THE ESSENTIALS 
 
 • OF 
 
 DESCRIPTIVE GEOMETRY 
 
 BY 
 
 F. G. HIGBEE, M. E. 
 
 i( 
 
 Professor and Head of Deparlmcnt of Descriptive Geometry and Drawing, 
 The State University of Iowa 
 
 SECOND EDITION REVISED 
 
 TOTAL ISSUE SEX THOUSAND 
 
 NEW YORK 
 
 JOHN WILEY & SONS, Inc. 
 
 London: CHAPMAN & HALL, Limited 
 
 1917 
 
£ngmeering 
 
 Copyright, 1915, 19171 
 
 BY 
 
 F. G. HIGBEE 
 
 Stanbopc iprcss 
 
 F. H. GILSON COMPANY 
 BOSTON, U.S.A. 
 
PREFACE 
 
 It has been the endeavor of the author in writing this text to 
 include only those portions of descriptive geometry which possess 
 industrial utility and which develop the quahties of mind so 
 essential in a draftsman. 
 
 First and foremost descriptive geometry should aim to teach 
 projection. A draftsman must be able both to read and to write 
 drawings with facility, and projection is the very grammar of 
 the language of the designer. Secondly, descriptive geometry 
 should aim to develop the ability to solve problems concerning 
 the relations of points, lines, and planes. These are, of course, 
 but elementary parts of all engineering structures as shown on 
 drawings and it is important that a draftsman be prepared to 
 solve problems relating to them directly on the drawing board. 
 Thirdly, and perhaps most important of all, descriptive geometry 
 should aim to promote the ability to analyse a problem into its 
 component parts, to reason from a given set of conditions to a 
 required set of conclusions, and to build up from the drawing 
 a mental picture of the object which is there represented, for 
 without the ability to analyse, to reason, and to visualize a 
 draftsman is lacking in the essential qualifications of his calHng. 
 
 For these reasons the subject has been discussed from the 
 point of view of a draftsman and the essential relations of points, 
 lines, and planes have been treated in the third quadrant. The 
 order of the material has been carefully considered and while 
 there is some departure from traditional arrangement it is be- 
 lieved that the selection of material and its arrangement will be 
 found both logical and conducive to a natural development of 
 the subject. Many problems of a carefully graded character 
 and of a practical "flavor" have been inserted at frequent inter- 
 vals in the belief that such work is invaluable in fixing principles 
 and promoting genuine interest. 
 
IV PREFACE 
 
 In the discussion on surfaces considerable variation may be 
 found from other works on the subject both in content and treat- 
 ment. It is believed, however, that the material included is 
 broad enough in character for practical purposes; and it is hoped 
 that the method of treatment, the character of the problems, 
 and the discussion on model making will stimulate interest in 
 this important and useful part of the subject. 
 
 In the preparation of this text the author has consulted and 
 acknowledges his indebtedness to the following standard texts: 
 Geometrie Descriptive by G. Monge; Theoretical and Practical 
 Graphics by F. N. Willson; Elements of Descriptive Geometry 
 by Albert E. Church; Elements of Descriptive Geometry by 
 C. W. MacCord; Engineering Drawing by Thos. E. French. 
 
 F. G. HIGBEE. 
 
 Iowa City, Ia., Feh. i, 1915. 
 
 PREFACE TO SECOND EDITION 
 
 The second edition of this book presents no departure from 
 the original endeavor of the author which was to discuss only 
 the essentials of descriptive geometry. 
 
 The material in the first edition has been corrected and re- 
 vised, some portions have been entirely rewritten, and at the 
 request of a number of teachers who are using the book as a 
 text a chapter on tangencies has been added. The changes 
 from the first edition are those which practical classroom use 
 has demonstrated as advisable. 
 
 The author wishes to acknowledge his indebtedness to his 
 colleagues in the department of Descriptive Geometry and 
 Drawing at the University of Iowa, to Dr. F. M. Comstock 
 and his associates at the Case School of Applied Science, to 
 Professor E. F. Chillman, Rensselaer Polytechnic Institute, and 
 to many others who have aided in the preparation of this edition 
 with helpful criticisms and suggestions. 
 
 F. G. HIGBEE. 
 
 Iowa City, Ia., March 15, 1917. 
 
CONTENTS 
 
 CHAPTER I Page 
 
 Orthographic Projection i 
 
 CHAPTER II 
 Profile Plane 1 1 
 
 CHAPTER HI 
 
 Assumption of Points and Lines 21 
 
 CHAPTER IV 
 Planes 27 
 
 CHAPTER V 
 Location of Points, Lines, and Planes 33 
 
 CHAPTER VI 
 
 Revolution of Points 38 
 
 CHAPTER VII 
 Problems on the Line 45 
 
 CHAPTER VIII 
 Problems on the Plane 52 
 
 CHAPTER IX 
 Problems on Angles 59 
 
 CHAPTER X 
 Problems on Points, Lines, and Planes 63 
 
 CHAPTER XI 
 Surfaces ^4 
 
 CHAPTER XII 
 
 Plane Surfaces 9^ 
 
 V 
 
VI CONTENTS 
 
 CHAPTER XIII Page 
 
 Cylindrical Surfaces loi 
 
 CHAPTER XIV 
 Conical Surfaces 115 
 
 CHAPTER XV 
 
 Intersection of Surfaces 131 
 
 CHAPTER XVI 
 Surfaces of Revolution 145 
 
 CHAPTER XVII 
 Warped Surfaces 155 
 
 CHAPTER XVIII 
 Tangent Planes and Lines 179 
 
 CHAPTER XIX 
 Model Making 193 
 
 CHAPTER XX 
 Appendix 211 
 
ESSENTIALS OF DESCRIPTIVE 
 GEOMETRY 
 
 CHAPTER I 
 ORTHOGRAPHIC PROJECTION 
 
 1. Descriptive Geometry is the science of graphic representa- 
 tion. It is the means by which objects are shown on drawings 
 and by which problems relating to these objects are solved. 
 
 2. When an object such as a machine, a bridge, a building, or 
 any elemental part of such an engineering structure, is designed 
 a drawing or set of drawings of it is made. Such a drawing is 
 not only invaluable to the designer in recording his ideas step by 
 step and in assisting him in his design but it is also indispensable 
 to the artisan who constructs the object. A drawing, as used in 
 the engineering sense, is a complete set of instructions from the 
 designer to the workman by means of which the workman may 
 reproduce, in exact shape and size and in material and finish, 
 the identical object which the designer represented on the 
 drawing. 
 
 The principles by which the shape and size of objects, and by 
 which problems relating to such representations are graphically 
 solved, are found in descriptive geometry. The art of putting 
 on the drawing such additional information as dimensions, shop 
 notes, finish, and other data regarding the construction of the 
 object represented, is not a part of descriptive geometry but is 
 included as a part of the art which is generally known as drafting. 
 
 3. There are two general systems by which graphic represen- 
 tations are made. One system has for its purpose the represen- 
 tation of objects as they appear, and is called scenographic pro- 
 jection. When an object is looked at from some particular 
 
2 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 point of view one can get an idea of its shape and size because at 
 least three faces of the object are seen, and the effect of hght and 
 shade serves to bring out the configuration, and because the eye 
 naturally compares and judges distances. If a representation 
 be made showing how the object appears to the observer, such a 
 drawing will be a scenographic projection, or, as it is more com- 
 monly called, a perspective drawing. 
 
 There are also a number of other forms of projection which 
 aim at the same result, — that is, to represent the object as it 
 appears, — ■ which may be called pseudo-perspective drawing. 
 Isometric, cabinet, and obhque projection are examples, but as 
 these are used as forms of perspective drawing and require 
 extensive treatment they will not be considered in this text. 
 
 4. The other system of projection has for its purpose not the 
 representation of objects as they appear, but the representation 
 of objects as they actually are. This form of representation is 
 called orthographic projection; and it is this form of projection 
 which is used in all engineering drawing where the purpose of the 
 drawing is to convey, from the draftsman to the workman, in- 
 formation and directions for building. 
 
 It is with orthographic projection and the use of orthographic 
 projection as a means of solving drawing-board problems graphi- 
 cally that this text has to do. 
 
 5. In perspective drawing the eye of the draftsman is assumed 
 to be at a definite distance from the object. Therefore the rays 
 of sight which travel from the eye to the corners and other parts 
 of the object converge to a point. If now a plane be set up be- 
 tween the eye and the object, as in Fig. i, and the points where 
 the lines of sight pierce this plane be found and joined, the result 
 will be a picture of the object observed. It is obvious that the 
 picture will be smaller than the object itself, because the plane, 
 which is called the picture plane, is between the eye and the 
 object. 
 
 It will be observed, then, that in perspective drawing the size 
 of the picture will vary inversely as the distance of the picture 
 plane from the object; when it is between the object and the 
 eye the picture will be smaller than the object, and when it is on 
 
ORTHOGRAPHIC PROJECTION 
 
 the far side of the object the picture will be larger. This is 
 due, of course, to the fact that the lines of sight converge to 
 the eye. 
 
 It will be observed that the relation of the lines in the object 
 to the corresponding lines in the picture is not the same for all 
 
 Eye^*^: 
 
 Fig. 
 
 lines. In other words, some lines are foreshortened more than 
 others owing to the position of the eye. 
 
 It will also be observed that in perspective more than one face 
 of the object is seen from one point of view. 
 
4 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 6. In engineering drawing — or in orthographic projection — 
 the eye of the draftsman is assumed to be at infinity. Therefore, 
 the Hnes of sight do not converge but are parallel. If now, as 
 
 
 
 \\ 
 
 
 
 
 
 \ 
 
 
 \ \ 
 
 Tofheeye "^ 
 
 < 
 
 
 \ 
 
 
 
 \ 
 
 Object 
 
 
 \ 
 
 \ 
 
 \ 
 
 
 
 Fig. 2. 
 
 in Fig. 2, a plane be set up between the eye and the object and 
 the points where the Hnes pierce this plane be found and joined 
 the result will be a picture of the object not as it appears but as 
 it actually is, and each hne of the picture will be the same length 
 as the corresponding line in the object. 
 
ORTHOGRAPHIC PROJECTION 5 
 
 It will be observed, therefore, that the size of the picture does 
 not vary with the position of the picture plane. No matter where 
 the plane is placed the size of the resulting picture will be the 
 same. Of course, in actual drawing, objects are drawn to scale. 
 That is, some part of an inch is taken on the drawing to represent 
 an inch on the object in order that large objects may be drawn 
 on convenient sized drawing sheets, but the reduction is the same 
 for all lines. If, for example, one line in the picture is drawn to 
 a scale of \" = i" all lines will be drawn to that same scale. 
 
 It will be observed, then, that the relation of lines in the object 
 to the corresponding lines in the picture is the same for all lines 
 and is the same no matter what distance the picture plane is from 
 the object. If the picture of a line be measured, therefore, it will 
 be found to equal in length the line itself. 
 
 It will also be observed that in orthographic projection only 
 one face of an object is visible from one point of view, and, 
 therefore, that to show more than one face more than one pic- 
 ture must be drawn and more than one point of view assumed. 
 
 7. In orthographic projection the picture planes are called 
 planes of projection and since more than one are required to show 
 three dimensions of an object one has been assumed vertical and 
 one has been assumed horizontal, because upon two planes an 
 object may in general be projected so as to give a complete idea 
 of its shape and size. The horizontal plane is called the horizon- 
 tal plane of projection and the view of the object obtained by 
 projecting it upon this plane is called the plan, or the plan view, 
 or the H projection. The vertical plane is called the vertical plane 
 of projection and the view obtained by projecting the object upon 
 this plane is called the elevation, or the front view, or the V pro- 
 jection. 
 
 8. Fig. 3 shows a picture of the two planes of projection 
 with an object ready to project upon them. The line where 
 the H and V planes intersect is called the ground line, — here- 
 after denoted as G. L., — and in drawings the ground line is the 
 line of reference which indicates the location of the object with 
 reference to the planes of projection. 
 
 To project the object in Fig. 4 upon the H plane, or to get a 
 
ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 Horizontal Plane 
 
 Ground Line 
 
 Vertical Plane 
 
 Fig. 
 
 plan view of it, the eye of the draftsman is supposed to be so far 
 
 above the H plane that the lines of sight are perpendicular to it. 
 
 Therefore, draw perpen- 
 diculars from each corner 
 of the object to the H 
 plane and join the points 
 where the perpendiculars 
 pierce H. To project 
 the object upon the V 
 plane, or to get an eleva- 
 tion of it, the eye of the 
 draftsman is assumed to 
 be in front of the V 
 plane and so far in front 
 that the hnes of sight 
 are perpendicular to it. 
 Therefore, draw perpen- 
 diculars from the object 
 
 to the V plane and join the points where the perpendiculars 
 
 pierce V. It will he 
 
 noted, therefore, that the 
 
 projection of a point 
 
 upon a plane is the foot 
 
 of a perpendicular from 
 
 the point to the plane. 
 
 It should be kept in 
 
 mind, however, that Fig. 
 
 4 is merely a picture, not 
 
 an actual representation, 
 
 showing how projections 
 
 are made. 
 
 9. Now it will be ob- 
 served that the two 
 
 planes of projection 
 
 stand at right angles to 
 
 each other, and that if the projections or views which are 
 
 shown on these planes are to be represented as they appear and 
 
 'H. Plane 
 
 Fig. 
 
ORTHOGRAPHIC PROJECTION 7 
 
 on one sheet of paper these two planes must be considered 
 as coinciding with the paper on which the drawing is made. 
 This is accompHshed as shown in Fig. 5 by revolving H 
 into coincidence with V. The H plane and the V plane are now 
 coinciding with the paper, but it will be noted that the relation 
 of the plan and the elevation with respect to each other and the 
 
 H. Plane 
 
 \ Plan 
 
 ~\ 
 
 Plan 
 
 Elevation 
 
 V. Plane 
 
 Elevation 
 
 Fig. s. 
 
 Fig. 6. 
 
 ground line has in no way been changed by revolving the H plane; 
 each remains the same distance from the ground Hne as before, 
 and it will be noted carefully that the plan and elevation of the 
 same point lie in the same perpendicular to the ground line. 
 
 In a drawing the planes of projection are not limited in extent, 
 so no boundary lines need be shown for them. A ground line, 
 as in Fig. 6, is drawn which indicates the position of the planes 
 
8 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 with respect to the object, — it being understood that the H 
 plane is above the V plane, — and the plan and elevation of the 
 object are drawn in the required position by locating them at the 
 proper distances from the G. L. Fig. 6 shows a plan and eleva- 
 tion of an object as usually drawn in descriptive geometry. 
 
 10. Now since the two planes of projection are indefinite in 
 extent, and stand at right angles to each other, it will be clear 
 
 that such a pair of 
 planes will divide 
 space into four quar- 
 ters as shown by the 
 pictorial drawing in 
 Fig. 7. These four 
 quarters are called 
 quadrants, or angles, 
 and for convenience 
 are designated as 
 First, Second, Third, 
 and Fourth Quad- 
 rants. Drawings may 
 be made with the ob- 
 ject in any of these 
 quadrants, but all 
 commercial drawing 
 is done either in the 
 third or first quad- 
 rant. The reason for 
 this will be evident 
 from a study of Fig. 
 7A which shows how the quadrants are located with respect to 
 the G. L. when the planes of projection coincide with the paper. 
 From this figure it will be seen that in third quadrant drawing 
 the plan is above the elevation; in first quadrant drawing the 
 plan is below the elevation; and in second and fourth quadrant 
 drawing the plan and elevation are on the same side of the ground 
 line, and therefore often interfere with each other. 
 Numbering the quadrants is done to afford a convenient way 
 
 
 V 
 
 \ 
 
 \ 
 \ 
 \ 
 \ 
 
 V 
 \ 
 \ 
 \ 
 
 
 H/ 
 
 A\ 
 
 /'^ 
 
 V 
 
 1 
 1 
 I 
 \ 
 
 \ 
 \ 
 
 \ 
 
 \ 
 
 \ 
 
 X 
 
 1 
 
 I 
 
 \ 
 
 \ 
 
 \ 
 \ 
 \ 
 \ 
 
 
 Fig. 7. 
 
ORTHOGRAPHIC PROJECTION 9 
 
 of indicating the location of an object with respect to the planes 
 of projection. When an object is said to be in the first quadrant 
 it means simply that it is above the H plane and in front of the 
 V plane; therefore its plan view will be below the elevation. 
 When an object is said to be in the third quadrant it means that 
 
 First Quadrant. Second Quadrant. Third Quadrant. Fourth Quadrant. 
 
 Plan below G. L. and H and V coinciding Plan above G. L. and H and V coinciding 
 elevation above G. L. above G. L. There- elevation below G. L. below G. L. There- 
 fore plan and eleva- fore both views be- 
 tion both above G. L. low G. L. 
 Fig. 7A. 
 
 it is below H and behind V; therefore its plan view will be 
 above its elevation. Also when a drawing is examined these rela- 
 tions serve to assist in visuahzing the object, and once the 
 "language" of projection and drawing is learned these details 
 are not thought of as such, and one reads the drawing much as 
 one reads a sentence without analysing it into subject, object, 
 verb, etc. 
 
 PROBLEMS IN PROJECTION 
 
 1. Draw the plan and elevation of a regular hexagonal prism whose base 
 is 2" in diameter and whose altitude is 3". 
 
 2. Draw the plan and elevation of a regular pentagonal pyramid whose 
 altitude is 3" and the sides of whose base are 2". 
 
 3. Draw the plan and elevation of a regular truncated hexagonal pyra- 
 mid whose lower base is 3" in diameter, whose upper base is 2" in diameter, 
 
lO ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 and whose height is 2" . Let the pyramid be in the first quadrant with its 
 lower base on H. 
 
 4. Draw the plan and elevation of a i" by 2" by 3" block having three 
 circular holes through it the i" way. The holes are \" in diameter and are 
 spaced with centers i" apart with the middle hole in the center of the 
 2" by i" face of the block. 
 
 5. Draw the plan and elevation of a cube with 2" edges. Let the cube 
 be in the first quadrant resting on H with its vertical faces inclining to V at 
 angles of 30 and 60 degrees. 
 
 6. Draw the plan and elevation of a cast iron plate 9" by 3' by 6'. In 
 the center of the 3' by 6' face is a hole 6" square and 4" deep; in the 9" by 6' 
 face are drilled two 3" holes through the plate 8" from each end and on the 
 center line. 
 
 7. A grain hopper in the shape of a regular truncated square pyramid 
 has an opening 4' by 4', and an outlet 12" by 12" which is 4' below the 
 opening. Draw the plan and elevation of the hopper when the opening 
 lies in the H plane with its edges incUning at angles of 45 degrees with the 
 G. L. 
 
 8. Draw the plan and elevation of a plain box 4" deep, 6" wide and 9" 
 long with its lid half raised. The box is built of |" material and the lid is 
 hinged along the 9" edge. 
 
 9. A hood is fastened to a wall. The opening of the hood is 4' by 4' and 
 its outlet is into a square pipe 6" in diameter also fastened to the wall 
 directly above the opening. The distance between the opening and outlet 
 is 3'. Draw the plan and elevation. 
 
 10. Draw the plan and elevation of an hexagonal pyramid whose base is 
 2" in diameter and whose altitude is 3". The apex of the pyramid is directly 
 above one corner of the base. 
 
 Note: Unless otherwise specified objects given in problems are to be 
 drawn in the third quadrant ; and when no distance from H and V is given 
 the views may be located at any convenient distance from the ground line. 
 It is well, however, to keep the distance between views relatively short so 
 that the eye may readily note which points are projections of each other; 
 in practical drafting only enough space is allowed between views to permit 
 of placing dimensions and to keep the views distinct. 
 
CHAPTER II 
 
 PROFILE PLANE 
 
 11. While in many cases a plan and elevation will show the 
 shape and size of an object it quite frequently happens that a 
 third view will be required for a complete representation. In 
 such a case the H and V planes of projection are not sufficient 
 and a third plane of projection must be added. The third view, 
 which is most commonly 
 used to supplement the 
 plan and elevation, is 
 called an end view, or a 
 side or end elevation, or 
 a profile projection; and 
 the plane upon which 
 this view is made is 
 called the profile plane, 
 or P plane. 
 
 12. In Fig. 8 is shown 
 a picture of the arrange- 
 ment of H, V, and P 
 for the third quadrant. 
 From this picture it will. 
 be clear that the profile 
 plane is perpendicular to the G. L., and therefore to both H 
 and V, and it may be located at any convenient point on the 
 G. L. and on cither or both sides of the object. In case it is 
 located on the right side of the object the view obtained will be 
 the right end view. It not infrequently occurs in drafting that 
 both end views are shown. 
 
 13. In Fig. 9 is shown how the H, V, and P planes, pic- 
 torially represented in Fig. 8, appear when they are revolved to 
 
 Fig. 8. 
 
12 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 coincide with the plane of the drawing. From this figure it will 
 be observed that the relations between H and V which have been 
 previously discussed are in no way altered, that the profile plane 
 takes a position to the right or left of the V plane, and that the 
 
 Fig. 9. 
 
 elevation and end view of the same point lie on fines which are 
 parallel to the G. L. It will also be observed that the end view 
 shows how far an object is below H, by means of the distance the 
 end view is below the G. L. ; and also shows how far an object is 
 behind V, by means of the distance the end view is to the right, 
 or left, of the profile ground fine. 
 
 There are, of course, two profile ground fines; one where the 
 P plane cuts V, and the other where it cuts H. In Fig. 9 it wiU 
 be noted that the ground fine between V and P is really an end 
 
PROFILE PLANE 
 
 13 
 
 view of the V plane, and the one between H and P is really an end 
 view of H. This latter coincides, after the planes are revolved, 
 with the G. L. for H and V. 
 
 In actual work the three planes are not limited in extent and 
 their location is determined only by the ground Unes. Fig. 10 
 
 Plan 
 
 Elevation 
 
 End View 
 
 Fig. 10. 
 
 shows how Fig. 9 appears as usually drawn with only the ground 
 lines showing the planes. 
 
 14. The arrangement of the three planes for all quadrants is 
 shown pictorially in Fig. 11, with the profile plane on the right. 
 Now it will be observed that if the profile plane be revolved so 
 that the portion for the third quadrant fall to the right of the 
 V plane the profile plane for the first quadrant will, of course, 
 revolve in the opposite way, and will be coincident with V. This 
 
14 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 is shown in Fig. 1 2 where a timber is partly in the first and partly 
 in the third quadrant. 
 
 In practical drafting, however, the arrangement of the three 
 planes for the first quadrant is as shown in Fig. 13 where the 
 
 Fig. II. 
 
 block shown at A is projected. The P plane is revolved out of 
 the way of the elevation in the same way as for the third quad- 
 rant, thus avoiding the confusion which might result from having 
 the end view superimposed upon the elevation. 
 
 One of the problems constantly before the draftsman for solu- 
 
PROFILE PLANE 
 
 15 
 
 Fig. 12. 
 
 \ \ 
 
 
 
 
 
 
 
 \ 
 
 
 
 
 
 \ 
 
 
 
 
 Fig. 13. 
 
 tion is the construction of the end views of an object which is 
 shown only in plan and elevation. Fig. 14 shows the plan and 
 elevation of an object of which it is desired to find both end 
 
 views. 
 
 The profile ground lines are assumed at any convenient point 
 along the given ground line as shown. Through the eleva- 
 
i6 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 tion of each point draw dotted lines parallel to the G. L. On 
 these dotted lines, the proper distance from the profile ground 
 lines, will lie the end view of each point. To find this location 
 
 lay off on the dotted lines distances to the right and left of the 
 profile ground line in V, the distance each point is back of V. 
 This may be done conveniently by drawing a 45 degree line as 
 shown and drawing perpendiculars. Since the distance each 
 
PROFILE PLANE 
 
 17 
 
 end view is from the profile ground line in V is equal to the dis- 
 tance it is back of V, points so located will be the proper distance 
 to the right or left of the profile G. L. 
 
 Fig. IS shows a similar problem solved in the first quadrant 
 and serves to show the arrangement of views for the first 
 quadrant. 
 
 By using similar methods of projection a third view of any 
 object may be obtained when two views are given. Thus: a 
 plan view may be found from the elevation and end view; or an 
 elevation may be found from the end view and plan. 
 
 PROBLEMS IN PROJECTION 
 
 II. Draw the plan, elevation, and right end view of a cube with edges 
 2" long. Let the upper face of the cube be parallel to and 1" below H with 
 
 :?! 
 
 -2M^ 
 
 n 
 
 Fig. 16. 
 
 Fig. 17. 
 
 its edges inclinmg 30 and 60 degrees to the G. L. Let the edge of the cube 
 nearest V be ^" behind V. 
 
 12. Draw three views of a block i" by 2" by 4". In the center of one 
 2" by 4" face is a \" square hole cut half way through the block, and in the 
 center of each i" by 2" face is a hole \" in diameter drilled i" deep. 
 
 13. Draw the plan, elevation, and left end view of a bar of iron 4" long, 
 
i8 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 2" wide, and \" thick which has four \" holes drilled through it on the center 
 line of the 2" by 4" face. The holes are spaced with their centers f " apart. 
 Let the face of the bar be \" below H and parallel to it, and let the center 
 line of this same face incline 60 degrees to the G. L. 
 
 14. Draw three views of a regular truncated hexagonal pyramid. Diam- 
 eter of large base 3"; diameter of small base i§"; distance between bases 
 2" . Use first quadrant projection and let the 3" base be in V. 
 
 Fig. 18. 
 
 -IM- 
 
 ^ 
 
 ^- 
 
 -VA- 
 
 -w 
 
 -2]4^ 
 
 « 
 
 IM— 
 
 
 - 
 
 -^f-^ 
 
 'Drill \' 
 
 
 
 
 
 
 ' 
 
 
 
 
 
 T- 
 
 
 
 
 
 
 
 
 
 ^^ 
 
 
 
 
 
 
 
 ' 
 
 
 
 Fig. 19. 
 
 15. A hopper has an opening in the floor 4' by 6'. Its 12" square outlet 
 is 5' below the floor with one edge directly under the center of one 6' edge 
 of the floor opening. Draw three views of the hopper. 
 
 16. Draw the plan and elevation and find the right end view of the 
 object shown in Fig. 16. 
 
 17. Draw the plan and elevation and find the left end view of the object 
 shown in Fig. 17. 
 
 18. Draw the two given views and find the plan of the object shown in 
 Fig. 18. 
 
 19. Draw the plan of the object whose end view and elevation are shown 
 in Fig. 19. 
 
 20. Draw the two given elevations and find the plan of the object shown 
 in Fig. 20. 
 
Fig. 20. 
 
 Fig. 21. 
 
 Fig. 22. 
 
 (19) 
 
20 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 DriU U" 
 
 Fig. 23. 
 
 Fig. 24. 
 
 21. Draw three views of the object shown in Fig. 21. 
 
 22. Draw a plan, elevation, and left end view of the object shown in 
 Fig. 22. 
 
 23. Draw the plan, elevation, and right end view of the object shown in 
 Fig. 23. 
 
 24. Draw a plan, elevation, and both end views of the object shown in 
 Fig. 24. 
 
CHAPTER III 
 
 ASSUMPTION OF POINTS AND LINES 
 
 15. In order to study the relation of various points and lines 
 which collectively form the representation of an object, and to 
 solve problems concerning these elements of the representation, 
 there must be a method of designating points and lines, and the 
 relation that these points 
 and Unes have to the 
 planes of projection must 
 be known. 
 
 In drafting a point is 
 almost never designated 
 by a letter, but in descrip- 
 tive geometry it has been 
 found convenient to indi- 
 cate points and lines by 
 this means. A point in 
 space is indicated by the 
 capital letter, as A, B, C, 
 etc., and its projections are 
 indicated by the corre- 
 sponding small letter, as 
 a, b, c, etc.; a, b, c, being 
 used to indicate the plan 
 view of the point and a', 
 b', d , etc., being used 
 to indicate its elevation 
 
 Fig. 25. 
 
 and ap, bp, Cp, etc., being used to indicate the end view. Thus: 
 the plan view of the point A is a, and its elevation is a', and 
 when the point A is referred to it means the actual point in space. 
 The Hne MN means a line in space whose plan view is mn, 
 whose elevation is m'n' and whose end view is mpUp. 
 
22 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 The relation that a' point or a line in space has to the planes of 
 projection may be discovered from its projections. In Fig. 25 
 is shown the plan and elevation of a hexagonal pyramid one of 
 whose edges is OM. Now the plan view of the point O is o and 
 its elevation is o', and an inspection of the figure will show that 
 the distance o is from the G. L. is the distance that O is from the 
 V plane; and the distance o' is from the G. L. is the distance 
 that O is from H. It is also to be seen that since o is above the 
 G. L. the point in space is behind V, and since o' is below 
 the G. L. the point in space is below H. Therefore, from the 
 drawing it may be found that the point O Hes in the third quad- 
 rant, and its distances from H and V may be measured. In like 
 manner the relation that M bears to H and V may be read from 
 the drawing. 
 
 16. To assume a point in space it is necessary only to assume 
 its two projections for, in general, a point in space is completely 
 located with reference to H and V when its plan and elevation 
 are shown. If, however, the problem is to assume a point in a 
 given quadrant at a given distance from H and V, the two views 
 must be located at the proper distances from the G. L. Thus, 
 in Fig. 25, O which is to Ue in the third quadrant 1" from H and 
 2" from V will be drawn as follows: Erect the dotted Hne 00' 
 perpendicular to the G. L.; on this hne will He o and o', the plan 
 and elevation of the required point. Since this required point 
 is the third quadrant it wih be below H and behind V; therefore, 
 the plan view o will be above the G. L. a distance equal to the 
 distance the point is behind V, or 2" . Therefore, locate o 2" 
 above the G. L. on the dotted perpendicular. Since the point 
 is i" below H, o' will lie in this same perpendicular 1" below the 
 G. L. 
 
 In like manner a point may be located in any of the quadrants 
 and at any specified distance from H and V. 
 
 17. To assume a line in space assume its plan and elevation 
 in any desired position. If two points of the line are given, the 
 plan and elevation of the line may be found by joining the plan 
 views of the two given points, and the elevation by joining the 
 elevations of the two given points. The plans and elevations 
 
ASSUMPTION OF POINTS AND LINES 
 
 23 
 
 of the given points will have to be located as indicated in 
 Article 16. 
 
 If the plan and elevation of a line are given and it is desired 
 to assume a point on the line it is necessary only to assume the 
 plan view of the point on the plan view of the line and the eleva- 
 tion of the point on the elevation of the line, and have these two 
 views lie in the same per- 
 pendicular to the G. L. 
 
 18. To assume two in- 
 tersecting lines assume 
 two lines which have a 
 common point. Fig. 26 
 shows AB and BC inter- 
 secting at the common 
 point B. 
 
 19. There are certain 
 relations which points 
 and lines have to the 
 planes of projection which 
 are apparent from a study 
 of the foregoing articles 
 and these relations may 
 be set forth as, — - 
 
 Observations. 
 
 a. The two views of a 
 point always lie in a per- 
 pendicular to the G. L. 
 
 h. The distance a point 
 is from H is equal to the 
 distance its elevation is from the G. L.; and the distance a 
 point is from V is equal to the distance its plan is from the G. L. 
 
 c. A point which lies in H will coincide with its own plan view 
 and its elevation will be in the G. L. Likewise a point in V will 
 coincide with its own elevation and its plan view will he in the 
 G. L. 
 
 d. A line which Hes in H or V will coincide with its view on H 
 or V and will have its other view in the G. L. 
 
 Fig. 26. 
 
24 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 e. When a line is parallel to H or V its view on H or V will be 
 equal in length to the line and its other view will be parallel to 
 the G. L. 
 
 A convenient method of visualizing any given problem in projection. The pencil 
 is held in the position of the given line and its projections on the cardboard 
 planes may be drawn. 
 
 A convenient method of showing the relation of the H, V, and P planes for the 
 third quadrant, and of the projections of any given line on the planes. 
 
 /. When a line is perpendicular to H or V its view on H or V 
 will be a point. The other view will be equal in length to the 
 line itself and perpendicular to the G. L, 
 
ASSUMPTION OF POINTS AND LINES 2 ; 
 
 g. When a line is parallel to the G. L. both views of the Kne 
 will be parallel to the G. L. 
 
 //. The projection of a Une on H or V will be equal in length to 
 the Hne itself when the Hne is parallel to H or V. 
 
 i. The projection of a Hne on H or V is either equal in length 
 to the line or shorter than the line. 
 
 j. A hne may He in one or in two or three quadrants, but never 
 in all four. 
 
 k. li Si point is on a Hne the plan view of the point wih He on 
 the plan view of the Hne, and the elevation of the point wiU He 
 on the elevation of the Hne. 
 
