generate a sur- 
 face of two parts which are on opposite sides of the vertex and 
 which are called the upper and lower sheets. 
 
 The intersection of the cone with a horizontal plane is usu- 
 ally taken as the base, but any plane section of the cone may be 
 considered the base. 
 
96 DESCRIPTIVE GEOMETRY 
 
 A right cone is one in which all the rectilinear elements make 
 the same angle with a straight line through the vertex called 
 the axis of the cone. 
 
 Any plane passing through the vertex and intersecting the 
 cone will cut rectilinear elements from the surface. 
 
 109. To represent the surface. A cone is usually represented 
 by the projections of its base, vertex and extreme elements. 
 
 In Fig. 62, AECG is the base and V the vertex of the cone. 
 The extreme elements in the horizontal view are tangent to the 
 base and pass through the horizontal projection of the vertex. 
 The extreme elements in the vertical view are drawn through 
 the extreme points of the vertical projection of the base and 
 through the vertical projection of the vertex. 
 
 110. To represent a rectilinear element of the surface. 
 Through any point M of the base, draw a straight line through 
 the vertex V ; this will be an element of the surface. 
 
 To represent a point of the surface. Locate a rectilinear ele- 
 ment, as MV, and take any point P of this element. 
 
 111. To represent a plane which contains a given point and is 
 tangent to a cone. 
 
 The tcmgent plane to a cone is tangent all along a rectilinear 
 element. 
 
 The intersection of the tangent plane with the plane of the base 
 is tangent to the base. 
 
 Let the cone be given as in Fig. 63, and let P be the given 
 point. 
 
 Since the required plane must contain a rectilinear element 
 of the cone, and therefore the vertex, a straight line joining the 
 vertex with the given point P will lie in the tangent plane and 
 pierce the plane of the base at M. A line through the point M 
 and tangent to the base of the cone is a horizontal of the re- 
 quired tangent plane. An element through the point of tan- 
 gency is the element of contact. This element, the horizon- 
 tal and the line YP all lie in the required tangent plane. 
 
CONES 
 
 97 
 
 There are as many solutions to the problem as there are tan- 
 gents to the base of the cone from the point in which the auxil- 
 iary line pierces the plane of the base. The problem cannot be 
 solved when the given point is inside the cone. 
 
 m' o' 
 
 Fig. 63. — Tangent plane to cone. 
 
 112. Problems. 
 
 1. A right cone and a point are given by their projections. Repre- 
 sent a plane which contains the point and is tangent to the cone. 
 
 2. A right cone with its base in a profile plane and a point are given 
 by their projections. Represent a plane which contains the point and 
 is tangent to the cone. 
 
 3. A right cone with its axis parallel to H and oblique to V and a 
 point are given by their projections. Represent a plane which contains 
 the point and is tangent to the cone. 
 
 4. Represent a plane which is tangent to a given oblique cone at a 
 point on the surface. 
 
 5. Represent a plane which is tangent to a given cone and parallel 
 to a given straight line. (A line through the vertex of the cone and 
 parallel to the given line will lie in the required tangent plane.) 
 
98 
 
 DESCRIPTIVE GEOMETRY 
 
 CONVOLUTES 
 
 113. A convolute is the path of a straight line moving along 
 and remaining tangent to a curve of double curvature such as 
 the helix. The surface, Fig. 64, is represented by drawing the 
 
 Fig. 64. — Convolute. 
 
 projections of the curvilinear directrix, an occasional element 
 and the intersection of the surface with a horizontal plane. If 
 the curvilinear directrix is a helix with axis perpendicular to 
 the horizontal plane, the base of the surface will be the in- 
 volute of the circle which represents the horizontal projection 
 of the helix. 
 
 This surface is of very little practical importance, so no fur- 
 ther discussion of it will be given. For a complete discussion 
 of it, see Salmon's Geometry of Three Dimensions, Art. 349. 
 
WABPED SURFACES 99 
 
 WARPED SURFACES 
 
 114. There is a great variety of warped surfaces, differing 
 from each other in their mode of generation and properties. 
 Only a few of the more common ones will be discussed. 
 
 There are several kinds of warped surfaces formed by mov- 
 ing a straight line to touch two other lines, straight or curved, 
 and keeping it parallel to a given plane, called a plane directer. 
 
 Warped surfaces are usually represented by drawing the 
 projections of one or more curves of the surface and a few of 
 the rectilinear elements. The projections of the directrices are 
 often given. 
 
 HYPERBOLOIDS OP REVOLUTION OF ONE SHEET 
 
 115. A hyperboloid of revolution of one sheet is the path of a 
 straight line revolving about another straight line as an axis, 
 the generatrix and axis not being in the same plane. 
 
 From the nature of the motion it is evident that no two posi- 
 tions of the generatrix can be brought into one plane without 
 distortion, therefore the surface is warped (Art. 95). 
 
 This is the only warped surface of revolution. 
 
 116. To represent the surface. In Fig. 65, MN is the axis and 
 AB the initial position of the generatrix. Each point of the 
 generatrix AB describes a horizontal circle. The point R, near- 
 est the axis, describes a circle of radius nr, called the circle of 
 the gorge. The point A describes a circle on a horizontal plane 
 usually considered the base of the surface. The horizontal 
 projection of the base and of the circle of the gorge form the 
 horizontal view of the surface. 
 
100 
 
 DESCRIPTIVE GEOMETRY 
 
 In each position of the genera- 
 trix AB there is a radius of the 
 circle of the gorge perpendicular 
 to it. Since the radius is parallel 
 to H, its horizontal projection will 
 be perpendicular to the horizon- 
 tal projection of the generatrix. 
 Hence the horizontal projections 
 of the elements of the surface are 
 tangent to the horizontal projec- 
 tion of the circle of the gorge. 
 The vertical projection of the cir- 
 cle of the gorge is a horizontal line 
 through / and the vertical pro- 
 jection of the base is a horizontal 
 line through a'. Therefore, to rep- 
 resent an element of the surface, 
 draw its horizontal projection cd 
 tangent to the horizontal projec- 
 tion of the circle of the gorge. The 
 vertical projection of C is c' and of 
 the point of tangency E is e'. 
 Hence c'e'd' is the vertical projec- 
 tion of the element. 
 The vertical projection of the surface is the hyperbola which 
 is the envelope of the vertical projections of the rectilinear 
 elements of the surface. Definite points on this hyperbola are 
 found by finding the points in which the rectilinear elements 
 pierce the meridian plane which is parallel to V. 
 
