B M SOI b5S 
 
 
 
 
 
 
 
 
 ■If!] 
 
 r ' 
 
 '/: 
 
Descriptive Geometry 
 
 for Students in 
 
 Engineering Science 
 
 and 
 
 Architecture 
 
 A CAREFULLY GRADED COURSE OF INSTRUCTION 
 
 BY 
 
 HENRY F. ARMSTRONG 
 
 Associate Professor of Descriptive Geometry and Drawing, McGill University 
 
 FIRST EDITION 
 
 FIRST THOUSAND 
 
 4 
 
 NEW YORK: 
 JOHN WILEY e^ SONS, Inc. 
 
 London: CHAPMAN i: HALL, Limited 
 
Copyright, igiS 
 
 BY 
 
 HENRY F. ARMSTRONG 
 
 THE SCIENTIFIC PRESS 
 
 ROBERT DRUMMOND AND COMPANY 
 
 BROOKLYN, N. Y. 
 
PREFACE 
 
 The writer offers the contents of this Text Book as the result of over 
 
 rfounrtht'^r"','"'""*^ "^ """"'p'^" ^^-^^'^y- ^-^ ^^ ^e 
 
 has found that a logical presentation of the subject, concerning itself largely 
 with the best sequence possible, and with the introduction, at an early 
 stage, of practical applications and well-graded exercises by which the first 
 rules and principles can be practised, is far more preferable and satisfactory 
 in Its results than the mode of procedure and division of the subject matter 
 commonly adopted in other text books on the subject 
 
 Difficult and complex phases of Descriptive Geometry and problems 
 beyond what the average student has time to assimilate during a limited 
 college course and beyond what is worth while in such a course, are avoided 
 because ol he tendency for students to become thereby discouraged and to 
 sX^LLtr^ "" ^*" ^'"""' "'^^™'- P™™ '° '^ - '"-esting 
 No pretense, therefore, has been made to place before students an 
 exhaustive treatise on the subject, but rather to deal with the essenti" 
 facts and methods readily arrived at and their useful application. To this 
 end, the student is frequently required to work, independently, exercises on 
 the problems of which illustrations are given and discussed, aiid himse to 
 propose questions or data for new exercises to be worked 
 
 It will be noticed that the study of the subject is conducted in such 
 a way that Part I of the Text Book may serve as an Introductory Course 
 more or less complete in itself, and may be sufficient and suitable for a 
 first term in a Science Course Curriculum, or for the upper forms in High 
 Schools; or. Parts I and II may be taken in the First Year and Part HI 
 VrJy I"" ' ^"^■""'•^' "^ ^°"^=" Course in Science, if in 
 undertaken '' ' " ""' ^'"''"^'"^ '"""'"'"' """= '"' ""^ "*°'» '" be 
 
 so liHle ""^ ""''^' •"' f"'' '"'''''''' '^' ™^Sinatixe faculty of the mind, 
 
 little exercised in the ordinary school curriculum, is afforded a .ood 
 
 traming. while the practice in grasping a collection of detailed conditions 
 
 the «ould-be engineer or architect in his future career. 
 Janu.lry, 11)15. 
 
CONTENTS 
 
 SEC. 
 
 1. The Projection Planes 
 
 2. Lines and Their Inclinations. 
 
 PAGE 
 
 4 
 
 PART ONE 
 
 CHAPTER I 
 
 CH.VPTER II 
 
 3. Plan and I^levation Fo-ms for Plane Rectilineal Figures g 
 
 4. Plan and Elevation Forms for the Circle 10 
 
 chapti:r III 
 
 5. Oblique Planes, Their Traces and Inclinations 13 
 
 6. Traces of Lines 17 
 
 CHAPri:R IV 
 
 7. Shadows of Lines ro 
 
 8. Shadows of Plane Figures 22 
 
 CHAPTER \' 
 
 (). The '■ Com[)ound Angle " for Lines 25 
 
 10. Projections of Plane Figures, Involving the Compound Angle 29 
 
 CHAPTER VI 
 
 11. Projeition of Simple Solids 31 
 
 I :;. Shadows oi Solitls and of Groups of Solids ;5 
 
 IWR'I" I'WO 
 
 CHAPTER \H 
 
 I ^ Rabattemcnt of Planes, and Projections of Figures Involving Its LTse 40 
 
 14. Points and Lines in Relation to Ohliiiue Planes 45 
 
 15. Intersection of Obli(iue Planes with Each Other, and of Lines with Oblique Planes 4S 
 
CONTENTS 
 
 CHAPTER VIII 
 
 SEC. 
 
 [6. Parallel Planes. 
 
 and Planes Dividing These Angles. . 56 
 
 17'. Dihedral Angles Contained by Oblique Intersecting Planes, 
 
 CHAPTER IX 
 
 x8. Projection of Rectilineal Angles, and of Figures and Solids Involving the Same 6^ 
 
 19. The Tetrahedron and the Oclahedron 
 
 CHAPTER X 
 
 69 
 
 20. Axometric and Isometric Projection ^^ 
 
 21. Axometric Projection, Continued 
 
 CHAPTER XI 
 
 22. Sections of Simple Solids... 
 
 23. Traces of Curved Surfaces. 
 
 PART THREE 
 
 CHAPTER XII 
 
 24. Tangent Planes to Cones and Cylinders ;.:' ^ "r^-" ' , " ^ '''' 
 
 25. Projection of Solids dependent on Tangent Planes to Right Circular Cones 
 
 CHAPTER XIII 
 
 26. Tangent Planes to a Sphere, and the Finding of an Oblique Plane with Given Inclinations 93 
 
 27. Tangent Planes Common to Three Given Spheres 
 
 CHAPTER XIV 
 
 Surfaces m 
 
 30. Surfaces of Revolution, and the Screw Thread 
 
 CHAPTER XV 
 
 31. Radial Projection. Perspective Projection . 
 
 32. Perspective Projection, continued 
 
c 
 
 /- 
 
 DESCRIPTIVE GEOMETRY 
 
 PART ONE 
 
 CHAPTER I 
 
 THE PROJECTION FLAXES 
 
 Section 1. In Descriptive Geometry the object is chiefly to prepare draw- 
 ings as follows: — 
 
 (a) Those which will display or describe by diferent views any object or 
 arrangement of lines or figures discussed; 
 
 {h) Those which will, by various analytical and constructive methods and 
 operations, discover or disclose facts as to shapes, inclinations, appear- 
 ances, sizes, etc.; and 
 
 (c) Those which will represent planes and how they may be disposed to one 
 another. 
 
 The views mentioned above in (a) are projections, and are made on what 
 are called planes of projection. The same projection planes, two in number, are 
 also made use of in the discussion of planes referred to in (c). lines being drawn 
 over the planes of projection and made to represent other planes in various alti- 
 tudes with respect to the projection planes. 
 
 The planes of projection are the Horizontal Plane and the Vertical Plane. 
 These are considered as being jlxed, and the lines, planes, figures or objects arc 
 considcM-ed as having a relation to them-near or otherwise as to distance, inclined 
 or otherwise as to attitude. 
 
 The drawings made either represent jjoints, fines, figures or objects hx views 
 thrown perpendicularly on to these planes of projection (the H.P. and the V P 
 as they arc commonly called), or they indicate the intersection of the planes of 
 projection by lines and planes. 
 
 ^ In order that the drawings ma>- be descriptive of what is under consideration 
 It IS usually necessary to have one drawin- of (he same thing on each of the planes 
 
2 DESCRIPTR^E GEOMETRY 
 
 of projection, so that, whatever it is, it may be studied from different points of view, 
 or better realized by the study of these drawings on the two planes of projection. 
 
 /;/ the case of projections it is advisable and convenient to place one drawing 
 representing the view projected perpendicularly on to the H.P. and another draw- 
 ing representing the view projected perpendicularly on to the V.P. These draw- 
 ings, projected perpendicularly from whatever is under discussion, will arrange 
 themselves perpendicularly opposite each other. 
 
 For convenience we may, while dealing with simple and elementary problems, 
 use one part of the drawing paper, or blackboard, upon which to make the draw- 
 ings or projections thrown on to the H.P. These drawings are called the hori- 
 zontal projections, or, more commonly, the plans. The other part of the drawing 
 paper, or blackboard, may be used for making the drawings or projections thrown 
 on to the V.P., called therefore the vertical projections, or, more commonly, the 
 elevations. 
 
 In order to make use of the drawing paper effectively in this way, a Hne is 
 drawn across it, and this is to represent the intersection line in which the V.P. 
 and the H.P. meet. We name this line XY, and usually the lower part of the 
 paper then represents the near part of the H.P., while the upper part represents 
 the upper part of the V.P. The line XY, however, must not be thought to limit 
 the extent of either of the planes, for it may sometimes be necessary to display 
 something which would not project on to the near part of the H.P., and then 
 the upper part of the paper would also represent the H.P., that is, the part of it 
 beyond the V.P.'s position. Similarly, the lower part of the paper may some- 
 times be required for projections on the V.P. from things placed below the level 
 of the H.P. 
 
 It must continually be remembered that the part of the drawing paper upon 
 which vertical projections or elevations are made is to be considered actually 
 vertical, and that the part of the paper used for the horizontal projections or 
 plans must always be considered horizontal. Sometimes it may be well to fold 
 the drawing paper along the XY line and then hold the V.P. part vertical, while 
 the H.P. part remains horizontal. Try this. 
 
 The student must learn as soon as possible to picture to himself, or arrange, 
 simple things in different positions and attitudes in respect to the two projection 
 planes arranged as above decided upon, one vertical and the other horizontal, 
 and imagine what should be the views projected to these planes and how the views 
 or projections will change with a change of attitude. 
 
 When it is realized that this process of projection is perpendicular to the pro- 
 jection planes and not like that of the camera, then it will be seen that the size 
 of the projection, plan or elevation, as the case may be, will not depend on distance. 
 
 The hand, resting on its edge on the table top, and arranged with the palm 
 vertical, might be used to represent roughly the V.P. of projection, and the table 
 top to represent the H.P. of projection. Now hold things with the other hand, 
 and consider what views or appearances (elevations and plans) you might get. 
 
THE PROJECTION PLAXES 3 
 
 projected perpendicularly on to the palm of the hand and on to the table top 
 respectively. The line across the table top where the vertical plane of the pahn 
 of the hand meets it, will be the XY line named above, and it will be seen that 
 any two drawings of an object which is held still, the one drawing on the table 
 top and the other on the palm of the hand, are perpendicularly opposite each other 
 across the XY line, when the hand is thrown back about the AT line as a hinge. 
 
 We have seen that there may be plans or horizontal projections and elevations 
 or vertical projections of points, lines, figures and objects, but it must be pointed 
 out that there is no such thing as the plan or the elevation of a plane. A plane, 
 however, other than the planes of projection, may be represented by a line or 
 lines drawn on the projection planes to mark its intersection with them. 
 
 EXERCISE I 
 
 1. Distinguish between attitude and position. 
 
 2. What are plans and elevations, and how are they obtained? 
 
 3. What is the XY? What is meant by H.P. and by V.P.? What is the proper relation 
 of the planes of projection to one another? 
 
 4. How can the same sheet of paper serve for both planes of projection? 
 
 5. How arc planes, other than the projection planes, represented? 
 
 6. How are a plan and an elevation of an object arranged so as to show that they are views 
 of the same object? 
 
 By further investigation on the same lines as suggested above, it will be seen 
 that the distances of an object held away from the planes of projection are the 
 same as the distances of the drawings or projections of it from XY. In other 
 words, the plan at 3" from XY will mean that the thing represented is 3" from the 
 V.P., and the elevation at 2" from AT will mean that the thing is 2" from the H.P. 
 
 When an object is on the near side of the V.P. it is said to be in front of the 
 V.P.. and its plan, on the paper, will be placed below the AT line. 
 
 EXERCISE II 
 
 I. Let the projection planes, for the purpose of this exercise, be represented by. say, a large 
 book, standing on the table, for the V.P., and by the table top for the H.P.. then arrange a book 
 or other rectangular object, in such altitudes, with respect to the projection planes, as to satisfy 
 the following descriptions: — 
 
 (i) The book, or tlat rectangular bo.\, placed horizontally with all its edges equally 
 inclined to the \'.P. 
 
 (2) The same, so thai its true form (rectangle) will appear in elevation. 
 
 (3) The same, so that its appearance will be a rhomboid on both planes. 
 
 14) The same, so that one edge is in the V.P. at, say, 45° to AT, and the surface of 
 it is at, say, 60° to the V.P. 
 
4 DESCRIPTIVE GEOMETRY 
 
 (5) The same, so that both projections are lines only, supposing the book to be very 
 
 thin. 
 
 (6) The same, so that its plan and elevation are both rectangles. 
 
 2. Make sketch or freehand drawings, on a paper or blackboard, not carefully measured 
 or scaled, as plans and elevations, to satisfy the descriptions above, and insert distances from 
 the planes of projection, wherever possible. 
 
 3. How are distances from the planes of projection shown, of any points in an object whose 
 plan and elevation are required or given? 
 
 LINES AND THEIR INCLINATIONS 
 
 Section 2. It will novv^ be seen that any object, such as, for instance, a book, 
 may have many different sets of drawings (a set = plan and elevation), differing 
 according to its attitude or arrangement in relation to the projection planes, to 
 represent it, and that, given any such set of drawings for the book, the book may 
 
 (i) 
 
 
 (ii) 
 
 
 
 (iii) 
 
 be' 
 
 
 (iv) 
 c b' 
 
 J a' 
 
 
 
 
 
 1 \ 
 
 
 \ \ 
 
 
 b'e 
 
 ( 
 
 :d:_ 
 
 — 
 
 1 \ 
 
 1 
 1 
 1 
 
 'c'd' 
 
 i\ \ 
 
 
 
 
 1 ! 1 1 
 1 1 1 1 
 
 
 
 
 d 
 
 
 1 
 
 d 
 
 ' /k ' 
 
 
 
 
 
 
 '/\l 
 
 id 
 
 
 
 
 
 
 
 
 ■■'-4:^' 
 
 
 J 
 
 c 
 
 b < 
 
 b 
 
 
 
 
 
 Fic 
 
 
 I. 
 
 
 
 be held in the proper way to give, or provide for, that set of drawings or pro- 
 jections. 
 
 In order to distinguish plans from elevations, it is customary, for purposes 
 of discussion, to use small letters a, b, c, etc., on the plans, and the same small 
 letters, with a short stroke or dash as a', h' , c' , etc., for the same points named 
 on the elevations. Capital letters A, B, C, etc., are used for the names of points 
 regardless of their projections. 
 
 Illustrations of projections^ or plans and elevations are here given, which 
 must be thoughtfully studied, while many facts are noted. 
 
 In Fig. I at (i) the point A, is represented at a and a' and has a distance above 
 the H.P. equal to twice that of its distance in front of the V.P. At (ii) the square 
 BCDE is arranged horizontally with edges BC and DE parallel to the V.P. The 
 distance of the elevation above XY shows the distance of the square above the 
 H.P. At (iii) the square still has its edges BC and DE parallel to the V.P. and 
 
LINES AND THEIR INCLINATIONS 5 
 
 its other edges still horizontal, i.e., parallel to the H.P.. but the square is inclined 
 to the H.P., and the angle the elevation line makes with XY is the angle the figure 
 makes with the H.P. The figure is, of course, still perpendicular to the V.P. 
 At (iv) the plan, the same shape and size as thef plan at (iii), has been so turned 
 that BC and DE are no longer parallel to the V.P., but as these edges are still 
 inclined the same amount as before, to the H.P., the difference of level of their 
 ends has not been altered, and so the elevation can be derived, partly, from the 
 elevation at (iii). Realize that the edges CD and BE are still horizontal, but 
 inclined to the V.P., the angle of inclination to the V'.P. being that which cd makes 
 with XY. 
 
 N.B. — A horizontal line may have any direction, for its plan, across the 
 paper. 
 
 Consider next the group of drawings in Fig. 2. At (i) a square is represented 
 horizontal, at a distance above the H.P. and a little in front of the V.P.; one 
 diagonal is perpendicular to the V.P. and is represented in elevation at b'd'. The 
 full length of the diagonal AC is shown in both the plan and the elevation. At 
 (ii) the elevation of (i) has been inclined to give an inclination of the figure to the 
 H.P., the point at b'd' still representing the horizontal diagonal whose plan bd 
 therefore shows true length; the plan of the other diagonal, however, is now short 
 of its true length, due to its inclination. This diagonal is said to he foreshortened, 
 that is, arranged so that the view of it does not show its true size. 
 
 Now, if without altering the size and shape of the plan of this square as seen in 
 Fig. 2 at (ii), it be moved so that the horizontal diagonal BD be at an inclination 
 to the V.P. represented by ^he angle which bd makes with AT, it will be seen 
 that the relative levels of the various points ABCD will remain as before, and 
 hence a new elevation for set 'iii) may be obtained from the plan as now arranged 
 
6 DESCRIPTIVE GEOMETRY 
 
 at (iii) and by making use of the levels from the elevation of (ii). The square 
 may now be described as having a diagonal horizontal and at an angle {a°) to the 
 V.P., and the figure as having an inclination (/3°) to the H.P. It will be seen 
 that in order to obtain set (iii) it is necessary to first make sets (i) and (ii). 
 
 From further inspection of Fig. i and Fig. 2 it will be seen that a line, in order 
 to show its true length in a projection, must be arranged parallel to the plane 
 upon which the projection is thrown; and, similarly with regard to plane figures, 
 i.e., for a figure to show its full size and shape in a projection, it must be arranged 
 parallel to the plane upon which the projection is made. 
 
 This matter is still further illustrated and discussed in Fig. 3, where it will 
 be seen that the plans in (i) and (ii) and (iii) remain full length because the line 
 is horizontal in each case, whereas the elevations vary, because the relation of the 
 fine to the V.P. varies. 
 
 (I) 
 
 (III) 
 
 ah 
 
 a' h' a' 
 
 b a b 
 
 ;b ^^ 
 
 I 
 
 Fig. 3. 
 
 In (iii) the plan and elevation are both full size. In (iv) the elevation is 
 full size because the line is still parallel to the V.P. as in (iii), and as shown by the 
 arrangement of its plan, now foreshortened, however, because the line is inchned 
 to the H.P. In (v) the plan is kept the same length as in (iv) and consequently 
 the levels for the ends of the elevation may be taken from the elevation in (iv). 
 At (vi) the line is vertical. 
 
 It will readily be seen that in Fig. 3 at (ii) the angle the line makes with the 
 V.P. is the angle that ab makes with XY, and that in (iv) .the angle the line makes 
 with the H.P. is the angle a'b' makes with XY. 
 
 Realize that although some of its projections are inclined to XY, the fine 
 AB itself is not inclined to XY. 
 
 If now it be required to discover the angles th^ line AB, in Fig. 3 as shown^ 
 at (v), makes with the projection planes, it is evident that the angle it makes with 
 
LIXES AND THEIR INX'LIXATIOXS 7 
 
 the H.P. is not what its elevation at (v) makes with XY, but what its elevation 
 at (iv) makes with AT, where the plan is arranged parallel to XY; and in order 
 to find what inclination the line AB has with the V.P. its elevation a'b' (v) must 
 be arranged horizontally as at (ii), and then its plan, the full length of the line, 
 will show with XY the angle required. 
 
 Let it be required to fmd the true lengths of any Hnes AB, Fig. 4. and CD, 
 Fig. 5, and also their inchnations to the planes of projection. Their plans and 
 elevations are shown at ab, a'b', Fig. 4 and at cd, c'd', Fig. 5. The line in each 
 case must be arranged parallel to the V.P. by swinging its plan to become parallel 
 to XY as at 0^2, Fig. 4, and at tJ.., Fig. 5, with the result, when the elevation 
 
 Y X 
 
 Fig. 4. 
 
 opposite this new arrangement of the plan in each case is set up. that the true 
 length d'b'2 and c'd' 2 respectively, is shown on the V.P., and the incUnation 0^, of 
 the line, to the H.P. may now be seen; also, the line in each case must be 
 arranged parallel to the H.P. by swinging its elevation parallel to XY, and 
 so providing for a plan, a-ib and cod in the cases being considered, that will 
 show true length and contain an angle, a°, with XY, equal to the angle the 
 hne makes with the V^P. 
 
 N.B. — The secondary positions for the lines in Figs. 4 and 5 are shown by 
 double lines to emphasize them. 
 
 EXERCISE III 
 
 I. Place the plan and elevation of the line AB, making the plan ab 2" long and the elevation 
 3" long. Find the true length of this line, and also its inclinations to the planes of projection. 
 Mark the angles as in Fig. 4 with arrow-headed arcs and use a. to indicate the angle with the V.P., 
 and ii for the angle with the H.P. 
 
 ^. The plan cd is 2" long. The line CD is 3" long. Draw an elevation for it, and show 
 what angles the line -CZ) makes with the planes of projection. 
 
8 DESCRIPTIVE GEOMETRY 
 
 3. The elevation of EF is 2" long, and is at 45° to XY . The line is at 30° to the H.P. Find 
 a plan for it, and also determine its true length. 
 
 4, 5 and 6. Find the true length of each of the given lines GH, JK and LM, and their indina- 
 tions to the planes of projection. 
 
CHAPTER II 
 
 PLAX AND ELEVATION FORMS FOR PLANE RECTILINEAL FIGURES 
 
 Section 3. Suppose it be required to lind the projections of any plane figure 
 inclined at a given angle to one of the projection i)lanes and having one of its 
 edges in that projection plane at a given inclination to the XV. Illustrations 
 of such a problem are shown in Fig. 6 and Fig. 7. 
 
 In Fig. 6 a square is chosen and is first placed, as at (i), in the V.P. Its plan 
 is therefore part of the XY line. In order that it may swing out at the required 
 
 Fig. 6. 
 
 Fig. 7. 
 
 angle, and yet leave one edge in the V.P., it must be arranged so that when the 
 line representing it in plan is moved to enclose the given angle a° with A'l'. one 
 edge of it, say AB, represented in elevation at a'b\ is in plan a point, as at ab, 
 and will remain as a point in AT. The shaded or "hatched" rectangle is now 
 the elevation shape required. This projection shape is the first thing to obtain 
 toward the solution of such a problem^as the above. 
 
 If now this elevation shape be rearranged so that the edge AB is at the given 
 angle to the AT, as at (ii), then it will be seen that a rhomboidal figure will be 
 obtained for plan. The distance of the plan line cd in (ii) from A'l' will be the same 
 as the distance it arrived at in (i\ since the elevation shape is the same in both 
 cases and therefore the relation of the figure to the V.P. is unchanged. 
 
 9 
 
10 DESCRIPTIVE GEOMETRY 
 
 In Fig. 7 the case of the regular pentagon should be studied. The pentagon, 
 as l.nally arranged at (ii), may be described as being inclined at (3° to the H.P. 
 and having one edge in the H.P. at a° to XY. The mode of procedure is similar 
 to that for the square in Fig. 6. 
 
 N.B. — It is advisable that the student, in order to comprehend clearly this 
 matter of projections, should frequently fold his drawing paper on the XY line 
 and stand that part of the paper with elevations on it, vertically, so that the planes 
 of projection are rightly related. He will then realize how the projections are 
 accounted for. 
 
 Before working Exercise IV on paper, it may be well for the student to cut a 
 square, and also a rectangle, out of a piece of cardboard, and practise placing 
 them according to the questions. Realize that the projections or views required 
 are thrown always perpendicularly on to the planes of projection. 
 
 EXERCISE IV 
 
 1. A rectangle 2" by f" is the plan for a square. Show an elevation for it, (a) when an edge 
 of the square is in the H.P. and perpendicular to XY, (b) when an edge in the H.P. is at 45° to XY. 
 
 2. A square of 2" edge is the plan for a rectangle having its long edges 3". Arrange the 
 square plan so that edges representing the short edges of the rectangle are at 30° to the V.P., 
 and then obtain an elevation. 
 
 3. Find the plan and elevation of a square, of 2" edge, inchned at 60° to the H.P and having 
 one edge horizontal at a distance of i" above the H.P. Place the square so that the horizontal 
 edges are parallel to the V.P. and in front of it. 
 
 4. Find the plan and elevation of a 2" square when one diagonal of it is horizontal and at 
 45° to the V.P., while the figure is inclined at 60° to the H.P. (This is the same as saying that 
 one diagonal being horizontal and at 45° to the V.P., the other is at 60° to the H.P.) 
 
 \ 5. The plan of a square is a rectangle 2§" by f ", with the long edges parallel to XY. 
 Find an elevation for this square. 
 
 6. A rectangle 2" by 3" has a long edge in the V.P. The figure is inclined to the V.P. at 
 60°. Show the elevation arranged so as to represent the rectangle with its long edges at 30° 
 to the H.P., and obtain a plan for it. 
 
 PLAN AND ELEVATION FORMS FOR THE CIRCLE 
 
 Section 4. A similar process to that for obtaining projections of the pentagon 
 in Fig. 7 is used for projections of the circle. 
 
 For the circle to appear circular in projection it must be placed parallel to 
 the projection plane. If it is horizontally arranged, its plan is a circle, and if it 
 is placed parallel to the V.P. its elevation is circular. 
 
 Hold some circular disc such as a coin or a piece of cardboard cut circular, 
 and it will be observed that, like any other plane figure, it can be represented by 
 a straight line on either one or the other, or on both at the same time, of the planes 
 
PLAX AND ELEVATION FORMS FOR THE CIRCLE 
 
 11 
 
 of projection; also, that when not {perpendicular to a projection plane, nor p>arallel 
 to it, the circle will appear in projection as an ellipse with the major axis equal 
 to the length of the diameter of the circle. This major axis represents the diameter 
 parallel to the projection plane upon which the ellipse appears. 
 