 I. When a Hne is parahel to H and oblique to V the elevation 
 of the Hne is parallel to the G. L. and the plan view makes an 
 angle with the G. L. equal to the angle the Hne makes with V. 
 
 In Hke manner when the Hne is parallel to V and incHned to H 
 its plan is parallel to the G. L. and its elevation makes an angle 
 with the G. L. equal to the angle the Hne makes with H. 
 
 m. When two Hues intersect they have a point in common. 
 Therefore, the plan views of these Hues will have a point in com- 
 mon, and the elevations will have a point in common. 
 
 n. Two Hnes which are parallel to each other in space wih have 
 plan views and elevations which are respectively parallel. 
 
 20. These observations should be verified by experiment. A 
 convenient method of doing this which at once serves to verify 
 the experiment and at the same time to assist the experimenter 
 to visuaHze the problem is shown in the illustrations. Two 
 pieces of cardboard are cut to form the planes of projection and 
 a pencil or a hat pin — which is easier to secure in place — is 
 held in the position the Hne occupies in space. The projections 
 of this position may be plotted on the cardboard and the planes 
 revolved, thus showing the relation of the plan and elevation of 
 the position to the G. L. These positions should, of course, 
 conform to the positions indicated in the observations. 
 
26 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 PROBLEMS IN LOCATION OF POINTS AND LINES 
 
 25. Locate a point Mi" behind V and 2" below H. 
 
 26. Locate a point N \" in front of V and \\" above H. 
 
 27. Locate a point O \" behind V and \\" above H. 
 
 28. Locate a point P 2" in front of V and i" below H. 
 
 29. Locate a point M in the third quadrant 2" from H and i" from V. 
 
 30. Locate a point N in the first quadrant i \" from H and 2" from V. 
 
 31. Locate a point O in the second quadrant 2" from H and V. 
 
 32. Locate a point P in the fourth quadrant i" from H and 2" from V. 
 II. Assume a line MN in the third quadrant parallel to H and i" below 
 
 H, and making an angle of 30 degrees with V. 
 
 34. Assume a line in the first quadrant parallel to V and 2" from it and 
 inclining to H at 60 degrees. 
 
 35. Assume a line in the third quadrant parallel to V and 2" from V, 
 and making an angle of 30 degrees with H. 
 
 36. Draw a line in the first quadrant parallel to the G. L. and i \" from 
 H and V. 
 
 37. Assume a fine with one end in the first and the other end in the third 
 quadrant. Draw an end view of this line. 
 
 38. Draw a line parallel to H and making an angle of 30 degrees to V in 
 both first and second quadrants. 
 
 39. Assume a point in the third quadrant 2" from H and i" from V. 
 Through this point draw two intersecting fines, one parallel to H and one 
 parallel to V. 
 
 40. Assume two intersecting lines in the first quadrant, one parallel to 
 H and one parallel to V and make them 3" long. Let the point of inter- 
 section be 2" from H and \" from V. 
 
 41. Assume a fine 3" long in the third quadrant parallel to the G. L. and 
 i" from H and V. Through the center of this line draw another line making 
 an angle of 45 degrees with both H and V. Draw an end view of the two 
 lines. 
 
 42. One line which inclines 45 degrees to both H and V is intersected by 
 another line which inclines 30 degrees to H and 60 degrees to V. The point 
 of intersection is 2" from H and V. Draw three views of the two lines in 
 the third quadrant. 
 
 43. Draw a line parallel to V i" behind it, and making 60 degrees with 
 H. Draw a second line parallel to this line \\" from it. 
 
 44. A line is 2" below H and parallel to H. One end is in V and the 
 other end is 3" behind V, and between these points the line is 4" long. 
 Through the middle point of this line draw a second line parallel to V. 
 
CHAPTER IV 
 PLANES 
 
 21. A plane is represented in descriptive geometry by its 
 traces. These traces are the lines in which the plane intersects 
 H, V, and P, and are designated as the H trace, the V trace, and 
 the P trace. 
 
 Since the G. L., which is common to both H and V, can meet 
 the plane in but one point it will be obvious that the H trace 
 and the V trace must meet each other at this point. In designat- 
 ing planes on a drawing this point is marked S, T or R, etc., 
 while the H trace is marked with the corresponding small letter 
 s, t, or r, etc.; and the V trace s', t', or r', etc.; and the P trace 
 ^p> *-p, or rp, etc. 
 
 In drafting, planes are usually so arranged as a part of the 
 object being drawn that they are represented by Hmiting hnes. 
 For example in a cube the faces are shown in plan and elevation 
 as squares and the traces of the planes of the faces are not needed. 
 But while the planes of any face of an object are usually not 
 needed to make the complete representation, yet they are useful 
 in solving drawing-board problems which arise in connection 
 with the drawing of the objects. They should, therefore, be 
 studied with the idea in mind that they are a means to an end 
 rather than an end in themselves. For this reason it will be 
 convenient to represent traces with a dashed hne arranged as 
 shown in the figures following. 
 
 22. Planes in descriptive geometry are considered to be 
 indefinite in extent. For this reason, then, it should be clear 
 that any plane extends through all four quadrants even though 
 its traces show it as being in only one. It is customary to show 
 the traces of a plane only in that quadrant where the problem 
 
 lies, yet if it were necessary these traces could be extended. 
 
 27 
 
28 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 In Fig. 27 is shown a plane in the first quadrant; in Fig. 28 a 
 plane in the second quadrant; in Fig. 29 a plane in the third 
 quadrant; and in Fig. 30 a plane in the fourth quadrant. Fig. 
 
 '/ / 
 
 / 
 
 I_ ' 
 
 T T 
 
 V 
 
 9 
 
 / 
 
 \ / \ 
 
 i 
 Fig. 27. Fig. 28. Fig. 29. Fig. 30. 
 
 V 
 
 Fig. 31. 
 
 31 shows a plane indefinite in extent and therefore in all four 
 quadrants: it is obvious from this figure that any pair of the 
 given H or V traces will locate this same plane. 
 
PLANES 29 
 
 23. Observations. 
 
 a. The traces of a plane which is parallel to the G. L. will 
 themselves be parallel to the G. L. Fig. 32 shows such a plane. 
 
 b. If two planes are parallel their respective traces will be 
 parallel. 
 
 c. If a plane is perpendicular to H, its V trace will be perpen- 
 dicular to the G. L. Likewise, when it is perpendicular to V its 
 H trace will be perpendicular to the G. L. 
 
 Fig. 32. 
 
 d. If a plane is perpendicular to H it will also be perpendicular 
 to any other plane whose H trace is perpendicular to its H trace. 
 Likewise a plane perpendicular to V will be perpendicular to any 
 other plane whose V trace is perpendicular to its V trace. 
 
 e. If the plan and elevation of a line are perpendicular respec- 
 tively to the H and V traces of a plane the line will be perpendicu- 
 lar to the plane. 
 
 /. If a line is perpendicular to a plane, the plan view of the 
 line will be perpendicular to the H trace of the plane and the 
 elevation of the line will be perpendicular to the V trace of 
 the plane. 
 
 g. A plane is determined by any two intersecting lines, any 
 three points, any two parallel lines, or a point and Hne. The 
 plane is, of course, in each case represented by its traces. 
 
 h. The plan view of the V trace of any plane is in the G. L.; 
 and the elevation of the H trace of any plane is also in the G. L. 
 
 i. If a line hes in a plane and is parallel to the H trace its plan 
 view will be parallel to that trace and its elevation will be parallel 
 
3° 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 to the G. L. Likewise, when the line is parallel to the V trace 
 its elevation will be parallel to that trace and its plan will be 
 parallel to the G. L. 
 
 These observations are collected here for convenience and 
 reference. No attempt has been made to prove the truth of the 
 assertions set forth because they are more or less axiomatic, and 
 both the proof and truth of these facts will be obvious as the 
 
 study of the subject 
 progresses, and may be 
 proved by experiment 
 as in Article 19. 
 
 24. To assume a 
 plane, assume any two 
 traces which are parallel 
 to the G. L., or which 
 intersect the G. L. at a 
 common point. 
 
 25. To assume a line 
 in a plane, assume either 
 view of the hne, as the 
 plan view ab in Fig. 33. 
 The plan view when ex- 
 tended cuts the G. L. at 
 m and cuts the H trace 
 of the plane at n. Since 
 n is on the H trace its 
 elevation will be on the 
 G. L. at n'; and since m 
 is on the G. L., m' — 
 
 Fig. S3- 
 
 since the line AB is to be in the plane T — must lie on the V 
 trace, m'n', therefore, is the elevation and mn the plan of MN 
 lying in the plane T. ab and a'b' are, of course, on mn and 
 m'n' and are the plan and elevation of a Hne AB lying in the 
 plane T. 
 
 26. To assume a point in a plane. — One method is to assume 
 a line in the plane — -as in the foregoing article — and on this 
 line assume the point. Another method more convenient to use 
 
PLANES 
 
 31 
 
 is to assume one view of the point — as the plan view a in Fig. 
 34 — and through this assumed plan view of the point draw the 
 plan mn of a line MN parallel to the H trace. Since this Kne 
 MN is parallel to the H trace its elevation will be parallel to the 
 
 Fig. 34. 
 
 G. L.; mn and m'n', then, are the plan and elevation of a line 
 MN lying in the plane T. Since a Hes on mn, a' will lie on 
 m'n', and the point A then, as represented by the plan a and 
 elevation a', will be a point in the plane T. 
 
 PROBLEMS ON PLANES 
 
 45. Assume a plane parallel to the G. L. so that its H trace is 2" back 
 of V and its V trace 3" above H. Show an end view of this plane. 
 
 46. A phme is perpendicular to H and inclined 30 degrees to V. Show 
 its three traces. 
 
 47. Draw a plane parallel to the plane in Problem 46. Draw the profile 
 traces 2" apart. 
 
32 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 48. Draw the H, V, and P traces of a plane in the third quadrant. Let 
 the H trace incline 60 degrees to the G. L. and the V trace 45 degrees to the 
 G. L. 
 
 49. Assume any line in the first quadrant oblique to H and V and show 
 the traces of a plane perpendicular to this line. 
 
 50. Assume any oblique plane in the third quadrant and show the traces 
 of a plane which will be perpendicular to this plane and to V. 
 
 51. Assume a plane whose H and V traces incline 60 degrees to the G. L. 
 In,this plane is cut a hole whose plan view is a i" square. Assume the loca- 
 tion of the plan view of the hole and show its plan, elevation, and end view. 
 
 52. Assume a plane whose H trace makes 45 degrees with the G. L. and 
 whose V trace makes 60 degrees with the G. L. In this plane is cut a hole 
 whose elevation is a circle i" in diameter. Find also the plan and end view 
 of this hole. The location of the elevation may be assumed. 
 
 53. A plane inclines 60 degrees to V and 30 degrees to H and is parallel 
 to the G. L. 3" from it. In this plane is cut a hole i" square; show its plan, 
 elevation, and end view. 
 
 54. A plane inclines 60 degrees to V and is perpendicular to H. In this 
 plane is cut a 1" circular hole with its center i^" below H, Show three 
 views of the plane with the hole cut in it. 
 
CHAPTER V 
 
 LOCATION OF POINTS, LINES, AND PLANES 
 
 27. Character of Lines. In order to facilitate the reading of 
 the drawings in the text all given and required lines will be drawn 
 heavy and all auxiliary and constructive lines will be drawn light. 
 When necessary the two views of a point will be connected by a 
 light dotted line perpendicular to the ground line, but wherever 
 possible the line is omitted to avoid the confusion of unnecessary 
 lines. The following chart shows the character and purpose of 
 
 Visible outline 
 
 Invisible outline 
 
 Trace of plane 
 
 Path of moving point 
 
 Joining projections of point 
 
 Center line 
 
 CHART OF LINES 
 
 each line used in the drawings and will be found convenient for 
 reference. 
 
 28. Method of Locating Points and Planes. — A point will be 
 located by giving its distances, or coordinates, in inches always 
 unless otherwise marked, from P, V, and H and in that order. 
 Thus: A is o; 2; 2 means that A in space is in the profile plane, 
 two inches from V, and two inches from H; or B is 3 ; 4; i means 
 that B is three inches from the profile plane, four inches from V, 
 and one inch from H. 
 
 Z2, 
 
34 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 In order to indicate on which side of the planes of projection a 
 point Hes, — or, in other words, in which quadrant it is, — plus 
 and minus signs will be used. Plus always means to the right 
 of P, in front of V, and above H, and unless otherwise marked 
 points will be understood to be plus. Thus A is 4; 3; 2 means 
 that A is four inches to the right of the profile plane, three inches 
 in front of V, and two inches above H. In other words, it is in 
 the first quadrant. 
 
 Minus always means to the left of P, behind V and below H. 
 Thus: B is —2; —3; —5 means that B is two inches to the 
 left of P, three inches behind V, and five inches below H. B, in 
 other words, is in the third quadrant. It will be convenient to 
 remember that a plus sign preceding the last two figures means 
 that the point is in the first quadrant. Thus M is —4; +2; +3 
 means M is in the first quadrant four inches to the left of P, two 
 inches in front of V, and three inches above H. N is 2 ; — i ; — 2 
 means that N is in the third quadrant two inches to the right of 
 P, one inch behind V, and two inches below H. 
 
 A line will be located by giving reference points for any two of 
 its points but it is to be understood that these two points do not 
 necessarily limit the extent of the line. Thus M is — 2 ; — 3 ; —4; 
 N is 3; —4; —2 means that the Kne MN is in the location indi- 
 cated by the location of M and N, but it is not necessarily limited 
 in length by M and N. 
 
 A plane will be located by giving first, the distance its point 
 on the ground line is from P; second, the angle its V trace 
 makes with the ground line; and third, the angle its H trace 
 makes with the ground line. The intersection of the traces 
 on the ground Hne is the vertex of these angles. To indicate 
 the direction of the traces plus and minus signs will be used as 
 follows: a plus sign preceding the angle given for the V trace 
 means that the V trace is below the ground line, and a minus 
 sign means that the V trace is above the ground line; a plus 
 sign preceding the angle given for the H trace means that the 
 H trace is below the ground line, and a minus sign means that 
 the H trace is above the ground line. JMeasurement of these 
 angles begins always on the ground line at the right of the 
 
LOCATION OF POINTS, LINES AND PLANES 
 
 35 
 
 vertex and plus angles for both traces are measured clockwise 
 while in inns angles for both traces arc measured contra-clock- 
 wise. The examples given below illustrate this method of locat- 
 ing planes. 
 
 /-60° 
 
 $=.0';-60; +45 
 
 S=0; + 45;-120' 
 
 S=0i-'-6(^+ao° 
 
 S=0;-150j+120' 
 
 Since planes are indefinite in extent the values of the angles 
 given in the problems often exceed i8o degrees, as T =+2", 
 — 300°, +300°. In such cases 180° may be subtracted from 
 the given values and the method given above will apply. 
 T =+2", —300°, +300° would then become T =+2", ~i2o°, 
 + 120°. 
 
 When a plane is parallel to the ground Hne the traces make 
 no angle with the ground line. Such planes are located by 
 giving first, the symbol 00 (which means that the plane is 
 parallel to the ground line and its traces meet at infinity); 
 second, the distance the V trace is from the ground Hne; and 
 third, the distance the H trace is from the ground Hne. A plus 
 
 T 
 
 S=oo; +l';'+Ji" 
 
 
 :s 
 
 i... 
 
 S = co:-i'f-l34" 
 
 S=oo:tlY-!/2" 
 
 sign preceding the distance given for the V trace means that 
 this trace is above H, and a minus sign means that this trace is 
 below H; a plus sign preceding the distance given for the H 
 trace means that this trace is in front of V, and a minus sign 
 means that this trace is behind V. The examples given above 
 
36 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 illustrate this method of locating planes parallel to the ground 
 line. 
 
 29. Accuracy and Checking. — In working the problems on 
 the drawing board it is reasonable to expect results within two 
 hundredths of an inch. That is, if a hne is actually two inches 
 long its length should be found within the limits one and ninety- 
 eight hundredths and two and two hundredths inches. Such a 
 degree of accuracy means, of course, great care in drawing and 
 plotting. An error made at the beginning of a problem is apt 
 to accumulate as the problem progresses, and the only way to 
 get close results is to check the problem step by step. 
 
 Checking means verifying a result by another method, and it 
 is important in drafting operations that this means of verifying 
 work be resorted to often. 
 
 PROBLEMS ON THE LOCATION OF POINTS, LINES, AND PLANES 
 
 55. A point is in the third quadrant 3" to the left of the profile plane, 2' 
 from H and 4" from V. Write these distances in the proper order, and with 
 the correct signs. Show also three views of the point. 
 
 56. The point M is 3" above H, 2" in front of V, and in the profile plane. 
 Show three views and give the coordinates of the point. 
 
 57. Locate the following points and indicate in what quadrant each is: 
 
 M = 0; 2"; 
 
 l" 
 
 N= -i"; 
 
 -i"; 2" 
 
 = -2"; 
 
 -ir; -2 
 
 
 2"; -i". 
 
 58. M = o; -2"; -i". N = +2; -i"; -§". = i"; -2"; -2". 
 From draw a line OP to the middle point of the line MN and give its 
 coordinates. 
 
 59. A = -3"; +2"; +1". B = -i"; +4"; +2". C = -I"; +1"; 
 + 1". Draw the plan, elevation, and end view of the triangle ABC. 
 
 60. M = +2; — i; —I. N = —2; +2; +2. Draw a plan, elevation, 
 and end view of this line. Use a profile plane at —3 for the end view. 
 
 61. A = o; —2; +1. B = o; +2; — i. Show a plan, elevation, and 
 end view of this line. 
 
 62. M = o; — 3; o. N = o; o; — 3. Show a plan, elevation, and end 
 view of this line and from the G. L. erect a perpendicular to it. Give the 
 coordinates of the point of intersection. 
 
 63- A= —3; —2; — i;B = — i;— 2; —4. With AB as one side con- 
 struct a square parallel to V and show three views of it. 
 
LOCATION OF POINTS, LINES, AND PLANES 37 
 
 64. M = o; -3; -2; N = +3; o; +2; O = +1; -i; o. Draw three 
 views of the triangle MNO. 
 
 65. Show three views of the solid — one of whose faces is ABCD — • 
 whose corners are given below: 
 
 A = o;-2;-2;B = +2;-2;-2;C = 4-2; -2; -4; D = o; -2; 4; 
 E = o; -4; -4; F = o; -4; -2; G = +2; -4; -2; H = +2; -4; -4- 
 
 66. A = +2; -3I; -2; B = +2; -i|; -2; C = +2; -2|; -3!; 
 0=+6; — 2|;— 2|. Draw three views of this pyramid considering ABC 
 as the base and O the apex. 
 
 67. A cube 2" on each edge has one corner at +2"; —3"; o. The edge, 
 of which this point is one end, slopes away from H at an angle of 30 degrees 
 to the left and is parallel to V. Draw three views of the cube in the third 
 quadrant. 
 
 68. Locate the three traces of the plane o; —30; +60. 
 
 69. Locate the three traces of the plane — 2 ; +60; —45. 
 
 70. Locate the three traces of the plane +1; +60; +30, 
 
 71. Locate the three traces of the plane o; —45; —75- 
 
 72. Locate the three traces of the plane o; —135; +45- 
 
 73. Locate the three traces of the plane o; —150; +105. 
 
 74. Locate the three traces of the plane — I ; —165; +150- 
 
 75. Locate the three traces of the plane + 2 ; —90; +90. 
 
 76. Locate the three traces of the plane —2; +90; —75. 
 
 77. Locate the three traces of the plane —3; +135; ~i35- 
 
 78. The plane S == o; —45; +60. Assume a line in this plane 2" above 
 and parallel to H. 
 
 79. The plane T = — 2; —135; +120. Assimie a line in this plane i" 
 behind and parallel to V. 
 
 80. The plane S = o; +60; —60. Assume a line in the plane 2" below 
 H and parallel to H. 
 
 81. The plane S= -2"; +120; —135. Assume a line in this plane 
 parallel to V and i" behind V. 
 
 82. Locate the planes 
 
 00 ; 
 
 -3; 
 
 -4 
 
 00 ; 
 
 -3; 
 
 00 
 
 00 ; 
 
 co; 
 
 +3 
 
 CO ; 
 
 co; 
 
 — 2 
 
 00 ; 
 
 + 2; 
 
 — I. 
 
 3"; 
 
 -4" 
 
 . Find the coordinates of a line 3" 
 
 83. The plane S = co ; +3 
 long in this plane 2" from H. 
 
 84. The plane T = co ; +2"; —3". Draw a hole i" square in this plane, 
 and show three views of it. 
 
CHAPTER VI 
 
 REVOLUTION OF POINTS 
 
 30. One of the most important operations in descriptive 
 geometry and the means by which a great many problems are 
 solved is the revolution of a point, or a Kne, from one position to 
 another. The actual laws governing such a movement are well 
 known, but to represent the change of position correctly, and to 
 visualize, or to get a mental picture of what is taking place dur- 
 ing the change of position, requires careful study. 
 
 Fig. 35. 
 
 31. When a point revolves about a line two things are ap- 
 parent at once: 
 
 1. That the point describes a circle in space. 
 
 2. That the plane of this circle is perpendicular to the line. 
 Put in other words, the revolution of any point about an axis 
 
 takes place in a plane perpendicular to the axis. 
 
 32. Fig. 35 shows the plan and elevation of a circle which 
 has been made by revolving the point A about the Hne MN which 
 lies in H. From this figure it will be seen that when A Hes in H, 
 
 38 
 
REVOLUTION OF POINTS 
 
 39 
 
 or the plane of the axis, it will be at Ah and that the distance from 
 Ah to the axis is equal to the radius of the circle. 
 
 In terms of the distances which are shown by the plan and 
 elevation of point A, this 
 radius is equal to the hypot- 
 enuse of a right triangle 
 whose base is equal to the 
 distance a is from the plan 
 of the axis, or ax, and whose 
 altitude is equal to the dis- 
 tance a' is below the G. L., 
 or the distance A is below 
 the plane of the axis, or a'g. 
 Since the plane of revolu- 
 tion is perpendicular to the 
 axis, A will fall on a per- 
 pendicular through a to mn, 
 either on the right or the left of MN, a distance from 
 MN equal to a'n', or the hypotenuse of a triangle whose 
 
 base is ax and whose altitude 
 is a'g. 
 
 33. Fig. 36 shows this same 
 problem worked out when 
 the axis MN lies in H but 
 oblique to V. In this case 
 Ah will be found on a per- 
 pendicular from a to mn 
 whose length is equal to the 
 hypotenuse of a triangle whose 
 base is ao, and whose altitude 
 is the distance from a' to the 
 G. L.,or a'g. 
 
 Fig. 37 shows a similar 
 Fig ^7 
 
 problem worked out with the 
 
 point revolved into V about an axis which lies in V. In this 
 
 case the point A falls on a perpendicular to m'n' whose length is 
 
 equal to the hypotenuse of a triangle whose altitude is equal to 
 
40 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 the distance a is from the G. L., and whose base is the dis- 
 tance a'o'. 
 
 Fig. 38. 
 
 34. A line may be revolved into H or V about an axis which 
 
 lies in H or V by revolving any two of its points. The hne and 
 
 ^ axis must, however, lie in 
 
 the same plane else one 
 point of the revolving line 
 would touch H or V before 
 the other. Fig. 38 shows a 
 line AB revolved into H at 
 AhBh about MN which 
 lies in H and is parallel to 
 AB. The location of the 
 points Ah and Bh may 
 be found as indicated in 
 Article 32. 
 
 35. Either a point or a 
 
 Fig. 39. _ ^ ^ 
 
 line may be revolved about 
 an axis into the plane of the axis when the plane does not coin- 
 cide with the H or V but is parallel to H or V. The method by 
 
REVOLUTION OF POINTS 
 
 41 
 
 which this may be done may be readily understood by con- 
 sidering that a secondary H or V plane passes through the axis. 
 The revolution, then, may be referred to this new H or V plane 
 and may be accomphshed as indicated in Article 32. 
 
 Fig. 39 shows the plan and elevation of a Une MN which is 
 to be revolved into the plane of the axis PQ. The revolution 
 may be accomplished as indicated above or by working the 
 
 Fig. 40. 
 
 problem without the use of the secondary H or V plane. In the 
 figure the hne MN has been revolved into the plane PQ directly, 
 M falls at a point whose elevation is m", m" being a distance 
 from the axis equal to the hypotenuse of a triangle whose base is 
 the distance of m' to the axis, or m'y', and whose altitude is the 
 distance M is from the plane of the axis, or mx. 
 
 36. To revolve a plane into H or V about the H or V trace. 
 Assume a point on the trace, as A on the trace t'T in Fig. 40. 
 
42 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 Revolve this point into H about tT as an axis as in Article 32. 
 Through this revolved position, Ah, and the point T on the G. L. 
 draw the Hne t'nT. This will be the revolved position of the V 
 trace as it Ues on H and t'nTt is the revolved position of the 
 plane T. 
 
 To revolve the plane into V a point on the H trace should be 
 taken and the V trace used as an axis. 
 
 Fig. 41. 
 
 37. To find the plan and elevation of a line in a plane when the 
 revolved position of the line and plane are given. 
 
 In Fig. 41 is shown a plane T and a Hne MN in the plane T as 
 they He revolved into the H plane at tTt'n and MhNh- To find 
 the actual position of the traces and the plan and elevation of 
 the line assume Ph, the revolved position of a point on the re- 
 volved V trace. P, since it is on the V trace, will lie in V and its 
 plan view therefore Hes on the G. L. Through Ph draw a per- 
 
REVOLUTION OF POINTS 43 
 
 pendicular to Tt, since Tt must be the axis of revolution, and 
 extend it to the G. L. at p. p, then, is the plan of a point P on 
 the V trace, and the elevation will be at p' which is below the 
 G. L. a distance equal to ox. To find ox lay off px equal to opn- 
 ox then is the base of the triangle whose hypotenuse is px, or 
 oPh- Draw t'T through T and p' and it will be the true position 
 of the V trace and tTt' will represent the true position of the 
 plane. 
 
 To find the plan and elevation of any line MN in this plane 
 whose revolved position is MrNh extend MhNh to Ah on the 
 H trace and Bh on the revolved V trace. When the plane is 
 revolved back to its true position tTt', Bh will fall at b' so that 
 Tb' is equal to BhT and Ah falls at Ah- A is shown in elevation 
 at a' on the G. L. and B is shown in plan at b, also on the G. L. 
 Anb is the plan and a'b' is the elevation of the required Hne 
 extended, and M and N may be located on this Hne. 
 
 PROBLEMS ON REVOLUTION 
 
 85.M = o;+i;o. N = +3; +4; o. = +1; +2; +2. Re- 
 volve O into H using MN as an axis. Find the length of the radius of 
 revolution. 
 
 86. A == o; o; +1. B = +3; o; +4. O = +2; -i; — i. Using AB 
 as an axis revolve O into V. Give the length of the radius of revolution. 
 
 87. M = o; — 2; — 2. N == +3; — i; — I. Revolve MN into H about 
 its own plan view. 
 
 88. A = — 3; o; o. B = +1; —.2; —2. Revolve AB into V about 
 a'b' as an axis. 
 
 89. A — +1; —2; — o. B = —2; — i; —3. Revolve AB into H about 
 ab. 
 
 90. The plane S = — 2; +30; —45. Revolve plane S into H about sS. 
 
 91. The plane T = +1; +120; —135. Revolve plane T into V about 
 its V trace. 
 
 92. The plane S = o; +45 ; —60. Revolve this plane into H and in 
 this position locate a 1" square hole in this plane. Show the plan, elevation, 
 and end view of the hole after the plane is returned to its original position. 
 
 93. The point T = o; o; o. The point t' = +3; o; —2. The point 
 ty = o; o; —3. t'Ttv is the revolved position of a plane T as it lies in V. 
 Show its true position. 
 
 94. The point S = o; o; o. The point s = +2; — 3; o. The point 
 s'h = —2; — 3; o. sSs'h is the position of a plane S as it Ues revolved into 
 H. Show it in its true position. 
 
44 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 95. The plane T = go ; — 2; — 3. The point O which lies in T lies at 
 the point o; x, — i. Find the plan and elevation of the point and revolve 
 into V about t'T. 
 
 96. Plane S = o; —60; +30. The point lies in this plane at —2; 
 — I ; X. Show a plan and elevation of this point and revolve it into H about 
 sS. 
 
CHAPTER VII 
 PROBLEMS ON THE LINE 
 
 38. Proposition i . Given two views of a line to find its length. 
 
 Discussion. Revolve the line parallel to any of the planes of 
 projection. In this position the line will be projected on the 
 plane to which it is parallel in its true length. Therefore, the 
 length of the projection is the length of the line itself. 
 
 Construction. First Method. Let it be required to find the 
 over-all length of the brace AB shown in the drawing of the 
 bracket in Fig. 42. Using the center line of the vertical post as 
 an axis turn the brace until it becomes parallel to the V plane 
 and takes the position shown by the light lines. Both A and B 
 will describe arcs of circles during the revolution, and since the 
 planes of these arcs are perpendicular to the axis of the revolu- 
 tion (Art. 32), the points will remain the same distance below 
 H. In the new position shown by a" and b" the Hne is parallel 
 to V and the distance a"b'' is the true length of the brace. 
 
 Construction. Second Method. In Fig. 43 the same prob- 
 lem is solved by revolving AB into H by using ab as an axis 
 of revolution. The revolved position of A is found at Ah by 
 Article 32. Ana is equal to the distance from a' to the G. L. and 
 is drawn perpendicular to ab. In like manner B is found at Br. 
 The hne AhBr being parallel to H is, of course, equal in length 
 to AB. 
 
 Corollary. Given the length of a line and one view of it to 
 construct the other view, when the position of some point on the 
 hne is known. 
 
 Construction. Let the problem be to make a drawing of the 
 hopper (Fig. 44) whose plan is abed — mnop. The length of 
 the edge BN is 12", and the plane of the opening ABCD Ues 
 4" below H. 
 
 45 
 
46 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 First draw the line a'b'c' 4" below the G. L.; this is the eleva- 
 tion of the opening ABCD. Now with b' as a center strike an 
 arc whose radius is 12"; somewhere in this arc will lie n", the 
 
 Fig. 42. 
 
 elevation of the revolved position of N. To find this point re- 
 volve BN parallel to G. L. at bn,. From n^ drop a perpendicu- 
 lar to intersect the arc; the point thus found is n", the elevation 
 
PROBLEMS ON THE LINE 
 
 47 
 
48 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 of the revolved position of N. BN may now be revolved to the 
 first position bn^ and n" will be shown at n'. The elevation of 
 the hopper may now be drawn as shown by the figure. 
 
 39. Special Cases, i. To find the length of a line which Hes 
 in the first and third quadrants. 
 
 2. To find the length of a Une which hes in the profile plane. 
 
 40. Proposition 2. Given two views of a line to find where it 
 pierces the planes of projection. 
 
 Discussion. The point where the line pierces H must he on 
 the line itself and in the H plane. Therefore, the projections of 
 the required piercing point will he on the projections of the given 
 
PROBLEMS ON THE LINE 
 
 49 
 
 line, and the elevation will lie on the G. L. If the elevation of 
 the line be extended to meet the G. L. this point will be the ele- 
 vation of the required piercing point; its plan view will He on the 
 
 Fig. 45. 
 
 plan view of the line and in a perpendicular from the elevation. 
 In a similar manner the V piercing point may be found. 
 
 Construction. In Fig. 45 let the problem be to find the plan 
 and elevation of the point where the center line OP of the timber 
 pierces H and V. By extending o'p' to the G. L. the point m' 
 
50 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 is found; m' is the elevation of the point where the line OP 
 pierces the H plane. The plan view of this H piercing point will 
 be on op where the perpendicular from m' intersects op at m; 
 since M is in H (as shown by the elevation m') and on OP (as m 
 and m' show) it will be the point where OP pierces H. 
 
 To find where OP pierces V, op is extended to the G. L. at n, 
 and n' is located on o'p' as shown. The point N, being in V and 
 on the line OP, will be the point where OP pierces V and n is the 
 plan and n' the elevation of this point. 
 
 In a similar manner the P piercing point, if it be required, may 
 be found. 
 
 41. Special Cases, i. Given the location of the H and V 
 piercing points of a line and the length of the line to draw its 
 plan and elevation. 
 
 2. Find the piercing points of a Hne which Hes in a profile 
 plane. 
 
 PROBLEMS ON THE LINE 
 
 97. Draw the plan and elevation of a triangular pyramid whose base is 
 A = o;-i;-i; B = +i;-3;-i; C = +1^; — 2; -i, and whose 
 apex isO=+i;— 2;— 4. Find the true length of OA. 
 
 98. The center line of a pipe intersects the floor of a room at the point 
 A = -36"; +48"; +0", and intersects the front wall at B = -36"; o"; 
 +36". Find how long the pipe is between these points. 
 