 To represent a point of the surface. Locate an element CD 
 and then take any point P of this element. 
 
 117. In Fig. 65, draw a line XY through the point R making 
 the same angle with H and having the same horizontal projec- 
 tion as the generatrix AB. For each point of XY there is a 
 corresponding point of AB which is the same distance from H 
 and also the same distance from MN. As AB and XY revolve 
 
 FiQ. 65. — Hyperholoid of rev- 
 olution of one sheet and tan- 
 gent plane. 
 
WARPED SURFACES 101 
 
 about MN as an axis, these two points will generate the same 
 circle and the two lines will therefore generate the same sur- 
 face. Hence it is evident that through every point of the sur- 
 face two rectilinear elements can be drawn. 
 
 118. To represent a plane which is tangent to a hyperboloid 
 of revolution of one sheet at a point of the surface. 
 
 Let the surface be given as in Fig. 65 and let P be the given 
 point (Art. 116). 
 
 Analysis. Draw the two rectilinear elements of the surface 
 which pass through the given point. Since the tangents to 
 these rectilinear elements coincide with the elements them- 
 selves, the elements determine the tangent plane to the surface 
 at the given point. Either of these rectilinear elements with 
 the tangent to the horizontal circle of the surface through the 
 given point will also represent the required tangent plane to 
 the surface. 
 
 Construction. Through P draw the rectilinear elements PC 
 and PO (Art. 116). These elements determine the required 
 tangent plane. OC is the line of intersection of this plane with 
 ihe plane of the base. Since the horizontal OC of this plane 
 intersects the circle of the base of the surface, it is evident that 
 the tangent plane is also an intersecting plane. 
 
 Since a rectilinear element of the surface, as KG, which 
 passes through any point K of the element OP does not inter- 
 sect the line OC, it does not lie in the tangent plane OPC. 
 Therefore OPC is not tangent to the surface at the point K. 
 Hence a tangent plane to a warped surface will contain a rec- 
 tilinear element of the surface but will be tangent to the surface 
 at only one point of that element. 
 
 HELICOIDS 
 
 119. A helicoid is the path of a straight line moving along 
 two helices having the same pitch and lying on concentric cylin- 
 ders of revolution. The generatrix makes a constant angle 
 with the plane of the bases of the cylinders and may or may not 
 intersect their common axis. 
 
102 DESCRIPTIVE GEOMETRY 
 
 From the nature of the motion it is evident that no two posi- 
 tions of the generatrix can be brought into one plane without 
 distortion, therefore the surface is warped (Art. 95). 
 
 If the generatrix does not cut the common axis of the cylin- 
 ders, it must remain a constant distance from it. 
 
 Each point of the generatrix will generate a helix having the 
 same axis and pitch as the directrices. In the particular case 
 where the generatrix is tangent to one of the directing helices, 
 the surface generated is a developable helicoid or convolute 
 which is a single curved surface. 
 
 When the generatrix is perpendicular to the axis, the surface 
 is a right helicoid, as in the square screw thread. When the 
 generatrix is at an acute angle with the axis, the surface is an 
 oblique helicoid, as in the triangular screw thread. 
 
 120. To represent the surface. A helicoid is represented by 
 showing the projections of its directrices, a few of its rec- 
 tilinear elements and its intersection with a horizontal plane. 
 
 In Fig. 66, let A, B, C be one of the helical directrices 
 
 and BBi the initial position of the generatrix. The generatrix 
 BBi makes the required angle oc with the plane of the bases of 
 the cylinders and is a constant distance d from the axis MN. 
 Any point Bj of BB^ can be selected as the point which is to 
 follow the other helical directrix. This directrix is then con- 
 structed having the same pitch and axis as the other directrix. 
 The horizontal projection of all the elements of the surface are 
 tangent to the circle which has d for a radius and mn for a cen- 
 ter. The vertical projection of any element CCi is found by 
 finding the vertical projections of the points C and C^ where the 
 element cuts the helical directrices and joining these points 
 
 with a straight line. Joining the points A, Bg, Cg, in 
 
 which these elements intersect the plane of the base gives the 
 base of the surface. 
 
 The helicoid can be constructed by first drawing the projec- 
 tions of both helical directrices. It is then necessary to draw 
 the rectilinear element in such a position that it will touch both 
 helices and make the required angle with the plane of the. base. 
 
WARPED SURFACES 
 
 103 
 
 This is accomplished by using any point on either directrix as 
 the vertex of a cone of revolution having its elements making 
 the required angle with the plane of the base. The point in 
 which the other directrix pierces this cone joined with the ver- 
 
 FiG. 66. — Helicoid and tangent plane. 
 
 tex of the cone gives one element of the helicoid. After this 
 element is drawn, the other elements can be drawn by the first 
 method given above. 
 
 A point of the surface is represented by first representing an 
 element of the surface and then taking a point of the element. 
 
 121. To represent a plane which is tangent to a helicoid at a 
 point of the surface. 
 
104 DESCRIPTIVE GEOMETRY 
 
 Let the surface be ^iven as in Fig. 66 and let Ci be the given 
 point, represented as in Art. 120. 
 
 Analysis, Construct a helix passing through the point Ci 
 and lying on the surface of the helicoid. The tangent to the 
 helix at the point C^ and the rectilinear element of the surface 
 through this point vrill determine the tangent plane. 
 
 Construction. The helix through the point C^ is constructed 
 by the method of Art. 93. The pitch of this helix must be the 
 
 same a§ the pitch of the helix ABC w^hich represents the 
 
 helicoid. The arc oc^ is the horizontal projection of part of this 
 helix. The straight line c^o^ is equal to the arc c^^o. Since the 
 point Oi is on the same level as the point 0, its vertical projec- 
 tion o\ is on a line through o' parallel to the ground line. Then 
 c^Oi and c\o\ are the projections of the tangent to the helix at 
 the point C^ (Art. 94). cc^ and dc\ are the projections of the 
 element of the helicoid passing through Ci (Art. 120). The 
 plane v^hich is determined by the two lines CiOi and CjC is the 
 required tangent plane. CgOg would be a horizontal of this plane. 
 
 The plane which is tangent to the helicoid at any point C^ is 
 not tangent to the surface at any other point as C, Fig. 66. 
 The plane CgCiOg is tangent to the surface at Cj. In like man- 
 ner, the plane of CCg and the tangent to the helix ABC at 
 
 C is tangent to the surface at the point C. These planes do not 
 coincide since the tangents to the helices at C and Ci are neither 
 parallel nor intersecting lines. Hence the plane which is tan- 
 gent to the helicoid at Ci is not tangent to the surface at any 
 other point. 
 