 Let it be required to find the projections of a circle inclined at, say, a' to the 
 H.P. and ha\'ing its horizontal diameter at. say. ^' to the V.P. 
 
 In Fig. 8 at (i) a circle is shown, horizontal, with a line for its elevation. If 
 the elevation line be rearrange^ at an inclination of a' to A' I' as at (u) it wiJI 
 represent the circle tilted or in^ned at a° to the H.P. The horizontal diameter 
 AB which is perpendicular to the V.P. will remain full length in plan at a2*2- while 
 the diameter at right angles to it. viz., CD, will now be foreshortened to C2d2- 
 
 (U) 
 
 (m) 
 
 Fig. 8. 
 
 A chord EF made parallel to AB wiU appear in ele\-ation as a pwint, and in 
 plan at €2/2- Other horizontal chords may be made use of, and eventually suffi- 
 cient points on the circimiference will be thus secured in the plan at u to insure 
 a good shape, when drawn by passing a freehand cur\-ed line through them, to 
 ser\-e as the edge of the circle when the circle is tilted or inclined to the H.P. 
 
 Xow if this plan, at (ii). be turned so that the horizontal diameter, a^^s, and 
 horizontal chords, are no longer j)erpendicular to the V.P.. but at 4" to the V-P. 
 ELS at <iii), then an elevation, in the shape of an ellipse instead oi 2 line, will be 
 btained. and this b done by deri\-ing the levels for the e! - us 
 
 points in the cur\-e. from the elevation line at (ii. Such .. . .-r, 
 
 Tiight also be obtained by making use of what is called a secondar\- ele\*atioo 
 3lane. represented as ha\ing its XV at any convenient distance, as at A'^Fi- This 
 
12 DESCRIPTIVE GEOMETRY 
 
 being settled, without moving the plan around, as was necessary at (iii), the ele- 
 vation (iv) is projected directly from the original plan at (ii), and heights for 
 the various points are transferred by measurement from the elevation at (ii). 
 
 N.B. — This secondary XY must make with the plan of the horizontal diam- 
 eter 02^2 at (ii) the same angle, /3°, as that contained between the XY and the plan 
 asbs at (iii). This will mean that X2Y2 makes with XY an angle 90° — /3°. 
 
 It will be seen that the solution of the problem requires a set of drawings 
 to be made, either such as (i), (ii) and (iii), or a set such as (i), (ii) and (iv). 
 
 EXERCISE V 
 
 1. Arrange a regular pentagon, of i|" edge, so that one edge is in the H.P. parallel to XY 
 and the figure is inclined to the H.P. at 45°. (Note. — The angle at the corner of a pentagon 
 is io8°.) 
 
 2. Incline a regular hexagon, of ij" edge, at 50° to theH.P., leaving one of its edges in the 
 H.P. Then turn the plan so that the edge in the H.P. is at 30° to XY, and show an elevation 
 for it when so placed. 
 
 3. Find a plan and an elevation for an equilateral triangle inchned at 60° to the H.P., and 
 having one edge, 2I", in the H.P. at 60° to the XY line. 
 
 4. Find the plan and elevation of a circle, 2^" diameter, when it is inclined at 50° to the 
 H.P. Then arrange the plan so that the horizontal diameter (the major axis of the ellipse found 
 for plan) is at 45° to the V.P., and obtain its elevation. 
 
 5. Find the plan of a circle 2^" diameter, inclined at 60° to the H.P., and find an elevation 
 on a secondary vertical plane, (with secondary XY), arranging it so that the horizontal diam- 
 eter of the circle makes an angle of 60° with the secc 'ary plane. 
 
 6. Find the plan and elevation of a circle, 2^" diameter, inclined at 70° to the V.P., and 
 having the diameter which is parallel to the V.P. at 45° to the H.P. 
 
CHAPTER III 
 
 OBLIQUE PLANES, THEIR TRACES ANT) INCITNATIONS 
 
 Section 5. It has been previously seen that in order that a line might be set 
 up at a given angle to the H.P., or in order to find the incUnation to the H.P. of 
 a given Hne, it is necessary to arrange it parallel to the V.P., or in it, and that then 
 the projection in the V.P. makes with XY the same angle as the line makes with 
 H.P. Also, that in order that a plane figure might be tilted or inclined to a definite 
 angle with the H.P., it is necessary to arrange it in a perpendicular attitude to the 
 V.P., that is, so that its appearance in elevation is a line only, and we have seen 
 that the angle this Une makes with the AT is the incHnation the plane figure 
 makes with the H.P. 
 
 Similarly, with respect to the inclination of a line or a plane figure to the V.P., 
 the line must be arranged parallel to, or in, the H.P., and the figure must be 
 arranged vertically, so that its plan is a straight line only, and the angle this 
 straight line then makes with A'F is the inclination of the given line or plane figure 
 to the V.P. 
 
 If, now, the plane of any figi \ that is, the imaginar}' plane of which the 
 figure is a part, be considered, it is clear that when such a plane is vertical it will 
 cut or intersect the V.P. in a vertical Hne, and at the same time cut or intersect 
 the H.P. in a line that makes with XY the angle that the said plane, arranged in 
 vertical attitude, makes with the V.P. (The room door opened will serve to 
 illustrate this.) 
 
 Also, it will be seen that the plane of a figure arranged perpendicularly to 
 the V.P., but incUned to the H.P., will cut or intersect the H.P. in a line perpen- 
 dicular to the XY, and will cut or intersect the V.P. in a hne which makes with 
 XY the angle the plane makes with the H.P. 
 
 Two such planes are represented in Fig. 9 at RST and LMX . The inter- 
 section line RS is made by the vertical plane RST meeting the V.P., and the inter- 
 section hne ST is made by the same plane meeting the H.P. So also the inter- 
 section Unes LM and MX arc made by the plane LMX , perpendicular to the \'.P., 
 but inchned to the H.P., meeting the planes of projection. 
 
 These intersection lines made by a plane meeting the planes of projection 
 are called the traces of the plane, and it is by traces that the attitude or arrange- 
 ment of any plane is expressed. 
 
 The traces found or marked in the V.P. are called \'ertical Traces, though 
 they are not necessarily vertical Unes, as, for instance, LM; and the traces on the 
 
 13 
 
14 
 
 DESCRIPTIVE GEOMETRY 
 
 H.P. are called Horizontal Traces, and these, of course, are always horizontal 
 lines. In mentioning these traces the initial letters only are used as V.T. and 
 H.T., instead of the whole words. 
 
 Planes having other attitudes than the two so far discussed, sometunes called 
 oblique planes, are shown at OPQ and VWZ. 
 
 The student must reaHze that we are not now deaHng with plans and eleva- 
 tions, but with intersections of the projection planes, made by planes of which there 
 can be no plans and elevations. These imagined planes are indicated by their 
 traces. 
 
 Reproduce on a paper with XY marked, these planes as in Fig. 9. Fold the 
 paper along the line XY so that the V.P. is vertical while the H.P. remains hori- 
 zontal, just as the projection planes are always supposed to be arranged in relation 
 to one another, and it will be realized that RS is really at right angles to ST, and 
 that LM is really at right angles to MN. In the other cases, also, the angle 
 
 Fig. 9. 
 
 which OP makes with PQ is much less than what it appears to be when the paper 
 is not so folded, and so also with VW and WZ. 
 
 In Fig. 9, if the Vertical Traces RS, LM, etc., be produced downward below 
 XY, the productions will show where the planes cut the V.P. below the level of 
 the H.P., and likewise, if the Horizontal Traces ST, MN, etc., be produced upward 
 beyond the XY line, it will be seen where the planes RST, LMN, etc., cut the 
 H.P. in that part of it beyond the XY where the V.P. passes through it. For 
 general purposes, however, it is sufficient to. represent the planes by two Hues 
 meeting in XY as in Fig. 9. 
 
 The student should now consider Fig. 10 and reaUze that whereas the plane 
 RST is perpendicular to the V.P., as seen by the fact that its H.T. is perpendicular 
 to XY, and shows its angle to the H.P. as a°, the line AB is parallel to the V.P., 
 its plan being parallel to XY, and therefore its elevation shows its inclination to 
 the H.P. as i8^ 
 
 The elevation a'b' , being arranged perpendicular to the trace RS, the angle 
 j3 is complement to the angle a, that is to say, if the plane is inclined to the H.P. 
 
 I 
 
OBLIQUE PLANES, THEIR TRACES AND INCLINATIONS 15 
 
 at 40°, then the line perpendicular to it is inclined at 50° to the H.P.; and as 
 ST shows the plane to be at 90° to the V.P.,'so, also, ab shows the line to be 0° 
 to the V.P., and it is conclusively evident that the line AB is perpendicular to the 
 plane RST. 
 
 In the case of the plane LAIN and the line CD, similar conclusions will be 
 arrived at, namely, that the line CD is perpendicular to the plane LMX, and that 
 the inclination of the line to the V.P. is complement to the inclination of the plane 
 to the V.P. 
 
 If the student will now take a pencil to serve for a line and arrange it per- 
 pendicularly to any oblique plane — the hand may be arranged to show the attitude 
 of the oblique plane — he will notice that the projection on to the H.P., i.e., the 
 plan of the line, is at right angles to the H.T. of the plane; and that the projection 
 of the line on to the V.P., i.e., the elevation of the line, is at the same time per- 
 
 du) 
 
 (IV) 
 
 Fig. ic. 
 
 pendicular to the V.T. of the plane. Illustrations of perpendicular lines to oblique 
 planes are shown at (iii) and (iv) in Fig. 10. 
 
 N.B. — The projections of the line need not cross the traces of the plane. 
 
 The method of linding the angles of inclination of any oblique plane whose 
 traces are given is now obvious, and will depend upon the facts noted above, 
 namely: — (i) A line perpendicular to an oblique plane can be shown by arranging 
 its elevation at right angles to the V.T. of the plane, and its plan at right angles 
 to the H.T. of the plane. (2) The inclination of the oblique plane to the V.P. 
 is complement to the inclination of the line perpendicular to it, to the V.P., and 
 the inclination of the oblique plane to the H.P. is complement to the inclination 
 of the line perpendicular to it. to the H.P. 
 
 That is to say, to find the inclinations to the planes of projection, made by 
 any plane whose traces are given, represent a line perpendicular to the plane by 
 arranging its elevation perpendicular to the V.T. and its plan perpendicular to 
 
16 
 
 DESCRIPTIVE GEOMETRY 
 
 the H.T., and find the inclinations of this line by the method shown in Fig. 4, 
 then subtract these inclinations from 90° in each case, and the inclinations of the 
 plane are obtained. 
 
 An illustration of the method is shown in Fig. 11, where, to find the inclina- 
 tions of the oblique plane RST a line perpendicular to it is represented at a'b', ab. 
 
 Fig. it. 
 
 The angle a° is the complement of the angle the fine AB makes with the H.P. 
 Therefore a° is the inclination of the given plane RST to the H.P.; and, for similar 
 reasons, the angle 0° is the inclination of the plane RST to the V.P 
 
 EXERCISE VI. 
 
 Find the inclinations to the planes of projection, of the planes represented at A, B and C, 
 mark with arrow-headed arcs the angles required, and say which angles they are. 
 
 N.B. — Take each of the planes A, B and C as a separate problem and work to a large scale. 
 
TRACES OF LINES 
 
 17 
 
 TRACES OF LINES 
 
 Sections. A line AB is represented by its projections at (i) in Fig. 12. It 
 will be observed that the A end of it is a i)oint in the V.P. at a\ and that the B end 
 of it is a point in the H.P. at b. Any other point in the line may be taken, as at 
 C, and it will readily be seen that C is at some distance in front of the V.P. and 
 at some distance, also, above the H.P. If the point C could be moved in the line 
 until it got down to the H.P. it would be at a definite place in the H.P., namely, 
 at b, and b' would be the elevation of that place. Similarly, if the point were 
 moved up the line until it met the V.P. it would arrive at a, a definite place in the 
 V.P. of which a is the plan. 
 
 Fig. 12 
 
 Now consider the points R and 5 at (ii). Let R be a. place or point on the 
 V.P. and 5 be another place or point on the H.P., and let it be required to find the 
 actual distance from R to 5. 
 
 R, being a point in the V.P., will have its plan directly opposite to it in XV 
 at r, and S, being a point in the H.P., will have its elevation perpendicularly oppo- 
 site to it on the V.P., in the AT line at s'. The double lines in the figure show 
 the plan and elevation for a line joining R to S, and its length may now be found 
 at Rs'y, which, of course, represents the actual distance from R to 5. 
 
 Another illustration is shown at (iii), where point .1/ is on the V.P. and A is 
 on the H.P. Following the same rule, the plan of point M is in A'l' at.;;/, and the 
 elevation of A is found in A'F at n'. The line joining ;;' to .1/ will give the ele- 
 vation of the line extending from point M on the V.P. to the point A on the H.P., 
 and the line joining m to N will give the plan of the same line. To obtain the 
 true length of the line whose projections arc so found, swing the plan line mA into 
 AI', and the line Mn'2, the length obtained, is the distance from .1/" to A. 
 
18 
 
 DESCRIPTIVE GEOMETRY 
 
 EXERCISE VII 
 
 1. ]\Iark a point R on the V.P. 2" above XY, and another point 5 on the H.P. perpendic- 
 larly opposite to it, at a distance of i^" from XY. Find the real distance from R to S. 
 
 2. Mark a point M on the V.P. at a distance of 2" below XY, and another point, TV, on 
 the H.P. i" in front of the V.P. and considerably to the right of M. Find the true distance 
 between the two points. 
 
 3. A point V on the V.P. is 2" above .YF. Find a point H on the H.P., i|" in front of 
 the V.P. and 3", real distance, from the point V. 
 
 In Fig. 13 the two projections of a line are given in each of five different 
 arrangements for it. If, in each case, the line be imagined to continue in its direction, 
 
 (i) . (ii) (iii) (iv) (v) 
 
 '/ ' ' ' ' "/ 1 1 \ 1 \ '; 1 \ 
 
 X i i 1 1 ! i \ \ /! 1 \ 
 
 II 1 1 1 / \ 1 III/ 1 1 \ ^ 1 
 
 \ 1 • 
 
 Fig. 13. 
 
 beyond the end of the given line, in other words, if_ the given line be produced, 
 then, if its direction brings it toward a plane of projection, the point it strikes in 
 the plane of projection, or the point of intersection it makes with that plane, is 
 called its trace on that plane. 
 
 The trace or intersection point is called the Horizontal Trace (H.T.) or the 
 Vertical Trace (V.T.) according to which plane it is found on. Thus at (i) the 
 line AB has an H.T., but not a V.T., since the line is parallel to the V.P. and there- 
 fore cannot meet it in any point. At (ii) there is a V.T. and no H.T. At (iii) 
 the line has both a V.T. and an H.T. So also with (iv) and (v). 
 
 The method of arriving at the traces of any line is no doubt apparent, namely : — 
 Through points in XY where the projections of the line, or the projections produced, 
 meet it, draw perpendiculars; then locate the H.T. in one of these where the plan 
 line meets it, and the V.T. in the other perpendicular where the elevation Hne 
 meets it. 
 
 If the given line has already met a plane of projection, as in (iv), before being 
 
TRACES OF LINES 
 
 19 
 
 [)roduced, then the place in that projection plane must be located and marked as 
 the trace of the line. 
 
 A special case presents itself when the projections of the line are both per- 
 pendicular to XY, as at ab, a'b' 
 
 in Fig. 14. To find the traces ^JaT-, 
 
 in this case, first swing the line 
 down to a horizontal attitude 
 so that a'ob' represents it in 
 elevation and a2b in plan. Pro- 
 duce the plan buj to c with its 
 elevation r'. It will now be 
 seen that if the line be length- 
 ened or produced sufficiently 
 to make it meet the V.P., its 
 elevation will have to be added 
 to, so that it will be from b' to 
 c'2, and this latter point will be 
 the V.T. of the given Hne. 
 
 Similarly, turn now the 
 plan ab to the position ab-j and 
 obtain the elevation a'b'2 for it 
 when the line is parallel to the 
 \'.P. Produce the elevation to 
 d', and find that the lengthened 
 line will need a proportionately lengthened plan 
 which is the H.T. of the given line. 
 
 Fig. 14. 
 
 Carry d by 
 
 arc to J2 
 
 KXKRCLSK VIII 
 
 nA 
 
 ^''" Kind and mark clearly the true lengths, the angles to the planes of pf^ction, and the traces 
 of thd lines whose projections are given at .1. B, C, etc. The lower linfes are, in all the cases, 
 
 a' 
 
 plans. Separate the dilTcrent cases so that no overlapping ol the solutions will lake place, and 
 work to a large scale. " "^ 
 
CHAPTER IV 
 
 SHADOWS OF LINES 
 
 Section 7. A ray of light may be represented by a line, and for our present 
 purposes, such a hne, with an arrow-head, will serve to indicate the direction of 
 parallel rays of light and will be represented by projections, plan and elevation, such 
 as may be seen at (i) in Fig. 15, the arrow-line below XY always being the plan. 
 
 Suppose a stick represented by a line AB he held or placed in various posi- 
 tions and attitudes in relation to the planes of projection as indicated by the pro- 
 
 CD 
 
 (II) (III) , (IV; 
 
 6n 
 
 T I 
 
 (VI) 
 
 (VII) 
 
 ! ab 
 
 abc 
 
 Fig. 15. 
 
 jections for it at (ii), (iii), etc., Fig. 15. A ray of light parallel to the obHque 
 line marked with arrow-heads at (i), striking the end B will be stopped there, 
 and instead of the ray reaching one of the planes of projection, the shadow of B 
 will be recorded on that plane. In other words, the shadow will be the H.T. or 
 the V.T. as the case may be, of the ray. See H and V in (ii) and (iii) respectively. 
 Do similarly, with regard to any other point, say C, in the hne representing the 
 stick at (iv) and at (v) . Also, other points in the Hne may be dealt with to prove that 
 a straight Kne will. have a straight Hne for its shadow on any plane it is cast upon. 
 
 Notice that when an end of the line or stick touches a plane, as in (vi) and 
 (vii), the shadow will start from that point, for, of course, no ray can reach the 
 plane at the point where the line touches the plane, and hence that point is part 
 of the shadow, and the shadow of any other point in the Hne wiU be found in a Hne 
 continuing across the plane from the touching point. 
 
 An examination of cases (v), Fig. 15, and (i). Fig. 16, wiH reveal the fact that 
 when a line is paraHel to a plane the shadow on that plane is equal in length to the 
 line itself, and paraUel to it. Hence in cases shown at (ii) and (iii) in Fig. 16 
 
 20 
 
SHADOWS OF LINES 
 
 21 
 
 this fact can be taken advantage of, and therefore the shadow of the point B having 
 been found, from it a parallel to the given line is drawn for so much of the shadow 
 as falls upon the plane to which the given line is parallel, and which catches the 
 shadow of B. From the point in XY where this shadow line meets XY, join up 
 to the shadow of A caught on the other projection plane. 
 
 In Fig. 1 6, (v), it will be seen to be necessary, where the line is an obHque 
 line, and part of its shadow is caught by the V.P., to first cast the shadow as if it 
 fell completely on the H.P., part of it beyond the V.P., and then to join up from 
 the point where it crosses AT to the shadow of its upper end on the V.P. 
 
 (II) 
 
 Three other cases are shown in Fig. 17, where the line is so arranged that it 
 is neither parallel to one plane nor to the other. In all three cases the end A 
 touches the V.P. and therefore that part of the shadow which falls on the V.P. 
 
22 
 
 DESCRIPTIVE GEOMETRY 
 
 must start from a\ while in (i) and (ii), for the same reason, that part of the shadow 
 faUing on the H.P. must start from b. The only way to obtain the direction of 
 the shadow as it crosses one plane or the other, is to find the shadow of some inter- 
 mediate point, chosen at convenience, such as M in (i), N, say, at one- third dis- 
 tance from A in (ii), and O, say, at half way between A and B in (iii). 
 
 The shadow of the portion BM in (i) is obviously hm2, and because a straight 
 line gives a straight line shadow on a plane, this shadow line must be continued 
 until it meets XY , from which point in XY the balance of the shadow can be 
 drawn to a . 
 
 EXERCISE IX 
 
 I. Find the shadows on the planes of projection for the lines at .4, B, C, D and E. The 
 directions for the rays are indicated by the arrow lines, the number of degrees near the arrow 
 line showing at what inclination the projection of the arrow line in each case must be made to 
 XY. In working the exercise, make the elevations of the given lines about 25" long, and arrange 
 the cases so as to avoid overlapping. 
 
 2. Make another set of lines arranged in various attitudes somewhat similar to those here 
 given, and find the shadows cast when the parallel rays of light are directed towards the planes 
 of projection, their elevations making 45° with XY and their plans 30^. 
 
 SHADOWS OF PLANE FIGURES 
 
 Section 8. The application of the method of finding shadows of lines to the 
 problem of finding shadows of plane figures is a simple step. Illustrations are 
 given in Fig. 18. 
 
 The rule for shadows of fines that are parallel to a projection plane upon 
 which they are cast will save a lot of work, as it will only be necessary, if that 
 rule is taken advantage of, to find the trace of a ray through one point (A) in 
 cases (i), (ii), (iv), (v) and (vi), and the H.T. of the ray through]^ and the V.T. 
 of the ray through B in case (iii). 
 
 Study of cases (iv) and (v) will show the obvious fact that a plane figure, 
 parallel to a plane, will cast its real shape, and size also, for its shadow, on that 
 
SHADOWS OF PLANE FIGURES 
 
 23 
 
 plane to which it is parallel; therefore in case (vi) it is necessary only to find the 
 trace of the ray through the centre A, and make a circle with this trace as centre 
 and radius the same as that of the circle casting the shadow. 
 
 Fir.. i8. 
 
 Further illustrations are shown in Fii^s. ig, 20 and 21, dealing with plant- 
 figures parallel to the H.P., but casting only part of the shadow, in each case, 
 on the H.P. 
 
 In Fig. 19 a square figure, parallel to the H.P., touches the \'.P. at one corner. 
 First find the H.T. of the ray through A at (72, and 
 with lines parallel to the edges of the figure, make 
 the shadow of the square on the H.P. so far as it 
 is cast upon that plane. Next find the V.T. of the 
 ray through B at b'2 and finish the outhne of the 
 shadow, which must, of course, run up to the point 
 where the figure touches the V.P. In Fig. 20 a 
 circle, parallel to theH.P., casts most of its shadow 
 on to the H.P., found by obtaining //, the H.T. of 
 the ray through the centre A, and using it as a 
 centre for a circle, or rather, the segment of a circle, 
 on the H.P. The rest of the shadow, namely, the 
 shadow of the segment BCD, will be caught by the j.i^ ip, 
 
 V.P. To obtain it, find V.T.'s for a number of rays 
 
 passing through points on the curve BDC. Some such points are marked at i, 2, 
 3 and 4 in the figure. A free-hand curve is drawn on the V.P. through the V.T.'s 
 of the rays passing into these points. 
 
 In Fig. 21 the H.T. of the ray through the centre .1, falls beyond the VF, 
 providing for only the smaller segment of a circle to be cast as shadow on the 
 
24 
 
 DESCRIPTIVE GEOMETRY 
 
 H.P. Use this point H as centre, as before, and describe the circular part of the 
 shadow appearing on the H.P. in front of the V.P., and then proceed to find the 
 shadow cast on the V.P. as in Fig. 20. 
 
 Fig. 21. 
 
 EXERCISE X 
 
 1. Find the shadows for the rectangle, the square and the circles shown at (a), {h), (c) and 
 (d). Keep them well apart from each other. 
 
 2. Find the shadow of the awning represented in plan and elevation at (e). 
 
 3. Arrange other cases for the square and the circle, varying the direction of the rays of light. 
 For example, let a horizontal circle, 2" diameter, touch the V.P. 3I" above the XY, and let the 
 rays have elevations at 45° and plans at 30° to XY. The result will be a shadow on the V.P. 
 elliptical in shape. 
 
 N.B. — The student is urged to verify these solutions by actual sunlight, and models held 
 in various relations to the planes of projection properly held at right angles to one another. 
 
CHAPTKR V 
 
 THE "COMPOUND ANGLE " FOR LINES 
 
 Section 9. Previously we have seen how to find the true length of a given 
 line, and also its inclinations to the projection planes, when the projections, plan 
 and elevation, are given; and it has been realized that the length made for 
 the projection of a given line will influence its inclination to the projection plane. 
 See Fig. 4. So, conversely, the projection length, for a given line, depends on its 
 real length and its inclination to the projection plane. Illustrate this fact by the 
 use of a stick or pencil to serve for a line. 
 
 Fig. 2: 
 
 It must also now be realized that the conditions of its length and of its inclina- 
 tion to a projection plane, establish the dilTerence of distance of the ends of the line 
 from that projection plane. 
 
 Thus in Fig. 22 at ^, a line arranged parallel to the V.P. will show its true 
 length, say, 2", and its true inclination to the H.P., say 45°, while its plan length, 
 depending on these, is shown by the arrow-ended line P below AT, and also the 
 difference of distance of its ends from the H.P. may be measured between the 
 levels of the dotted lines passing horizontally through the ends of the elevation, 
 and marked by another arrow-headed line, D. 
 