 99. The center line of a timber is so located that one end is at the point 
 M = o; +36"; +6", and the other end is at the point N = -48"; -36"; 
 
 — 36". Find how long the timber is. 
 
 100. A brace rests against a floor at o; —24"; o, and against the wall at 
 
 — 36"; o; —48". Find the true length of the brace (center line only con- 
 sidered) between these points. 
 
 loi. The center hne of a tunnel is 500' long. One portal lies at the point 
 o; — o'; —40', and from this point the tunnel runs north 60 degrees east on 
 a rising 6 per cent grade. Show its plan and elevation and find the actual 
 distance between portals. 
 
 Note. In all land measurement distances are measured on the horizon- 
 tal. Thus, the length of the tunnel given as 500' is reaUy the length of its 
 H projection or plan view, and the distance asked for is the actual linear 
 distance from one portal to the other. 
 
 102. A drain starts at a point -48"; -36"; o, and runs north 45 degrees 
 east for 96" on a falling 10 per cent grade. How many inches of drain pipe 
 win be required? 
 
PROBLEMS ON THE LINE 51 
 
 103. A prop 10' long rests on a floor 6' in front of wall. Find where it 
 will rest against the wall when its plan view inclines 60 degrees to the wall, 
 
 104. A pipe lies in a side wall so that it slopes 30 degrees to the horizontal 
 and goes through the ceiUng 6' back of the front wall. Find where it 
 pierces the front wall. 
 
 105. A tunnel starts from the bottom of a shaft at o; —100'; —65', and 
 runs south 60 degrees east on a rising 15 per cent grade. Find the point 
 where the tunnel will reach the surface and how long it will actually be. 
 
 106. A brace rests against a wall 60" above the ground, and rests against 
 the ground 60" in front of the wall. The brace is 108" long. Show three 
 views of it. 
 
 107. How many feet of pipe will be required to join an opening in a floor 
 18" back of the front wall of a building with an outlet in the left side wall 
 60" below the floor and 36" back of the front wall? The opening in the 
 floor is 48" from the side wall. 
 
CHAPTER VIII 
 PROBLEMS ON THE PLANE 
 
 42. Proposition 3. To find the traces of a plane when two 
 views of any of its lines are given. 
 
 Discussion. Since the trace of a plane is the line in which the 
 plane cuts H or V it must contain the points in which all lines 
 contained in the plane cut H or V. Therefore, if the H and V 
 piercing points of the given lines be found these will determine 
 the traces of the plane of the lines. 
 
 Construction. Let it be required in Fig. 46 to find the traces 
 of the plane of that side of the hood given by the two lines AB 
 and BC. By Proposition 2 find where AB and BC pierce H and 
 V; AB piercing H at M and V at N; BC piercing H at and V 
 at P. Join M and O; this will give the H trace tT. Join N and 
 P; this gives the V trace t'T. The plane of the two lines then 
 is tTt'. The traces must of course meet on the G. L. 
 
 43. Special Cases. Find the traces of the plane of two lines: 
 
 1. When one line is parallel to H. 
 
 2. When one fine is parallel to H and the other parallel to V. 
 
 3. When one line is parallel to the G. L. 
 
 4. When one fine is parallel to the G. L. and the other is 
 
 parallel to H or V. 
 
 5. When the two lines are given by their plans and end views. 
 Note. In working the above special cases it should be 
 
 remembered that when a fine lies in a plane and is parallel to 
 H or V, it will be parallel to the H or V trace of the plane. See 
 observations on planes, Article 23. 
 
 44. Proposition 4. Given the plan and elevation of a point 
 to pass through it a plane parallel to two lines whose plans and 
 elevations are given. 
 
 52 
 
PROBLEMS ON THE PLANE 
 
 S3 
 
 Discussion. If through the given point two lines be drawn 
 parallel to the two given Hues, the plane of these two Unes will 
 be the required plane. 
 
 Construction. Let the problem be to cut off the trough, 
 Fig. 47, one of whose edges is NP, by a plane through P parallel 
 to MN and NO. Through p draw pq and pr parallel to the plan 
 of the two given Hues, mn and no. Also draw p'q' and p'r' 
 parallel to m'n' and n'o'. Find now where these lines, PQ and 
 PR, which are parallel to the given Hnes, pierce H and V. PQ 
 pierces H at X and V at q'. PR pierces V and y'. Draw q'y' 
 and extend it to T; also draw Tx. These are the H and V 
 
54 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 traces respectively, and since PR is parallel to H the trace tT 
 should be parallel to pr. The plane T then is the required plane 
 through P parallel to MN and MO. 
 
 45. Special Cases. Pass a plane through a point parallel to 
 two given lines: 
 
 1. When one of the lines is parallel to the G. L. 
 
 2. When one of the lines is parallel to the H or V. 
 
 3. When one line is parallel to H and the other is parallel to V. 
 
 4. Pass a plane through one Une parallel to another. 
 
 46. Proposition 5. Given the plan and elevation of a point 
 to pass through it a plane parallel to a given plane. 
 
 Discussion. Since the required plane is to be parallel to the 
 given plane the corresponding traces of these two planes will be 
 
PROBLEMS ON THE PLANE 
 
 55 
 
 parallel. The direction of the required traces is, therefore, 
 known. To find their location a line may be drawn through the 
 given point parallel to either trace of the given plane and its H 
 
 or V piercing point found. With one point on either trace and 
 the direction of the traces known the plane may be located. 
 
 Construction. In Fig. 48 is shown the plan and elevation of 
 an hexagonal pyramid whose base lies in the plane T. The prob- 
 
56 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 lem is to cut off the prism by a plane parallel to T through the 
 point O. 
 
 Through draw a line OP parallel to t'T. This line pierces 
 H at P. Through P draw sS parallel to tT and through S draw 
 s'S parallel to t'T; sSs' is the required plane through O parallel 
 to the plane T. 
 
 Fig. 49. 
 
 47. Special Case. Pass a plane through a point parallel to a 
 plane which is parallel to the G. L. 
 
 48. Proposition 6. Given the plan and elevation of a Une to 
 draw through a point, whose plan and elevation are given, a 
 plane perpendicular to the line. 
 
 Discussion. When a line is perpendicular to a plane the plan 
 and elevation of the Hne are perpendicular to the corresponding 
 traces of the plane. Thus the direction of the traces of the 
 required plane is known. If then a Hne be drawn through the 
 
PROBLEMS ON THE PLANE 57 
 
 given point parallel to the direction of either of these traces this 
 line will be in the required plane. By finding its H or V piercing 
 point the location of one point on one of the traces may be found 
 and the plane located. 
 
 Construction. Let the problem be to find the traces of a plane 
 through the point O which will cut a section perpendicular to 
 MN, the axis of the pyramid in Fig. 49. The H trace of the 
 required plane will be perpendicular to the plan view of the axis, 
 and the V trace will be perpendicular to the elevation. There- 
 fore, through 0, the given point, draw a fine parallel to the 
 direction of the required H trace; its plan view will be op and 
 its elevation o'p'. This line OP is a line of the required plane 
 since it contains a point in the required plane and is parallel to 
 one of the traces of that plane. It pierces V at p' and if through 
 p' r'R be drawn perpendicular to m'n' the elevation of the axis, 
 it will be the V trace of the required plane. The H trace is Rr 
 which is perpendicular to the plan view of the axis. rRr', then, 
 is the required plane. 
 
 49. Special Cases, i. Construct this same problem when the 
 point does not lie on the line to which the plane is to be per- 
 pendicular. 
 
 2. Find the location of a plane which is parallel to a given 
 plane at a given distance from it. 
 
 PROBLEMS ON THE PLANE 
 
 108. The point A = o; — .?; —3; the point B = +1; — i; — i; the 
 point C = -\-2; —3; — 2. Find the traces of the plane of the triangle ABC. 
 
 109. The point A = -|-i; — i; —3; the point B = +1; —2; — i; the 
 point M = -I-3; — i; —3; the point N = -)-3; —2; — i. Pass a plane 
 through AB and MN and show its traces. 
 
 no. The line AB is parallel to the G. L. i" below H and 2" back of V. 
 The Hne MN is parallel to the G. L. i" above H and 2" in front of V. Find 
 the traces of the plane of these two lines. 
 
 111. The point M = — i; +2; +2; the point N = —2; — i; — i; the 
 point O = — 2; -\-2; — i; the point P = o; — i ; — 2. Pass a plane through 
 P parallel to the lines MN and ON. 
 
 112. The point A = o; — i; o; the point B = -\-2; — i; —3; the point 
 C = -\-2; —3; —3; the point = —2; —3; — i. Pass a plane through 
 O parallel to AB and BC. 
 
58 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 113. The plane T = 00 ; —2; —4; the point = —3; —2; —3. Pass 
 a plane through O parallel to plane T. 
 
 114. The plane S = o; +60; —60; the point = +2; — i; —3. Pass 
 a plane through O parallel to plane S. 
 
 115. The point M = — i; —2; — i; the point N = —2; —3; —2. 
 Pass a plane through M perpendicular to the line MN. 
 
 116. The point A = — i; —3; — 4; the point B = —3; — i; — i. Pass 
 plane through AB perpendicular to it at its middle point, 
 
 117. The point M = o; o; o; the point N = — i; —3; —3; the point 
 O = —3; — i; —2. Pass a plane through O perpendicular to MN. 
 
CHAPTER IX 
 PROBLEMS ON ANGLES 
 
 50. Proposition 7. Given two views of an angle to find its 
 true size. 
 
 Discussion. If the plane of the angle be revolved into co- 
 incidence with H or V the angle in this position will be shown in 
 its true size. Therefore, find the plane of the angle, and using 
 either the H or the V trace as an axis revolve the angle into H or 
 V. When the angle coincides with H or V it will be shown in 
 its true size. 
 
 Construction. In Fig. 50 let the problem be to find the true 
 size of all the angles between the edges of the section of the tri- 
 angular prism; or, in other words, the shape and size of the 
 triangle ABC. By Proposition 3 find the plane of the triangle; 
 its traces are rR and r'R. With the V trace r'R, as an axis, 
 revolve the triangle into V by the method in Article 36. When 
 coinciding with V the triangle will occupy the position AyByCv, 
 and as this is its true shape and size the edges and the angles may 
 be measured with scale and protractor. 
 
 51. Corollary. Given two views of one side of an angle, the 
 traces of its plane, and its size, to construct the views of the 
 other side. 
 
 Discussion. If the given side be revolved into H or V about 
 the trace of the given plane, the angle may be constructed in this 
 position in its true size. If now the plane be revolved back to 
 its original position the two views of the required angle may be 
 found. 
 
 Construction. Let the problem be, in Fig. 51, to construct 
 a rectangular opening in the plane R with MN as one edge and 
 the other edge one-half as long. By Article 34 revolve MN about 
 rR into H. In this position MN will lie at MhNh and the plane 
 R will lie at rRr'n. With MrNh as one side construct the rec- 
 
 59 
 
6o 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 tangle MrNhOhPh making MrPh equal to ^ MhNh- This 
 rectangle is the true shape and size of the required rectangular 
 opening. 
 
 Now revolve rRr'n back to its first position rRr' and with it 
 revolve the rectangle MhNhOhPh- To find the plan of P draw 
 
 Fig. so. 
 
 the diagonal NrPh and let it intersect the trace rR at a. Now 
 when the plane R returns to its first position Nh revolves to n 
 and the diagonal aBn revolves to ab. Since Ph is on this 
 diagonal p, the required plan view of P will be found at the 
 point where a perpendicular from Ph to rR intersects ab. To 
 find the elevation of P draw first a'b', a' being on the G. L. since 
 
PROBLEMS ON ANGLES 
 
 6i 
 
 A is on the trace rR, and b' being on r'R. p', the elevation of 
 P, will be found on a'b' in a perpendicular to the G. L. from p. 
 In like fashion, O may be found and the plan and elevation of the 
 required rectangular opening constructed. 
 
 Fig. si. 
 
 52. Corollary. Given a line and a point to draw through the 
 point a line making a given angle with the given line. 
 
 Discussion. If a plane be passed through the given point and 
 line it will be the plane of the required angle. If this plane be 
 revolved into H or V the true relation of the point to the line 
 will be shown when they are in the new position. The given 
 angle may then be constructed. If now the plane be revolved 
 
62 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 back to its original position the plan and elevation of the required 
 angle may be found. 
 
 Construction. Let the construction be made according to 
 the discussion. 
 
 53. Special Cases, i. To find the size of an angle when one 
 side is parallel to the G. L. 
 
 2. To draw two views of the bisector of an angle. 
 
 PROBLEMS ON ANGLES 
 
 118. Given the triangle ABC, A = o; —3; —2; B = — i; — i; — i; 
 C = —2: —2; —3, to find its true shape and size. 
 
 119. The point A = o; o; —3; the point B = +1; o; — i; the point 
 C = +2; o, —2; the point O = — i; —3; — i. Find the true shape and 
 size of each of the faces of the triangular pyramid whose base is ABC and 
 upon whose apex is O. 
 
 120. The point M = o; —3; — i; the point N = o; — i; —3; the point 
 0=— 2;— i;— I. Find the true shape and size of the angle MNO. 
 
 121. The plane S = o; —60; +45. The point O in this plane lies at 
 + 2; +1; +x. With O as one corner construct an equilateral triangular 
 hole in the plane S with sides i" long. Let the base of the triangular hole 
 be parallel to the H trace of the plane. 
 
 122. The plane T = —3; —45; +60. A Hne lies in this plane at M = 
 — 5; — x; —I and N = —6; — 2;y. With MN as one side construct a 
 square hole in plane T. 
 
 123. The plane ,S = o; +45; —60. The point O = +2; — i; x. 
 Through the point O, which Hes in plane S, draw a hne in plane S perpen- 
 dicular to the H trace of S and find the angle it makes with the V trace. 
 
 124. M = —3; —2; —2. N == o; — i; — i. O = —2; — i; —3. 
 Draw a line through making an angle of 30 degrees with MN. (Two 
 solutions possible.) 
 
 125. M = -3; -i; -I. N = o;-2;-3. = -2; -4; -i. 
 Through O draw a line perpendicular to MN. 
 
 126. A = -i; +2; +2. B=-i;+3;+4. O = o; +1; +1. 
 Through draw a hne perpendicular to AB. 
 
 127. A = o; +1; +4. B = o; — i; —2. C = +2; —3; — i. Show 
 three views of the bisector of the angle ABC. 
 
 128. An observer stands in a hghthouse tower at A = +1000'; +300'; 
 + 100', and at 10 o'clock observes a ship N. 30° E. from him at an angle of 
 depression of 30 degrees. One hour later the same ship is N. 60° W. from 
 him at an angle of depression of 15 degrees. Assuming the ship keeps the 
 same course and speed how will it bear from him and what wUl be its angle 
 of depression at 12 o'clock? 
 
CHAPTER X 
 PROBLEMS ON POINTS, LINES, AND PLANES 
 
 54. Proposition 8. To draw the plan and elevation of the 
 intersection of two given planes. 
 
 Discussion. Since the line of intersection contains all of the 
 points common to both planes, its location will be determined 
 if any two of these common points be located. If, then, a line 
 
 Fig. 52. 
 
 be drawn from the intersection of the H traces to the intersection 
 of the V traces, the plan and elevation of this line show the 
 required intersection. 
 
 Construction. Let T and S, Fig. 52, be the given planes 
 whose intersection is required. The H traces intersect at a point 
 whose plan is a and whose elevation is a'. The V traces intersect 
 at a point whose plan is b and whose elevation is b'. ab, then, 
 is the plan and a'b' is the elevation of the required intersection 
 AB. 
 
 63 
 
64 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 In case the traces of the given planes do not intersect con- 
 veniently on the drawing, Fig. 53 shows a method of finding 
 the required intersection without using the intersection of the 
 traces. T and S are again the given planes, Rr is an auxiliary 
 plane parallel to V, which cuts the line AB from the plane S, and 
 the line BC from the plane T, thus giving a point B common to 
 both planes. In hke manner a second point N on the required 
 
 Fig. S3. 
 
 intersection may be found by passing an auxiliary plane U 
 through S and T parallel to H. B and N when joined give the 
 plan view bn and the elevation b'n' of the required intersection 
 of the planes S and T. 
 
 55. Special Cases, i. Find the intersection of two planes 
 when one pair of traces are parallel. 
 
 2. Find the intersection of two planes which are parallel to 
 the G. L. 
 
 56. Proposition 9. Given a plan and elevation of a line to 
 find where it pierces a given plane. 
 
PROBLEMS ON POINTS, LINES, AND PLANES 
 
 65 
 
 Discussion. If any plane be passed through the given line it 
 will contain all points of that line. It will, therefore, contain 
 the required point. Since the required point Hes in the given 
 plane and in the auxihary plane it will lie on their hne of inter- 
 section. If then the line of intersection be found and the point 
 
 Fig. 54. 
 
 common to it and the given line be located it will be the required 
 piercing point. 
 
 Construction. In Fig. 54 is shown the plan and elevation of 
 a timber whose center line is OP. The problem is to find the 
 point where this line OP pierces the plane tTt'. Pass the auxil- 
 
66 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 iary plane R through OP. Since the required piercing point lies 
 in the plane R and in the given plane T it will lie on their inter- 
 section MN. (See Proposition 8.) The required point lies on 
 MN and OP; therefore, it will be at the intersection of these two 
 lines or at Q. q then is the plan and q' is the elevation of the 
 required point. 
 
 It should be noted that any auxiliary plane through OP will 
 serve, but, for convenience in drawing, one which is perpendicular 
 to H or V should be used. The points where the edges of the 
 timber pierce T may be found in a similar manner and when 
 joined give the plan and elevation of the section cut from the 
 timber by plane T. 
 
 57. Special Cases. Find where a line pierces a plane: 
 
 1. When the hue is parallel to the G. L. 
 
 2. When the plane is parallel to the G. L. 
 
 3. When the plane is parallel to the G. L. and the Hne lies in 
 
 a profile plane. 
 
 4. When the plan and elevation of the line are parallel respec- 
 
 tively to the H and V traces of the given plane. 
 
 58. Proposition 10. Given the plan and elevation of a point 
 to find its shortest distance from a given plane. 
 
 Discussion. The shortest distance from a point to a plane is 
 the perpendicular distance from the point to the plane. If the 
 perpendicular be drawn from the point to the given plane and 
 the point where the perpendicular pierces the plane be found, 
 the distance between these two points will be the required 
 shortest distance. 
 
 Construction. In Fig. 55 the problem is to find the altitude 
 of the pyramid whose base lies in the plane T and whose apex is 
 at O. From O draw a Une perpendicular to the plane T. (See 
 Observation/, Article 23.) Find where this line pierces plane T 
 by Article 56. p is the plan and p' the elevation of this piercing 
 point. Then op is the plan and o'p' is the elevation of the 
 shortest distance from the apex of the pyramid to the plane of 
 its base. Its true length may be found by Article 38, as shown 
 at o"p'. 
 
 It should be noted that P, which is the foot of the perpendicu- 
 
PROBLEMS ON POINTS, LINES, AND PLANES 
 
 67 
 
 lar drawn from O to the plane T, is the projection of the point 
 O on the obHque plane T. 
 
 Fig. 55. 
 
 59. Corollary. Given a plane, the plan and elevation of the 
 projection of a point upon this plane, and its distance from the 
 plane to find the plan and elevation of the point. 
 
 Discussion. If a perpendicular be erected to the given plane 
 through this given projection of the point, the required point 
 
68 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 will lie in this line. To find its location a point on the perpen- 
 dicular must be found at the given distance from the plane. 
 
 Construction. Let the construction be made in accordance 
 with the suggestions in the discussion. 
 
 60. Special Cases, i. Find the shortest distance from a point 
 to a plane when the plane is parallel to the G. L. 
 
 2. Given two parallel planes to find the shortest distance 
 between them. 
 
 PROBLEMS 
 
 129. The plane S = o; —45; +60. The plane T = +3; —300; +300. 
 Find the true length of that portion of the intersection of planes S and T 
 which lies between H and V. 
 
 130. The plane S = 00 ; —3; — i. The plane T = qo ; — i; —3. Find 
 the plan and elevation of the intersection of these planes, 
 
 131. The plane S = o; +60; —60. The plane T = 00; —4; —3. Find 
 the plan and elevation of the intersection of planes S and T, 
 
 132. The plane S = o; —90; —75. The plane T = —2; —75; —6c. 
 Find the plane and elevation of the intersections of S and T. 
 
 133. The plane T = —2; —go; —300. The plane S = +1; —90; 
 -I-300. Find the plan and elevation of the intersection of planes S and T. 
 
 134. The plane S = —2; +45; —60. The point M = —2; — i; —5. 
 The point N = +2; —4; — i. Find where the line MN pierces the plane S. 
 
 135. The plane S = co ; —4; —2. The point M = —3; — i; —4. 
 The point N = o; —3; — i. Find where MN pierces plane S. 
 
 136. The plane S = o; +75°; —30°. The line MN is in the third 
 quadrant parallel to the G. L. 2" from H and V. Find where the line MN 
 pierces plane S. 
 
 137. The point M = o; — i; — 2. The point N = —3; — i; —4. The 
 plane S = — 3; —60°; —60°. Find where MN pierces plane S. 
 
 138. The plane S = o; +135°; —105°. The point O = — i; —3; —2. 
 Find the shortest distance from to plane S. 
 
 139. The plane R = 00 ; —3; —2. The point O = — o; —3; —4. 
 Find the shortest distance from O to the plane R and show its plan and 
 elevation. 
 
 140. The plane S = o; +60°; —45°. Pass a plane parallel to plane S 
 and 2" from it. 
 
 141. The plane S = o; +75°; —60°. The plane R = — 2; +75°; —60°. 
 Find the shortest distance between the planes. 
 
 142. The plane S = o; -45°; +60°. The point M = -2"; -2"; x. 
 M is one corner of a regular square pyramid whose base lies in plane S. The 
 base of the pyramid is a 2" square and the altitude is 3". Draw a plan, 
 
PROBLEMS ON POINTS, LINES, AND PLANES 69 
 
 elevation, and end view of the pyramid when one edge of its base is parallel 
 tosS. 
 
 143. The plane R = o; +45"; -60°. The point is +3"; -y"; -2". 
 This point is the center of a i" square which lies in plane R with one edge 
 making an angle of 30 degrees with Rr. Draw the plan, elevation, and left 
 end view of a cube, one of whose faces is this square. 
 
 144. A point of light is located at +3; +6; +3. A ray of light from 
 this point strikes a mirror in H at the point o; +2; o and is reflected to V. 
 Find the coordinates of the point where the reflected ray strikes V. (It 
 will be remembered that the angles of incidence and reflection are equal.) 
 
 145. A ray of light comes from the point +6; +5; +4, and strikes a 
 mirror at the point +3; +3; +2. The plane of the mirror inclines 60 de- 
 grees to V and is perpendicular to H. Find where the reflected ray strikes V. 
 
 61. Proposition ii. Given the plan and elevation of a line to 
 find the plan and elevation of its projection upon any given plane. 
 
 Discussion. The projection of any point upon any plane is 
 the foot of a perpendicular from the point of the plane. If 
 perpendiculars be drawn from any two points of the given line 
 to the given auxihary plane and the points where these perpen- 
 diculars pierce this plane be found, these piercing points when 
 joined will be the required projection of the Kne. 
 
 Construction. The problem in Fig. 56 is to find the plan and 
 elevation of the projection of the face MNOP of the casting on 
 the plane R to which this face is parallel. From each corner of 
 the face draw a perpendicular to the plane R and find where 
 these perpendiculars pierce the plane. The plan view of the 
 projection on R is abed — these points being the plan view of 
 the piercing points of the perpendiculars — and the elevation is 
 a'b'c'd'. AhBhChDh shows the face in its true size after it has 
 been revolved into H about rR as an axis, and set to one side 
 of the actual plane of revolution so as to avoid confusion in the 
 drawing. 
 
 62. Special Cases. To project a Une upon any plane: 
 
 1. When the Hne is parallel to the G. L. 
 
 2. When the plane is parallel to the G. L. 
 
 3. When the plan and elevation of the given line are parallel 
 
 to the corresponding traces of the given plane. 
 
 63. Proposition 12. Given the plan and elevation of a line 
 and of a point, to find the shortest distance between them. 
 
70 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 Discussion: First Method. If the line and the point be 
 revolved into H or V about the trace of the plane which contains 
 
 Ch 
 
 Fig. 56. 
 
 them they will be shown in this revolved position in their true 
 relation. If now from the revolved position of the given point 
 a perpendicular be drawn to the revolved position of the line, 
 
PROBLEMS ON POINTS, LINES, AND PLANES 7 1 
 
 this perpendicular will be the measure of the shortest distance 
 from the point to the given line. If a plan and elevation of this 
 distance be required the point must be revolved back to the first 
 position, and with it the perpendicular. 
 
 Construction. Let the construction be made in accordance 
 with the discussion. 
 
 Discussion: Second Method. It not infrequently happens 
 that the relation of the given point to the given line is such as 
 to make the construction by the first method awkward. In 
 such cases it will be found more convenient to pass a plane 
 through the point perpendicular to the line, and by finding where 
 the given line pierces this plane a point will be found which, when 
 joined to the given point, will be the shortest distance from the 
 given point to the given hne. 
 
 Ccnstruction. Fig. 57 shows such a case. The problem is to 
 nnd the length of the center Hne required for a timber to reach 
 from point A to the second timber whose center line is OQ. 
 Through A the plane R is passed perpendicular to OQ. (Article 
 48.) The hne OQ pierces this plane at B. (Article 56.) AB 
 then is the required shortest distance and abi shows its true 
 length. 
 
 64. Proposition 13. Given the plans and elevations of two 
 Hnes not in the same plane to find the plan and elevation of the 
 shortest Hne which can be drawn between them. 
 
 Discussion. If through one of the given lines a plane be 
 passed parallel to the other, and if upon this plane the second 
 Hne be projected, this projection will intersect the first Hne at a 
 point which will be one end of the shortest connecting Hne. If 
 now at this point a line be erected perpendicular to the plane it 
 will intersect the second line at a point which will be the other 
 end of the shortest connecting Hne. If the length of this line be 
 required it may be found as in Article 38. 
 
 Construction. Fig. 58 shows lwo timbers lying in different 
 planes. The problem is to find the shortest third timber to 
 join them — center Hnes only being considered. Through PQ 
 pass a plane parallel to AB. This plane is rRr'. Upon plane 
 R project AB. (Article 61.) This projection on R is shown by 
 
72 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 Fig. .t;7. 
 
PROBLEMS ON POINTS, LINES, AND PLANES 
 
 73 
 
 mn and mV. Now at O, where PQ and MN intersect, erect a 
 perpendicular to the plane R. This perpendicular intersects 
 AB at C. CO then is the center Hne of the shortest timber that 
 can connect PQ and AB. Its true length may be found as 
 in Article 38. 
 
 65. Proposition 14. Given the plan and elevation of a line 
 to fmd the true size of the angle it makes with a given plane. 
 
 Discussion. The angle a line makes with a plane is equal to 
 the angle a line makes with its projection on that plane. If, 
 therefore, the given line be projected on the given plane and the 
 true size of the angle between the given Hne and the projection 
 on the given plane be measured it will be the required angle the 
 given hne makes with the given plane. 
 
74 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 Construction. In Fig. 59 the problem is to find the angle 
 between the plane S and the center Hne NQ of a tunnel. By 
 
 Fig. 59. 
 
 Article 56, find where the center Kne NQ pierces the plane S. 
 This point is M, and it is the vertex of the required angle. From 
 any other point N of NQ draw a line perpendicular to the plane 
 
PROBLEMS ON POINTS, LINES, AND PLANES 
 
 75 
 
 S and find where it pierces the plane S. This point is P, 
 shown in plan at p and in elevation at p', and MP is the 
 projection of MN upon S. NMP, then, is the angle the line 
 NQ makes with the plane S. Its true size is shown at NhMhPh- 
 (Article 50.) 
 
 Further Discussion. In case only the size of the angle the line 
 makes with the plane is required a shorter method may be used. 
 
 Fig. 60. 
 
 In the problem just solved it will be observed that since the angle 
 MPN is a right angle — NP being perpendicular to MP — the 
 angle PNM is the complement of the required angle. If, there- 
 fore, from any point of the given hne a perpendicular be drawn 
 to the given plane the angle between the perpendicular and the 
 given line will be the complement of the required angle. The 
 true size of this complement may be found and subtracted from 
 90 degrees thus giving the size of the required angle. 
 
76 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 Construction. Fig. 60 shows the method of finding the size 
 of the angle between the plane S and the Hne OQ. The perpen- 
 dicular from O to S pierces V at P; OQ pierces V at Q. r'R is 
 the V trace of the plane of the angle QOP and its true size may 
 be found as shown. Subtracting q'Ovp' from 90 degrees gives 
 NOvp', the size of the angle OQ makes with S. 
 
 66. Special Cases, i. Find the angle a given line makes with 
 H, V, or P. 
 
 2. Given the plan and elevation of a line and of its projection 
 on some plane to locate the traces of the plane. 
 
 PROBLEMS 
 
 146. The plane T = o; +60; -60. The point M = o; -i; -i|. 
 The point N = +2; — i ; — i. Find the plan and elevation of the line MN 
 when projected upon plane T. 
 
 147. The plane S = ^; -3"; +4". The line MN is parallel to the 
 G. L. 1" from H and V in the third quadrant. Find the plan and elevation 
 of MN when projected upon plane S. 
 
 148. The point M = o; o; o. The point N = +3; "3; — 3- The 
 point O = +2; —3; —I. Find the shortest distance from O to MN and 
 show its plan and elevation. 
 
 149. A tunnel runs from a point N = o; — 6; o S. 60° W. on a falling 
 6 per cent grade. The bottom of a shaft is located at a point — 3 ; — 2 ; — 10. 
 Find the actual length of the shortest tunnel which will connect the bottom 
 of the shaft and the tunnel and show its plan and elevation. 
 
 ■ 150. A drain runs N. 30° E. from a point o; o; — i on a falling 10 per cent 
 grade. Show the plan and elevation and find the length of the shortest 
 connection between the drain and the point + 1 ; — 2 ; o. 
 
 151. A steam pipe runs through the points M = o; —3; — i and 
 N = +3; —6; —I. A second pipe runs through the points A = — i; — i; 
 —4 and B = +2; — i; —2. Find the shortest connection between the 
 pipes and give its true length. Use center lines only. 
 
 152. A tunnel joins the bottoms of two shafts which are located at 
 o; —8; —50 and +50; —25; —75. A second tunnel starts at o; o; —75 
 and runs N. 45° E. on a rising 20 per cent grade. Draw a plan and elevation 
 of the shortest connecting tunnel which can join these two and give its 
 actual length. 
 
 153. The plane of an ore body is located at o; +75°; —60°. This plane 
 is cut by a tunnel at the point O = + 2 ; -3 ; x. The tunnel runs S. 60° E. 
 through this point on a rising 10 per cent grade. Find the angle the tunnel 
 makes with the plane of the ore. 
 
PROBLEMS ON POINTS, LINES, AND PLANES 77 
 
 154. The plane of a body of ore is located at S = o; +75°; —60°. A 
 shaft is sunk at a point +3; — 3; o. Find the angle the shaft will make 
 with the plane of the ore and how deep the shaft will have to be in order to 
 reach the ore. 
 
 155. A brace 14' long meets the ground 8' in front of a wall and rests 
 against the waU 6' above the ground. Find the angles the brace makes 
 with the ground and the wall. 
 
 67. Proposition 15. Given the angles a line makes with H 
 and V, and the plan and elevation of some point on the line, to 
 draw the plan and elevation of the line. 
 
 Discussion. If a line of any assumed length be drawn through 
 the point parallel to H and making the required angle with V the 
 plan view of the line will be equal in length to the plan view of 
 the required line. Also, if a line be drawn through the given 
 point parallel to V and making the required angle with H the 
 elevation of this line will be equal in length to the elevation of 
 the required line. Having the length of the two views and the 
 location of one point the line may be constructed. 
 
 Construction. In Fig. 61, O is the bottom of a shaft from 
 which runs a tunnel with a given grade and direction. The 
 grade is equivalent to the angle it makes with H and its direction 
 is equivalent to the angle it makes with V. 
 
 Draw a line through O parallel to H making the given direc- 
 tion with V. Assume OQ as the length of the tunnel, o'q", 
 then, is the length of the elevation of the required line. Likewise 
 draw through O a line parallel to V and making the given grade 
 with H, and make it equal in length to OQ. op^, then, is the 
 length of the plan view required. Revolve these two lines until 
 P and Q coincide at N ; on and o'n' then will be the required plan 
 and elevation of the center line of the tunnel. 
 
 Note. The sum of the angles given must not exceed 90 
 degrees or the problem becomes impossible. 
 