 HYPERBOLIC PARABOLOIDS 
 
 122. A hyperbolic paraboloid is the path of a straight line 
 moving along two straight lines not in the same plane, and re- 
 maining parallel to a given plane. 
 
 The moving line is called the generatrix, the two fixed 
 straight line the directrices and the given plane the plane di- 
 rector. Any position of the generatrix is called an element of 
 the surface. 
 
WARPED SURFACES 105 
 
 This surface has a second rectilinear generation in which any 
 two rectilinear elements of the first generation may be taken as 
 directrices and a plane which is parallel to the first directrices 
 as a plane directer. It follows that through any point of a 
 hyperbolic paraboloid, two rectilinear elements can always be 
 drawn. 
 
 Any intersection of the surface by a plane, which is not a 
 straight line intersection, is a hyperbola or a parabola, hence 
 the name of the surface. 
 
 It is evident from the nature of the motion of the generating 
 line that the surface is warped. 
 
 123. To represent the surface. The surface is usually rep- 
 resented by the projections of the two rectilinear directrices, 
 the plane directer and some of the rectilinear elements. 
 
 To represent a rectilinear element of the surface. Let AB 
 and CD, Fig. 67, be the rectilinear directrices and let V be the 
 plane directer. Through any point M on the directrix AB, 
 draw MN parallel to the plane directer V and intersecting the 
 other directrix CD at N. mn and mV are the horizontal and 
 vertical projections respectively of a rectilinear element of the 
 surface. 
 
 To represent a point of the surface. Eepresent a rectilinear 
 element, as MN, Fig. 67, then take any point of this element as 
 P. p and p' are the projections of the required point. 
 
 124. To represent a plane which is tangent to the surface at 
 a point of the surface. 
 
 Let 0, Fig. 67, be the given point. 
 
 Analysis. Through the given point pass a plane parallel to 
 the elements of each generation. These planes will intersect 
 the surface in the two rectilinear elements which determine the 
 tangent plane at the given point. 
 
 Construction. The plane through parallel to the elements 
 of the first generation will cut the element VW from the sur- 
 face. Through draw OG parallel to CD, and OU parallel to 
 AB. These lines will determine a plane parallel to the elements 
 of the second generation. This plane will intersect the element 
 
106 
 
 DESCRIPTIVE GEOMETRY 
 
 XY in Z. The line joining Z with will be an element of the 
 second generation. OW and OZ will therefore determine the 
 required tangent plane to the surface at the point 0. 
 
 /' 
 
 Fig. 67. — Hyperholic paraboloid and tangent plane. 
 
 CONOIDS 
 
 125. A conoid is the path of a straight line moving along two 
 other lines, one straight and the other curved, and remaining 
 parallel to a given plane. 
 
 To represent the surface. A conoid is usually represented by 
 the projections of the directrices, the plane directer, and some 
 of the rectilinear elements. 
 
WARPED SURFACES 
 
 107 
 
 To represent a rectilinear element of the surface. Let the 
 circle ABCD, Fig. 68, and the straight line MN, be the given 
 
 directrices, and let V be the plane 
 directer. Through any point E on 
 the curved directrix draw EM par- 
 allel to the plane directer V, and 
 intersecting the rectilinear direc- 
 trix at M. em is the horizontal 
 and e'm' the vertical projection of 
 a rectilinear element of the sur- 
 face. 
 
 To represent a point of the sur- 
 face. Represent a rectilinear ele- 
 ment, as EM, Fig. 68, and then 
 take any point of this element, as 
 0. and o' are the projections of 
 a point of the surface. 
 
 126. To represent a plane which 
 is tangent to the surface at a given 
 point of the surface. Let 0, Fig. 
 68, be the given point. The tan- 
 gent plane must contain the rectilinear element EM through the 
 given point. It must also contain the line OP tangent to the 
 elliptical section at the same point. EM and OP will therefore 
 represent the required tangent plane. 
 
 -Conoid and tangent 
 plane. 
 
 CYLINDROmS 
 
 127. A cylindroid is the path of a straight line moving along 
 two curves, and remaining parallel to a given plane. 
 
 To represent the surface. A cylindroid is usually repre- 
 sented by the projections of the two curved directrices, the 
 plane directer, and some of the rectilinear elements. 
 
 To represent a rectilinear element of the surface. Let ABC 
 and DEG, Fig. 69, be l^he curved directrices, and let H be the 
 
108 
 
 DESCRIPTIVE GEOMETRY 
 
 plane directer. Through any point B of one directrix draw a 
 straight line parallel to the plane directer H and intersecting 
 the other directrix at E. h'e' is the vertical and te the horizon- 
 tal projection of a rectilinear element of the surface. 
 
 Fig. 69. — Cylindroid and tangent plane. 
 
 To represent a point of the surface. Represent any element, 
 as BE, Fig. 69, and then take any point of this element, o and 
 o' are the projections of a point of the surface. 
 
 128. To represent a plane which is tangent to the surface at a 
 given point of the surface. Let P, Fig. 69, be the given point. 
 The tangent plane must contain the rectilinear element CP 
 through this point and the tangent to a plane section of the 
 surface at P. NPM is a section of the surface by a plane per- 
 pendicular to V. pk is the horizontal and p'k' the vertical pro- 
 jection of the tangent to this curve at P. Then CP and PK 
 represent the tangent plane to the surface at the point P. 
 
DOUBLE CURVED SURFACES OF REVOLUTION 109 
 
 DOUBLE CURVED SURFACES 
 
 (Only double curved surfaces of revolution will he discussed.) 
 
 129. Some of the simple double curved surfaces of revolution 
 are the following : 
 
 A sphere is the path of a circle revolving about its diameter 
 as an axis. 
 
 A torus or anchor ring is the path of a circle revolving about 
 a straight line which lies in the plane of the circle but does not 
 cut the circumference. 
 
 An ellipsoid of revolution is the path of an ellipse revolving 
 about either axis. "When the ellipse is revolved about its major 
 axis, a prolate ellipsoid is generated; when about its minor 
 axis, an oblate ellipsoid. These surfaces are sometimes called 
 spheroids. 
 
 A paraboloid of revolution is the path of a parabola revolving 
 about its axis. 
 
 A hyperboloid of revolution of two sheets is the path of a 
 hyperbola revolving about the axis which passes through the 
 foci. 
 