 A line of the same length arranged at, say, 30° to the V.P., will show that angle 
 between its plan and XY, and also will show its true length in the plan if the line 
 be placed as a parallel to the H.P., as in Fig. 22 at 5, with similar results as to 
 projection length, E, on the other plane, and difference of distance of its ends, D. 
 from that other plane. 
 
26 
 
 DESCRIPTIVE GEOMETRY 
 
 Note that, because, while the length of the hne remains the same at 5 as at ^, 
 the angle is changed, therefore the length of projection is changed, and also the 
 difference of distance of its ends from the projection plane is changed. 
 
 Hence it will be seen that there are four facts, which must be borne in mind, 
 about any given hne, namely: — (a) Length; (b) Inclination — attitude in relation 
 to a projection plane; (c) Projection length, and (d) Diference of distance of its 
 ends from the projection plane. Given any two of these facts the others may 
 be found. 
 
 In further illustration of this, let a Hne, say 2" long, have its lower end fixed 
 in XY, and be inclined at, say, 45° to the H.P. By placing it in the V.P., 
 as in a previous lesson, the Hne 
 will show its true length for 
 elevation and its true inclination 
 to the H.P. in its projection on 
 the V.P. It win be foreshort- -^ 
 ened in plan as seen in Fig. 23 
 at A to an extent depending on 
 its true length, 2", and its in- 
 clination, 45°, to the H.P. In 
 other words, the plan length 
 
 proper to this^line is found, viz., ab, and also the level for the free end of the line 
 is found to be at a height H above the level of the lower fixed end. Again, if a 
 line, say 2" long, be arranged to lie in the H.P. at, say, 30° to the V.P., with 
 one end fixed in the XY as at B in the figure, then the elevation length, a'b', and 
 the proper distance, Z>, from the V.P., of the free end, are determined. 
 
 Fig. 23. 
 
 EXERCISE XI 
 
 A; 
 
 Find the other projection for each of the given five lines, one projection of which is given, 
 together with a description as follows: — 
 
 AB is 25" long and one end of it is in the H.P. 
 CD is inclined to the H.P. at 30°. 
 EF is at 60° to the \'.P. 
 GH is at 45° to the H.P. 
 JK is at 60° to the V.P. 
 
THE "COMPOUND ANGLE" FOR LINES 
 
 27 
 
 By combining or compounding the results we have obtained at A and at B 
 in Fig. 23 we shall arrive at a method for the solution to the problem of finding 
 the plan and elevation, properly placed in relation to each other and to XV, of 
 any given line at given angles to the planes of projection. 
 
 Thus, as an illustration, let the line be 2" long, with one end tlxed in XY, 
 and let the angle it makes with the H.P. be 45°, and that with the V.P. 30^. 
 
 In Fig. 24 at (i) the height or level for the upper end of the 2" line inclined 
 at 45° to the H.P. is marked by the dotted line hh, and the elevation length for a 
 2" line at 30° to the V.P. is obtained and marked e'e'. This line is then moved 
 by an arc into its proper position, namely, between the levels proper for a 2" line 
 at 45° to the H.P., i.e., between XY and the parallel ////. Now as the projections 
 of any line are necessarily perpendicularly opposite each other across XY, it follows 
 that the plan for this line at A must be opposite its elevation, and as it must also 
 
 (in 
 
 (III) 
 
 Fio. 24. 
 
 be limited to the space between XY and the parallel dd due to the 30'' angle, a 
 perpendicular from the free end, that is, the upper end of the elevation, drawn to 
 this parallel dd, will decide its plan. 
 
 At (ii) the same result is arrived at by obtaining the plan length, marked pp, 
 and moving it by an arc into its position due to the 30^ angle, and from this plan 
 the elevation, which must be opposite to it, is found. 
 
 If, instead of the arrangement of the line by which its projections meet in a 
 point in AT as in Fig. 24 at (i) and (ii), an arrangement of it is desired, so that its 
 upper end is, say, in the V.P. and its lower end is, say, in the H.P., then the solu- 
 tion will be obtained most readily by working it out as at (iii), in the following 
 order: — (a) plan length due to 45°; (b) distance of lower end from AT due to 30"; 
 (c) plan put into position, and (d) elevation found. 
 
 This problem, discussed in Fig. 24, is sometimes spoken of as the problem of 
 the Compound Angle. 
 
 In Fig. 25 are shown two common cases of error made by beginners in their 
 attempts to solve this problem without properly understanding what they are 
 
28 DESCRIPTIVE GEOMETRY 
 
 doing. The student is urged to make frequent use of some commonplace model 
 of the planes of projection, arranged at right angles to one another, and to mark 
 upon them projections of lines and to study these. Then open out the 90° angle 
 between the planes, and so represent the flat paper upon which the projections 
 are to be made. Eventually the student will be readily able to "read" sets of 
 
 Fig. 25. 
 
 projections and also to imagine or picture them, for cases described, without having 
 to VianHlp a moHel. 
 
 to handle a model 
 
 EXERCISE XII 
 
 Find projections for the following lines: — 
 
 (i) AB, 2\" long, inclined at 45° to the H.P. and at 30° to the V.P. 
 
 (2) CD, 2\" long, inclined at 50° to the H.P. and at 20° to the V.P. 
 
 In both the above cases show two different ways of dealing with the construction. 
 
 (3) EF, 2\" long, at 40° to the H.P. and at 50° to the V.P. 
 
 (4) Find projections for a 2\" line, at 40° to the H.P. and at 30° to the V.P., by a method 
 
 providing for one end of the Hne to be in the H.P. and the other in the V.P. 
 
 (5) Propose other cases, varying the position of, say, the lower end of the line. 
 
 N.B. — When proposing new cases for oblique lines, the student should realize that the sum 
 of the two angles for the same line must not exceed 90°, and that in the case of the sum of the 
 two angles being the limit of 90°, the plan and elevation will both appear as perpendicular to XY 
 
 It will now be recognized that upon this problem depends a method of finding 
 or arranging traces for any oblique plane whose angles with the projection planes 
 are given. For instance, if it be required to find the traces for a plane which is 
 inclined at 50° to the H.P. and at 70° to the V.P., since the traces required are 
 perpendicular to the projections of a line that is at right angles to the plane, it 
 will be seen that a Hne at 40° to the H.P. and at 20° to the V.P. will be a perpen- 
 dicular to this plane; therefore, find the projections of this 40°, 20° line, and then 
 make traces for the required 50°, 70° plane, perpendicular to these projections. 
 
PROJECTIONS OF PLANE FIGURES 
 
 29 
 
 EXERCISE Xlir 
 
 Find the traces for the following oblique planes: — 
 
 (a) 60° to the H.P. and 50^ to the V.P. 
 
 (b) 45° to the H.P. and 65° to the \'.P. 
 
 (c) 30° to the H.P. and 60° to the V.P. 
 
 N.B. — Here it should be noticed, in proposing additional exercises, that the sum of the two 
 angles for an oblique plane cannot be less than qo°, and in the case of the sum being equal to 90" 
 the traces for the plane will run parallel to the XV line. 
 
 PROJECTIOXS OF PL.VNE FIGURE.S, IXVOLVIXO TflE COMPOFXI) AXGLE 
 
 Section 10. Illustration may now be given of the application of the com- 
 pound angle problem to cases of plane figures in which it is involved. And first 
 let it be realized, experimentally, that when two lines meet at right angles, if one 
 of them is in a plane of projection or parallel to it, the projection on that plane, 
 
 Fig. 26, 
 
 Fig. 27. 
 
 of the second line, will be perpendicular to the projection of the first Une. Con- 
 versel}-, if one of them has to be inclined to the planes of projection, the arrange- 
 ment of the other must follow it at right angles to the projection of the first, on 
 the plane to which the second one is parallel. Consider, for example, that if one 
 diagonal of a square be horizontal and the other be at 45° to the H.P. and 30° 
 to the V.P., then it is impossible to say that the horizontal one makes any par- 
 ticular angle with the V.P., for its plan simply follows at right angles the plan of 
 the inclined diagonal. It is only possible therefore to say of it that it is horizontal. 
 Study the cases shown in Figs. 26 and 27. The plan and elevation in Fig. 26 
 have been found for a square having one edge, AB, in the H.P., and an adjacent 
 edge, BC, inclined at 50° to the H.P. and at 30° to the V.P.; and in Fig. 27 the 
 plan and elevation have been found for a square when one diagonal. AR, is hori- 
 
30 DESCRIPTIVE GEOMETRY 
 
 zontal and the other diagonal is at a compound angle— 50° to the H.P. and 30° to 
 the V.P. 
 
 The first step is to find the plan shaj £, as at (i), quite irrespective of the com- 
 pounding of angles for the incHned edge or diagonal. The next step is to find the 
 compound an^le for the obhque line— the edge BC in Fig. 26 and the diagonal 
 CD in Fig. 27. For this, any length of Hne may be taken to serve the purpose, 
 because all that is required is to obtain the direction of the plan, marked by the 
 arrow line in each case. Now, either fit the plan shape' to this arrow line, as at 
 J2C2, Fig. 26, or make C2d2 parallel to it as in Fig. 27. The elevation, in each case, 
 is the last thing to find, and It may be derived from the newly arranged plan, and 
 the first elevation which provides levels. 
 
 Care must be taken, that, in working out the compound angle, the same 
 length of line must be used (any length will do) to be placed at 30° to XY, as that 
 which has been placed at 45° to XF. 
 
 EXERCISE XIV 
 
 1. Find the plan and elevation of a rectangle, 2" by 3", when a short edge is in the H.P. 
 and the long edges are inclined at 40° to the H.P. and at 25° to the V.P. 
 
 2. Find the plan and elevation of a square 2" edge, when one diagonal is horizontal and the 
 other is at 50° to the H.P. and at 25° to the V.P. 
 
 3. Find the plan and elevation for a circle 2" diameter, when the diameter. perpendicular 
 to the horizontal diameter (there is always a horizontal one) is at 55° to the H.P. and at 20° to 
 the V.P. 
 
 4. Propose, and work out, other problems of plane figures to involve the compound angle, 
 and realize that questions like Nos. 5 and 6, which follow, do not involve the problem of finding 
 the compound angle. 
 
 5. Find the plan and elevationwaf a rectangle having its long edges horizontal and inclined 
 at 30° to the V.P., and.its short edges at 50° to the H.P. 
 
 6. Find the plan and elevation of a square when c ne diagonal is horizontal and at 50° to the 
 V.P. and the other diagonal is at 25° to the V.P. 
 
( HAPTKR \l 
 
 PROJECTIOX OF SIMI'LK SOLIDS 
 
 SectiOx\ 11. The student will now be able to deal with easy problems 
 involving the use of simple solids, and as the solids used will be familiar tj-pes, the 
 attention can be chiefly given to the descriptions with regard to position or place, 
 and with regard to attitude while in any particular position. The student is 
 advised to make use of small cardboard boxes or models of wood, and continu- 
 
 FlG. 28. 
 
 Fig. 29. 
 
 Fig. 30. 
 
 ally to remember that we are to represent these solids by projecting their appear- 
 ances perpendicularly on to the two planes of projection. Therefore he should 
 study the possible attitudes or arrangements of the solids in relation to a vertica- 
 plane and a horizontal plane held or placed conveniently for experimenting purl 
 poses. 
 
 In the illustrations Figs. 28, 29 and 30. the order of procedure may be studied 
 for finding plans and elevations of such solids referred to above. 
 
 In Fig. 28 is represented a cube having one of its faces in the H.P. and another 
 face at 30° to the V.P. The dotted line in the elevation represents an edge 
 not seen, since the student is supposed to look toward the projection plane with 
 the solid between him and the projection plane. This applies to both planes, 
 hence the dotted lines appearing both in plans and in elevations in Figs. 29 and 30. 
 Note that the same edges are not represented by dotted lines in both projections. 
 
 31 
 
32 
 
 DESCRIPTIVE GEOMETRY 
 
 In Fig. 29 the cube has one edge in the H.P. at 30° to XY, and a face adjacent, 
 that is, to which that edge belongs, is at 30° to the H.P. There must evidently 
 be a preliminary set of projections at (i) before the plan and elevation required 
 can be found at (ii). This also applies to Fig. 30, where the cube has one edge of a 
 face in the H.P. and another edge of the same face is at 60° to the H.P. and at 
 25° to the V.P. The preUminary plan shape and elevation for levels are shown 
 at (i), and the compound angle problem for the edge BC is worked out as in Fig. 26, 
 the arrow line showing the direction for &2C2 in the required plan. 
 
 In the illustration Fig. 31 a regular hexagonal prism is arranged so that a 
 short edge is in the H.P. and the long edges are at a compound angle, viz., 45° 
 to the H.P. and 20° to the V.P. The upper end of the solid is turned away from 
 
 Fig. 31, 
 
 the V.P., and is therefore fully visible as a foreshortened hexagon, while dotted 
 lines have to be used for part of the outline of the lower end. In working the 
 compound angle problem it is only necessary to find the plan direction, marked 
 with the arrow line in the figure, as the elevation will naturally assume its proper 
 direction. 
 
 In Fig. 32 a right cylinder is arranged so that its axis is at 50° to the H.P. 
 and at 15° to the V.P. This means that the circular end will be at 40° to the 
 H.P. Its plan should be found by the method previously explained. The com- 
 pound angle for any length of line RR is found in such a way (see Fig. 24, (iii)) 
 that when the arrow line, which indicates the direction for the plan of the axis, 
 is obtained, and the plan of the axis of the cylinder is made parallel to it, the lower 
 end of the cylinder is turned away from the V.P. and appears in the vertical pro- 
 jection in full view, therefore is clear-Hned all around. The upper end will be 
 partly outlined by a dotted line. The elevation of the axis should be drawn, 
 
PROJECTION OF SIMPLE SOLIDS 
 
 33 
 
 and, of course, will not be at 50'^ to XY. When the elevation ellipses have been 
 correctly drawn, the width of the elevation of the cylinder, that is, at right angles 
 to the axis, will be the same as that of the plan. 
 
 Fig. 3j. 
 
 In Fig. 33 a regular pentagonal pyramid is made to rest with one of its tri- 
 angular faces in the H.P. and the axis of the solid — the line from its apex to the 
 centre of its base — is at 30° to the V.P. This means that the compound angle 
 
 Fig. 
 
 must be found for the axis, since there are two angles to be taken into consideration, 
 namely, the angle a° which it makes with the H.P.. and the 30^ angle it makes with 
 the \'.P. In the fmal position, therefore, the plan oi the axis is made parallel to 
 the arrow line obtained by compounding these angles. 
 
34 
 
 DESCRIPTIVE GEOMETRY 
 
 In Fig. 34 there is no compound angle problem required to be worked out. 
 A right circular cone is represented with its axis at 45° to the H.P., and the hori- 
 zontal diameter of its base is at 50° to the V.P. Notice that in drawing the lines 
 for the sides of the incHned cone, where an ellipse is part of the drawing, these side 
 lines are tangent to the curve of the ellipse, and do not run into the ends of the 
 major axis of the ellipse. It will be advisable to draw the line ee the full length 
 
 Fig. 34. 
 
 of the diameter of the base, at right angles to the elevation obtained of the axis 
 of the cone. In finding the plans and elevations of circles inclined, not less than 
 eight points in the curve should be found. 
 
 EXERCISE XV 
 
 1. Find the plan and elevation for a cube, 2" edge, which has one edge in the H.P. at 60° 
 to XY, and a face adjacent to that edge at 30° to the H.P. 
 
 2. A pentagonal prism has one short edge, ij" long, in the H.P. The long edges, 3" long, 
 are at 45° to the H.P. and at 25° to the V.P. Find the plan and elevation. 
 
 3. A hexagonal pyramid, base edge i", axis 3", has one triangular face in the H.P., and 
 the horizontal diagonal of the base is at 45° to the V.P. Find the plan and elevation. 
 
 4. A right circular cone, diameter of base 2", axis 3", lies on its side on the H.P., with its 
 apex therefore in the H.P. Find the projections for it when its axis is at 30° to the V.P. 
 
 5. Propose and work out other problems, for example, on the square prism, the cylinder, 
 the pentagonal pyramid and the cube, after the manner of those already given, some to involve 
 the problem of the compound angle and some not. 
 
 N.B. — This suggestion that the student himself shall be required to prepare in writing the 
 exact description and data for additional exercises, and then work them out, is very important, 
 for it is a good test of the student's grasp of the subject so far as the ground has been covered, 
 and his powers of concentration. 
 
SHADOWS OF SOLIDS AND OF GROUPS OF SOLIDS 
 
 35 
 
 SHADOWS OF SOLIDS AND OF GROUPS OF SOLIDS 
 
 Section 12. The more difficult cases of shadow problems, involving the use 
 of solids, may now be undertaken. Let the rays of light be parallel as previously, 
 and directed toward the planes of projection, the direction to be shown b)- the 
 plan and elevation of an arrow line. 
 
 Before commencing to work out the solutions for such shadow problems 
 as now to be dealt with, the student must endeavor to picture or imagine the solid 
 or group represented by the plan and elevation, or described verbally, with the 
 light directed toward it, and then try to decide what edges will cause or are likely 
 to cause the outline or contour of the shadow which will be cast. 
 
 Fig. 35- 
 
 The student will not be able to make very satisfactory progress until he can 
 readily conceive mental pictures of what he has to deal with, realizing the relation 
 of what he is picturing to the vertical and horizontal planes of projection, and 
 judging what the projections and other results might be like. Therefore it is 
 important that the student should take advantage of every opportunity which 
 presents itself to train himself in these matters. 
 
 In Fig. 35 it will be seen that a cube is represented, and by studying it as 
 suggested above, it will be concluded that the light will fall on three of its faces, 
 and that two vertical edges, those at .1 and at C, and four horizontal edges, the 
 upper ones AB and BC, and the lower ones AD and DC, are the edges which will 
 cast shadows to give the shadow shape. Deal with them in the order suggested 
 by the numbers i to 6 in the figure. Notice that 3 and part of 4 will be parallel 
 to DC and DA respective!} . 
 
 In Fig. 36 it will be seen necessary to cast the shadow of the apex of the pyra- 
 mid on to the H.P. beyond the A'F line first, in order to draw from that point, 
 
36 
 
 DESCRIPTIVE GEOMETRY 
 
 on the H.P. the Unes forming the contour for that part of the shadow falhng upon 
 the H.P. in front of the V.P. (see Fig. i6, (v)). These contour Hnes are then con- 
 tinued up the V.P. from where they break at XY, to the actual shadow, on the 
 V.P., of the apex. Notice that as two triangular faces of the pyramid, visible 
 
 Fig. 36. 
 
 Fig. 37. 
 
 in plan, have no light falling on them, they are shaded. They are said to be in 
 shade, as distinguished from the cast shadow just found. In the case of the cone 
 at Fig. 37, the part in shade is determined by the radii that are perpendicular to 
 the tangent hnes serving as contour or boundary Unes for the shadow on the H.P. 
 
 In Fig. 38 it must first be realized that the outhne of the shadow of the circular 
 block there represented will be caused by the vertical hnes on its sides found in 
 plan at A and at B, the upper rim ACB, and the lower run ADB. The shadows 
 of the vertical lines at A and B should first be found, then, by taking points on 
 
SHADOWS OF SOLIDS AND OF GROUPS OF SOLIDS 37 
 
 the rim lines mentioned, the curves can be obtained. The part of the lower rim 
 casting its shadow on the H.P. will give a circular shadow with its centre at e. 
 Only this, and the straight-Hne parts of the shadow have been found in Fig. 38; 
 the remainder has been purposely left unfinished in order to emphasize the impor- 
 tance of first obtaining correctly the parts that are here found. In each of the 
 cases, Figs. 35 to 38, the part of the surface in shade and visible in the elevation 
 is shaded with hatching to indicate that fact. 
 
 Fig. 39 represents a flat block or slab resting on a vertical cylinder. First 
 find the complete shadow of the group as it is cast on the planes of projection, 
 and, in doing this, after drawing the shadow outlines on the H.P. and starting 
 
 Fig. 39. 
 
 Fig. 40. 
 
 them vertically up the V.P., find the outline of that part of the shadow which 
 falls on the V.P. The best order in which the parts of the outhne are obtained 
 is indicated by the numbers. Notice that 5 will be made parallel to 4. and 6 made 
 parallel to 3, for reasons that are obvious. 
 
 After this shadow on the planes of projection is found, then find the shadow 
 cast on the near or front half of the cylinder, by a portion of the upper solid. 
 The group is reproduced in Fig. 40 in order to avoid confusion of lines, and to 
 show only the shade and shadow seen on the near part of the cylindrical solid. 
 The line AB should first be drawn, perpendicular to the plan direction of the rays. 
 Then from B a vertical line on the near or front surface of the cvlinder in elevation 
 
38 
 
 DESCRIPTIVE GEOMETRY 
 
 divides the light from the shade. Points in the order i to 8 should then be decided 
 on, and shadows of these points cast on the surface of the cyhnder. Freehand 
 curves will then mark the outline of the shadow cast. Notice that point 2 is the 
 one that casts a shadow on the outside generator whose plan is C. 
 
 In Fig. 41 a group consisting of a pyramid standing on a square block is repre- 
 sented. The shadow caught by the top surface of the block will be obtained by 
 supposing the surface of it extended sufficiently to hold the whole shadow. This 
 will give lines i and 2. The points a and h must now follow the shadows of the 
 
 Fig. 
 
 edges they are on, to az and ^2. From ^2 the shadow of the slant edge of the pyra- 
 mid continues on the H.P., parallel to 2 on the upper surface of the block, until 
 it reaches the XY , when it runs up the V.P. to the shadow of the apex, on the V.P., 
 and from ao a line on the V.P. to the shadow of the apex, completes the contour 
 or outline of the shadow cast. 
 
 EXERCISE XVI 
 
 1. Work out carefully all the cases dealt with in Figs. 35 to 41, making much larger drawings. 
 
 2. Find shadows on the H.P., the V.P. and on the under solid, for a group consisting of a 
 circular slab 2" diameter, f" thick, resting centrally on a cube, if" edge. The cube rests on 
 the H.P. with its vertical faces equally inclined to the V.P., and one vertical edge \" from the 
 V.P. The rays of light have plans at 60° to XY, and elevations at 45° to XY. 
 
 3. A square block 2" edge, and 1" thick, rests with a square face centrally on a circular 
 cylinder, if" diameter, 2\" high, standing on the H.P. The horizontal edges of the square 
 
SHADOWS OF SOLIDS AND OF GROUPS OF SOLIDS 
 
 39 
 
 block are equally inclined to the \-.P. A short edge of the upper sol.d is m the \ .P. The real 
 inclinations of the rays are 50° to the H.i>. and 30° to the X.V. Find the shade and shadows 
 
 4 \ right circular cone, 2" high, 2" diameter of base, stands vertically on the centre of a 
 square block, i" thick, sides 3"- The block has one rectangular face parallel to the \ .1 ^and 
 i'' from it. The rays of light have plans at 45° to XY and elevations at 30 to A Y. I-md the 
 
 shade and shadows. r wu , 
 
 5. Prepare other exercises, using your best judgment, and work them out for further 
 
 practice. 
 
PART TWO 
 
 CHAPTER VII 
 
 RABATTEMENT OF PLANES, AND PROJECTIONS OF FIGURES 
 INVOLVING ITS USE 
 
 Section 13, The process known as rabattement is that by which a plane is 
 made to swing on one of its traces, as on a hinge, until whatever is represented as 
 being in the plane, whether it be point, line or figure, is turned into one of the 
 planes of projection, and its true form is made evident, also its relation to the trace of 
 
 Fig. 42. 
 
 the plane. In Fig. 42 the plane RST, arranged perpendicularly to the V.P., has 
 in it a point P. If the plane is made to swing about its trace 5r as a hinge line 
 until it is turned over into the H.P., either on the one side or on the other, it carries 
 the point with it, and the point's rabattement is said to be at P2 or at P3. If D be 
 a point perpendicularly opposite P, and in the H.T. or hinge line, then DP may 
 be considered as the radius line for an arc in which the point P moves. This 
 radius length, as shown in the figure at (ii), is the hypotenuse of a right-angled 
 triangle having for its base the perpendicular distance from the plan p to the 
 
 40 
 
RABATTEMENT OF PLANES 
 
 41 
 
 trace at d, and for its height the distance from A'l' to p' . The right angle at c in 
 the figure should be specially noted. 
 
 In Fig. 43 an oblique plane is represented at RST and a point P in it with 
 plan p and elevation p' . The construction for rabattement is identical with that 
 of Fig. 42, and the point when rabatted appears in the H.P. at p2 or at pi. It 
 -hould be noticed that whereas in Fig. 42 the true inclination between the two 
 trace lines RS and ST is a right angle, in Fig. 43 the true inclination between 
 the traces of the plane is the angle between ST and a line from S through />3 or 
 through p2. This angle may be greater or less than a right angle according to 
 the arrangement of the traces of the oblique plane given. 
 
 In Fig. 44 is shown the rabattement of a plane carrying with it a line which 
 has previously been placed in the plane and arranged so as to be at 30° to the 
 H.P. To arrange the line in the plane it must first be placed in the V.P. with 
 
 Fig. 43. 
 
 one end of it at any convenient point hh' , the other end of it in A'l' at c. The plan 
 he is then moved, by an arc with h as centre, to bring the lower end of the line 
 into the H.T. of the plane at a. The projections of the line are ah, a'b\ and its 
 rabattement is at ah-i or at ah?.. 
 