 68. Proposition 16. Given two planes to find the angle be- 
 tween them. 
 
 Discussion. The angle between the planes is measured in a 
 plane perpendicular to both planes or to their line of intersection. 
 If, then, a plane be passed through any point of a line common 
 
78 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 to both planes and the intersection of this plane with both of the 
 others be found the angle thus formed will be the required angle 
 between the given planes, and its size may be determined. 
 
 Construction. In Fig. 62 is shown a plan and elevation of a 
 trough. The problem is to find the angle between ilB sides. 
 
 A 
 
 1^ 
 
 
 / 
 
 9" 
 
 :^„ 
 
 Fig. 61. 
 
 Assume the point on the line common to the planes of the two 
 sides, and through this point, by Article 48, pass a plane perpen- 
 dicular to the line. This plane is rRr'. Now by Article 54 find 
 the intersection of R and S, and R and T. This is OP in the first 
 case and OQ in the second, and POQ is the required angle. Its 
 true size, pOnq, may be found by Article 50. 
 
 69. Special Cases, i. Find the angle a plane makes with H 
 and V. 
 
PROBLEMS ON POINTS, LINES, AND PLANES 
 
 79 
 
 2. Find the angle between two planes which are parallel to 
 the G. L. 
 
 70. Proposition 17. Given one trace of a plane and the angle 
 the plane makes with the corresponding plane of projection to 
 find the other trace. 
 
 Fig. 62. 
 
 Discussion. The vertex of the given angle will lie on the given 
 trace and the angle will lie in a plane perpendicular to the given 
 trace. This angle may, therefore, be constructed and since one 
 
8o 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 side of it lies in the required plane the piercing point of this side 
 will lie on the required trace. 
 
 Construction. Let the problem be, Fig. 63, to find the plane 
 of the ore body whose strike is the Hne sS (the "strike" is the 
 
 s 
 
 / 
 
 Fig. 63. 
 
 direction of any horizontal line in the plane) and whose dip is 
 60 degrees (the "dip" is the inclination of the plane to the hori- 
 zontal). Lay off an angle of 60 degrees with its vertex o on sS 
 
PROBLEMS ON POINTS, LINES, AND PLANES 
 
 8l 
 
 and one side OP perpendicular to sS. Pnop will be the revolved 
 position of the given angle the ore body makes with H, and Ph 
 will be the revolved position of the point where OP pierces V. 
 Since the line OP must lie in the plane of the ore revolve the angle 
 till it is perpendicular to H, and Ph will fall at a point shown in 
 plan and in elevation at p and p'. P, then, is a point in s'S, 
 the required trace. 
 
 71. Special Case. Given the angles a plane makes with H 
 and V and the location of one point in the plane to find the traces 
 of the plane. 
 
 Discussion. If through the given point a line be drawn 
 perpendicular to the required plane this Hne will make with H 
 and V angles equal to the complements of the angles the plane 
 makes with H and V. If, therefore, a Une be drawn through the 
 given point making with H and V complements of the given 
 angles, this Kne will be perpendicular to the required plane, and 
 the plane may then be located. 
 
82 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 Note. The sum of the given angles must not be less than 90 
 degrees else the problem will be impossible. 
 
 Construction. Let O be the given point in Fig. 64 and let it 
 be required to find the traces of the plane through O which makes 
 60 degrees with H and 45 degrees with V. 
 
 By Article 67, draw a line through O making an angle of 30 
 degrees with H and 45 degrees with V. Now, by Article 48, pass 
 through O a plane perpendicular to the line. sSs' is such a plane 
 and since it is perpendicular to a Hne making 30 degrees with H 
 and 45 degrees with V the plane itself makes 60 degrees with H 
 and 45 degrees with V. S, then, is the required plane. 
 
 PROBLEMS 
 
 156. Through the point —6; —4; —3 draw a line 3" long which will 
 make angles of 30 degrees with H and V. 
 
 157. The bottom of a shaft is located at o; —6; —10. Through this 
 point a tunnel is driven making an angle of 60 degrees with V and 15 degrees 
 with H. Give the bearing of the tunnel. 
 
 158. A guy wire is anchored at o; — 3; o and is fastened to the top of a 
 stack at +3; +1; x. Find how high the stack is when the wire makes an 
 angle of 60 degrees with the ground. Find also the angle the wire makes 
 with V. 
 
 159. Find the angle between the end and side of a floor hopper; size of 
 opening in floor 6' by 8', size of outlet 12" by 12", depth of hopper 3'. 
 
 160. Find the angles between the sides of a hood which fits in the corner 
 of a room. The opening of the hood is 24" square and its outlet is 6" square. 
 The distance between opening and outlet is 18". 
 
 161. Three shafts are sunk to the plane of an ore body. One is located 
 at the point —3; —4; —10; a second is located at o; — i; —18; the third 
 is located at +1; —6; —24. Find the angle the plane of the ore makes 
 with H and V. 
 
 162. A V-shaped flume runs N. 30° E. on a falling 10 per cent grade. The 
 bottom of the flume runs through the point — 2; o; —2 and the angle be- 
 tween the sides of the flume is 30 degrees. Draw a plan and elevation of 
 the flume assuming it to be made up of a 2" by 10" piece fastened to a 2" by 
 8" piece. 
 
 163. An ore body dips 60 degrees and strikes S. 15° W. from the point 
 — 2; —6; —10. Show its outcrop and find the angle it makes with the 
 vertical plane. 
 
 164. The bottom of a wefl is at o; —6; —18. The plane of a stratum 
 of clay through this point dips 60 degrees and makes an angle of 75 degrees 
 with V. Find the traces of this plane. 
 
PROBLEMS ON POINTS, LINES, AND PLANES 83 
 
 165. The outcrop of a stratum of stone runs N. 60° W. from the point 
 +6; —4; o and dips 15 degrees. Find the angle the stratum makes with V. 
 
 166. The wedge-shaped prow of a snow plough makes angles of 60 de- 
 grees with H and V, the planes of the faces inclining so as to meet. Find 
 the angle at which a plate will have to be bent in order to form a shoe over 
 the intersection of these planes. 
 
CHAPTER XI 
 SURFACES 
 
 72. A surface is the area made by a line moving according to 
 some law. 
 
 In descriptive geometry surfaces are classified according to the 
 character of the moving Hne, called the generatrix, and the law 
 controlling its motion. 
 
 73. The generatrix may be either a straight or a curved Hne 
 and its motion may be controlled in one of several ways. It 
 may be made to move so as to touch other lines either straight 
 or curved, called directrices, and to remain parallel to a plane, 
 called a plane director; or it may be made to revolve about 
 another line, called the axis of revolution; or its motion may be 
 controlled in other ways. All surfaces, however, are generated 
 by a line whose motion is controlled in some specified way. 
 
 74. Surfaces thus generated are divided into two general 
 classes: ruled surfaces and double curved surfaces. A ruled sur- 
 face is generated by the motion of a straight line. A double 
 curved surface is generated by the motion of a curved line. It 
 is obvious, therefore, that a straight edge may be made to coin- 
 cide with a ruled surface while it will coincide with a double 
 curved surface in one point only; this fact offers a convenient 
 method of identifying surfaces. 
 
 75. There are three kinds of ruled surfaces: Plane, single 
 curved, and warped. 
 
 A plane surface is generated by a hne moving so as to con- 
 stantly touch straight line directrices which either intersect or 
 are parallel. Thus, in Fig. 65, the hne MN is the generatrix, 
 and the traces of the plane, sS and s'S, are the intersecting direc- 
 trices. The hne MN, then, moves so as to constantly touch 
 these two lines, thus generating the plane sSs'. In further dis- 
 cussions of plane surfaces it will be understood that surfaces 
 
 84 
 
SURFACES 
 
 85 
 
 with plane faces are meant; the intersections of the plane faces, 
 or planes, form the edges of the surfaces which hereafter will be 
 called plane surfaces. 
 
 Any combination of intersecting planes, usually three or more, 
 forms a plane surface. Many of these combinations have 
 geometrical names such as prisms, pyramids, etc., with any 
 number of faces from three upward. Some of these surfaces 
 have commercial names which identify the uses to which they 
 are put, as, for example, the hopper which geometrically is a 
 
 Fig. 65. 
 
 truncated rectangular pyramid. It is with such surfaces as have 
 commercial uses that this text has to do rather than with those 
 which possess mathematical interest only. 
 
 76. There are three kinds of single curved surfaces: cylin- 
 drical, conical, and convolute. All of these surfaces are ruled 
 and the ruhngs — or the positions of the generating hne — are 
 called elements. 
 
 A cylindrical surface is generated by a straight line moving 
 so as to always remain parallel to its first position and to con- 
 stantly touch a plane curve. ^ When this plane curve is closed 
 the surface is called a cylinder. In Fig. 66 the generating line 
 
 ' A plane curve is one which lies in a plane, as a circle; a space curve is one 
 which does not lie in a plane, as a helix. 
 
86 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 BX moves so that it remains parallel to its first position, BX, 
 and constantly touches the curve ABC, thus generating a cylin- 
 drical surface. In Fig. 67 the directrix is taken as a circle in H 
 
 and since this is a closed curve 
 the resulting surface is called a 
 cyhnder. It will be noted that 
 the elements of a cyhnder are 
 parallel to each other. A conical 
 surface is a surface generated by 
 a straight Hne so moving that 
 one point remains stationary 
 while the line constantly moves 
 along a plane curve. This sta- 
 tionary point is called the apex, 
 and it is obvious that since 
 the generating line extends beyond the apex there will be 
 generated two surfaces, called nappes, one on each side of the 
 
 Fig. 68. 
 
 apex. In discussions of conical surfaces in this text only one 
 nappe will be considered, however. When the plane curve 
 
SURFACES 
 
 87 
 
 directrix is a closed curve the surface is called a cone. In Fig. 68 
 the generatrix moves so that it goes through the point O, the apex, 
 and constantly touches the curve 
 XY, thus generating a conical sur- 
 face. In Fig. 69 the curved direc- 
 trix is a circle and the resulting 
 surface is, therefore, a cone. 
 
 A convolute surface is generated 
 by a straight line moving so as to 
 be constantly tangent to a space 
 curve. In Fig. 70 the space curve 
 
 1-2-3 II is the curvilinear 
 
 directrix to which the elements are 
 constantly tangent, and a portion 
 of the surface is shown by the 
 positions of the generating Une at i 
 
 Fig. 69. 
 
 2A, 3B, 7F. The 
 
 most important convolute surface 
 commercially is the helical con- 
 volute, so-called because its cur- 
 vilinear directrix is a helix, and 
 since this surface is a special case 
 of helicoid it will be considered 
 in detail in connection with that 
 surface. 
 
 77. A warped surface is a ruled 
 surface generated by a straight 
 hne so moving that the elements 
 of the resulting surface neither 
 intersect nor are parallel. Thus 
 in Fig. 71 the generating line AX 
 so moves that the elements of the 
 resulting surface neither intersect 
 nor are parallel. In this surface 
 the lines AB and CD are recti- 
 linear directrices and the genera- 
 trix, AX, constantly touches them; the H plane is the plane 
 director as all of the positions of AX remain parallel to this plane. 
 
 Fig. 70. 
 
88 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 There are innumerable warped surfaces since there are 
 ^ innumerable combinations 
 
 of generatrices, directrices, 
 and plane directors but 
 comparatively few of them 
 are important commer- 
 cially, and these will be 
 discussed in detail in a 
 later chapter. 
 
 78. A surface of revolu- 
 tion is generated by the 
 motion of a line about 
 another line called the axis 
 of revolution. The gen- 
 erating line may be either 
 straight or curved ; in case 
 the generatrix is a straight 
 
 
 ii 
 
 \x 
 
 
 afy^ 1 
 
 5^1 
 
 Ic' 
 
 
 A 
 
 
 
 
 
 
 
 
 
 "^ \ 
 
 
 \ \ 
 
 
 ^! \ 
 
 d' 
 
 Fig. 71. 
 
 Hne the resulting surface will be a ruled surface of revolution; if 
 the generatrix is a curved line 
 the resulting surface will be a 
 double curved surface of revo- 
 lution. In Fig. 72 the circle 
 whose center is at O is re- 
 volved about the axis of 
 revolution AB. The resulting 
 surface is a double curved 
 surface of revolution called an 
 annular torus. In the figure 
 one-quarter of the surface has 
 been cut away in order to show 
 the curvihnear generatrix. 
 Whether the generatrix is 
 straight or curved all surfaces 
 of revolution have a circular 
 section in planes perpendic- 
 ular to the axis of revolution. 
 
 79. Developable surfaces are those which may be rolled out 
 
SURFACES 
 
 89 
 
 on a plane surface so that all the elements will coincide with the 
 plane. In Fig. 73 is shown a cone lying in a plane so that the 
 element OX coincides with plane. If the cone be now rolled 
 out the curve of the base will rectify itself on the line xx as the 
 elements of the surface successively come in contact with the 
 plane. The result, oxx, is called a development of the cone. 
 The only developable surfaces are the single curved surfaces as 
 
 Fig. 73. 
 
 these obviously are the only surfaces whose elements may be 
 made to he in a plane without changing their relation to each 
 other. In practice, however, a number of warped and double 
 curved surfaces are considered developable ; this is made possible 
 by approximate methods of development which give patterns 
 which conform closely enough to the original surface for practical 
 purposes. The bottoms of hemispherical water tanks are good 
 examples of this fact. 
 
90 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 CLASSIFICATION OF SURFACES 
 
 I. Ruled. Ruled 
 
 surfaces 
 be 
 
 may 
 
 fi. Plane surfaces / J"^'"^, 
 such IS 1 Pyramids 
 
 I, Polyhedrons 
 
 Single curved f Cylinders 
 surfaces such i Cones 
 as I Convolutes 
 
 (Cylindroids 
 Conoids 
 Helicoids, 
 etc. 
 
 Surfaces 
 are 
 either 
 
 2. Double Curved. 
 Double curved 
 
 surfaces 
 be 
 
 may 
 
 fi. Surf aces of Rev- 
 olution such 
 as 
 
 Surfaces of 
 Double Curv- 
 ature not 
 surfaces of 
 revolution 
 such as 
 
 ' Spheres 
 Ellipsoids 
 Paraboloids 
 Hyperboloids 
 Tori 
 
 See page 155 
 for a com- 
 plete clas- 
 sification 
 
 See complete 
 classifica- 
 tion which 
 follows 
 
 Ellipsoids with a variable 
 
 generatrix 
 Elliptical paraboloids 
 Elliptical hj^erboloids 
 ^ Serpentines 
 
 Surfaces of 
 Revolution 
 may be 
 either 
 
 Ruled sur- 
 faces such as 
 
 Single curved surfaces 
 as 
 
 2. Warped surface as 
 
 Double curved sur- 
 faces as 
 
 Rt. Cir. 
 
 Cylinder 
 
 Rt. Cir. 
 
 Cone 
 
 The hyperboloid 
 
 of revolution 
 
 of I sheet. 
 
 Only case. 
 
 Sphere 
 
 Ellipsoid 
 
 Paraboloid 
 Hyperboloid of 2 sheets 
 
 ; Prolate 
 I Oblate 
 
CHAPTER XII 
 PLANE SURFACES 
 
 80. A plane surface is represented by the plan and elevation 
 of any one of its sections and of its edges. While the drawing 
 may be made by making this section in any plane it is usually 
 taken for convenience in H or V; in whatever plane the section 
 lies it is called a base. 
 
 When the base lies in H or V it will be apparent that it is 
 formed by the intersections of the traces of the planes of the 
 faces, the edges of the base in reahty being portions of the traces. 
 The edges of the surface are the intersections of the planes of 
 the faces. Thus, in Fig. 74, the planes R, S, and T form the 
 faces of the surface, a triangular prism, and the intersections of 
 these planes, XA, YC, ZB, form the edges of the surface; the 
 triangle XYZ, made by the intersections of the V traces of these 
 planes, forms the V base; the triangle ABC made by the inter- 
 sections of the H traces forms the H base. 
 
 81. To assume a point on a plane surface any line as MN, 
 Fig. 74, may be assumed in the plane R, S, or T by Article 25, 
 then any point as on this line will he on the surface of the 
 prism. 
 
 82. In case the given base of the surface does not lie in H or 
 V the problem may be solved either by extending the surface 
 so as to obtain a base in either H or V, or in a plane parallel to 
 H or V, or by using the plane of the given base as an auxiUary 
 plane of projection. As a rule the problem may be solved more 
 readily by finding a base in H or V. 
 
 Note. A section is the intersection obtained by cutting a 
 surface by a plane. If the plane of the section is perpendicular 
 to the axis of the surface the section is called a right section; 
 other sections are, in general, oblique sections. 
 
 91 
 
92 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 83. Proposition 18. Given the size and location of the right 
 section of any prism to find the plan and elevation of the surface. 
 
 Fig. 74. 
 
 Discussion. Since the right section lies in a plane perpendicu- 
 lar to the axis the edges of the surface will be perpendicular to 
 this plane. If, then, through the corners of the given right sec- 
 tion lines be drawn perpendicular to the plane of this section 
 these hues will be the edges of the surface and the points where 
 
PLANE SURFACES 
 
 93 
 
 these edges pierce H and V will determine the corners of the 
 bases in H and V. 
 
 't 
 
 Fig. 75 
 
 Construction. In Fig. 75, T is the plane of the given right 
 section whose center is at the given point O. The section is to 
 be a square of given dimensions. To find the plan and elevation 
 
94 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 of this section O is revolved into V at Ov; around this position 
 of O the square may be constructed in its true size and revolved 
 back to its true position in plane T. (Article 37.) Through 
 the corners of the square in its true position draw the lines which 
 form the edges of the surface. Their plans and elevations will 
 be perpendicular to the corresponding traces of the plane T, as 
 ab and a'b'. Extend these Hnes until they pierce H, — in this 
 case, — and join the piercing points. The area thus enclosed 
 will be the base of the surface in H. In the figure this area has 
 been cross hatched, thus representing the object to be a solid. 
 
 84. Proposition 19. Given the plan and elevation of a plane 
 surface to find the true shape and size of its right section. 
 
 Discussion. Pass a plane perpendicular to the axis of the 
 surface; this will be the plane of the right section. Find where 
 the edges of the surface pierce this plane; these points when 
 joined will determine the right section. Revolve this area into 
 H or V about the H or V trace of the section plane; this will 
 give the true shape and size of the required section. 
 
 Construction. The figure is to be drawn according to the 
 above discussion. 
 
 85. Proposition 20. To find the section cut from any plane 
 surface by any plane. 
 
 Discussion : First Method. Find the intersection of the plane 
 of each face of the surface with the given cutting plane. Since 
 each of the lines thus found is common to both the surface and 
 the cutting plane the area enclosed by these intersections will 
 be the section cut from the surface by the plane. 
 
 Construction. In Fig. 76 the surface is assumed to be a 
 rectangular pyramid cut by the plane sSs'. The plane R is the 
 left face of the soHd and cuts the plane S in the fine AB ; also the 
 plane T, which is the plane of the right face of the solid, inter- 
 sects the plane S in the line CD. In like manner the lines AD 
 and BC may be found. Since these four lines are common to 
 both the surface and the plane S the area they enclose will be the 
 section cut from the pyramid by the cutting plane S. If the true 
 size of this area is desired it may be found by revolving ABCD 
 into H or V about the corresponding trace Ss or Ss'. 
 
PLANE SURFACES 
 
 95 
 
 Discussion: Second Method. Find where each edge of the 
 surface pierces the cutting plane. Since these points all lie in 
 one plane and are common to the surface and the cutting plane 
 they will, when joined, form the required intersection. 
 
 U^ 
 
 Fig. 76. 
 
 Construction. In Fig. 77, is the apex of a solid hexagonal 
 pyramid whose base is in a plane parallel to H. By Article 56 
 find where each edge pierces the cutting plane S, as at X, Y, etc. 
 Join these points in order and the area thus enclosed will be the 
 section cut from the surface by the plane S. If the true size of 
 
96 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 the section is required it may be found by revolving it into H or 
 V. In this figure the section is revolved into H about the trace 
 sS (Article 36), and its true size is shown by the hatched area 
 at XhYh, etc. 
 
 Fig. 77. 
 
 86. Proposition 21. To develop any plane surface. 
 
 Discussion. Find the true size of each face of the surface and 
 lay these faces out on a plane with their common edges coin- 
 ciding. 
 
PLANE SURFACES 
 
 97 
 
 Construction. In Fig. 78 is shown a hood such as might be 
 used over a forge. To make a development of it first find the 
 true size of the edge OA. At any convenient place on the paper 
 lay off OA equal to a'o", then with A as a center and AD as a 
 radius strike an arc and with O as a center and OD equal to OA 
 as a radius strike a second arc intersecting the first at D. The 
 area AOD is the true size of the back of the hood. In similar 
 fashion using OD as a common edge construct the area OCD 
 
 Fig. 78. 
 
 equal to the side of the hood, and continuing this process each 
 face of the hood may be laid out. The lips of the hood, or the 
 portions lying below the line ABCD are laid off by constructing 
 rectangles with one side equal to AD, DC, CB, and AB and the 
 other side equal to the depth of the lip. These rectangles are 
 of course joined to the hood along the lines AD, DC, etc. 
 
 In case the edges of the surface to be developed do not meet 
 in a point, as in Fig. 78, or are not parallel, as in the case of a 
 prism, the development may be made as shown in Fig. 79. 
 The method used is the same in principle as that used in Fig. 78, 
 modified slightly to suit the altered conditions. 
 
 Construction. Lay off the area BCHG making BC = be, 
 CH = c'h', BG = b'g' and GH = gh. This will be the back of 
 
98 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 the piece. Now with C as a center and a radius equal to c'e' 
 strike an arc, and with H as a center and h'e' as a radius strike 
 a second arc intersecting the first at E. This point E is the 
 position of E in the development of the face CDEH. Now to 
 find the position of D strike arcs with C as a center and c'd' as 
 
 iG 
 
 a radius, and with E as a center and e'd' as a radius. These arcs 
 intersect at D in the figure and thus the side CDEH may be 
 developed by dividing it into triangles and locating the corners 
 as shown. The layout for the remaining portion of the pieces 
 will be apparent from an inspection of the figure. 
 
PLANE SURFACES 99 
 
 PROBLEMS ON PLANE SURFACES 
 
 Special Proposition. Given the axis, the apex, and the size and shape 
 of the right section at a given point on the axis, to find the plan and eleva- 
 tion of a pyramid. Write the discussion and draw a figure demonstrating 
 the above proposition. 
 
 167. The center line of a timber 4" square pierces H 8" in front of V 
 and incHnes 45 degrees to H and 30 degrees to V. Draw a plan, elevation, 
 and end view of the timber. 
 
 168. The center line of a timber 4" square pierces H at O (o"; —6"; o") 
 and pierces V at Q (18"; o"; -12"). Draw a plan and elevation of the 
 timber and show its intersection with H and V. 
 
 169. The point O is o"; -6"; o". The point Q is -8"; o"; -8". The 
 line OQ is the center line of a rectangular pipe i" by 3" in section. Draw a 
 plan and elevation of the pipe and find the true size of the hole it makes in 
 passing through the profile plane located at —5"; 90; 90. 
 
 170. The point is -2"; -4"; o". The point Q is 2"; o"; -4". The 
 line OQ is the center line of an hexagonal prism whose base in H is a regular 
 hexagon with sides i" long. Draw a plan, elevation, and end view of the 
 prism and show its base in V. 
 
 171. Make a drawing showing how to trim the ends of a 2" by 4" stick 
 so that it will fit against V making an angle of 45 degrees, will fit against H 
 making an angle of 30 degrees, and will have a center line 18" long between 
 these points. 
 
 172. The plane T is o"; -60; +60. The point O in this plane is -3"; 
 x"; —3" and is the center of a regular hexagon with i" sides which lie in 
 plane T. Draw a plan and elevation of the prism perpendicular to V which 
 could cut this hexagon from the plane. 
 
 173. The point is o"; -3"; -3". The point Q is 6"; -i"; o". 
 The hne OQ is the center line of an hexagonal prism whose base in P is a 
 regular hexagon with i" sides and center at O. Draw a plan, elevation, 
 and end view of this prism and find the true shape and size of its right section. 
 
 174. The base of a pyramid is a hexagon with li" sides in H and its 
 center at the point o"; -3"; o". The apex of the pyramid is at Q (-4"; 
 — i"; —3"). Draw a plan and elevation of this pyramid and find its right 
 section at a point 2" from the apex. 
 
 175. The plane T is o"; -60; +45. The point O is in this plane at 
 _4". -2"; x". This point O is the center of a rectangle 2" by 3" whose 
 sides incline at 45 degrees to the H trace of plane T. The rectangle is the 
 base of a quadrilateral pyramid whose altitude is 3". Show three views 
 of the pyramid. 
 
 176. A trough is built up of a 2" by 8" piece nailed to two 2" by 6" 
 pieces. Draw the plan, elevation, and end view of a section 18" long which 
 slopes 30 degrees to H and V. 
 
lOO ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 177. Make a development of a hopper whose opening is a rectangle 18" 
 by 36" and whose 12" by 12" outlet is formed by cutting the hopper by a 
 plane inclining 30 degrees to the horizontal. The distance between the 
 centers of the openings is 18". 
 
 178. Make the pattern for a chute 12" square in section to connect the 
 opening in a bin wall 48" above the floor with an opening in the floor 36" 
 from the bin wall. 
 
 179. An air duct of 6" by 8" pipe follows the line of intersection of the 
 walls and ceiling of a room with its 8" side against the ceiling. Lay out 
 patterns for two-piece elbow needed in the corners. 
 
 180. Lay out the pattern for a pipe to connect a 12" square hole in a 
 floor with a bin wall. The center Hne of the pipe incHnes 30 degrees to the 
 floor, 45 degrees to the bin waU, and is 72" long. 
 
 181. Design the transition piece to connect a 12" square air duct with a 
 4" square air duct. The two ducts lie on the floor of a bmlding against the 
 wall with their ends 6" apart. 
 
 182. Design a ventilating hood for the corner of a room. The opening 
 of the hood is to be 18" by 24", the outlet is to be in the corner into a 9" by 
 12" pipe, and the hood is to be 16" deep with 4" lips. 
 
 183. A 12" square ventilating pipe runs along the intersection of the 
 ceiUng and front waU of a room. Design the elbow which wiU take the pipe 
 down the side wall from the corner at an angle of 30 degrees with the ceiling. 
 
 184. Make the patterns for an elbow turning an angle of 105 degrees 
 using 4" by 6" pipe with the plane of the angle in the direction of the long 
 dimensions of the pipe. 
 
 185. Make the patterns for an elbow turning an angle of 60 degrees 
 using 4" by 6" pipe with the plane of the angle in the direction of the short 
 dimension of the pipe. 
 
 186. A water duct runs down the side of a building at an angle of 15 de- 
 grees with the horizontal and turns around the corner of the building. Lay 
 out patterns for an elbow to make the turn using 3" square pipe. 
 
CHAPTER XIII 
 CYLINDRICAL SURFACES 
 
 87. A cylindrical surface is represented by the plan and 
 elevation of some curve of the surface and of the limiting elements 
 of the surface. For convenience this curve is usually taken in 
 H or V and is therefore the locus of the points in which the 
 elements of the surface pierce H or V. 
 
 88. In a cylinder this curve is called a base and for conven- 
 ience it is taken as a rule either in H or V, or in a plane parallel 
 to H or V. Since the base of a cyHnder must be a closed curve, 
 cylinders will be either circular or elliptical in section, or the 
 section will be some irregular closed plane curve such as, for 
 example, would be formed by cutting a corrugated tin pipe. 
 When a cylinder with a circular right section is inchned to H or 
 V its bases in those planes will be elhpses; when the cylinder is 
 elliptical in right section and is inclined to H or V its bases will, 
 in general, be elhpses, although if the angle of the inchnation be 
 properly adjusted they may become circles. Thus, in Fig. 80 
 is a cyhnder with a circular base in H. This cyhnder must be 
 elliptical in section. In Fig. 81 is a cylinder with an elhptical 
 base in V. This cyhnder may be either circular or elliptical in 
 section, according to the angle its axis makes with V. 
 
 89. CyUnders are defined by their bases. Thus: a cyHnder 
 with a circular base; a cyHnder with an elliptical base, etc. A 
 rigJit cylinder is one whose elements are perpendicular to the 
 plane of its base. Such cyHnders may be either circular or 
 elliptical in section and are generally named accordingly, as a 
 right circular cyHnder, or a right elliptical cyHnder. 
 
 90. To assume a point on the surface of a cylinder it is neces- 
 sary only to assume an element of the surface and on this element 
 locate the point. In Fig. 80, OP is the axis of a cyHnder whose 
 base is a circle in H. Assume the element XY parallel, of course, 
 
I02 
 
 f'SSEN'HALS- OF DESCRIPTIVE GEOMETRY 
 
 to the axis and piercing H in the base at Y. Any point on this 
 element will be a point on the surface of the cylinder. The 
 element projected at x'y' has two plan views, one at xy and the 
 other at qr. Either of these is correct as one is on one side of 
 the cylinder and the other is directly opposite on the other side. 
 This same thing is shown in Fig. 8i. If the point Q were given 
 
 Fig. 8o. 
 
 Fig. 8i. 
 
 by its plan view only, two V projections would be possible, one 
 at q' and one at x', since the element through Q has two possible 
 V projections. In given problems it will always be stated on 
 which element to take the point. 
 
 91. Proposition 22. Given the axis and right section of any 
 cylinder to find its plan and elevation. 
 
 Discussion: First Method. Revolve the axis parallel to H 
 or V. In this position draw the right section. Through the 
 ends of the axes of the curve of the right section draw elements; 
 these limiting elements will pierce H or V at the ends of the axes 
 of the base. 
 
 Construction. In Fig. 82, OQ is the given axis and the given 
 right section is a circle whose diameter is dc long. Revolve OQ 
 
CYLINDRICAL SURFACES 
 
 103 
 
 parallel to V; in this position it is projected at oq^ and o'q". 
 At any convenient point P place the given right section. Its 
 elevation will be a straight line, a"b", since the axis is parallel to 
 V; and its plan view will be an ellipse whose major axis will be 
 
 d,c, = a"b" and whose minor axis will be a,b„ or the projection 
 of a"b" on H. Through A and B draw lines parallel to OQ in 
 its revolved position; these will be limiting elements and will 
 pierce H at y, and x,, the ends of the major axis of the base in H. 
 Likewise draw the limiting elements through D and C; these 
 
I04 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 pierce H at n^ and m^ respectively and locate the ends of the 
 minor axis of the base in H. Now revolve the axis back to its 
 given position OQ and locate the true position of the axes of the 
 base. Since the angle that OQ makes with H remains' constant 
 the axes of the base remain constant and in the same relation to 
 OQ. Therefore, x^ will revolve to x, y^ to y, n^ to n, and m^ 
 to m. Construct the ellipse of the base and draw the projec- 
 tions of the Hmiting elements. 
 
 Discussion: Second Method. Pass a plane perpendicular to 
 the given axis; this plane contains the given right section. Con- 
 struct the projections of the given right section, draw the 
 Hmiting elements, and find where they pierce H or V. These 
 points will give the axes of the base in H or V. 
 
 Construction. Through any convenient point P, Fig. St,, pass 
 a plane T perpendicular to the given axis OQ. (Article 48.) 
 Revolve the point P into H about Tt. With this point P as a 
 center draw the given right section in its true size. When this 
 circle is revolved to its proper position in plane T it will be pro- 
 jected as an ellipse and the diameters AC and BD will become 
 the major and minor axes of this ellipse as it is projected in H. 
 Find the projections of A, B, C, and D by Article 37 and through 
 these points draw the projections of elements of the surface. 
 These elements pierce H at yx and mn, which points are re- 
 spectively the ends of the major and minor axes of the base. 
 Draw the ellipse determined by these points and find the pro- 
 jections of the Hmiting elements. 
 
 92. Proposition 23. To find the curve cut from any given 
 cylinder by any given plane. 
 
 Discussion. Pass planes through the cylinder parallel to its 
 axis. These auxiliary planes cut elements from the surface of 
 the cyHnder and lines from the cutting plane. These Hnes inter- 
 sect the elements in points common to both the cylinder and the 
 cutting plane. If a sufficient number of auxiHary planes be 
 used enough points will be obtained to plot the curve of inter- 
 section. 
 