 These surfaces of revolution are usually represented with the 
 axis perpendicular to the horizontal plane. The surface is rep- 
 resented in the horizontal view by its intersection with a hori- 
 zontal plane or by the projection of its largest horizontal circle. 
 The surface is represented in the vertical view by the projection 
 of the meridian curve parallel to the vertical plane. 
 
 130. To represent a point on a surface of revolution. 
 
 Let the surface be given with the axis perpendicular to the 
 horizontal plane. 
 
 Assume the horizontal projection of the point any place on 
 the horizontal projection of the surface. Through this point, 
 draw a circle with center at the horizontal projection of the 
 axis. This is the horizontal projection of a circle lying on the 
 surface. By projection, the vertical projection of the circle, 
 which is a straight line parallel to the ground line, is found. 
 The vertical projection of the point is on the vertical projec- 
 
110 DESCRIPTIVE GEOMETRY 
 
 tion of the circle. These projections represent a point on the 
 surface. 
 
 If the vertical projection of the point is assumed, the vertical 
 projection of the circle must be drawn first, and then the hori- 
 zontal projection of the circle and point found. 
 
 131. Problems. 
 
 1. Represent a point on the surface of a sphere. 
 
 2. Represent a point on the surface of a prolate ellipsoid. 
 
 3. Represent a point on the surface of an oblate ellipsoid. 
 
 4. Represent a point on the surface of a paraboloid of revolution. 
 6. Represent a point on the surface of a torus. 
 
 132. Tangent planes to double curved surfaces of revolution. 
 
 The tangent plane to a surface at a point on the surface is de- 
 termined by two straight lines tangent at this point to two lines 
 of the surface. The simplest curves passing through the point 
 on a surface of revolution are usually the meridian line and the 
 circle which is cut from the surface by a plane perpendicular 
 to the axis of the surface. The tangent plane, therefore, is 
 determined by two straight lines, one tangent to the circular 
 section at the given point and the other tangent to the meridian 
 section at that point. 
 
 133. To represent a plane which is tangent to an ellipsoid of 
 revolution at a point of the surface. 
 
 Let the surface be given as in Fig. 70, and let P be the given 
 point represented by the method of Art. 130. 
 
 Analysis. A tangent to the meridian curve at the given point 
 and a tangent to the circle of the surface at this point will rep- 
 resent the tangent plane (Art. 132). 
 
 Construction, prx is the horizontal and r'x' the vertical pro- 
 jection of the circle of the surface through the given point P. 
 pm is the horizontal projection and p'm' the vertical projection 
 of the tangent to the circle at the point P. ap is the horizontal 
 view of the meridian section through P. If the meridian curve 
 through P is revolved about the axis of the surface until the 
 plane of the curve is parallel to V, it will have the ellipse rVi/ 
 for its vertical projection, P moving to X. At x' draw icV tan- 
 
DOUBLE CURVED SURFACES OF REVOLUTION 
 
 111 
 
 gent to the ellipse xVy', Then x'a! is the vertical projection of 
 the revolved position of the tangent to the meridian section at 
 P. In the counter revolution of the meridian plane, X moves 
 
 Fig. 70. — Elli'^soid, of revolution and tan- 
 gent 'plane, 
 
 to P and the point A, in the axis, remains fixed. Then ap is the 
 horizontal and a'p' the vertical projection of the tangent to the 
 meridian curve at P. PM and PA represent the required plane 
 which is tangent to the surface at the point P. 
 
112 DESCRIPTIVE GEOMETRY 
 
 134. Problems. 
 
 1. Represent a plane which is tangent to a sphere, (a) at a point of 
 the surface; (b) and which contains a point outside the surface. 
 
 2. Represent a plane which is tangent to a prolate ellipsoid, (a) at a 
 point of the surface; (b) and which contains a point outside the surface. 
 
 3. Represent a plane which is tangent to a paraboloid of revolution, 
 (a) at a point of the surface; (b) and which contains a point outside 
 the surface. 
 
 4. Represent a plane which is tangent to a torus, (a) at a point of 
 the surface; (b) and which contains a point outside the surface. 
 
 135. To represent a plane which contains a given straight line 
 and is tangent to a sphere. 
 
 Let MN, Fig. 71, be the given straight line and let C be the 
 center of the given sphere. 
 
 Analysis. Assume that the required plane is drawn through 
 the given line and tangent to the sphere. Now if an auxiliary- 
 plane be passed through the center of the sphere and perpen- 
 dicular to the line MN, it will cut a point from MN, a great 
 circle from the sphere and a line from the tangent plane which 
 will pass through the point on MN and be tangent to the great 
 circle cut from the sphere. Therefore, to make the construc- 
 tion for the tangent plane, pass a plane through the center of 
 the sphere and perpendicular to the line MN. From the point 
 in which this plane cuts MN, draw a tangent to the great circle 
 cut from the sphere. This tangent and the line MN determine 
 the required tangent plane. 
 
 CoTistruction. HH and FF represent a plane which contains 
 the center of the sphere and is perpendicular to the liue MN 
 (Art. 50). D is the point in which the line MN pierces this 
 plane. When the plane of HH and FF, which contains a great 
 circle of the sphere and the point D, is revolved about HH as an 
 axis until it is parallel to H, the point D moves to Dg and the 
 great circle has the same horizontal projection as the sphere, 
 c^g^z is the horizontal projection of the revolved position of a 
 tangent from D to the great circle. By counter revolution the 
 
DOUBLE CURVED SURFACES OF REVOLUTION 
 
 113 
 
 projections de and d'e' of the required tangent are found. 
 DE and MN represent the required tangent plane. 
 
 Then 
 
 Fig. 71. — Plane containing line and tangent to sphere. 
 
 136. Problems. 
 
 1. Represent a plane which contains a line parallel to V and is tan- 
 gent to a given sphere. 
 
 2. Represent a plane which contains a line parallel to H and is tan- 
 gent to a given sphere. 
 
 3. Represent a plane which contains a line parallel to G. L. and is 
 tangent to a given sphere. 
 
 4. Represent a plane which contains a line in a profile plane and is 
 tangent to a given sphere. 
 
 PROBLEMS 
 
 1. To a given oblique cone with base in a horizontal plane, pass a 
 tangent plane which makes a given angle with H. 
 
 2. To a given oblique cone with base on a plane parallel to V, pass a 
 tangent plane which makes a given angle with V. 
 
114 DESCRIPTIVE GEOMETRY 
 
 3. To a given oblique cylinder with base in a horizontal plane, pass a 
 tangent plane which makes a given angle with H. 
 
 4. To a given oblique cylinder with base on a plane parallel to V, 
 pass a tangent plane which makes a given angle with V. 
 
