The D. Van Noftrand Company
intend this book to be sold to the Public at the advertised price, and supply it to the Trade on terms which will not allow of reduction.
I THE THEORY
OF
ENGINEERING DEAWING
BY
ALPHONSE A. ADLER, B.S., M.E.
Member American Society of Mechanical Engineers; Instructor in Mechanical Drawing and Designing, Polytechnic Institute, Brooklyn, N. Y.
NEW YORK
D. VAN NOSTRAND COMPANY
25 PAEK PLACE
1912
Copyright, 1912
BY
D. VAN NOSTRAND COMPANY
THE SCIENTIFIC PRESS DNUMMOND AND
BROOKLYN, N. Y,
PREFACE
ALTHOUGH the subject matter of this volume is, in large measure, identical with that of many treatises on descriptive geometry, the author has called it " Theory of Engineering Drawing," believing that this title indicates better than could any other, the ultimate purpose of the book. That texts on descriptive geometry appear with some degree of frequency, with but few, if any, additions to the theory, indicates that teachers are aware of certain weaknesses in existing methods of presenting the subject. It is precisely these weakness that the present work aims to correct.
The author emphasizes the fact that the student is concerned with the representation on a plane of objects in space of three dimensions. The analysis, important as it is, has for its primary purpose the development of methods for such representation and the interpretation of the resulting drawings. It is nowhere regarded as an end in itself. The number of fundamental prin- ciples has been reduced to a minimum; indeed it will be found that the entire text is based on the problem of finding the piercing point of a given line on a given surface, and a few additional operations. The accepted method of presenting the subject, is to start with a set of definitions, to consider in detail the ortho- graphic projection of a point, and then, on the foundation thus laid, to build the theory of the projection of lines, surfaces, and solids. Logical and beautiful as this systematic developmeen may be, it nevertheless presents certain inherent difficulties, chief of which is that the student is confronted at the outset with that most abstract of all abstractions, the mathematical point. In this volume the order of presentation is reversed and the reader is asked to consider first some concrete object, a box, for instance, the study of which furnishes material of use in the later discussion of its bounding surfaces and lines.
The " Theory of Engineering Drawing " is divided into four
iii
^59666
iv PREFACE
parts. Part I treats of oblique projection, orthographic projec- tion, and a special case of the latter, axonometric projection. The student is advised to give special attention to the classification at the end of this section, because it gives a complete outline of the entire subject. Part II contains a variety of problems of such nature as to be easily understood by those whose training has not extended to the more highly specialized branches of com- mercial or engineering practice. Part III considers convergent projective line drawing, more familiar under the name of perspec- tive. Part IV has to do with the pictorial effects of illumination, since a knowledge of shades and shadows is frequently required in the preparation of complicated drawings.
No claim is made to originality of subject matter, but it is not possible to acknowledge indebtedness to individual writers, for the topics discussed have been widely studied, and an historical review is here out of place. The author wishes, however, to express his sense of obligation to Professor William J. Berry of the Depart- ment of Mathematics in the Polytechnic Institute of Brooklyn for his criticism of Chapters IX and X, and other assistance, and to Mr. Ernest J. Streubel, M.A., of the Department of English for his untiring efforts in preparing the manuscript for the press.
POLYTECHNIC INSTITUTE OF BROOKLYN, October, 1912.
CONTENTS
PART I
THE PRINCIPLES OF PARALLEL PROJECT ING-LINE
DRAWING
CHAPTER I INTRODUCTORY
ART. PAGE
101. Nature of drawing 3
102. Science and art of drawing 4
103. Magnitude of objects 4
104. Commercial application of drawing 4
CHAPTER II OBLIQUE PROJECTION
201. Nature of oblique projection 6
202. Oblique projection of lines parallel to the plane of projection 7
203. Oblique projection considered as a shadow 8
204. Oblique projection of lines perpendicular to the plane of projection . 9
205. Oblique projection of the combination of parallel and perpendicular
lines to the plane of projection 10
206. Oblique projection of circles 11
207. Oblique projection of inclined lines and angles 12
208. Representation of visible and invisible lines 13
209. Drawings to scale 14
210. Examples of oblique projection 14
211. Distortion of oblique projection 20
212. Commercial application of oblique projection 21
CHAPTER III ORTHOGRAPHl'c PROJECTION
301. Nature of orthographic projection 25
302. Theory of orthographic projection 26
303. Revolution of the horizontal plane 27
v
vi CONTENTS
ART. PAGE
304. Position of the eye 27
305. Relation of size of object to size of projection 28
306. Location of object with respect to the planes of projection 28
307. Location of projections with respect to each other 29
308. Dimensions on a projection 29
309. Comparison between oblique and orthographic projection 29
310. Orthographic projection considered as a shadow 30
311. Profile plane 30
312. Location of profiles 31
313. Section plane 33
314. Supplementary plane 34
315. Angles of projection 36
316. Location of observer in constructing projections 36
317. Application of angles of projections to drawing 37
318. Commercial application of orthographic projection 38
CHAPTER IV AXONOMETRIC PROJECTION
401. Nature of isometric projections 45
402. Theory of isometric projection 46
403. Isometric projection and isometric drawing 47
404. Direction of axes 48
405. Isometric projection of circles 48
406. Isometric projection of inclined lines and angles 49
407. Isometric graduation of a circle 49
408. Examples of isometric drawing 51
409. Dimetric projection and dime trie drawing 54
410. Trimetric projection and trimetric drawing 55
411. Axonometric projection and axonometric drawing 56
412. Commercial application of axonometric projection 56
413. Classification of projections 57
PART II
GEOMETRICAL PROBLEMS IN ORTHOGRAPHIC* PROJECTION
CHAPTER V REPRESENTATION OF LINES AND POINTS
501. Introductory 61
502. Representation of the line 62
503. Line fixed in space by its projections 64
504. Orthographic representation of a line 65
505. Transfer of diagrams from orthographic to oblique projection, 66
CONTENTS vii
ART. PAGE
506. Piercing points of lines on the principal planes 66
507. Nomenclature of projections 68
508. Representation of points 68
509. Points lying in the principal planes 69
510. Mechanical representation of the principal planes 69
511. Lines parallel to the planes of projection 70
512. Lines lying in the planes of projection 71
513. Lines perpendicular to the planes of projection 72
514. Lines in all angles 72
515. Lines with coincident projections 74
516. Points in all angles 75
517. Points with coincident projections 75
518. Lines in profile planes 75
CHAPTER VI REPRESENTATION OF PLANES
601. Traces of planes parallel to the principal planes 80
602. Traces of planes parallel to the ground line , 80
603. Traces of planes perpendicular to one of the principal planes 82
604. Traces of planes perpendicular to both principal planes 83
605. Traces of planes inclined to both principal planes 83
606. Traces of planes intersecting the ground line 84
607. Plane fixed in space by its traces 84
608. Transfer of diagrams from orthographic to oblique projection 84
609. Traces of planes in all angles 86
610. Projecting plane of lines 86
CHAPTER VII ELEMENTARY CONSIDERATIONS OF LINES AND PLANES
701. Projection of lines parallel in space 89
702. Projection of lines intersecting in space 89
703. Projection of lines not intersecting in space 90
704. Projection of lines in oblique planes 91
705. Projection of lines parallel to the principal planes and lying in an
oblique plane 92
706. Projection of lines perpendicular to given planes 95
707. Revolution of a point about a line 96
CHAPTER VIII
PROBLEMS INVOLVING THE POINT, THE LINE, AND THE
PLANE
801. Introductory 99
802. Solution of problems 99
803. Problem 1. To draw a line through a given point parallel to a given
line. . . 100
viii CONTENTS
804. Problem 2. To draw a line intersecting a given line at a giver^ point .. 100
805. Problem 3. To find where a given line pierces the principal
planes ................................................ 101
806. Problem 4. To pass an oblique plane through a given oblique line. . 102
807. Special cases of the preceding problem ......................... 102
808. Problem 5. To pass an oblique plane through a given point ....... 103
809. Problem 6. To find the intersection of two planes, oblique to each
other and to the principal planes ........................... 104
810. Special case of the preceding problem ........................... 104
811. Problem 7. To find the corresponding projection of a given point
lying in a given oblique plane, when one of its projections is given ................................................... 104
812. Special case of the preceding problem .......................... 105
813. Problem 8. To draw a plane which contains a given point and is
parallel to a given plane .................................. 106
814. Problem 9. To draw a line perpendicular to a given plane through
a given point ............................................ 107
815. Special case of the preceding problem .......................... 108
816. Problem 10. To draw a plane through a given point perpendicular to
a given line ............................................. 108
817. Problem 11. To pass a plane through three given points not in the
same straight line ........................................ 109
818. Problem 12. To revolve a given point, not in the principal planes,
about a line lying in one of the principal planes ................ 110
819. Problem 13. To find the true distance between two points in space
as given by their projections. First method. Case 1 ........ Ill
820. Case 2 ............................... ..................... 112
821. Problem 13. To find the true distance between two points in spdce
as given by their projections. Second method. Case 1 . .. 113
822. Case 2 ..................................................... 113
823. Problem 14. To find where a given line pierces a given plane ...... 114
824. Problem 15. To find the distance of a given point from a given
plane ................................................... 115
825. Problem 16. To find the distance from a given point to a given line. ... 115
826. Problem 17. To find the angle between two given intersecting
lines ................................................... 116
827. Problem 18. To find the angle between two given planes ......... 117
828. Problem 19. To find the angle between a given plane and one of
the principal planes ...................................... 118
829. Problem 20. To draw a plane parallel to a given plane at a given
distance from it ......................................... 119
830. Problem 21. To project a given line on a given plane ............ 120
831. Problem 22. To find the angle between a given line and a given
plane .................................................. 121
832. Problem 23. To find the shortest distance between a pair of skew
lines ................................................... 122
833. Application to other problems ................................. 125
CONTENTS ix
i
ART. PAGE
834. Problem 24. Through a given point, draw a line of a given length,
making given angles with the planes of projection 125
835. Problem 25. Through a given point, draw a plane, making given
angles writh the principal planes 127
836. Problem 26. Through a given line, in a given plane, draw another
line, intersecting it at a given angle 129
837. Problem 27. Through a given line, in a given plane, pass another
plane making a given angle with the given plane 130
838. Problem 28. To construct the projections of a circle lying in a given
oblique plane, of a given diameter, its centre in the plane being known 131
CHAPTER IX
*
CLASSIFICATION OF LINES
901. Introductory 144
902. Straight line 144
903. Singly curved line 144
904. Representation of straight and singly curved lines 144
905. Circle 145
906. Ellipse 146
907. Parabola 147
908. Hyperbola 147
909. Cycloid 148
910. Epicycloid 149
911. Hypocycloid 150
912. Spiral 151
913. Doubly curved line 151
914. Representation of doubly curved lines 151
915. Helix 152
916. Classification of lines 154
917. Tangent 154
918. Construction of a tangent 154
919. To find the point of tangency 155
920. Direction of a curve 156
921. Angle between curves 156
922. Intersection of lines 156
923. Order of contact of tangents 157
924. Osculating circle 158
925. Osculating plane 158
926. Point of inflexion. Inflexional tangent 159
927. Normal 159
928. Rectification 159
929. Involute and Evolute 160
930. Involute of the circle. . . 161
CONTENTS
CHAPTER X CLASSIFICATION OF SURFACES
ART. PAGE
1001. Introductory 165
1002. Plane surface 165
1003. Conical surface 166
1004. Cone 166
1005. Representation of the cone 167
1006. To assume an element on the surface of a cone 168
1007. To assume a point on the surface of a cone 168
1008. Cylindrical surface 169
1009. Cylinder 169
1010V Representation of the cylinder 170
1011. To assume an element on the surface of a cylinder 170
1012. To assume a point on the surface of a cylinder 171
1013. Convolute surface 171
1014. Oblique helicoidal screw surface 173
1015. Right helicoidal screw surface 174
1016. Warped surface 174
1017. Tangent plane 175
1018. Normal plane 175
1019. Singly curved surface 176
1020. Doubly curved surface 176
1021. Singly curved surface of revolution 176
1022. Doubly curved surface of revolution 176
1023. Revolution of a skew line 177
1024. Meridian plane and meridian line 177
1025. Surfaces of revolution having a common axis 177
1026. Representation of the doubly curved surface of revolution 178
1027. To assume a point on a doubly curved surface of revolution 178
1028. Developable surface 179
1029. Ruled surface 179
1030. Asymptotic surface 179
1031. Classification of surfaces , 180
CHAPTER XI
INTERSECTIONS OF SURFACES BY PLANES, AND THEIR DEVELOPMENT
1101. Introductory 184
1102. Lines of intersection of solids by planes 185
1103. Development of surfaces 185
1104. Developable surfaces 185
1105. Problem 1. To find the line of intersection of the surfaces of a
right octagonal prism with a plane inclined to its axis 186
CONTENTS xi
j
ART. PAGE
1106. Problem 2. To find the developed surfaces in the preceding prob-
lem 187
1107. Problem 3. To find the line of intersection of the surface of a right
circular cylinder with a plane inclined to its axis 188
1108. Problem 4. To find the developed surface in the preceding problem . 189
1109. Application of cylindrical surfaces 189
1110. Problem 5. To find the line of intersection of the surfaces of a
right octagonal pyramid with a plane inclined 'to its axis. . . 190
1111. Problem 6. To find the developed surf aces in the preceding problem 191
1112. Problem 7. To find the line of intersection of the surface of a right
circular cone with a plane inclined to its axis 192
11 13. Problem 8. To find the developed surface in the preceding problem . 193
1114. Application of conical surfaces 194
1115. Problem 9. To find the line of intersection of a doubly curved
surface of revolution with a plane inclined to its axis 194
1116. Problem 10. To find the line of intersection of a bell-surface with a
plane 195
1117. Development by triangulation 196
1118. Problem 11. To develop the surfaces of an oblique hexagonal
pyramid 196
1119. Problem 12. To develop the surface of an oblique cone 197
1120. Problem 13. To develop the surface of an oblique cylinder 198
1121. Transition pieces 199
1122. Problem 14. To develop the surface of a transition piece connect-
ing a circular opening with a square opening 200
1123. Problem 15. To develop the surface of a transition piece connect-
ing two elliptical openings whose major axes are at right angles
to each other 201
1124. Development of doubly curved surfaces by approximation 202
1125. Problem 16. To develop the surface of a sphere by the gore
method 203
1126. Problem 17. To develop the surface of a sphere by the zone
method 204
1127. Problem 18. To develop a doubly curved surface of revolution by
the gore method 205
CHAPTER XII
INTERSECTIONS OF SURFACES WITH EACH OTHER AND THEIR DEVELOPMENT
1201. Introductory 210
1202. Problem 1. To find the line of intersection of the surfaces of two
prisms 211
1203. Problem 2. To find the developments in the preceding problem. . -. 211
1204. Problem 3. To find the line of intersection of two cylindrical sur-
faces of revolution whose axes intersect at a right angle .... 212
xii CONTENTS
ART. PAGE
1205. Problem 4. To find the developments in the preceding problem . . . . 213
1206. Problem 5. To find the line of intersection of two cylindrical
surfaces of revolution whose axes intersect at any angle. . 213
1207. Problem 6. To find the developments in the preceding problem. . . 214
1208. Application of intersecting cylindrical surfaces to pipes 214
1209. Problem 7. To find the line of intersection of two cylindrical sur-
faces whose axes do not intersect 215
1210. Problem 8. To find the developments in the preceding problem. . . 215
1211. Intersection of conical surfaces 216
1212. Problem 9. To find the line of intersection of the surfaces of two
cones whose bases may be made to lie in the same plane, and whose altitudes differ 216
1213. Problem 10. To find the line of intersection of the surfaces of two
cones whose bases may be made to lie in the same plane, and whose altitudes are equal 218
1214. Problem 11. To find the line of intersection of the surfaces of two
cones whose bases lie in different planes 219
1215. Types of lines of intersection for surfaces of cones 221
1216. Problem 12. To find the line of intersection of the surfaces of a
cone and a cylinder of revolution when their axes intersect at
a right angle 222
1217. Problem 13. To find the line of intersection of the surfaces of a cone
and a cylinder of revolution when their axes intersect at any angle 222
1218. Problem 14. To find the line of intersection of the surfaces of an
oblique cone and a right cylinder 223
1219. Problem 15. To find the developments in the preceding problem . 224
1220. Problem 16. To find the line of intersection of the surfaces of an
oblique cone and a sphere 224
1221. Problem 17. To find the line of intersection of the surfaces of a
cylinder and a sphere 225
1222. Problem 18. To find the line of intersection of two doubly curved
surfaces of revolution whose axes intersect 226
1223. Commercial application of methods 226
CONTENTS xiii
PART III
THE PRINCIPLES OF CONVERGENT PROJECTING-LINE
DRAWING
CHAPTER XIII PERSPECTIVE PROJECTION
ART. PAGE
1301. Introductory 235
1302. Scenographic projection 235
1303. Linear perspective 236
1304. Visual rays and visual angle 236
1305. Vanishing point 237
1306. Theory of perspective projection 237
1307. Aerial perspective 237
1308. Location of picture plane 237
1309. Perspective of a line 238
1310. Perspectives of lines perpendicular to the horizontal plane 239
1311. Perspectives of lines parallel to both principal planes 239
1312. Perspectives of lines perpendicular to the picture plane 240
1313. Perspectives of parallel lines, inclined to the picture plane 240
1314. Horizon 241
1315. Perspective of a point 242
1316. Indefinite perspective of a line 242
1317. Problem 1. To find the perspective of a cube by means of the
piercing points of the visual rays on the picture plane 244
1318. Perspectives of intersecting lines 245
1319. Perpendicular and diagonal 245
1320. To find the perspective of a point by the method of perpendiculars
and diagonals 246
1321. To find the perspective of a line by the method of perpendiculars
and diagonals 248
1322. Revolution of the horizontal plane 249
1323. To find the perspective of a point when the horizontal plane is
revolved 249
1324. To find the perspective of a line when the horizontal plane is
revolved 250
1325. Location of diagonal vanishing points 251
1326. Problem 2. To find the perspective of a cube by the method of
perpendiculars and diagonals 251
1327. Problem 3. To find the perspective of a hexagonal prism 253
1328. Problem 4. To find the perspective of a pyramid superimposed on
a square base 254
1329. Problem 5. To find the perspective of an arch 254
1330. Problem 6. To find the perspective of a building 256
1331. Commercial application of perspective 258
1332. Classification of projections . 260
xiv CONTENTS
PART IV
PICTORIAL EFFECTS OF ILLUMINATION CHAPTER XIV
PICTORIAL EFFECTS OF ILLUMONATION IN ORTHO- GRAPHIC PROJECTION
ART. PAGE
1401. Introductory 265
1402. Line shading applied to straight lines 265
1403. Line shading applied to curved lines 263
1404. Line shading applied to sections 267
1405. Line shading applied to convex surfaces 267
1406. Line shading applied to concave surfaces 268
1407. Line shading applied to plane surfaces 268
1408. Physiological effect of light 268
1409. Conventional direction of light rays 269
1410. Shade and shadow 269
1411. Umbra and penumbra 269
1412. Application of the physical principles of light to drawing 270
1413. Shadows of lines 270
1414. Problem 1. To find the shadow cast by a cube which rests on a
plane 271
1415. Problem 2. To find the shadow cast by a pyramid, in the principal
planes 272
1416. Problem 3. To find the shade and shadow cast by an octagonal
prism having a superimposed octagonal cap 273
1417. Problem 4. To find the shade and shadow cast by a superimposed
circular cap on a cylinder 275
1418. High-light 275
1419. Incident, and reflected rays 276
1420. Problem 5. To find the high light on a sphere 276
1421. Multiple high lights 277
1422. High lights on cylindrical or conical surfaces 277
1423. Aerial effect of illumination 277
1424. Graduation of shade -. 278
1425. Shading rules 278
1426. Examples of graduated shades 279
CHAPTER XV
PICTORIAL EFFECTS OF ILLUMINATION IN PERSPECTIVE PROJECTION
1501. Introductory 282
1502. Problem 1. To draw the perspective of a rectangular prism and its
shadow on the horizontal plane 282
1503. General method of finding the perspective of a shadow 284
CONTENTS xv
{
ABT. PAGE
1504. Perspectives of parallel rays of light 284
1505. Perspective of the intersection of the visual plane on the plane
receiving the shadow 285
1506. Application of the general method of finding the perspective of a
shadow 285
1507. Problem 2. To draw the perspective of an obelisk with its shade
and shadow 286
1508. Commercial application of the pictorial effects of illumination in
perspective 289
PART I
PART I
PRINCIPLES OF PARALLEL PROJECTING-LINE
DRAWING
CHAPTER I
INTRODUCTORY
101. Nature of drawing. Drawing has for its purpose the exact graphic representation of objects in - space. The first essential is to have an idea, and then a desire to express it. Ideas may be expressed in words, in pictures, or in a combination of both words and pictures. If words alone are sufficient to express the idea, then language becomes the vehicle of its transmission. When the idea relates to some material object, however, a drawing alone, without additional information may satisfy its accurate conveyance. Further, some special cases require for their expres- sion a combination of both language and drawing.
Consider, for purposes of illustration, a maple block, 2 inches thick, 4 inches wide and 12 inches long. It is easy to conceive this block of wood, and the mere statement, alone, specifies the object more or less completely. On the other hand, the modern news- paper printing press can not be completely described by language alone. Anyone who has ever seen such a press in operation, would soon realize that the intricate mechanism could not be described in words, so as to make it intelligible to another without the use of a drawing. Even if a drawing is employed in this latter case, the desired idea may not be adequately presented, since a circular shaft is drawn in exactly the same way, whether it be made of wood, brass, or steel. Appended notes, in such cases, inform the constructor of the material to use. From the foregoing, it is evident, that drawing cannot become a uni- versal language in engineering, unless the appended descriptions and specifications have the same meaning to all.
4 PARALLEL PROJECTING-LINE DRAWING
102. Science and art of drawing. Drawing is both a science and an art. The science affects such matters as the proper arrangement of views and the manner of their presentation. Those who are familiar with the mode of representation used, will obtain the idea the maker desired to express. It is a science, because the facts can be assimilated, classified, and presented in a more or less logical order. In this book, the science of drawing will engage most of the attention; only such of the artistic side is included as adds to the ease of the interpretation of the drawing.
The art lies in the skilful application of the scientific prin- ciples involved to a definite purpose. It embraces such topics as the thickness or weight of lines, whether the outline alone is to be drawn, or whether the object is to be colored and shaded so as to give it the same appearance that it has in nature.
103. Magnitude of objects. Objects visible to the eye are, of necessity, solids, and therefore require the three principal dimensions to indicate their magnitude — length, breadth, and thickness. If the observer places himself in the proper position while viewing an object before him, the object impresses itself on him as a whole, and a mental estimate is made from the one position of the observer as to its form and magnitude. Naturally, the first task will be to represent an object in a single view, show- ing it in three dimensions, as a solid.