 An illustration of the use of this method is given in Fig. 45, where it is required 
 to find the plan of a square when it is inclined to the H.P. at 45° and one edge 
 of it is inclined to the same projection plane at 20°. On the rabattement ahi of 
 the line AB the figure, true size and shape, is constructed at c-zdzCi- Any ix)int 
 in this rabattement figure will move across the H.T. at right angles until it 
 reaches its plan position. Thus c-z and d-^ will be carried across to their plans 
 c and (/ respectively. 
 
 In order to get the plan of E from Ci, the line f.x/j is produced to the H.T. 
 or hinge line at/, an immovable point, from which a line through d is drawn until 
 a point opposite c-z is found at c. Parallels to cd and dc will complete the plan 
 required of the square. 
 
42 
 
 DESCRIPTIVE GEOMETRY 
 
 When the figure is required to be found in an obhque plane, then the elevation 
 obtainable will be a figure as at Fig. 46. In Fig. 45 since the plan only was called 
 for, the plane of the figure, RST, was arranged perpendicular to the V.P., and 
 this would give a line only for elevation, if it were needed. 
 
 In Fig. 46 the plan and elevation are obtained of a regular pentagon placed 
 in a given oblique plane RST and having one of its edges at a given angle, say 20°, 
 to the H.P. The line ab, a'b' is first found in place, as in the case of Fig. 45, and 
 then rabatted over to a2& (see Fig. 43). On this latter is constructed the pentagon. 
 In constructing this pentagon, the angles of it at c and d are drawn, each 108°, 
 and de and eg made equal to cd. The fifth corner of the pentagon is then obtained 
 by intersecting diagonals from c and d made parallel to the sides de and eg 
 
 Fig. 46. 
 
 respectively. To obtain the plan of the edge whose rabattement is de, produce de 
 to the H.T. at / and from that point draw a line through the plan of D till a point 
 opposite e is found. For the remaining two corners of the plan draw diagonal 
 lines parallel to the plans of the edges already found. The elevation of the point 
 E may be obtained by drawing a line through the elevation of D from the elevation 
 in XY of the point F. 
 
 The use of rabattement is again illustrated in Fig. 47, where two lines AB 
 and BC enclose an angle the size of which it is required to find. As any two lines 
 which meet each other have a common plane whose traces will contain the traces 
 of the lines, it is only necessary to find the H.T. of each of the hnes at R and S 
 respectively, and the fine passing through these is the H.T. of the plane of the 
 lines. Use this H.T. as a hinge line about which to rabat the plane. Set up 
 the usual right-angled triangle whose hypotenuse can be used to swing down B 
 to its rabattement 62. This point joined to the traces of the lines will give the 
 
RABATTEMENT OF PLANES 
 
 43 
 
 rabattement of the angle, showing its true size. This is marked in the figure 
 by the arrowed arc. 
 
 EXERCISK X\II 
 
 1. Find the plan of a regular pentagon, i^" edge, whose plane is at 60' to the H.P. and which 
 has an edge at 30° to the H.P. 
 
 2. I'"ind the plan and elevation of a regular hexagon, i" edge, when it lies in an oblique 
 plane with \'.T. at 60° to XY and H.T. at 45° to XY, and when one edge of the tigure is at 
 20° to the H.P. 
 
 3. Find the true inclinations between the lines meeting as at .1, B and C. 
 
 4. I'ind the true inclination between the H.T. and the \'.T. of any given oblique plar 
 
 By experimenting with two sticks of pencil, realize that a line, inclined to a 
 plane, makes an angle with it which is the complement of the angle the same line 
 makes with a perpendicular to the plane from any point in the inclined line. 
 
 Thus, in Fig. 48 the line AB is evidently inclined to the plane RST. If a 
 perpendicular to the plane RST be drawn from A by making its elevation per- 
 pendicular to RS and its plan perpendicular to ST, there will be contained by 
 
44 
 
 DESCRIPTIVE GEOMETRY 
 
 these two lines, AB and AC, an angle which is the complement of the angle which 
 AB makes with RST. Find this angle by the method employed in Fig. 47. The 
 angle indicated by the arrow-headed arc in Fig. 48 is the required angle the Hne 
 AB makes with the oblique plane RST. 
 
 Fig. 48. 
 
 EXERCISE XVIII 
 
 Find the inclination to the given oblique plane, of the line AB in each of the cases shown 
 a.t A, B and C. 
 
POINTS AND LINES IN RELATION TO OBLIQUE PLANES 45 
 
 POINTS AND LINES IX RKLATION' TO OBLIQUE PLANES 
 
 Sectiox 14. Realize by experiment the following facts: — 
 (i) Parallel lines must be projected as parallel lines, i.e., their plans will be 
 parallel, and their elevations will be parallel, to each other. 
 
 (2) A horizontal line on an inclined plane must be parallel to any other hori- 
 zontal line on the same plane and therefore parallel to the H.T. of that plane. 
 
 (3) A line on an oblique plane must have its V.T., if it has one, in the V.T. 
 of that oblique plane. 
 
 (4) A line on an oblique plane and parallel to the V.P. is parallel to the \'.T. 
 of that plane. 
 
 (5) A line inclined to both planes of projection will have its traces in the traces 
 of any plane containing it. 
 
 Fig. 51. 
 
 Thus the plan ab in Fig. 49 is made parallel to ST to represent a horizontal 
 line in the plane of which ST is the H.T., and when AB is produced it will meet 
 the y.V. in a point cc' in the V.T. of the plane RST in which it Hes. and the ele- 
 vation a'b' of the horizontal line can then be determined. This consideration 
 helps us to fmd projections of definite points on oblique planes. For example. 
 in Fig. 50 a point on the plane LMX has its plan given at a, and there arc shown 
 three different ways of obtaining its elevation: — i, by a horizontal line; 2, by a line 
 parallel to the V.P., and 3, hy an oblique line, the plan of the line in each case 
 passing through the point, and the elevation of the line being obtained. 
 
 In Fig. 51 is shown how to locate, in plan and elevation, any point on a given 
 plane and having given distances from the planes of projection, viz., say 15" from 
 
46 
 
 DESCRIPTIVE GEOMETRY 
 
 the V.P. and i" above the H.P. The solution lines may be drawn in the order 
 indicated, resulting in first the elevation and then the plan. 
 
 EXERCISE XIX 
 
 I. Find the other projection, in each < -e, for the point in the given plane, one projection 
 of the point being given. 
 
 2. Draw traces for any plane, and find the plan and elevation of a point on it f " in front of 
 the V.P. and il" above the H.P. 
 
 In Fig. 52 is shown how to set up, in plan and elevation, a perpendicular to a 
 given plane from any point in it. It will be remembered that a perpendicular 
 to a plane has its projections perpendicular to the traces of the plane. Let aa' be 
 the point. First make a perpendicular of any convenient length, hmited in the 
 
 Fig. 52. 
 
 figure at the arrow-head. Then turn this line so as to arrange it parallel to one 
 of the planes of projection, thus showing its true length. Cut off the required 
 part on this true-length line, and by a parallel to XY the length of one of the pro- 
 jections is determined. A perpendicular across XY will define the other projection. 
 
POINTS AND LINES IN RELATION TO OBLIQUE PLANES 47 
 
 EXERCISE XX 
 
 1. Find the projections of a 2" line perpendicular to a given plane and starting from it in 
 point i" from both planes of projection. Let the plane be inclined at 40° to the H.P. and 65° 
 to the V.P. 
 
 2. The traces of a plane are V.T. at 30° to XV and H.T. at 45° to AT. A line perpen- 
 dicular to this plane has a plan 2" long. Find the projections of a 2" portion of the line. 
 
 3. Work out on a larger scale the following cases: — 
 
 i. Find plan and elevation of a point on the H.T. of this plane, 2" from the AT, 
 
 and set up a perpendicular line from it, 2" long, 
 ii. Find a point on this plane i\" above the H.P. and i" in front of the \".P. 
 iii. Find a 2" line perpendicular to this plane from the point C in it. 
 iv. Find the elevation of AB and its true length. It lies in the given plane. 
 V. Find the plan of D which lies in the given oblique plane. Given d'. 
 vi. Find the plan of KF which is perpendicular to the given plane, and obtain its 
 
 true length, 
 vii. Find the plan of the line (HI which lies in the given oblique plane, 
 viii. Find the elevation of the line JK which is in the given plane, and show its inclina- 
 tions to the H.P. 
 ix. The plan is given of a triangle lying in the oblique plane RST. Find its eleva- 
 tion then by rabattement of each corner in turn, on to the H.P., show the 
 true shape and size of the triangle. 
 
 4. Find the true inclination between the traces of each of the planes at iii. iv and v. 
 
48 
 
 DESCRIPTIVE GEOMETRY 
 
 INTERSECTION OF OBLIQUE PLANES WITH EACH OTHER, AND OF LINES 
 WITH OBLIQUE PLANES 
 
 Section 15, If any free line be taken, represented, for instance, by a pencil 
 for purposes of experiment, it will be realized that any number of planes may 
 contain it, and the condition of this will be, that the trace lines of such planes 
 pass through or contain the trace points of the line. 
 
 In the illustration in Fig. 53 there are six such planes shown. Two of them 
 are specially important and useful, owing to the fact that they can be drawn or 
 
 X y 
 
 yV 
 
 :^' I 3 
 
 i\ 
 
 ! .1 
 
 ]^-^^4i.T. 
 
 Fig. 53. 
 
 represented without the necessity of first discovering the trace points of the given 
 line. They are Nos. 2 and 4 — the one, a plane perpendicular to the V.P., and 
 the other a vertical plane or plane perpendicular to the H.P. 
 
 The next thing to realize is that since planes intersect each other in straight 
 lines, it will easily be possible to show the plan and elevation of an intersection 
 line by joining the point on the V.P. common to the Vertical Traces of the two 
 planes (i.e., where the two V.T.'s meet), to the point on the H.P. where the planes' 
 H.T.'s meet. In Fig. 54 an illustration is given where plane RST intersects plane 
 LMN in line AB. The plan of the intersection line is at ah and its elevation 
 is a'h'. 
 
INTERSECTION OF OBLIQUE PLANES WITH EACH OTHER 49 
 
 Further realize that a free Unc directed to any plane, or passing through it, 
 will h^e an intersection point on that plane, and this will be a po.nt ,n the mter- 
 
 Fig. 54. 
 
 rzr:ri:T.t=:i:n/:r;rL='r.v" 
 
 Fio. 55- 
 
 A plane /,.1/.V containing AB intersects ^r in the line passing through V and H 
 and shown in plan and elevati.in at vH and I .1/. 
 
50 
 
 DESCRIPTIVE GEOMETRY 
 
 When the plan of the line AB, namely, ab, is produced to meet the plan of 
 this intersection line, namely, vH, the plan of the intersection point is found at 
 P, and P', the elevation of it, can be at once obtained. 
 
 Be careful to note that if a vertical plane is used in which to contain the given 
 line, then the elevation of the intersection will meet, or cross, the elevation of the 
 line, or the line produced, at the elevation of the intersection point required, and 
 that if a plane perpendicular to the V.P. is used to contain the line, it will be the 
 plan of the intersection line that will meet the plan' of the given line in the plan 
 of the intersection point required. 
 
 EXERCISE XXI 
 
 I. Find the intersection, showing it by plan and elevation, of the line AB with the plane 
 /?5r in each of the cases ^, 5 and C. 
 
 Y 1 
 
 2. Find the distance of the point A from the plane RST at D. 
 
 3. Find the projections of the intersection line in each of the cases at E, F, G and H where 
 planes are arranged so as to intersect each other. Work to a large scale. 
 
CHAPTER VIII 
 PARALLEL PLANES 
 
 Section 16. It needs no demonstration to realize that planes parallel to one' 
 another will meet the planes of projection in parallel trace lines. 
 
 In Fig. 56 at (i) two vertical planes are shown, and it is evident that the dis- 
 tance between these planes is not the distance between their Vertical Traces, 
 which may be, for these planes, a varying distance apart according to the angle 
 
 Fic. 56. 
 
 at which they meet the V.P., but the distance between them is the distance between 
 their Horizontal Traces, because they meet the H.P. perpendicularly. 
 
 In the case at (ii) the distance between the planes is the same as that between 
 their V.T.'s, because in this case the planes meet the V.P. perpendicularly. 
 
 At (iii) the parallel planes are oblique, and the method of Imding the distance 
 between them is to start a jcrpe ndicular line from som£_4iiiLnt in one of them, 
 say, PQZ. and find where it intersects the other plane RST, then fmd the length 
 of the line from one to the other. To do this, take any point in the H.T. of the 
 plane PQZ, as at aa', and from it set up a perpendicular, marked with arrow-heads 
 in the figure. Now consider this line to be in a vertical plane L.l/.V, and find the 
 
52 
 
 DESCRIPTIVE GEOMETRY 
 
 intersection of LMN with RST, giving a line whose elevation is VH and which 
 contains point P shown at p'p. Find the true length of AP at a'p'2' This is the 
 distance between the two given planes. 
 
 EXERCISE XXII 
 
 Find the true distance between the parallel planes A and B, also between C and D. 
 
 It was previously seen that a horizontal line in a plane was parallel to the 
 H.T. of that plane, and' now it will be seen that any horizontal line parallel to a 
 plane is parallel to the H.T. of that plane. And, as parallel lines have plans parallel 
 to one another, it will be realized that when two planes are parallel to one another, 
 
 Fig. 57. 
 
 the H.T. of one of the planes is parallel to any horizontal line on the other plane. 
 Hence the solution given in Fig, 57 for findin g a plane parallel to ?i given on^^ 
 and at a given distance from it. 
 
PARALLEL PLANES 
 
 53 
 
 From any point A in the given plane RST set up a perpendicular AB and cut 
 ofT a part, AP, say, i" long. Through P draw a horizontal hne parallel to the 
 plane RST by making its plan pq parallel to ST and its elevation p'q' parallel to 
 XY. The V.T. of this line is at q\ which is also a point in the V.T. of the required 
 parallel plane. Draw LM parallel to RS, and 3/-V parallel to ST. It should 
 be noticed that the plane LMX has its traces perpendicular to the projections of 
 the line AB, and that this plane has been passed through a certain definite point 
 P, marked in the line AB. 
 
 So, likewise, if it be required to pass a plane through any given point P, Fig. 58, 
 and to arrange it perpendicularly to any given line, AB, it is only necessary to 
 make a second hne, CP, through P, with its plan perpendicular to the plan of the 
 given line AB, and to represent a horizontal line, in this way, on the required 
 plane. This line CP will have its V.T. at V, which is a point in the V.T. of the 
 required plane RST. RS will, of course, be made perpendicular to the elevation 
 a'b' of the given line. 
 
 A problem illustrating the application of two or three recently discussed 
 methods is worked out in Fig. 59, where it is required to lind the centre of the 
 sphere, which has upon its surface four points whose projections are given. The 
 four points are shown in plan and elevation at aa, bb', cc' and dd' . Any point 
 on the surface of a sphere, joined to any other point on the surface by a straight 
 line, will give a chord of the sphere, and a plane bisecting this chord will cut the 
 sphere into two equal parts, that is, will pass through the centre of the sphere. 
 Notice also another fact, namely, that if three points only be chosen as points on 
 the surface of a sphere, the size of the sphere, upon the surface of which these 
 points may lie, may be any size provided that it is not less in diameter than the 
 circle upon the circumference of which these points may have place. Con- 
 sequently, in order to limit the size of the sphere to be dealt with, a fourth point is 
 necessary, and all four of the points must be made use of in the solution. 
 
 By joining the given points, several chords may be obtained, as shown in plan 
 and elevation. Fig. 59, and, according to the argument above, if planes are made 
 
54 
 
 DESCRIPTIVE GEOMETRY 
 
 to bisect them at right angles, each plane so found will pass through the centre 
 of the sphere. Thus, by joining ab and a'b' the chord AB \s represented, and 
 from the centre point of it, pp', a horizontal line PQ is drawn as a horizontal line 
 on the plane perpendicular to AB, giving the point q' as its V.T. through which 
 the V.T. of the AB plane may be drawn perpendicular to a'h'. Its H.T. is then 
 drawn perpendicular to ah. So again in the two other cases, giving the AC plane 
 and the CD plane. The plan and elevation of the intersection of plane CD with 
 
 Fig. 59. 
 
 plane ^C is next found, and then the intersection of plane AB with one of these 
 (conveniently AC m this. solution), and the plans of the intersections obtained 
 cross each other in the plan of the centre of the sphere. Likewise the elevations 
 of the intersections cross each other in the elevation of the centre of the sphere. 
 
 The radius of the sphere may now be found, if desired, by obtaining the 
 true length of the line joining this centre point to any of the given points A, B, C 
 or D. The sphere may then be represented by circles, one for plan and one for 
 elevation, with the points found, as centres. 
 
PARALLEL PLANES 
 
 55 
 
 EXERCISE XXIII 
 
 I. Find the traces of a plane parallel to, and i^" distant from, a given plane shown at A, 
 whose V.T. is at 60° to XY and H.T. at 30° to XV. 
 
 2. Find the traces of a plane parallel to the given one shown at B and i" perpendicular 
 distance from it. 
 
 3. Kind a plane perpendicular to the given line shown at C. and passing through the given 
 point P. 
 
 T/ 
 1 »(" I 
 
 ! 1 ' »»' 
 
 \ - 1 1 11 
 
 1 1 T^' ! i 
 
 •«' j ! Ill 
 
 1 lit''' '1 ' t/'' 
 A' ' II' ! 1 1 1 V 
 
 k i '1 
 
 j |v id it. I 1 
 
 1 • 1 \ 1 j 
 
 1 1 1 ''* 
 
 id 1 • ! 
 
 4. Find the centre of the sphere which has on its surface the four given points shown at 
 A BCD. 
 
 5 Find the plan and elevation of the sphere which has on its surface the four given points 
 EFGU 
 
56 
 
 DESCRIPTIVE GEOMETRY 
 
 DIHEDRAL ANGLES CONTAINED BY OBLIQUE INTERSECTING PLANES, 
 AND PLANES DIVIDING THESE ANGLES 
 
 Section 17. In considering dihedral angles, or angles contained by planes 
 as they incline to one another, the illustration at (i), Fig. 60, shows two planes 
 RST and LMN meeting each other in a vertical line at H, and meeting the H.P. 
 at right angles with the result that the angle and its supplement, named the dihedral 
 angles contained by these two planes, are equalled by the angles MHS and SEN 
 on the H.P. 
 
 Similarly, at (ii), the two planes RST and LMN meet the V.P. at right angles, 
 and the dihedral angles contained by them are evident at MVS and SVL, because 
 
 Fig. 60. 
 
 they are equalled by the rectilineal angles contained by the traces or intersections 
 RS and LM on the V.P., that is, on the plane perpendicular to them. 
 
 At (iii) the planes are so arranged that they are not perpendicular to one 
 of the planes of projection, and consequently a third plane, other than one of the 
 planes of projection which served the purpose in cases (i) and (ii), must be made 
 use of, in order that the intersections by the two given planes made on this third 
 plane may be found, and the angles by the intersections on it measured. 
 
 Because at (iii) the planes are so arranged as to intersect each other in a hori- 
 zontal line not perpendicular to the vertical plane of projection, the third plane, 
 with its H.T. marked 3, is arranged so that it cuts the given planes at right angles. 
 It cuts the intersection line of the given planes at P, and when this point is rabatted 
 onto the H.P. at P2 and joined to a and b where the third plane's H.T. crosses 
 the H.T.'s of the given planes, the angle aP^b and its supplement are obtained, 
 to which the dihedral angles required are equal. 
 
DIHEDRAL ANGLES-OBLIQUE INTERSECTING PLANES 57 
 
 In Fig. 6 1 the two planes are so arranged that their common intersection 
 line is inclined from the point v' where the vertical traces cross, to the point H 
 where the horizontal traces cross. In this case a third plane, perpendicular to the 
 two given ones, will therefore be inclined, and, being at right angles to the intersec- 
 tion line VH, its H.T. marked 3, must, of course, be made at right angles to the 
 plan, vH, of the intersection line. 
 
 By turning vH, with centre v, into the XF, and carrying with it the point c, 
 the real inclinations of the intersection line and of the third plane may be repre- 
 sented on the V.P., where the line I'P is a view of the intersection line showing 
 
 
 »\ 
 
 r 
 
 ^ 
 
 / 
 
 
 X / 
 
 
 
 X 
 
 .u\ 
 
 .y 
 
 '^^^\. 
 
 \^ 
 
 \ 
 
 
 / 
 
 h 
 
 1 
 1 
 1 
 
 
 
 
 x*^ 
 
 XT/ 
 
 7 ^^ 
 / ^ 
 
 1 
 
 1 
 
 
 Fig. 61. 
 
 its inclination to the H.P. and the line at right angles to it through P shows the 
 inclination of the plane 3. Point P marks the level at which they intersect. From 
 the point P, where the inclined intersection line passes through the inclined plane 3, 
 the distance, indicated by a bracket, down the plane 3 to the H.P. is then taken 
 and transferred to cPi. Join Pi to a and to h, and these lines, which are the 
 rabattements of intersections made by RST and LM'S , respectively, with plane 
 3, as in case (iii), Fig. 60, will give angles to which the dihedral angles required 
 are equal, namely, aP-ih and its supplement. 
 
 In working exercises, it is advisable, until the student is familiar with each 
 step, to make hair lines and attach names to all the various lines and points whi!<^ 
 the solution is in progress. 
 
58 
 
 DESCRIPTIVE GEOMETRY 
 
 (Notice that the third plane's H.T., marked 3, and arranged at any convenient 
 place perpendicular to the plan vH of the intersection line, may need to be produced 
 beyond XY in order to meet the H.T. of LMN at h, as in Fig. 62.) 
 
 EXERCISE XXIV 
 
 Find and mark the angles contained by the planes represented in pairs at (i), (2), (3) and 
 (4). Make the drawings very much larger than what is shown here. 
 
 By experiment with an open book, it will be realized that any dihedral angle, 
 contained by the covers of the book, may be divided by arranging a leaf or leaves 
 of the book, so that from one cover to a leaf is a dihedral angle, and from that 
 leaf to another, separated from it, or to the other cover, will be another dihedral 
 angle. The leaf, representing a plane, divides the dihedral angle between the 
 covers which represent other planes, and will be seen to have the same intersec- 
 tion line as the covers have, namely, the hinge edge of the book. 
 
 Now, let it be required to find the traces of planes which will bisect the 
 dihedral angles contained by any two planes whose traces are given. It will be 
 clear that such traces will pass through the traces of the intersection line made by 
 the given planes, since they must contain the same intersection line. 
 
 In Fig. 62 the planes RST and LMN are the given planes with their inter- 
 section Hne passing through V and H. After proceeding as in the case explained 
 in Fig. 61, the rabatted intersections oi RST and LMN with plane 3 will give the 
 angles aP2b and aPod marked by arrow-headed arcs, showing the sizes of the 
 dihedral angles contained by the given planes. 
 
 Now, bisectors of these angles, marked in the figure by straight lines with arrow- 
 heads, will serve as rabatted intersections made with plane 3 by planes bisecting 
 the dihedral angles, and, of course, these intersection lines will have their H.T.'s 
 in the H.T. of plane 3 at e and/ respectively, and also it will be realized that these 
 
DIHEDRAL AXGLES-OBLIQUE INTERSECTING PLANES 
 
 59 
 
 H.T.'s will be in the H.T.'s of the required bisecting planes. Since the bisecting 
 planes contain the intersection line passing through V and //, these points, V and 
 //, are in the traces of the bisecting planes required. The result is. that the line 
 through II and c is the H.T., and from where this meets XV a line through -/ is 
 the V.T., of one of the required planes, while /// is the H.T., and a line through 
 v' to meet JH in XY, is the V.T. of the other required plane. 
 
 If, as is the case in this figure, there is not room enough to lind the meeting 
 of the H.T. and V.T. on XY for the last plane, then a point xx' should be chosen 
 
 Fig. 62. 
 
 on the common intersection line, and a horizontal line on the required plane should 
 be drawn to meet the \'.P. at kk'. This latter point on the \'.P. is in the V.T. 
 required, which is obtained by joining v' to k'. 
 
 If the H.T. of this last plane, marked /F, is parallel to AT, of course the \'.T. 
 of it, passing through ?' \\i\\ also be parallel to A'l^. 
 
 Also, if ^he bisector Pyf runs nearly parallel to H.T. 3. so that there is not 
 room enough to find/, then the direction of the line/// may be obtained by making 
 use of an inclined line in the plane required. 
 
60 
 
 DESCRIPTIVE GEOMETRY 
 
 EXERCISE XXV 
 
 Find the traces of planes which bisect the dihedral angles contained by the pairs of planes 
 given at A and at B. 
 
 It will now be seen, that, by supposing the rabattement of the angle between 
 two planes, and working conversely to the method in Fig. 6i, it will be possible 
 to find the traces of a second plane at a given angle to a first one, whose traces 
 may be given, and intersecting it in a line of given inclination to the H.P. 
 
 Fig. 63. 
 
 For illustration, let the plane RST, Fig. 63, be given, and let it be required j 
 to find the traces of another which will cut into this one in a line of 30° to the! 
 H.P., and make a dihedral angle of, say, 70° with this given plane. 
 