 5. Find a plane which contains the ground line and is tangent to a 
 given sphere. Find the angles which the plane makes with the planes 
 of projection. 
 
 6. Pass a plane tangent to a sphere and parallel to a given plane. 
 
 7. Inscribe a sphere in a given tetrahedron. 
 
 8. Having given the axis of a cylinder of revolution and a tangent 
 plane, draw the projections of the element of contact. (The tangent 
 plane must be parallel to the axis. ) 
 
 9. Through a given point, pass a plane tangent to a cylinder of rev- 
 olution, having given the axis and radius of the cylinder, without find- 
 ing the projections of the cylinder. 
 
 10. Having given the radius and the axis of a cylinder of revolution, 
 to pass a plane tangent to the cylinder and parallel to a given straight 
 line, without finding the projections of the cylinder. 
 
 11. A circle which lies in a plane parallel with H is the base of a 
 cone. The vertex of the cone is in G. L. Through some other point on 
 Gr. L., pass a plane tangent to the cone. Show the angles which the 
 plane makes with the planes of projection. 
 
 12. Find a common normal to two cylinders of revolution. 
 
 13. To two given cylinders, pass tangent planes which are parallel 
 to each other. 
 
 14. To a given cone and cylinder, pass tangent planes which are par- 
 allel to each other. 
 
 15. Two cones with their bases on H have the same altitude; pass a 
 plane tangent to each so that the planes will be parallel. 
 
 16. Through a given point, pass a plane tangent to two given spheres, 
 
 17. Pass a plane tangent to two given spheres and parallel to a given 
 straight line. 
 
 18. Pass a plane tangent to three given spheres. 
 
 19. Through a given point, pass a plane which is equidistant from 
 three given spheres. 
 
 20. Through a given straight line, pass a plane which is equidistant 
 from two given spheres. 
 
 21. Pass a plane which is equidistant from four given spheres. 
 
 22. Find a common tangent plane to a sphere and a cylinder of rev- 
 olution. 
 
 23. Find a common tangent plane to a sphere and a cone of revolution. 
 
CHAPTER Y 
 
 PLANE SECTIONS AND DEVELOPMENTS OF CURVED 
 
 SURFACES 
 
 137. The intersection of an oblique plane with a curved sur- 
 face which is ruled can be found by finding the points in which 
 the rectilinear elements of the surface pierce the given oblique 
 plane. These piercing points can be found by the usual method 
 of finding where a straight line pierces a plane or by using a 
 view of the surface taken on an auxiliary plane of projection as 
 described in Art. 61. This auxiliary plane is at right angles to 
 a horizontal or a frontal of the given oblique plane. A line 
 joining these piercing points in the proper order is the re- 
 quired intersection. The rectilinear elements should be taken 
 near enough together to give the line of intersection as accu- 
 rately as desired. 
 
 The intersection of an oblique plane with a double curved 
 surface is found by using a system of auxiliary planes. Each 
 of the auxiliary planes will cut a straight line from the oblique 
 plane and a curved line from the double curved surface and 
 these lines will intersect in points of the required line of inter- 
 section. The auxiliary planes should be so taken that they will 
 cut simple curves from the double curved surface and these 
 curves should be in simple positions with reference to the planes 
 of projection. 
 
 138. To find the intersection of a right circular cylinder with 
 a given oblique plane and to develop the surface of the cylinder. 
 
 Let the cylinder be given as in Fig. 72 and let AB and AC 
 represent the oblique plane. 
 
 Analysis. Find the points in which the rectilinear elements 
 of the cylinder intersect the given oblique plane. A curved line 
 
116 
 
 DESCRIPTIVE GEOMETRY 
 
 joining these points in the proper order is the required line of 
 intersection. 
 
 Construction. Since the elements of the cylinder are perpen- 
 dicular to H, the horizontal projecting plane of the element 
 
 Fig. 72. — Plane section and development of right cylinder. 
 
 through D may be the horizontal projecting plane of some other 
 element as the one through E. The line of intersection of this 
 projecting plane with the plane of AB and AC cuts the elements 
 D and E at X and Y, points on the required line of intersection. 
 As many other points as are necessary to determine an accurate 
 curve can be found in the same way. 
 
 139. To draw a straight line tangent to the curve of intersec- 
 tion at any point. Select the point at which the tangent is to 
 be drawn and then represent a plane tangent to the cylinder at 
 this point (Art. 106). The intersection of this plane with the 
 cutting plane is the required tangent line (Art. 98). In Fig. 72, 
 MN is the tangent to the curve of intersection at the point G. 
 
PLANE SECTIONS AND DEVELOPMENTS OF SURFACES 117 
 
 140. To develop the surface of the cylinder showing the curve 
 of intersection with the oblique plane. If the cylinder be rolled 
 on a tangent plane until each element has come into the plane, 
 the base will develop into the the straight line h^kz. The dis- 
 tance k202=ko, Ozd^^od, etc. In the development, the elements 
 will be perpendicular to kzk^ since they are perpendicular to the 
 plane of the base. To get the line of intersection on the de- 
 velopment, lay off kzZz^k'z'y 6L^x.^=dJx\ etc. A smooth curve 
 passing through the points z^^ x^, 2/2? z^ is the development of the 
 line of intersection. 
 
 141. To find the intersection of an oblique cylinder with a 
 given oblique plane and to develop the surface of the cylinder. 
 
 Let the cylinder be given as in Fig. 73 and let AB and AC 
 represent the oblique plane. 
 
 The plane is taken perpendicular to the elements of the cylin- 
 der. This is not necessary in order to get the intersection but 
 it is necessary for the development. 
 
 Analysis. Find the points in which the rectilinear elements 
 of the cylinder intersect the given oblique plane. A curved line 
 joining these points in the proper order is the required line of 
 intersection. 
 
 By following the above analysis, the intersection XYZG is 
 found. 
 
 142. To draw a straight line tangent to the curve of intersec- 
 tion at any point. EG and EN represent the tangent plane to 
 the cylinder at the point G (Art. 106). The intersection of this 
 plane with the plane of AB and AC is the line MN. MN is the 
 required tangent to the curve of intersection at the point G. 
 