104. Commercial application of drawing. It must be remembered that the function of drawing is graphically to present an idea on a flat surface — like a sheet of paper for instance — so as to take the place of the object in space. The reader's imagination supplies such deficiency as is caused by the absence of the actual object. It is, therefore, necessary to study the various underlying principles of drawing, and, then, apply them as daily experience dictates to be the most direct and accurate way of their presentation. In any case, only one interpretation of a drawing should be possible, and if there is a possibility of ambiguity arising, then a note should be made on the drawing calling attention to the desired interpretation.
INTRODUCTORY
QUESTIONS ON CHAPTER I
1. In what ways may ideas be transmitted to others?
2. What topics are embraced in the science of drawing?
3. What topics are included in the art of drawing?
4. How many principal dimensions are required to express the magni-
tude of objects? What are they?
5. What is the function of drawing?
6. Is the reader's imagination called upon when interpreting a drawing?
Why?
CHAPTER II OBLIQUE PROJECTION
201. Nature of oblique projection. Suppose it is desired to draw a box, 6" wide, 12" long and 4" high, made of wood \" thick. Fig. 1 shows this drawn in oblique projection. The method of making the drawing will first be shown and then the theory on which it is based will be developed. A rectangle abed, 4"X6", is laid out, the 6" side being horizontal and the 4" side' being vertical. From three corners of the rectangle,
FIG. 1.
lines ae, bf, and eg are drawn, making, in this case, an angle of 30° with the horizontal. The length 12" is laid off on an inclined line, as eg. The extreme limiting lines of the box are then fixed by the addition of two lines ef (horizontal) and fg (vertical). The thickness of the wood is represented, and the dimensions showing that it is \" thick indicate the direction in which they are laid off. The reason for the presence of such other additional lines, is that they show the actual construction.
The sloping lines in Fig. 1 could be drawn at any angle other
6
OBLIQUE PROJECTION
than 30°. In Fig. 2, the same box is drawn with a 60° inclina-
tion. It will be seen, in this latter case, that the inside bottom
of the box is also shown
prominently. It is customary
in the application of this type
of drawing, to use either 30°,
45° or 60° for the slope, as these
lines can be easily drawn with
the standard triangles used in >
the drafting room.
202. Oblique projection of lines parallel to the plane of projection. In developing the theory, let XX and YY, Fig. 3, be two planes at right angles to each other. Also, let ABCD be a thin rectangular plate, the plane of FIG. 2.
which is parallel to the plane XX.
Suppose the eye is looking in the direction Aa, inclined* to the plane XX. Where this line of sight from the point A on the
FIG. 3.
object appears to pierce or impinge on the plane XX, locate the point a. From the point B, assume that the eye is again directed
* The ray must not be perpendicular, as this makes it an orthographic projection. The ray cannot be parallel to the plane of projection, because it will never meet it, and, hence, cannot result in a projection.
8 PARALLEL PROJECTING-LINE DRAWING
toward the plane XX, in a line that is parallel to Aa; this second piercing point for the point B in space will appear at b. Similarly, from the points C and D on the object, the piercing points ron [the plane will be c and d, Cc and Dd being parallel to Aa.
On the plane XX, join the points abed. To an observer, the figure abed will give the same mental impression as will the object ABCD. In other words, abed is a drawing of the thin plate ABCD. ABCD is the object in space; abed is the corresponding oblique projection of ABCD. The plane XX is the plane of projection; Aa, Bb, Cc, and Dd, are the projecting lines, making any angle with the plane of projection other than at right angles or parallel thereto. The plane YY serves the purpose of throwing the plane XX into stronger relief and has nothing to do with the projection.
It will be observed that the figure whose corners are the points ABCDabcd is an oblique rectangular prism, the opposite faces of which are parallel because the edges have been made parallel by construction. From the geometry, all parallel plane sections of the prism are equal, hence abed is equal to ABCD, because the plane of the object ABCD was originally assumed parallel to XX, and the projection abed lies in the plane XX. As a corollary, the distance of the object from the plane of pro- jection does not influence the size of the projection, so long as the plane of the object is continually parallel to the plane of projection.
Indeed, any line, whether straight or curved, when parallel to the plane of projection has its projection equal to the line itself. This is so because the curved line may be considered as made up of an infinite number of very short straight lines.
203. Oblique projection considered as a shadow. Another way of looking at the projection shown in Fig. 3 is to assume that light comes in parallel lines, oblique to the plane of pro- jection. If the object is interposed in these parallel rays, then abed is the shadow of A^CD in space, and thus presents an entirely different standpoint from which to consider the nature of a projection. Both give identical results, and the latter is here introduced merely to reenforce the understanding of the nature of the operation.
OBLIQUE PROJECTION
9
204. Oblique projection of lines perpendicular to the plane of projection. Let XX, Fig. 4, be a transparent plane surface, seen edgewise, and ab, an arrow perpendicular to XX, the end a of the arrow lying in the plane. Suppose the eye is located at r so that the ray of light rb makes an angle of 45° with the plane XX. If all rays of light from points on ab are
x FIG. 4.
FIG. 5.
parallel to rb, they will pierce the plane XX in a series of points, and ac, then, will become the projection of ab on the plane XX. To an observer standing in the proper position, looking along lines parallel to rb, ac will give the same mental impression as the actual arrow ab in space; and, therefore, ac is the projection of ab on the plane XX. The extremities of the line ab are hence projected as two distinct points a and c. Any intermediate point, as d, will be projected as e on the projection ac.
From the geometry, it may be noticed that ac is equal to ab, because cab is a right angled triangle, and the angle acb equals the angle cba, due to the adoption of the 45° ray. Also, any limited portion of a line, perpendicular to the plane of projection, is projected as a line equal to it in length. The triangle cab may be rotated about ab as an axis, so that c describes a circle in the plane XX and thus ac will always remain equal to ab. This means that the rotation merely corresponds to a new position of the eye, the inclination of the ray always remaining 45° with the plane XX.
10
PARALLEL PROJECTING-LINE DRAWING
The foregoing method of representing 45° rays is again shown as an oblique projection in Fig. 5. Two positions of tiie ray are indicated as rb and sb; the corresponding projections are ac and ad. Hence, in constructing oblique projections, the lines that are parallel to the plane of projection are drawn with their true relation to each other. The lines that are perpendicular to the plane of projection are drawn as an inclined line of a length 'equal to tke line itself and making any angle with the horizontal,
FIG. 6.
at pleasure. Here, again, the plane YY is added. The line of intersection of XX and YY is perpendicular to the plane of the paper, and is shown as a sloping line, because the two planes themselves are pictured in oblique projection.*
205. Oblique projection of the combination of parallel and perpendicular lines to the plane of projection. Fig. 6 shows a box and its projection, pictorially indicating all the mental steps required in the construction of an oblique pro- jection. The object (a box) A is shown as an oblique projection;
* Compare this with Figs. 1 and 2. The front face of the box is shown as it actually appears, because it is parallel to the plane of projection (or paper). The length of the box is perpendicular to the plane of the paper and is projected as a sloping line.
OBLIQUE PROJECTION
11
its projection B on the plane of projection appears very much distorted. This distortion of the projection is due to its being an oblique projection, initially, which is then again shown in oblique projection. From what precedes, the reader should find no difficulty in tracing out the construction. Attention may again be called to the fact that the extremities of the lines per- pendicular to the plane of projection are projected as two distinct points.
206. Oblique projection of circles.* When the plane of a circle is parallel to the plane of projection, it is drawn with a compass in the ordinary way, because the projection is equal to the circle itself (202). When the plane of the circle is per- pendicular to the plane of projection, however, it is shown as an ellipse. Both cases will be illustrated by Fig. 7, which shows a cube in oblique projection. In the face abed, the circle is shown as such, because the plane of the circle is par- allel to the plane of pro- jection, which, in this case, is the plane of the paper. It will be noticed that the circle FlG- 7-
in abed is tangent at points
midway between the extremities of the lines. If similar points of tangency are laid off in the faces aefb and fbcg and a smooth curve be drawn through these points, the result will be an ellipse; this ellipse is, therefore, the oblique projection of a circle, whose plane is perpendicular to the plane of projection. The additional lines in Fig. 7 show how four additional points may be located on the required ellipse.
It may be shown that a circle is projected as an ellipse in all cases except when its plane is parallel to the plane of projection, or, when its plane is chosen parallel to the projecting lines. In the latter case, it is a line of a length equal to the diameter of the
*When projecting circles in perpendicular planes, the 30° slope offers an advantage because the ellipse is easily approximated. See Art. 405.
PARALLEL PROJECTING-LINE DRAWING
circle. The reason for this will become evident later in the- subject. (It is of insufficient import at present to dwell on it at length.)
207. Oblique projection of inclined lines and angles.
At times, lines must be drawn that are neither parallel nor perpendicular to the plane of projection. A reference to Fig. 8 will show how this is done. It is desired to locate a hole in a cube whose edge measures 12". The hole is to be placed in the side bfgc, 8" back from the point c and then 4" up to the point h. To bring this about, lay off ck = 8" and kh (vertically)
= 4" and, then h is the 12" - required point. Also, hkc
is the oblique projection of a right angled tri- angle, whose plane is per- pendicular to the plane of projection. Suppose, further, it is desired to locate the point m on the face abfe, 1" to the right of the point a and 5" back. The dimensions show how this is done. Again, man is the ob- lique projection of aright angled triangle, whose
legs are 5" and 1". This method of laying off points is virtually a method of offsets.* The point m is offset a distance of 5" from ab; likewise h is offset 4" from eg. If it be required to lay off the diagonal of a cube, it is accomplished by making three offsets from a given point. For instance, consider the diagonal ce. If c is the starting point, draw eg perpendicular to thft plane (shown as an inclined line), then fg, vertically upward, and finally fe, horizontally to the left; therefore, ce is the diagonal of the cube, if eg = gf = fe.f
* This method is of importance in that branch of mathematics known as Vector Analysis. Vectors are best drawn in space by means of oblique projection.
t If two given lines are parallel in space, their oblique projections are parallel under any conditions. The projecting lines from the extremities of
OBLIQUE PROJECTION
13
A word may be said in reference to round holes appearing in the oblique faces of a cube. As has been shown, circles are here represented as ellipses (206), but if it were desired to cut an elliptical hole at either h or m, then their projections would not give a clear idea of the fact. Such cases, when they occur, must be covered by a note to that effect; an arrow from the note pointing to the hole would then indicate, unmistakably, that the hole is to be drilled (for a round hole), otherwise its shape should be called for in any way that is definite. It seems, therefore, that oblique projection cannot fulfill the needs of commercial drawing in every respect; and, indeed, this is true. Other methods also have certain advantages and will be treated subsequently 4
208. Representation of visible and invisible lines. While viewing an object, the observer finds that some lines on the object are visible. These lines are drawn in full on the projection. There are, however, other lines, invisible from the point of view chosen and these, when added, are shown dotted. Fig. 9 shows all the visible and invisible lines on a hollow circular cylinder. Dimensions are appended and the cylinder shaded so that no question should arise as to its identity. It can be observed that the drawing is clear in so far as it shows that the hole goes entirely through the cylin- der. Were the dotted lines FIG. 9. omitted, one could not tell
whether the hole went entirely through, or only part way through. Hence, dotted lines may add to the clearness of a drawing; in such cases they should be added. At times, how- ever, their addition may lead to confusion; and, then, only the
the given lines determine planes that cut the plane of projection in lines which are the projections of the given lines. The projecting planes from the given lines are parallel, and, hence, their projections are parallel, since it is the case of two parallel planes cut by a third plane.
t The student will obtain many suggestions by copying such simple illustrations as Figs. 1, 2, 7, 8 and 9.
14 PARALLEL PROJECTING-LINE DRAWING
more important dotted lines added, and such others, considered unnecessary, should be omitted. Practice varies in this latter respect and the judgment of the draftsman comes into play at this point; ability to interpret the drawing rapidly and accurately is the point at issue.
209. Drawings to scale. Objects of considerable size cannot be conveniently represented in their full size. The shape is maintained, however, by reducing the length of each a definite proportion of its original length, or, in other words, by drawing to scale. Thus, if the drawing is one-half the size of the object, the scale is 6" = 1 ft. and is so indicated on the drawing by a note to that effect. The scales in common use are 12" = 1 ft. or full size; 6" = 1 ft. or half-size; 3" = 1 ft. or quarter size; 1J"; 1"
f"; 4"; f"; i"; A"; i"; A"; A"; etc. = i ft. The smaller
sizes are used for very large work and vice versa. In railway work, scales like 100' = 1 inch, or 10000' = ! inch are common. In watch mechanism, scales like 48" = I ft. or " four times actual size " or even larger are used, since, otherwise, the drawings would be too small for the efficient use of the workman. Irre- spective of the scale used, the actual dimensions are put on the drawing and the scale is indicated on the drawing by a note to that effect.
If some dimensions are not laid out to the scale adopted, the drawing may create a wrong impression on the reader and this should be avoided if possible. When changes in dimen- sions occur after the completion of a drawing and it is im- practicable to make the change, the dimension may be underlined and marked conveniently near it N T S, meaning " not to scale."
210. Examples of oblique projection. Fig. 10 shows a square block with a hole in its centre. The dimension lines indicate the size and the method of making the drawing when the planes of the circles are chosen parallel to the plane of the paper (plane of projection). The circle in the visible face is drawn with a compass to the desired scale. The circle in the invisible face (invisible from the point of view chosen) is drawn to the same radius, but, its centre is laid off on an inclined line, a distance back of the visible circle, equal to the thickness of the block. The circle in the front face is evidently not in the same plane
OBLIQUE PROJECTION
15
as that in the distant face. The line joining their centres is thus perpendicular to the plane of projection, and is, hence, laid off as an inclined line.
FIG. 10.
FIG. 11.
»
When the plane of the circles is made perpendicular to the plane of projection, the circles are projected as ellipses. Fig. 11 shows how the block of Fig. 10 is drawn when such is the
16
PARALLEL PROJECTING-LINE DRAWING
case. It is to be observed, that the bounding square is to be drawn first and then the ellipse (projection of the circle) is inscribed. When the circles in both faces are to be shown, the bounding
FIG. 13.
square must be replaced by a bounding rectangular prism. This rectangular prism is easily laid out and the ellipses are inserted in the proper faces. The method of using bounding figures of
OBLIQUE PROJECTION
17
simple shape is of considerable importance when applying the foregoing principles to oblique projection.
FIG. 14.
FIG. 15.
Fig. 12 is another illustration of an object, differing from Figs. 10 and 11 in so far as the hole does not go entirely through
18
PARALLEL PROJECTING-LINE DRAWING
from face to face. The centres for the different circles are found on the axis. The distances between the centres, measured on the inclined line, is equal to the distances between the planes of the corresponding circles. Since the circles are drawn as such, the planes must, therefore, be parallel to the plane of projection. The axis of the hole is perpendicular to the plaite of projection, and, hence, is projected as an inclined line.
The object in Fig. 12 is also shown with the planes of the circles perpendicular to the plane of projection in Fig. 13. Every step of the construction is indicated in the figure and the series of bounding prisms about the cylinders is also shown.
FIG. 16.
A somewhat different example, showing the necessity of bounding figures, is given in Fig. 14. The bounding figure, shown in Fig. 15, is laid out as given and it becomes a simple matter to insert the object subsequently. It should be noted how the rectangular projection becomes tangent to the cylinder and that the only way to be certain of the accuracy of the drawing, is to use these bounding figures and make mental record of the relative location of the lines that make up the drawing.
If objects are to be drawn whose lines are inclined to each other, the principles so far developed offer simple methods for their presentation. Fig. 16 shows a tetrahedron with a bounding
OBLIQUE PROJECTION
19
rectangular prism. The apex is located on the top face and its position is determined from the geometric principles imposed. Solids, as represented in the text-books on geometry, are drawn •in this way. Some confusion may be avoided by observing that angles are only preserved in their true relation in the planes parallel to the plane of projection (207).
The concluding example of this series is given in Fig. 17. It is known as a bell-crank and has circles shown in two planes at right angles to each other, The example furnishes the clue
FIG. 17.
to constructing any object, however complicated it may be. Base lines ab and be are first laid out to the required dimensions and to the desired scale. In this example, the base lines are chosen parallel to the plane of projection, and hence are projected as a right angle, true to dimensions. The thickness of the two lower cylinders is laid off as an inclined line from each side of the base line and the circles are then drawn. The upper circles (shown as ellipses) may need some mention. A bounding rec- tangular prism is first drawn, half of which is laid off on each side of the base line (true in this case but it may vary in others).
20
PARALLEL PROJECTING-LINE DRAWING
The circles and inclined lines are filled in after the guiding details are correctly located. This drawing may present some difficulty at first, but a trial at its reproduction will reveal no new prin- ciples, only an extreme application.
On completion of the drawing, the bounding figure may be removed if its usefulness is at an end. When inclined lines appear frequently on the drawing, the bounding figures can be made to serve as dimension lines, and so help in the interpre- tation. The draftsman must determine what is best in each case, remembering, always, that the drawing must be clear not only to himself, but to others who may have occasion to read it.
211. Distortion of oblique projection. A view of a com- pleted machine suffers considerable distortion when drawn in
\x FIG. 18.
FIG. 19.
oblique projection because the eye cannot be placed in any one posi- tion, whereby it can view the drawing in the manner the projec- tion was made. * To overcome this difficulty to some extent and to avoid bringing the distortion forcibly to the attention of the observer, the projecting lines can be so chosen that the per- pendicular to the plane of projection is projected as a shorter line than the perpendicular itself. Fig. 18 shows this in con- struction. XX is a vertical transparent plate, similar to that shown in Fig. 4. The ray rb makes an angle with XX greater than 45°, and, by inspection, it is seen that the projection of ab on XX is ac, which is shorter than ab, the perpendicular.
The application of the foregoing reduces simply to this: All lines and curves parallel to the plane of projection are shown * This condition is satisfied in Perspective Projections.
OBLIQUE PROJECTION 21
exactly the same as in oblique projection with 45° ray inclination. The lines that are perpendicular to the plane of projection are reduced to i, |, J, etc. of their original length and reduced or increased to the scale adopted in making the drawing. This mode of representation* is suitable for making catalogue cuts and the like. It gives a sense of depth without very noticeable distortion, due to two causes: the impossible location of the eye while viewing the drawing, and, the knowledge of the apparent decrease in size of objects as they recede from the eye. A single illustration is shown in Fig. 19.
212. Commercial application of oblique projection.
Oblique projection is useful in so far as it presents the three dimensions in a single view. When curves are a part of the outline of the object, it is desirable to make the plane of the curve parallel to the plane of projection, thereby making the projection equal to the actual curve and also economizing time in making the drawing. Sometimes it is not possible to carry this out completely. Fig. 17, already quoted, shows an example of this kind. It is quite natural to make the drawing as shown, because the planes of most of the circles are parallel to the plane of projection, leaving, thereby, only one end of the bell-crank to be projected with ellipses.
Oblique projections, in general, are perhaps the simplest types of drawings that can be made, if the objects are of com- paratively simple shape. They carry with them the further advantage that even the uninitiated are able to read them, when the objects are not unusually intricate. The making of oblique projections is simple, but, at the same time, they call on the imagination to some extent for their interpretation. This is largely due to the fact that the eye changes its position for each point projected, and that no one position of the eye will properly place the observer with respect to the object.
The application of oblique projection to the making of drawings for solid geometry is already known to the student and the resulting clarity has been noticed. Other types of pro- jections have certain advantages which will be considered in due order.
* This type of projection has been called Pseudo Perspective by Dr. MacCord in his Descriptive Geometry.
PARALLEL PROJECTING-LINE DRAWING
The convenience of oblique projection to the laying out of piping diagrams is worthy of mention. Steam and water pipes, plumbing, etc., when laid out this way, result in an exceedingly readable drawing.
QUESTIONS ON CHAPTER II
1. What is an oblique projection?
2. What is a. plane of projection?
3. What is a projecting line?
4. Prove that when a rectangle is parallel to the plane of projection the
projection of the rectangle is equal to the rectangle itself.
5. Does the distance of the object from the plane have any influence on
the size of the projection? Why?
6. Prove that any line, whether straight or curved, is projected in its
true form when it is parallel to the plane of projection.
7. Show under what conditions a projection may be considered as a
shadow.
8. Prove that when a line is perpendicular to the plane of projection,
it is projected as a line of equal length, when the projecting rays make an angle of 45° with the plane of projection. Use a diagram.
9. Prove that any limited portion of a line is projected as a line of
equal length, when the line is perpendicular to the plane of projec- tion and the projecting lines make an angle of 45° with the plane of projection.
10. Show how a perpendicular may be projected as a longer or a shorter
line, if the angle of the projecting lines differs from 45°.
11. Why can not the projecting lines be selected parallel to the plane of
projection?
12. Show that when a line is parallel to the plane of projection, it is
projected as a. line of equal length, irrespective of the angle of the projecting lines, provided the projecting lines are inclined to the plane of projection.
13. Why may the slope of the projection of a line perpendicular to the
plane of projection be drawn at any angle?
14. Prove that when two lines are parallel to each other and also to the
plane of projection, their projections are parallel.
15. Prove that when two lines are perpendicular to the plane of pro-
jection, their projections are parallel to each other.
16. Draw two rectangular planes at right angles to each other so that the
edges of the planes are parallel or perpendicular to the plane of the paper (or projection).
17. Draw a cube in oblique projection and show which lines are assumed
parallel to the plane of projection and which lines are perpendicular to the plane of projection.
18. Draw a cube in oblique projection and show how the circles are
inserted in each of the visible faces.
OBLIQUE PROJECTION
23
19. Show how angles are laid off on the face of a cube in oblique pro-
jection.
20. Under what conditions is the angular relation between lines pre-
served?
21. Draw a line that is neither parallel nor perpendicular to the plane
of projection. (Use the cube, in projection, as a bounding figure.)
22. Prove that any two lines in space are projected as parallels when
they themselves are parallel.
23. Under what conditions will the oblique projection of a line be a
point?
24. How are visible and invisible lines represented on a drawing?
25. What is meant by drawing to scale?
26. Why is it desirable to have all parts of the same object drawn to
true scale?
v.
FIG. 2A.
FIG. 2B.
27. What considerations govern the choice of the scale to be used on a
drawing?
28. Show how the distortion of an oblique projection may be reduced by
changing the angle of the projecting lines.
29. Why is it impossible to locate the eye in one position and view the
projection in the manner in which it was made?
30. Draw a rectangular box with a hinged cover, in oblique projection,
and show the cover partly raised.
31. Draw a lever, having a round hole on one end so as to fit over a
shaft. Have the plane of the circles parallel to the plane of projection.
32. Draw an oblong block, 2"x3"x6" long, in oblique projection,
having a 1" hole in its centre, 4" deep.
33. Draw a cylindrical shaft, 6" in diameter and 18" long, in oblique
projection, having a rectangular hole, 2"X3"x5" deep, from each end. Lay out to a scale of 3" = 1 ft. and affix all dimensions.
24 PARALLEL PROJECTING-LINE DRAWING
34. Draw a triangular prism, in oblique projection, showing how the
bounding figure is used.