DIHEDRAL ANGLES-OBLIQUE INTERSECTING PLANES 61 
 
 From any point V in RS draw a line on the V.P. at 30^ to XY. With :•, the 
 plan of V, as centre, bring the plan of this line around by an arc until its lower end 
 is at // in ST. Then, at any convenient place, cross this plan of the intersection 
 line by the H.T. of plane 3, previously made use of in Figs. 60 and 61, and carry 
 the point c into XY so as to show from the point there obtained the inclination 
 of plane 3 and its intersection of the given intersection line at P. The bracket 
 line, now measured off at cPj, will give a i)oint which when joined to a will show 
 the rabattement of intersection of the given plane RST with the plane 3. Apply 
 to this rabattement the angle 70° and so obtain the rabatted intersection Pzb of 
 the required plane with plane 3, and b, a point in its H.T. 
 
 By joining H and b and producing the line to XY the H.T. of the required 
 plane is found, and its V.T. may then be found by drawing a line, from this point 
 in XY, through v'. 
 
 EXERCISE XXVI 
 
 Find the traces of a plane which makes with the given plane an angle of 65° and intersects 
 it in a line inclined at 30° to the H.P. Two cases, .4 and B. 
 
CHAPTER IX 
 
 PROJECTION OF RECTILINEAL ANGLES, AND OF FIGURES AND SOLIDS 
 INVOLVING THE SAME 
 
 Section 18. Two inclined lines meeting each other at a point above the H.P. 
 may be made to form any angle as they meet, and if, say, two sticks of pencil, 
 to serve for lines, be used experimentally, it may be seen that the projections of 
 the two lines and of the angle they contain will vary according to how the pair 
 of lines is disposed in relation to the planes of projection. Notice that the plane 
 
 Fig. 64. 
 
 of the two lines, that is, the plane in which both have place, varies in its attitude, 
 with the variation of the angle formed by the two lines, while the lines may still 
 remain each at its own particular angle to the H.P. 
 
 As illustration, let there be two lines A and B. Let A be inclined at, say, 
 35° to the H.P., and B have an inclination of, say, 50° to the H.P. It must first 
 be realized that the greatest possible angle that can be contained by these lines 
 is 95°, and in order that this may be so, the plane of the lines would have to be 
 vertical. Such an arrangement of them is shown in Fig. 64, where the plans of 
 the lines are parts of the XY line. Realize that the lengths of their plans will 
 not vary so long as the lines are not increased in length and their angles with the 
 H.P. are not changed. The plans, however, may be moved the one toward the 
 other so that instead of the distance between their lower ends or H.T.'s being as 
 great as from a to ^ in the figure, it may be reduced to, say, the distance a to b2, 
 the line B being moved so as to have its H.T. at bo. The angle contained by the 
 
 62 
 
PROJECTION OF RECTILINEAL ANGLES 
 
 63 
 
 two lines A and B is now reduced, since the distance a to bo, subtending the angle, 
 is reduced, but not the lengths of the lines. The new elevation of B is now at 
 B2, and the angle contained by A and B has been reduced to an angle of which 
 the plan and elevation are indicated by the arrow-headed arcs. The true size 
 of this angle can readily be obtained by rabattement about ab^ as a hinge line, abz 
 being the H.T. of the plane of the two lines as now arranged. 
 
 As a specific case, let it be required to find plan and elevation for an angle 
 contained by a pair of lines of any length when one of them is at 30° to the H.P. 
 and the other is at 45° to the H.P., and the angle contained by them is 80°. 
 
 By placing both lines in the V.P. as in Fig, 65-, (i), at A and B, starting them 
 both from the same point and producing them downward to their H.T.'s at a and b 
 respectively, their plans, proper for the particular angles they make with the H.P., 
 are found. Allowing the 30° line A to remain in the V.P., the 45° line B must 
 be moved toward A in order to reduce the angle from what it is at present (105°) 
 
 Fic. 65. 
 
 to 80°. This 80° will be subtended by a line, across the H.P. between the H.T.'s 
 of the lines, the length of which must be found, therefore with c' as centre move 
 B toward A until 80° is enclosed as shown by the arrow-headed arc. This gives 
 the distance abo required to subtend 80° when the lines are A and B. Now, with c 
 as centre bring the plan cb round by an arc until the H.T. of B is at its proper dis- 
 tance from the H.T. of A. This will require the use of an intersecting arc with 
 a as centre and (7/)2 as radius. The new place for the H.T. of B is ^3- Join it to c 
 for the new position of plan, and then obtain the elevation of the line as shown. 
 By joining a to b^ the H.T. of the plane of the two lines is obtained, and the V.T. 
 of the plane of the two lines will be dc\ 
 
 In Fig. 65 at (ii) is shown a particular case which is very useful. The case 
 is that of the right angle (90°) contained by two lines, the sum of whose angles 
 to the H.P. is less than 90°. The two lines are first placed as at A and B in the 
 V.P., with their plans therefore in AT. If, now, A be allowed to remain in the 
 V.P. wath ca as its plan, and B is to be placed so as to contain with .1 an angle 
 
64 DESCRIPTIVE GEOMETRY 
 
 of 90°, then the elevation of B will be a perpendicular to the elevation A, because, 
 as we have seen previously, when a right angle is contained by two lines, and one 
 of them is in a projection plane or parallel to it, then the other has its projection 
 on the same plane perpendicular to the projection of the first. Therefore c'h' is 
 at once obtained as the elevation of the B line, and its plan can then be found 
 by making a perpendicular from h' and intersecting it by an arc with c as centre, 
 which carries down the plan length ch into its new place. Notice that by joining 
 a toh the H.T. of the plane of the two lines is obtained, and, of course, the V.T. 
 of the same plane is ac', coinciding with the elevation ac' of the line A which is 
 in the V.P. 
 
 N.B. — When the sum of the angles made with the H.P. by the given Hnes 
 is over 90°, the angle contained by them must be less than 90°. Consequently, 
 for large contained angles the inclinations of the lines to the H.P. must be small 
 accordingly. 
 
 EXERCISE XXVII 
 
 1. Find the plan and elevation of an angle of 45° contained by two lines when one of them 
 is in the V.P. and inclined to the H.P. at 35°, and the other is at 50° to the H.P. 
 
 2. Find the plan and elevation of two lines, one at 25° to the H.P., and in the V.P., and the 
 other at 45° to the H.P., when they contain an angle of 100°. 
 
 3. Find the plan and elevation of two lines, one at 45° to the H.P. and in the V.P., and 
 the other at 30° to the H.P., when they contain an angle of 45°. 
 
 4. Find the plan and elevation of an angle of 90° when contained by two lines, one at 35° 
 to the H.P., and the other at 20° to the H.P. 
 
 N.B. — In each case mark which is the plan of the contained angle, and which is the ele- 
 vation of it. 
 
 An application of the problem above discussed may now be made to the finding 
 of the projections of any rectilineal figure, triangle, square or polygon, when two 
 adjacent sides, or a diagonal and an adjacent side, or two diagonals, are given 
 at inclinations to the H.P. Instead of inchnations for edges, etc., levels may 
 be given for corners, and the inclinations of edges or diagonals ascertained 
 accordingly. 
 
 In Fig. 66 is represented in plan and elevation a square having one edge at 
 25° to the H.P. and in the V.P., and an adjacent edge at 40° to the H.P. As the 
 angle contained by the two Unes is 90°, the 40° line can be found in its proper place 
 in relation to the 25° one, by the method of Fig. 65 (ii). The edge-length of the 
 square must be marked off as c'a' on the true length of the 25° line and as c'h'2 on 
 the true length of the 40° line, a'c' gives ac as its plan, and from c'h'2 may be 
 obtained c'h' and ch, the elevation and the plan respectively, of the 40° edge of 
 the square. Opposite edges of the figure being parallels, their projections may 
 be obtained by making them parallel to those already obtained. The square 
 
PROJECTIOX OF RECTILINEAL AXGLES 
 
 65 
 
 thus found may now be considered as the face of a cube, and it will be seen that 
 by drawing a line through ////i, which points are the H.T.'s of the two edges of 
 the square, the H.T. of the plane of the square is obtained, and Ila'c' will be the 
 V.T. of the plane of the square. A perpendicular to this plane from, say, the 
 point A may be made by the method explained in Fig. 52. If this perpendicular 
 be made equal in length to the edge of the square, then it will be seen that the 
 cube can be represented by drawing other perpendiculars with projections equal 
 in length to the projections of this one and parallel to it; and by joining the upper 
 ends of them the drawing will be completed. To correctly represent this cube 
 as a solid the three edges in plan from the corner d should be made dotted lines, 
 and the edge a'c' in the elevation should also be made a dotted line. 
 
 Fig. 66. 
 
 In Fig. 67 the plan and elevation of a regular pentagon (whose angle is 108°) 
 is found, one edge being at 25° to the H.P. and an adjacent edge at 35° to the 
 H.P. Following the construction shown in Fig. 65, (i), the inclined lines are found 
 in position at ac, ch for plans, and ac', c'b' for elevations. Point C is then rabatted 
 to Co, where the angle 108° appears at its true size, and the figure is formed with 
 this angle for one of its corners. From the figure so formed in rabattement the 
 corner points are carried over the H.T. line ab perpendicularly to the plan, and 
 then from the plan the elevation is obtained. Advantage is taken of the fact 
 that there is a diagonal of the pentagon parallel to each side of it. 
 
 The plan and elevation of the pentagon having been completed, the centre 
 of it may be obtained by directing lines from the centres of any two sides to oppo- 
 site corners and letting them intersect as at dd' . 
 
 The pentagon might be considered to be the base of a pyramid, and then a 
 perpendicular from dd' may be found, of a given length, to serve as the axis of the 
 solid, by the method of Fig. 52. The upper end of this axis, i.e., the apex of the 
 pyramid, may then be joined by straight lines to the corners of the pentagon to 
 
66 
 
 DESCRIPTIVE GEOMETRY 
 
 complete the representation of the solid. Care must be taken to make certain 
 lines dotted lines to represent them as being out of view, in the plan and in the 
 elevation. In Fig. 67 the apex pp' is found, but the solid is not shown. 
 
 EXERCISE XXVIII 
 
 1. Find the plan and elevation of a regular pentagon i\" edge, when one edge is at 20° to 
 the H.P. and an adjacent edge is at 35° to the H.P. 
 
 2. Find the plan and elevation of a right hexagonal pyramid when one edge of the base is 
 at 15° to the H.P. and an adjacent edge of the base is at 40° to the H.P. The length of t*he 
 base edge is i", and of the axis 3". 
 
 3. Find the plan and elevation of a cube 2" edge, when one edge is at 25** to the H.P. and 
 an adjacent edge is at 50° to the H.P. 
 
 4. Find the plan and elevation of a pentagon, 1" edge, when three consecutive corners of 
 it are at levels \" , l" and if", above the H.P. 
 
 THE TETRAHEDRON AND THE OCTAHEDRON 
 
 Section 19. Of the " regular " soHds, we have already dealt with the cube, 
 and we shall now proceed to consider two others, the tetrahedron, with four equal 
 faces, each an equilateral triangle, and the octahedron, which has eight equal faces, 
 each being an equilateral triangle. 
 
THE TETRAHEDRON AND THE OCTAHEDRON 
 
 67 
 
 The appearance and characteristics of these solids may be judged from the 
 representations of them in Fig. 68 at (i) and (ii). 
 
 The tetrahedron has a corner perpendicularly opposite the centre of each 
 face; while the perpendicular distance of a corner from the opposite face is the 
 height of the solid. The simplest way by which to find* the height of any tetra- 
 hedron is to place it on the H.P., as shown at (iii), and then, since all of its edges 
 are equal, and the edge ^-B is parallel to the V.P., a'h' shows true length, its plan 
 being at ah. The height of the solid is therefore found to be b'c'. 
 
 The octahedron might be looked upon as two square pyramids, base to base, 
 the height of each being equal to half the diagonal of its base. In other words, 
 
 Fig. 68. 
 
 the solid has three diagonals at right angles to one another, and each one therefore 
 at right angles to the plane of the two others. 
 
 The simplest way to arrange this solid is shown at (iv) in Fig. 68, where the 
 full height in the elevation is made equal to a diagonal o^ the square representing 
 the solid in plan. 
 
 At (v) the solid is shown cast over on to one of its faces, and in plan a dotted 
 triangle for the under face, equilateral, and another, clear-lined for the upper face 
 will be necessary. All the sloping edges are seen in plan and form a regular he.xagon. 
 
 In Fig. 69 an octahedron, of given edge ah, is arranged so that one diagonal 
 is in the V.P. at if to the H.P., and another diagonal is at 45° to the H.P. 
 
68 
 
 DESCRIPTIVE GEOMETRY 
 
 From c' the half diagonal, obtained from the square on ah, is marked off on 
 each of the inclined lines. When the plans and elevations of these have been 
 determined, they are extended to make the complete diagonals. The third diagonal 
 is then found, as a perpendicular to the plane of the two already found. Par- 
 ticular care must be taken in clear-lining the projections, so that the correct use 
 of dotted lines is made for the edges not in view. It should be recognized that 
 dd' is the near and upper end of the third diagonal. 
 
 Fig. 69. 
 
 EXERCISE XXIX 
 
 1. Find the projections of a tetrahedron, 2" edge, when one edge is at 25° to the H.P. and 
 another edge is at 45° to the H.P. 
 
 2. Find the projections of an octahedron, \\" edge, when one diagonal is in the V.P. at 
 20° to the H.P., and another diagonal is at 35° to the H.P. 
 
 3. Find the projections of an octahedron \\" edge, when one edge is in the V.P. at 20° to 
 the H.P. and a diagonal, adjacent to it, is at 40° to the H.P. 
 
 4. Find the plan of an octahedron, \\" edge, with one face in the H.P., and show an ele- 
 vation of it, derived from this plan, on a plane to which no edge of the solid is parallel. 
 
CHAPTER X 
 
 AXO.METRIC AND ISOMKTRIC PROJKCTIOX 
 
 Section 20. Realize experimentally or otherwise the following facts: — 
 
 1. When two inclined lines meet, there is a plane of these lines which has its 
 H.T. passing through the H.T.'s of the two lines; 
 
 2. When a line is perpendicular to a plane, the plan of the line and the H.T. 
 of the plane are at right angles to one another; 
 
 3. Three lines may meet in a point, and each one be perpendicular to both 
 of the others, that is, to the plane of the others; and 
 
 4. Three such lines, each perpendicular to the others, may either be equally 
 inclined to the H.P. or all differently inclined. 
 
 Fig. 70. 
 
 Let there be three lines, .1. B and C, meeting in one pi)int. and containing 
 right angles with each other. If they be equally inclined t > the H.P., the three 
 right angles will appear in plan equal to one another, and the plans of the right 
 angles will be 1 20° in each case, as at (i) in Fig. 70. 
 
 Now, if the point where A, B and C meet each other be at some distance 
 above the H.P., then some point in one of them, say, // in .1. will be the H.T. of 
 line .1, and the H.T. of B and the H.T. of C can then be placed similarly in the 
 lines B and C at the same distance from the point where they all meet, as seen 
 at (ii). 
 
70 
 
 DESCRIPTIVE GEOMETRY 
 
 By drawing lines through these H.T.'s we have the H.T.'s of the planes of 
 pairs of lines, and it will be seen that as the hnes A and B contain a right angle, 
 they will, when rabatted, expose that right angle, which is, of course, the angle 
 of a semicircle; therefore, taking advantage of this fact, on HH make a semicircle, 
 and by a perpendicular across the trace HH from the plan of the point where A 
 and B meet, to the circumference of the semicircle, we obtain the rabattement of 
 A and B as seen at ^2 and Bo, and so, true lengths are now shown for these lines 
 A and B. 
 
 Next consider the case when the lines A, B and C are at different angles to 
 the H.P., and let A be at 25° and B be at 45°, then by the method explained and 
 illustrated in Fig. 65, (ii), the plans of A and B are found and their H.T.'s are 
 at HH in Fig. 71. Through HH draw the H.T. of the plane of A and B, and the 
 
 Fig. 71 
 
 plan of C will be perpendicular to this H.T. Hne. This is shown by the arrow line 
 in the figure. The H.T. of this arrow Hne can now be found by drawing the H.T. 
 of the plane of B and C perpendicular to the plan A, or by drawing the H.T. of 
 the plane of A and C perpendicular to plan B. This H.T. of C is indicated at HC. 
 
 From point HC to the point where the three hnes meet is the plan of C, and 
 this plan may be swung into XY and then the elevation of C, on the V.P., may be 
 drawn, to show the angle, marked in the figure with an arrow-headed arc, just 
 in the way that the lines A and B showed their inclinations, 25° and 45° respectively, 
 when originally placed in the V.P. 
 
 As in the previous case. Fig. 70, (ii), since each pair of lines contains a right 
 angle, true lengths for ^, B and C may readily be arrived at by rabatting them 
 into semicircles, as shown at A2, B2 and C2, Fig. 72. Upon these rabattements 
 any true length measurements may be made as at ^2 and then carried perpendicu- 
 larly across to the plan line A Sit d. 
 
AXOMETRIC AND ISOMETRIC PROJECTION 
 
 71 
 
 This arrangement of three lines, A, B and C, provides a very useful means of 
 making single projections or plans which show three dimensions to defmite scales, 
 and because of the three lines which serve as axes of direction for measurable 
 distances, the method is known as Axometric Projection. 
 
 It will be recognized that the " scale " of measurement along each of the 
 three lines, A, B and C, or axes of projection, as they are called, in Fig. 70 at (ii), 
 is the same, because the axes are equally inclined to the projection plane. This 
 is therefore referred to as Isometric — a special case of axometric projection. The 
 scales of measurement along the axes A, B and C in Figs. 71 and 72, however, are 
 all different from one another, so that, for instance, 2" on the axis A is represented 
 by a longer line than 2" on the axis B. 
 
 As an illustration of the use of this method, let it be required to find the axo- 
 metric projection of a skeleton cube, each face of which will appear as a foreshort- 
 
 FiG. 72. 
 
 ened view of Fig. 73, (,i). There will be twelve bars of square section, each one 
 represented by three lines, as suggested at (ii). There are three directions for the 
 edges of a cube, four edges to each direction, therefore the three axes of projection 
 may each contain an edge. Let two of the axes be at 25° and 45° respectively, 
 to the projection plane. 
 
 First find the axes .1, B and C and rabattements of them at .!_', B2 and €2- 
 On these rabattements mark oft" measurements taken from the edges of the square 
 at (i), and transfer them to the axes by perpendiculars to the trace lines. Then 
 by parallels the three upper faces of the cube, all foreshortened in appearance, 
 may be found. Three of the bars are by this time represented, each by three lines. 
 By the careful use of parallels each of the remaining nine bars may be found. All 
 the twelve bars have square ends which may be shown as at fii) in the figure. 
 
72 
 
 DESCRIPTIVE GEOMETRY 
 
 The drawing here shown is not completed, but should be carried out on a large 
 scale and completed by the student. 
 
 (II) 
 
 Fig. 73. 
 
 EXERCISE XXX 
 
 1. Find the projection of a 2" cube when 
 two of the axes of projection are inclined at 25° 
 and 40° respectively to the projection plane. 
 
 2. Find the inclination to the projection 
 plane of the third axis in i. 
 
 3. Find by axometric projection a skeleton 
 cube 3I" outside, 2\" inside, measurement. One 
 axis is at 30° and another at 40° to the projection 
 plane. 
 
 4. Find in isometric projection the two parts, 
 separated as in the illustration herewith, of a 
 mortise and tenon joint. Use dotted lines to 
 represent those lines not in view. 
 
AXOMETRIC PROJECTION, CONTINUED 
 
 73 
 
 AXOMETRIC PROJECTION, Continued 
 
 Section 21. Since " scale," with respect to drawings, is the ratio of projection 
 length to real length, it will be seen that the axes in axometric projection may be 
 arranged in place when scales for any two of them are given, by setting up two 
 lines at inclinations to the H.P. that will give those scales, and, having arranged 
 their plans correctly, placing the third axis in proper relation to these. The scale 
 of the third may then be obtained also. 
 
 Thus, suppose the scales for A and B are to be f and -^ respectively, the incli- 
 nations of A and B to the H.P. may be obtained as in Fig. 74 at (i), where ab, 
 equal to 4 divisions, is represented in plan by ac equal to 3 of the same divisions; 
 
 Fig. 74. 
 
 that is, ac is f full length, and the inclination for this scale of f is the angle bac. 
 Again, ad, equal to 6 divisions, is represented in plan by ac, equal to 5 divisions, 
 i.e., ae is i full length and the angle for the axis which will have a scale of ,^ will 
 be the angle dac. 
 
 Employing the method as shown in Fig. 71, set up at Fig. 74, (ii), the angle 
 dae at a and the angle bac at /3, and proceed to find axes A and B with their H.T.'s 
 at HH. The third axes, C, will be found and its H.T. obtained as in Fig. 71. By 
 swinging the plan of that portion of it above the H.P. into AT, the plan length 
 RS and the real length TS are obtained, which will give the scale for the third axis 
 
 RS 
 
 C as 
 
 TS' 
 
 From what was seen previously (see Note in Section 18), the contained angle 
 between the axes of projection being 90°, the inclinations of them to the projection 
 plane must be small, so that the sum of them is less than 00°, and this limits the 
 
74 
 
 DESCRIPTIVE GEOMETRY 
 
 use of axometric projection to large scales only, if the drawings are to serve the 
 purpose they are intended to serve. 
 
 Suppose the arrangement of the axes of projection to be given in plan as at 
 (i), Fig, 75, and let it be required to find the projection of some solid having its 
 edges or other lines in the directions of the axes given. First choose any point 
 in one of the lines, say, H, in axis A at (i) , and proceed to find points in B and C 
 that are at the same level, by lines from H perpendicular to C and B respectively. 
 Then by rabatting the right angle contained by axes A and B, also by A and C, 
 into semicircles, true lengths are obtained at ^2, Bo and C2, and the measurements 
 of the cube or other solid may be marked on these and transferred to A, B and C 
 respectively, as in Fig. 73 at (iii). 
 
 (TI) 
 
 Fig. 75- 
 
 If the incHnations of these given axes. Fig. 75, are required to be found, it 
 will be seen that a vertical plane containing one of them, say, axis A , will cut the 
 plane of the others in the Hne RT, see (ii), and that HR is at right angles to RT, 
 therefore the vertical semicircle on the line HT for diameter, rabatted as at HR2T, 
 will show the angle the axis A makes with the projection plane, namely, the angle 
 R2HR. This estabhshes the height of R above the projection plane. Hence, 
 if the H.T.'s of B and C be brought around into the line HRT, their angles will 
 Hkewise be seen. 
 
 A further illustration is given in Axometric Projection in Fig. 76. Let it be 
 required to find the projection of a pentagonal pyramid when one edge of the base 
 is at 25° to the projection plane and the axis of the solid is at 40° to the same plane. 
 
 First find the three axes of projection. As the axis of the sohd has to be in the 
 40° direction, the base must be placed in the plane of the two others. The base, 
 not being right angular, must be placed in a rectangle as at (ii). As AB at (ii) 
 
AXOMETRIC PROJFXTIOX, CONTIXUED 
 
 75 
 
 contains an edge of the base, the Hne AB must be fitted on to the 25" axis of pro- 
 jection. Therefore, place it in rabattement at /l2i^2and draw the complete figure 
 contained in the rectangle BoAoCo, and proceed to carry it over to its position 
 in plan. Having completed the base in plan, find its centre C, and from that 
 centre draw a line in the direction of the 40° axis. Find, by the proper method, 
 the projection length of the axis at coa^ or c-zb-y, according to its length, and then 
 mark this off on the axis line of the pyramid starting from c, thus giving the apex 
 at (/ or h. Join the apex to the corners of the base and fmish as usual with dotted 
 lines for some edges as the case may require. 
 
 Fig. 76. 
 
 If a circular hole has to be represented, in a block, for instance, in axomctric 
 projection, points should be obtained in the circumference by using diagonals and 
 parallels to the sides of a square made to contain the circle, the sides of the square 
 being made to follow axis directions in the projection. 
 
 EXERCISK XXXI 
 
 1. The plans of the three axes of projection enclose angles of iio^. 120° and i.^o"". Find 
 the inclinations of the three axes to the projection plane. 
 
 2. The scales of two of the axes for a projection are J^ and j^. Find the projection of the axes 
 and represent the scale of the third axis. 
 
 3. P'ind by axomctric projection the plan of a regular hexagonal pyramid, when the axis 
 of the pyramid, 3" long, is at 30° to the projection plane, and one of the edges of the base, i" 
 long, is at 45° to the projection plane. 
 
CHAPTER XI 
 
 SECTIONS OF SIMPLE SOLIDS 
 
 Section 22. Sections of solids are made by passing planes through them. 
 By rabatting these planes, carrying with them the outline points of the section 
 shape, the true form of the particular section may be obtained. For convenience, 
 the section plane is usually arranged either perpendicularly to the H.P. or per- 
 pendicularly to the V.P. 
 
 Fig. 77 
 
 Illustrations of the process are shown in several figures now to be considered. 
 
 In Fig. 77 a cube has a vertical section plane RST passing through it. Hori- 
 zontal edges are cut by it at a, b and c, while at d two of the inclined edges are 
 cut. Realize that three inclined faces are cut and one vertical one. By using 
 S as a centre and making rabattement of all the points in which the plane cuts 
 the edges, the true form is obtained on the V.P. It is covered with hatching or 
 shading in the figure. 
 