 143. To develop the surface of the cylinder. Since the sec- 
 tion plane is perpendicular to the elements of the cylinder, it is 
 evident that the line of intersection will be a straight line in the 
 
 development. The true size of the right section is x"\f'z" 
 
 This is found by revolving the section until it is parallel to the 
 horizontal plane about AC as an axis. In the development lay 
 off x^Vz^^x'^', y2^2=y"^'\ etc., along the straight line w^w^. On 
 a line perpendicular to W2W2 at the point x^, lay off X20i=x^o; 
 
118 
 
 DESCRIPTIVE GEOMETRY 
 
 x^o being the true length of the element from the plane of the 
 section to the plane of the base. In a similar manner, lay off 
 the elements at 2/2? ^2? ^tc. A smooth curve through the points 
 ^2, 2^2j ^2j is the development of the base. 
 
 Fig. 73. — Plane section and development of oblique cylinder. 
 
 144. To find the intersection of an oblique cone with a given 
 oblique plane and to develop the surface of the cone. 
 
 Let the cone be given as in Fig. 74 and let AB and AC rep- 
 resent the oblique plane. 
 
 Analysis. Find the points in which the rectilinear elements 
 of the cone intersect the given oblique plane. A curved line 
 joining these points in the proper order is the required line of 
 intersection. 
 
 By following the above analysis, the intersection XYZG is 
 found. 
 
 145. To draw a straight line tangent to the curve of intersec- 
 tion at any point. KG and KN represent the tangent plane to 
 the cone at the point G (Art. 111). The intersection of this 
 
PLANE SECTIONS AND DEVELOPMENTS OF SURFACES 
 
 119 
 
 plane with the plane of AB and AC is the line MN. MN is the 
 required tangent to the curve of intersection at the point G. 
 
 146. To develop the surface of the cone showing the curve of 
 intersection with the oblique plane. Draw the elements V^D, 
 
 Fig. 74. — Plane section and development of oblique cone. 
 
 ViE, ViO, etc. Then lay off the true length v\d\ of V^D along 
 any straight line as ^2^2- With the point V2 as a center and a 
 radius equal to the true length v\e\ of the element V^E, strike 
 an indefinite arc. With the point (^2 ^^ ^ center and a radius 
 equal to the rectified arc DE, strike an arc cutting the first arc 
 at ^2. Then ^2^2 is the element V^E laid off on the development. 
 By using Vg as a center and a radius YjO and 63 as a center and 
 a radius equal to the rectified arc EO, the point O2 is located. 
 
120 DESCRIPTIVE GEOMETRY 
 
 By continuing this process, the development as shown in Fig. 74 
 is found. Points on the line of intersection are located on the 
 development by laying off V2X2=v\x\, V2y2=v\y\, etc. A 
 
 smooth curve through the points Xz, 2/2, jS2j ^2 is the curve 
 
 of intersection laid off on the development. The greater the 
 number of elements developed, the greater will be the accuracy 
 of the development. 
 
 147. To find the intersection of a warped surface with a given 
 oblique plane. 
 
 Analysis. Since a warped surface is a ruled surface, it is 
 only necessary to find the points in which the rectilinear ele- 
 ments of the surface pierce the given oblique plane. A line 
 joining these points is the required intersection. 
 
 148. To draw a straight line tangent to the curve of intersec- 
 tion at any point. Select the point of the curve at which the 
 tangent is to be drawn. The intersection of the tangent plane 
 to the surface at this point with the plane of the section is the 
 required tangent line. 
 
 A warped surface cannot be developed. 
 
 149. To find the intersection of any surface of revolution with 
 a given oblique plane. 
 
 Let the surface be given as in Fig. 75 and let AB and AC rep- 
 resent the oblique plane. 
 
 Analysis. Cut the surface and plane by a system of jauxiliary 
 planes perpendicular to the axis of the surface. The auxiliary 
 planes will cut circles from the surface which will intersect the 
 lines cut from the given plane in points of the required curve 
 of intersection. 
 
 Construction. Since the method is the same for all surfaces 
 of revolution, let the construction be made for the torus as given 
 in Fig. 75. The planes, T, Tj, Tg, etc., are parallel with H and 
 cut lines from the plane of AB and AC which are parallel to a 
 horizontal of this plane, and intersect the circles cut from the 
 torus in the points X, Xj ; Y, Y^ ; etc., points of the required 
 curve of intersection. By continuing in this manner enough 
 
PLANE SECTIONS OF SURFACES 
 
 121 
 
 points can be found to determine the curve of intersection of 
 the given oblique plane with the torus. 
 
 150. To determine particular points of the curve of intersec- 
 tion. To find the points where the curve of intersection touches 
 the highest and lowest circles of the torus, use the auxiliary 
 
 Fig. 75. — Plane section of surface of revolution. 
 
 planes which contain these circles. The vertical projections 
 of these circles are straight lines which form part of the 
 outline of the torus in the vertical view. To find the points 
 where the curve of intersection touches the circular part of the 
 outline of the surface in the vertical view, use the auxiliary- 
 plane Tj which contains the axis and is parallel to V. To find 
 
122 DESCRIPTIVE GEOMETRY 
 
 the points where the curve of intersection touches the circles 
 which represent the torus in the horizontal view, use the auxil- 
 iary plane Tg which contains the center of the torus and is par- 
 allel to H. 
 
 151. To draw a straight line tangent to the curve of intersec- 
 tion at a given point. Select any point as G at which the tan- 
 gent is to be drawn. The intersection MN of the tangent plane 
 to the torus at this point with the plane of AB and AC is the 
 required tangent line. 
 
 152. To find the line of intersection of any surface with a 
 given oblique plane by using an auxiliary plane of projection. 
 
 Fig. 76 represents an oblique cone cut by the plane of HH and 
 FF. The auxiliary plane of projection is taken at right angles 
 to the horizontal HH of the given plane. In the auxiliary view, 
 the given plane is represented by the line x^z^ and the points in 
 which the elements of the cone pierce this plane by the points 
 ^1, 2/i) ^1' By projection the horizontal view xyz and the ver- 
 tical view x'y'z' of the intersection are found. 
 
 In order for the auxiliary view of the cutting plane to show 
 as a straight line,, the auxiliary plane of projection should be 
 taken at right angles to a horizontal or frontal of the given 
 plane. 
 
PLANE SECTIONS OP SURFACES 
 
 123 
 
 -'#.". 
 
 
 F^G. 76. — 'Plane section of oMique cone by using an 
 auxiliary plane of projection. 
 
124 DESCRIPTIVE GEOMETRY 
 
 PROBLEMS 
 
 In each of the following problems draw a straight line tangent 
 to the curve of intersection at some point not on an extreme 
 element of the surface. 
 