35. Draw a hexagonal prism, in oblique projection, and show all the
invisible (dotted) lines on it.
36. Draw a hexagonal pyramid in oblique projection.
37. Make a material list for the box shown in Fig. 2A.
38. Draw the circular cylinder, 4" diameter and 6" long, shown in
Fig. 2B. On the surface of this cylinder, two semi-circular grooves are cut, as shown by the dimensions. Make two drawings in oblique projection, one showing the plane of the circles parallel to the plane of projection and the other, with the plane of the circles perpendicular to the plane of projection.
NOTE. — For additional drawing exercises see examples in Chapters III and IV.
CHAPTER III ORTHOGRAPHIC PROJECTION
301. Nature of orthographic projection. Take, for example, a box 6"X12"X4" high, made of wood, \" thick. This box is shown orthographically in Fig. 20, and requires two distinct views to illustrate it properly. The upper view, or elevation, shows the side of the box whose outside dimensions are 4"X12", while the thickness of the wood is indicated by the
!&
|
0 o |
o o |
|
FIG. 20.
FIG. 21.
dotted lines. The lower view is called the plan, and is obtained by looking down into the inside of the box. It is thus to be remembered that the two views are due to two distinct directions of vision on the part of the observer.
As another example of this mode of representation, consider the object shown in Fig. 21. It is here a rectangular plate, 5"X3" and I" thick, with a square hole in its centre. The metal around the square hole projects \" above the surface of the plate. In addition, there are four bolt holes which enable the part to be secured to a machine with bolts. As before, two views are shown with the necessary dimensions for con- struction.
25
26 PARALLEL PROJECTING-LINE DRAWING
302. Theory of orthographic projection. Let Fig. 22 represent an oblique projection of two plane surfaces HH and W, at right angles to each other. For mechanical operations to be performed later, it is assumed that they are hinged at their intersection so that both planes may be made to lie as one flat surface, instead of two separate surfaces at right angles to each other. The plane HH, shown horizontally, is the hori- zontal plane of projection; that shown vertically, is the vertical plane of projection; their intersection is called the ground line. The two planes, taken together, are known as the principal planes
FIG. 22.
of projection. The object is the 4"X6"X12" box chosen as an illustration in Fig. 20. The drawing on the horizontal plane is the horizontal projection; while that on the vertical plane is the -vertical projection.
The method of constructing the projection consists of dropping perpendiculars from the object upon the planes of projection. Thus, in other words, the projecting lines are perpendicular to the plane of projection. To illustrate: The box is so located in space that the bottom of it is parallel to the horizontal plane (Fig. 22) and the 4"X12" side is parallel to the vertical plane. From the points A, B, C, and D, perpendiculars are drawn to the vertical plane and the points, where these perpendiculars pierce or impinge on the plane, are marked a', b'; c', and d', to
ORTHOGRAPHIC PROJECTION 27
correspond with the similarly lettered points on the object. By joining these points with straight lines, to correspond with the lines on the object, the vertical projection is completed, when the dotted lines showing the inside of the box are added.
Turning to the projection on the horizontal plane, it is seen that A, B, E, and F are the corners of the box in space, and that perpendiculars from these points to the horizontal plane determine a, b, e, and f as the horizontal projection. It is assumed that the observer is looking down on the horizontal plane and therefore sees the inside of the box; these lines are hence shown in full, although the projecting perpendiculars are omitted so as to avoid too many lines in the construction.
303. Revolution of the horizontal plane. It is manifestly impracticable to carry two planes at right angles to each other, each containing one projection of an object. A more con- venient way is to represent both projections on a single plane surface, so that such drawings can be represented on a flat sheet of paper. The evident expedient, in this case, is to revolve the horizontal plane about the ground line as an axis, until it coincides with the plane of the vertical plane. The conventional direction of rotation is shown by the arrow in Fig. 22, and to accomplish this coincidence, a 90° revolution is required. In passing, it may be well to note that it makes no difference whether the horizontal plane is revolved as suggested, or whether the vertical plane is revolved in the opposite direction into coin- cidence with the horizontal plane. Both accomplish t>e same purpose, and hence either method will answer the requirements.
304. Position of the eye. The perpendicular projecting lines drawn to the planes of projection correspond with a line of sight that coincides with these perpendiculars. Each point found en the projection, corresponds to a new position of the eye. All projecting lines to one plane are then evidently parallel because they are all perpendicular to the same plane. As two projec- tions are required, two general directions of vision are necessary. That for the horizontal projection requires the eye above that plane, continually directed perpendicularly against it; thus the eye is continually shifting in position, although the direction of vision is fixed. Also, the vertical projection requires that the eye be directed perpendicularly against it, but in this case, the
28
PARALLEL PROJECTING-LINE DRAWING
line of sight is perpendicular to that required for the horizontal projection.
305. Relation of size of object to size of projection.
The object is projected on the planes by lines perpendicular to it. If the plane of the object is parallel to the plane of pro- jection, then the projection is equal to the object in magnitude. This is true because the projecting lines form a right prism and
FIG. 23.
all the parallel plane sections are the same (compare with 202). Fig. 23 gives the construction of the projection in Fig. 21. ABCDa'b'c'd' is such a right prism because the plane of the object is parallel to the plane of projection and the projecting lines are perpendicular to the plane of projection.
306. Location of object with respect to the planes of projection. For purposes of drawing, the location of the object to the planes of projection is absolutely immaterial. In fact,
ORTHOGRAPHIC PROJECTION 29
the draftsman intuitively makes the projections and puts corre- sponding projections as close as is necessary to economize room on the sheet.
307. Location of projections with respect to each other.
In Figs. 20 and 21, the vertical projection is placed directly above the horizontal projection. Reference to Fig. 23 will show why such is the case. When the horizontal plane is revolved into coincidence with the vertical plane, the point d will describe the arc of a circle dd" * which is a quadrant; d" is the ultimate position of the point d after revolution, and must be on a line d'd" which is perpendicular to the original position of the hori- zontal plane. So, too, every point of the horizontal projection is located directly under the corresponding point in the vertical projection, and the scheme for finding its position is identical to that for finding d at d".
308. Dimensions on a projection. When the principal planes of the object are turned so that they are parallel to the planes of projection, then the edges will, in the main, be perpendicular to the planes of projection. In Fig. 23, DF is one edge of the object and it is perpendicular to the vertical plane of projection W. The projection of this line is d', because the projecting perpendicular from any point on DF will coincide with DF itself. The result of this is that the thickness of the object is not shown when the length and breadth are shown, or, in other words, only two of the three principal dimensions are shown in a single view. Thus, another view is required to show the thickness. If DC be considered a length, and DF a thickness, the horizontal pro^ jection shows both as dc and df. The vertical projection does not show the thickness DF as it is perpendicular to the vertical plane of projection. Hence, in reading orthographic projections, both views must be interpreted simultaneously, as each shows but two of the three principal dimensions and only one of the three is common to both projections.
309. Comparison between oblique and orthographic projections. It is of interest here to show wherein the ortho- graphic projection differs from the oblique. When the plane
* d' is read d prime; d" is read d second; d"' is read d third; and so on.
30
PARALLEL PROJECTING-LINE DRAWING
of the object is parallel to the plane of projection, the projection on that plane is equal to the object, whether it is projected orthographically or obliquely. When a line is perpendicular to the plane of projection its extremities have two distinct pro- jections in oblique projection, but only one in orthographic pro- jection. This latter statement means simply that if the projecting lines instead of being oblique to the plane of projection, gradually assume the perpendicular position, the two projections of the extremities of any line approach each other until they coincide when the projecting lines are perpendicular. Therefore, in orthographic projection, the third dimension vanishes and a new view must be made in addition to the other, in order to represent a solid.
310. Orthographic projection considered a shadow. The
horizontal and the vertical projections may be considered as shadows on their respective planes. The source of light must be such that the rays emanate in parallel lines, and are directed perpendicularly to the planes of projection. -Evidently, the two views are due to two distinct positions of the source of light, one whose rays are perpendicular to the horizontal plane while casting the horizontal shadow, and the other, whose rays are perpendicular to the vertical plane while casting the vertical shadow.
311. Profile plane. Let A, in Fig. 24, be the horizontal projection and B, the vertical projection of an object. The
two views are identical, and to one unfamiliar with the object, they are indefinite, as it is im- possible to tell whether they are projections of a cylinder or of a prism. By the addition of either view C or D, it is at once appar- ent that the object in question
is a circular cylinder, a hole FIG. 24. running part way through it
and with one end square.
Fig. 25 shows how this profile is made. As customary, the horizontal and vertical planes are present and the projection on these planes should now require no further mention. A
ORTHOGRAPHIC PROJECTION
31
profile plane (or end plane as it may be called) is shown on the far side of the object and is a plane that is perpendicular to both the horizontal and vertical planes (like the two adjacent
FIG. 25.
walls and the floor of a room meeting in one corner). A series of perpendiculars is dropped from the object upon this profile
FIG. 26.
plane, as shown by the dotted lines, and thus the side view is determined.
312. Location of profiles. If the profile view is to show the object as seen from the left side, it is put on the left side of the drawing, and vice versa. Fig. 24 shows two profile views located in accordance with this direction. Either views B and C
32
PARALLEL PROJECTING-LINE DRAWING
or B and D completely represent the object. In this case, although this is not always so, the horizontal projection is not essential.
Fig. 26 gives still another illustration of an object that is not as symmetrical as that immediately preceding. The illus- tration is chosen to show exactly how the profile planes are revolved into the vertical plane, if the vertical plane be assumed as the plane of the paper. A is the horizontal and B the vertical projection of the object. C and D are two profiles, drawn against the vertical projection, whereas E is a profile drawn against the horizontal projection. Fig. 27 shows a plan view of the vertical and two profile planes. In reading this drawing, the
FIG. 27.
horizontal plane is the plane of the paper, while the vertical plane is seen on edge and is shown as W, as are also the left and right profile planes indicated respectively as LL and RR.
In making the projection on the horizontal plane, the object is above the plane and the projecting perpendiculars are dropped from points on the object to the horizontal plane, which in this case is the plane of the paper. The construction of the vertical projection (that on W) is indicated by the arrow A. The arrangement here shown corresponds to the views A and B in Fig. 26.
When making the profile projections, the planes are assumed as transparent, and are located between the object and the observer. As the observer traces the outline on these profile planes, point by point, each ray being perpendicular to the plane, the resultant picture so drawn becomes the required projection. If, then, the planes LL and RR be revolved in the direction of the
ORTHOGRAPHIC PROJECTION
33
arrows until they coincide with the vertical plane, and then the vertical plane be further revolved into the plane of the paper, the final result will be that of Fig. 26 with view E omitted.
View E is a profile drawn against the horizontal projection and is shown on the left because it is the projection on the profile plane LL. It has been revolved into the horizontal plane, by revolving the profile plane so that the upper part of the plane moves toward the object into coincidence with the horizontal plane.
Fig. 26 has more views than are necessary to illustrate the object completely. In practice, all would not be drawn, their presence here is necessary only to show the method.
313. Section plane. The addition of dotted lines to the drawing of complicated objects is unsatisfactory at times on account of the resultant confusion of lines. This difficulty can be overcome by cutting the object by planes, known as section planes. The solid material when so exposed is sectioned or cross- hatched by drawing a series of equidistant lines over the exposed area. A convenient mnemonic in this connection is to assume that the cut is made by a saw and that the resultant tooth marks represent the sec- tion lines. Fig. 28 shows what
is known as a stuffing box on a steam engine. This is a special case where but one projection is shown in section and one profile. The left-hand view might have been shown as an outside view, but the interior lines would then have been shown dotted. As it is, the object is cut by the plane ab and this half portion is shown to the left, sectioned of course, because the cut is not actual.
Another example is seen in Fig. 29, where a fly-wheel is represented in much the same way as in the illustration in Fig. 28. It differs somewhat from that immediately preceding in so far as the two views do not have the theoretical relation. Were the wheel actually cut by the plane ab then the arms (spokes) shown in the profile would have to be sectioned. As
FIG. 28.
34 PARALLEL PROJECTING-LINE DRAWING
shown, however, the arms appear in full as though the section plane passed through the wheel a short distance ahead of the spokes. The convention is introduced for a double purpose: In the first place it avoids peculiar projections as that for the plane cd for instance, where the spokes would be foreshortened because they incline to the plane of projection. In the second place, the sectioning of the spokes is the conventional method of showing a band wheel,* that is, a wheel with a solid web, or, in other words, without spokes. Hence, it appears that although it may not seem like a rational method of drawing, still the attending advantages are such as make it a general custom. The mechanics who use the drawings understand this, and therefore it becomes common practice.
FIG. 29.
\
Many more examples could be added, but they would be too complicated to be of illustrative value. It may be said, that, in some cases, six or more sections may be made to illustrate the object completely. They are located anywhere on the draw- ing and properly indicated, similar to cd in Fig. 29. It may also be mentioned, that in cases like that of the fly-wheel, the shaft is not cut by the section plane but is shown in full as it appears in Fig. 29.
314. Supplementary plane. Fig. 30 shows a Y fitting used in pipe work for conveying steam, water, etc., and consists of a hollow cylindrical shell terminating in two flanges, one at
* Fig. 43 is an example of a band wheel.
ORTHOGRAPHIC PROJECTION
35
each end. From this shell there emerges another shell (in this
case, smaller in diameter), also terminating in a flange. A is the
Y fitting proper; B is the end view of one flange, showing the
bolt holes for fastening to a
mating flange on the next
piece of pipe, not shown. The
view B shows only the one
flange that is represented by
a circle, because the profile
plane is chosen so as to be
parallel to that flange. If the
flange at C be projected on
this same profile plane, it FIG. 30.
would appear as an ellipse,
and, as such, could not be drawn with the same facility as a
circle, Here, then, is an opportunity to locate another plane,
FIG. 31.
called a supplementary plane, parallel to the flange at C. The projection of C on this supplementary plane will be a circle and is therefore readily drawn with a compass.
36
PARALLEL PROJECTING-LINE DRAWING
Only one-half of the actual circle may be shown if desired so as to save time, space, or both time and space. The bolt holes are shown in the supplementary view at C just the same as in view B. Either plane may be considered supplementary to the other; on neither plane is the entire projection made, because the object is of so simple a character. To avoid the possibility of any error arising, the supplementary project' en is, if it is at all possible, located near the part to be illustrated. If, for any reason, this view cannot be so located, a note indicating the proper position of the view is added to the drawing.
315. Angles of projection. Up to the present point, no attention has been devoted to the angles in which the pro-
|
0 |
i i ^N |
0 |
0 |
1 1 1 1 |
C |
|
|
r^ |
o |
— 1 |
||||
|
1 |
||
|
1 |
||
|
0 |
I i |
o |
|
1 1 |
1st Angle 2nd Angle 3rd Angle FIG. 32.
4th Angle
jections were made. As will soon appear, the examples so far chosen were all in the first angle. Fig. 31 shows two planes HH and W, intersecting at right angles to each other. The planes form four dihedral angles, numbered consecutively in a counter-clockwise manner as indicated. The same object is shown in all four angles, as are also the projections on the planes of projection, thus making four distinct projections.
316. Location of observer in constructing projections.
The eye is always located above the horizontal plane in making any horizontal projection. That is, for objects in the first and second angles, the object is between the plane and the observer ; for objects in the third and fourth angles the plane is between
ORTHOGRAPHIC PROJECTION 37
the object and the observer. While constructing the vertical projections the eye is always located in front of the vertical plane. That is, for objects in the first or fourth angles, the object is between the plane and the observer ; but in the second and third angles, the plane is between the object and the observer. This latter means simply that the observer stands to the right of this vertical plane W and views it so that the line of sight is always perpendicular to the plane of projection.
317. Application of angles of projection to drawing.
If the horizontal plane be revolved about the ground line XY as indicated by the arrows, until it coincides with the vertical plane of projection, it will be seen that that portion of the hori- zontal plane in front of the vertical plane will fall below the ground line, whereas that portion of the horizontal plane behind the vertical plane will rise above the ground line.
Supposing that in each position of the object in all four angles, the projections were made by dropping the customary perpendiculars to the plane of projection, and in addition, that the revolution of the horizontal plane is accomplished, then the resultant state of affairs will be as shown in Fig. 32. For purposes of illustration, the ground line XY is drawn although it is never used in the actual drawing of objects.* Also, the object has been purposely so located with respect to the planes that the second and fourth angle projections overlap. Mani- festly the second and fourth angles cannot be used in drawing if we wish to be technically correct. It may be possible so to locate the object in the second and fourth angles, by simply changing the distance from one plane or the other, that the two projections do not conflict, but a little study will show that the case falls either under first or third angle projection, depending upon whether the vertical projection is above the horizontal projection, or, below it.
It will be seen that in the first angle of projection, the plan is below and the elevation is above; whereas in the third angle of projection the condition is reversed, that is, the plan is above and the elevation is below. Strictly speaking, the profile, section and supplementary planes, have nothing to do with the angle
* When lines, points, and planes are to be represented orthographically, the ground line becomes a necessary adjunct.
38
PARALLEL PEOJECTING-LINE DRAWING
of projection, but it is quite possible to take a single projection with its profile, and locate it so that it corresponds to a third angle projection. Thus, there appears a certain looseness in the application of these principles. In general, the third angle of projection is used more than any other * as the larger number
|
o |
-1 — |
i |
0 |
1st Angle
3rd Angle
FIG. 33.
of mechanics are familiar with the reading of drawings in this ftngle. Fig. 33 shows the same object in the first and third angles of projection. A profile, or end view, is also attached to each, thus making a complete, though simple illustration.
318. Commercial application of orthographic projection.
Orthographic projection is by far the most important method of making drawings for engineering purposes. Other types of drawing have certain advantages, but, in general, they are limited to showing simple objects, made up principally of straight lines.
Some experience is required in reading orthographic projections because two or more views must be interpreted simultaneously. This experience is readily acquired by practice, both in making drawings and in reading the drawings of others.
To compensate for the more difficult interpretation of this type of drawing, there are inherent advantages, which permit the representation of any object, if it has some well defined shape. By the aid of sections, profiles, and supplementary planes, any side of a regular body can be illustrated at will, and further than this, the curves are shown with such peculiarity as characterizes them. Bodies, such as a lump of coal, or a spade
* The third angle of projection should b6 used in preference to the first because the profile, section, and supplementary planes conform to third angle projection.
ORTHOGRAPHIC PROJECTION
39
full of earth, are considered shapeless, and are never used in engineering construction. Even these can be represented ortho- graphically, however, although it is quite difficult to draw on the imagination in such cases.
QUESTIONS ON CHAPTER III
1. What are the principal planes of projection? Name them.
2. What is the ground line?
3. What angles do the orthographic projecting lines make with the
plane of projection?
FIG. 3A.
4. What is the horizontal projection?
5. What is the vertical projection?
6. Why is the horizontal plane revolved 90° after making the pro-
jections?
FIG. ZB.
7. Could the vertical plane be revolved instead of the horizontal plane?
8. Make a sketch of the planes of projection and show by arrow how
the revolution of the planes is accomplished.
9. How is the eye of the observer directed in making an orthographic
projection?
40
PARALLEL PROJECTING-LINE DRAWING
10. What is the general angular relation between the projecting lines
to the horizontal plane and the vertical plane of projection?
11. Why is the projection of the same size as the object?
12. Why is the ground line omitted in making orthographic projections?
13. Why are corresponding projections located directly over each other?
Show by diagram.
14. Why does an orthographic projection show only two of the three
principal dimensions of the object?
"1 "
FIG. 3C.
15. Why must the two views of an orthographic projection be interpreted
simultaneously?
16. Compare the direction of the projecting lines of oblique projection
with those of orthographic projection.
17. Show how the source of light must be located in order that the
shadow should correspond to a true projection.
FIG. 3D.
18. To cast the horizontal and vertical shadows, is it necessary to have
two distinct positions for the source of light?
19. What is a profile plane?
20. How is the profile plane located with respect to the principal planes
of projection?
21. How is the profile plane revolved into the plane of the drawing?
Show by diagram.
ORTHOGRAPHIC PROJECTION
41
22. How are the profiles located with respect to the main projection?
Show by some simple sketch.
23. What is a section plane?
24. When is a section plane desirable?
25. How is the section constructed?
TIG. 3F.
26. How is the section located with respect to the main projection?
27. Show a simple case where only one projection and one section com-
pletely determine an object.
FIG. 3F.
28. What is a supplementary plane?
29. When is it desirable to use a supplementary plane?
30. How is the supplementary projection located with respect to the
main projection?
42
PARALLEL PROJECTING-LINE DRAWING
31. Show why the profile plane is a special case of the supplementary
plane.
32. Make a diagram showing the four angles of projection and show
how they are numbered.
|
o |
o |
|
|
o |
0 |
|
FIG. 3#.
FIG. 3G.
FIG. 37.
33. How is the revolution of the planes accomplished so as to bring
them into coincidence?
34. How is the observer located in making first angle projections?
ORTHOGRAPHIC PROJECTION
43
35. How is the observer located in making second angle projections?
36. How is the observer located in making third angle projections?
37. How is the observer located in making fourth angle projections?
38. Why are only the first and third angles of projection used in
drawing?
39. How does the third angle differ from the first angle projection?
Show by sketch.
40. Why is the third angle to be preferred?
FIG. 3J.
41. Why is it more difficult to read orthographic projections than oblique
projections?
42. What distinct advantages has orthographic projection over oblique
projection?
FIG.
FIG. 3L.
43. Make a complete working drawing of 3-A and show one view in
section. Assume suitable dimensions.
NOTE: A working drawing is a drawing, completely dimensioned, with all necessary views for construction purposes.
44. Make a complete working drawing of 3-B and show one view in
section. Assume suitable dimensions.
45. Make a complete working drawing of 3-C and show one view in
section. Assume suitable dimensions.
46. Make a complete working drawing of 3-D and show one. view in
section. Assume suitable dimensions.
44 PARALLEL PROJECTING-LINE DRAWING
47. Make a complete working drawing of 3-E and show the profile on
the left, to the top view. Assume suitable dimensions.
48. Make a complete working drawing of 3-F and show the profile on
the left, to the top view. Assume suitable dimensions.
49. Make a complete working drawing of 3-G and show the lower view
as a first angle projection. Assume suitable dimensions.
50. Rearrange 3-G and show three views in third angle projection.
51. Make a section of 3-H through the web and observe that the web
is not sectioned. Assume suitable dimensions.
52. Make a supplementary view of the 45° ell shown in 3-1. Assume
suitable dimensions.
53. Make a supplementary view of 3-J. Assume suitable dimensions.
54. Make three views of 3-K in third angle projection. Assume suitable
dimensions.
55. Make three views of 3-L in third angle projection. Assume suitable
dimensions.
CHAPTER IV AXONOMETRIC PROJECTION
401. Nature of isometric projections. Consider three lines intersecting at a point and make the angles between each pair of lines equal to 120°; the three angles will then total 360° which is the total angle about a point. If on one of these lines, or lines parallel thereto, lengths are laid off, on the other line, or lines parallel thereto, breadths are laid off, and on the remaining line, or lines parallel thereto, thicknesses are laid off, it seems quite reasonable that a method may be devised whereby the three principal dimensions can be plotted so as to represent objects in a single view. This method when carried out to completion results in an isometric projection.