 In Fig. 78 a square pyramid has a section plane RST perpendicular to the 
 V.P., passing through it. The slant edges of the pyramid are all cut in points 
 whose elevations are marked a'b'c' and d'. By using S as centre and rabatting 
 all these points on to the H.P. the true form is obtained. Notice that it is neces- 
 sary to have the plans of the four points. These are readily found in the case of 
 a and of d, but as the projections of the edges in which B and C occur are so nearly 
 
 76 
 
SECTIONS OF SIMPLE SOLIDS 
 
 77 
 
 perpendicular to XV these points are carried by level lines to the slant edge in 
 which a occurs and from their plans in the plan of this edge they are carried by 
 arcs to their proper places in the plans of the edges they belong to, and thence 
 by perpendiculars across the H.T. line to their rabattements. The true form 
 may now be drawn and is marked in the figure by hatching. 
 
 In Fig. 79 a right circular cylinder has a section plane perpendicular to the 
 V.P., passing through it in such a way as to cut the top horizontal surface and cut 
 also a considerable amount of its cur\'ed vertical surface. Since the section shape 
 of a right circular cylinder by a plane inclined to its axis is an ellipse, the result in 
 this case will be part of an ellipse, as the shape of the section. Points in the curved 
 outline of the true form are obtained by choosing straight generator lines, as they 
 are called, on the curved surface of the cylinder, marked in this example at a, b 
 
 c, d, c, etc. Portions of the elevations of these generators are shown in order to 
 obtain the elevations of the intersection points to be rabattcd. In the figure 
 the shape of section is shown covered with hatching. 
 
 In Fig. 80 a right circular cone has a section plane RST passing through it, 
 and since this plane cuts all the generators of the curved surface it produces an 
 ellipse for the shape of the section. This shape is rabatted on to the H.P. by 
 using 5 as centre. Generators, taken in such places as will give a good distribu- 
 tion of points in the curve, are drawn at a, b, c, d, e, etc. These in elevation are 
 seen to be cut by the section plane, and the points in which they are cut are carried 
 over by arcs to the H.P. It is necessary to locate them in plan, which is straight- 
 forward work except in the case of generator bb'. The point in this generator 
 must be carried horizontally to the generator c, and its plan carried back to the 
 generator b by an arc as seen in the figure. 
 
78 
 
 DESCRIPTIVE GEOMETRY 
 
 Fig. 8 1 shows a right circular cone with a section plane RST passing through 
 it so as to give the hyperbola as a section. The plane for the hyperbola is parallel 
 to the plane of two generator lines of the cone, whereas the plane for a parabola 
 has only one generator line of the cone parallel to it. The two generators parallel 
 to the plane of the hyperbola in this case are shown at a and b, and their Horizontal 
 Traces are at ////. The curve of the hyperbola is shown in plan at cc with its true 
 form by rabattement at C2C2. The H.T.'s of this curve are at gg. 
 
 Now, if the line cc which lies on the curved surface of the cone be continually 
 produced in the same plane it becomes straight, at infinity, on opposite sides 
 of the cone, and since the plane of the parallel generator lines A and B is at such 
 
 Fig. 81. 
 
 a small distance from the plane of the hyperbola, the tangents to the hyperbola 
 lines, at infinity, and the generator hnes near them are pairs of parallels. These 
 tangents and generators on the surface of the cone at infinity will meet the cir- 
 cumference of the circular section of the cone at infinity, at right angles, and the 
 H.T.'s of the planes of the pairs of parallels at infinity will be perpendiculars to 
 the generators whose plans are a and b respectively. 
 
 Consequently, on the Horizontal Plane of projection, from points // perpen- 
 diculars to the plans a and b will pass through the H.T.'s of the tangents at infinity 
 to the hyperbola, which tangent lines also He in the plane of the hyperbola. There- 
 fore HH are the H.T.'s of the tangents at infinity to the hyperbola, and the plans 
 of them are the lines dd, made parallel in plan to the generators a and b. 
 
TRACES OF CURVED SURFACES 79 
 
 The tangents at infinity to the h\perbola are called the Asymptotes, and meet 
 each other at an angle. In Fig. 8i they are rabatted on to the H.P. and show the 
 angle contained, which is marked by an arrow-headed arc. 
 
 Strictly speaking, the asymptotes, lying outside the two branches of a hyper- 
 bola, which are sections of two equal cones united at their ape.xes and having the 
 same axis line, are lines which never really meet the curves of the hv-perbola. 
 
 exp:rcise XXXII 
 
 1. A square prism, 15" edge of end, 2§" long, has a long edge in the H.P, at 30° to XV, 
 and the face adjacent to it is at 30° to the H.P. Find the true form of section by a vertical plane 
 at 45° to the \'.P., passing through the soh'd and cutting its axis i" from one end. Two 
 solutions. 
 
 2. Find the true form of section of a right hexagonal pyramid, base in H.P. with one diag- 
 onal of base parallel to A'F, by a plane perpendicular to V'.P., at 45° to the H.P., and passing 
 through a point in the axis |" from the base. Base edge i". Axis 22". 
 
 3. Find the true form of a parabola by a plane passing through a cone, when its base is 2" 
 diameter, and height of apex of the cone above the base 25". The plane cuts the axis in the 
 centre. 
 
 4. Find the asymptotes and show the angle contained by them in the case of a hyperbola 
 made by a plane at 15° tp the axis of a cone, cutting the axis in a point i" from the apex. The 
 angle of the cone at the apex is 40°. 
 
 TRACES OF CURVED .SURFACES 
 
 Section 23. When a right circular cylinder is inclined to a projection plane, 
 its curved surface, produced to meet the plane, will meet it in the curve of an 
 ellipse, for the projection plane might be considered as a section plane passing 
 through the circular cylinder at an angle to its axis. This applies also to the 
 curved surface of a cone. The line in which the curved surface of the cone meets 
 the projection plane will be an ellipse, if the cone is a right circular one, i.e., one 
 that has a circular section in any plane at right angles to its a.xis. 
 
 The line of intersection in which the curved surface of a cylinder or of a cone, 
 whether its axis be inclined or not. will meet the projection plane, if the curved 
 surface is produced to do so, is called the trace of the curved surface, and may be 
 the Vertical trace or the Horizontal trace according to whether it is found on the 
 V.P. or on the H.P. Illustrations are shown in Figs. 82. and 83. 
 
 It will be seen that in the case of the right circular cylinder, the H.T. is an 
 ellipse with its minor axis equal to the diameter of the cylinder, whereas in the 
 case of the right circular cone the ellipse for H.T. has a minor axis which varies 
 in length with the inclination of the cone or with the distance of its apex from the 
 projection plane. The method of obtaining the trace is to choose any number of 
 generator lines and produce them until their traces are found, then to draw, by 
 freehand, the curve through these trace points. 
 
80 
 
 DESCRIPTIVE GEOMETRY 
 
 A further illustration is shown in Fig. 84, where it is important to realize 
 that the generators whose plans are at aa have their elevations at a'a', and that 
 
 Fig. 82. 
 
 Fig. S3. 
 
 these are best located by recognizing the fact that they pass through the ends 
 of the horizontal diameter of the end of the cylinder. The topmost generator 
 
 Fig. 84. 
 
 line, having its plan at h, has its elevation at //, which is not the topmost line of 
 the elevation. Other generators, chosen at convenience for the purpose of secur- 
 
TRACES OF CURVED SURFACES 81 
 
 ing points in the trace, such as those marked cc-z in the figure, should meet the end 
 of the cyUnder in pairs, at the ends of horizontal chords, for then the elevation 
 marked c'2 can the more certainly be located by making use of the horizontal 
 chord through the end of the one marked cc' . 
 
 N.B. — The projections for the circular ends of the solids should be carefully 
 found by the method previously discussed. (See Section 4, Fig. 8.) 
 
 EXERCISE XXXIII 
 
 1. Find the H.T. of the curved surface of a right circular cylinder, i\" diameter, when its 
 axis is at 35° to the H.P. and the plan of it is at 25' to XY . 
 
 2. Find the H.T. of the curved surface of a right circular cone, whose base is i\" diameter 
 and axis ih" long. Let the axis be parallel to the V.P., and inclined to the H.P. at 45°. 
 
 3. Show the plan and elevation of a cone whose H.T. is a circle 2^" diameter, and whose 
 axis is 3" long and inclined at 40° to the H.P. and 40° to the V.P. Show the V.T. of its curved 
 surface when the centre of the circular H.T. is i\" from XY. 
 
 4. Find the shadow of a sphere of 2" diameter, having its centre 2" above the H.P. and 2" 
 in front of the \'.P., when the rays of parallel light are inclined at 45° to the H.P. and 30° to 
 the V.P. 
 
 N.B. — The shadow problem here given involves the method discussed in this section. 
 
PART THREE 
 
 CHAPTER XII 
 
 - TANGENT PLANES TO CONES AND CYLINDERS 
 
 Section 24. Since the generator of a cylinder, and also of a cone, is a 
 straight line, it is evident that a plane, tangent to or touching the curved surface 
 
 Fig. 85. 
 
 of one or of the other, will touch it along a straight hne, and, in the case of the 
 cone, will include its apex. Such a tangent plane may be shown, as usually planes 
 are shown, by its traces, and its traces will be tangential to the curved traces of 
 the curved surface of the solid. 
 
 In the illustration, Fig. 85, a right circular cone, that is, one whose sec- 
 tion perpendicular to its axis is circular, stands vertically on the H.P., conse- 
 
 82 
 
TANGENT PLANES TO CONES AND CYLINDERS 
 
 83 
 
 quently, the circle, which is the circumference of its base, is the H.T. of its cur\^ed 
 surface. A plane, tangent to this cone, will therefore have its H T. tangent to 
 this circular H.T.. and this is shown at ST. As the apex aa' is in the tangent 
 plane, a horizontal line may be drawn through it, lying in the tangent plane. 
 This will have its plan parallel to ST, and its elevation will give its V.T. at V. 
 RS, the V.T. of the plane, may then be drawn. 
 
 It will be realized next, that, as all the generators of the vertical cone meet 
 the H.P. at the same incUnation, therefore the generator whose plan is ab. is at 
 the same inclination as the generator whose elevation is a'c', and whose inclination 
 is indicated by an arrow-headed arc in the figure; and, since the plane is at the 
 same inclination as a line in it perpendicular to its trace, and AB is perpendicular 
 to the trace ST. the tangent plane RST is at the angle indicated at c'. 
 
 Fig. 86. 
 
 The converse of this problem is specially useful. For instance, if it be 
 required to fmd the traces of a plane of given inclination to the H.P. having its 
 H.T. in a given direction, it is only necessary to set up a right circular cone whose 
 generators are at the given inclination for the required plane, and then make the 
 H.T. of the plane tangent to the circular trace, the base of the cone is the H.P.. 
 and proceed as before to find the V.T. of the plane. 
 
 In Fig. 86 the V.T. of a plane is given at RS. Suppose the plane, whose V.T. 
 is RS, to be inclined at 6o° to the H.P.,, and let it be required to find its H.T. 
 Then, because any point in this V.T. is in the plane and may be taken as the 
 apex of a right circular vertical cone to which the plane is tangent, from any such 
 point A draw the line a'c' in the V.P. at 6o° to the H.P. The circumference 
 of the circle drawn, with a as centre and ac' as radius, is the H.T. of the 6o° cone, 
 and the point A, its apex, is contained in every tangent plane to the cone. There- 
 fore from the point, namely 5. in .VI', where the V.T. line meets it, draw the H.T. 
 
84 
 
 DESCRIPTIVE GEOMETRY 
 
 line, ^ST, of the required 60' plane, making it tangent to the circular trace of the 
 cone. The line SN shows the H.T. of another plane, having the same V.T., 
 and tangent to the same 60° cone, therefore also at 60° to the H.P. There are, 
 therefore, two planes, RST and RSN, satisfying the conditions. 
 
 In Fig. 87 a line AB is given, and it is shown how a plane of any particular 
 inclination to the H.P., greater than the inclination of the given line to the H.P., 
 may be found, when it has to contain the given Kne. For the solution, any point 
 cc' in the given line may be taken, to serve as the apex of a right circular vertical 
 cone. 
 
 Fig. 87. 
 
 The cone may be represented by drawing the line c'd' to make with XY 
 an angle, a, equal to the given inclination of the required plane, and with 
 cd, its plan, as radius, drawing the circular H.T. of its curved surface. Next, 
 find the H.T. of the given line AB a.t H, and through H draw tangent lines to 
 the circle. These are the horizontal traces of two tangent planes to the cone, 
 each containing the given line. Horizontal Hnes on these planes, drawn through 
 the apex C, will give points V and Vi in the Vertical Traces required to more com- 
 pletely determine the planes. HeV is one of the planes. The other plane, whose 
 H.T. is Hf, gives an opportunity of showing how to obtain the V.T. without first 
 making Hf meet XY. Thus, from any point g, in Hf, draw the plan of a line through 
 some point in the line ^^— through the apex C will do— and produce this plan 
 
TAXGEXT PLANES TO COXES AXD CYLIXDERS 
 
 85 
 
 and its elevation till its V.T. is found at V2. Join V1V2 and this line is part of 
 the Ventcal Trace of the second plane, part of whose H.T. is the line II f. 
 
 EXERCISE XXXIV 
 
 1. Find a plane at 50° to the H.P. containing ihe given point .1 at (i). Let the H.T. of the 
 plane be at 45° to the A' I'. 
 
 2. Find a plane at 65^ to the H.P. and containing the given line AB at (iij. 
 
 3. The \'.T. of a plane is given at (iii). Show the H.T. of it when the plane is at 45'' to the 
 H.P. Two solutions. 
 
 
 (i 
 
 
 
 (ii) /; 
 
 (iii) 
 
 
 
 
 . Ta' 
 
 / 
 
 
 
 
 4 
 
 
 d( 
 
 1" 
 
 ^^'^ 
 
 
 X 
 
 ' 
 
 
 
 :!(i^\ 
 
 Y 
 
 
 -'' 
 
 
 a\ 
 
 \»/ 
 
 
 
 
 ' ia 
 
 \ 
 
 
 
 
 
 
 
 ^ 
 
 
 
 By experiment with a cylindrical object — a roll of paper will do — it should 
 be realized that if the cylinder is inclined, there may be any number of planes 
 tangent to it, of angles to the H.P. not less than the angle of inclination of the 
 axis of the cylinder to the H.P. 
 
 Thus, in Fig. 88 at (i) the plane RST is tangent to the inclined cylinder and 
 is at the same inclination to the H.P. as that of the cylinder. Any plane of greater 
 inclination may be made tangent to the same cylinder, until one is reached that 
 is vertical, such as that whose H.T. is at LL. If the cylinder were inclined to the 
 V.P. also, then this plane LL would have a V.T. perpendicular to the A' I'. 
 
 As previously noted, the condition for a plane to be tangent to a cylinder 
 is, that it shall contain a straight-line generator of the curved surface, and as all 
 such generators have their H.T.'s in the cur\'e of the ellipse serving for the H.T. 
 of the cylinder, the H.T. of any tangent plane to the cylinder must be a tangent 
 line to the curved H.T. of the surface of the cylinder. 
 
 Now let ah. a'h\ at (ii), be the projections of a line parallel to the generators 
 of the given cylinder, and from some point in it, serving as the apex of a cone, 
 arrange a right circular vertical cone whose generators are at, say a° to the H.P. 
 Two a° tangent planes may now be represented by their H.T.'s at ac and ad, 
 to contain the line AB; and because parallel planes have their H.T.'s parallel, 
 and AB is a parallel to all the tangent planes to the given cylinder, being parallel 
 
86 
 
 DESCRIPTIVE GEOMETRY 
 
 to all its generators, there will be four a° planes, two parallel to plane ac and two 
 parallel to ad, and they will be shown by their H.T.'s, tangent to the curved H.T. 
 
 Fig. 88. 
 
 of the cylinder, as at e, f, g and h. Each of these four planes contains one 
 generator of the cylinder, and therefore, since the generators of the cylinder are 
 
 Fig. 89. 
 
 parallel to the V.P., the V.T.'s of these planes will be parallel to the elevations of 
 the generators of the cylinder. Therefore, by producing the H.T. marked g 
 
TAXGEXT PLANES TO CONES AND CYLINDERS 
 
 87 
 
 to XY, the V.T. for it may be drawn as shown, parallel to the generators of the 
 cylinder. So, also, with the V.T.'s of the other planes. X.B. The ellipse for H.T. 
 should be obtained by the method shown in Section 23. 
 
 In Fig. 89 is shown a right circular cylinder whose axis is inclined to both 
 planes of projection. Its H.T., an ellipse, is also shown. Suppose a tangent plane 
 
 Fig. 90. 
 
 to this cylinder has ST for its H.T., obtained as in the previous case; then because 
 the point H is the H.T. of a generator of the cylinder, and //' is the elevation of 
 //, the generator having // for its H.T. can be located, as shown in the figure at 
 GG', and produced until its V.T. is found at V. Now draw VSR, which is the 
 V.T. of the tangent plane. So proceed with others. 
 
 In Fig. 90, a right circular cone is represented in plan and elevation, with 
 axis inclined to both planes of projection. Its H.T., the large ellipse, is also shown. 
 
88 DESCRIPTIVE GEOMETRY 
 
 Let it be required to find the traces of planes tangent to this inclined cone and 
 having a given inclination to the H.P. Since all planes, tangent to the given cone, 
 must pass through the apex of it, and a plane of a particular inclination, say a°, 
 must be tangent to a vertical right circular cone whose generators are at that 
 particular inclination, it is necessary to make use of the apex of the given inclined 
 cone to serve as the apex of such a vertical one. The H.T. of the vertical cone 
 referred to is the circle H in the figure. 
 
 It should now be realized that a plane tangent to both cones at the same time, 
 will have its H.T. as a common tangent to the H.T.'s of the two cones, and will 
 contain the common apex A. The H.T. of such a plane is shown at ST, and to 
 obtain the point V in its V.T., a horizontal line, on the tangent plane, may be 
 drawn through the apex aa'. The line SV is the V.T. required. 
 
 In the case illustrated in Fig. 90, three other tangent planes of the same 
 inclination may be found. If the H.T. of one of them is parallel, or nearly parallel, 
 to the XY, or at such a small angle with it that there is not room for the point 
 to be located in XY, where the H.T. and the V.T. meet, then two inclined Hues 
 on the plane required may be drawn through the apex and through any convenient 
 points c and d in the H.T., and their V.T.'s found at Vi and V2. The line drawn 
 through these V.T. points is the V.T. of the tangent plane whose H.T. is the 
 line upon which the points c and d were chosen. 
 
 EXERCISE XXXV 
 
 1. Find the four tangent planes, each at 65° to the H.P., to a right circular cylinder whose 
 axis is at 40° to the H.P. and at 25° to the V.P. Diameter of cylinder i|". Let the cylinder 
 rest on a point in the H.P. 2" from XY. 
 
 2. Find the traces of planes inclined to the H.P. at 65° and tangent to a right circular cone 
 of iV' base, and ih" axis, when the axis is at 40° to the H.P. and the plan of the axis is at ■ 
 45° to XY. Let the cone rest on the H.P. in a point 15" from the XY. 
 
 PROJECTION OF SOLIDS DEPENDENT ON TANGENT PLANES TO 
 RIGHT CIRCULAR CONES 
 
 Section 25. In finding the projections of prisms, cubes and pyramids when 
 different faces of the same soHd are at different inclinations to the same projec- 
 tion plane, it is necessary to make use of the plane tangent to a right circular cone. 
 Illustrations are given and explained in Figs. 91 and 92. 
 
 In Fig. 91 the method is shown for finding the plan of a cube, or of a right 
 prism, when one face is at, say, 40° to the H.P. and another face is at, say, 70° 
 to the H.P. 
 
 RST is the 40° plane in which one of the faces will be found, and is arranged 
 perpendicularly to the V.P. for greater convenience. The line AB, shown in 
 
PROJECTION OF SOLIDS DEPENDENT ON TANGENT PLANES 89 
 
 plan at ab and in elevation at a'b', is an edge of the cube perpendicular to the 40 "" 
 face and therefore may serve as an edge of the 70° face. 
 
 The 70° plane, to include this edge of the 70° face, is found, according to 
 the method recently explained, by the use of a cone of 70° with its apex in the line 
 AB or that line produced. This 70° plane, thus containing AB, has its H.T. 
 at ////. 
 
 Fig. 91. 
 
 The point A is common to both planes, the 40° and the 70°; so, also, is the 
 point // where the two planes' H.T.'s meet. Join // to a therefore, and so obtain 
 the plan of the intersection of the 40° plane with the 70° plane. This line will con- 
 tain the edge of the cube common to the two faces whose inclinations are given. 
 Rabatte//a to ha2, and using ao as the comer for a square, mark off J2C2 on the 
 rabatted line, and also make d-ido perpendicular to it. These measurements, in 
 
90 
 
 DESCRIPTIVE GEOMETRY 
 
 the case of a cube, will be the same as that of a'b' previously chosen as the length 
 of an edge of the cube, or they should be made the same as the edges of the 
 40° face of the prism, if it be a prism that has to be projected. 
 
 Fig. 02. 
 
 It will readily be seen that from this rabattement of the two edges the plans 
 ac and ad may be obtained. Three edges are now found in plan, namely AB^ 
 
PROJECTION OF SOLIDS DEPEXDEXT OX TAXGEXT PLAXES 91 
 
 AC and AD, and parallels to these will be necessary in order to complete the 
 plan of the cube. 
 
 Of the two faces found in the figure, the face dace is a 40° face, and the face 
 bacf is a 70° face. 
 
 In Fig, 92 is shown the case of a pyramid in which the mjth ;d requires 
 the employment of two right circular cones, since the triangular faces of the solid 
 are inclined to the plane of the base. 
 
 Let it be required to find the plan of a square pyramid when the base is inclined 
 at, say, 45° to the H.P. and one triangular face is at, say, 65° to the H.P. The base 
 plane is marked RST in the figure, and is purposely arranged perpendicularly 
 to the V.P. in order to simplify the work. At (i) in the figure the square pyramid 
 is set up in plan and elevation in such a way as to find the angle, a'^, of the vertical 
 cone to which each of the triangular faces of the solid is tangent. Each face is 
 tangent on a generator which is at the same time the middle line of a triangular 
 face and meets the middle point of a base edge of the pyramid, as at d in be. At 
 any convenient place on the plane i^^T at (ii) set up the right cone with base angle 
 equal to that in (i), namely a°. Then find its horizontal trace, the large ellipse, 
 on the H.P. A tangent plane to this inclined cone, with its inclination accord- 
 ing to that given in the problem for a triangular face of the pyramid, viz., 65°, 
 is now obtained, with its H.T. at HH2. This trace is made tangent to the large 
 ellipse and at the same time tangent to the circular trace of a vertical cone of 65° 
 from the same apex as that of the inclined cone, viz., the point aa'. The V.T. 
 of this plane is not needed, so is therefore neglected. 
 
 The generator ae, a'e\ of the inclined cone, which at the same time lies in the 
 65° plane is now drawn, e being its H.T. Its elevation ae crosses RS, giving the 
 elevation d' , of the point in which it passes through the base plane. The line ad in 
 (ii) corresponds to the ad in (i), and by drawing a line through d and the point // 
 where the H.T.'s of the 45° and 65° planes intersect, the plan of the intersection 
 of the two planes, base plane and face plane, is obtained. The rabattement of thij 
 intersection line is ///. On this line /7/mark off, on either side of the rabattement 
 of point d, marked d-i, the measurements of dh, dc from the base edge in (i), and 
 construct the square in rabattement. Bring this up to its proper place in plan, 
 where it will appear foreshortened, and the edges of it will be tangents to the small 
 ellipse, the plan of the base of the inclined cone. Join the corners of the square 
 to the apex a, and the plan will be completed when such edges as are not in view 
 are shown by dotted lines. In the figure the plan of the solid is not completed, 
 in order to save confusion of lines. 
 
 N.B. — In the figure the construction work for finding the ellipses is reduced 
 or omitted as far as possible to save confusion of lines. In finding the plan of the 
 base of the inclined cone, however, there is shown a method of finding plans of 
 horizontal chords of it. Point g is the elevation of such a chord. Its plan can 
 be found by making a semicifcle as shown, and a perpendicular from g in the 
 diameter of it. to the circumference, thus giving half the chord length. 
 
92 ' DESCRIPTIVE GEOMETRY 
 
 EXERCISE XXXVI 
 
 1. Work out completely the problem in Fig. gi, and obtain an elevation of the solid on a 
 vertical projection plane whose XY is at 45° to the XY given. Let the length of the edge of the 
 cube be 2". 
 
 2. Work out completely the problem in Fig. 92, making the base edge of the pyramid 2" 
 and its height 3I", with the triangular face at 70° to the H.P. instead of at 65°. 
 
 3. Find the plan and elevation of a square prism, when the square end of it is at 45° to the 
 H.P. and a rectangular face is at 70° to the H.P. Short edges il", long edges 2^". 
 
 4. Find the plan of a regular pentagonal pyramid, i" edge of base, height 3", when the base 
 is incHned to the H.P. at 40°, and one triangular face is at 65° to the H.P. 
 
CHAPTER XIII 
 
 TANGENT PLANES TO A SPHERE, AND THE FINDING OF AN OBLIQUE PLANE 
 WITH GIVEN INCLINATIONS 
 
 Section 26. In discussing tangent planes to spheres it must chiefly be noticed 
 that the radius of the sphere, from the tangent point on its surface, is a perpen- 
 dicular to the tangent plane, and its projections will therefore be perpendicular to 
 the traces of the tangent plane. 
 
 yiC: 93- 
 
 ru-.. 94. 
 