 1. The base of a right cylinder is an ellipse in a horizontal plane. 
 Find the intersection of this cylinder with a given oblique plane. 
 
 2. Cut a circle with diameter equal to the major axis of the ellipse 
 from the cylinder in problem 1. 
 
 3. The axis of a right circular cylinder makes 60° with H and is 
 oblique to V. Find the intersection of the cylinder with a horizontal 
 plane. 
 
 4. The base of an oblique cylinder is a circle in a horizontal plane. 
 Cut a circle from the cylinder by a plane which is not parallel to the 
 base. 
 
 5. From a right circular cone with base in a horizontal plane, cut an 
 ellipse and show the true size of the section. 
 
 6. From a right circular cone with base in a horizontal plane, cut a 
 parabola and show the true size of the section. 
 
 7. From a right circular cone with base in a horizontal plane, cut a 
 hyperbola and show the true size of the section. 
 
 8. An oblique cone has a circle in a horizontal plane for a base. Cut 
 a circle of given radius from this cone by a plane which is not parallel 
 to the base. 
 
 9. Is the shortest path around a cone, following the surface, starting 
 at a point and returning to the same point, a plane curve? 
 
 10. Find the intersection of the hyperboloid of revolution of one 
 sheet by a plane. 
 
 11. Find the intersection of a helicoid by a plane. 
 
 12. Find the intersection of a hyperbolic paraboloid by a plane. 
 
 13. Find the intersection of a conoid by an oblique plane. 
 
 14. Find the intersection of a cylindroid by an oblique plane. 
 
 15. Find the intersection of a sphere by an oblique plane. 
 
 16. Find the intersection of an oblate ellipsoid of revolution by a 
 plane. 
 
 17. Find the intersection of a prolate ellipsoid of revolution by a 
 given oblique plane. 
 
 18. Find the intersection of a torus by a given oblique plane. 
 
CHAPTER VI 
 INTERSECTIONS OF CURVED SURFACES 
 
 153. To find the line of intersection of any two surfaces. In 
 
 general, pass a system of auxiliary planes through the surfaces. 
 The planes will cut lines from each surface and the intersections 
 of these lines will be points on the required line of intersection. 
 The auxiliary planes should be passed so that they will cut the 
 simplest lines possible from the given surfaces. 
 
 Sometimes the intersection can be more easily found by using 
 a system of auxiliary spheres in place of the system of planes. 
 For example, if the axes of two surfaces of revolution intersect, 
 a system of spheres having their centers at the intersection of 
 the axes and radii of different lengths, will cut circles from each 
 of the original surfaces. The intersections of these circles are 
 points on the required line of intersection. 
 
 In finding the line of intersection of two curved surfaces, the 
 points where this line touches the extreme elements of both 
 surfaces should be found. Any one of these points can usually 
 be found by taking an auxiliary plane which contains the ex- 
 treme element through that point. In some cases the curve of 
 intersection does not touch the extreme elements of the surface. 
 
 154. To find the line of intersection of two oblique cylinders. 
 Let the cylinders be given as in Fig. 77. 
 
 Analysis. Pass a system of auxiliary planes parallel to the 
 elements of both cylinders. These planes will cut straight line 
 elements from each cylinder which will intersect in points of 
 the required curve of intersection. 
 
 Construction. Through any point A draw two lines, one par- 
 allel to the elements of one cylinder and the other parallel to 
 the elements of the other cylinder. The plane of these lines 
 
126 
 
 DESCRIPTIVE GEOMETRY 
 
 cuts the plane of the bases of the cylinders in the line tt. Since 
 the auxiliary planes are parallel, each of them will cut a line 
 from the plane of the bases which is parallel to tt. The plane 
 Tg, which is tangent to one of the cylinders, cuts the element 
 
 Fig. 77. — Intersection of two cylinders. 
 
 BBi from that cylinder and the elements CCi and DD^ from the, 
 other cylinder. These elements intersect in the points X and T; 
 points of the required line of intersection. Other points on the 
 line of intersection are found in the same way. A curved line 
 joining in the proper order the points thus found is the required 
 line of intersection of the cylinders. 
 
 155. To draw a tangent to the curve of intersection at any 
 point. Select the point at which the tangent is to be drawn and 
 
INTERSECTIONS OF CURVED SURFACES 127 
 
 pass a plane tangent to each cylinder at this point. The inter- 
 section of these planes is the required tangent to the curve. 
 
 156. To determine which part of the curve of intersection is 
 visible. A point on the curve of intersection is visible in the 
 horizontal view if it lies on an element of each cylinder which 
 comes from the visible part of the base of the cylinder in that 
 view. In like manner, a point is visible in the vertical view 
 when it lies on an element of each cylinder which comes from 
 the visible part of the base of the cylinder in that view. 
 
 157. To find the line of intersection of a cone and cylinder. 
 Let the surfaces be given as in Fig. 78. 
 
 Analysis. Pass a system of auxiliary planes through the ver- 
 tex of the cone and parallel with the elements of the cylinder. 
 These planes will cut from each surface straight line elements 
 which will intersect in points of the required line of intersec- 
 tion. A line through the vertex of the cone and parallel with 
 the elements of the cylinder will lie in all of the auxiliary 
 planes. 
 
 Construction. The line Y^X, through the vertex of the cone 
 and parallel with the elements of the cylinder, lies in all tly' 
 auxiliary planes and pierces the plane of the bases at X, a point 
 common to all of the lines which these planes cut from the plane 
 of the bases. Through x, draw the lines xt, xt^, etc., cutting the 
 bases of both surfaces. From the points where these lines cross 
 the bases draw elements of the surfaces. These elements will 
 intersect in points of the required line of intersection. In this 
 manner the curve of intersection as shown in Fig. 78'is found. 
 
 The visible part of the curve is determined and the tangent 
 is drawn in the same manner as in Arts. 155 and 156. 
 
 15& To determine in advance the nature of the curves in 
 which cylinders and cones intersect. In the system of auxiliary 
 planes used in Fig. 78, there are two T and Tg, which are tan- 
 gent to the cone, xt and xt^ are the intersections of these 
 planes with the plane of the bases of the surfaces. The plane T 
 is tangent to the cone on the back side and the plane Tg is tan- 
 gent to the cone on the front side. Both of these tangent planes 
 
128 
 
 DESCRIPTIVE GEOMETRY 
 
 cut the cylinder. It is evident then that the cylinder extends 
 beyond the cone on both sides ; the cone running into the cylin- 
 der from one side and out on the other side, forming two dis- 
 tinct curves of intersection, as is shown in the figure. If one of 
 
 Fig. 78. — Intersection of cone and cylinder. 
 
 these tangent planes had cut the base of the cylinder while the 
 other did not, there would have been one continuous curve of 
 intersection. If the cylinder had been between these planes 
 and not cut by either of them, there would again have been two 
 distinct curves of intersection, where the cylinder runs into the 
 cone from one side and where it runs out on the other. In the 
 
INTERSECTIONS OF CURVED SURFACES 129 
 
 case where the planes are tangent to both cone and cylinder, 
 the two curves of intersection just touch each other. 
 