 Construction. When the cyHnder is perpendicular to H or V. 
 In Fig. 84 the given circular cyHnder with its base in H is cut by 
 
CYLINDRICAL SURFACES 
 
 105 
 
 a plane T. Through the axis perpendicular to plane T pass 
 auxiliary plane R. This plane cuts two elements from the cylin- 
 der at D and C, and a Hne DC from the plane T. This line in- 
 
 FlG. 83. 
 
 tersects the two elements at D and C, two points of the required 
 curve. In Hke manner the plane U cuts two elements at A and B 
 and a Hne AB from the i)lane T. This line intersects the two ele- 
 
[o6 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 ments at A and B. From an inspection of the drawing it will be 
 seen that C is the point where the plane T enters the cylinder and 
 
 Fig. 84. 
 
 the point D is where the plane leaves the cylinder; therefore C and 
 D are the two points on the curve of intersection farthest apart, 
 or the ends of the major axis of the curve. Since the auxiliary 
 
CYLINDRICAL SURFACES 107 
 
 plane U contains the line AB and is perpendicular to the plane R 
 the two points A and B will be the ends of the minor axis of the 
 curve of intersection. Now if AB and CD be revolved into V 
 about Tt', the true size of the curve of intersection may be 
 found as shown. If the projections of this curve of intersection 
 be required they may be found by revolving the ellipse AvByCvDy 
 to its original position, or by passing more auxihary planes 
 parallel to U, as W, and locating points as in plane U. 
 
 Construction, When the cylinder is oblique to H or V. In 
 Fig. 85 is shown a cyhnder obHque to H with circular base in H. 
 This cylinder is cut by the plane T as in the case just discussed. 
 Pass planes, such as R, U, W, through the cylinder parallel to 
 its axis. Each plane, except the plane tangent along the Hmit- 
 ing elements, cuts two elements from the cyhnder and a straight 
 Hne from the cutting plane T. These lines are of course parallel 
 as shown at AM, BN, CQ, and intersect the elements of the 
 cyhnder in points of the curve of intersection, as A, B, C, and 
 D. If a number of auxihary planes are passed a sufficient num- 
 ber of points will be determined to plot the curve of intersec- 
 tion. In finding the true size of this curve the points may be 
 revolved into V or H about the corresponding trace. In this 
 case the curve is revolved into V about the V trace taking the 
 position AvByCvDy. 
 
 93. It must be remembered in working problems of this 
 character that the more points there are found on the curve 
 the more accurate will be the final result. In the figures given 
 as illustrations only a few points have been located in order to 
 avoid confusion in the drawings. 
 
 It must also be remembered that only by solving the problem 
 in an orderly fashion can accurate results be secured. It is not 
 wise to attempt short cuts, and in the end it will be found most 
 satisfactory to pass the auxihary planes one at a time and to 
 determine all of the points in each plane before passing to the 
 next. Any attempt to pass all of the auxihary planes, then to 
 draw all of the elements these planes cut from the cylinder, and 
 then to locate all of the lines cut from the cutting planes results 
 either in throwing the problem into a hopeless confusion of lines, 
 
Io8 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 Fig. 85. 
 
CYLINDRICAL SURFACES 
 
 109 
 
 tiG. 86. 
 
no 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 or else requires more painstaking work to avoid this tangle than 
 the method suggested. This condition is true of all problems 
 of this character. 
 
 94. Proposition 24. Given any cylinder to find its right 
 section. 
 
 Discussion. Pass a plane through the cylinder perpendicular 
 to its axis and find the curve of intersection; this will be a right 
 section of the cyUnder. 
 
 Construction. If both the true size and the projections of 
 the right section are required the problem may be best solved 
 by Proposition 23. If, however, only the true size of the right 
 section is required the solution in Fig. 86 will be found more 
 simple. Let OQ be the axis of the given cyhnder and T the 
 perpendicular plane containing the right section. The axis OQ 
 pierces the plane T (Article 56) at P; this will be the center of the 
 right section. Now draw a fine AB through P parallel to the H 
 trace of T; this fine will be the longest Hne which can be drawn 
 between two elements of the surface; hence it 
 is the major axis of the curve of intersection. 
 To find the minor axis draw CD through P 
 perpendicular to the trace tT. Revolve AB and 
 CD into V about the V trace to AyBv and 
 CyDy; with these Hnes as axes construct the 
 ellipse which will be the right section of the 
 cylinder. 
 
 95. It will have been noted in connection with 
 Article 92 that the intersection of a right 
 cylinder with an obhque plane is a curve which 
 is symmetrical about its center lines. This fact is utilized in 
 constructing cylindrical elbows so that the pipe section may 
 remain uniform while the elbow turns an angle. 
 
 In Fig. 87 is shown the plan and elevation of a cyhnder cut by 
 an obhque plane. Since this curve of intersection is symmetri- 
 cal the upper part of the cyhnder may be turned through 180 
 degrees to the position shown in Fig. 88 and the two parts of the 
 cyhnder will still join on the same curve of intersection while the 
 axis turns an angle of 90 degrees. Fig. 89 shows how this same 
 
 Fig. 87. 
 
CYLINDRICAL SURFACES 
 
 III 
 
 principle may be applied to elbows with more than one section. 
 If the problem be to draw a four section elbow turning an angle 
 of 6 degrees, Fig. 90, draw two hnes 
 making the given angle with each other, 
 as AO and DO, and then divide the angle 
 by drawing BO and CO so that the angle 
 
 BOA: 
 
 3 
 
 Find the points i, 2, 3, and 4 all equi- 
 distant from O and with these points as 
 centers lay off on AO, BO, CO, and DO the diameters of the 
 
 right sections of each part of the 
 elbow. The limiting elements 
 may be drawn through the ends of 
 the diameter of the right section. 
 96. Proposition 25. To de- 
 velop any cylinder. 
 
 Discussion. Since the ele- 
 ments of a cylinder are perpen- 
 dicular to the plane of the right 
 section the curve of the right sec- 
 tion will roll out in the develop- 
 ment into a straight hne to which 
 all the elements of the surface are 
 perpendicular. Therefore, lay off 
 a line equal in length to the right 
 section and erect perpendiculars 
 to this Kne. If the distances be- 
 tween these perpendiculars be 
 made to equal the distances be- 
 tween the elements along the curve 
 of the right section and the perpen- 
 diculars be made equal in length 
 to the elements the result will be 
 ^^*^- ^9- a development of the cylinder. 
 
 Construction. In Fig. 91, the cyhnder taken is part of an 
 elbow with a 90 degree turn. With the bow dividers set off on 
 
112 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 the base the distances 1-2, 2-3, 3-4, etc., equal and small enough 
 so that the chord of the arc is equal, practically, to the arc itself, 
 
 and through these points 
 draw elements of the cyHn- 
 der. The last step, 11-12, 
 may not be equal to the 
 others but this does not 
 matter. Now draw a hne 
 i-i (it will be convenient 
 to draw this Hne with the 
 T square a continuation 
 of the projection of the 
 base of the cylinder) and 
 divide it as shown so that 
 the divisions are equal to 
 the distances laid off on 
 the base of the cylinder. 
 Through these points erect perpendiculars equal in length to 
 the elements. Through the points a, b, c, d, etc., thus found 
 draw a curve. The area bounded by the points a-i-i-a 
 
 Fig. 90. 
 
 1 •: 3 4 5 7 8 y 10 11 12 11 10 !J 8 7 6 5 -1 3 3 1 
 
 Fig. 91. 
 
 is the developed cylinder. It will be obvious from the figure 
 that since the development is symmetrical on each side of the 
 center line 1 2-I only one-half of the base need be divided in order 
 to develop the whole cylinder. 
 
CYLINDRICAL SURFACES I13 
 
 PROBLEMS ON CYLINDERS 
 
 Special Proposition. Given a right circular cylinder parallel to the 
 ground line and an oblique plane to find the curve cut from the cylinder by 
 the plane. Write the discussion and draw a figure demonstrating the above 
 proposition. 
 
 187. The axis of a cyhnder pierces H at a point i\" in front of V. A 
 second point on this line, 4" from the first point, is 3" in front of V and 2\" 
 above H. A right section of this cylinder is a circle 2" in diameter. Draw 
 a plan and elevation of this cylinder and show the true shape and size of its 
 base in H. 
 
 188. The point is o"; - 2"; o" . The point P is 4"; -4"; -3". The 
 line OP is the axis of a cylinder whose right section is a circle 2" in diameter. 
 Draw a plan and elevation of this cylinder and find its base in H. 
 
 189. The axis of a cylinder pierces H at a point 2" back of V and pierces 
 V at a point 3" below H. Between these two piercing points the axis is 6" 
 long. Draw a plan and elevation of the cylinder showing its base in H as a 
 circle i \" in diameter and find its base in V. 
 
 190. The axis of a cylinder pierces V 4" below H, and inclines 30 degrees 
 to V and 45 degrees to H. Draw a plan and elevation of the cylinder and 
 find its base in H when its base in V is a circle 2" in diameter. 
 
 191. A 3" circular air pipe crosses the corner of a room. Its axis cuts 
 the side wall 5" from the corner and the front wall 6" from the corner; both 
 of these points are 8" below the ceiling. Show the true size of the holes 
 which will have to be cut for the pipe. 
 
 192. The axis of a right circular cylinder 3" in diameter pierces H 2" 
 back of V and is perpendicular to H. Find the traces of a plane which will 
 show the angle at which this cylinder has to be cut in order to get an ellipse 
 whose axes are 4" and 3". 
 
 193. The point O is o"; -o"; -if". The point P is o"; -4"; -if'. 
 This line is the axis of a right circular cylinder whose diameter is 2\". The 
 cylinder is cut by a plane located at 4"; 135; 240. Show a plan and eleva- 
 tion of the curve of intersection and its true size. 
 
 194. The point O is o"; -2"; -o"; the point P is o"; -2"; -4". 
 The plane T is —4"; +30; —45. Find the true shape and size of the curve 
 cut from the cylinder whose axis is OP and whose diameter is 2" . 
 
 195. Draw the plan and elevation of a cylinder whose base is a 2" circle 
 in H with its center 2" below the G. L., and whose axis is inclined so that its 
 plan makes 45 degrees with the G. L. and its elevation makes 60 degrees 
 with the G. L. Cut this cylinder by a plane which makes an angle of 75 
 degrees with the axis. Find the plan and elevation of the curve of inter- 
 section and its true size. 
 
 196. The point is o"; - 3"; o" ; the point Q is 3"; -5"; -4". The 
 line OQ is the a.xis of a cylinder whose base in H is a circle 2\" in diameter. 
 
114 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 The plane T is 4"; —45; +45. Draw a plan and elevation of the inter- 
 section of the plane and cylinder and find its true size. 
 
 197. The plane S is o; —45; —60. The point O lies in this plane at 
 — i"; —2"; x", and is the center of a 2" circle whose plane is S. Draw the 
 plan and elevation of the cyHnder of which this circle is a right section and 
 find one of its bases in H or V. 
 
 198. The point O is -2"; -3"; o". The point Q is 2"; o"; -5". The 
 line joining OQ is the axis of a circular cylinder whose diameter is 2". Draw 
 the plan and elevation of the cylinder showing its bases in H and V. 
 
 199. The axis of a cylinder whose right section is a 2" circle pierces H 3" 
 back of V and pierces V 4" below H. Between these two points the axis is 
 7" long. Draw a plan and elevation of the cyHnder and find its base in P. 
 
 200. A right circular cylinder whose diameter is 2|" is cut at an angle 
 of 60 degrees to its axis. Make a pattern for this cylinder assuming the 
 longest element of the cylinder to be 4". 
 
 201. A right elliptical cyhnder is cut off by a plane inclining to the axis 
 at such an angle that the section cut is a circle. The right section of the 
 cylinder is an ellipse 2" by 3" and its longest element is 5". Lay out a 
 pattern for this cylinder. 
 
 202. Make a development of the air pipe in Problem 191 and lay out on 
 the development the line of intersection between the pipe and the walls. 
 
 203. Make the patterns for a four piece elbow of 6" pipe. The elbow 
 turns an angle of 90 degrees, with a radius of center line 12" long. 
 
 204. The axis of a 3" circular cylinder pierces H 4" back of V and pierces 
 V 3" below H; between these two piercing points the axis is 5" long. Make 
 a pattern of the cylinder between its H and V bases. 
 
 205. A hne of 8" pipe slopes at the rate of 4' in every 10' and runs due 
 northeast. At a point o"; —6"; o" the pipe meets the horizontal floor it is 
 to drain. Make a pattern showing how the last section of pipe must be cut 
 to fit into the floor. 
 
 206. Make the patterns for a five piece elbow of 8" pipe. The pipe 
 turns an angle of 105 degrees, with a radius of 16". 
 
CHAPTER XIV 
 CONICAL SURFACES 
 
 97. A conical surface is represented by the projections of some 
 curve of its surface and the Umiting elements which meet at the 
 apex. As in the case of cylinders this curve is usually taken in 
 H or V and is the locus of the piercing points of the elements. 
 
 98. A cone is represented by the projections of any one of its 
 sections and the limiting elements. For convenience the section 
 taken is usually the base in H or V, and the projections of the 
 limiting elements are found by joining the limiting points in the 
 base with the projections of the apex. Thus, in Fig. 92, the base 
 is an ellipse in H whose center is at 0, and the apex of the cone 
 is given at Q. To find the projections of the hmiting elements 
 for the plan views of the cone tangents are drawn from q to the 
 ellipse; and to find the projections of the Hmiting elements for 
 the elevation of the cone Hnes are drawn from q' to the extreme 
 points of the elevation of the base. It will be obvious that the 
 limiting elements of the elevation are not projections of the 
 Hmiting elements of the plan view. 
 
 99. Cones are identified by their bases. Thus: a cone with 
 a circular base or a cone with an elHptical base, etc. As in the 
 case of cyHnders a cone with a circular base wiH be elHptical in 
 section when its axis is incHned to the plane of the base, and a 
 cone with an ehiptical base may be either elliptical or circular 
 in section according to the angle of inclination. 
 
 A right cone is one whose axis is perpendicular to the plane 
 of the base. Right cones may be either circular or elliptical in 
 section and are named accordingly, as: a right circular cone, or 
 a right elHptical cone. 
 
 100. To assume a point on the surface of the cone either 
 projection of a point may be assumed on "the corresponding pro- 
 
ii6 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 jection of some element and the other projection of the point 
 found by locating it on the other projection of the same element. 
 Thus, in Fig. 92, the projection y' is assumed on the projection 
 of the element q'r' and the projection y then will be found on qr. 
 It will be noted, as in the case of cylinders, that there are two 
 elements of the cone whose elevations coincide with this line, one 
 being QR and the other QP, so that it is possible to have two 
 points on the cone whose elevations coincide at x', one being Y 
 
 Fig. 92. 
 
 Fig. 93. 
 
 and the other X. In given problems the data are usually given 
 which determine which point is intended, otherwise two solutions 
 to such problems are possible. 
 
 In Fig. 93 a similar problem is shown. In this case the plan 
 view of the point is assumed and its elevation located at one of 
 the points x' or y', depending on which side of the cone the point 
 is to be taken. 
 
 1 01. Proposition 26. Given the axis of a cone, the size and 
 location of its right section to find the plan and elevation of the 
 cone. 
 
 Discussion. If the cone be revolved parallel to a plane of 
 projection, the given right section may be drawn in its given 
 location. If, then, elements be drawn through this section from 
 
CONICAL SURFACES 
 
 117 
 
 the apex the base of the cone in H or V may be found by finding 
 where the elements pierce H or V. 
 
 Construction. In Fig. 94, OQ is the given axis, and P the 
 given point on this axis where it pierces the given right section. 
 The right section in this construction is taken as a circle whose 
 diameter is equal to x"y"- Revolve OQ parallel to V; in this 
 
 q' q" 
 
 Fig. 94. 
 
 position it is projected at oq^ and o'q" and P falls at p^ and p". 
 Draw x"y" equal to the given diameter of the right section per- 
 pendicular to o'q" ; this line will be the elevation of the right sec- 
 tion, since OQ is parallel to V, and the ellipse x^y^ will be the plan 
 view. Now through x" and y" draw the elements projected 
 on V at q"d" and q"a", and on H at q^d^ and q^a^. These 
 elements pierce H at a^ and d^, which points will be the ends of 
 the major axis of the base in H. To fmd the minor axis of the 
 
Il8 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 base bisect a^d^ by the line whose direction is b^c^. The length 
 of the Hne is as yet unknown but whatever this may be, since it 
 is perpendicular to V, the hne will be projected on V at the point 
 b"c". Join b"c" with q"; this will be the elevation of the two 
 elements which pierce H at the ends of the minor axis. Now 
 these elements cut the curve of right section in its revolved 
 position at M and N, and if through these points elements be 
 drawn they will pierce H at c, and b^, or the ends of the minor 
 axis. The axis may now be revolved to its original position, and 
 the true position of these axes located. The curve in the base 
 may be constructed and the limiting elements drawn. 
 
 102. Proposition 27. To find the curve cut from any given 
 cone by any given plane. 
 
 Discussion. Through the apex of the cone pass auxiliary 
 planes intersecting the cone and the given plane. These planes 
 cut elements from the cone and Knes from the given plane. 
 Since these elements and hnes he in common planes they will 
 intersect in points which wih be common to both the cone and 
 the given plane. If a sufficient number of auxihary planes be 
 passed enough points may be found to plot the curve of inter- 
 section. 
 
 Construction. When the axis of the cone is perpendicular to 
 H or V and its base is a circle. In Fig. 95, sSs' is the given plane 
 cutting the right circular cone whose apex is at Q. Through the 
 apex Q pass the plane R perpendicular to Ss. This plane cuts 
 from the plane S the hne AB, which hne pierces the surface of the 
 cone at A and B. Since this hne is the longest line which can be 
 drawn common to both the cone and the plane S it will be the 
 major axis of the curve of intersection. Now through P, the 
 middle point of AB, draw CO in plane S and perpendicular to 
 AB. This hne CO pierces the surface of the cone at the points 
 C and D. To find C and D pass a plane uUu' through the 
 apex Q and the hne CO. This plane cuts two elements from the 
 cone which intersect the hne CO at the points C and D where 
 CO cuts the surface of the cone. Since it is the shortest hne 
 which can be drawn common to both cone and plane S, and 
 since it is perpendicular to AB, CD will be the minor axis of the 
 
CONICAL SURFACES 
 
 119 
 
 curve of intersection. To find its true size revolve the axes of 
 the curve into V about s'S and construct the ellipse AvBvCvDy. 
 If the projections of this curve of intersection are required they 
 may be found by passing other auxihary planes, such as R, 
 through the apex of the cone and finding points by the method 
 used to find points A and B. 
 
 Fig. 95. 
 
 Construction. When the axis of the cone is oblique to H and 
 V, the base being any plane curve. 
 
 In Fig. 96, S is the given plane cutting the cone whose apex 
 is Q. Pass auxiliary planes R, U, V, etc., through Q perpendicu- 
 
I20 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 lar to H. The auxiliary plane R cuts from the cone one element 
 and from the plane S the line OM. This line OM intersects this 
 element at F, one point on the curve of intersection. In like 
 manner the plane U determines the point E of the curve. Now 
 
 Fig. 96. 
 
 in problems of this character it generally happens that the 
 auxiliary planes run off the paper, as, for example, V, making it 
 impossible to find points common to the V traces, as m' and n'. 
 To avoid the difficulty which arises the following method of 
 finding the intersections of S with the cutting plane is convenient. 
 
CONICAL SURFACES 121 
 
 Since the auxiliary planes all pass through the apex of the 
 cone and are perpendicular to H, they will have a line in com- 
 mon which passes through Q perpendicular to H. Now this 
 line pierces the given plane S in a point which is common to S 
 and all the auxiliary planes. This point, therefore, is common 
 to all the lines cut from S by the auxiliary planes. Through this 
 point and the intersections of the H traces lines of intersection 
 between the planes may be drawn without using the V traces. 
 This point in the given case is O. To find it locate a point O 
 in the plane S, having its plan view at q (Article 26). Since the 
 line common to all the auxiliary planes is perpendicular to H 
 through the apex its plan view will coincide with q. 
 
 Now the auxiliary plane V cuts from the plane S the Hne OK 
 and from the cone two elements; this line and the elements 
 intersect at B and D on the required curve. In like manner 
 other points may be determined and the curve of intersection 
 drawn. To find the true shape and size of the curve it is neces- 
 sary to revolve it into H or V about the corresponding trace and 
 to locate a number of points, as AyByCv, etc. 
 
 103. Proposition 28. Given any cone to find the shape and 
 size of its right section at any given point. 
 
 Discussion. If the cone be revolved parallel to H or V the 
 right section may be drawn through the given point. In this 
 position the axes of the right section will be projected on H or 
 V in their true size. 
 
 Construction. In Fig. 97, the given cone has a circular base 
 in V and an apex at O. About the center of the base revolve 
 the cone parallel to H, the projections of the apex falling at o^ 
 and o". Now find OQ, the axis of the cone, by bisecting the 
 angle at the apex and on this line locate the given point P, at 
 which the section is to be taken. Through P, pass a plane R 
 perpendicular to the revolved position of OQ. This plane con- 
 tains the revolved position of the right section, and the points 
 C and D, where the limiting elements pierce R, are the ends 
 of the minor axis. Since CD is parallel to H and the axes are 
 perpendicular, the major axis will be projected on H as a point 
 at p^ and on V as a line through p" perpendicular to c"d"o 
 
122 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 Now through p^ draw the projection of the elements which pass 
 through the ends of the major axis; both of these elements are 
 projected in H at o^q^ and pierce V at x" and y". o"x" and 
 o"y" are the elevations of the elements which pass through the 
 ends of the major axis at a" and b". a"b" and c"d" are ele- 
 
 r 
 
 Fig. 97. 
 
 vations, then, of the axes of the right section at P, and by 
 revolving them into V about r'R the true size of the curve may 
 be drawn at AvByCvDy. 
 
 104. If the base of the cone is an ellipse whose major axis 
 coincides in direction with the projection of the axis of the cone, 
 this same method may be used to find the right section. In such 
 a case, however, it is possible for the section to be a circle and if 
 the cone inclines so that the plane R cuts a Kne c^d^ from the 
 cone equal to the hne a"b" such will be the case. 
 
 When this method will not apply the right section may be 
 
CONICAL SURFACES 1 23 
 
 found as in Article 102 by taking the cutting plane perpen- 
 dicular to the axis of the cone. 
 
 105. As in the case of cylinders the fact that an oblique plane 
 cuts a symmetrical curve from a right cone may be utilized in 
 
 Fig. 98. Fig. 99. 
 
 constructing conical elbows. Fig. 98 shows a typical case when 
 the top of a given cone is cut off, turned through 180 degrees, 
 
 Fig. 100. Fig. ioi. 
 
 and set back. The axes of the two sections not only turn an 
 angle but they no longer meet. The reason for the latter point 
 will be clear from an inspection of Fig. 99. 
 
124 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 Fig. loi shows an elbow made up of sections of the same 
 cone, the cone being divided as shown in Fig. loo. Since the 
 axis of the cone does not pierce the plane of each joint in the 
 center of the plane of the joint the axes will have to be offset as 
 shown. 
 
 A simple method of drawing such elbows is shown in Fig. 103, 
 
 which is a design for a ship's ventilator of three conical and one 
 
 cylindrical sections, whose axes are given by the broken line 
 
 M 
 
 Fig. 102. 
 
 ABCDE. The problem is to find the size of the right cones 
 which will fit together with plane joints. 
 
 The principle by which the problem may be worked may be 
 stated as: 
 
 106. Proposition 29. To find the size of a cone or cy Under 
 which will fit another cone or cylinder with a plane joint, the 
 axes and the size of one cone or cyUnder being given. 
 
 Discussion. Since an oblique plane cuts an ellipse from both 
 a right cone and a right cyHnder, the problem is to find where to 
 place this plane so that the sections cut from the surface will be 
 equal. In Fig. 102 the right cyHnder whose axis is AO is to form 
 an elbow with a right cone whose axis is BO, and whose right 
 section at B is the circle shown. Now if at O, the intersection of 
 the axes, a sphere be drawn to which the given cylinder is tan- 
 gent, and a cone be drawn with BO as an axis tangent to the 
 sphere, this cone will join the cylinder in an ellipse whose plane 
 
CONICAL SURFACES 
 
 125 
 
 is shown as the joint in the elbow, MN. Since the cone and the 
 cylinder are each tangent to a common sphere they will have one 
 curve of intersection in common. 
 
 Construction. To apply this principle to the drawing of the 
 ventilator cowl made up of conical sections, spheres must be 
 drawn at B, C, and D as shown in Fig. 103. The diameter of 
 the sphere at B must be equal to the diameter of the cylinder; 
 the diameter of the spheres at C and D are to be assumed so that 
 
 Fig. 103. 
 
 the increase in size in the sections of the cowl is fairly uniform. 
 Tangent to these spheres are drawn the limiting elements of the 
 cones and where these pairs of limiting elements intersect will be 
 the joints. The opening at E is usually given so that the limiting 
 elements of the last section are drawn from the ends of the given 
 diameter of the opening at E to the sphere whose center is at D. 
 
 107. Proposition 30. To develop any given cone. 
 
 Discussion. Draw lines from a common point — the apex — 
 equal in length to the elements of the cone, and spaced according 
 to their distances apart on some curve of the surface. 
 
126 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 Construction. When the given cone is a right circular cone. 
 In Fig. 104 the base of the given cone is a circle in H, the apex is 
 at O, and the cone has been cut off along the plane of a'g'. Since 
 the cone is right all the elements from the apex to the base are 
 
 Fig. 104. 
 
 equal in length; therefore, with o' as a center and o'l' as a 
 radius describe the arc i^^i6, etc., and at any convenient 
 point draw the Hne o'gi equal to o'g'i'. This Hne will be the 
 first element in the development. Now divide the circle of the 
 base into a convenient number of parts and step these off on 
 
CONICAL SURFACES 
 
 127 
 
 the developed base as i, 2, 3, 4, 5, 6, etc. Find where each of the 
 elements pierces the plane through a'g' and find the true distance 
 from O to each point. Lay this distance off on the development 
 
 Fig. 105. 
 
 as at g, f, e, d, etc. In the figure only a few points have been 
 worked out to avoid the confusion of lines but in practice the 
 elements should be spaced about as shown by the points i, 2, 
 3, 4, etc. 
 
128 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 Since the development will be symmetrical about a center line 
 Oi6 the two parts may be constructed simultaneously. 
 
 Construction. Method of Triangulation. When the given 
 cone is oblique. In Fig. 105 the given cone is taken with its 
 axis parallel to V with its base in H and its apex at O. Lay 
 off o'8 equal to element 08. Now with the true length of the 
 arc 7-8 as a radius and 8 as a center strike an arc, and with o' 
 as a center and the true length of O7 as a radius strike a second 
 arc intersecting the first at 7. o'7 then is the position in the 
 development of the element O7. In practice the distance 87 
 must be made small enough so that the triangle 0^87 will lie 
 on the surface of the cone; in the figure these lines are taken 
 far apart to avoid confusion. 
 
 In like fashion with 0^7 as one side construct triangle 0^76 
 equal to the triangle O76 on the cone, and proceeding this way 
 the entire surface of the cone may be developed by dividing it up 
 into small triangles and laying these out in their true size and 
 relation. 
 
 This method is the method by which most developments are 
 made. It will apply to the development of any object whose 
 surface is capable of being divided into triangles. 
 
 PROBLEMS ON CONES 
 
 Special Proposition. Given a cone and a line to find where the line cuts 
 the surface of the cone. Write the discussion and draw a figure demon- 
 strating the above proposition. 
 
 207. The axis of a cone is 3 J" long. It pierces H i\" in front of V and 
 inclines 45 degrees to H. The apex of the cone is 3" in front of \. A right 
 section of the cone 2" from the apex is a circle 1" in diameter. Draw a plan 
 and elevation of the cone and find its base in H. 
 
 208. The point O is o"; -3"; o". The point Q is -3"; -5"; -4". 
 The line OQ is the axis of a cone whose right section 4" from the apex Q is a 
 circle ih" in diameter. Draw the plan and elevation of the cone and show 
 its base in H. 
 
 209. The axis of a cone pierces H 3" in front of V. The apex of the 
 cone is 3" above H and 2" behind V and the axis from the apex to the H 
 piercing point is 7" long. The base of the cone in H is a circle 2\" in 
 diameter. Find the base of the cone in V. 
 
 210. The point O is -3"; -3"; o". The point Q is 2"; -3"; -4". 
 The H base of this cone whose axis is OQ is a circle 2\" in diameter. Find 
 the base of the cone in P. 
 
CONICAL SURFACES 1 29 
 
 211. The point O is -2"; o"; -3". The point Q is 2"; -4"; -3". 
 TIic base of tlie cone whose axis is OQ is a i" circle in P = c"; go°; 90°. 
 Find the base of this same cone in V when Q is the apex. 
 
 212. A conical lamp shade whose height is 8" is hung from the center 
 of a 9' by 12' room with its apex 7' 6" above the floor. The base of the 
 shade is 1 2" . Find the lighted areas on the walls and floor of the room made 
 by the cone of light reflected from the shade. 
 
 213. The axis of a right cone is perpendicular to H 25" behind V; the 
 base of this cone in H is a 3" circle and its altitude is 4". Cut this cone by 
 three planes: one at an angle to its axis cutting all of the elements; one 
 parallel to the axis; and one parallel to one element of the cone. Show the 
 true shape and size of each of these curves of intersection and investigate 
 its properties. 
 
 214. Draw a plan and elevation of the cone in Problem 213 and cut the 
 cone by any plane oblique to both H and V. Show the true shape and size 
 of the curve of intersection. 
 
 215. ThepointOiso"; -o"; -21". The point Q is o"; -2!"; -21". 
 The base of the cone whose axis is OQ is a circle 2\" in diameter in V. The 
 cone is cut by plane T = 3I"; —30; —60. Find the true shape and size of 
 the curve of intersection and its plan and elevation. 
 
 216. The point O is 3!"; -2\"\ o" . The point Q is if"; -4!"; -3I". 
 The base of this cone is a circle 3" in diameter in H. The cone is cut by a 
 plane which is o" ; +30; —60. Find the true shape and size of the curve 
 of intersection. 
 
 217. The point O is 31"; -2^"; -o". The point Q is o"; -2!"; -2!". 
 The base in H of this cone is a circle 2\" in diameter. The plane T is 
 o"; —60; +45 and cuts the cone. Find the true shape and size of the 
 curve of intersection. 
 
 218. Draw the plan, elevation, and left end view of a right cone whose 
 base is a \\" circle lying in plane T — sl"; 60; 30; with its center at 
 O — 3f"; if"; x" and whose altitude is 3". 
 
 219. A cone of revolution is located in the third quadrant with its 
 base in plane T —2\"\ +60; +30 with its center at the point O —5"; x"; 
 — 1 1". The base is a circle 2" in diameter and the altitude is 3I". Draw 
 three views of the cone. 
 
 220. Lay out the patterns for a conical hopper. Diameter of large 
 opening 12"; diameter of small opening 4"; distance between openings 6". 
 
 221. A 20" smoke stack passes through a roof which inclines 30 degrees 
 with the horizontal. A conical canopy forms the flashing between the roof 
 and stack. Make a pattern of tliis canopy with the shortest clement 6" 
 long. 
 
 222. Lay out a pattern for the cone in Problem 216 between the base 
 and the intersection with the plane. 
 
130 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 223. The point is o"; 3"; o". The point Q is o"; -2; +4". The 
 line OQ is the axis of a cone whose base in H is a 3" circle. Make a pattern 
 for that portion of the cone lying between H and V. 
 
 224. Make a pattern for a sink strainer. This is in the form of one 
 quarter of a cone whose opening is a 16" circle and whose height is 8". 
 
 225. Make the patterns for a conical elbow of three sections to turn an 
 angle of 90 degrees. Diameter of large opening 10"; diameter of small 
 opening 4"; radius of center line 12". 
 
 226. Design a conical elbow of four sections for a 120 degree turn to 
 connect a pipe of 12" diameter with another pipe of 6" diameter, radius of 
 center line 20". 
 
CHAPTER XV 
 INTERSECTION OF SURFACES 
 
 1 08. Proposition 31. Given two plane surfaces to find their 
 lines of intersection. 
 
 Discussion: First Method. Find where the edges of one 
 surface pierce the planes of the faces of the other surface. These 
 points when joined in proper order determine the line of inter- 
 section. 
 
 Construction. In Fig. 106 the right hexagonal prism is 
 intersected by a rectangular prism whose edges are obHque to H 
 and V. Find where these edges pierce the faces of the hexagonal 
 prism and join the points in the proper order. In the given 
 case the rectangular prism pierces the hexagonal prism, thus 
 giving two intersections: the entering one at ABCD and the de- 
 parting one at EFGH. The prisms have been assumed as solids 
 in the drawing. 
 
 Discussion. Second Method. Find the intersections of the 
 planes of the faces. These lines will be the edges of the inter- 
 section of the two surfaces. 
 