Let, in Fig. 34, OA, OB, and OC be the three lines, drawn as directed, so that the angles between them are 120°. Suppose it is desired to draw a box
4"X6"X12", made of wood *>
\" thick. Lay off 12" on OB (to any convenient scale), 6" on OA and 4" on OC. From A, draw a line AD parallel to OB and AE par- allel to OC; also, from C draw CE, parallel to OA and CF, parallel to OB. The thickness of the wood is to be added and the direction in which this is laid off is indicated by the direction
of the corresponding dimension line. In the drawing, the line OC was purposely chosen vertical so that OA and OB may be readily drawn with a 60° triangle. The other lines, added to Indicate the construction, are self-explanatory.
FIG. 34.
46
PARALLEL PROJECTING-LINE DRAWING
This, then, is an isometric projection, and, as may be noted, is a rapid method of representing objects in a single view. Com- parison may be made with Figs. 1 and 20, which show the same box in oblique and orthographic projection, respectively.
402. Theory of isometric projection. Let Fig. 35, represent a cube shown on the left side of a transparent plane W. If the observer, located on the right, projects this cube ortho- graphically (projecting lines perpendicular to the plane) on the plane W, and at the same time has the cube turned so that each of the three visible faces is projected equally on the plane, the
resultant projection is an isometric projection of the cube on the plane W.
The illustration on the right of Fig. 35 shows how this cube appears when orthographically projected. OA, OB, and OC are called the isometric axes. As each face of the cube is initially equal to the other faces, and as each edge is also equal, then, with equal inclination of the three faces, their projections are equal. Hence, the three angles are each equal to 120° and the three isometric axes are equal in length. To use these axes for drawing purposes merely requires that all dimensions of one kind (lengths, for instance) be laid off on any one line, or lines parallel thereto, and that this process be observed for the three principal dimensions.
Isometric projection, is, therefore, a special case of ortho- graphic projection; because, in its conception, the principal planes
AXOXOMETRIC PROJECTION
47
of the object are inclined to the plane of projection. As solids are thereby represented in a single view, only one projection is necessary.
403. Isometric projection and isometric drawing. When a line is inclined to the plane of projection, the orthographic projection of the line is shorter than that of the line itself, for, if the degree of inclination continue, the line will eventually become perpendicular to the plane and will then be projected as a point. Thus, in Fig. 35, OA, OB, and OC are drawn shorter than the actual edges of the cube, and the projection on the right is a true isometric projection. In the application of this mode of pro-
FIG. 36
jection, however, it is easier to lay off the actual distance (01 any proportion of it) rather than this foreshortened distance, If commercial scales are used in laying out the drawing instead of the true isometric projections, then it is called an Isometric Drawing.
The distinction between the two is a very fine one, since, if the ratio of foreshortening * be used as a scale to which the drawing is made, then it is possible by a simple statement, to change from isometric drawing to isometric projection. The commercial name will be followed and they will be called isometric drawings, always bearing in mind that the distinction means little.
* This ratio of the actual dimension to the true projected dimension is 100 : 83 and may be computed by trigonometry.
48
PARALLEL PROJECTING-LINE DRAWING
404. Direction of axes. It is usual to assume one of the axes as either horizontal, or vertical, as under these circum- stances, the other two can be drawn with the 60° triangles which are standard appliances in the drawing room. Fig. 36 shows a wood block in several positions, each of which is an isometric drawing; here, the location of the observer with respect to the block is at once apparent.
405. Isometric projection of circles. As no line is shown in its true length when it lies in the faces of the cube and is projected isometrically, then, also, no curve that lies in these faces can be shown in its true length and, therefore, in its true shape. This is due to the foreshortening caused by the inclina-
FIG. 37
tion of the plane of the three principal dimensions to the plane of projection.
Fig. 37 shows two cubes, the one on the left appears as a single square because it is an orthographic projection and this plane was parallel to the plane of projection. On the right, an iso- metric drawing is shown. In the orthographic projection on the left, a circle is inscribed in the face of the cube. The circle, so drawn, is tangent to the sides of the square at points midway between the extremities of the lines. When this square and its inscribed circle is shown isometrically, the points of tangency do not change, but, as the square is projected as a rhombus, the circle is projected as an ellipse, and is a smooth curve that is tangent at mid-points of the sides of the rhombus. A rapid,
AXONOMETRIC PROJECTION
49
though approximate method of drawing an ellipse * is shown in the upper face of the isometric cube. The major and the minor axes of the ellipse can be laid off accurately by drawing the diagonals eg and fh (the notation being alike in both views) ; a tangent to the circle is perpendicular to the radius, and, for points on the diagonals, this tangent is shown as mn (all four are alike because they are equal). Showing mn isometrically means that the gm = gn in one view is equal to the gm = gn in the other view; the hm = hn in one view is equal to the hm = hn in the other view, and so on, for the four possible tangents at the diagonals.
406. Isometric projection of inclined lines and angles.
Suppose it is desired to locate the centre of a hole in one of the
faces of a cube as at h (Fig. 38).
The hole is 12" back from the
point f, on the line fm, and 6" up
from the point m; hfm is the
isometric representation of a right
triangle whose legs are 6" and
12". It will be observed that
none of the right angles of the
cube are shown as such, therefore,
the angle hfm is not the true angle
corresponding to these dimensions;
and, hence, it cannot be measured
with a protractor in the ordinary
way.
The point k is located in a similar manner on the top of the cube, while ank is the isometric drawing of a right triangle whose legs are 9" and 13". A diagonal ed of the cube is also shown.
The general method of drawing any line in space is to plot the three components of the line. For instance, the point d is located with respect to the point e, by first laying off ef, then fg, and finally gd.
407. Isometric graduation of a circle, t If the plane of a circle is parallel to the plane of projection, the projection is
* The exact method of constructing an ellipse will be found in any text- book on geometry.
f This method is also applicable to oblique projection.
FIG. 38.
50 PARALLEL PROJECTING-LINE DRAWING
equal to the circle itself and, hence, it can be drawn with a compass to the required radius. Any angle is then shown in its true size and thus the protractor can be applied in its graduation.
When circles are shown isometrically, however, their pro- jections are ellipses, and the protractor graduation is applicable no longer. In the upper face of the cube, shown isometrically in Fig. 39, the circle is shown as an ellipse whose major and minor axes are respectively horizontal and vertical. If, on the major axis ab, a semicircle is drawn, and graduated by laying
FIG. 39.
off angles at 30° intervals, the points CDEF and G are obtained making six angles at 30°, or a total of 180°, the angular measure of a semicircle. If, then, the plane of the circle is rotated about ab as an axis, until the point E coincides with e, each point of division on the semicircle will find itself on the similarly lettered point of the ellipse; because, then, the plane of the semicircle coincides with the plane of the upper face of the cube. Therefore, aoc, cod, doe, etc., are 30° angles, shown isometrically.
In the side face of the same cube are shown two methods of laying off 45° angles. The fact that both methods locate the
AXONOMETRIC PROJECTION
51
same points tends to show that either method, alone, is applicable. The lettering is such that the construction should be clear without further explanation.
408. Examples of isometric drawing. As an example, the steps in the drawing of a wooden horse, shown in Fig. 40 and used in the building trades for supporting platforms and the like, will be followed. The making and the subsequent interpre-
FIG. 40.
tation of drawings of this kind is facilitated by the introduction of bounding figures of simple shape. In the example chosen, the horse is bounded by a rectangular prism. The attached dimensions show the necessary slopes and may be introduced so as to replace the bounding figure. There is nothing new in this drawing and therefore the description will not be needlessly exhaustive. The bounding figure, when appended to a drawing like that of Fig. 40, helps to emphasize the slopes of the various lines. In general, the bounding figure should be removed on completion of the drawing.
52
PARALLEL PROJECTING-LINE DRAWING
Fig. 41 shows a toothed wheel. The plane of the circles corresponds to the plane of the top face of a cube. To construct it, it is necessary to lay out the prism first and then to insert the ellipses. The graduation of the circle is similar to that indicated in Art. 407.
TIG. 41.
A lever is shown in Fig. 42. The centre line ab lies in the top face of the lever proper. Circumscribing prisms determine the ends of the lever and also the projecting cylin- drical ends.
FIG. 42.
A gas-engine fly-wheel is illustrated in Fig. 43. Here the fly-wheel is shown so that the planes of the circles correspond to one of the side faces of a cube. The wheel has a solid web (without spokes) and a quarter of it is removed and shown in section.
AXONOMETRIC PROJECTION 53
PARALLEL PROJECTIXG-LINE DRAWING
Fig. 44 represents a bell-crank. Again the scheme of using base-lines in connection with circumscribing prisms is shown. This view should be compared with that given in Fig. 17.
409. Dimetric projection and dimetric drawing. Let Fig. 45 be the isometric projection of a cube, with the invisible edges shown dotted. A disturbing symmetry of the lines is at once apparent. This objection becomes serious when applied to drawing if the objects are cubical, or nearly so.
The foregoing difficulty may be partially overcome by turning the cube so that only two faces are projected equally and the remaining
face may be larger or smaller at pleasure. Fig. 46 shows this condition represented. The angle aoc is larger than either the angles aob or cob, the latter two (aob and cob) being
FIG. 45.
equal. The faces A and B are projected equally, whereas C is smaller in this particular case, though it need not be. The illustration as shown on the right of Fig. 46 is a dimetric pro- jection and, as such, the same scale is applied to the axes oa and
AXONOMETRIC PROJECTION
55
oc because they are projected equally. The axis ob is longer, since the corresponding edge is more nearly parallel to the plane of projection W than either oa or oc. Hence, to be theoretically correct, a different scale must be used on the axis ob.
When dimetric projection is to be commercially applied, confusion may result from the use of two distinct scales. If one scale is used on all three axes the combination becomes a dimetric drawing. Hence, a dimetric drawing differs from an isometric drawing, in so far, as two of three angles are equal to each other for the dimetric drawing; whereas, in isometric drawing, all three are equal, that is, the axes are 120° apart.
What is true of the direction of the axes in isometric drawing is equally true here, that is, the angular relation between the axes, alone, determines the type of projection, the direction of any one is entirely arbitrary.
Dimetric drawings do not entirely remove the objectionable symmetry, yet they find some use in practice, although they present no distinct advantage over any other type.
410. Trimetric projection and trimetric drawing. Fig. 47 shows a cube which is held in such a position that the orthographic projection of it will result in the unequal projection of the three visible faces. This, then, becomes a trimetric projection.
To make true projections, three different scales must be used. This, of course, is objectionable and, hence, recourse is had to a trimetric draw- ing. A trimetric drawing, therefore, requires three axes. The angles be- tween the axes differ, but no one
angle can be a right angle in any case.* The same scale is applied to all three. Also, the axes may have any direction,
* This evidently makes it an "oblique projection. It is impossible to project a cube orthographically to produce this; since, if two axes are parallel to the plane of projection, the third axis must be perpendicular and is hence projected as a point. The oblique projection of a cube, which is turned as it would be in isometric drawing, presents no new feature since it results, generally speaking, in a trimetric projection.
FIG. 47.
56 PARALLEL PROJECTING-LINE DRAWING
so long as attention is paid to the angular relation be- tween them.
If isometric drawings introduce the disturbing symmetry, then the trimetric is to be recommended, unless one of the other types of drawing is found to be more suitable. No examples are given in this connection, because the application of dimetric and trimetric drawings involve no new features. It is only to be remembered, that artistic taste may dictate the direction of the axes, so as to present the best view of the object to be illustrated.
411. Axonometric projection and axonometric drawing.
Isometric, dimetric, and trimetric projection form a group which may be conveniently styled as axonometric projections. All three are the result of the orthographic projection of a cube, so that the three principal axes are projected in a manner as already indicated. Axonometric projections are therefore a special case of orthographic projection, but their advantages are sufficiently prominent to warrant separate classification.
For isometric projection, one scale is used throughout; for dimetric projection, two separate scales are used; and for tri- metric projection, three distinct scales are used. When applying axonometric projections to drawing, the same scale is used on all axes, and the group then represents a series which may be called axonometric drawing.
The distinction between isometric projection and isometric drawing has been pointed out (Art. 403). It becomes more prominent, however, in dimetric and trimetric projection.
412. Commercial application of axonometric projection.
For objects of simple shape, with few curves, isometric drawings serve a useful purpose, because they are easily made and are easily read by those unfamiliar with drawing in general. When curves are frequent and it is desirable to picture objects in a single view, oblique projections offer advantages over isometric drawings because it may be possible to make the planes of the curves parallel to the plane of projection. The curves will then be projected as they actually appear. When curves appear in many planes, then orthographic projections will answer require- ments best, but, as already mentioned, their reading is more difficult, due to the simultaneous interpretation of two or more views.
AXONOMETRIC PROJECTION 57
Before dismissal of the subject of projections in general, attention is called to Fig. 48 which is a peculiar application of isometric drawing. It is frequently used as an example of an optical illusion. By concentrating the vision at the centre of the picture, there seems to be a sudden change from six to seven cubes, or vice versa, depend- ing upon whether the central corner be regarded as a projecting or a depressed corner. This is due to the fact that all of the cubes are shown of the same size, a condition which is contrary to our everyday experience. As objects recede from the eye, they appear smaller; and
in isometric projection there is no correction for this. In fact, all the projections so far considered, draw on the imagination for their interpretation,* and, therefore, they cannot present a per- fectly natural appearance.
413. Classification of projections. All projections having parallel projecting lines may be classified according to the method by which they are made. This classification furnishes a useful survey of the entire subject and also serves to emphasize the distinction between the different methods.
It will be found, on analysis, that if the object be conceived in space with its parallel projecting lines, the oblique projections result when the plane of projection cuts the projecting lines obliquely. When the plane of projection cuts the projecting lines at a right angle, then the orthographic series of projections arise. If, still further, the projecting lines, coincide with some of the principal lines on the object, and the plane of projection is at right angles to the projecting lines, then the two-dimension orthographic projections result, and these are commonly called mechanical drawings. If the projecting lines do not coincide with some of the principal lines on the object, then the axono- metric series of projections follow.
"The projections that overcome these objections are known as Per- spective Projections. In these, the projecting lines converge to a point at which the observer is supposed to be located. Photographs are perspectives in a broad sense.
58
PARALLEL PROJECTING-LINE DRAWING
CLASSIFICATION OF PROJECTIONS HAVING PARALLEL PROJECTING LINES
All lines projected equal in length. Hence, same scale used on all.
Lines parallel to plane of projection drawn to equal length. Lines per- pendicular to plane, projected as shorter lines to di- minish distortion.
Principal lines of object parallel or perpendicular to planes of projec- tion. Sometimes called O r t h o - graphic projec- tion.
Isometric. All three axes of cube projected equally, hence, axes are 120° apart and equal in length. Applied as Iso- metric Drawing.
Dimetric. Two of the three axes projected equally, hence, only two angles equal. Ap- plied as Dimetric Drawing.
Trimetric. Axes projected unequ- ally, hence, all three angles differ. Applied as Tri- metric Drawing.
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with plane of |
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jecting lines |
projection. |
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inclined to |
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plane of pro- |
Inclination of |
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jection but |
projecting |
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parallel to |
lines greater |
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than 45° |
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with plane of |
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projection. |
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Mechanical |
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Drawing. |
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showing two |
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principal di- |
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mensions on . |
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lines perpen- |
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dicularto |
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the plane of |
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PROJEC- TIONS |
Axonometric. |
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projection, |
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showing |
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Curved J Not used in Engineering drawing. Surfaces: |
AXONOMETRIC PROJECTION 59
QUESTIONS OX CHAPTER IV
1. What are the isometric axes?
2. How are they obtained?
3. What is the angular relation between the pairs of axes?
4. Show why the isometric projection is a special case of orthographic.
5. What is the distinction between isometric projection and isometric
drawing?
6. What direction do the axes have for their convenient application
to drawing?
7. How is a circle projected isometrically?
8. Show the approximate method of drawing the isometric circle.
9. How are inclined lines laid off isometrically?
10. How are angles laid off isometrically?
11. Show that the laying off of an inclined line is accomplished by laying
off the components of the line.
12. How is the isometric circle graduated in the top face of the cube?
13. Show how the graduation is accomplished in one of the side faces
of the cube.
14. What is a dimetric projection?
15. What is a dimetric drawing?
16. What angular relation exists between the axes of a dimetric pro-
jection?
17. What is a trimetric projection?
18. What is a trimetric drawing?
19. What angular relation exists between the axes of a trimetric drawing?
20. Show why the trimetric drawing eliminates the disturbing symmetry
of an isometric drawing.
21. Why cannot the angle between one pair of axes be a right angle?
22. What are axonometric projections?
23. Why do projections with parallel projecting lines draw on the imag-
ination for their interpretation?
24. Draw the object of Question 33 in Chapter 2 in isometric drawing.
Use bounding figure.
25. Draw a triangular prism in isometric drawing. . Use a bounding
figure.
26. Make an isometric drawing of a hexagonal prism. Use a bounding
figure.
27. Make an isometric drawing of a hexagonal pyramid. Use a bounding
figure.
28. Make an isometric drawing of 3-A (Question in Chapter 3).
29. Make an isometric drawing of 3-B.
30. Make an isometric drawing of 3-C.
31. Make an isometric drawing of 3-D.
32. Make an isometric drawing of 3-F.
33. Make an isometric drawing of 3-G.
34. Make an isometric drawing of 3-H.
60. PARALLEL PROJECTING-LINE DRAWING
35. Make an isometric drawing of 3-1.
36. Make an isometric drawing of 3-J.
37. Make an isometric drawing of 3-K.
38. Make an isometric drawing of 3-L.
39. Make a complete classification of all projections having parallel
projecting lines.
PART II
GEOMETRICAL PROBLEMS IN ORTHOGRAPHIC PROJECTION
CHAPTER V REPRESENTATION OF LINES AND POINTS
501. Introductory. Material objects are bounded by sur- faces, which may be plane or curved in any conceivable way. The surfaces themselves are limited by lines, the forms of which may be straight or curved. Still further, these lines terminate in points. Thus, a solid is really made up of surfaces, lines and points, in their infinite number of combinations; and these may be considered as the mathematical elements that make up the solid. The mathematical elements must be considered as concepts as they have no material existence and, hence, are purely imagin- ative; that is, a surface has no thickness, therefore, it has no volume. A line or a point is a still further reduction along this line of reasoning. The usefulness of these concepts must be admitted, however, in view of the fact that they play such an important role in the conception of objects.
In general, the outline of any object is found by intuitively locating certain points, and joining the points by proper- lines; the lines, when taken in their proper order determine certain surfaces, and the space included between them forms the solid (the material object) in question. Hence, to view material objects analytically, the nature of their mathematical elements must be known.
In Part II of this book, the graphical representation of mathe- matical concepts engages the attention. Whether the treatment of the subject be from the viewpoint of mathematics or of drawing, entirely depends upon the ultimate use. In the two chapters to
61
62 GEOMETRICAL PROBLEMS IN PROJECTION
be presented (V and VI), the method of representing lines, points and planes, orthographically, will be discussed. So far, attention has been directed to the representation of material objects, as common every day experience renders such objects familiar to us.
Before proceeding with the subject in all its detail, familiarity must be gained with certain fundamental operations on lines, points and planes, as well as with their graphical representation on two assumed planes called the planes of projection. The fundamental operations are grouped and studied without apparent reference to future application. It is desirable to do this so as to avoid frequent interruptions in the chain of reasoning when applying the operations to the solution of problems.
The student will save considerable time if he is well versed in the fundamental operations. In view of this, many questions are given at the end of the chapters so that his grasp of the sub- ject may be tested from time to time. It may be needless to say that the solution of subsequent problems is utterly impossible without this thorough grounding. Frequent sketches should be made, representing lines, points and planes in positions other than shown in this book. These sketches should be made both in orthographic and in oblique projection. By this means, the student will increase his experience in the subject, much more than is possible by all the reading he might do. The proof of one's ability always lies in the correct execution of the ideas presented. The subject under consideration is a graphical one, and, as such, drawing forms the test mentioned. It ha§ been considered necessary to caution the student so as to avoid the complications that will result later, due to insufficient preparation. Only in this way will the subject become of interest, to say nothing of its importance in subsequent commercial applications.
502. Representation of the line. Let Fig. 49 be an oblique projection of two planes intersecting at right angles to each other. The plane W is called the vertical plane of projection and HH is called the horizontal plane of projection; these planes intersect in a line XY which is termed the ground line. The two planes taken as a 'whole are known as the principal planes; and by their intersection, they form four angles, numbered as shown in the figure. In what immediately follows, attention
REPRESENTATION OF LINES AND POINTS
63
will be concentrated on operations in the first angle of projection, and later, extended to all four angles.
Suppose a cube ABCDEFGH is located in the first angle of projection, so that one face lies wholly in the horizontal plane and another face lies wholly in the vertical plane. The two faces then lying in the principal planes, intersect in a line which coin- cides with the ground line for the conditions assumed. Suppose, further, that it is ultimately desired orthographically to represent the diagonal AG of the cube. For the present, the reasoning will be carried out by the aid of the oblique projection.
The construction of the horizontal projection consists in drop- ping upon the horizontal plane, perpendiculars from points on
FIG. 49.
the line. For the line AG, the projecting perpendicular from the point A is AE, and E is then the horizontal projection of the point A in space.* As for the point G on the line, that already lies in the horizontal plane and is its own horizontal projection. Thus, two horizontal projections of two points on the line are established, and, hence, the horizontal projection of the line is determined by joining these two projections. This is true, because all the perpendiculars from the various points on the line lie in a plane, which is virtually the horizontal projecting plane of the line. It cuts the horizontal plane of projection in a line EG, which is the horizontal projection of the line AG in space.
* It is to be noted that the projection of a point is found at the place where the projecting line pierces the plane of projection.
64
GEOMETRICAL PROBLEMS IN PROJECTION
Putting this in another form, the projection EG gives the same mental impression to an abserver viewing the horizontal plane from above as does the line AG itself; in fact, EG is a drawing of the line AG in space.
Directing attention for a moment to the vertical plane, it is found that the construction of the vertical projection consists in dropping a series of perpendiculars from the line AG to that plane. For point G, the perpendicular to the vertical plane is GH, and H is thus the vertical projection of G. The point A lies in the vertical plane and, hence, is its own vertical projection. A line joining A and H is the vertical projection of AG. Again, a plane passed through AG, perpendicular to the vertical plane,
FIG. 49.
will cut from it the line AH, which is the vertical projection as has been determined. Thus, AH is a drawing of AG because it conveys the same mental impression to an observer who views it in the way the projection was made.