 The same method may be used to determine the nature of the 
 curve of intersection of two cylinders or of two cones. 
 
 159. To find the line of intersection of two cones. 
 
 Analysis. Pass a system of auxiliary planes through the ver- 
 tices of both cones. These planes will cut straight lines from 
 each cone which will intersect in points of the required line of 
 intersection. A line joining the vertices of the cones will lie in 
 all of the auxiliary planes and will pierce the plane of the bases 
 in a point common to all the lines which these planes cut from 
 the plane of the bases. 
 
 Let the construction be made for finding the curve of inter- 
 section. Also determine which part of the curve is visible and 
 draw a tangent to the curve by methods previously given. 
 
 160. To find the line of intersection of a sphere and a cone or 
 a sphere and a cylinder. 
 
 Analysis 1, If the cone or cylinder has a circle for a base, 
 the auxiliary planes can be passed parallel to the base. Each 
 plane will cut a circle from the sphere and a circle from the 
 cone or cylinder. The intersections of these circles are points 
 on the required line of intersection. 
 
 Analysis 2. Take the horizontal or vertical projecting planes 
 of the rectilinear elements of the cone or cylinder. These planes 
 will cut circles from the sphere which will intersect the ele- 
 ments cut from the cone or cylinder in points of the required 
 line of intersection. To find these points, it may be necessary 
 to revolve the projecting planes until they are parallel to one 
 of the planes of projection. 
 
 Let the construction be made in accordance with the above 
 analysis. 
 
 161. To draw a straight line tangent to the curve of intersec- 
 tion at any point. The required tangent will be the line of in- 
 tersection of two planes, one tangent to the cone or cylinder 
 and the other tangent to the sphere at the given point. 
 
130 DESCRIPTIVE GEOMETRY 
 
 PROBLEMS 
 
 1. Find the intersection of an oblique cone and a right circular cylin- 
 der. The bases of both surfaces are circles on a horizontal plane. 
 
 2. Find the intersection of a right circular cone with a sphere. The 
 base of the cone is on a horizontal plane and the center of the sphere 
 not on the axis of the cone. 
 
 3. Find the intersection of a right circular cylinder with a right 
 circular cone. The axes of both surfaces are perpendicular to H but 
 do not coincide. 
 
 4. Find the intersection of a sphere with a right circular cylinder. 
 The axis of the cylinder is perpendicular to H but does not pass 
 through the center of the sphere. 
 
 5. Find the intersection of a right circular cone with a right circular 
 cylinder. The base of the cone is on a horizontal plane and the axis 
 of the cylinder is parallel to the ground line. Draw a tangent to the 
 curve of intersection. 
 
 6. Find the intersection of a triangular prism with a sphere. 
 
 7. Find the line of intersection of a cone with a triangular prism. 
 
 8. Find the line of intersection of the two given oblique cylinders, 
 which have their bases in a horizontal plane, and draw a tangent to the 
 curve at some point of the curve not on an extreme element of either 
 surface. 
 
 Cylinder A < 
 
 Center of lower base, A(3i, X, — 4) 
 Center of upper base, 0(— 3^, —If, 0) 
 Major axis is 3^" long and is parallel to V 
 Minor axis is 2^" long 
 
 r Center of lower base, B(— 3^, Y, — 4) 
 Cy'linder B J Center of upper base, C(i, — 4i, 0) 
 1 Radius of base If" 
 
 Note. — Take the origin at the center of a ll''xl5" sheet, with the 
 ground line parallel to the longer side and passing through the center. 
 X may vary from — 1^" to —ZV and Y may vary from —2" to —3". 
 
 9. Find the line of intersection of the given oblique cylinder and the 
 given oblique cone, which have their bases in a horizontal plane, and 
 draw a tangent to the curve at some point of the curve not on an ex- 
 treme element of either surface. 
 
PROBLEMS 131 
 
 {Center of the lower base, A(i, X, — 3^) 
 Center of the upper base, C(— 3, — li, 0) 
 Radius of the base, li" 
 
 r Center of the base, B(— 3i, Y. — 3i) 
 Cone B J Vertex, Y^dh ~5i, li) 
 I Radius of the base, If" 
 
 Note.— Take the origin at the center of a ll"xl5'' sheet, with the 
 ground line parallel to the longer side and passing through the center. 
 X may vary from —If' to — 3i" and Y may vary from —2" to — 3J". 
 
 10. Find the line of intersection of two oblique cones, which have 
 their bases in a horizontal plane, and draw a line tangent to the curve 
 of intersection at some point of the curve not on an extreme element 
 of either surface. 
 
 r Center of the base, A(4i, X, — 4f) 
 Cone A J Vertex of the cone, V^(l, — 2|, —2) 
 Radius of the base, 1^" 
 
 
 r Center of the base, B(0, Y, — 4|) 
 Cone B < Vertex of the cone, V^CB^, — 5i, 0) 
 (^Radius of the base. If" 
 
 Note. — Take the origin at the center of a ll"xl5" sheet with the 
 ground line parallel to the longer side and passing through the center. 
 X may vary from — If" to — 3i" and Y may vary from — 2" to — 3". 
 
 11. Find the line of intersection of the two given cones, which have 
 their bases in a horizontal plane, and draw a line tangent to the curve 
 of intersection at some point of the curve not on an extreme element of 
 either surface. 
 
 r Center of the base, A(0, — 3i, — 6f ) 
 Cone A J Vertex of the cone, VJO, —31, 0) 
 [^ Radius of the base, 3" 
 
 /Center of the base, B(i, —4, — 6f) 
 Vertex of the cone, V^C— 3i, —3, — li) 
 Radius of the base, 2J" 
 
 
 Take origin at the center of a ll"xl5" sheet, with the ground line 
 passing through the center and parallel to the shorter side of the sheet. 
 
 12. Find the line of intersection of a torus and an oblique cylinder. 
 
VB (7!*' 
 
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