 Construction. In Fig. 107 a rectangular hopper, base in 
 H 1-2-3-4 and apex 0, intersects a rectangular pipe. The face 
 of the hopper, 1-4-O, lies in the plane rRr'; and this plane 
 intersects U — the plane of the upper face of the pipe — in the 
 line PQ. The portion of PQ between the edges O4 and Oi — or 
 AD — is one side of the opening in the pipe made by the hopper. 
 In like manner CB may be obtained by finding the intersection 
 of plane U with plane vVv'; and with these four points, A, B, C, 
 D, the intersection may be drawn. 
 
 109. Proposition 32. To find the intersection of two given 
 cylinders. 
 
 Discussion. Pass auxiliary planes through the cylinders cut- 
 ting elements from each. These elements, being in a common 
 
 131 
 
132 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 plane, intersect in points common to both cylinders, or points 
 on the required curve of intersection. 
 
 Fig. io6. 
 
 Construction: First Case. When the bases of the given 
 cylinders lie in the same plane. 
 
 In Fig. 1 08 one cyHnder is perpendicular to H and the other 
 obHque to H and V; both have circular bases in H. Pass plane 
 
INTERSECTION OF SURFACES 
 
 133 
 
 Fig. 107. 
 
134 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 R through both cylinders cutting elements from each. These 
 elements intersect at A, B, C, and D, as shown. If enough auxil- 
 iary planes similar to R are passed a sufficient number of points 
 may be obtained to draw a curve of intersection. 
 
 Fig. io8. 
 
 Construction: Second Case. When the bases of the given 
 cylinders are in different planes. 
 
 In Fig. 109 the larger cylinder has a circular base in H and is 
 parallel to V; the smaller has a circular base in V and is parallel 
 to H. Pass auxihary planes, as S, through the cyhnders parallel 
 to the axis of each and cutting elements from each. These ele- 
 
INTERSECTION OF SURFACES 
 
 135 
 
 ments intersect in points of the curve of intersection, as D and 
 B. To find the direction of the auxihary planes pass any plane 
 through the axis of one cylinder parallel to the axis of the other. 
 If the auxihary planes be drawn parallel to this plane they will 
 
 Fig. 109. 
 
 cut elements from the cyhnders since they are parallel to the 
 axis of each .cylinder. 
 
 no. Something of the nature of the curve of intersection may 
 be determined in advance. Thus, in Fig. 109, plane R is ob- 
 
136 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 viously the first auxiliary plane which can cut elements from the 
 smaller cylinder, and the plane T is obviously the last auxiliary 
 plane which can cut the smaller cylinder. Therefore, since all 
 of the elements of the smaller cylinder pierce the surface of the 
 larger one, it becomes apparent that the smaller one completely 
 enters the larger. On the other hand, if the plane rRr' failed to 
 cut elements from the larger cylinder, then it would be apparent 
 that all of the elements of the smaller cyhnder did not pierce the 
 larger, and the resulting curve would be more like the one in 
 Fig. 108. By the use of limiting planes in this fashion the 
 character of the required intersection may be somewhat prede- 
 termined and the problem attacked with greater appreciation. 
 
 111. The visible and invisible portions of the curve of inter- 
 section should be shown by using solid and dotted lines. To 
 determine which points are visible it must be remembered that 
 surfaces are not transparent, nor are they solids — unless so 
 specified — and that when one looks into the end of a cone or a 
 cyhnder the base is not considered an obstruction to the view. 
 A line which enters a surface is visible only up to the entering 
 point, invisible within the surface, and visible again upon leaving 
 the surface. The simplest method of locating visible points on 
 a curve of intersection of two surfaces is to locate first the visi- 
 ble elements of the surface and of course points on these ele- 
 ments are also visible. It is not necessary to actually draw these 
 elements but with a little practice and study one can readily 
 locate the visible and invisible points by inspection. 
 
 112. Proposition 33. Given two cones to find their curve of 
 intersection. 
 
 Discussion. Pass auxiliary planes through the apices of the 
 two cones cutting elements from each. These elements being 
 in the same plane will intersect each other in points common to 
 both surfaces, or points on the curve of intersection. 
 
 Construction. When the bases of the given cones are in the 
 same plane, as in Fig. no. 
 
 Draw a Une through the apices of the cones and find where it 
 pierces H and V. These two points will be common to all the 
 auxiHary planes. In the case given in Fig. no this line is par- 
 
INTERSECTION OF SURFACES 
 
 137 
 
 Fig. I 10. 
 
138 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 allel to H so that all the H traces of the auxiliary planes will be 
 parallel to it. The plane S cuts one element from the right 
 hand cone and two elements from the other; the elements in- 
 tersect at B and D, two points on the curve of intersection. In 
 Kke manner other points may be located but all the points 
 common to both surfaces will lie in auxiHary planes between 
 planes S and T. 
 
 Fig. III. 
 
 Construction. When the bases of the given cones are in 
 different planes. 
 
 In Fig. Ill two right cones, one with its base in H and the 
 other with its base in P, intersect. As before, the apices are 
 joined; this Hne OQ is common to all the auxiliary planes and 
 its H piercing point, O, is common, therefore, to all the H traces 
 and its P piercing point, Q, to all the P traces. As before, locate 
 
INTERSECTION OF SURFACES 
 
 139 
 
 the limiting planes, S and R, and find the points common to both 
 surfaces. 
 
 113. Proposition 34. Given a cone and a cyhnder to find 
 their intersection. 
 
 Discussion. Pass planes through the apex of the cone parallel 
 to the elements of the cylinder. These planes will cut elements 
 
 Fig. 112. 
 
 from each surface, and since these elements are in a common 
 plane they will intersect in points common to both surfaces. 
 
 Construction. When the bases of the given surface are in 
 the same plane. 
 
 In Fig. 112 the given cone and cylinder have circular bases in 
 V and are obHque to both H and V. Through O, the apex of 
 
I40 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 the cone, draw a line OQ parallel to the axis of the cylinder. 
 This line will be common to all the auxiliary planes. 
 
 The first plane whose V trace passes through Q and which 
 cuts both the cone and cylinder is R. Plane R cuts one element 
 from the cone and two elements from the cylinder; these elements 
 
 Fig. 113. 
 
 intersect at D and C which are two points on the required inter- 
 section. The last plane which cuts both surfaces is plane S. 
 This plane cuts only one element from the cylinder and two from 
 the cone; these elements intersect at A and B which are two 
 more points on the required curve. Between these limiting 
 
INTERSECTION OF SURFACES 141 
 
 planes other auxiliary planes may be passed in sufficient number 
 to completely locate the curve. 
 
 Construction. When the bases of the given surfaces are in 
 different planes. 
 
 In Fig. 113 the base of the cylinder is in P and its elements are 
 parallel to the G. L., and the base of the cone is in H and its axis 
 is perpendicular to H. The construction will be obvious from a 
 study of the drawing. 
 
 PROBLEMS ON THE INTERSECTION OF PLANE SURFACES 
 
 227. The axis of an hexagonal prism whose sides are 2" wide pierces V 2" 
 below H, and is perpendicular to V. The axis of a square prism whose 
 sides are i" wide pierces P 2" below H and 2" back of V, and is parallel 
 to the G. L. The diagonals of the right section of the square prism are 
 perpendicular and parallel to H. Draw a plan and elevation of the two 
 prisms showing their intersection. 
 
 228. The base of a pyramid is parallel to and \" below H. The base is 
 2" by 3" with its 3" edge parallel to V i" back of V. The apex of the 
 pyramid is 2" back of V, 2\" below H, and in a profile plane i" to the 
 right of the right edge of the base. The axis of a 1" by i" prism pierces 
 H i}^" to the right of the profile plane of the apex, 2" back of V, incHnes 
 45 degrees to H, and is parallel to V. Draw a plan and elevation showing 
 the intersection. 
 
 229. The point M is -2|"; o"; -2". The point N is i|"; -4"; -2". 
 The line MN is the axis of a prism whose base in V is a rectangle 2" by i". 
 The point is 2>\" \ -2"; o". The point Q is o"; -5^'; -3I". The line 
 OQ is the axis of an hexagonal pyramid whose base in H has i" sides. Draw 
 a plan and elevation showing the intersection. 
 
 230. The point A (|"; —4"; —2\") is the apex of a pyramid whose base 
 is CDEF. The point C is -2H"; -\h"\o. The point D is -\h"\ 
 -2"; o". The point E is -A"; -|"; o". The point F is -2"; o"; o". 
 The point B (—2"; —2 A"; —lii") is the apex of a second pyramid whose 
 base is MNOP. The point M is U"; o"; o". The point O is 2tV'; -li"; 
 o". The point N is U"; - 1§"; o". The point P is 2^"; o"; >". Draw a 
 plan and elevation of these two pyramids showing their intersection. Make 
 a development of the pyramid A-CDEF with this line of intersection traced 
 on it. 
 
 231. The axis of a 6" square timber pierces V at o"; o"; — 18", is parallel 
 to H, and inclines 60 degrees to V. The axis of a second timber of the same 
 size pierces H at +6"; —18'; o" , is parallel to V, and inclines 60 degrees to 
 H. Draw a plan and elevation of these two timbers and show also a plan and 
 elevation of the shortest 6" square timber which can be fitted between them. 
 
142 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 PROBLEMS ON THE INTERSECTION OF CYLINDERS 
 
 232. The axis of a cylinder whose base in H is a 2" circle pierces H at 
 o; — i|";o". A second point on this axis is at 2"; — sii"; —2". The 
 axis of a second cyUnder whose base in H is a ij" circle pierces H at 2|"; 
 — 1|"; o". A second point on the axis of this cylinder is at i"; —41"; —2%". 
 Draw a plan and elevation of these two intersecting cylinders and show the 
 plan and elevation of their curve of intersection. 
 
 233. The axis of a cylinder whose base in H is a 2" circle pierces H at 
 o"; — ii";o". A second point on this axis is at 2"; —2^,"; —2". The 
 axis of a second cylinder whose base in H is a if" circle pierces H at 3^"; 
 
 — i"; o". A second point on this axis is at 2"; —2 J"; —2\". Show the 
 plan and elevation of the curve of intersection of these two cylinders. 
 
 234. The axis of a cylinder whose base in V is a 2|" circle pierces V at 
 o"; o"; — 3I". The axis of a second cylinder whose base is a i|" circle 
 pierces H at x\"; —4"; o". These two axes intersect at the point 2f"; 
 
 — 2|"; —2 A". Draw a plan and elevation of the curve of intersection. 
 
 235. The axis of a cyhnder whose base in V is a lA" circle pierces V at 
 o";o"; —I J"; a second point on this axis is at 2"; —3^"; —2\". The axis 
 of a second cylinder whose base in H is a 1 1" circle pierces H at the point 
 2f"; — i; o". A second point on this axis is at o"; — 3I"; —42"- Draw 
 the plan and elevation of the two cylinders showing their curve of inter- 
 section. 
 
 236. The axis of a 12" pipe lies parallel to H and 36" below H; it in- 
 clines 60 degrees to V. The axis of a second 12" pipe is parallel to the G. L. 
 24" back of V and 60" below H. Draw the plan and elevation of these two 
 pipes and make the pattern for the shortest 8" pipe which will connect them. 
 
 PROBLEMS ON THE INTERSECTION OF CONES 
 
 237. The center of the base of a cone is at o"; —3"; o". The base is a 
 circle 3" in diameter. The apex of the cone is at 42"; —55"; —55". The 
 center of the base of a second cone is at 3I"; — iJ"; o"; the base is a circle 
 2" in diameter. The apex of this cone is at 2"; — Sj"; —4". Draw a plan 
 and elevation of the two cones showing their curve of intersection. 
 
 238. The center of the base of a cone is at o"; — 3M"; — if". The base 
 is a 3" circle whose plane is parallel to V. The apex of this cone is at 4^"; 
 
 — tb"; — 5f". The base of a second cone is at 3"; —^W; —2". The 
 base of this cone is a 2\" circle whose plane is parallel to V. The apex is at 
 the point o"; —A"; — 4f". Draw a plan and elevation of these two cones 
 and show their curve of intersection. 
 
 239. The base of a cone in V has its center at the point o" ; o" ; —\\"\ the 
 diameter of the base is 2\" . The apex of this cone is at 2 J"; —3!"; — 2f". 
 The base of a second cone in H has its center at 3"; — li"; o"; the base is 
 
INTERSECTION OF SURFACES 143 
 
 2" in diameter. The apex of this cone is at f"; —if"; —3!". Draw a 
 plan and elevation of the cones showing their curve of intersection. 
 
 240. The base of a cone in H has its center at o"; —if"; o"; the diam- 
 eter of the base is 2 1". The apex of this cone is at 4I"; —3!"; —45". The 
 base of a second cone in V is at 4I"; o"; — if"; the base is 2 J" in diameter. 
 The apex of this cone is at — |"; —5"; — 2I". Show a plan and elevation 
 of the two cones and find their curve of intersection. 
 
 241. The center of the base of a right cone is 2f" behind V, and its apex 
 is 2)i" below H. The cone is perpendicular to H and its base in H is 3I". 
 A second right cone, which is perpendicular to V, has the center of its base 
 i\" below H in V, and its apex is 42" behind V. The base of the second 
 cone is a circle 2\" in diameter. Draw a plan, elevation, and end view of 
 the two cones, and show three views of their curve of intersection. 
 
 242. The center of a 2 J" circle whose plane is parallel to V is at o"; 
 
 — 3"; —2". This circle is the base of a right cone whose apex is at o"; 
 
 — h"'-, —2". The center of the base of a second cone is at 2^^"; —if"; 
 
 — 2^"; the apex of this cone is at — 2|"; — 2|"; — f". The base of this 
 second cone is a 2\" circle lying in the plane i|"; 270; 300. Show the plan 
 and elevation of these cones and their curve of intersection. 
 
 PROBLEMS ON THE INTERSECTION OF CYLINDERS AND 
 
 CONES 
 
 243. The center of the base of a cylinder lies in H at the point o"; 
 
 — 2"; o"; the base is a circle 3" in diameter. A second point on the axis 
 of the cylinder is at 3"; —32"; — if"- The center of the base of a cone lies 
 in H at the point 4I"; — ig"; o"; the diameter of this base is 2". The 
 apex of this cone is at f"; — 4I"; — 3i". Draw a plan and elevation of the 
 cone and cylinder showing the curve of intersection. 
 
 244. The base of a cylinder has its center in H at o"; —i\"; o"; the 
 diameter of the base is 2" . A second point on the axis of this cylinder is at 
 2"; —35"; —2". The center of the base of a cone lies in H at the point 
 3"; — 1 2"; o"; the diameter of this base is if". The apex of the cone is at 
 \"; —35"; — 3f". Draw the plan and elevation of the cyHnder and cone 
 and show their curve of intersection. 
 
 245. The base of a cylinder has its center in V at o"; o"; — 15"; the 
 diameter of this base is i \" . A second point on the axis of the cylinder is at 
 2"; — 3x'V"; — 2i". The center of the base of a cone Kes in Hat 3 f"; — itV; 
 o"; the diameter of the base is 2\" . The apex of the cone is at the point 
 
 — I"; —2"; —3 1". Draw a plan and elevation of the two surfaces and show 
 the curve of intersection. 
 
 246. A cylinder 3" in diameter lies parallel to H and V with its axis 2" 
 back of V and 3" below H. The apex of a right cone coincides with this 
 axis of the cylinder and the projections of the axis of the cone inclines 60 de- 
 
144 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 grees to GL. A right section of the cone 2" from the apex is 2" in diameter. 
 Draw a plan, elevation, and end view of the cone and cylinder and show 
 their curve of intersection. 
 
 247. The axis of a right cone pierces V i§" above H and inclines 45 de- 
 grees to V; the apex is 4" above H. Between these two points the axis of 
 the cone is 4" long, and a right section of the cone 3" from the apex is a circle 
 1 3" in diameter. At a point 2" from the apex the axis of the cone is cut 
 by the axis of a right circular cylinder 2" in diameter whose axis is parallel 
 to H and V. Draw three views of these two surfaces showing the curve of 
 intersection. 
 
CHAPTER XVI 
 
 SURFACES OF REVOLUTION 
 
 114. A Surface of Revolution is represented in a drawing by 
 the projections of its meridian curves, or sections. These me- 
 ridian curves are always taken 
 in planes through the axis 
 parallel to the planes of pro- 
 jection. Thus, in Fig. 114, 
 the plan view of the sphere is 
 the projection of the curve 
 cut from the sphere by the 
 meridian plane AB; and the 
 elevation of the sphere is the 
 elevation of the curve cut 
 from the sphere by the me- 
 ridian plane CD. 
 
 115. To assume a point on 
 the surface of a surface of 
 revolution one projection may 
 be assumed, as x in Fig. 
 114, and the other projection 
 found by passing a meridian 
 plane MN through the as- 
 sumed point and revolving 
 this plane into coincidence 
 with the meridian plane CD. 
 In this position the point X 
 will be projected horizontally 
 
 at x^, and vertically at x" on the meridian curve. The plane 
 MN may now be revolved back to its original position and the 
 projection of point X on the surface will be found at x and x'. 
 
 145 
 
 — B 
 
 Fig. 114. 
 
146 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 It will be noted that the point projected on H at x may have 
 two elevations and which of these is to be taken depends upon 
 which side of the sphere the point is assumed originally. 
 
 116. In finding the intersections cut from surfaces of revolu- 
 tion by other surfaces the problem will be greatly simplified if 
 the following rule is observed: 
 
 Pass the auxiliary planes through the given surface in such a 
 direction that only the simplest and most easily located curves and 
 elements of the surfaces are cut out. 
 
 Since surfaces of revolution have circular sections in planes 
 perpendicular to the axes of revolution it will be obvious that, 
 where expedient, the cutting planes should be passed in this 
 direction. 
 
 117. Proposition 35. Given any surface of revolution to find 
 its intersection with any plane. 
 
 Discussion. Pass auxiliary planes through the surface per- 
 pendicular to the axes of revolution. These planes cut circles 
 from the surface and Hnes from the given plane. Since each 
 circle and Hne lies in a common plane they will intersect in points 
 common to the given surfaces, hence on the required intersection. 
 
 Construction. In Fig. 115 the given surface is an ellipsoid 
 with its axis of revolution perpendicular to H. It is cut by the 
 plane sSs'. To find the required intersection pass planes through 
 the plane and ellipsoid perpendicular to the axis or parallel to H, 
 as R. The plane R cuts the line MN from the plane S and the 
 circle, whose diameter is AB, from the ellipsoid. This line inter- 
 sects the circle at X and Y, two points on the required curve. 
 In Hke manner plane V cuts the circle whose diameter is CD from 
 the surface and from the plane cuts the line OP. These intersect 
 at O and P, two more points on the required curve; continuing 
 this process enough points may be found to locate the curve. If 
 the true size of the curve be required it may be found by re- 
 volving it into H or V about the corresponding trace of plane S. 
 
 118. Proposition 36. Given any surface of revolution to find 
 its intersection with any cylinder. 
 
 Discussion. Pass auxiliary planes through the surfaces so 
 that circles are cut from the surface of revolution and elements 
 
SURFACES OF REVOLUTION 
 
 147 
 
 from the cylinder. These elements cut the circles in points 
 common to both surfaces and are points of the required inter- 
 section. 
 
 Construction. In Fig. 116 the cy Under has a circular base in 
 H and intersects a semicircular pipe bend (annular torus). The 
 auxiliary plane R cuts from the surface of the torus a circle which 
 
148 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 is the elevation of the surface, and from the cylinder the limiting 
 elements of the elevation. The elements cut this circle at A and 
 F, or two points on the required curve of intersection. In like 
 manner auxiliary plane S cuts a smaller circle from the torus and 
 
 Fig. ii6. 
 
 two elements from the cylinder. These intersect at E and B, 
 two more points on the curve. Auxiliary plane T locates points 
 D and C, and in this manner the entire curve may be located. 
 
 119. Proposition 37. Given any surface of revolution to find 
 its intersection with a cone. 
 
SURFACES OF REVOLUTION 
 
 149 
 
 Discussion. See Article 116. 
 
 Construction. In Fig. 117 is a cylinder of revolution with a 
 hemispherical top, intersected by a cone with a circular base. 
 From a study of the figure it will appear that the auxiliary 
 planes should be passed parallel to H, for in this direction each 
 
 Fig. 117. 
 
 plane cuts a circle from each surface. Thus, plane R cuts from 
 the cone a circle whose radius is y'l and from the other surface a 
 circle whose radius is x'2. These two circles intersect at A and 
 B two points on the required curve. By passing other planes 
 parallel to R the curve of intersection may be completely deter- 
 mined. 
 
ISO 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 120. Proposition 38. Given any two surfaces of revolution 
 whose axes intersect to find the curve of intersection of the 
 surface. 
 
 Fig. 118. 
 
 Discussion. If, with the intersection of the axes of the given 
 surfaces as a center, auxihary spheres be passed through the 
 surfaces they will cut circles from each surface. These circles 
 intersect in points on the curve of intersection. 
 
SURFACES OF REVOLUTION 
 
 151 
 
 Construction. In Fig. 118 the sphere is cut by a cone. The 
 axes intersect at O. With this point as a center describe a sphere 
 as ABC. This sphere cuts from the given sphere a circle verti- 
 cally projected as a Hne at a'b' and from the cone another circle 
 vertically projected as a line at m'n'. These 
 two circles intersect at two points, both of 
 which are vertically projected at x' and 
 horizontally projected on the plan view of 
 circle AB at x and y. By passing other 
 spheres and proceeding in the same manner 
 the curve of intersection may be completely 
 determined. 
 
 121. Proposition 39. To develop any sur- 
 face of revolution. 
 
 Discussion. Theoretically, surfaces of 
 revolution are not capable of development 
 but, practically, such surfaces are developed 
 by methods which give approximations, 
 which are near enough for commercial pur- 
 poses. Examples of this are shown in the 
 following developments of the sphere. In 
 the first method the surface of the sphere 
 is divided by meridian planes into gores as 
 shown in Fig. 119. In the second method, 
 Fig. 120, the surface of the 
 sphere is divided into zones 
 by making each zone a 
 frustum of some cone which 
 nearly coincides with the 
 surface of the sphere. Of- 
 ten these two methods are 
 combined in laying out the bottoms of hemispherical tanks. 
 
 Construction. Gore Method. Divide the surface into gores 
 by meridian planes, keeping in mind the fact that the narrower 
 the gores are made the more nearly will the development ap- 
 proach the true surface of a sphere. In Fig. 119 the hemisphere 
 is divided into 16 equal gores. To lay out each gore divide the 
 
152 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 sphere as shown in the figure by horizontal planes 1-2-3-4, etc.. 
 and lay off the distances 1-2; 2-3; 3-4, etc., in the line 1-7, 
 
 thus making the line 1-7 equal to I the great circle of the sphere. 
 Through the point i, 2, 3, 4, etc., thus located draw horizontal 
 Hnes; these will be the developed circles cut by the horizontal 
 
SURFACES OF REVOLUTION 1 53 
 
 planes through i, 2, 3, 4, etc. On each of these horizontal hnes 
 lay off the distances between the meridian planes. Thus, at the 
 point I lay off ab equal to the arc ab; at point 2 lay off the 
 length of arc between the meridian planes in the plane 2. 
 Sixteen gores, arranged as shown in the figure, developed by 
 this method will give an approximate pattern of the hemisphere. 
 
 Conp+.ruction. Zone Method. Divide the surface of the 
 hemisphere in Fig. 120 into zones by making each zone the frus- 
 tum of a cone whose apex is on the axis of the hemisphere and 
 whose elements are tangent to the hemisphere. Develop these 
 frusta by any convenient method of laying out a cone and 
 arrange them as shown ; the result will approximate very closely 
 a development of a sphere. 
 
 As in the gore method, it should be remembered that the more 
 zones used the nearer the result will approach the surface of a 
 hemisphere. 
 
 PROBLEMS ON SURFACES OF REVOLUTION 
 
 248. An ellipsoid of revolution is generated by revolving an ellipse 2" by 
 3" about its major axis. Cut this surface by a plane oblique to both H and 
 V and find the plan and elevation of the curve of intersection and its true 
 size. 
 
 249. An annular torus is generated by revolving a i" circle about an 
 axis lying in the plane of this circle and I5" from its center. Cut this torus 
 by a plane oblique to both H and V and find the plan, elevation, and true 
 size of the curve of intersection. 
 
 250. The center of a hemisphere is at o"; —3"; o". The center of the 
 base of a cylinder is at 5"; —3"; o". The radius of the hemisphere is 1^", 
 and the base of the cylinder is an ellipse lying in H with its 2" major axis 
 perpendicular to the G. L. and its i|" minor axis parallel to the G. L. The 
 axis of the cyUnder is parallel to V and inclines 15 degrees to H. Draw a 
 plan and elevation of these two surfaces showing the curve of intersection. 
 
 251. A tank with a hemispherical bottom is drained by a pipe which 
 runs off at an angle of 30 degrees to the horizontal. The tank is 72" in 
 diameter, the pipe is 6" in diameter, and the axis of the pipe pierces the 
 bottom of the tank at a point on its surface directly below the center 
 of its hemispherical bottom. Draw a plan and elevation of the tank and 
 pipe showing the curve of the joint between them. 
 
 252. The axis of an ellipsoid of revolution is perpendicular to H 3" 
 behind V. The surface is generated by the revolution of an ellipse 3" by 2", 
 
154 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 the 2" dimension being parallel to H and the upper end of the 3" axis being 
 1 1" below H. The axis of a right cone is parallel to the G. L. 3" below H 
 and 3" behind V. The apex of the cone is 2" to the right of the axis of the 
 ellipsoid and the center of the base Ues in a profile plane 2" to the left of the 
 axis. The base of the cone is a circle 2\" in diameter lying in this profile 
 plane. Draw a plan and elevation of these two surfaces and show their 
 curve of intersection. 
 
 253. In the Pierce-Arrow automobile the headlights — which are pa- 
 raboloids of revolution — are set into the mud guards of the front wheels. 
 Assuming the necessary dimensions find the shape of the hole which will 
 have to be cut into the cylindrical mud guard to receive the lamp. 
 
 254. Make the patterns for the bottom of a tank which is a hemisphere 
 of 42" radius. 
 
 255. Show how to cut the tin to cover a dome which is a semi-ellipsoid 
 in form with axes 84" by 60". 
 
 256. Draw three views of a U. S. Standard hexagonal nut for a 2" bolt. 
 The nut has a spherical chamfer. 
 
CHAPTER XVII 
 WARPED SURFACES 
 
 122. A Warped Surface is represented by the projections of 
 its directrices and a sufficient number of positions of its genera- 
 trix to show the character of the surface, and to indicate how the 
 generating Hne moves. Thus, in Fig. 71, the projections of the 
 directrices AB and CD are shown and a sufficient number of 
 positions of the generatrix AX are given to indicate the character 
 of the surface and that AX remains parallel to the H plane, which 
 in this case is the plane director. 
 
 123. In addition to the warped surfaces which are unclassified 
 there are three general kinds of warped surfaces : 
 
 1. Warped surfaces with two Hnear directrices and a plane 
 director, 
 
 2. Helicoidal surfaces. 
 
 3. Warped surfaces with three linear directrices. 
 
 Some of these surfaces have commercial applications and such 
 as these will be studied in detail but the majority of warped 
 surfaces, owing to their complex nature and undevelopable 
 character, are not used in practical work and possess only mathe- 
 matical interest. The following table gives a classification of 
 warped surfaces with the names of some of the more important 
 ones. 
 
 CLASSIFICATION OF WARPED SURFACES 
 
 Warped surfaces 
 with two linear 
 directrices and 
 a plane director 
 
 Linear directrices, both Hyperbolic paraboloid 
 straight lines 
 
 Linear directrices, one Conoid 
 straight and one curved 
 
 , Linear directrices, both Cylindroid 
 I curved lines 
 
156 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 2. Warped surfaces 
 with a helical 
 directrix 
 
 Right helicoid Generatrix perpendicular to the axis 
 
 ObUque hehcoid Generatrix oblique to axis 
 
 Convolute (de- Not a warped surface but a single 
 velopable heli- curved surface to which the heli- 
 coid) coid approaches as a limit 
 
 Warped surfaces 
 with three linear 
 directrices 
 
 1. Linear directrices, 
 
 all straight lines 
 
 2. Two linear direc- 
 
 trices, one straight 
 and one curved 
 
 3. One linear directrix 
 
 straight and two 
 curved 
 
 4. Linear directrices, 
 
 all curved lines 
 
 Hyperboloid of one nappe, 
 a surface of revolution 
 
 Warped arch 
 Warped cone 
 
 Also 
 
 124. To assume a point on a warped surface one of two 
 
 methods may be used. In case it is possible to draw the pro- 
 jection of an element through the assumed projection of the 
 point, as in Fig. 121, this may be done and the other projection 
 of the point may be found on the corresponding projection of 
 the element. In Fig. 121, x' was assumed and through it was 
 drawn the projection p'q' of an element of the surface passing 
 through that point. The H projection of this element is, of 
 course, pq, and on this projection the other projection x may 
 be located. 
 
 In Fig. 122 this method is not applicable since through the 
 assumed projection of the point x it is not convenient to draw the 
 projection of the element containing it. To find x' in this case, 
 pass a plane R through the point perpendicular to H. This plane 
 cuts from the warped surface the line OPQ, and obviously x' will 
 be found located on its projection o'p'q'. 
 
 125. Of warped surfaces having a plane director Fig. 71 is an 
 example. This surface has two straight Une directrices and is 
 a common example of warping. This surface is called an hyper- 
 bolic paraboloid because one set of planes will cut hyperbolas 
 
WARPED SURFACES 
 
 157 
 
 Fig. 122. 
 
158 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 from the surface while another set perpendicular to the first set 
 will cut parabolas. (See Fig. 123.) As practical examples of 
 
 this surface the pilot on a 
 locomotive and the bow of 
 a ship may be cited. 
 
 If one directrix is a curved 
 line and one remains a 
 straight line the surface be- 
 comes a conoid. An ex- 
 ample of such a surface 
 may be found in chisels 
 Fig. 123. - An hyperbolic paraboloid. Note shaped from round bar stock 
 
 that planes parallel to the base of the and in the bowS of ships. 
 model will cut out hyperbolas, and that jrjg ^^i shows an oblique 
 planes parallel to its ends will cut out . , , _,. , 
 
 parabolas. ^onoid, and Fig. 1 24 shows a 
 
 string model of a right conoid. 
 
 Where both directrices become curved lines the surface 
 becomes a cylindroid. 
 Fig. 122 shows such a 
 surface and illustrates 
 the skewed arch, one of 
 the useful applications 
 of the surface. The 
 cyHndroid is also made 
 use of in buildings for 
 connecting arched pas- 
 sageways at different ele- 
 vations. 
 
 Many interesting ap- 
 plications of these and 
 other warped surfaces 
 may be seen in the 
 bodies, mud guards, and 
 other portions of modern 
 automobiles. 
 
 126. Proposition 40. Given the directrices and the plane 
 director to draw the plan and elevation of a warped surface. 
 
 Fig. 124. — A string model of a right conoid. 
 The base of this surface is a circle which is also 
 the curved directrix; the bar at the top is the 
 straight line directrix. The plane director is 
 parallel to the ends of the box. 
 
WARPED SURFACES 1 59 
 
 Discussion. The directrices may both be straight lines, or 
 may both be curved lines, or one may be straight and one curved. 
 The character of the surface will, of course, be determined by 
 the character of these directrices as described in Article 125, but 
 the method of drawing will be the same for all cases. 
 
 Construction. In Fig. 122 let AB and CD be the two given 
 directrices, in this case both being curvilinear. Let the H plane 
 be the plane director. From any point. A, on one directrix 
 draw a line, as AC, parallel to the plane director, H. This 
 line intersects the other given directrix at C. Therefore, since 
 AC touches both directrices and is parallel to the plane director 
 it will be an element of the surface. In like manner other 
 elements may be drawn and the plan and elevation outlined. 
 In commercial drafting only the limiting elements, or those ele- 
 ments which define the contour of the surface, are shown. 
 
 In Fig. 121, the oblique conoid, the method of construction 
 is the same. In this case the curvilinear directrix is the circle 
 MN and the rectilinear directrix is AB, and the plane director 
 is an oblique plane rRr'. The elements must, of course, touch 
 both MN and AB and be parallel to plane R. 
 
 127. A helicoidal surface is generated by a line moving uni- 
 formly about and along an axis, making a constant angle with 
 it. If the generatrix is of definite length it is obvious that 
 during the generation of such a surface one end of the genera- 
 trix — • that touching the axis — will describe a straight line, 
 while the other end will follow a path in space which is a curve 
 of double curvature called a helix. It must be apparent that all 
 points between these extreme points of the generatrix will also 
 describe helices of sizes which will vary according to the position 
 of the point on the generatrix. The surface of the helicoid, then, 
 may be said to be made up of a series of concentric hehces, and 
 in order to study such a surface properly the nature of the helix 
 must first be considered. 
 