503. Line fixed in space by its projections. The location of the principal planes is entirely arbitrary, as is also the line in Question; but, when both are once assumed, the line is fixed in space by its projections on the principal planes. If a plane be passed through the horizontal projection EG (Fig. 49), perpen- dicular to the horizontal plane, it will contain the line AG in space, since the method is just the reverse of that employed in finding the projection. Similarly, a plane through the vertical projection, perpendicular to the vertical plane, will also contain
REPRESENTATION OF LINES AND POINTS
65
the line AG. It naturally follows that the line AG is the inter- section of the horizontal and vertical projecting planes and, therefore, the line is absolutely fixed, with reference to the prin- cipal planes, by its projections on those planes.
504. Orthographic representation of a line. The line that has been considered so far is again represented in the left-hand view of Fig. 50, in oblique projection. The edges of the cube have been omitted here in order to concentrate attention to the line in question. AG is that line, as before, EG is its horizontal projection and AH is its vertical projection.*
Suppose that the plane HH is revolved in the direction of the arrows, 90° from its present position, until it coincides with the
FIG. 50.
vertical plane W. The view on the right of Fig. 50 shows the resultant state of affairs in orthographic projection. AH is the vertical projection and EG is the horizontal projection. AE in one view is the equivalent of AE in the other; HG and EH in one view are the equivalents of HG and EH in the other. Both views represent the same line AG in space. At first sight, it may appear that the oblique projection is sufficiently clear, and such is the case; but, in the solution of problems, the orthographic projection as shown on the right possesses many advantages. In due time, this mode of representation will be considered, alone, without the oblique projection.
* The projections of a line may be considered as shadows on their respective planes. The light comes in parallel rays, perpendicularly directed to the planes of projection. See also Arts. 203 and 310 in this connection.
66
GEOMETRICAL PROBLEMS IN PROJECTION
505. Transfer of diagrams from orthographic to oblique projection. It is desirable to know how to transfer diagrams from one kind of projection to the other. If the orthographic projection appears confusing, the transfer to oblique projection may be of service. On the right of Fig. 50, is given the ortho- graphic projection of the line AG. To construct the oblique projection from this, draw the principal planes as shown in the left-hand diagram. From any point E on the ground line in the oblique projection, lay off EA, vertically, equal to EA in the orthographic projection. On the sloping line (ground line), lay off EH equal to EH in the orthographic projection. Then draw HG in the horizontal plane, equal to HG in the orthographic projection. A line joining A and G in the oblique projection
FIG. 50.
gives the actual line in space; hence, in the oblique projection, the actual line and both of its projections are shown. In the orthographic projection, only the projections are given; the line itself must be imagined. Compare the method of constructing the oblique projection with Art. 207 and note the similarity.
506. Piercing points of lines on the principal planes.
If AG (Fig. 50) be considered as a limited portion of line indefi- nitely extended in both directions, then A and G are the piercing points on the vertical and horizontal planes respectively. On observation, it will be found that the vertical piercing point lies on the vertical projection of the line, and also on a perpendicular to the ground line XY from the point where the horizontal pro- jection intersects the ground line; hence, it is at their intersection.
REPRESENTATION OF LINES AND POINTS
67
Likewise, the horizontal piercing point lies on the horizontal projection of the line and also on the perpendicular erected at the intersection of the vertical projection with the ground line; hence, again, it is at the intersection of these two lines.
In Fig. 51 let ab be the horizontal projection of a line AB in space, and a'b' be the corresponding vertical projection. The line, if extended, would pierce the horizontal plane at c and the vertical plane at d'. This line is shown as an oblique projection in Fig. 52. All the lines that are required in the mental process are shown in this latter view. The actual construction in ortho- graphic projection is given in Fig. 51. To locate the vertical piercing point prolong both projections, and at the point where the horizontal projection intersects the ground line, erect a per-
FIG. 51.
FIG. 52.
pendicular until it intersects the prolongation of the vertical projection. This locates the vertical piercing point. To find the horizontal piercing point, prolong the vertical projection until it intersects the ground line, and at this point, erect a per- pendicular to the ground line. Then, the point at which this perpendicular intersects the prolongation of the horizontal pro- pection is the horizontal piercing point.
A convenient way of looking at cases of this kind is to assume that XY is the edge of the horizontal plane while viewing the verti- cal projection, therefore, AB must pierce the horizontal plane somewhere in a line perpendicular to the vertical plane at the point c'. In viewing the horizontal plane, XY now represents the vertical plane on edge. Here, again, the line AB must pierce
68
GEOMETRICAL PROBLEMS IN PROJECTION
the vertical plane somewhere in a line perpendicular to the hori- zontal plane at the point d.
507. Nomenclature of projections. In what follows, the actual object will be designated by the capital letters, as the line AB for instance. The horizontal projections will be indicated by the small letters as the line ab, and the vertical projections by the small prime letters as the line a'b'. It should always be remembered that in orthographic projection, the projections alone are given, the actual object is to be imagined.
508. Representation of points. A single point in space is located with respect to the principal planes as shown in Fig. 53. A is the actual point while a is its horizontal, and a' its vertical projection. The distance of A above the horizontal plane is equal to the length of its projecting perpendicular Aa and this is equal to a'o because Aa and a'o are both perpendicular to the hori-
FIG. 53.
zontal plane HH. Also, Aa' and ao are perpendicular to the verti- cal plane, therefore, the figure Aaoa' is a rectangle, whose opposite sides are necessarily equal. Hence, also, the distance of A from the vertical plane is equal to the length of its projecting perpen- dicular Aa' which also equals ao. Performing the usual revolu- tion of the horizontal plane, a will reach a" on a line a'a" which is perpendicular to the ground line XY. It has been shown that a'o is perpendicular to XY and it only remains to prove that oa" is a continuation of a'o. This must be so, because a revolves about the ground line as an axis, in a plane determined by the two intersecting lines a'o and ao. This plane cuts from the ver-
REPRESENTATION OF LINES AND POINTS 69
tical plane the line a'a" which is perpendicular to XY because a portion (a'o) of it is perpendicular. As the point a revolves about XY as an axis, it describes a circle, whose radius is oa, and hence oa" must equal oa.
In projection,* this is shown on the right-hand diagram of Fig. 53. Both figures are lettered to correspond as far as con- sistent. The actual point A is omitted, however, in the right- hand diagram because the very object of this scheme of repre- sentation is to locate the point from two arbitrary planes (prin- cipal planes), solely by their projections on those planes. This latter is an exceedingly important fact and should always be borne in mind.
509. Points lying in the principal planes. If a point lies in one of the principal planes, it is its own projection in that plane and its corresponding projection
lies in the ground line. Fig. 54 shows such cases in projection. A is a point \a $ Cc>
lying in the vertical plane, at a distance a'a above the horizontal plane; its ver- [b
tical projection is a' and its correspond- -pio. 54.
ing horizontal projection is a. B is
another point, lying in the horizontal plane, at a distance bb' from the vertical plane; b is its horizontal and b' the correspond- ing vertical projection.
If a point lies in both planes, the point coincides with both of its projections and must therefore be in the ground line. C is such a point, and its two projections are indicated by cc', both letters being affixed to the one point. The use of these cases will appear as the subject develops.
510. Mechanical representation of the principal planes.
For the time being, the reader may find it desirable to construct two planes f so that lines and points may be actually represented
* Hereafter, orthographic projection will usually be designated simply as "in projection."
t For classroom work, a more serviceable device can be made of hinged screens, constructed of a fine mesh wire. Wires can be easily inserted to represent lines and the projections drawn with chalk. The revolution of the planes can be accomplished by properly hinging the planes so that they can be made to lie approximately flat.
70
GEOMETRICAL PROBLEMS IN PROJECTION
FIG. 55.
with reference to them. If two cards be slit as shown in Fig. 55, they can be put together so as to represent two planes at right
angles to eaj&h other. Lines may be represented by use of match-sticks and points by pin-heads, the pin being so inserted as to represent its projecting perpendicular to one plane. The idea is recommended until the student be- comes familiar with the involved opera- tions. As soon as this familiarity is obtained, the cards should be dispensed with, and the operations reasoned out [in space, as far as possible, without the use of any diagrams.
611. Lines parallel to the planes of projection. Assume a Ime parallel to the horizontal plane. Evidently this line cannot pierce the horizontal plane on account of the parallelism. It will pierce the vertical plane, however, if it is inclined to that plane .
FIG. 56.
In Fig. 56 this line is shown both in oblique and orthographic projection. The piercing point on the vertical plane is found by erecting a perpendicular at the point where the horizontal projection intersects the ground line. It will also lie on the ver- tical projection of the line. As the line is parallel to the horizontal plane, its vertical projection is parallel to the ground line. The piercing point will therefore lie at the intersection of the two lines; that is, of the perpendicular from the ground line and the vertical projection of the line. In both views of the figure, AB is the given line and b' is its piercing point.
If, on the other hand, the line is parallel to the vertical plane
REPRESENTATION OF LINES AND POINTS
71
(Fig. 57) and inclined to the horizontal plane, it will pierce the horizontal plane at some one point. Its horizontal projection is now parallel to the ground line XY. The horizontal piercing
FIG. 57.
point is at b, as shown, and is found in much the same way as that in the illustration immediately preceding.
A case where the line is parallel to both planes is shown in
FIG. 58.
Fig. 58. This line cannot pierce either plane and, therefore, both its projections must be parallel to the ground line as depicted.
512. Lines lying in the planes of projection. If a line lies in the plane of projection, it is its own projection in that plane, and its corresponding projection lies in the ground line. When the line lies in both planes of projection it must therefore coincide with the ground line. Fig. 59 shows in projection the three cases possible. The first is a line lying in the horizontal plane, ab is its horizontal projection and
72 GEOMETEICAL PROBLEMS IN PROJECTION
a'b' is its corresponding vertical projection. The second is a
line lying in the vertical plane, c'd' is its vertical projection and cd is its corresponding horizontal projection. Y The third case shows a line in both planes and its horizontal and vertical projections coincide with the ground FIG 59. line and also the line itself; the
coincident projections are indicated
as shown at ee' and ff, read ef and e'f.
513. Lines perpendicular to the planes of projection.
If a line is perpendicular to the horizontal plane, its projection on that plane is a point, because both projecting perpendiculars from the extremities must coincide with the perpendicular line. The vertical projection, however, shows the line in its true length, perpendicular to the ground line, as such a line is 'parallel to the
|
f '"' tf TT "Vl * V rr -1 |
V c 2 ^« |
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X c 1 n 4 |
3 |
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Ut |
|
|
,1 |
V |
|
FIG. 60. |
vertical plane. Fig. 60 shows this in projection as AB, and, in addition a profile plane is added which indicates the fact more clearly. CD is a line perpendicular to the vertical plane with one extremity of the line in that plane. In the profile plane, the lines AB and CD appear to intersect, but this is not necessarily the case. The construction of the projections of the line CD is identical with that immediately preceding.
514. Lines in all angles. So far, the discussion of lines and points in space has been limited entirely to the first angle. If a line is indefinite in extent, it may pass through several of the four angles. A case in each angle will be taken and the salient features of its projections will be pointed out.
Fig. 61 shows a line passing through the first angle. It con- tinues beyond into the second and fourth angles. In projection,
REPRESENTATION OF LINES AND POINTS
73
the condition is shown on the right. The projections ab and a'b' show those of the limited position that traverses the first angle.
The horizontal
The dotted extensions show the projection of the indefinite line, and are continued, at pleasure, to any extent.
A line in the second angle is shown in Fig. 62. projection is ab and the ver- tical projection is a'b'. If the horizontal plane be re- volved into coincidence with the vertical plane, the view in projection will show that, in this particular case, the projections of the lines cross pIG 62
each other. Only the limited
portion in the second angle is shown, although the line may be indefinitely extended in both directions.
A third angle line is illustrated in Fig. 63. It may be observed,
FIG. 63.
in comparison with a line in the first angle, that the horizontal and vertical projections are interchanged for the limited portion
74
GEOMETRICAL PROBLEMS IN PROJECTION
of the line shown. In other words the horizontal projection of the line is above the ground line and the vertical projection is below it. The case may be contrasted with Fig. 61.
The last case is shown in the fourth angle and Fig. 64 depicts
FIG. 64.
this condition. Both projections now cross each other below the ground line. Contrast this with Fig. 62. The similarity of the foregoing and Arts. 315, 316 and 317 may be noted.
515. Lines with coincident projections. Lines, both of whose projections lie in the ground line have been previously considered (511). If a line passes through the ground line from
66'
FIG. 65.
the second to the fourth angle, so that any point on the line is equidistant from both planes, then the two projections of the line will be coincident. Fig. 65 illustrates such a condition. The line is not indeterminate by having coincident projections, however, because the oblique projections can be constructed
REPRESENTATION OF LINES AND POINTS 75
from the orthographic representation. The projection on the profile plane shows that this line bisects the second and fourth angles.
If the line passes through the ground line, from the second to the fourth angles, but is not equidistant from the principal planes, then both projections will pass through the same point on the ground line.
516. Points in all angles. Fig. 66 shows four points A, B, C, and D, lying, one in each angle. The necessary construction lines are shown. On the right, the condition is depicted in projection, the number close to the projection indicating the angle in which the point is located.
Observation will show that the first and third angle projec-
t !'
H 2 |3
I !
-' fr
FIG. 66.
tions are similar in general appearance, but with projections interchanged. The same is true of the second and fourth angles; although, in the latter cases, both projections fall to one or the other side of the ground line.
517. Points with coincident projections. It may be
further observed in Fig. 66 that in the second and fourth angles, the projection b, b' and d, d' may be coincident. This simply means that the points are equidistant from the planes of projection. The case of the point in the ground line has been noted (508), such points lie in no particular angle, unless a new set of principal planes be introduced. It can then be considered under any case at pleasure, depending upon the location of the prin- cipal planes.
518. Lines in profile planes. A line may be located so that both of its projections are perpendicular to the ground line.
76
GEOMETRICAL PROBLEMS IN PROJECTION
The projections must therefore be coincident,* since they pass through the same point on the ground line. Fig. 67 shows an example of this kind. Although the actual line in space can be determined from its horizontal projection ab and its vertical
FIG. 67.
projection a'b', still this is only true because a limited portion of the line was chosen for the projections. The profile shown on the extreme right clearly indicates the condition. The location of the profile with respect to the projection should also be noted. The view, is that obtained by looking from right to left, and is therefore located on the right side of the projection. The number- ing on the angles should also help the interpretation.
D
H
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C 1« |
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b', |
d'l |
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X. |
/ |
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a' |
1 |
2 3 |
4 |
|
a |
,, .. |
„' |
FIG. GS.
Fig. 68 shows a profile (on the left-hand diagram) of one line in each angle. The diagram on the right shows the lines in projection. The numbers indicate the angle in which the line is located.
* Compare the coincident projections in this case with those of Art. 515.
REPRESENTATION OF LINES AND POINTS 77
QUESTIONS ON CHAPTER V
1. Discuss the point, the line, and the surface, and show how the material
object is made up of them.
2. What are the mathematical elements of a material object?
3. Why are the mathematical elements considered as concepts?
4. How is the outline of a material object determined?
5. What is meant by the graphical representation of mathematical con-
cepts?
6. What are the principal planes of projection?
7. What is the ground line?
8. How many dihedral angles are formed by the principal planes and
how are they numbered? Make a diagram.
9. How is a line orthographically projected on the principal planes?
10. How many projections are required to fix the line with reference to
the principal planes? Why?
11. Show how one of the principal planes is revolved so as to represent a
line in orthographic projection.
12. Assume a line in the first angle in orthographic projection and show
how the transfer is made to an oblique projection.
13. Under what conditions will a line pierce a plane of projection?
14. Assume a line that is inclined to both planes of projection and show
how the piercing points are determined orthographically. Give the reasoning of the operation.
15. Do the orthographic projections of a line represent the actual line in
space? Why?
16. Show a point in oblique projection and also the projecting lines to the
principal planes. Draw the corresponding diagram in orthographic projection showing clearly how one of the principal planes is re- volved.
17. Draw, in projection, a point lying in the horizontal plane; a point
lying in the vertical plane; a point lying in both planes. Observe nomenclature in indicating the points.
18. Indicate in what angle the points shown d
in Fig. 5-A are located.
19. Draw a line parallel to the horizontal t<*
plane but inclined to the vertical plane in orthographic projection. Transfer
diagram to oblique projection. ! , [b>
20. Draw a line parallel to the vertical plane ;6
but inclined to the horizontal plane in 'J'J
orthographic projection. Transfer dia- FIG. 5-A.
gram to oblique projection.
21. Draw a line parallel to both principal planes in orthographic pro-
jection. Transfer diagram to oblique projection.
22. When a line is parallel to both principal planes is it parallel to the
ground line? Why?
78
GEOMETRICAL PROBLEMS IN PROJECTION
23. In Fig. 5-B, make the oblique projection of the line represented.
24. Draw the projections of a line lying in the horizontal plane.
25. Draw the projections of a line lying in the vertical plane.
26. Draw the projections of a line lying in the ground line, observing the
nomenclature in the representation.
a
N,
FIG. 5-B.
FIG. 5-C.
27. In Fig. 5-C give the location of the lines represented by the ortho-
graphic projections. Construct the corresponding oblique projec- tions.
28. Show two lines, one perpendicular to each of the planes and also
draw a profile plane for each indicating the advantageous use in such cases.
29. When a line is perpendicular to the horizontal plane, how is it pro-
jected on that plane?
30. When a line is perpendicular to the vertical plane, why is the horizon-
tal projection equal to it in length?
31. Draw a line in the second angle, in orthographic projection. Transfer
the diagram to oblique projection.
32. Draw a line in the third angle, in orthographic projection. Transfer
the diagram to oblique projection.
33. Draw a line in the fourth angle, in orthographic projection. Transfer
the diagram to oblique projection.
34. In Fig. 5-D, specify in which angles each of the lines are situated.
g'
|
Y ! |
i r Y |
|
! i i ! i— ~J |
FIG. 5-D.
35. Draw lines similar to 5-D in orthographic projection and transfer
the diagrams to oblique projection.
36. Make the orthographic projection of a line with coincident projections
and show by a profile plane what it means. Take the case where the line passes through one point on the ground line and is per- pendicular to it, and the other case where it is inclined to the ground line through one point.
REPRESENTATION OF LINES AND POINTS 79
37. Locate a point in each angle and observe the method of indicating
them.
38. Make an oblique projection of the points given in Question 37.
39. Make an orthographic projection of points with coincident pro-
jections and show under what conditions they become coincident.
40. In what angles are coincident projections of points possible? Show
by profile.
41. Make the oblique projections of the lines shown in Fig. 68 of the
text. Use arrows to indicate the lines.
42. Why is it advantageous to use a profile plane, when the lines are
indefinite in extent, and lie in the profile plane?
CHAPTER VI REPRESENTATION OF PLANES
601. Traces of planes parallel to the principal planes.
Let Fig. 69 represent the two principal planes by HH and W intersecting in the ground line XY. Let, also, RR be another plane passing through the first and fourth angles and parallel to W. The plane RR intersects the horizontal plane HH in a line tr, which is called the trace of the plane RR. As RR is parallel
FIG. 69.
to W, the case is that of two parallel planes cut by a third plane, and, from solid geometry, it is known that their intersections are parallel. If the horizontal plane is revolved, by the usual method, into coincidence with the vertical plane, the resulting diagram as shown on the right will be the orthographic represen- tation of the trace of a plane which is parallel to the vertical plane.
If, as in Fig. 70, the plane is parallel to the horizontal plane, the condition of two parallel planes cut by a third plane again presents itself, the vertical plane of projection now being the cutting plane. The trace t'r' is then above the ground line, and, as before, parallel to it.
602. Traces of planes parallel to the ground line. When a plane is parallel to the ground line, and inclined to both planes
80
REPRESENTATION OF PLANES
81
of projection, it must intersect the principal planes. The line of intersection on each principal plane will be parallel to the ground
FIG. 70.
line because the given plane is parallel to the ground line and, hence, cannot intersect it. Fig. 71 shows this condition in oblique
R
FIG. 71.
and orthographic projection, in which tr is the horizontal trace and t'r' is the vertical trace.
A special case of this occurs if the plane passes through the ground line. Both traces then coincide with the ground line and the orthographic repre- sentation becomes indeterminate unless the profile plane is attached.
Fig. 72 is a profile and shows several planes passing through the ground line, each of which is now determined. It may be possible, however, to introduce a new hori- zontal plane H'H', parallel to the principal horizontal plane. Such an artifice will bring the case into that immediately preceding. It may then be shown as an ordinary
H
IT'
FIG. 72.
82
GEOMETRICAL PROBLEMS IN PROJECTION
orthographic projection, just as though the original horizontal plane were absent. The same would still be true if a new ver- tical plane were added instead of the horizontal plane, or, if an entirely new set of principal planes were chosen so that the new principal planes would be parallel to the old ones.
603. Traces of planes perpendicular to one of the prin= cipal planes. Fig. 73 shows a plane that is perpendicular to
the horizontal plane but inclined to the vertical plane. It may be imagined as a door in a wall of a room. The angle with the vertical plane (or wall in this case) can be changed at will by swinging it about the hinges, yet its plane always remains per-
FIG. 74.
pendicular to the horizontal plane (floor of the room). The inter- section with the vertical plane is perpendicular to the horizontal plane because it is the intersection of two planes (the given plane and the vertical plane), each of which is perpendicular to the horizontal. As a consequence the vertical trace is perpen- dicular to the ground line, because, when a line is perpendicular
REPRESENTATION OF PLANES
83
to a plane it is perpendicular to every line through its foot (from geometry). The orthographic representation of three distinct planes is shown on the right; the cases selected all show TV perpendicular to the ground line.
What is true regarding planes perpendicular to the horizontal plane is equally true for planes similarly related with respect to the vertical plane. Fig. 74 gives an illustration of such a case. Here, the horizontal trace is perpendicular to the ground line but the vertical trace may make any angle, at will, with the ground line.
604. Traces of planes perpendicular to both principal planes. When a plane is perpendicular to both principal planes,
-Y
FIG. 75.
its two traces are perpendicular to the ground line. Such a plane is a profile plane and -is shown in Fig. 75.
605. Traces of planes inclined to both principal planes.
It may be inferred from the preceding, that, if the plane is
FIG. 76.
inclined to both principal planes, neither trace can be perpen- dicular to the ground line. Fig. 76 shows such a case, the
84
GEOMETRICAL PROBLEMS IN PROJECTION
orthographic representation of which in projection is shown on the right.
Fig. 77 shows two cards, each slit half way, as indicated. These can be fitted together to represent the principal planes.
If at any point, a and a', on XY slits tt and ft' be made, it will be found on assembling the y cards that a third card can be
7^- Y inserted.
/( In the upper card, another
slit sV may be made, through the point a', with its direction parallel to tt. As in the previous
FIG. 77.
instance, a card can again be inserted. This case is of in- terest, however, because both traces become coincident on the revolution of the horizontal plane, and fall as one straight line tTt', as shown in projection in Fig. 77.
606. Traces of planes intersecting the ground line. It
must have been observed in the cases where the given plane is inclined to the ground line that both traces pass through the same point on the ground line. This becomes further evident when it is considered that the ground line can intersect a plane at but one point, if at all. This one point lies in the ground line, and, hence, it has coincident projections (509).