 In the illustration. Fig. 93, a plane is shown tangent to a given sphere the 
 centre of which is aa'. The radius, which is perpendicular to the plane, is also 
 shown in elevation and plan, a'p' and up. A great circle of the sphere, that is, 
 one whose plane passes through the centre of the sphere, is shown in plan by the 
 (lotted line be and in elevation by the circle b'c'. It is perpendicular to the tan- 
 gent plane, and has for its radius the line AP. 
 
 In Fig. 94, any point on the surface of the sphere is shown in plan at p. It 
 is required to find the tangent plane to the sphere at this point. Any number 
 of points on the surface of the sphere, and at the same level, would be in the cir- 
 cumference of a horizontal circle whose eU^vation would appear as a straight line 
 
 93 
 
94 
 
 DESCRIPTIVE GEOMETRY 
 
 a'b'. Consequently, such a horizontal circle is represented and p' found in its 
 elevation. 
 
 Similarly, if the point be thought of as being on the circumference of a vertical 
 circle whose plan would be a straight line through p parallel to XY, a circle in 
 elevation would need to be drawn in order to locate the elevation of the point. 
 A tangent plane to the sphere at P may now be obtained by drawing a horizontal 
 line through P in the direction of the H.T. required, i.e., perpendicular in plan, 
 to the plan of the radius from P. Thus will be obtained a point V in the V.T. 
 
 Fig. 95. 
 
 required, and the traces of the plane may now be drawn perpendicular to the 
 radius from P. 
 
 In Fig. 95 the same problem as in Fig. 94 is solved by the employment of a 
 vertical plane containing the radius from the given point and therefore at right 
 angles to the plane required. This vertical plane has HH for its H.T. By rabatting 
 it on to the H.P., the centre of the sphere is carried over to C2, and a rabatted great 
 circle of the sphere is made on the H.P., shown in the figure covered with hatching. 
 The point P will now appear rabatted to p2, and the rabattement of the radius is 
 seen at c-ypi- At right angles to this draw poh as the rabattement of the intersec- 
 tion of the required plane with the vertical plane employed whose H.T. is the 
 line HH. This gives /?, a point in the H.T. of the required plane. Through h 
 draw the H.T. marked ST, and through 5 draw RS at right angles to the elevation 
 
TANGENT PLANES TO A SPHERE 
 
 95 
 
 of the radius. RST is the plane required. The height of /»' above XY is taken 
 from the rabattement, as indicated by the bracket line. This construction, Fig. 
 95, will be necessary, used conversely, when the trace of a plane tangent to a sphere 
 is given, and it is required to find the other trace of the plane and the projections 
 of the tangent point. 
 
 EXKRCISK XXXVII 
 
 1. Find the planes tangent to the sphere whose projections are at .1, each containing a 
 point on its surface, for which />' is the elevation. 
 
 2. A sphere, 2" diameter, touches both planes of projection. Find the two planes, each 
 touching the sphere in a point 15" above the H.P. and i§" from the V.P. 
 
 3. A sphere is given at B, and the H.T. of a plane tangent to it. Find the V'.T. and also 
 show the plan and elevation of the tangent [)oint. 
 
 4. Find the traces of a plane tangent to the sphere whose projections are at C, and make 
 it perpendicular to the line AB, whose projections are given. Mark the plan and elevation 
 of the tangent point. 
 
 5. Find the traces of three planes equally inclined to each other and all at 60^ to the H.T. 
 Let ihcm be tangent to a sphere of 2" diameter, resting on the H.P. with centre i^" from the \'.P. 
 Find also the inclination between any two of the tangent planes. 
 
 By e.xperiment and examination it should be realized that a sphere may be 
 enveloped by a right circular cone, having for its apex any point outside the 
 sphere, and that if two cones envelope the same sphere, a tangent plane to the 
 sphere may contain both apexes; also, that any two unequal spheres may be en- 
 veloped by a cone, the apex of which will lie in the line passing through their centres. 
 The tangent lines on the surfaces of such spheres as are enveloped by cones are 
 circumferences of circles, smaller than great circles of the spheres, and perpendicular 
 to the axis of the cone in each case. 
 
96 
 
 DESCRIPTIVE GEOMETRY 
 
 Fig. 96 is an illustration of the use of cones enveloping a single sphere, and 
 is a second method for obtaining the traces of an oblique plane whose angles 
 with the planes of projection are given. The previously discussed and more com- 
 monly used method was explained in Part I, Section 9, and was shown to depend 
 on the projections of a Hne set up at angles complementary to those of the required 
 plane, the traces of the required plane being then drawn perpendicular to the pro- 
 jections of the line. 
 
 In the method now discussed, a sphere with its centre in the XY, is represented 
 by the circle with centre at c. This circle therefore serves for both plan and 
 elevation of the sphere. 
 
 Fig. 96. 
 
 Suppose the required plane is to have angles of 60° to the H.P. and 45° to 
 the V.P., then a vertical cone to envelope the sphere, with its generators at 60"*, 
 to the H.P. and its apex therefore at V in the V.P., will have part of its circular 
 H.T. represented by the arc de, and any plane, whose H.T. is a tangent to this 
 arc, and whose V.T. contains the apex V, is therefore a plane at 60° to the H.P. 
 Next, represent another enveloping cone, about the same sphere, with its axis 
 perpendicular to the V.P., and its apex therefore at the point H in the H.P., and 
 make its generators at 45° to the V.P. Part of its circular V.T. will be the arc 
 fg. Any plane whose V.T. is a tangent to this arc and whose H.T. passes through 
 
TANGENT PLANES COMMON TO THREE GIVEN SPHERES 97 
 
 ^ is a plane at 45° to the V.P. because it is tangent to a right cone perpendicular 
 to the V.P. 
 
 We now have two cones enveloping the same sphere, and a plane containing 
 both apexes, and tangent to the sphere, will contain a generator of each cone. 
 Such a plane may now be represented by drawing its H.T. through H, tangent to 
 the circular H.T. of the 60° cone and its \'.T. through V, tangent to the circular 
 V.T. of the 45° cone. These traces, being traces of the same plane, meet in the 
 XY at R. 
 
 EXERCISE XXXVIII 
 
 Find, by the method employing enveloping or tangent conea/to a sphere, the traces of 
 the following oblique plane: / 
 
 {a) At 70° to the H.P. and at 55° to the \'.P. 
 {b) At 35° to the H.P. and at 75° to the V.P. 
 {c) At 65° to the H.P. and at 50° to the V.P. 
 
 TANGENT PLANES COMMON TO THREE GIVfeN SPHERES 
 
 Section 27. In Fig. 97 is illustrated the method of finding tangent planes 
 common to three unequal spheres, and the projections of the tangent points on 
 their surfaces. 
 
 The three spheres .4, B and C, are resting on the H.P., so that enveloping 
 cones of any two of them have their apexes in the H.P. The enveloping cone 
 of .1 and B has its apex at Hi, and one enveloping .1 and C has its apex at H-z- 
 The line through Hi and II2 is therefore the H.T. of a plane tangent to the three 
 spheres, and by passing a vertical plane, perpendicular to this tangent plane, 
 through the centre of one of the spheres, say, C, and rabatting as in Fig. 95. the 
 tangent point cc' is obtained. The V.T .of the tangent plane will be perpen- 
 dicular to the elevation of the radius which ends in c' . 
 
 Just as the radius to point C is perpendicular to the tangent plane found, 
 so, in like manner, the radius lines to the tangent points on the surfaces of A and 
 B will be perpendicular to the plane, therefore plans of these radius lines made 
 perpendicular to the H.T., and elevations of them made perpendicular to the V.T. 
 may be drawn. These must be limited at aa and bb' by generator lines on the 
 surfaces of the cones, one from H2 through cc' to limit the radius line from the 
 centre of sphere A, and another from this point aa' found on sphere A, drawn to 
 III, to limit the radius of the sphere B at bb'. The elevations of the points may be 
 obtained by carrying perpendiculars across A'l' from the plans, instead of using 
 elevations of generator lines. Both ways are shown in the figure. One checks 
 the accuracv of the other. 
 
98 
 
 DESCRIPTIVE GEOMETRY 
 
 EXERCISE XXXIX 
 
 1. Find two tangent planes to three spheres, the planes not to pass between any of the 
 spheres, and mark the projections of the tangent points on their surfaces. The spheres have 
 their centres all at the same level, and these centres are the corners of an equilateral triangle 
 2" side, with no side parallel to the V.P. Diameters of the spheres if", i\" and f " respectively 
 
 2. Three spheres rest on the H.P. and touch each other. Their centres are at difTerent 
 distances from the V.P. Their diameters are if", |" and f" respectively. Find the inclined 
 plane tangent to all three spheres and mark the projections of the tangent points. 
 
 Fig. 97. 
 
 When the centres of three unequal spheres are at unequal distances from 
 the planes of projection and the spheres are not all resting on the H.P., then the 
 line joining the apex points of the enveloping cones will be oblique, and the problem 
 becomes more involved. If, however, the line joining the apex points be considered 
 in relation to one sphere only, then it becomes a matter of finding the planes con- 
 taining this line and tangent to the one sphere. It will be seen that the tangent 
 planes found will be tangent to the other spheres, the apex of whose common 
 enveloping cone is a point in the line. The tangent points, also, having been 
 obtained on the one sphere, the tangent points on the others may be found as in 
 the case considered in Fig. 97. 
 
 Let ah, a'h' in Fig. 98 be an inclined line, such as that referred to above, and 
 cc' the centre of a sphere whose projections are given. Let it be required to find 
 
TANGENT PLANES COMMON TO THREE OIVEN SPHERES 
 
 99 
 
 the tangent planes to this sphere, which contain the Hne AB. The order of 
 procedure might be as follows: — (i) Find the traces, II and V, of the given line. 
 These will be in the traces of the planes required. (2) Find a plane, RST, passing 
 through the centre of the sphere and perpendicular to the line AB. (3) By the 
 
 Fig. 98. 
 
 help of a plane LMX, containing the line -1^, and intersecting plane RST, find 
 the point of intersection which AB makes with RST at />/>'. (4) Rabatte the point 
 P about 5rto p2, and also rabatte the centre C to C2, and draw the rabatted great 
 circle of the sphere made by the intersection of the sphere by plane RST passing 
 through its centre. (5) From p-^ draw the rabatted tangent lines to this great 
 
100 DESCRIPTIVE GEOMETRY 
 
 circle and find their H.T.'s at Ih and Ho. (6) Join Hi and H2 to p and so obtain 
 the plans of two lines which are tangent to the sphere and have a common point 
 P in the given line AB. (7) The planes required, tangent to the sphere, will each 
 contain one of these tangent lines and the given Hne, therefore draw the traces 
 of the required planes by making their H.T's at HiH and H2H respectively. Their 
 V.T.'s will pass through the V.T. of the given line AB, viz., V. (8) The projec- 
 tions of the tangent points may be obtained by passing them over from their 
 rabattements at J2 and co to their plans at d and e respectively, and then obtain- 
 ino; their elevations in the usual way. 
 
 EXERCISE XL 
 
 1. Find the traces of the tangent planes to the sphere, of 2" diameter, which has its centre 
 i\" from both planes of projection, and contains a line whose H.T. is 2" from XY and 2" from 
 the plan of the centre of the sphere, while its V.T. is 3" above XY and if" from the elevation 
 of the centre of the sphere. Both these given traces are to the right of the centre of the sphere. 
 Also, mark the projections of the tangent points on the surface of the sphere. 
 
 2. Arrange three spheres of difTerent sizes and not far away from each other, placing them 
 at different levels. Find two planes tangent to all the spheres and mark the tangent points on 
 the spheres. 
 
CHAPTER XIV 
 
 SIMPLE CASES OF IXTERPEXETRATIOXS OF SOLIDS, AND THE DEVELOPMENTS 
 
 OF SURFACES 
 
 Section 28. The matter of finding intelligently the intersection of the 
 surface or surfaces of one solid with the surface or surfaces of another, depends 
 
 IlG. yy. 
 
 chiefly upon recognizing the use of sectional planes or of surfaces which will con- 
 tain elements or generators of both of the given surfaces, meeting each other in 
 points common to both given surfaces. The solids should be so arranged as to 
 provide for the use of planes which will readily give plans and elevations of the 
 generators or elements, and their intersection points common to both surfaces. 
 In Fig. 99 is shown a square prism penetrating a cylinder. In making the 
 
 101 
 
102 
 
 DESCRIPTIVE GEOMETRY 
 
 projections of the solids, or, rather, such parts of them as may be necessary, 
 start by making the square which will serve as the rabattement of the end of 
 the square prism. 
 
 Inspection of the figure will disclose the following facts:— (i), that a vertical 
 section plane containing the axis of the cylinder contains two vertical generators 
 of the cylinder, and may also contain the edges A and B of the prism which has 
 been placed parallel to the V.P.; (2) that the elevations of the points of intersec- 
 tion of these two edges with the generators of the surface of the cylinder are a'a' 
 and b'b' respectively; (3) that similarly c'c' are the elevations of the intersections 
 of the edge C with the surface of the cylinder; (4) that for other points, common 
 to both surfaces, straight lines, of any convenient number, and placed conveniently 
 
 (i.e., not necessarily at equal distances from each other) to run parallel to the V.P., 
 are drawn on the surfaces of the prism at i, 2, 3, 4, first in plan and then in eleva- 
 tion, by taking advantage of the rabatted square end; (5) that vertical generator 
 lines on the surface of the cylinder, and in the same vertical section planes as the 
 lines on the prism, pass through the points where these lines i, 2, 3 and 4 cut the 
 curved surface of the cylinder. These are then projected, resulting in the eleva- 
 tions of points, common to both surfaces, and shown at d', e', f, g' , etc.; (6) that 
 these elevation points must be joined by freehand curved lines to indicate the 
 elevation of the common section Hues. 
 
 In Fig. 100 is shown what is called the development of the surfaces of the prism 
 of Fig. 99, in order to show what the section lines on those surfaces appear like. 
 The long lines marked ACBDA are set up at distances from each other equal 
 to the width of the faces of the prism. Then the straight lines which were drawn 
 
SIMPLE CASES OF IXTERPEXETRATIOXS OF SOLIDS 103 
 
 upon those faces in Fig. 99 are drawn at their proper distances from each other, 
 taken from the rabatted end and numbered. The distances of the points of inter- 
 section on all these lines and upon the edges ACBD are now taken from the eleva- 
 tion in the figure, and joined by freehand curves, thus marking the exact way in 
 which each face has been penetrated by the surface of the cylinder. 
 
 In a similar way the curved surface of the cylinder may be " develoi)cd " 
 and the intersection lines upon its surface shown. The circumference of the end 
 
 Fig. ioi. 
 
 is laid out as a straight line, and pcri)cndiculars are made from it to represent 
 generators. These are cut at heights taken from the elevation in the figure and 
 then the points so found are joined by curved lines, showing the e.xact way in 
 which the surface has been cut. 
 
 A further illustration of development is given in Fig. loi, where it will be 
 seen that the circular base of the solid is laid out as well as the cun'ed surface, 
 and the intersection lines made on them are displayed. For convenience the 
 curved surface is divided into twelve equal parts. One-twelfth of the circum- 
 ference of the base, that between 7 and S in the plan. is. by an approximate 
 
104 DESCRIPTIVE GEOMETRY 
 
 method, found to be equal to the tangent marked 76. The method is, to make 
 the hne from point 7 through the centre to a, equal to three times the radius of the 
 circle, and from a to draw ab through the other end of the 30° arc to the tangent 
 line from 7. Since all the division lines of the curved surface are equal, from c in 
 the elevation, as centre, draw the arc i, 2, 3, etc., to i, marking on it the divisions, 
 equal to the line jb in the plan drawing. The intersections of the lines on the 
 surface of the cone are now carried across, by parallels to XY, to the line ci, 
 and from that line carried by arcs with c as centre to their development. Between 
 the points 2 and 3 the circumference of the base is cut by the plane RST, and at 
 this point, therefore, in the development, draw a tangent circle, so that the 
 intersection of the base from this point may be marked on the circle representing 
 it in the development. 
 
 EXERCISE XLI 
 
 1. Find the elevation of the intersection lines, made by the surfaces meeting each other 
 of a square prism, 2" short edge, standing vertically with a vertical face at 30° to the V.P., and 
 another square prism, i|" short edge, with long edges horizontal and parallel to the V.P. A 
 face of this prism is at 20° to the H.P., so that both of the diagonals of the end of it are inclined, 
 the one at 65° and the other at 25°, to the H.P. Arrange it so that three of the vertical edges of 
 the first prism pass -through the second prism, and one of the vertical edges of the first passes 
 through the axis of the second. (N.B. — In working out the result, it must be seen that the vertical 
 edges of the first prism penetrate faces of the second prism in points through which should be 
 drawn, on the faces of the second prism, hnes parallel to the long edges. The elevations of 
 these should then be obtained.) 
 
 2. A right cylinder, 2" diameter, has its axis vertical, and a square prism, with one diagonal 
 of end horizontal and at 70° to the V.P., has its axis inclined at 20° to the H.P. The axis of 
 the prism passes through the axis of the cylinder. The edge of end of the prism is ij" long. 
 (N.B. — It is suggested that in setting up these solids the prism should be arranged correctly 
 first, in relation to the planes of projection, and the cylinder placed in position after this has 
 been done.) 
 
 3. Two right circular cylinders interpenetrate each other. Let one be placed with its axis 
 vertical, its diameter being 2", and let the other, with diameter if", have its axis inclined at 
 20° to the H.P. and parallel to the V.P., is" farther away from the V.P. than the axis of the 
 other cylinder. Show the elevation of the intersection curve made by their surfaces. Make 
 also a development of the curved surface of the 2" cylinder to show the intersection line upon 
 it. 
 
 MORE DIFFICULT CASES OF INTERPENETRATION OF SOLIDS, AND THE 
 PROJECTION OF THE INTERSECTIONS OF THEIR SURFACES 
 
 Section 29. When the intersecting surfaces are those of the cone and the 
 sphere, then all that is necessary, in order to obtain the projection of points 
 common to both surfaces, is to make use of planes which give circular sections of 
 the cone, and pass through the sphere. If the circular sections of the sphere, 
 
MORE DIFFICULT CAbES OF INTEKPEXETRATIOX OF SOLIDS lu5 
 
 by these planes, intersect the circular sections of the cone by the same planes, 
 then the points of intersection are points on both surfaces. 
 
 Another method of determining the intersection of the surfaces of these soHds. 
 is by the use of intersecting spherical surfaces instead of intersecting planes, and 
 depends on the following facts: — 
 
 (i) That a sphere whose centre is in the axis of a right circular cone will 
 intersect the surface of that cone, if it intersects at all. in a circle perpendicular 
 to the axis of the cone; 
 
 Fig. io: 
 
 Fig. 103. 
 
 (2) That if a sphere intersects another sphere the intersection of their sur- 
 faces is the circumference of a circle, the plane of which is perpendicular to the line 
 joining their centres; and 
 
 (3) That if a sphere, with its centre in the axis of a right circular cone inter- 
 penetrates at the same time another sphere, then the intersection of the circular 
 sections it makes with both solids will be points common to the surfaces of both 
 solids. 
 
 Illustrations of the use of both methods referred to are given in Figs. 102 
 and 103. In Fig. 102, a right circular cone with its axis vertical, and a sphere 
 
106 
 
 DESCRIPTIVE GEOMETRY 
 
 whose centre is cc', are arranged so that their surfaces intersect. The centre 
 of the sphere is not necessarily at the same distance from the V.P. as the axis of 
 the cone. By the use of horizontal section planes a number of points will be 
 found in plan, where the circular sections of the solids intersect. Their elevations 
 are readily discovered, and by joining the points by freehand curves, the projec- 
 tions of the section line or lines, are obtained. 
 
 In Fig. 103, concentric sectional spheres, having their common centre at the 
 apex of the cone, are represented a.t A, B, C, etc. Since the centre of the given 
 
 Fig. 104. 
 
 sphere is at the same distance from the V.P. as the axis of the given cone, the 
 sections the given cone and sphere make with the concentric spheres will appear in 
 elevation as straight lines, intersecting each other in the elevations of points com- 
 mon to their surfaces. These points may readily be found in plan in the plans 
 of the circular sections of the cone. For example, the cone intersects the sphere 
 A in the circle whose elevation is a'a', and the given sphere intersects the sphere 
 A in the circle whose elevation is a' 2a' 2. Where these cross each other is the 
 point a'z, which is the elevation for two points a^ and as in plan. 
 
MORE DIFFICULT CASES OF INTERPEXETRATIOX OF SOLIDS 107 
 
 In Fig. 104 is shown how best to deal with the problem of linding the plan 
 and elevation of the intersection of the surface of a cone with the surface of a 
 cylinder, when the two solids interpenetrate. 
 
 The c>linder is placed with its a.xis parallel to the V.P., and a secondary ele- 
 vation plane is represented, perpendicular to the ordinary V.P., with its XY 
 marked X2Y2 in the figure. Upon this secondary vertical plane' is drawn an 
 end view of the cylinder, and another view of the cone. 
 
 Inclined planes perpendicular to this secondary vertical plane, and con- 
 taining generators of the cylinder and of the cone, are represented with their H.T.'s 
 at A, B, C, etc., and their V.T.'s, passing through the secondary elevation of the 
 apex of the cone at c". One generator of the cone and two generators of the 
 cylinder are contained in the plane A . The intersections made by the generators 
 result in two points whose elevations are a and a'2. Plane B contains two genera- 
 tors of each solid and consequently four points common to the two surfaces are 
 found; and so on. After sufTicient points are found the freehand curves may be 
 drawn through them. In the figure, the elevation has been completed, in order 
 to emphasize the fact that it is better to work it out first. The plan may now 
 jjc obtained, either by projecting from the elevations of the points on to the plans 
 of the generators of the cone, or by making plans of the generators of the cylinder 
 to cross the plans of the generators of the cone in the points required. This latter 
 method will be necessary in the case of the plans to be marked a and a-z. 
 
 EXERCISt: XLII 
 
 1. Find plan and elevation of the intersection of the surfaces of a right circular cone and a 
 sphere, by using the method involving spherical section surfaces. The cone has a base 2 J" 
 diameter, in the H.P. and a height 2^". The sphere, diameter 2", rests on the H.P., and has 
 its centre in the curved surface of the cone. 
 
 2. Find, by the method of horizontal section planes, the plant and elevation of the inter- 
 section line, when a cone with 25" circle for its H.T., and with an axis 2§" long from the centre 
 of its H.T. to its ape.x, inclined at 70° to the H.P., interpenetrates with a sphere, 2" diameter, 
 resting on the H.P. and having its centre in the shortest line from the apex to the H.P.. on the 
 surface of the cone. 
 
 3. A right circular cylinder. 2" diameter, lies with one of its generators in the H.P., parallel 
 to the \'.P. A cone, whose H.T. is a circle 3" diameter, has the centre of this circle in the 
 plan of the axis of the cylinder. The axis of the cone is 4" long, and has a plan J" long at 45° 
 to XY. Find the plan and elevation of the intersection of the surfaces of the two solids. 
 
 To obtain the projections of the surface intersections when inclined cones and 
 cylinders, not necessarily right circular ones, interpenetrate, it is necessary to make 
 use of section planes which will have in them straight generator lines on both sur- 
 faces. Hence, when the surfaces of cones are cut by those of other cones, or of cylin- 
 ders, the apex points of the cones must be in the section planes made use of; and in 
 
108 
 
 DESCRIPTIVE GEOMETRY 
 
 the case of cylinders, their axes must be parallel to the section planes. Thus, in Fig. 
 105, where two cylinders are represented whose given H.T.'s are circles and whose 
 axes are directed as indicated by the plans and elevations of generators of their 
 
 Fig. 105. 
 
 surfaces, it is necessary to take any generator of one cylinder, or a parallel to it 
 as ab, a'b' , and from some point in that line to draw another line, say ac, a'c', 
 parallel to the generators of the second cylinder. The H.T.'s of these two lines 
 are h and c respectively, and the plane of the two lines has the line HH for its H.T. 
 All the generators of both surfaces are parallel to this plane, and hence any plane 
 
MORE DIFFICULT CASES OF IXTERPENETRATIOX OF SOLIDS 109 
 
 made parallel to this one, and having its H.T. crossing the circular H.T.'s of the 
 given cyhnders will therefore contain generators of the surfaces of both cylinders. 
 The points in which these generators cut each other are points in the intersection 
 line of the surfaces, and the plan and elevation of it can in this way be found, as 
 shown. It will be found an advantage to arrange tangent planes to commence 
 with. These in Fig. 105, are marked i and 2 respectively, and it will be there seen 
 that a strip on each of the cylinders is not pierced by the surface of the other, and 
 the curves will be turned back from these strips accordingly. 
 
 Fio. 106. 
 
 In Fig. 106 two cones are represented. A hne through the ape.\ points 
 will be contained in the planes which have generators of both surfaces. Hence 
 the V.T. and the H.T. of this line must be found, as at V and Z7, and the traces of 
 planes passed through them. Those planes, such as VRII, which contain genera- 
 tors of both cones, will give points common to both surfaces. A number of such 
 points must be found in order to obtain the projections of the cur\-e or curves. 
 
 In Fig. 107, a cylinder and a cone are represented, and here it will be seen 
 that since the apex of the cone must be in all the section planes that are of sersice 
 in finding the intersection, and these section planes must contain generators of 
 
no 
 
 DESCRIPTIVE GEOMETRY 
 
 the cylinder also, therefore they must contain a line, through the apex point, 
 parallel to the generators of the cylinder. Having drawn this line and obtained 
 its traces, as at V and H, it is only necessary then to draw traces of section planes 
 through V and H which will cross the H.T. of the cylinder and the V.T. of the 
 
 Fig. 107. 
 
 cone, as the plane VRH does, and obtain plans and elevations of generators of 
 
 both surfaces, as they cross each other, thus finding projections of section points. 
 