 128. The helix is a space curve generated by moving a point 
 around and along an axis at a uniform rate. The distance from 
 the axis may be constant giving a cylindrical helix (Fig. 125), or 
 it may vary uniformly giving a conical helix (Fig. 126). In 
 
i6o 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 either of these cases the distance the generating point moves in 
 the direction of the axis during one revolution is called the pitch. 
 Practical examples of such curves may be found in the center 
 lines of cylindrical and conical springs (Figs. 127 and 128). 
 
 Fig. 125. 
 
 Fig. 126. 
 
 129. To construct a cylindrical helix it is necessary to know 
 the pitch and diameter of the cylinder on which the helix may be 
 wrapped. In Fig. 129 the pitch and diameter of the cylinder are 
 given. The plan view of the curve will be, of course, a circle and 
 
WARPED SURFACES l6l 
 
 may be drawn at once. To find the elevation, divide the pitch 
 distance into any number of convenient parts, say 24, and divide 
 the circle of the H projection into the same number of parts. 
 Let the generating point start in H at point o. In moving from 
 o to I the point makes one twenty-fourth of a complete turn 
 
 about the axis, so it therefore advances in the direction of the 
 axis one twenty-fourth of the pitch; i' then is the elevation of 
 its second position. In like manner all twenty-four positions 
 may be located and the curve plotted. 
 
 In drawing a conical helix the same principle applies except 
 
l62 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 that it must be remembered that the H projection will be a 
 spiral and not a circle and that, since the helix is on the surface 
 of a cone, the distance of the generating point from the axis 
 varies directly with the distance from its first position. Fig. 126 
 
 Fig. 128. 
 
 Fig. 129. 
 
 shows the construction of a conical hehx and the method of 
 finding its projections. 
 
 130. To draw a tangent to a helix. If in any given helix the 
 generating point starts from H, as in Fig. 125, it will be found 
 that the distance from any position of the point on the curve 
 back to H along the path of the helix is equal to the distance from 
 
WARPED SURFACES I 63 
 
 that same point back to H along the path of the tangent to the 
 helix at that point. This is true because the curve and its tan- 
 gent make the same angle with H, and it may be proved by 
 wrapping a string around a cylinder in the form of a helix. Now 
 if a portion of the string be unwound from the cylinder it will be 
 seen that the unwrapped portion is tangent to the portion of the 
 hehx still wrapped on the cyHnder, and that the end of the string 
 always remains in the same plane. This fact is made use of in 
 drawing a tangent to a helix. In Fig. 125 draw px perpendicular 
 to op, and make px equal to p-1. This line will be the H pro- 
 jection of the tangent to the helix and x will be the point where 
 the tangent pierces H, because px is the plan of a tangent Hne 
 equal in length to the length of the helix from H to P. p'x' then 
 will be the elevation of the tangent and will be tangent to the 
 elevation of the helix. 
 
 131. Proposition 41. Given the axis, the generatrix, and the 
 pitch to draw the plan and elevation of a helicoidal surface. 
 
 Discussion. A helicoidal surface may be best represented by 
 showing a plan and elevation of its axis, its base in H or V, and 
 a sufficient number of elements to define the contour of the sur- 
 face. For use in solving problems in connection with this surface 
 it is convenient to have also the plan and elevation of the helix 
 described by the end of the generating line. 
 
 Construction. In Fig. 130 let AB be the generating line in its 
 first position. Let the axis be the line through A perpendicular 
 to H, and let the distance that the point B rises during one 
 revolution of AB about the axis be equal to twice the distance 
 from 12' to the G. L. Draw the heHx described by point B 
 according to the directions in Article 129. As the point B takes 
 up the position on this helix indicated by the points i, 2, 3, etc., 
 the point A will remain in contact with the axis and will rise equal 
 distances, as shown by a'l, a'o, a'3, etc. By joining the successive 
 positions of A and B the positions of the generating line as it 
 moves about the axis may be shown. If the points where these 
 several positions of the generating line pierce H be found, as at 
 b, bi, bo, hz, etc., the curve of the base may be located. From an 
 inspection of the drawing it will be seen that the intersection of 
 
164 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 Fig. 130. 
 
WARPED SURFACES 
 
 165 
 
 the surface with H gives a curve called the Spiral of Archimedes. 
 In the drawing only a little over one-half of one revolution of the 
 line AB about the axis has been shown. 
 
 132. There are two general kinds of helicoidal surfaces both 
 of which have many practical uses. When the generatrix is in- 
 clined to the axis, as in Fig. 
 130, the resulting helicoid is 
 called an oblique helicoid; and 
 when the generatrix is per- 
 pendicular to the axis, as in 
 Fig. 134, the surface is called 
 a right helicoid. In its prac- 
 tical applications the heli- 
 coidal surface is not indefinite 
 in extent but is limited usually 
 by cylinders. Fig. 131 shows 
 a plan and elevation of an 
 oblique helicoidal surface in 
 the form of a screw thread. 
 Fig. 132 shows a right heli- 
 coidal surface such as is used 
 in cams. Practical examples 
 of helicoids may be found in 
 so-called "spiral" stairways, 
 screw conveyors, propellers, 
 screw threads, twist drills, etc. 
 
 133. Proposition 42. Given 
 the axis, the pitch, and gen- 
 eratrix to draw the plan and 
 elevation of a right helicoid. ^^*^- ^3i- 
 
 Discussion. Since nearly all right helicoids are necessarily 
 limited in extent for practical purposes the method by which they 
 are drawn may be shown by a practical illustration. The 
 principle involved will be, of course, the same for any similar 
 surface. 
 
 Construction. Let OP be the given axis in Fig. 133, and let 
 Oi be the given generatrix which advances the given pitch dis- 
 
i66 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 tance during one revolution. The point i in revolving about the 
 
 axis describes a circle whose plan view is shown at i 13 1, 
 
 and moves up or down the given pitch distance. Therefore, 
 divide the circle into 24 equal parts and the pitch distance into 
 
 Fig. 132 
 
 the same number of parts. As the point moves from i it will 
 also advance so that it will take up the position 2, 3, 4, etc., 
 thus describing a hehx; and at the same time O moves along 
 the axis so that the motion of the hne Oi generates a right 
 
WARPED SURFACES 
 
 167 
 
 helicoid. In the given figure the square iXYZ, of which the 
 line Oi is one side, has been revolved, thus cutting in the cyl- 
 inder a square thread, whose faces are right helicoids. 
 
 Fig. 134 shows a right helicoidal surface in the form usually 
 
1 68 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 used for conveyors. The flights are right helicoidal surfaces and 
 are mounted on a shaft. When the conveyor is enclosed in a 
 hollow cyhnder and revolved it is capable of delivering material 
 along the length of the enclosing casing. 
 
 134. When the axis of a helicoid becomes a cylinder the 
 surface becomes a convolute, and for this reason the convolute, 
 which is a single curved surface, is often called the developable 
 helicoid. The helicoid is the Hmiting form of the convolute, 
 
 for as the cylinder, about 
 which the directing helix 
 is wrapped, approaches a 
 straight line so does the 
 convolute surface approach 
 the hehcoid as a limit. A 
 hehcal convolute, then, is 
 generated by a Hne moving 
 tangent to a given heHx. 
 (See Fig. 135.) 
 
 135. Proposition 43. 
 Given the directing helix to 
 
 Fig. 135. -a string model of Tdevebpable draw the plan and elevation 
 
 helicoid. The strings are each tangent to of a COnvolute surface, 
 the heli.x of which they form the envelope, DiscuSsion. A COnvolute 
 
 and which may be seen in the illustration. . ,, i i •., 
 
 IS usually represented by its 
 base and the plan and elevation of the directing helix, and a 
 number of elements. Since the surface is generated by a line 
 moving tangent to the directing hehx, it becomes necessary to 
 draw a number of tangents to the helix and find where they 
 pierce H or V to get the base. 
 
 Construction. In Fig. 136, MN is the axis of the given helix 
 ABCD. By Article 130 draw a number of tangents to the helix 
 and find where these tangents pierce H at i, 2, 3, 4, etc. The 
 curve through these points will be the base in H. 
 
 136. In Fig. 137 is shown the plan and elevation of a con- 
 veyor, whose flights are made developable hehcoids. To draw 
 the plan and elevation of such a surface the diameter and pitch 
 of the smaller helix — the directrix — must be known, and also 
 
WARPED SURFACES 
 
 169 
 
 the length of the generatrix or the outside diameter of the con- 
 veyor flights. 
 
 Given these data the drawing is made as follows: 
 Tangent to the directing helix at B draw AB. A is the point 
 where the line, which is a position of the generatrix parallel to 
 
 Fig. 136. 
 
 V, pierces the limiting outside cyUnder; A, then, Is a point on 
 the Umiting outside helix. With A as one point construct a helix 
 with a pitch equal to the directing helix on the limiting outside 
 cylinder. If lines be drawn by connecting the points in order, 
 as A and B, etc., they will be elements of the surface. It 
 should be noted that with these same given data another con- 
 
lyo 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 volute may be constructed with AC as the generating line, in 
 which case the surface will slope in the opposite way to the sur- 
 face shown. 
 
 137. To develop a convolute. One flight, or turn, of a con- 
 volute such as shown in Fig. 137 becomes a portion of a fiat 
 
 DEVELOPMENT OF ONE FLIGHT 
 
 Fig. 137. 
 
 ring when developed. The radius R, of the inner circle of the 
 ring, is equal to the Hne ed since the tangents have a constant 
 slope. This length is obtained by drawing a perpendicular b'd 
 to the tangent a'b', and from the point e where a'b' cuts the 
 elevation of the Hmiting element of the inner cylinder a hori- 
 
WARPED SURFACES 171 
 
 zontal line ed. The length of this line ed is the graphical solu- 
 tion of the equation r sec- • 6, in which r is the radius of the inner 
 cylinder and d is the angle the tangent AB makes with H. The 
 radius R', of the outer circle of the developed ring, is found by 
 drawing AB tangent to the developed inner helix. Since AB is 
 the true length of an element the point A will therefore be a 
 point on the circumference of the outer circle, and through this 
 point this circle may be drawn. The length of this edge of the 
 ring will be the developed length of the outer helix and may be 
 found by solving the equation Vp- -|- 2 iri''^, in which P is the 
 pitch of the helix and r' is the radius of the plan view of the helix. 
 This length may be found graphically by laying off a right tri- 
 angle whose base is equal to the circumference of the plan view of 
 the helix and whose altitude is equal to the pitch. The hypothe- 
 nuse of this triangle is equal in length to the helix for one turn. 
 The length of the inner edge of the ring may be found in a similar 
 manner. 
 
 138. The helix is used extensively in the generation of other 
 surfaces of commercial importance. Of these, screw threads 
 and the so-called " spiral springs " are the best examples. Coiled 
 springs made up of rectangular stock are right helicoids while 
 those made up of round stock, such as the serpentine shown in 
 Fig. 128, are surfaces of double curvature. 
 
 139. Of warped surfaces of the third class, — those having 
 three linear directrices, — comparatively few are used in practical 
 work. These directrices may all be curved lines, or one may be 
 curved and two straight, or two may be curved and one straight, 
 or all may be straight. 
 
 The pecuharity of this last surface — which is called the hyper- 
 boloid of revolution of one nappe — is that it is the only sur- 
 face of revolution which is warped. It may be generated by 
 revolving a straight Hne about another not in the same plane 
 with it. This surface is a surface of revolution, and, since it is 
 possible to have the generatrix slope two ways and generate the 
 same surface, the surface is doubly ruled. (See Fig. 138.) It is 
 difi&cult to imagine this surface as being also generated by a line 
 moving so as to touch three other straight lines but that it can 
 
172 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 be done can be proved. If any three positions of the generating 
 Kne as it revolves about the axis be taken as directrices, and a 
 position of the generatrix of the opposite slope be moved so as 
 to touch them, it will follow the surface. 
 
 The surface is used to some extent in transmitting motion from 
 one shaft to another not in the same plane with it. The principle 
 
 by which this is accom- 
 plished is shown in Fig. 
 
 139- 
 
 If the line AB be revolved 
 about each shaft it will gen- 
 erate two hyperboloids hav- 
 ing a common element, or 
 AB. Naturally, then, when 
 these hyperboloids roll on 
 their respective axes they 
 will constantly be in contact 
 and therefore one may be 
 made to drive the other. 
 
 140. Another warped sur- 
 face of this class having two 
 curved directrices and one 
 
 Fig. 138. — A model of an hyperboloid of straight directrix is some- 
 one nappe showing by the position of the ^-j^ies met with in arches. 
 
 strings how the surface may be doubly t^ 1 
 
 j^lg^j If the two curves are semi- 
 
 circular, as in Fig. 140, and 
 lie in parallel planes, and if the straight line directrix MN 
 lies in a plane perpendicular to the planes of the curved 
 directrices through their centers the surface becomes a cow's 
 horn arch. 
 
 A similar surface of this same general character is the warped 
 cone, shown in Fig. 141. In this surface the curvilinear direc- 
 trices do not necessarily lie in parallel planes nor is the rectiUnear 
 directrix in a plane perpendicular to them. A common applica- 
 tion of this surface is shown in Fig. 141, where the warped cone 
 is used as a hood. The method of construction will be obvious 
 from an inspection of the figure. 
 
WARPED SURFACES 
 
 173 
 
 Fig. 139. 
 
174 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
WARPED SURFACES 
 
 175 
 
 141. Proposition 44. To develop a warped surface. 
 Discussion. While warped surfaces may not be truly devel- 
 oped, patterns for some warped surfaces may be laid out which, 
 
 Fig. 141, 
 
 when formed, approximate the original surface closely enough 
 for commercial purposes. An example of this is shown in the 
 development of the warped cone in Fig. 142. 
 
176 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
WARPED SURFACES 177 
 
 Construction. Make the development by the method of tri- 
 angulation as explained in Article 107. 
 
 142. There are many other forms of warped surfaces than 
 those which are here considered ; but as they are seldom encoun- 
 tered in practical work and possess, therefore, little interest to 
 the engineer or architect, it has been thought well to omit them. 
 The principles by which warped surfaces in general are repre- 
 sented and by which problems relating to such surfaces are solved 
 have been discussed in detail, and no peculiar difficulties should 
 be encountered in solving problems relating to the more obscure 
 surfaces omitted from consideration here. 
 
 PROBLEMS ON WARPED SURFACES 
 
 257. The plane T is o"; 90; 45. The point lies in plane T at i|"; 
 i^"; o" and is the center of a 2" circle also in the plane T. The plane S 
 is 5"; 90; 150. The point Q lies in the H trace of plane S in the profile 
 plane 3 J"; 90; 90 and is the center of a 3" circle lying in plane S. With 
 these two circles as directrices and the V plane as plane director draw a plan 
 and elevation of a cylindroid. 
 
 258. The plane S is o"; 90; 30. The point is in the H trace of plane S 
 and in the profile plane 2|"; 90; 90. This point is the center of a 2\" 
 circle lying in plane S. The line MN is perpendicular to V at the point 
 ^\"\ o"; — li"- With the circle and line as directrices and the V plane as 
 plane director draw a plan, elevation, and end view of a conoid. 
 
 259. The point A is o"; -3"; o". The point B is 3"; o"; -3". The 
 point C is 4"; o"; -\V'. The point D is 5^"; -3i"; o". With AB and 
 CD as directrices, and with the H plane as plane director draw the plan, 
 elevation, and end view of the hyperbolic paraboloid. 
 
 260. A cyhnder 2§" in diameter has cut on it a single square thread whose 
 pitch is I". Draw a plan and elevation. 
 
 261. A 4" screw has 3 standard V-threads per inch. Draw a plan and 
 elevation of the screw. 
 
 262. A bundle conveyor passes from the top floor of a building to the 
 wrapping counter in the basement. The distance between floors is 10' and 
 in passing from one floor to the floor below the bundle makes 3 turns. The 
 conveyor is in the form of a right helicoid 5' in diameter and mounted on a 
 2' shaft. Draw a plan and elevation of the conveyor. 
 
 263. Draw the plan and elevation of a screw conveyor 14" in diameter 
 mounted on a 3" shaft. The conveyor is capable of delivering 7500 cubic 
 feet per hour when revolving 100 revolutions per minute. 
 
178 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 264. Draw the plan and elevation of a helical spring made of |" square 
 stock. Inside diameter of spring 2\" \ length 6"; pitch i". 
 
 265. Draw the plan and elevation of the spring in Problem 264 made up 
 of I" round stock. 
 
 266. A conical spring is 6" in diameter at its base and 2" in diameter at 
 its top. It is made up of 4 turns of i" square stock. Draw the plan and 
 elevation of the spring. 
 
 267. Draw a plan and elevation and give the length of the stock required 
 to make the spring in Problem 266 out of f" round stock. 
 
 268. Draw the plan and elevation of a convolute conveyor whose diam- 
 eter is 18", whose pitch is 16", and whose shaft is 6" in diameter. 
 
 269. Draw a plan and elevation of a wood screw which is formed by two 
 convolute surfaces. Outside diameter 4", diameter at the root of the 
 thread 2", pitch 2" . 
 
 270. A railway which runs due north and south crosses over a street 
 which runs north 75 degrees east on a cow's horn arch. The openings in the 
 arch are circles 40' in diameter lying in parallel planes 80' apart. Draw a 
 plan, elevation and end view of the arch. 
 
 271. A reducing hood whose opening is a circle 36" in diameter and whose 
 outlet is a circle 8" in diameter is fastened to a wall so that the circular open- 
 ings are tangent to the wall. These tangent points are in a line which is 
 vertical and which is the straight Hne directrix of the surface. The length 
 of the hood along the wall is 42", the plane of the opening inclines 45 degrees 
 to the wall, and the plane of the outlet is perpendicular to the wall. Draw a 
 plan and elevation of the hood and make a pattern for it. 
 
 272. A reducing hood has two circular openings whose centers are in a 
 line perpendicular to H, and 36" apart. The outlet of the hood is a circle 
 18" in diameter whose plane is perpendicular to this line; and the inlet is a 
 circle whose diameter is 48" and whose plane inchnes 30 degrees to this 
 line. Draw a plan and elevation of this hood and make a pattern for it. 
 
 273. Make a pattern for a convolute conveyor flight whose diameter is 
 12", pitch 8", diameter of shaft 3". 
 
 274. Two shafts, one vertical and one inclined 45 degrees in a vertical 
 plane, are 2" in diameter, and their center lines are 7" apart. Draw the plan, 
 elevation, and end view of two hyperboloids of revolution which will operate 
 on these shafts in the ratio of 4 to 3. 
 
CHAPTER XVIII 
 TANGENT PLANES AND LINES 
 
 143. A plane is tangent to a ruled surface when it contains 
 one and only one element of the surface. The element common 
 to the tangent plane and the surface is called the element of 
 contact. 
 
 144. A plane is tangent to a double curved surface when it 
 contains one and only one point of the surface. The point 
 common to the tangent plane and the surface is called the 
 point of contact, or the point of tangency. 
 
 145. A line is tangent to a curve when it contains one and 
 only one point of the curve. The point common to the curve 
 and the line is called the point of tangency. 
 
 146. A plane which is tangent to a surface contains all of the 
 lines tangent to curves of the surface at the point where these 
 curves touch the tangent plane. In a single curved surface 
 such as a cone, for example, every line tangent to a curve on the 
 surface of the cone will lie in the plane tangent to the cone along 
 the element which passes through the point of tangency. Like- 
 wise in a double curved surface such as a sphere every circle 
 which passes through the point of contact will have a tangent 
 Hne which will lie in the plane tangent to the sphere at that 
 point. 
 
 147. Proposition 45. Given any single curved surface to 
 find the plane tangent to it at a given point on the surface. 
 
 Discussion. If an element of the surface be drawn through 
 the given point it will be the element of contact. Since this 
 line lies in the tangent plane it will pierce H and V in the corre- 
 sponding traces of that plane. The tangent plane will also con- 
 tain any tangent to any curve of the surface at the point where 
 this curve passes through the element of contact. Since the 
 
 179 
 
i8o 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 base of the given surface is such a curve it will be obvious that 
 the trace of the required plane will be tangent to that base. 
 
 Construction. Let the given surface in Fig. 143 be a cylinder 
 with its base in V and the given point be O. Draw the element 
 OX through O. This will be the element of contact and since 
 it pierces V at X this point will be on the V trace of the required 
 
 Fig. 143- 
 
 tangent plane. Through X draw Tt' tangent to the base of the 
 cylinder. This will be the V trace of the required tangent plane. 
 The H trace may be found by locating Q, the point where the 
 element of contact pierces H, and through this point and T 
 drawing the H trace Tt. Since Q is very often remote or inac- 
 cessible a more convenient way of finding the H trace is to 
 draw through O a line parallel to the V trace, as the line OP. 
 
TANGENT PLANES AND LINES l8l 
 
 Since OP is parallel to a line of the plane through a point in the 
 plane it will itself lie in the required plane and will, therefore, 
 pierce H in a point on the required H trace. P is this point 
 and Tt may be drawn through P and T. 
 
 148. Proposition 46. Given any single curved surface to find 
 a plane tangent to it through any given point not on its surface. 
 
 Discussion. If through the given point without the surface 
 a line be drawn parallel to the element of contact, or through 
 some point of the element of contact, this Hne will he in the 
 tangent plane and will pierce H and V in the corresponding 
 traces of the required plane. Through these piercing points 
 the traces may then be drawn as in the previous proposition. 
 
 Construction. When the given surface is a cyHnder. Draw 
 a line through the given point parallel to the elements of the 
 cylinder. This Hne will He in the tangent plane and will pierce 
 H and V in the required H and V traces of the tangent plane. 
 With these points known the traces may be located as in Propo- 
 sition 45. Let the drawing be made in accordance with these 
 suggestions. 
 
 Construction. When the given surface is a cone as in Fig. 
 144. Let be the given point. Join O with the apex Q. This 
 line OQ must He in the required tangent plane since it contains 
 O, the given point, and Q, the apex through which aU tangent 
 planes must pass. OQ pierces H at P and pierces V at M. 
 Through P draw Tt tangent to the base of the cone; this wiD 
 be the H trace of the required tangent plane. Through Q 
 draw a line QN paraUel to Tt; this wiU be a Hne in the required 
 plane. It pierces V at N. Through N and M draw t'T, the V 
 trace of the required plane. If the traces meet at the ground 
 line the work is checked. 
 
 Construction. When the given surface is a convolute. 
 Through the given point draw a Hne making the same angle 
 with H, or V, as the elements of the convolute make with H, or 
 V. Let this Hne generate a right cone whose base will be a 
 circle lying in the plane of the base of the given convolute. If 
 now a plane be passed tangent to this cone it will contain the 
 given point, and if this same plane be made tangent to the con- 
 
l82 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 volute it will be the required plane. To fit these conditions one 
 trace of the plane must be tangent both to the base of the cone 
 and to the base of the convolute. The other trace may be lo- 
 
 FlG. 144. 
 
 cated as in the foregoing propositions. Let the construction 
 be made in accordance with these suggestions. 
 
 149. Proposition 47. Given any single curved surface to 
 find a plane tangent to it which is parallel to any given line. 
 
 Discussion. Since the plane is to be tangent to the given 
 surface it must, therefore, contain an element of the surface. 
 Also since it is to be parallel to a given line it must contain a 
 line parallel to the given line. Since the two lines may be made 
 to intersect they will determine the tangent plane as indicated 
 in the following constructions. 
 
 Construction, ^^^^en the given surface is a cylinder. Through 
 any point on the given line draw a line parallel to the elements 
 
TANGENT PLANES AND LINES 
 
 183 
 
 of the cylinder. The plane of these two lines will be parallel to 
 the required tangent plane since this plane contains two Knes par- 
 allel to two lines of the required plane. Parallel to this plane and 
 tangent to the given cyhnder the required plane may be drawn. 
 Let the drawing be made in accordance with these suggestions. 
 Construction. When the given surface is a cone. Through 
 
 the apex of the cone draw a Hne parallel to the given line. 
 Through this Hne and tangent to the cone pass a plane. This 
 plane will be the required plane since it is tangent to the cone 
 and contains a line parallel to the given line. Let the drawing 
 be made in accordance with these suggestions. 
 
 Construction. Let the given surface be a convolute as in 
 Fig. 145. PQ is the given line to which the tangent plane is to 
 
1 84 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 be parallel. Assume the point O on this line and through O 
 draw OM parallel to BN, one of the elements of the convolute. 
 Let the line OM generate a right cone whose base is a circle 
 in V. Now tangent to the cone thus generated and through 
 the line PQ pass the plane R. This plane will be parallel to the 
 required plane since it contains the given line and also a line 
 parallel to the element of contact on the convolute. Since the 
 elements of the cone generated by OM make the same angle 
 with V that the elements of the convolute do it is evident that 
 the element of this cone along which rRr' is tangent will be 
 parallel to the element of contact of the convolute. Therefore, 
 if tTt' be drawn parallel to rRr' and tangent to the convolute, 
 it will be the required tangent plane. 
 
 150. Proposition 48. Given any single curved surface and 
 its intersection with a plane to find the line tangent to this curve 
 at a given point. 
 
 Discussion. Since the curve of intersection Hes in a plane 
 the tangent line will also lie in this plane. This required line 
 will also lie in the plane which is tangent to the given surface 
 at the given point. If, therefore, a plane be passed tangent to 
 the given surface it will cut from the plane in which the given 
 intersection lies a line which will be tangent to the given curve 
 at the given point. 
 
 Construction. Let the given surface be a cone as in Fig. 146 
 with the given curve lying in plane sSs'. Let O be the point 
 at which it is required to draw a tangent to the curve of inter- 
 section. Pass the plane rRr' tangent to the cone at the point 
 O (Proposition 45). Find the intersection of plane R with 
 plane S (Proposition 8). This line is MN and it is tangent to 
 the given curve at O. 
 
 151. This same method will apply to any problem wherein 
 it is required to find the line tangent to any curve cut from a 
 surface by a plane. Stated in general terms the method may 
 be given as follows: 
 
 Pass a plane tangent to the surface at the given point. The 
 intersection of this tangent plane with the plane cutting the given 
 curve will always be tangent to this curve at the given point. 
 
TANGENT PLANES AND LINES 
 
 l8S 
 
 152. Proposition 49. Given any warped surface to find the 
 plane tangent to it at a given point on the surface. 
 
 Discussion. The required tangent plane will contain the ele- 
 ment of the surface through the point (the element of contact) 
 
 Fig. 146. 
 
 and the line tangent to any curve of the surface passing through 
 the given point. These two lines will determine the required 
 plane. 
 
 Construction. Let the given surface be an obhque helicoid 
 as in Fig. 147 and let the point at which the tangent plane is to 
 be drawn be A. Through A draw the element A2; this will be 
 one line of the required plane. At A draw also AM tangent to 
 the helix passing through A. (Art. 130.) This line will also lie 
 
1 86 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 in the tangent plane. Through AM and A 2 pass the plane tTt' 
 which will be the required tangent plane. 
 
 It should be kept clearly in mind that a plane tangent to a 
 warped surface at one point may cut the surface at a point very 
 
 near this point. And also that while the element of contact 
 A2 Hes in the required tangent plane this plane is 7iot tangent all 
 along this element. From this it should be clear that in Fig. 147 
 tT is not tangent to the H base of the helicoid, and that the 
 
TANGENT PLANES AND LINES 187 
 
 tangent plane at any point will contain only lines tangent to 
 curves of the surface through that point. 
 
 153. Proposition 50. Given any warped surface to find the 
 plane tangent to it through any given point not on its surface. 
 
 Discussion. In general an indefinite number of planes may 
 be passed through a pomt not on the surface tangent to a warped 
 surface since an indefinite number of lines may be drawn through 
 the given point tangent to curves of the surface. Any one of 
 these tangent lines with the element which passes through the 
 point of tangency will determine a plane which will contain the 
 given point and be tangent to the given surface. 
 
 Construction. To find any one of these tangent planes pass 
 a plane through the given point cutting a curve from the surface. 
 Through the point draw a tangent to this curve. Through the 
 point of contact of this tangent line draw an element of the 
 given surface. The tangent line and this element will deter- 
 mine a plane which will meet the required conditions. Let the 
 drawing be made in accordance with these suggestions. 
 
 154. Proposition 51. Given any warped surface to find the 
 plane tangent to it which is parallel to a given line. 
 
 Discussion. If a plane be passed through the given surface 
 parallel to the given line it will cut from the surface a curve to 
 which it will be possible to draw a tangent which will be parallel 
 to the given line. This tangent together with the element of 
 the given surface through the point of contact of this tangent 
 will determine a plane which will meet the required conditions. 
 Since an mdefinite number of planes may be passed through the 
 surface parallel to the given lines there will be an equal number 
 of possible tangent planes. 
 
 Construction. Let the construction be made in accordance 
 with the given discussion. 
 
 155. Proposition 52. Given any double curved surface to 
 find the plane tangent to it at a point on the surface. 
 
 Discussion. The plane tangent to a double curved surface 
 will contain all the lines tangent to curves of the surface passing 
 through the given point. If any two of these tangent Hnes be 
 found they will determine the tangent plane at that point. 
 
l88 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 Construction. Let the given surface be an annular torus as 
 in Fig. 148 and let the given point on its surface be O. Through 
 O draw the Une OQ tangent to the horizontal circle on the 
 surface of the torus passing through O. Also draw the line OM 
 tangent to that circle of the surface which lies in the meridian 
 
 Fig. 1 48. 
 
 plane through O. (This tangent may be found by revolving O 
 into the meridian plane parallel to V. In this position of O the 
 tangent OP may be drawn and then revolved back to its true 
 position with P, on the axis of revolution, remaining sta- 
 tionary.) These two tangent lines intersecting at the given 
 
TANGENT PLANES AND LINES 189 
 
 point O determine the required plane and the traces may be 
 found by locating the piercing points of these Unes, at M, Q, 
 and N. 
 
 156. Proposition 53. Given any double curved surface to 
 find the plane tangent to it through any given point not on the 
 surface. 
 
 Discussion. If the given point be taken as the apex of a 
 cone whose elements are all tangent to curves lying on the 
 given surface it will be apparent that any plane tangent to this 
 cone will also be tangent to the given surface. Such a plane, 
 since it contains the given point and is tangent to the given 
 surface, meets the required conditions. From this it may be 
 seen that an indefinite number of planes may be passed through 
 a point tangent to a surface. 
 
 157. Proposition 54. Given any double curved surface to 
 find the plane tangent to it and parallel to any given hne. 
 
 Discussion. If a series of lines be drawn tangent to curves 
 lying on the given surface and parallel to the given line these 
 Hnes wiU be the elements of a cylindrical surface since they are 
 all parallel. Any plane then tangent to this cylindrical surface 
 will also be tangent to the given surface and parallel to the given 
 Hne. Since there are an infinite number of such planes there 
 are an indefinite number of planes which will meet the required 
 conditions. 
 
 158. Proposition 55. Given any two surfaces and their curve 
 of intersection to find a Hne tangent to this curve at a given 
 point. 
 
 Discussion. Since the Hne is tangent to a curve of each 
 surface it must He in a plane tangent to the surface at the given 
 point. If, therefore, planes be passed tangent to the given sur- 
 faces at the given point and their Hne of intersection found this 
 Hne will be tangent to the curve of intersection of the two sur- 
 faces at the given point. 
 
 Construction. Let the given surfaces be two cylinders as 
 shown in Fig. 149 and let O be the point on their curve of inter- 
 section at which a tangent is to be drawn. Through O pass a 
 plane tangent to each cyHnder (Proposition 45). Plane sSs' is 
 
190 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 Fig. 149. 
 
TANGENT PLANES AND LINES 
 
 191 
 
 tangent to the right hand cylinder and plane tTt' is tangent to 
 the left hand cylinder. Find the intersection of these two 
 tangent planes (Proposition 8). One pair of traces intersect at 
 N and the other pair at M thus gi\'ing MN as the line of inter- 
 section. When extended MN passes through the given point 
 and the Hne MO is tangent to the curve of intersection at the 
 given point. 
 
 PROBLEMS ON TANGENT PLANES AND TANGENT LINES 
 
 275. Assume a cylinder with a circular base in H and with its axis oblique 
 to H and piercing V. Pass a plane tangent to this cylinder at any assumed 
 point. Is the V trace of the tangent plane tangent to the V base of the 
 cylinder? 
 
 276. Assume a cone whose base is a circle in V and whose axis is oblique 
 to H and V. Pass a plane tangent to this cone at any assumed point on 
 its surface. 
 
 277. Assume a cylinder with its base in 4 and its axis oblique to H and 
 V. Through any point assumed outside the surface of the cylinder pass a 
 plane tangent to the surface. Give both solutions. 
 