607. Plane fixed in space by its traces. Two intersect- ing lines determine a plane (from solid geometry). Hence, the two traces of a plane fix a plane with reference to the principal planes because the traces meet at the same point on the ground line. If the diagram is such that the traces do not intersect in the ground line within the limits of the drawing, it is assumed that they will do so if sufficiently produced. The limiting posi- tion of a plane whose traces cannot be made to intersect the ground line is evidently a plane parallel to it. This plane is then parallel to the ground line and its traces must also be parallel to it (602).
608. Transfer of diagrams from orthographic to oblique projection. Let the right-hand diagram of Fig.
REPRESENTATION OF PLANES
85
78 represent a plane in orthographic projection. Through any point S, on the ground line, pass a profile plane sSs', inter- secting the two traces tT and Tt' at points o and p. To transfer the orthographic to the oblique projection, lay off the principal planes HH and W as shown, intersecting in the ground line
FIG. 78.
XY. Lay off any point S on the ground line in the oblique pro- jection and then make So and Sp of the orthographic equal to the similarly lettered lines in the oblique projection. The profile plane sSs' is therefore determined in the oblique projection. Make TS of one diagram equal to the TS of the other, and com-
FIG. 79.
plete by drawing the traces through T, o and T, p. To increase the clarity of the diagram, a rectangular plane may be shown as though it passes through the principal planes. Fig. 78 has this plane added.
A case where the two traces are coincident is shown in Fig. 79. Two profile planes are required, but only one trace of each
86 GEOMETRICAL PROBLEMS IN PROJECTION
is needed. All the necessary construction lines are shown in this diagram.
609. Traces of planes in all angles. As planes are indef- inite in extent, so are their traces; and, therefore, the traces are .not limited to any one angle. In the discussion of most problems, it may be possible to choose the principal planes so as to limit the discussion to one angle — usually the first. The advantage to be gained thereby is the greater clarity of the diagram, as then the number of construction lines is reduced to a minimum. Third angle projection may also be used, but the transfer to oblique projection is undesirable. Second and fourth angles are avoided because the projections overlap (317, 514, 516). -
Fig. 80 shows the complete traces of a given plane T. The
\
_,. x-r/' ~\7~ ^^ \.* * / *
A'
/ X T V XT/ V T 1-
X \ ^ ~/i 2 8\ /4\ ^
\ A t\'t r*
FIG. 80. FIG. 81.
full lines indicate the traces in the first angle; the dotted lines show the continuation in the remaining three angles. Fig. 81 represents each quadrant separately for the same plane that is shown in Fig. 80. The appended numbers indicate the angle to which the given plane is limited.
610. Projecting plane of lines. It is now evident that the finding of the projection of a line is nothing more nor less than the finding of the trace of its projecting piano. The two perpendiculars from a line to the plane of projection are necessarily parallel and therefore determine a plane. This plane is the projecting plane of the line and its intersection (or trace) with the plane of projection contains the projection of the line.
Manifestly, any number of lines contained in this projecting plane would have the same projection on any one of the principal planes, and that projection therefore does not fix the line in space.
KEPRESENTATION OF PLANES
87
If the projection on the corresponding plane of projection be taken into consideration, the projecting planes will intersect, and this
FIG. 82.
intersection will be the given line in space, the point in question.
Fig. 82 illustrates
QUESTIONS ON CHAPTER VI
1. What is the trace of a plane?
2. Draw a plane parallel to the vertical plane passing through the first
and fourth angles, and show the resulting trace. Make diagram in oblique and orthographic projection.
3. Take the same plane of Question 2 and show it passing through the
second and third angles.
4. Show how a plane is represented when it is parallel to the horizontal
plane and passes through the first and second angles. Make diagram in oblique and orthographic projection.
5. Take the same plane of Question 4 and show the trace of the plane
when it passes through the third and fourth angles.
6. Show how a plane is represented when it is parallel to the ground
line and passes through the first angle.
7. How is a plane represented when it is parallel to the ground line and
passes through the second angle? Third angle? Fourth Angle?
8. Show how a plane passing through the ground line is indeterminate
in orthographic projection.
9. When a plane passes through the ground line show how the profile
plane might be used to advantage in representing the plane.
10. When a plane passes through the ground line, show how an auxiliary
principal plane may be used to obtain determinate traces.
11. When a plane is perpendicular to the horizontal plane and inclined to
the vertical plane, show how this is represented orthographically. Make, also, the oblique projection of it.
88 GEOMETRICAL PROBLEMS IN PROJECTION
12. When a plane is perpendicular to the vertical plane and inclined to
the horizontal plane show how this is represented orthographically. Make, also, the oblique projection of it.
13. When a plane is perpendicular to both principal planes draw the
orthographic traces of it.
14. Why is the ground line perpendicular to both traces in Question 13?
Is this plane a profile plane?
15. When a plane is inclined to the ground line show how the traces are
represented.
16. Why do both traces of a plane intersect the ground line at a point
when the plane is inclined to it?
17. Why do the traces fix the plane with reference to the principal planes?
FIG. 6-A.
18. When a plane is parallel to the ground line, why are the traces of the
plane parallel to it?
19. Show an oblique plane in all four angles of projection (use only the
limited portion in one angle and use the same plane as in the illustration).
20. In what angles are the planes whose traces are shown in Fig. 6-A?
t'.>
.-. /
,x-
t FIG. 6-B. FIG. 6-C.
21. Show how a line in space and its projecting perpendiculars determine
a plane which is the projecting plane of the line. Is the projection of the line the trace of the projecting plane of the line?
22. Is it possible to have two separate lines whose projections are coin-
cident on one plane? How are such lines determined?
23. Given the traces of a plane in orthographic projection as shown in
Fig. 6-B construct the oblique projection of it.
24. Construct the oblique projection of a plane having coincident projec- tions (Fig. 6-C).
CHAPTER VII ELEMENTARY CONSIDERATIONS OF LINES AND PLANES
701. Projection of lines parallel in space. When two lines in space are parallel, their projecting planes are parallel, and their intersection with any third plane will result in parallel lines. If this third plane be a plane of projection, then the traces of the two projecting planes will result in parallel projections.
Fig. 83 shows two lines, AB and CD in space. The lines are
FIG. 83.
FIG. 84.
shown by their horizontal projections ab and cd, which are parallel to each other, and by their vertical projections a'b' and c'd', which are also parallel to each other. A perfectly general case is represented pictorially on a single plane of projection in Fig. 84.
If two
702. Projection of lines intersecting in space.
lines in space intersect, their projections intersect, because the two lines in space must meet in a point. Further, the projection of this point must be common to the projections of the lines.
In Fig. 85 two such lines, AB and CB are shown, represented, as usual, by their pro- jections. O is the intersection in space, FIG. 85. indicated by its horizontal and vertical projections o and o',
90
GEOMETRICAL PROBLEMS IN PROJECTION
respectively. EF and GH (Fig. 86) are two other lines, chosen to show how in the horizontal plane of projection, the two pro- jections may coincide, because the plane of the two lines happens also to be the horizontal projecting plane. The case is not inde- terminate, however, as the vertical projection locates the point M in space. The reverse of this is also true, that is, the vertical
FIG. 86.
FIG. 87.
FIG. 88.
instead of the horizontal projections may be coincident. General cases of the above are represented pictorially in Figs. 87 and 88. Should the horizontal and the vertical projections be coin- cident, the lines do not intersect but are themselves coincident in space and thus form only one line.
703. Projection of lines not intersecting in space.*
There are two possible cases of lines that do not intersect in space.
FIG. 90.
The case in which the lines are parallel to each other has previously been discussed (701). If the two lines cannot be made to lie in the same plane, they will pass each other without intersecting. Hence, if in one plane of projection, the projections intersect, they cannot do so in the corresponding projection.
This fact is depicted in Fig. 89. AB and CD are the two * Called skew lines.
CONSIDERATIONS OF LINES AND PLANES
91
lines in space. Two distinct points E and F, on the lines are shown in the horizontal projection as e and f; their vertical projections are, however, coincident. Similarly, G and H are also two distinct points on the lines, shown as g' and h' in the vertical projection, and coincident, as g and h in the horizontal projection. The pictorial representation is given in Fig. 90.
704. Projection of lines in oblique planes. When a third plane is inclined to the principal planes, it cuts them in lines of intersection, known as traces (601). Any line, when inclined to the principal planes will pierce them in a point. Hence, if a plane is to contain a given line, the piercing points of the line must lie in the traces of the plane. Viewing this in another way, a plane may be passed through two parallel or two inter- secting lines. On the resulting plane, any number of lines may be drawn, intersecting the given pair. Hence, an inclined line must pass through the trace, if it is contained in the plane.*
Fig. 91 shows a plane tit' indicated by its horizontal trace tT and its vertical trace TV. It is required to draw a line AB in this plane. Suppose the hori- zontal piercing point is assumed at b, its corresponding vertical projection will lie in the ground line (509). Also, if its vertical piercing point is assumed at a', its corresponding horizontal pro- jection will be a. With two horizontal projections of given points on a line and two vertical
projections, the direction of the line is determined, for, if the horizontal and vertical projecting planes be erected, their inter- section determines the given line (610).f
As a check on the correctness of the above, another line CD may be assumed. If the two lines lie in the same plane and are not parallel, they must intersect. This point of intersection
* When the plane is parallel to the ground line, a line in this plane parallel to the principal planes cannot pierce in the traces of the plane. See Art. 602.
f It must be remembered that the principal planes must be at right angles to each other to determine this intersection. In the revolved position, the planes would not intersect in the required line.
pIG
92
GEOMETRICAL PROBLEMS IN PROJECTION
is M shown horizontally projected at m and vertically, at m'. The line joining the two is perpendicular to the ground line (508). This is illustrated in oblique projection in Fig. 92.
It should be here noted that if the line is to be contained by the plane, only the direction of one projection can be assumed, and that the corresponding projection must be found by the principles so far developed.
A converse of this problem is to draw a plane so that it shall contain a given line. As an unlimited number of planes can be passed through any given line, the direction of the traces is not fixed. Suppose AB (Fig. 91) is the given line, then through the horizontal piercing point b draw any trace tT, and from T where this line intersects the ground line, draw Tt', through the vertical
FIG. 92.
piercing point a'. The point T may be located anywhere along XY. All of these planes will contain the line AB, if their traces pass through the piercing points of the line.
Still another feature of Fig. 91 is the fact that from it can be proved that two intersecting lines determine a plane. If AB and CD be the two lines intersecting at M, their horizontal piercing points are b and c and their vertical piercing points are a! and d' respectively. Two points fix the direction of a line, and, hence, the direction of the traces is fixed; tT is drawn through cb and Tt' is drawn through a'd'. The check lies in the fact that both traces intersect at one point T on the ground line (606).
705. Projection of lines parallel to the principal planes and lying in an oblique plane. If a line is horizontal, it is parallel to the horizontal plane, and its vertical projection must be
CONSIDERATIONS OF LINES AND PLANES
93
parallel to the ground line (511). If this line lies in an oblique plane, it can have only one piercing point and that with the vertical plane, since it is parallel to the horizontal. Thus, in Fig. 93, let tit' be the given oblique plane, represented, of course, by its traces. The given line is AB and a'b' is its vertical projection, the piercing point being at a'. The corresponding horizontal projection of a' is a. When a line is horizontal, as in this case, it may be considered as being cut from the plane fit by a hori- zontal plane, and, as such, must be parallel to the principal horizontal plane. These two parallel horizontal planes are cut by the given oblique plane fit' and, from geometry, their lines of intersection are parallel. Hence, the horizontal projection
. FIG. 93.
of the horizontal line must be parallel to the horizontal trace, of the given plane because parallel lines in space have parallel projections. Accordingly, from a, draw ab, parallel to Tt, and the horizontal line AB is thus shown by its projections.
A line, parallel to the vertical plane, drawn in an oblique plane follows the same analysis, and differs only in an interchange of the operation. That is to say, the horizontal projection is then parallel to the ground line and pierces in a point on the horizontal trace of the given plane; its vertical projection must be parallel to the vertical trace of the given plane. In Fig. 93, CD is a line parallel to the vertical plane and cd is the horizontal projection, piercing the horizontal plane at c vertically projected at c'; c'd' is, therefore, the required vertical projection.
94
GEOMETRICAL PROBLEMS IN PROJECTION
A check on the problem lies in the fact that these two lines must intersect because they lie in the same plane by hypothesis. The point of intersection M is shown as m in the horizontal
FIG. 94.,
projection, and as m' in the vertical projection. The line joining these points is perpendicular to the ground line. The oblique projection of this is shown in Fig. 94.
Fig. 95 is a still further step in this problem. Three lines
FIG. 95.
AB, CD, and EF are shown, all being in the plane tTt'. AB is parallel to the vertical plane, CD is parallel to the horizontal plane and EF is any other line. It will be observed that the three lines intersect so as to form a triangle MNO shown by having the area shaded in both its projections.
CONSIDERATIONS OF LINES AND PLANES
95
706. Projections of lines perpendicular to given planes.
If a line is perpendicular to a given plane, the projections of the line are perpendicular to the corresponding traces of the plane.
Let, in Fig. 96, LL be any plane and MM any other plane intersecting it in the trace tr. Also, let AB be any line, perpen- dicular to MM, and ab be the projection of AB on LL. It is desired to show that the projection ab is perpendicular to the trace tr. Any plane through AB is perpendicular to the plane MM, because it contains a line perpendicular to the plane by hypothesis. Also, any plane through Bb, a perpendicular to the plane LL, is perpendicular to LL. Hence, any plane contain-
FIG. 96.
ing both lines (it can do so because they intersect at B) will be perpendicular both to LL and MM. As the plane through ABb is perpendicular to the two planes LL and MM, it is also perpendicular to a line common to the two planes, such as tr. Thus, ab is perpendicular to tr. In fact, any line perpendicular to a plane will have its projection on any other plane perpendicular to the trace of the plane, because, instead of LL being assumed as the plane of projection and BA a perpendicular to another plane MM, the conditions may be reversed and MM be assumed as the plane of projection and Bb the perpendicular.
The converse of this is also true. If Ao be assumed the pro- jection of Bb on MM, a plane passed through the line Ao per- pendicular to the trace tr will contain the lines AB and Bb, where
96
GEOMETEICAL PROBLEMS IN PROJECTION
Bb is the given line and AB then the projecting perpendicular to the plane MM.
707. Revolution of a point about a line. Frequently it is desirable to know the relation of a point with respect to a line, for, if the line and point are given by their projections, the true relation may not be apparent. To do this, revolve the point about the line so that a plane through the point and the line will either coincide or be parallel to the plane of projection, then they will be projected in their true relation to each other. The actual distance between the point and the line is then shown as the perpendicular distance from the point to the line.
Consider the diagram in Fig. 97* and assume that the point A is to be revolved about the point B. The projection of A on
B « a"
FIG. 97/
the line Oa" is a and, to an observer looking down from above the line Oa", the apparent distance between A and B is aB. Oa' is the apparent distance of A from B to an observer, looking orthographically, from the right at a plane perpendicular to the plane of the paper through Oa'. Neither projection gives the true relation between A and B from a single projection. If the point A is revolved about B as an axis, with BA as a radius, until it coincides with the line Oa", A will either be found at a" or at a'", depending upon the direction of rotation. During the revolution, A always remains in the plane of the paper and de- scribes a circle, the plane of which is perpendicular to the axis, through B, the centre of the circle.
* This diagram is a profile plane of the given point and of the principal
CONSIDERATIONS OF LINES AND PLANES
97
A more general case is shown in Fig. 98 where BB is an axis lying in a plane, and A is any point in space not in the plane containing BB. If A be revolved about BB as an axis, it will describe a circle, the plane of which will be perpendicular to the axis. In other words, A will fall somewhere on a line Ba" perpendicular to BB. The line Ba" must be perpendicular to BB because it is the trace of a perpendicular plane (706). This point is a" and Ba" is equal to the radius BA. Contrast this
FIG. 98.
with a, the orthographic projection of A on the plane containing BB. Evidently, then, a is at a lesser distance from BB than a". Indeed, BA equals Ba" and is equal to the hypothenuse of a right triangle, whose base is the perpendicular distance of the projection of the point from the axis, and whose altitude is the distance of the point above the plane containing the line. The angle AaB is a right angle, because a is the orthographic projection of A.
QUESTIONS ON CHAPTER VII
1. If two lines in space are parallel, prove that their projections on any
plane are parallel.
2. When two lines in space intersect, prove that their projections, on
any plane, intersect.
3. Draw two lines in space that are not parallel and still do not intersect.
4. Show a case of two non-intersecting lines whose horizontal projections
are parallel to each other and whose vertical projections intersect. Show also, that the horizontal projecting planes of these lines are parallel.
98 'GEOMETRICAL PROBLEMS IN PROJECTION
5. Make an oblique projection of the lines considered in Question 4.
6. Prove that when a line lies in a plane it must pass through the traces
of the plane.
7. Given one projection of a line in a plane, find the corresponding pro-
• jection.
8. Given a plane, draw intersecting lines in the plane and show by the
construction that the point of intersection satisfies the ortho- graphic representation of a point.
9. Show how two intersecting lines determine a plane by aid of an
oblique projection.
10. In a given oblique plane, draw a line parallel to the horizontal plane
and show by the construction that this line pierces the vertical plane only. Give reasons for the construction.
11. In Question 9, draw another line parallel to the vertical plane and
show that this second line intersects the first in a point.
12. Given an oblique plane, draw three lines; one parallel to the hori-
zontal plane, one parallel to the vertical plane and the last inclined to both planes. Show, by the construction, that the three lines form a triangle (or meet in a point in an exceptional case).
13. Prove that when a line is perpendicular to a plane the projection of
this line on any other plane is perpendicular to the trace of the plane. Show the general case and also an example in orthographic projection.
14. A line lies in a given plane and a point is situated outside of the
plane. Show how the point is revolved about the line until it is contained in the plane.
15. Prove that the point, while revolving about the line, describes a
circle the plane of which is perpendicular to the axis (the line about which it revolves).
CHAPTER VIII
V
PROBLEMS INVOLVING THE POINT, THE LINE, AND THE PLANE
801. Introductory. A thorough knowledge of the preceding three chapters is necessary in order to apply the principles, there developed, in the solution of certain problems. The commercial application of these problems frequently calls for extended knowl- edge in special fields of engineering, and for this reason, the application, in general, has been avoided.
Countless problems of a commercial nature may be used as illustrations. All of these indicate, in various ways, the im- portance of the subject. In general, the commercial problem may always be reduced* to one containing the mathematical essentials (reduced to points, lines and planes). The solution, then, may be accomplished by the methods to be shown sub- sequently.
802. Solution of problems. In the solution of the following problems, three distinct steps may be noted: the statement, the analysis, and the construction.
The statement of the problem gives a clear account of what is to be done and includes the necessary data.
The analysis entails a review of the principles involved, and proceeds, logically, from the given data to the required conclusion. On completion of the analysis* the problem is solved to all intents and purposes.
The construction is the graphical presentation of the analysis. It is by means of a drawing and its description that the giver data is associated with its solution. It may be emphasized again that the drawings are made orthographically, and that the actual points, lines and planes are to be imagined.
By a slight change in the assumed data, the resultant con- struction may appear widely different from the. diagrams in the book. Here, then, is an opportunity to make several con- structions for the several assumptions, and to prove that all
99
100 GEOMETRICAL PROBLEMS IN PROJECTION
follow the general analysis. The simpler constructions might be taken and transformed from orthographic to oblique projec- tion; this will show the projections as well as the actual points, lines and planes in space. By performing this transformation (from orthographic to oblique projection), the student will soon be able to picture the entire problem in space, without recourse to any diagrams.
To bring forcibly to the student's attention the difference between the analysis and the construction, it may be well to note that the analysis gives the reasoning in its most general terms, while the construction is specific, in so far as it takes the assumed data and gives the solution for that particular case only.
803. Problem 1. To draw a line through a given point, parallel to a given line.
Analysis. If two lines in space are parallel, their projecting planes are parallel and their intersection with the principal planes are parallel (701). Hence, through the projections of the given point, draw lines parallel to the projections of the given lines. Construction. Let AB, Fig. 99, be the given line in space, d, represented by its horizontal pro- jection ab and its vertical projection a'b'. Further, let G be the given point, similarly represented by its horizontal projection g and its ver- tical projection g'. Through the horizontal projection g, draw cd parallel to ab and through g', draw FlG 99> c'd' parallel to a'b'.
As the length of the line is not
specified, any line that satisfies the condition of parallelism is permissible. Therefore, CD is the line in space that is parallel to AB through the point G. A pictorial representation of this is shown in Fig. 84.
804. Problem 2. To draw a line intersecting a given line at a given point.
Analysis. If two lines in space intersect, they intersect in a point that is common to the two lines. Therefore, their pro- jecting planes will intersect in a line which is the projecting line of the given point (702). Hence, through the projections of the
THE POINT, THE LINE, AND THE PLANE
101
FIG. 100.
given point, draw any lines, intersecting the projections of the given lines.
Construction. Let AB, Fig. 100, be the given line, shown horizontally projected as ab and ver- tically projected as a'b'. Let, also, G be the given point, situated on the line AB. As no direction is specified for the intersecting line, draw any line cd through the horizontal projec- tion g and this will be the horizontal projection of the required line. Similarly, any other line c'd' through g' will be the vertical projection of the required line. Hence, CD and
AB are two lines in space, intersecting at the point G. The pictorial representation of this case is depicted in Figs. 87 and 88.
805. Problem 3. To find where a given line pierces the principal planes.
Analysis. If a line is oblique to the principal planes, it will pierce each of these in a point, the corresponding projection of which will be in the ground line. Hence, a piercing point in any principal plane must be on the projection of the line in that plane. It must also be on a perpendicular erected at the point where the corresponding projection crosses the ground line. Therefore, the required piercing point is at their intersection. Construction. Let AB, Fig. 101, be a limited portion of an indefinite line, shown by its horizontal projection ab and its vertical projection a'b'. The portion chosen AB will not pierce the principal planes, but its continua- tion, in both directions, will. Prolong the vertical projection a'b' to c' and at c', erect a perpendicular to XY, the ground line, and continue it, until it intersects the prolonga- tion of ab at c. This will be the horizontal piercing point. In the same way, prolong ab to d, at d, erect a perpendicular to the ground line as dd', the intersection of which with the prolonga- tion of a'b', at d' will give the vertical piercing point. Hence,
FIG. 101.
102
GEOMETRICAL PROBLEMS IN PROJECTION
if CD be considered as the line, it will pierce the horizontal plane of projection at c and the vertical plane of projection at d'. A pic- torial representation in oblique projection is shown in Fig. 52.
806. Problem 4. To pass an oblique plane, through a given oblique line.
Analysis. If a plane is oblique to the principal planes, it must intersect the ground line at a point (606); and if it is to contain a line, the piercing points of the line must lie in the traces of the plane (704). Therefore, to draw an oblique plane containing a given oblique line, join the piercing points of the line with any point of the ground line and the resulting lines will be the traces of the required plane.