 Other simple cases of interpenetration, not requiring for their solution any 
 
 new principle, or serious change of method, will doubtless suggest themselves. 
 
 EXERCISE XLIII 
 
 Work to completion, in plan and elevation, the cases of interpenetration shown in Figs. 106 
 and 107. 
 
SURFACES OF REVOLUTION, AND THE SCREW THREAD U; 
 
 SURFACES OK RE\OLLTIOX, AND THE SCREW THREAD 
 
 Sectiox 30. Thus far we have dealt with three different surfaces of revolu- 
 tion, the cylinder, the cone and the sphere. The first is formed by revolving a 
 straight line round an axis of rotation while maintaining a parallel attitude in 
 relation to it. The second is formed by revolving a straight line round an axis 
 of rotation, the revolving line having one point of it in the axis of rotation. The 
 third is formed by revolving a semicircle, making use of its diameter as the axis 
 of rotation. Others will now be considered, namely, the hyperboloid and the 
 helicoid. 
 
 The hyperboloid of revolution is obtained as the result of a straight line revolv- 
 ing about an axis of rotation, but not in the same plane with it. Each point in 
 the revolving line travels in the circumference of a circle for its locus, the plane 
 of which is at right angles to the axis. This will be noted in Fig. io8, where c, 
 c'c' is the axis of rotation, and ah. ah' is the line revolving. .\ny points taken 
 in this line revolve in paths which are the circumferences of circles whose planes 
 are horizontal, the axis being vertical, and on joining the elevations of the extreme 
 limits to which these points move from one side to the other, the freehand cur\-e 
 gives the projection of the contour of the surface of revolution generated, namely, 
 the hyperboloid of revolution. 
 
 The helicoid. or surface of revolution generated by a straight line moving 
 so that any point in it travels in a helicoidal line or helLx, as a spiral, while the 
 
112 
 
 DESCRIPTIVE GEOMETRY 
 
 line is not in the same plane with the axis, but is constantly maintaining the same 
 attitude in relation to it, is used in the making of screw-threads on bolts, etc. 
 
 Fig. iio. 
 
 In projecting screw-threads, the curved edge lines of the threads are the 
 paths of points in the generating lines of hehcoids, and may be thought of as lines 
 
SURFACES OF REVOLUTION, AND THE SCREW THREAD 113 
 
 drawn on the surfaces of cylinders or drums. The development of such a cylin- 
 drical surface of a bolt, showing the thread line, would give that line as a straight 
 one, as ac in the Fig. 109. The Hne ab is the length of the circumference of the 
 circular section, and the distance be represents the pitch or distance travelled in 
 the direction of the axis during one revolution, ac is the straight line represent- 
 ing the helicoidal line developed as from the surface of a cylinder. These curved 
 helicoidal lines may be projected on to the V.P., and in this way a screw-thread is 
 represented. 
 
 As in illustration, in Fig. no, let ABB2 be the triangular section, made 
 by a plane which has in it the axis of the bolt, of a screw-thread to be pro- 
 jected, when the large circle in the plan, represents the section of the bolt. From 
 B to B2 is the pitch of the thread. Two helicoidal lines must be found, one, on 
 the outside cylindrical surface, and passing through the point A, and the other 
 on an inner cylinder or drum upon which is situated the point B. Notice that 
 the pitch is the same for both lines. 
 
 For convenience, 30° divisions of the circular plan are made, therefore twelve 
 in all, and for each 30° division over which the line moves it rises ,Vt of the pitch. 
 For a right-hand screw the thread runs upwards in the direction of the arrow- 
 headed line. 
 
 For a thread of square section there will be four helicoidal lines to draw. In 
 such a case if the pitch be f ", the edge of the square section will be |". 
 
 EXERCISE XLI\' 
 
 1. Find the elevation of a screw-thread on a vertical bolt 2^" diameter, when the section 
 of the thread by a plane including the a.xis, is an equilateral triangle. The thread to be single, 
 and pitch |". 
 
 2. Find the projection of a bolt screw-thread with square section. Diameter of bolt 3", 
 pitch f . 
 
 3. Find the surfaces of revolution generated by the edges of a cube made to rotate on one 
 of its solid diagonals, i.e., on a diagonal passing through its centre. Edge 2". 
 
CHAPTER XV 
 
 RADIAL PROJECTION. PERSPECTIVE PROJECTION 
 
 Section 31. Things are made visible to us by means of light, and in order 
 to obtain, in the eye, a picture of anything, rays of light, which travel in straight 
 lines, must proceed from the thing looked at, to the eye. The retina, at the back 
 of the eye, is the picture surface, a curved one, receiving the projection or view 
 of the object. This projection or view is obtained by a process called Radial 
 Projection. It is the same as that by which a plane surface, the plate or flat 
 film, in a camera, receives a picture. 
 
 In Orthographical Projection, by which we have obtained plans and eleva- 
 tions of things, the projectors are at right angles to the projection plane. In 
 Radial Projection the projectors are inchned to the projection plane. Radial 
 Projection is used in different ways for the projection of maps, etc., but is most 
 commonly used, by Architects and others, in what is called Perspective Projection. 
 
 In Perspective Projection a single projection plane is used, and is situated 
 between the object and the eye. It can be seen through, by the spectator, as 
 he looks at the object. The view or projection of the object is caught by this 
 projection plane, or Picture Plane as it is called, which is, in attitude, at right 
 angles to the direction of vision. 
 
 Since, in a natural way, and ordinarily, a person's sight is directed, as he stands, 
 in a horizontal direction, the Picture Plane, in Perspective Projection, is arranged 
 as a vertical plane, with a point, perpendicularly opposite the eye, marked upon 
 it, as the Centre of Vision. 
 
 The eye-level exactly agrees with what one sees in the distance as the hori- 
 zon, and is represented on the Picture Plane by a horizontal Hne, in which the 
 Centre of Vision point is situated. This line, representing the eye-level and at 
 the same time the horizon, is called the Horizon Line, and parallel to it, on 
 the Picture Plane, another horizontal Hne is drawn, at a distance below the 
 Horizon Line equal to the distance of the spectator's eye from the ground and is 
 consequently called the Ground Line. 
 
 The distance from the Horizon Line to the Ground Line is decided upon 
 for each case, and should be such as will best serve the purpose m viewing the 
 object to be represented. Drawings, of course, are necessarily made strictly to 
 some chosen scale — |" to the foot, ^" to the foot, half size, etc., as may be suitable. 
 
 The distance between the object and the spectator is variable, and may be 
 any distance chosen. The appearance of the object depends very much upon the 
 
 114 
 
RADIAL PROJECTION. PERSPECTIVE PROJECTION 115 
 
 distance chosen. This distance being settled, it is then necessary to fix the posi- 
 tion, between the object and the spectator, of the Picture Plane, and upon this 
 will depend the size of the drawing or projection of the object. 
 
 Let the student imagine the vertical transparent Picture Plane, between 
 his eye and the object, and he will realize that, as the plane is placed nearer to the 
 object the view of it which he could trace on the plane will be larger, and as he 
 brings the Picture Plane towards the eye, the view of the object, traced on the 
 plane, will be smaller. 
 
 X Plan of Picture Piano 
 
 Xz 
 
 \<-V 
 
 S*Plan of Spectator's Eye 
 
 Fig. III. 
 
 This and other matters will be better understood if reference is made to Fig. 
 Ill, where the proper arrangement for perspective projection is shown in plan. 
 It will be seen that the spectator is at a distance OS from the object, and that 
 A'l', representing the vertical picture plane, at right angles to the direction of 
 vision rei)resentcd by the direction of the line OS, is placed not far away from 
 the object in relation to the distance of the spectator from it, and consequently, 
 the size of the picture of the object obtained by the interception of the rays, from 
 the object to the eye. by the plane at .VI'. will be fairly large. On the other 
 
116 DESCRIPTIVE GEOMETRY 
 
 hand, if the picture plane be placed so as to have its plan at X2Y2, that is, rather 
 close to the eye in relation to the distance of the object from it, then the picture 
 obtained by the interception of the rays from the object as they proceed to the 
 eye, will be correspondingly small. The point C is the plan of what was spoken 
 of as the Centre of Vision, and is commonly marked CV. 
 
 If a vertical plane be imagined as containing the horizontal line of sight 
 or direction of vision, the line SC, then anything to the left of this vertical plane, 
 as, for instance, the corner A of the object, is said to be to the left of the spectator, 
 and anything to the right of this same vertical plane, as 5, is said to be to the right 
 of the spectator. The whole object might be to the left, or to the right, of the 
 spectator, and of course that would mean that the plan of it would have to be 
 placed to the left, or right, of the line 5C, accordingly. 
 
 In order to make a perspective projection or drawing, the essential things are: — 
 
 (i) The placing of the spectator S, and the Une of sight SC. 
 
 (2) The placing of the object, represented by plan, at its proper distance 
 
 from the spectator, measured in the direction of the line of sight, and 
 at its proper distance to the right or left of the spectator. 
 
 (3) The placing of the picture plane, represented by XY, and 
 
 (4) The decision as to the height of the eye above the horizontal plane of 
 
 the ground, on which, or in relation to which, the object is placed. 
 
 To represent the two lines Ground Line and Horizon Line, two lines, so 
 named, are drawn at their proper distances apart, across the paper parallel to XY, 
 and at any convenient place between the point 5 and the line XY, as in Fig. 112. 
 
 Before proceeding to discuss specific cases of perspective projection, certain 
 facts which have a bearing on the intelligent understanding of the method adopted, 
 must be realized. These are as follows: — 
 
 (i) All parallel lines receding from the spectator appear to converge, 
 regardless of attitude or direction. 
 
 (2) All horizontal lines regardless of level, when receding from the spectator 
 
 approach the horizon, and when produced to infinity lose themselves 
 on the horizon in points commonly spoken of as vanishing points, and 
 
 (3) Any number of parallel horizontal lines, regardless of levels or direction, 
 
 have the same vanishing point in the horizon. 
 
 Consequently, therefore, lines on the ground plane, or other horizontal 
 lines below the level of the eye, if receding from the spectator, will be represented 
 by lines rising in the projection from their near ends towards the eye-level line 
 which represents the horizon, and, similarly, horizontal receding lines above the 
 level of the eye will be represented by lines drawn downwards from their near ends 
 towards the eye-level line where their vanishing points are. 
 
 Consequently, also, horizontal lines that are perpendicular to the projection 
 plane, that is, in the same direction as the line of sight, wherever they are, will 
 
RADIAL PROJECTION. PERSPECTIVE PROJECTION 117 
 
 have their vanishing point at C, or CA' ., in the horizon Hne, that is, at the Centre 
 of Vision. 
 
 An illustration is shown in Fig. 112, where a vertical shaft standing centrally 
 on a square block is represented. The plan of this group is placed a little to the 
 
 Fig. 112. 
 
 left of the spectator, measured to the left of the line of sight, and is at a distance, 
 measured along the line of sight, from the spectator whose position is marked 5. 
 The picture plane, that is, the projection plane, is placed vertically at AT, which 
 is its plan, and is then represented again, lower down on the paper, with .VI' or 
 Ground Line parallel to the original position and the eye-level line or Horizon 
 
118 DESCRIPTIVE GEOMETRY 
 
 Line represented at the height decided upon for the eye. In the figure the 
 elevation of the group is shown, exactly opposite the plan of it, on this plane, and 
 in this way the true heights on the picture plane, where horizontal lines through 
 the points of the group, and perpendicular to the plane, meet it, are obtained. 
 
 To obtain the Perspective Projection it is now only necessary to represent 
 all the horizontal lines that are really perpendicular to the Picture Plane, as pro- 
 ceeding to the Centre of Vision, C.V., and drop perpendiculars to them from 
 the points in the plan XY of the picture plane, where rays from the points, to the 
 eye, are intercepted by the picture plane. In the figure most of these perpen- 
 diculars are only started, as dotted lines, not carried all the way, in order to avoid 
 confusion. 
 
 Fig. 112 illustrates what is often spoken of as Parallel Perspective, since the 
 lines necessary for it are either parallel to the picture plane or parallel to the Line 
 of Sight. 
 
 N B. — Since the view or projection is not influenced, except in size, by the 
 distance of the picture plane from the spectator, and the things represented are 
 often large things drawn to a small scale, the picture plane is arranged as if near 
 the object in order to get a reasonably large drawing for the projection. 
 
 EXERCISE XLV 
 
 1. Find the perspective projection of a box with a lid. Its measurements are 2 ft. X3 ft. 
 and it is i| ft. high. Thicknesses may be omitted. Let the picture plane be 10 ft. from the 
 spectator and the box 2 ft. beyond the plane, or 2 ft. " into the picture " as it is said, and the 
 nearest corner 4 ft. to the left. Let one end be parallel to the picture plane and the front of 
 the box be on the near side. Let the lid be opened to an angle of 45°. 
 
 2. Make a perspective projection of a double cross, that is having four arms at right angles 
 to each other, with a pedestal and shaft similar to those in Fig. 112. Let the spectator's eye 
 be 5 ft. above the ground and the total height of the group 10 ft., the pedestal being i^ ft. thick 
 and 4 ft. square, one edge of the square being parallel to the picture plane and 3 ft. into the 
 picture (i.e., beyond the picture plane). Make the shaft and crossbars i ft. square in section, 
 each arm being i| ft. long, and the distance from pedestal to crossbars 5 ft. Let the nearest 
 corner of the pedestal be 3 ft. to the right of the spectator and the distance of the picture plane 
 from the spectator 12 ft. 
 
 3. Make a perspective projection of any familiar object such as a book, a hut, a boat-house, 
 a table, or any simple thing arranged in parallel perspective. 
 
 PERSPECTIVE PROJECTION, Continued 
 
 Section 32. We have seen that the vanishing point, in the horizon, for hori- 
 zontal lines perpendicular to the picture plane, is the same as that for the line of 
 sight or line from the eye perpendicular to the picture plane, and marked C.V.^ 
 the central vanishing point, or Centre of Vision. 
 
 We shall now consider the determination of vanishing points for lines that 
 are not perpendicular to the picture plane. For a horizontal line, the vanish- 
 
PERSPECTIVE PROJECTION, CONTINUED 
 
 119 
 
 ing point must be in the horizon, and must, for our purposes of projection, be 
 represented by a point on the horizon line drawn on the picture plane. This point, 
 the given line's vanishing point, will be the same as for a line proceeding from 
 the eye of the spectator parallel to the given line. Hence the method shown in 
 Fig. 113, namely, from S draw a line parallel to the plan of the given horizontal 
 line, ab, to meet XV, the projection plane, in a point whose plan is V, and whose 
 place in the Horizon Line on the picture plane is V.P. 
 
 Since the line ab and the parallel to it SV are in the same plane, an oblique 
 plane, it will be realized that the trace, on the picture plane, of the line ab, viz., 
 
 xi' Plan of Pi Mro riano V Y 
 
 Horizon Line / tV 
 
 I .. I 
 
 
 
 X 
 
 Fig. ii.v 
 
 the point whose plan is v, joined to the trace of the line 5F at the point whose 
 plan is r, will give the intersection of the oblique plane with the picture plane. 
 This is shown as the line, across the picture plane, marked v'V , and so it will 
 be realized that the perspective projection of the line ab, a line in the oblique 
 plane, as seen from the point 5, a point in the same plane, by radial projection in 
 that oblique plane, must be a foreshortened line lying in the intersection line 
 v'V in the projection plane. The plans of the rays are directed from a and b to 
 S in the drawing, and are seen to meet the plan of the projection plane, AT. at 
 points from which vertical projectors are drawn to determine the perspective 
 projection ah' . The point in the Horizon Line marked \" is the vanishing point 
 
120 
 
 DESCRIPTIVE GEOMETRY 
 
 for the line AB and for all other lines parallel to it>-It is therefore also marked 
 V.P. A second hne, horizontal, and immediately above the hne AB, is shown 
 in projection at a'2^'2- Its vertical trace in the picture plane is y'2. The inter- 
 section, of the plane of this second line and the eye point, with the picture plane, 
 is the trace line v'2V', and since, as before, the rays of light from thi§ upper line 
 
 Fig. 114. 
 
 travel to the eye in the obhque plane whose intersection with the picture plane 
 is v'oV, the representation of the line in perspective is inclined, and is found 
 to be a'2&'2. 
 
 In Fig. 114, an illustration is given involving the use of vanishing points other 
 than the central vanishing point. These are marked V.Pi and V.Po. 
 
 It will be noticed that as there are no lines in the object perpendicular to the 
 Picture Plane, the C.V. does not come into use as a vanishing point in this case. 
 
PERSPECTIVE PROJECTION, CONTINUED 121 
 
 The nearest corner of the cottage is at a certain distance into the picture, 
 i.e., beyond the Picture Plane, and also at a distance to the left of the spectator, 
 that is, to the left of the central line of vision, or Line of Sight. 
 
 The plan of the cottage is arranged so that the length way of it is at an 
 angle of 40° to the Picture Plane. At a convenient place to the left, on the Ground 
 Line as base, is erected an end elevation, from which levels for the vertical 
 traces of lines, such as that for the roof ridge, may be found, from which to draw 
 vanishing lines to the vanishing points. These vanishing lines on the Picture 
 Plane, as was seen above, are intersections of the Picture Plane by oblique planes, 
 each containing the eye and some line, such as the ridge line of the roof, and there- 
 fore containing the projection of it, by radial projection from the said line to 
 the eye. In Fig. 114, the drawing has been made as economically as possible, 
 in order to save confusion of lines. It will be readily seen that by continuing 
 in the same manner, windows and door ways, etc., might be added. 
 
 The student will doubtless realize that, if it were necessary to find vanishing 
 points for oblique Hnes, these would be found in the vertical traces, on the Picture 
 Plane, of vertical planes through the eye, parallel to such oblique lines. For the 
 purposes of our present study, however, sufficient has been undertaken. 
 
 EXERCISE XLVI 
 
 1. Work out to a scale, such as |" to the ft., a plan and elevation for a structure such as that 
 given in the Fig. 1 14, where the sizes are 15 ft. by 26 ft., with roof plan ig ft. by 30 ft. The nearest 
 corner of the roof is 4 ft. into the picture and 3 ft, to the left of the spectator. The distance 
 of the spectator is 40 ft., that is, 36 ft. from the picture plane. Suitable heights and details should 
 be chosen. Height of the eye 6 ft. 
 
 2. Numerous problems will readily suggest themselves, such as books lying unevenly upon 
 one another, pieces of furniture, buildings, etc. 
 
iNDi:x 
 
 Asymptotes. 70 
 Attitude. 2 
 Axes of projection. 71 
 Axis, of cone, 34 
 
 of cylinder. 32 
 
 of rotation, 11 1 
 Axometric projection, 71 
 
 Centre of vision, (W., 114 
 Chords, of circles, 1 1 
 
 of the sphere, 53 
 Circles, projected, 11 
 
 as great circles of spheres, 04 
 as H.T.'s of right vertical cones, 83 
 as sections of right cones and cylin- 
 ders, 70 
 as sections of si)heres with spheres, 105 
 shadows of, 2^^ 
 ("omplement angle, 14 
 Compound angle, 27 
 Cone, projection of. 34 
 sections of, 7S 
 shadow of, 36 
 tangent planes to. S2, 8q 
 I races of. 70 
 Cones enveloping spheres, g6. 97 
 Cube, projected under ditlerenl conditions. 
 
 I'igs. 28, 20. 30, 45. 66, 73, Qi 
 Curved surfaces, traces of, 7q 
 Cylinders, projections of. ,^2 
 sections of. 77 
 shadows of. 37 
 tangent planes to. S5 
 traces of. 71) 
 
 Develo{)ments of surfaces. io-\ 103 
 
 Dihedral angles. 56 
 
 Distances from planes of projection, 4 
 
 Elevations, 2 
 
 Kllil)se, the foreshortened a|jpearance of a 
 circle, 1 1 
 the section of a cylinder or cone, 77 
 the trace of a cylinder or of a cone. 80 
 
 Eye level. 116 
 
 Foreshortened appearance. 5. 6 
 
 Generator lines. 7 
 Ground line. 1 14 
 
 Helicoid. 1 1 1 
 
 Helix. 1 1 1 
 
 Horizon line. 1 14 
 
 Horizontal diameter of inclined circle 
 lines on oblique plane. 45 
 plane of projection. H.I*.. 
 
 Hyperbola. 78 
 
 Hyperboloid of revolution. 1 1 1 
 
 Inclination, 5 
 
 of lines to oblique planes. 44 
 
 of lines to planes of projection. 7 
 
 of planes to each other, 60 
 
 Interpenetrations of solids, loi. 105. io6, laS 
 
 Intersection, of lines with planes. 18. 4q 
 
 of planes with each other, 48 
 
 Isometric projection, 71 
 
 l.ine. generator, of a cylinder or cone. 77. 7g. 01 
 of intersection. AT. 2 
 of intersection of oblique planes. 48 
 of intersection of surfaces of interpcne 
 
 trating solids. 105. 107 
 representing ray of light. 20, 35. 1 14 
 shadow of. :o 
 tangent to great cirrle of a sphere. o<) 
 
 l.M 
 
124 
 
 INDEX 
 
 Lines, as edges and diagonals of figures, 4, 5 I Projection plane, I 
 
 at inclinations to both planes of pro- Pyramids, projection of, 33, 65, 91 
 
 jection, 27 j sections of, 76 
 
 inclinations, attitudes, and true lengths shadows of, 35, 3S 
 
 of, 6, 7 
 inclined to oblique planes, 44 
 inclined to one another and projected, Rabattement, of great circle of a sphere, 94 
 
 43, 62 
 in planes, 45 
 
 perpendicular to planes, 46 
 projection lengths of, 25 
 
 Notation adopted for naming points and their 
 projections, etc., 4 
 
 Oblique planes, found, angles given, 28, q6 
 angles contained by, 57 
 rabatted, 41, 42 
 represented by lines, i, 14 
 
 Octahedron, 66 
 
 Orthographical or perpendicular projection, i, 
 114 
 
 Parabola, 78 
 
 Parallel perspective, 118 
 
 planes, 51 
 Penetration, see Interpenet rations. 
 Perspective projection, 114 
 Picture plane, 114 
 Pitch, 113 
 
 IMane figures, projection forms of, q, 10 
 projection of, 20 
 shadows of, 22 
 Planes, oblic}ue, 13 
 
 angles contained by, 56 
 intersection of, 48 
 
 parallel, 51 
 
 rabattement of, 40 
 Plans, 2 
 Points, at distances from planes of projection, 4 
 
 on oblique planes, 45 
 
 on planes of projection, 17 
 
 on surface of sphere, 54, 03 
 
 planes passing through, 53 
 
 tangent, on spheres by planes, gS 
 Prisms, 32, 88 
 
 development of surfaces of, 102 
 
 interpenetration of, 101 
 
 of planes, 40 
 
 of right angles in axometric 
 projection, 74 
 
 of sections of solids, 76 
 
 to determine angles between 
 lines, 42 
 
 to determine dihedral angles, 56 
 Radial projection, 114 
 Rays of light, 20, 35, 114 
 Revolution, surfaces of, iii 
 
 Scale, 73 
 
 Screw-threads, 112 
 Secondary elevation, of a figure, 11 
 of solids, 107 
 Secondary XY line, 11, 107 
 Sections of solids, with curved surfaces, 77, 78 
 
 with plane surfaces, 76 
 Shade, 36 
 
 Shadow of sphere, 81 
 Shadows, of figures, 22 
 of lines, 20 
 of solids, 35 
 Skeleton cube, 71 
 Solids, in perspective, 114, 120 
 
 projection of, 31, 88 
 Sphere, found, when points on its- surface are 
 given, 53 
 great circle of, 94 
 interpenetrating with other spheres 
 
 or with cones, 105 
 shadow of, 81 
 tangent planes to, 93, 97 
 to find points on surface of, 93 
 with enveloping cones, 96, 98 
 Spiral, 1 1 1 
 
 Supplement angle, 56 
 Surfaces of revolution, iii 
 
 Tangents at inlinity to the hyperbola, 78 
 Tangent planes, to cones, 82 
 
 to cylinders, 8^ 
 
INDEX 
 
 V2'i 
 
 Tangent planes, to spheres, 03 
 Tetrahedron, 66 
 
 height of, 67 
 Traces, of curved surfaces, 79 
 of lines, iS 
 of planes, i,^ 
 True forms, of figures by rabattement. 47 
 
 of sections of solids. 76 
 True inclinations, of lines with each other by XY line, 2 
 rabattement of plane containing them. 42 A'sKj. or secondary AT line, 11, 107 
 
 ! True inclinations, of lines to planes. 44 
 True lengths of lines, 7 
 
 Vertical plane of projection, V.P.. i 
 Vertical transparent picture plane in per- 
 sfK'Ctive projection. 115 
 
14 DAY USE 
 
 KBTORN TO DBSK BROM WH.CH BORKOWBO 
 
 I OAN OEPT. 
 
 Renewed booUs are s,-— --d—»"- 
 
 LD 21-100m-6,'56 
 (B9311sl0)476 
 
 General Library . 
 Universiry of California 
 Berkeley 
 
YD 05053 
 
 -f\' 
 
 UNIVERSITY OF CALIFORNIA UBRARY