 278. Assume a right circular cylinder with its axis parallel to the GL. 
 Pass a plane tangent to this cylinder through a point assumed outside its 
 surface. Give both solutions. 
 
 279. Assume a right circular cone with its base in H. Assume a line 
 MN. Find the plane tangent to the cone which will be parallel to MN. 
 (There are some positions of the line MN which would make the problem 
 impossible. What are they?) 
 
 280. At some point on the curve of intersection in Problem 193 draw a 
 line tangent to the curve. 
 
 281. Assume a point on the curve of intersection in Problem 215 and 
 at this point draw a line tangent to the curve. 
 
 282. The axis of an oblique helicoid is perpendicular to H 3" in front 
 of V. The generating line inclines 60 degrees to H and pierces H 2" from 
 the foot of the axis in its initial position. The pitch is 3". Pass a plane 
 tangent to the helicoid at a point which the initial piercing point of the 
 generatrix would assume after one-eighth of a turn. 
 
 283. Pass a plane tangent to an assumed point on the surface of the 
 conoid in Problem 25(8. 
 
 284. Given a sphere to pass a plane tangent to it at a given point on its 
 surface. Work this problem by an original method. (Hint — a plane 
 tangent to a sphere will be perpendicular to the radius at the point of 
 contact.) 
 
 285. Assume an ellipsoid of revolution and a point on its surface. Pass 
 a plane tangent to the ellipsoid at the given point. 
 
192 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 286. Assume an hyperboloid of revolution similar to the one shown in 
 Fig. 138 and pass a plane tangent to it at any point on its surface. 
 
 287. Assume a point on the curve of intersection of any of the single 
 curved surface problems in Chapter XV and pass a line tangent to the 
 curve at this point. 
 
 288. Assume a point of the curve of intersection of any of the problems 
 in Chapter XVI and find the line tangent to the curve at this point. 
 
CHAPTER XIX 
 
 MODEL MAKING 
 
 159. One of the most interesting and practical means of 
 studying the development of surfaces and the laying out of 
 patterns for them consists in the construction of actual scale 
 models. This not only affords practice in the use of drawings 
 and gives experience in constructing the object represented in 
 the drawing, but it is also an excellent means of showing how 
 curves of intersection appear when laid on a pattern as well as 
 
 Fig. 150. — A paper model of a blast furnace piping problem constructed to scale 
 from ordinary detail paper and mounted on a wooden base. 
 
 when they are an actual part of the object on which they lie. 
 While such model making must of course be done to scale yet 
 the experience acquired serves to illustrate many of the practical 
 considerations which must be kept in mind when laying out 
 sheet metal work. 
 
 160. Practical Hints. In making paper models such as the 
 one shown in Fig. 150 a good grade of detail paper will give satis- 
 factory results. After the developments of the several surfaces 
 have been accurately made they may be cut out — leaving an 
 allowance along the proper edges of the pattern for lap — and 
 glued together, thus forming the actual surface. 
 
 193 
 
194 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 Fig. 151. — Illustration showing two methods of making joints in paper model 
 making. The method at the right gives better results for joints along straight 
 lines while the one at the left is better for curved joints. 
 
 Fig. 152. — Patterns for the parts of the model shown in Fig. 155 before being 
 glued together Note the method for making the joints. 
 
MODEL MAKING 
 
 195 
 
 Two convenient methods for making the joint along an element 
 of the surface are shown in Fig. 151. The joint in the cylinder 
 on the right is overlapped and on the one side of the pattern 
 tongues are cut to fit into the slits on the other side of the pattern. 
 This method has been found to give good results and is some- 
 what easier than the one shown on the cylinder at the left in 
 Fig. 151. For joints of two intersecting surfaces it will be found 
 more convenient to cut tongues on both patterns, the tongues on 
 one side being arranged to fit into spaces between the tongues 
 on the other as shown in the cylinder 
 at the left in Fig. 151. It requires 
 a little care and foresight to get the 
 tongues and spaces to knit but the 
 resulting joint is so much easier to 
 form and is so much more secure 
 that the extra trouble is worth the 
 time it takes. 
 
 Both library paste and glue are 
 good " solder " for the joints. 
 Library paste is pleasanter to work 
 with and takes less time to dry but 
 it seems to hold less well than glue. 
 A glued joint has been found to last 
 indefinitely but joints fastened with 
 paste are apt to break open in time. 
 
 161. The tendency of surfaces 
 such as cones and cylinders when 
 formed from detail paper patterns is to be flat along the 
 joints. This is due to the extra stiffness at these points caused 
 by overlapping edges. An easy method of overcoming this and 
 securing rigidity in the model is to insert cardboard reinforce- 
 ments as shown in Fig. 153. These cardboard reinforcements 
 may be made in the form of right sections and glued in place. 
 The model will thus be made to keep its proper shape and will 
 be much stiflfer. 
 
 162. Naturally a model such as the one shown in Fig. 154 will 
 get more or less soiled from handling during its construction and 
 
 Fig. 153. — A view of the model 
 shown in Fig. 154 cut open to 
 show the method of securing 
 stiffness and true form by in- 
 serting cardboard forms. 
 
196 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 its appearance may be greatly improved if upon completion it 
 is given a coat of bronze or silver paint. This covers the glue 
 marks and gives the model a fresh and finished appearance at a 
 trifling cost. 
 
 Fig. 154. — A paper model. A close 
 examination of the opening in the 
 cowl will show the method used 
 to make the joints. 
 
 Fig. 155. — Another paper 
 model. 
 
 Fig. 156. — A collection of paper models made by the methods just described. 
 \\'ith the exception of two or three all of these models were made by students 
 during one course. 
 
 The models which are used here for illustration, both the 
 finished models and the ones which have been cut open to show 
 methods of construction, were made by students in the regular 
 required course in descriptive geometry at the University of 
 
MODEL MAKING 
 
 197 
 
 Iowa. This fact is mentioned to impress upon students who use 
 this text that work of this character and quahty is quite within 
 their capacity. 
 
 GENERAL PROBLEMS 
 
 289 302. Draw a plan and elevation of the object shown in the drawing 
 and lay out patterns for each part. After the patterns have been laid out 
 to the same scale as the drawing, they are to be fitted together and glued tc 
 form a model of the object. 
 
 Problem 289. Hopper 
 
 
 A 
 
 B 
 
 C 
 
 D 
 
 E 
 
 F 
 
 G 
 
 a 
 
 12" 
 
 12" 
 
 2" 
 
 5" 
 
 6" 
 
 5" 
 
 2" 
 
 b 
 
 36" 
 
 18" 
 
 6" 
 
 0" 
 
 18" 
 
 0" 
 
 6" 
 
 c 
 
 36" 
 
 18" 
 
 6" 
 
 0" 
 
 18" 
 
 IS" 
 
 6" 
 
 d 
 
 36" 
 
 18" 
 
 6" 
 
 6" 
 
 18" 
 
 0" 
 
 6" 
 
 e 
 
 30" 
 
 12" 
 
 4" 
 
 4" 
 
 12" 
 
 9" 
 
 12" 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
igS ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 
 
 
 
 
 -i 
 
 
 k 
 
 
 
 ; 
 
 
 
 B 
 
 
 
 
 
 
 y 
 
 f 
 
 < — C > 
 
 
 
 
 
 ^"^^^^ 1 
 
 D 
 
 
 ^^^^^^ 
 
 
 
 
 ^-__^ 
 
 
 F 
 
 
 < — G — > 
 
 
 
 
 ^-L- 
 
 
 
 
 ' 
 
 II 
 
 
 
 Problem 290. Rectangular Tapered Connecting Pipe 
 
 
 A 
 
 B 
 
 C 
 
 D 
 
 E 
 
 F 
 
 G 
 
 H 
 
 L 
 
 K 
 
 a 
 
 12" 
 
 6" 
 
 6" 
 
 12" 
 
 6" 
 
 6" 
 
 6" 
 
 24" 
 
 6" 
 
 3" 
 
 b 
 
 12" 
 
 3" 
 
 6" 
 
 9" 
 
 8" 
 
 6" 
 
 5" 
 
 36" 
 
 6" 
 
 4^" 
 
 c 
 
 12" 
 
 4" 
 
 6" 
 
 9" 
 
 8" 
 
 6" 
 
 5" 
 
 36" 
 
 6" 
 
 4" 
 
 d 
 
 12" 
 
 6" 
 
 6" 
 
 12" 
 
 6" 
 
 6" 
 
 6" 
 
 24" 
 
 6" 
 
 6" 
 
 e 
 
 10" 
 
 10" 
 
 6" 
 
 6" 
 
 3" 
 
 3" 
 
 6" 
 
 20" 
 
 6" 
 
 0" 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
MODEL MAKING 
 
 199 
 
 Problem 291. Twisted Connecting Pipe 
 
 
 A 
 
 B 
 
 C 
 
 D 
 
 E 
 
 F 
 
 G 
 
 a 
 
 12" 
 
 6" 
 
 12" 
 
 6" 
 
 3" 
 
 3" 
 
 22" 
 
 b 
 
 18" 
 
 12" 
 
 20" 
 
 9" 
 
 6" 
 
 6" 
 
 30" 
 
 c 
 
 8" 
 
 8" 
 
 12" 
 
 6" 
 
 4" 
 
 4" 
 
 24" 
 
 d 
 
 16" 
 
 8" 
 
 16" 
 
 8" 
 
 6" 
 
 6" 
 
 30" 
 
 e 
 
 . 8" 
 
 4" 
 
 8" 
 
 4" 
 
 6" 
 
 6" 
 
 24" 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
200 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 
 
 
 
 
 <r A 
 
 ;> 
 
 
 ^ 
 
 
 D 
 
 ^^ 
 
 
 
 /> 
 
 ^ 
 
 X 
 
 \ 
 
 V 
 
 
 
 / 
 
 ( 
 
 1 
 
 
 
 / 
 
 
 
 ■" 
 
 
 
 
 
 
 ^ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Problem 292. Conical Connection for Offset Cylinders 
 
 
 A 
 
 B 
 
 C 
 
 D 
 
 E 
 
 X 
 
 Notes 
 
 a 
 
 8" 
 
 16" 
 
 8" 
 
 8" 
 
 14" 
 
 60 
 
 
 b 
 
 12" 
 
 18" 
 
 12" 
 
 10" 
 
 12" 
 
 45 
 
 
 c 
 
 8" 
 
 16" 
 
 8" 
 
 8" 
 
 14" 
 
 30 
 
 
 d 
 
 12" 
 
 18" 
 
 12" 
 
 10" 
 
 12" 
 
 30 
 
 
 e 
 
 8" 
 
 14" 
 
 10" 
 
 8" 
 
 18" 
 
 60 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
MODEL MAKING 
 
 20I 
 
 E 
 -L 
 
 ^^<^ 
 
 ) i_ 
 
 Problem 293. Hopper for Pipe 
 
 
 A 
 
 B 
 
 C 
 
 D 
 
 E 
 
 F 
 
 G 
 
 X 
 
 Pipe 
 Section 
 
 a 
 
 12" 
 
 6" 
 
 6" 
 
 10" 
 
 3" 
 
 5" 
 
 10" 
 
 90 
 
 Rect. 
 
 b 
 
 15" 
 
 15" 
 
 10" 
 
 18" 
 
 5" 
 
 9" 
 
 12" 
 
 90 
 
 Circle 
 
 c 
 
 6" 
 
 6" 
 
 6" 
 
 10" 
 
 3" 
 
 s" 
 
 10" 
 
 90 
 
 Circle 
 
 d 
 
 15" 
 
 15" 
 
 10" 
 
 18" 
 
 5" 
 
 9" 
 
 12" 
 
 120 
 
 Circle 
 
 e 
 
 9" 
 
 9" 
 
 6" 
 
 10" 
 
 3" 
 
 5" 
 
 10" 
 
 75 
 
 Circle 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Note: In all of the above problems the axis of the hopper remains vertical. 
 
2C2 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 Problem 294. Conical Hopper on Cylindrical Pipe 
 
 
 A 
 
 B 
 
 C 
 
 D 
 
 E 
 
 X 
 
 Y 
 
 a 
 
 12" 
 
 12" 
 
 14" 
 
 12" 
 
 12" 
 
 90 
 
 90 
 
 b 
 
 12" 
 
 12" 
 
 12" 
 
 8" 
 
 8" 
 
 90 
 
 90 
 
 c 
 
 18" 
 
 12" 
 
 20" 
 
 12" 
 
 18" 
 
 75 
 
 90 
 
 d 
 
 12" 
 
 12" 
 
 12" 
 
 8" 
 
 8" 
 
 90 
 
 75 
 
 e 
 
 12" 
 
 12" 
 
 20" 
 
 12" 
 
 18" 
 
 90 
 
 90 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
MODEL MAKING 
 
 203 
 
 Problem 295. Intersection of Cylindrical Pipes 
 
 
 A 
 
 B 
 
 C 
 
 D 
 
 E 
 
 X 
 
 Y 
 
 a 
 
 12" 
 
 12" 
 
 16" 
 
 6" 
 
 6" 
 
 45 
 
 90 
 
 b 
 
 iS" 
 
 18" 
 
 20" 
 
 10" 
 
 10" 
 
 60 
 
 90 
 
 c 
 
 10" 
 
 10" 
 
 20" 
 
 6" 
 
 6" 
 
 45 
 
 90 
 
 d 
 
 12" 
 
 12" 
 
 20" 
 
 9" 
 
 9" 
 
 60 
 
 90 
 
 e 
 
 18" 
 
 18" 
 
 20" 
 
 10" 
 
 10" 
 
 105 
 
 90 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
204 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 Problem 296. Reducing Breeching 
 
 
 A 
 
 B 
 
 C 
 
 D 
 
 E 
 
 F 
 
 G 
 
 H 
 
 X 
 
 Y 
 
 a 
 
 6" 
 
 6" 
 
 14" 
 
 8" 
 
 8" 
 
 8" 
 
 6" 
 
 6" 
 
 45 
 
 45 
 
 b 
 
 6" 
 
 8" 
 
 18" 
 
 12" 
 
 18" 
 
 10" 
 
 6" 
 
 10" 
 
 60 
 
 45 
 
 c 
 
 8" 
 
 6" 
 
 16" 
 
 8" 
 
 8" 
 
 8" 
 
 6" 
 
 4" 
 
 60 
 
 45 
 
 d 
 
 8" 
 
 8" 
 
 18" 
 
 12" 
 
 12" 
 
 12" 
 
 8" 
 
 8" 
 
 30 
 
 30 
 
 e 
 
 6" 
 
 6" 
 
 16" 
 
 10" 
 
 10" 
 
 10" 
 
 8" 
 
 8" 
 
 30 
 
 60 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
MODEL MAKING 
 
 20.^ 
 
 Problem 297. Connecting Rod End 
 
 
 A 
 
 B 
 
 C 
 
 D 
 
 E 
 
 D' 
 
 R 
 
 a 
 
 3" 
 
 6" 
 
 12" 
 
 il" 
 
 8" 
 
 21" 
 
 li" 
 
 b 
 
 4" 
 
 8" 
 
 12" 
 
 4" 
 
 10" 
 
 5" 
 
 3" 
 
 c 
 
 3" 
 
 6" 
 
 12" 
 
 -4" 
 
 8" 
 
 3§" 
 
 2" 
 
 d 
 
 4" 
 
 8" 
 
 10" 
 
 2" 
 
 10" 
 
 3" 
 
 4" 
 
 e 
 
 3" 
 
 6" 
 
 8" 
 
 ir' 
 
 6" 
 
 3" 
 
 2" 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Note: Make model of rectangular part only. 
 
2o6 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 Problem 2c 
 
 Cowl for Ship's Ventilator 
 
 
 A 
 
 B 
 
 C 
 
 D 
 
 E 
 
 -y- Number 
 of Sections 
 
 a 
 
 12" 
 
 6" 
 
 18" 
 
 8" 
 
 14" 
 
 I OS 
 
 Five 
 
 b 
 
 18" 
 
 9" 
 
 24" 
 
 9" 
 
 18" 
 
 los 
 
 Five 
 
 c 
 
 18" 
 
 9" 
 
 24" 
 
 9" 
 
 14" 
 
 105 
 
 Five 
 
 d 
 
 12" 
 
 6" 
 
 18" 
 
 8" 
 
 14" 
 
 90 
 
 Two 
 
 e 
 
 14" 
 
 7" 
 
 20" 
 
 9" 
 
 12" 
 
 105 
 
 Five 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
MODEL MAKING 
 
 207 
 
 Problem 299. Transition Piece 
 
 
 A 
 
 B 
 
 C 
 
 D 
 
 E 
 
 F 
 
 G 
 
 H 
 
 a 
 
 12" 
 
 4" 
 
 8" 
 
 4" 
 
 6" 
 
 12" 
 
 6" 
 
 6" 
 
 b 
 
 12" 
 
 4" 
 
 8" 
 
 4" 
 
 6" 
 
 12" 
 
 3" 
 
 6" 
 
 c 
 
 12" 
 
 4" 
 
 8" 
 
 4" 
 
 6" 
 
 8" 
 
 4" 
 
 3 
 
 d 
 
 12" 
 
 4" 
 
 8" 
 
 4" 
 
 6" 
 
 12" 
 
 3" 
 
 3" 
 
 e 
 
 12" 
 
 4" 
 
 8" 
 
 4" 
 
 4" 
 
 12" 
 
 6" 
 
 10" 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
2o8 
 
 ESSENTIALS OE DESCRIPTIVE GEOMETRY 
 
 -;;- 
 
 Problem 300. Ventilator Cowl with Y-Connection and Square Hood 
 
 
 A 
 
 B 
 
 6" 
 6" 
 6" 
 
 C 
 
 D 
 
 E 
 
 F 
 
 G 
 
 H 
 
 K 
 
 L 
 
 4" 
 4" 
 6" 
 
 M 
 
 18" 
 18" 
 18" 
 
 N 
 
 4" 
 4" 
 6" 
 
 
 
 8" 
 8" 
 8" 
 
 X 
 
 45 
 60 
 60 
 
 Y 
 
 45 
 45 
 60 
 
 Z 
 
 
 
 a 
 
 32" 
 
 12" 
 
 26" 
 
 18" 
 
 16" 
 
 12" 
 
 18" 
 
 8" 
 
 105 
 
 b 
 
 32" 
 
 12" 
 
 26" 
 
 18" 
 
 16" 
 
 12" 
 
 18" 
 
 8" 
 
 105 
 
 c 
 
 36" 
 
 12" 
 
 30" 
 
 14" 
 
 18" 
 
 10" 
 
 12" 
 
 12" 
 
 105 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
MODEL MAKING 
 
 209 
 
 Problem 301. Square Hood with Y-Connection, Elbow, and 
 Tr-ansition Piece 
 
 
 A 
 
 B 
 
 6" 
 6" 
 6" 
 
 C 
 
 D 
 
 E 
 
 F 
 
 G 
 
 8" 
 8" 
 
 8" 
 
 H 
 
 K 
 
 L 
 
 8" 
 8" 
 8" 
 
 M 
 
 8" 
 8" 
 
 8" 
 
 N 
 
 
 
 8" 
 8" 
 8" 
 
 P 
 
 8" 
 6" 
 
 8" 
 
 X 
 
 Y 
 
 45 
 45 
 60 
 
 Z 
 
 60 
 
 45 
 45 
 
 a 
 
 32" 
 
 12" 
 
 26" 
 
 iS" 
 
 16" 
 
 18" 
 
 12" 
 
 18" 
 
 45 
 
 b 
 
 32" 
 
 XT." 
 
 26" 
 
 18" 
 
 16" 
 
 20" 
 
 10" 
 
 18" 
 
 60 
 
 c 
 
 32" 
 
 12" 
 
 26" 
 
 18" 
 
 16" 
 
 20" 
 
 10" 
 
 18" 
 
 75 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
2IO 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 O 
 
 u 
 
CHAPTER XX 
 
 APPENDIX 
 
 The following geometrical constructions will be found useful 
 in drawing many of the problems in this text. While it is under- 
 stood that many of them are well known problems in plane 
 geometry it has been thought well to include them here for the 
 convenience of the student. 
 
 It is to be understood that no attempt has been made to prove 
 any of the constructions, and also that the particular construc- 
 tion given may not be the only 
 method of arriving at the same 
 result. 
 
 The attempt has been made to 
 draw the figure in such a way 
 that the construction will be 
 more or less obvious, and there- 
 fore the explanation has been 
 made as brief as possible. The 
 methods here given have been 
 found by experience to be the 
 most practical constructions for 
 this kind of work. 
 
 Given the length of one side to construct a regular pentagon. 
 
 Let AB be the given side. With AB as a radius and B as a 
 center draw the arc NPAQC. With A as a center and AB as the 
 radius draw the arc MPBQE. With P as a center and the same 
 radius draw the arc MAOBN. Draw the hues PQ, MC, NE, 
 thus locating E and C. With E and C as centers and AB as 
 radius draw arcs intersecting at D. ABDCE, then, is the re- 
 quired regular pentagon. 
 
212 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 Given the length of one side to construct a regular hexagon. 
 
 Let AB be the given side. With A and B as centers and AB 
 as radii draw arcs intersecting at O. This will be the center of 
 the hexagon. With O as a center and a radius equal to OA draw 
 the circumscribing circle. Lay off chords on this circle equal in 
 length to AB, thus locating points C, D, E, and F. ABCDEF 
 is the required hexagon. 
 
 E _ P D 
 
 Given the distance " across the flats " to construct a regular 
 hexagon. 
 
 Let MN and RS be two hues representing the sides of a hexa- 
 gon. Draw PQ perpendicular to MN and RS, and bisect it by 
 the perpendicular TL. The point O is the center of the inscribed 
 
 circle of the hexagon. Draw 
 OV making an angle of 60 
 degrees with OQ. Draw also 
 the inscribed circle VQP with 
 O as a center and OQ as a 
 radius. At V draw AF tan- 
 gent to OV; AF is the length 
 of one side of the required 
 hexagon which may now be 
 constructed by any convenient 
 method. 
 
 Given the length of one side to construct a regular polygon 
 of any given number of sides. 
 
 Let AB be the given side and let it be required to draw a poly- 
 gon with seven sides. With A as a center and a radius equal to 
 AB draw the arc BOGP. Divide this arc into seven equal parts 
 
APPENDIX 
 
 213 
 
 and through the second point, G, draw the line AG. This will 
 be the side of the polygon adjacent to AB. Bisect AB and AG 
 by perpendiculars meeting at O. This point O will be the center 
 of the circumscribing circle. Draw this circle and lay off on it 
 seven chords each equal to AB, thus locating the points C, D, E, 
 and F. ABCDEFG, 
 then, is the required 
 polygon of seven sides. 
 
 To draw a tangent 
 to a circle from a given 
 point without the cir- 
 cumference. 
 
 Let MANB be the 
 given circle whose cen- 
 ter is at O, and let P 
 
 be the given point from which the tangent is to be drawn. Draw 
 OP and upon OP as a diameter describe the circle OABP cutting 
 the given circle at the points A and B. Draw PA and PB ; these 
 will be the required tangents from P and A and B will be the 
 tangent points. 
 
 To draw a line tangent to two given circles. 
 
 Let MANB be one circle and CD the other circle to which the 
 Hne is to be drawn tangent. Draw PO connecting the centers. 
 Lay off NI equal to PD and describe the circle HIJK. This 
 circle is equal in radius to the difference between the radii of the 
 two given circles. 
 
 Now by the previous problem draw the tangents PJ and PH 
 
214 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 from P to the circle HIJK. Extend OH to cut the given circle 
 at A and extend OJ to cut at B. A and B will be the points of 
 
 tangency and AC and BD drawn 
 from these points parallel to 
 HP and JB will be the two lines 
 which may be drawn tangent to 
 the two given circles. 
 
 To draw an ellipse by tram- 
 mels, given the axes. 
 
 Let MN be the major axis and 
 OP the minor axis of the ellipse. 
 Cut a strip of paper as shown and 
 lay off AC equal to half of MN 
 and lay off BC equal to half of OP. Now keeping the point A 
 always on the axis OP and the point B always on the axis MN 
 move the trammel about and mark the various positions of C. 
 A sufficient number of points may be thus marked to enable the 
 curve to be accurately drawn 
 in with the irregular curve as 
 a guide. 
 
 To locate the foci of the 
 ellipse. 
 
 With P as a center and 
 half of MN as a radius draw 
 arcs cutting MN at F and 
 F', the required foci. 
 
 To draw a tangent to an 
 ellipse at a point on the 
 curve. 
 
 Let L be the given point. 
 Draw LF and LF', the focal 
 radii. Bisect the exterior 
 angle between LF and LF' and draw the 
 required tangent. 
 
 To draw an ellipse, given the axes. Second Method. 
 Draw the circle upon MN, the major axis; and draw also the 
 circle upon PQ, the minor axis. Draw oi, 02, 03, etc. From a, 
 
APPENDIX 
 
 215 
 
 b, c, etc., draw horizontal lines; from i, 2, 3, etc., draw vertical 
 lines. Locate the intersections of these lines as at i', 2', 3', etc., 
 and find a sufficient number so that a curve may be drawn 
 through them as shown. This curve will be the required ellipse. 
 
 To draw a tangent to an ellipse from a point without the curve. 
 
 Locate F and F', the foci of the given ellipse. Let L be the 
 given point at which the tangent is to be drawn. Draw LF, and 
 with LF as a radius and L as a center describe the arc FR. With 
 F' as a center and MN as a radius describe an arc cutting the 
 arc FR at R, and draw RF'. The point where this line RF' cuts 
 the elHpse or K is the point of tangency and LK, then, is the 
 required tangent. 
 
 To draw a parabola when its axis, its vertex, and a point on 
 the curve are given. 
 
 Let AB be the axis, A the vertex, and D a point on the curve. 
 Draw DC and AC and divide each into the same number of equal 
 parts, thus locating points i, 2, etc., and a, b, c, etc. Draw Ai, 
 A2, A3, A4, A5, A6, A7, and find where each of these fines is 
 intersected by fines through a, b, c, d, e, f, g drawn parallel to 
 AB. The intersections thus located wiU be points on the re- 
 quired parabola, and when a suflicient number of them have 
 been found the curve may be drawn with the irregular curve. 
 
2l6 
 
 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 To draw a tangent to a parabola at a point on the curve. 
 
 Let L be the given point. Draw LP perpendicular to AB and 
 make NA equal to PA. Draw NL; this will be the required 
 tangent to the curve at L. 
 
 To draw a tangent to a parabola from a point not on the 
 curve. 
 
 Let CD be the directrix of the given parabola whose focus is 
 F, and let P be the point from which the Hne is to be drawn tan- 
 
 C 
 
 
 C 1 2 3 4 5 
 
 D 
 5 
 4 
 3 
 
 2 
 
 1 
 
 E 
 
 a\^^ 
 
 
 Fb^ 
 
 Q- 
 
 
 
 gent to the curve. With P as a center and a radius Pf describe 
 the arc cutting the directrix at K and L. From these points 
 draw Hues LM and KN perpendicular to CD cutting the parab- 
 ola at M and N, M and N, then, are the tangent points, and 
 lines drawn from P to these points will be tangent to the curve. 
 
 To draw an hyperbola, given the major axis and one point on 
 the curve. 
 
 Let AB be the given axis and D the given point. Draw the 
 rectangle BCDE, and divide CD and ED into any number of 
 equal parts, in this Q^se six. Draw A5, A4, A3, A2, Ai, also 
 B5, B4, B3, B2, Bi. At the intersection of each of these lines 
 from A with the corresponding lines from B will be located a 
 point on the curve, and by finding a sufficient number of points 
 the curve may be drawn as shown. 
 
APPENDIX 
 
 217 
 
 To draw a tangent to an hyperbola at a given point on the curve. 
 
 Let AB be the major axis of the hyperbola and L the given 
 
 point at which the tangent is to be drawn. Draw LM perpen- 
 
 dicular to AB. On OM as a diameter describe the circle 
 cutting the major auxiliary circle at K. Draw KN perpen- 
 dicular to AB. Draw NL, the required tangent. 
 
 To rectify a given arc. An approximate method. 
 
 Let AB be the given arc. Divide it as shown into a number 
 of equal parts. Let the distances into which the arc is divided, 
 as Ai, 12, 23, 3B, be made small enough so that the chord sub- 
 tending this small arc is 
 practically equal to the 
 arc itself. With the bow 
 dividers lay off on the 
 given straight line these 
 distances. The distance 
 AB on the straight line, 
 then, will be equal to 
 the distance AB on the 
 arc. 
 
 If it be desired to lay off on a given arc any given length the 
 same method may be used. The whole point of this method 
 
2l8 ESSENTIALS OF DESCRIPTIVE GEOMETRY 
 
 is in keeping the distances so small that there is no difference, 
 practically, between the arc and its subtending chord. 
 
 Other methods of laying off arcs onto straight lines may be 
 used, but the above is by far the most convenient for the 
 draftsman. 
 
 To measure an angle by its natural tangent. 
 
 Let the given angle to be measured be ACB. Its natural 
 
 AB 
 
 tangent is — — . Divide, therefore, AB by BC and look up the 
 BC 
 
 value of the angle corresponding to this quotient in any rehable 
 
 ^A table of natural tangents. The 
 result will be the value of the 
 angle in degrees and minutes. 
 
 To lay off a given angle by its 
 natural tangent. 
 
 Lay off B any convenient dis- 
 tance from C, say 3", and erect 
 a perpendicular BA. To find the length of BA look up the value 
 of the natural tangent of the given angle in any reliable table of 
 natural tangents, multiply it by three, and lay off this distance 
 on BA, thus locating A. Connect A with C and BCA then will 
 be the required angle. 
 
 Note. Since tables of natural tangents are computed on a 
 base of i the value given for the natural tangent must always be 
 multiplied by the length of the base when it is taken greater than 
 unity. 
 
Wiley Special Subject Catalogues 
 
 For convenience a list of the Wiley Special Subject Catalogues, 
 envelope size, has been printed. These are arranged in groups 
 — each catalogue having a key symbol. (See Special Subject 
 List Below). To obtain any of these catalogues, send a 
 postal using the key symbols of the Catalogues desired. 
 
 1 — Agriculture. Animal Husbandry. Dairying. Industrial 
 Canning and Preserving. 
 
 2 — ^Architecture. Building. Masonry. 
 
 3 — Business Administration and Management. Law. 
 
 Industrial Processes: Canning and Preserving; Oil and Gas 
 Production; Paint; Printing; Sugar Manufacture; Textile. 
 
 CHEMISTRY 
 
 4a General; Analytical, Qualitative and Quantitative; Inorganic; 
 Organic. 
 
 4b Electro- and Physical; Food and Water; Industrial; Medical 
 and Pharmaceutical; Sugar. 
 
 CIVIL ENGINEERING 
 
 5a Unclassified and Structural Engineering. 
 
 5b Materials and Mechanics of Construction, including; Cement 
 and Concrete; Excavation and Earthwork; Foundations; 
 Masonry. 
 
 5c Railroads; Surveying. 
 
 5d Dams; Hydraulic Engineering; Pumping and Hydraulics; Irri- 
 gation Engineering; River and Harbor Engineering; Water 
 Supply. 
 
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CIVIL ENGnSTEEKlNG— Continued 
 5e Highways; Municipal Engineering; Sanitary Engineering; 
 Water Supply. Forestry. Horticulture, Botany and 
 Landscape Gardening. 
 
 6 — Design. Decoration. Drawing: General; Descriptive 
 Geometry; Kinematics; Mechanical. 
 
 ELECTRICAL ENGINEERING— PHYSICS 
 
 7 — General and Unclassified; Batteries; Central Station Practice; 
 Distribution and Transmission; Dynamo-Electro Machinery; 
 Electro-Chemistry and Metallurgy; Measuring Instruments 
 and Miscellaneous Apparatus. 
 
 8 — Astronomy. Meteorology. Explosives. Marine and 
 Naval Engineering. Military. Miscellaneous Books. 
 
 MATHEMATICS 
 9 — General; Algebra; Analytic and Plane Geometry; Calculus; 
 Trigonometry; Vector Analysis. 
 
 MECHANICAL ENGINEERING 
 10a General and Unclassified; Foundry Practice; Shop Practice. 
 10b Gas Power and Internal Combustion Engines; Heating and 
 
 Ventilation; Refrigeration. 
 10c Machine Design and Mechanism; Power Transmission; Steam 
 
 Power and Power Plants; Thermodynamics and Heat Power. 
 1 1 — Mechanics. 
 
 12 — Medicine. Pharmacy. Medical and Pharmaceutical Chem- 
 istry. Sanitary Science and Engineering. Bacteriology and 
 
 Biology. 
 
 MINING ENGINEERING 
 
 13 — General; Assaying; Excavation, Earthwork, Tunneling, Etc.; 
 Explosives; Geology; Metallurgy; Mineralogy; Prospecting; 
 Ventilation. 
 
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