Construction. Let AB, Fig. 102, be the given line. This
FIG. 102.
FIG. 103.
line pierces the horizontal plane at a and the vertical plane at b'. Assume any point T on the ground line XY, and join T with a and also with b'. Ta and Tb' are the required traces, indicated, as usual, by tTt'. Thus, the plane T contains the line AB. The construction in oblique projection is given in Fig. 103.
807. Special cases of the preceding problem. If the
line is parallel to both planes of projection, the traces of the plane will be parallel to the ground line (602), and the profile plane may be advantageously used in the drawing.
If the line is parallel to only one of the principal planes, join the one piercing point with any point on the ground line, which results in one trace of the required plane. Through the point on the ground line, draw the corresponding trace, parallel to the
THE POINT, THE LINE, AND THE PLANE
103
projection of the line in the plane containing this trace. The case is evidently that considered in Art. 705.
808. Problem 5. To pass an oblique plane, through a given point.
Analysis. If the oblique plane is to contain a given point, it will also contain a line through the point. Hence, through the given point, draw an oblique line and find the piercing points of this line on the principal planes. Join these piercing points with the ground line and the result will be the required traces of the plane.
Construction. Let C, Fig. 104, be the required point, and let AB be a line drawn through this point. AB pierces the hori-
FIG. 104.
FIG. 105.
b H
zontal plane at a and the vertical plane at b'. Join a and b' with any assumed point T and tTt' will be the required traces. The plane T, contains the point C, because it contains a line AB through the point C. Fig. 105 shows this pictorially.
NOTE. An infinite number of planes may be passed througri the point, hence the point T was assumed. It might have been assumed on the opposite side of the point and would still have contained the given point, or the auxiliary line through the point.
Also, it is possible to draw a line through the given point parallel to the ground line. A profile construction will then be of service (602).
Again, a perpendicular line may be drawn through the given point and a plane be passed so that it is perpendicular to one or both planes of projection (603, 604).
104
GEOMETRICAL PROBLEMS IN PROJECTION
809. Problem 6. To find the intersection of two planes, oblique to each other and to the principal planes.
Analysis. If two planes are oblique to each other, they intersect in a line. Any line in a plane must pass through the traces of the plane. As the line of intersection is common to the two planes, it must pass through the traces of both planes and hence it passes through the intersection of these traces.
Construction. Let T and S, Fig. 106, be the given planes. The horizontal piercing point of their line of intersection is at b,
FIG. 106.
FIG. 107.
vertically projected at b'; the vertical piercing point is at a', horizontally projected at a. Join ab and aV, as they are the projections of the line AB, which is the intersection of the planes T and S. In oblique projection, this appears as shown in Fig. 107.
810. Special case of the preced= ing problem. If the two planes are chosen so that the traces in one plane do not intersect within the limits of the drawing, then draw an auxiliary plane R (Fig. 108) and find the inter- section, AB as shown. From c' draw c'd', parallel to a'b' and from c, draw cd, parallel to ab. The line of inter- section of the given planes is thus deter- mined.
FIG. 108.
811. Problem 7. To find the corresponding projection of a given point lying in a given oblique plane, when one of its pro- jections is given.
Analysis. If a line lies in a given plane and also contains
THE POINT, THE LINE, AND THE PLANE
105
a given point, the projections of this line will also contain the projections of the given point. Hence, through the projection of the given point, draw the projection of a line lying in the given plane. Then find the corresponding projection of the line. The required projection of the given point will lie on the inter- section of a perpendicular to the ground line, through the given projection of the point with the corresponding projection of the, line.
Construction. Let tTt', Fig. 109, be the traces of the given plane and c the horizontal projection of the given point. Draw ab, the horizontal projection of a line in the plane, through c the horizontal projection of the given point. The horizontal
FIG. 109.
FIG. 110.
piercing point of the line AB is in the trace Tt at a, and its cor- responding projection lies in the ground line at a'. Further, the vertical piercing point lies on a perpendicular to the ground line from the point b and also in the trace Tt', hence, it is at b' and a'b' is thus the corresponding projection of the line ab. The required projection of the given point lies on a line through c, perpendicular to the ground line, and also on a'b'; hence, it is at their intersection c'. The point C, in space, is con- tained in the plane T, and c and c', are corresponding pro- jections. The oblique projection of this problem is given in Fig. 110.
812. Special case of the preceding problem. The point in the above problem was purposely chosen in the first angle, in order to obtain a simple case. It may be located any where ,.
106 GEOMETRICAL PROBLEMS IX PROJECTION
however, because the planes are indefinite in extent. For instance, in Fig. Ill the vertical projection is selected below the ground line. However, a single projection does not locate a point in space. It may be assumed as lying either in the third or fourth angles (515). Subsequent operations are dependent upon the
angle in which the point is chosen. Assume, for instance, that the point is in the fourth angle; the traces of the given plane must then also be assumed as being in the fourth angle. Thus, with the given plane T, Tt is the horizontal trace, and Tt'" is the vertical trace (609). In completing pIG in the construction by the usual method,
let c' be the assumed vertical pro- jection, and through it, draw a'b' as the vertical projection of the assumed line through the given point and lying in the given plane. This line pierces the horizontal plane at b and the vertical plane at a'. Hence, ab and a'b' are the corresponding projections of the line AB in space which is situated in the fourth angle. The corresponding horizontal projection of c' is c, and thus the point C in space is determined.
Had the point been assumed in the third angle, then the traces Tt" and Tt'" would have been the ones to use, Tt" being the horizontal, and Tt'" the vertical trace. The construction would then, in general, be the same as the previous.
813. Problem 8. To draw a plane which contains a given point and is parallel to a given plane.
Analysis. The traces of the required plane must be parallel to the traces of the given plane. A line may be drawn through the given point parallel to an assumed line in the given plane. This line will then pierce the principal planes in the traces of the required plane. Hence, in the given plane, draw any line. Through the given point, draw a line parallel to it, and find the piercing points of this line on the principal planes. Through these piercing points draw the traces of the required plane parallel to the corresponding traces of the given plane. The plane so drawn is parallel to the given plane.
Construction. Let T, Fig. 112 be the given plane, and G
THE POINT, THE LINE, AND THE PLANE
107
the given point. In the plane T, draw any line CD as shown by cd and c'd' its horizontal and vertical projections respectively. Through G, draw AB, parallel to CD and ab and a'b' will be the projections of this line. The piercing points are a and b' on the horizontal and vertical planes respectively. Draw b'S parallel to t'T and aS parallel to tT, then sSs' will be the traces
FIG. 113.
of the required plane, parallel to the plane T and containing a given point G. A check on the accuracy of the construction is furnished by having the two traces meet at S. Also, only one plane will satisfy these conditions because the point S cannot be selected at random. This construction is represented picto- rially in Fig. 113.
814. Problem 9. To draw a line perpendicular to a given plane through a given point.
Analysis. If a line is perpendicular to a given plane the projections of the line are perpendicular to the corresponding traces of the plane. Hence, draw through each projection of the given point, a line perpendicular to the corresponding trace (706).
Construction. Let T, Fig. 114, be the given plane, and C, the given point. Through c, the horizontal projection of the given point, draw ab, perpendicular to Tt; and through c', the vertical pro- jection of the given point, draw a'b', perpendicular to Tt'.
FIG. 114.
108
GEOMETRICAL PROBLEMS IN PROJECTION
Thus, AB is perpendicular to the plane T. of this problem is given in Fig. 115.
An oblique projection
815. Special case of the preceding problem. If the
point is chosen in the third angle, then it must be observed that Tt" is the horizontal trace (Fig. 116) and Tt'" is the vertical trace. Again, the line AB is drawn perpendicular to T, by making a'b', through c', perpendicular to TV", and ab, through c, perpendicular to Tt".
816. Problem 10. To draw a plane through a given point perpendicular to a given line.
Analysis. The traces of the required plane must be perpen-
a' >*
FIG. 115.
FIG. 116.
dicular to the corresponding projections of the given line (706), Through one projection of the given point, draw an auxiliary line parallel to the trace; the corresponding projection of this line will be parallel to the ground line, because it is a line parallel to that plane in which its projection is parallel to the trace (705). Find the piercing point of this auxiliary line, and through this point, draw a line perpendicular to the corre- sponding projection of the given line. Where this intersects the ground line, draw another line, perpendicular to the corresponding projection of the given line. The traces are thus determined.
Construction. Let, in Fig. 117, AB be the given line, and C the given point. For convenience, assume a horizontal line in the plane as the auxiliary line. Then, through c, draw cd,
THE POINT, THE LINE, AND THE PLANE
109
perpendicular to ab, and through c', draw c'd', parallel to the ground line. The piercing point of this auxiliary line is d', and only one point on either trace is required. Hence, through d', draw TV, perpendicular to a'b', and from T, draw Tt, per- pendicular to ab. T is, therefore, the required plane, and Tt must be parallel to cd because both must be perpendicular to ab. Fig. 118 shows a pictorial representation of the same problem.
NOTE. Instead of having assumed a line parallel to the horizontal plane, a line parallel to the vertical plane might have been assumed. In the latter case, the vertical projection would have been perpendicular to the vertical projection of the line,
FIG. 118.
and the horizontal projection would therefore have been parallel to the ground line. Also, a point in the horizontal plane would have fixed the traces, instead of a point in the vertical plane as shown in the problem.
817. Problem 11. To pass a plane through three given points not in the same straight line.
Analysis. If two of the points be joined by a line, a plane may be passed through this line and revolved so that it contains the third point. In this position, the plane will contain a line joining the third point with any point on the first line. Hence, join two points by a line, and from any point on this line, draw another line, through the remaining point. Find the piercing points of these two lines, and thus establish the traces of the required plane.
110
GEOMETRICAL PROBLEMS IN PROJECTION
Construction. Let AB and C, in Fig. 119, be the three given points. Join A and B and find where this line pierces the prin- cipal planes at d and e'. Assume any point H, on the line AB, and join H and C; this line pierces the principal planes at g and f. Join dg and f e' and the traces obtained are those of the required plane. A check on the accuracy of the work is furnished by the fact that both traces must meet at one point on the ground line, as shown at T. Hence, the plane T contains the points A, B and C. Fig. 120 is an oblique projection of this problem.
NOTE. In the construction of this and other problems, it may be desirable . to work the problem backwards in order to
FIG. 119.
FIG. 120.
obtain a simpler drawing. It is quite difficult to select three points of a plane, at random, so that the traces of the plane shall meet the ground line within the limits of the drawing. In working the problem backwards, the traces are first assumed, then, any two distinct lines are drawn in the plane, and, finally, three points are selected on the two assumed lines of the plane. It is good practice, however, to assume three random points and proceed with the problem in the regular way. Under these conditions, the piercing points are liable to be in any angle, and, as such, furnish practice in angles other than the first.
818. Problem 12. To revolve a given point, not in the principal planes, about a line lying in one of the principal planes.
Analysis. If a point revolves about a line, it describes a circle, the plane of which is perpendicular to the axis of revolu-
THE POINT, THE LINE, AND THE PLANE 111
tion. As the given line is the axis, the point will fall somewhere in the trace of a plane, through the point, perpendicular to the axis. The radius of the circle is the perpendicular distance from the point to the line; and is equal to the hypothenuse of a tri- angle, whose base is the distance from the projection of the point to line in the plane, and whose altitude is the distance of the corresponding projection of the point from the plane containing the line (707).
Construction. Assume that the given line AB, Fig. 121, lies in the horizontal plane and therefore is its own projection, ab, in that plane; its corresponding projection is a'b' and lies in the ground line. Also, let C be the ,
given point, shown by its projections c CK
and c'. Through c, draw cp, perpendic- ular to the line ab; cp is then the trace v «_' &' of the plane of the revolving point and
the revolved position of the point will fall
somewhere along this line. The radius of
p\
\\
the circle is found by making an auxiliary \c"
view, in which c'o is the distance of the pIG 12i.
point above the horizontal plane, and
oq = cp is the distance of the horizontal projection of the given point from the axis. Hence, c'q is the radius of the circle, and, therefore, lay off pc" = c'q. The revolved position of the point C in space is, therefore, c". The distance pc" might have been revolved in the opposite direction and thus would have fallen on the opposite side of the axis. This is immaterial, however, as the point is always revolved so as to make a clear diagram. NOTE. In the construction of this problem, the line was assumed as lying in the horizontal plane. It might have been assumed as lying in the vertical plane, and, in this case, .the operations would have been identical, the only difference being that the trace of the plane containing the path of the point would then lie in the vertical plane. The auxiliary diagram for deter- mining the radius of the circle may be constructed below the ground line, or, if desirable, in an entirely separate diagram.
819. Problem 13. To find the true distance between two
points in space as given by their projections. First method.
Analysis. The true distance is equal to the length of the
112
GEOMETRICAL PROBLEMS IN PROJECTION
line joining the two points. If, then, one projecting plane of the line be revolved until it is parallel to the corresponding plane of projection, the line will be shown in its true length on the plane to which it is parallel.
Construction. Case 1. When both points are above the plane of projection. — Let AB, in Fig. 122, be the given line. For convenience, revolve the horizontal projecting plane about the projecting perpendicular from the point A on the line. The point B will describe a circle, the plane of which is perpendicular to the axis about which it revolves. As the plane of the circle (or arc) is parallel to the horizontal plane, it is projected as the arc be. The corresponding projection is bV, because its
&' c
FIG. 122.
FIG. 123.
plane is perpendicular to the vertical plane. In the position ac, the projecting plane of AB is parallel to the vertical plane, and c' is the vertical projection of the point b, when so revolved. Hence, a'c' is the true length of the line AB in space.
820. Case 2. When the points are situated on opposite sides of the principal plane. — Let AB, in Fig. 123, be the given line. Revolve the horizontal projecting plane of the line about the horizontal projecting perpendicular from the point A on the line. The point B will describe an arc which is horizontally pro- jected as be and vertically projected as b'c'. The ultimate position of b' after revolution is at c', whereas a' remains fixed. Hence, a'c' is the true length of the line AB.
THE POINT, THE LINE, AND THE PLANE
113
821. Problem 13. To find the true distance between two points in space as given by their projections. Second method.
Analysis. The true distance is equal to the length of a line joining the two points. If a plane be passed through the line and revolved about the trace into one of the principal planes, the distance between the points will remain unchanged, and in its revolved position, the line will be shown in its true length.
Construction. Case 1. When both points are above the plane of projection. — Let A and B Fig. 124 be the two points in question. For convenience, use the hori- zontal projecting plane of the line as the revolving plane; its trace on the horizontal plane is ab, which is also the projection of the line. The points A and B, while revolving about the line ab, will describe circles, the planes of which are perpendic- ular to the axis, and, hence, in their re- volved position, will lie along lines aa" and bb". In this case, the distance of the projections of the points from the axis is zero, because the projecting plane of the line joining the points was used. The
altitudes of the triangles are the distances of the points above the horizontal plane, and, hence, they are also the hypothenuses of the triangles which are the radii of the circles.* Therefore, lay off aa" = a'o and bb// = b/p along lines aa" and bb", which are perpendicular to ab. Hence, a"b" is the true distance between the points A and B in space.
822. Case 2. When the points are situated on opposite
sides of the principal plane. — Let AB, Fig. 125, be the two points. From the -y projections it will be seen that A is in the first angle and B is in the fourth angle. If, as in the previous case, the horizontal projecting plane is used as the revolving plane then the horizontal FlG 125 projection ab is the trace of the revolv-
ing plane as before. The point A falls
to a" where aa" = a'o. Similarly, B falls to b" where bb" = b'p, * See the somewhat similar case in Problem 12.
b"
A
114
GEOMETRICAL PROBLEMS IN PROJECTION
but, as must be noticed, this point falls on that side of ab, oppo- site to the point a". A little reflection will show that such must be the case, but it may be brought out by what follows : The line AB pierces the horizontal plane at m and this point must remain fixed during the revolution. That it does, is shown by the fact that the revolved position of the line a"b" passes through this point.
NOTE. In both cases, the vertical projecting plane might have been used as the revolving plane and after the operation is performed, the true length of the line is again obtained. It must of necessity be equal to that given by the method here indicated.
823. Problem 14. To find where a given line pierces a given plane.
Analysis. If an auxiliary plane be passed through the
FIG. 126.
FIG. 127.
given line, so that it intersects the given plane, it will cut from it a line that contains the given point. The given point must also lie on the given line, hence it lies on their intersection.
Construction. Let T, Fig. 126, be the given plane and AB the given line. For convenience, use the horizontal projecting plane of the line as the auxiliary plane; the horizontal trace is cb and the vertical trace is cc'. The auxiliary plane cuts from the plane T the line CD. The vertical projection is shown as c'd' and the horizontal projection cd is contained in the trace of the horizontal projecting plane, because that plane was pur-
THE POINT, THE LINE, AND THE PLANE
115
posely taken as the cutting plane through the line. It is only in the vertical projection that the intersection m' is determined; its horizontal projection m is indeterminate in that plane (except from the fact that it is a corresponding projection) since the given line and the line of intersection have the same horizontal projecting plane. An oblique projection of this problem is given in Fig. 127.
824. Problem 15. To find the distance of a given point from a given plane.
Analysis. The perpendicular distance from the given point to the given plane is the required distance. Hence, draw a per- pendicular from the given point to the given plane, and find where this perpendicular pierces the .given plane. If the line joining the given point and the piercing point be revolved into one of the planes of projection, the line will be shown in its true length.
Construction. Let T, Fig. 128, be the given plane, and A, the given point. From A, draw AB, perpendicular to the plane T; the pro- jections of AB are therefore perpendicu- lar to the traces of the plane. If the horizontal projecting plane of the line AB be used as the auxiliary plane, it cuts the given plane in the line CD, and pierces it at the point B. Revolve the projecting plane of AB about its horizontal trace ab into the horizontal plane of projection. A will fall to a", where aa" = a'o, and B will fall to b", where bb" = b'p. Therefore, a"b" is the distance from the point A to the plane T.
FIG. 128.
825. Problem 16. To find the distance from a given point to a given line.
Analysis. Through the given point pass a plane perpen- dicular to the given line. The distance between the piercing point of the given line on this plane and the given point is the required distance. Join these two points by a line and revolve this line into one of the planes of projection; the line will then be seen in its true length.
116
GEOMETRICAL PROBLEMS IN PROJECTION
FIG. 129.
Construction. Let AB, Fig. 129, be the given line, and G, the given point. Through G, draw a plane perpendicular to AB (816), by drawing gc perpendicular to ab, and g'c', parallel to
the ground line; the piercing point of this line on the vertical plane is c'. Hence, the traces Tt' and Tt, perpendicular to a'b' and ab, respectively, through the point c', will be the traces of the required plane. AB pierces this plane at B, found by using the horizontal projecting plane of AB as a cutting plane, this cutting plane inter- secting in a line DE. The point B must be on both AB and DE. The projected distance, then, is the distance between the points B and G ; the true distance is found
by revolving BG into the horizontal plane. The latter operation is accomplished by using the horizontal projecting plane of BG and revolving -it about its trace bg; G falls to g", where gg" = g'p, and B falls to b", where bb" = b'o. Thus, g"b" is the true distance between the point G and the line AB.
826. Problem 17. To find the angle between two given intersecting lines.
Analysis. If a plane be passed through these lines and revolved into one of the planes of projection, the angle will be shown in its true size. Hence, find the piercing points of the given lines on one of the planes of projection; the line joining these piercing points will be the trace of the plane containing the lines. Revolve into that plane and the revolved position of the two lines shows the true angle.
Construction. Let AB and AC, Fig. 130, be the two given lines intersecting at A. These lines pierce the horizontal plane of projection at b and c and be is the trace of a plane containing the two lines. If A be considered as a point revolving about the. line be, it then describes a circle, the plane of which is per- pendicular to be and the point A will coincide with the horizontal plane somewhere along the line oa". The radius of the circle
THE POINT, THE LINE, AND THE PLANE 117
described by A is equal to the hypothenuse of a right triangle, where the distance ao, the projec- tion of a from the axis, is the base, and a'p, the distance of the point above the plane, is the altitude. This is shown in the triangle a'pq, where pq is equal to ao and therefore a'q is the required radius. Hence, make oa" = a'q, and a" is the re- volved position of the point A in space. The piercing points b and c of the given lines do not change
their relative positions. Thus a"b
... , , ... FIG. 130.
and a c are the revolved position
of the given lines, and the angle ba"c is the true angle.
827. Problem 18. To find the angle between two given planes.
Analysis. If a plane be passed perpendicular to the line of intersection of the two given planes, it will cut a line from each plane, the included angle of which will be the true angle. Re- volve this plane, containing the lines, about its trace on the principal plane, until it coincides with that plane, and the angle will be shown in its true size.
Construction. Let T and S, Fig. 131, be the two given planes, intersecting, as shown, in the line AB. Construct a sup- plementary plane to the right of the main diagram. H'H' is the new horizontal plane, shown as a line parallel to ab. The line AB pierces the vertical plane at a distance a'a above the horizontal plane. Accordingly, a'a is laid off on V'V, perpendicu- lar to H'H' ; it also pierces the horizontal plane at b shown in both views. The supplementary view shows ba' in its true relation to the horizontal plane, and is nothing more or less than a side view of the horizontal projecting plane of the line AB. If, in this supplementary view, a perpendicular plane cd be drawn, it will intersect the line AB in c, and the horizontal plane in the trace dfe shown on end. The lettering in both views is such that similar letters indicate similar points. Hence, dfe is the trace of the plane as shown in the main diagram, and ec and dc are the two lines cut from the planes T and S by the plane
118
GEOMETRICAL PROBLEMS IN PROJECTION
cde. When the plane cde is revolved into coincidence with the horizontal plane, c falls to c" in the supplementary view and is projected back to the main diagram as c". Therefore, ec"d is the true angle between the planes, because e and d remain fixed in the revolution.
828. Problem 19. To find the angle between a given plane and one of the principal planes.
Analysis. If an auxiliary plane be passed through the given plane and the principal plane so that the auxiliary plane is per-
FIG. 131.
pendicular to the intersection of the given plane and the prin- cipal plane, it will cut from each a line, the included angle of which will be the true angle. If, then, this auxiliary plane be revolved into the principal plane, the angle will be shown in its true size.
Construction. Let T, Fig. 132, be the given plane. The angle that this plane makes with the horizontal plane is to be determined. Draw the auxiliary plane R, so that its horizontal trace is perpendicular to the horizontal trace of the given plane; the vertical trace of the auxiliary plane must as a consequence be perpendicular to the ground line as Rr'. A triangle rRr' is
THE POINT, THE LINE, AND THE PLANE
119
cut by the auxiliary plane from the given plane and the two principal planes. If this triangle be revolved into the horizontal plane, about rR as an axis, the point r' will fall to r" with Rr' as a radius. Also, the angle rRr" must be a right angle, because it is cut from the principal planes, which are at right angles to each other. Hence, Rrr" is the angle which the plane T makes with the horizontal plane of projection.
The construction for obtaining the angle with the vertical plane is identical, and is shown on the right-hand side, with plane S as the given plane. All construction lines are added and no comment should be necessary.
NOTE. The similarity of Probs. 18 and 19 should be noted. In Prob. 19 the horizontal trace is the intersection of the given
'
s/«
FIG. 132.
plane and the horizontal plane; hence, the auxiliary plane is passed perpendicular to the trace. A similar reasoning applies to the vertical trace.
829. Problem 20. To draw a plane parallel to a given plane, at a given distance from it.
Analysis. The required distance between the two planes is the perpendicular distance, and the resulting traces must be parallel to the traces of the given plane. If a plane be passed perpendicular to either trace, it will cut from the principal planes and the given plane a right angled triangle, the hypothenuse of which will be the line cut from the given plane. If, further, the triangle be revolved into the plane containing the trace and the required distance between the planes be laid off perpendicular to the hypothenuse cut from the given plane, it will establish a point on the hypothenuse of the required plane. A line parallel
120 GEOMETRICAL PROBLEMS IN PROJECTION
to the hypothenuse through the established point will give the revolved position of a triangle cut from the required plane. On the counter revolution, this triangle will determine a point in each plane, through which the required traces must pass. Hence, if lines be drawn through the points so found, parallel to the traces of the given plane, the traces of the required plane are established.
Construction. Let T, Fig. 133, be the given plane, and r"g the required distance between the parallel planes. Pass a plane
rOr', perpendicular to the hori- zontal trace Tt; its vertical trace Or is, therefore, perpendicular to the ground line. The re- volved position of the triangle cut from the two principal planes and the given plane is rOr". Lay off r"g, perpendicular to rr", and equal to the required distance between the planes. Draw uu" parallel to rr" and
FlG j33 the triangle cut from the prin-
cipal planes and the required plane is obtained. On counter revolution, u" becomes u' and u remains fixed; Su' and Su, parallel respectively to Tr' and 1*r, are the required traces. Therefore, the distance between the planes T and S is equal to r"g.
830. Problem 21. To project a given line on a given plane.
Analysis. If perpendiculars be dropped from the given line upon the given plane, the points, so found, are the projec- tions of the corresponding points on the line. Hence, a line joining the projections on the given plane is the required pro- jection of the line on that plane.
Construction. Let T, Fig. 134, be the given plane, and AB the given line. From A, draw a perpendicular to the plane T; its horizontal projection is ac and its vertical projection is a'c'. To find where AC pierces the given plane, use the hori- zontal projecting plane of AC as the cutting plane; FE is the line so cut, and C is the resultant piercing point. Thus, C is the projection of A on the plane T. A construction, similar in
THE POINT, THE LINE, AND THE PLANE
121
detail, will show that D is the projection of B on the plane T. Hence, CD is the projection of AB on the plane T.
831. Problem 22. To find the angle between a given line and a given plane.
Analysis. The angle made by a given line and a given plane is the same as the angle made by the given line and its projection
FIG. 134.
on that plane. If from any point on the given line, another line be drawn parallel to the projection on the given plane, it will also be the required angle. The projection of the given line on the given plane is perpendicular to a projecting perpendicular from the given line to the given plane. Hence, any line, parallel to the projection and lying in the projecting plane of the given line to the given plane is also perpendicular to this projecting perpendicular. Therefore, pass a plane through the given line and the projecting perpendicular from the given line to the given plane. Revolve this plane into one of the planes of pro-
122
GEOMETRICAL PROBLEMS IN PROJECTION
jection, and, from any point on the line, draw a perpendicular to the projecting perpendicular. The angle between this line and the given line is the required angle.
Construction. Let T, Fig. 135, be the given plane, and AB the given line. From B, draw a perpendicular to the plane T, by making the projections respectively perpendicular to the traces of the given plane. Find the piercing points e and f, on the horizontal plane, of the given line and this perpendicular. Revolve the plane containing the lines BE and BF; B falls to b", on a line b"p, perpendicular to ef. The distance b"p is equal to the hypothenuse of a right triangle, where bp is the base and
FIG. 135.
b'o is the altitude; b'oq is such a triangle, where oq = bp. Hence, b"p is laid off equal to b'q. If from any point d, a line dc be drawn, perpendicualr to b"f, then b"dc is the required angle, as dc is parallel to the projection of AB on the plane T in its revolved position.
832. Problem 23. To find the shortest distance between a pair of skew * lines.
Analysis. The required line is the perpendicular distance between the two lines. If through one of the given lines, another line be drawn parallel to the other given line, the intersecting lines will establish a plane which is parallel to one of the given lines.
* Skew lines are lines which are not parallel and which do not intersect.
THE POINT, THE LINE, AND THE PLANE
123
The length of a perpendicular from any point on the one given line to the plane containing the other given line, is the required distance.
Construction. Let AB and CD, Fig. 136, be the two given
FIG. 136.
lines. Through any point O, on AB, draw FE, parallel to CD, and determine the piercing points of the lines AB and FE; a, e and f, b' are these piercing points, and, as such, determine the plane T. CD is then parallel to the plane T. From any point G, on CD, draw GH, perpendicular to the plane T, and
124
GEOMETRICAL PROBLEMS IN PROJECTION
find its piercing point on that plane. This point is H, found by drawing gh perpendicular to Tt and g'h' perpendicular to Tt'; the horizontal projecting plane cuts from the plane T, a line MN, on which is found H, the piercing point. GH is there- fore the required distance, but to find its true length, revolve
FIG. .136.
the horizontal projecting plane of GH into the horizontal plane. On revolution, H falls to h", where hh" is equal to the distance h' above the ground line; and, similarly, G falls to g". There- fore, h"g" is the true distance between the lines AB and CD.
THE POINT, THE LINE, AND THE PLANE
125
NOTE. In order to find the point on. each of the lines at which this perpendicular may be drawn, project the one given line on the plane containing the other. Where the projections cross, the point will be found.
ADDITIONAL CONSTRUCTIONS
833. Application to other problems. The foregoing prob- lems may be combined so as to form additional ones. In such cases, the analysis is apt to be rather long, and in the remaining few problems it has been omitted. The construction of the
FIG. 137.
problem might be followed by the student and then an analysis worked up for the particular problem afterward.
834. Problem 24. Through a given point, draw a line of a given length, making given angles with the planes of projection.
Construction. Consider the problem solved, and let AB, Fig. 137, be the required line, through the point A. The con- struction will first be shown and then the final position of the line AB will be analytically considered. From a', draw a'c' of a length 1, making the angle a with the ground line, and draw ac, parallel to the ground line. The horizontal projecting plane of AB has been revolved about the horizontal projecting perpen- dicular, until it is parallel to the vertical plane, and, therefore
126
GEOMETRICAL PROBLEMS IN PROJECTION
a'c' is shown in its true length and inclination to the horizontal plane. On the counter-revolution of the projecting plane of AB, it will be observed that the point A remains fixed, because it lies in the axis; B describes a circle, however, whose plane is parallel to the horizontal plane, and is, therefore, projected as the arc cb, while its vertical projection (or trace, if it be consid- ered as a plane instead of the moving point) is b'c', a line parallel to the ground line. The angle that the line AB makes with the horizontal plane is now fixed, but the point B is not finally located as the remaining condition of making the angle ji with the hori- zontal plane is yet conditional.
In order to lay off the angle that the line makes with the
FIG. 137.
vertical plane, draw a line ad, through a, making the angle (} with the ground line, and of a length 1. From a', draw a'd', so that d' is located from its corresponding projection d. Draw bd, through d, parallel to the ground line, and where its inter- section with the arc cb, locates b, the final position of the hori- zontal projection of the actual line. The corresponding pro- jection b', may be located by drawing the arc dV, and finding where it intersects the line b'c', through c', parallel to the ground line. The points b and b' should be corresponding projections, if the construction has been carried out accurately. In reviewing the latter process, it is found that the vertical projecting plane
THE POINT, THE LINE, AND THE PLANE
127
of the line has been revolved about the vertical projecting per- pendicular through A, until it was parallel to the horizontal plane. The horizontal projection is then ad, and this is shown in its true length and inclination to the vertical plane.
It may also be noted that the process of finding the ultimate position of the line is simply to note the projections of the path of the moving point, when the projecting planes of the line are revolved. That is, when the horizontal projecting plane of the line is revolved, the line makes a constant angle with the hori- zontal plane; and the path of the moving point is indicated by its projections. Similarly, when the vertical projecting plane of the line is revolved, the line makes a constant angle with the vertical plane; and the path of the moving point is again given
b
1 <4 a',
6'
FIG. 138.
by its projections. Where the paths intersect, on the proper planes, it is evidently the condition that satisfies the problem. There are four possible solutions for any single point in space, and they are shown in Fig. 138. Each is of the required length, and makes the required angles with the planes of projection. The student may try to work out the construction in each case and show that it is true.
835. Problem 25. Through a given point, draw a plane, making given angles with the principal planes.
Construction. Prior to the solution of this problem, it is desirable to investigate the property of a line from any point on the ground line, perpendicular to the plane. If the angles that this line makes with the principal planes can be determined, the actual construction of it, in projection, is then similar to the preceding problem. To draw the required plane, hence, resolves itself simply into passing a plane through a given point, per- pendicular to a given line (816).
128
GEOMETRICAL PROBLEMS IN PROJECTION
Let, in Fig. 139, T be the required plane, making the required angles a with the horizontal plane, and g with the vertical plane. To find the angle made* with the horizontal plane, pass a plane perpendicular to the horizontal trace as AOB, through any as- sumed point O on the ground line; OA is4 thus perpendicular to Tt and OB is perpendicular to the ground line. This plane cuts from the plane T, a line AB, and the angle BAO is the required angle a. Similarly, pass a plane COD, through O perpen- dicular to the vertical trace, then CO is perpendicular to Tt and OD is perpendicular to the ground line; DCO is the required angle g, which the plane T makes with the vertical plane. The planes AOB and COD intersect in a line OP. OP is perpendicular
FIG. 139.
to the plane T because each plane AOB and COD is perpendicular to the plane T (since they are each perpendicular to a trace, which is a line in the plane), and, hence, the line common to the two planes (OP) must be perpendicular to the plane. The angles OPC and OPA are, therefore, right angles, and, as a result, angle POA = 90°-«, and angle POC = 90°-g. Hence, to draw a perpendicular to the required plane, draw a line making angles witH the principal planes equal to the complements (90° vininus the angle) of the corresponding angles. It is evident that any line will do, as all • such lines, when measured in the same way, will be parallel, hence, it is not necessary, although convenient, that this line should pass through the ground line. To complete the problem, le't AB, Fig. 140, be a line making
THE POINT, THE LINE, AND THE PLANE
129
angles 90° -a with the horizontal plane, and 90° -& with the vertical plane. Through G, the given point, draw a perpendicular plane to AB, and the resultant plane T is. the required plane
FIG. 140.
making an angle a, with the horizontal plane, and an angle (i, with the vertical plane.
NOTE. As there are four solutions to the problem of drawing a line making given angles with the principal planes, there are also four solutions to this problem. The student may ghow these cases and check the accuracy by finding the angle between a given plane and the principal planes (828).
836. Problem 26. Through a given line, in a given plane, draw another line, intersecting it at a given point, and at a given angle.
Construction. Let AB, Fig. 141, be the given line, T the given plane and G the given point. Revolve the limited portion of the plane tTt' into coincidence with the
horizontal plane. Tt remains fixed but Tt' revolves to Tt". To find the direction of Tt", consider any point b' on the original position of the trace Tt'. The distance Tb' must equal Tb" as
130
GEOMETRICAL PROBLEMS IN PROJECTION
this length does not change on revolution; since Tt is the axis of revolution, the point B describes a circle, the plane of which is perpendicular to the axis, and, therefore, b" must also lie on a line bb", from b, perpendicular to the trace Tt.
In the revolved position of the plane, the point a remains unchanged, while B goes to b". Hence, ab" is the revolved position of the given line AB. The given point G moves to g", on a line gg", perpendicular to Tt and must also be on the line ab". Through g", draw the line cd", making the required angle a with it. On counter-revolution, c remains fixed, and d" moves to d'. Thus CD is the required line, making an angle a with another line AB, and lying in the given plane T.
FIG. 142.
837. Problem 27. Through a given line in a given plane, pass another plane, making a given angle with the given plane.
Construction. Let T, Fig. 142, be the given plane, AB the given line in that plane, and a the required angle between the planes. Construct a supplementary view of the line AB; H'H' is the new horizontal plane, the inclination of ab' is shown by the similar letters on both diagrams. The distance of b' above the horizontal plane must also equal the distance b' above H'H' and so on. Through h, in the supplementary view, draw a plane hf perpendicular to ab'. Revolve h about f to g and locate g as shown on the horizontal projection ab, of the main
THE POINT, THE LINE, AND THE PLANE
131
diagram. The line eg is cut from the plane T, by the auxiliary plane egf, hence, lay off the angle a as shown. This gives the direction gf of the line cut from the required plane. The point f lies on gf and also on ef .which is perpendicular to ab. Hence, join af and produce to S; join S and b' and thus estab-
FIG. 143.
lish the plane S., The plane S passes through the given line AB and makes an angle a with the given plane T (827).
838. Problem 28. To construct the projections of a circle lying in a given oblique plane, of a given diameter, its centre in the given plane being known.
Construction. Let T, Fig. 143, be the given plane, and G
132
GEOMETRICAL PROBLEMS IN PROJECTION
the given point lying in the plane T. When g', is assumed, for instance, g is found by drawing a horizontal line g'a' and then ga is its corresponding projection; g can therefore be determined as shown. When the plane of a circle is inclined to a plane of
FIG. 143.
projection, it is projected as an ellipse. An ellipse is determined, and can be constructed, when its major and minor axes are given.* To construct the horizontal projection, revolve the plane T about Tt, until it coincides with the horizontal plane as It". The direction of Tt" is found by drawing aa" perpendicular to Tt and laying off Ta' = Ta" (836). The centre of the circle in
* For methods of constructing the ellipse see Art. 906.
THE POINT, THE LINE, AND THE PLANE 133
its revolved position is found by drawing a"g" parallel to Tt and gg" perpendicular to Tt; g" is this revolved position. With the given radius, draw c"d"e"f", the circle of the given diameter. Join e"d" and f"c"; prolong these lines to o and q, and also draw g"p parallel to these. Thus, three parallel lines in the revolved position of the plane T, are established and on counter revolution, they will remain parallel. The direction gp, of one of them is known, hence, make fq and eo parallel to gp. The points cdef are the corresponding positions of c"d"e"f" and determine the horizontal projection of the circle. The line ec remains equal to e"c" but fd is shorter than f"d", hence, the major and minor axes of an ellipse (projection of the circle) are determined. The ellipse may now be drawn by any convenient method and the horizontal projection of the circle in the plane T will be complete.
By revolving the plane T into coincidence with the vertical plane, Tt'" is found to be the revolved position of the trace Tt and g"', the revolved position of the centre. The construction is identical with the construction of the horizontal projection and will become apparent, on inspection, as the necessary lines are shown indicating the mode of procedure. As a result, k'l'm'n' determine the major and minor axes of the ellipse, which is the vertical projection of the circle in the plane T.
NOTE. As a check on the accuracy of the work, tangents may be drawn in one projection and the corresponding projection must be tangent at the corresponding point of tangency.
QUESTIONS ON CHAPTER VIII
1. Mention the three distinct steps into which the solution of a problem
may be divided.
2. What is the statement of a problem?
3. What is the analysis of a problem?
4. What is the construction of a, problem?
5. What type of projection is generally used in the construction of a
problem?
Note. In the following problems, the construction, except in a few isolated cases, is to be entirely limited to the first angle of pro- jection.
6. Draw a line through a given point, parallel to a given line. Give
analysis and construction.
134 GEOMETRICAL PROBLEMS IN PROJECTION
7. Draw a line intersecting a given line at a given point. Give analysis
and construction.
8. Find where a given oblique line pierces the planes of projection.
Give analysis and construction.
9. Transfer the diagram of Question 8 to oblique projection.
10. Pass an oblique plane through a given oblique line. Give analysis
and construction.
11. Make an oblique projection of the diagram in Question 10.
12. Pass a plane obliquely through .the principal planes and through a
line parallel to the ground line. Give analysis and construction. Hint: Use a profile plane in the construction.
13. Transfer the diagram of Question 12 to an oblique projection.
14. Pass an oblique plane through a line which is parallel to the hori-
zontal plane but inclined to the vertical plane. Give analysis and construction.
15. Transfer the diagram of Question 14 to an oblique projection.
16. Pass an oblique plane through a line which is parallel to the vertical
plane but inclined to the horizontal plane. Give analysis and construction.
17. Transfer the diagram of Question 16 to an oblique projection.
18. Pass an oblique plane through a given point. Give analysis and
construction.
19. Make an oblique projection of the diagram in Question 18.
20. Find the intersection of two planes, oblique to each other and to the
principal planes. Give analysis and construction.
21. Transfer the diagram of Question 20 to an oblique projection.
22. Find the intersection of two planes, oblique to each other and to the
principal planes. Take the case where the traces do not intersect on one of the principal planes. Give analysis and construction.
23. Find the corresponding projection of a given point lying in a given
oblique plane, when one projection is given. Give analysis and construction.
24. Transfer the diagram of Question 23 to an oblique projection.
25. Draw a- plane which contains a given point and is parallel to a given
plane. Give analysis and construction.
26. Transfer the diagram of Question 25 to an oblique projection.
27. Draw a line through a given point, perpendicular to a given plane.
Give analysis and construction.
28. Transfer the diagram of Question 27 to an oblique projection.
29. Draw a plane through a given point, perpendicular to a given line.
Give analysis and construction.
30. Transfer the diagram of Question 29 to an oblique projection.
31. Pass a plane through three given points, not in the same straight
line. Give analysis and construction.
32. Transfer the diagram of Question 31 to an oblique projection.
33. Revolve a given point, not in the principal planes, about a line
lying in one of the principal planes. Give analysis and construc- tion.
THE POINT, THE LINE, AND THE PLANE 135
34. Transfer the diagram of Question 33 to an oblique projection and
show also the plane of the revolving point.
35. Find the true distance between two points in space, when both
points are in the first angle of projection. Use the method of revolving the projecting plane of the line until it is parallel to one plane of projection. Give analysis and construction.
36. Make an oblique projection of the diagram in Question 35 and show
the projecting plane of the line.
37. Find the true distance between two points in space, when one point
is in the first angle and the other is in the fourth angle. Use the method of revolving the projecting plane of the line until it is parallel to one plane of projection. Give analysis and construction.
38. Make an oblique projection of the diagram in Question 37 and show
the projecting plane of the line.
39. Find the true distance between two points in space, when both
points are in the first angle of projection. Use the method of revolving the projecting plane of the line into one of the planes of projection. Give analysis and construction.
40. Transfer the diagram of Question 39. to an oblique projection and
show the projecting plane of the line and the path described by the revolving points.
41. Find the true distance between two points in space, when one point
is in the first angle and the other is in the fourth angle of pro- jection. Use the method of revolving the projecting plane of the line into one of the planes of projection. Give analysis and construction.
42. Transfer the diagram of Question 39 to an oblique projection and
show the projecting plane of the line and the path described by the revolving points.
43. Find where a given line pierces a given plane. Give analysis and
construction.
44. Transfer the diagram of Question 43 to an oblique projection.
45. Find the distance of a given point from a given plane. Give analysis
and construction.
46. Transfer the diagram of Question 45 to an oblique projection. Omit
the portion of the construction requiring the revolution of the points.
47. Find the distance from a given point to a given line. Give analysis
and construction.
48. Transfer the diagram of Question 47 to an oblique projection. Omit
the portion of the construction requiring the revolution of the points.
49. Find the angle between two given intersecting lines. Give analysis
and construction.
50. Make an oblique projection of the diagram in Question 49.
51. Find the angle between two given planes. Give analysis and con-
struction.
52. Make an oblique projection of the diagram in Question 51.
136 GEOMETRICAL PROBLEMS IN PROJECTION
53. Find the angle between a given plane and the horizontal plane of
projection. Give analysis and construction.
54. Transfer the diagram of Question 53 to an oblique projection.
55. Find the angle between a given plane and the vertical plane of pro-
jection. Give analysis and construction.
56. Transfer the diagram of Question 55 to an oblique projection.
57. Draw a plane parallel to a given plane at a given distance from it.
Give analysis and construction.
58. Project a given line on a given plane. Give analysis and construction.
59. Make an oblique projection of the diagram in Question 58.
60. Find the angle between a given line and a given plane. Give analysis
and construction.
61. Make an oblique projection of the diagram in Question 60.
62. Find the shortest distance between a pair of skew lines. Give
analysis and construction.
63. Make an oblique projection of the diagram in Question 62.
64. Through a given point, draw a line of a given length, making given
angles with the planes of projection.
65. Show the construction for the three remaining cases of the problem
in Question 64.
66. Through a given point, draw a plane, making given angles with the
principal planes.
67. Prove that the construction in Question 66 is correct by finding
the angles that the given plane makes with the principal planes. (Note that there are four possible cases of this problem.)
68. Show the construction for the three remaining cases of the problem
in Question 66.
69. Prove that the constructions in Question 68 are correct by finding
the angles that the given plane makes with the principal planes.
70. Through a given line, in a given plane, draw another line intersecting
it at a given point, and at a given angle.
71. Transfer the diagram of Question 70 to an oblique projection.
72. Through a given line in a given plane, pass another plane, making
given angles with the given plane.
73. Make an oblique projection of the diagram in Question 72.
74. Construct the projections of a circle lying in a given oblique plane,
the diameter and its centre in the given plane being known.
Note. The following exercises embrace operations in all four angles.
75. Given a line the first angle and a point in the second angle, draw
a line through the given point, parallel to the given line.
76. Given a line in the first angle and a point in the third angle, draw
a line through the given point, parallel to the given line.
77. Given a line in the first angle and a point in the fourth angle, draw
a line through the given point, parallel to the given line.
78. Given a line in the second angle and a point in the third angle,
draw a line through the given point, parallel to the given line.
THE POINT, THE LINE, AND THE PLANE 137
79. Given a line in the third angle and a point in the fourth angle, draw
a line through the given point, parallel to the given line.
80. Given a line in the fourth angle and a point in the second angle,
draw a line through the given point, parallel to the given line.
81. Draw two intersecting lines in the second angle .
82. Draw two intersecting lines in the third angle.
83. Draw two intersecting lines in the fourth angle.
84. Find where a given line pierces the principal planes when the limited
portion of the line is in the second angle.
85. Find where a given line pierces the principal planes when the limited
portion of the line is in the third angle.
86. Find where a given line pierces the principal planes when the limited
portion of the line is in the fourth angle.
87. Show the second angle traces of a plane passed through a second
angle oblique line.
88. Show the third angle traces of a plane passed through a third angle
oblique line.
89. Show the fourth angle traces of a plane passed through a fourth
angle oblique line.
90. Given a line in the second angle and parallel to the ground line,
pass an oblique plane through it. Use a profile plane as part of the construction.
91. Given a line in the third angle and parallel to the ground line,
pass an oblique plane through it. Use a profile plane as a part of the construction.
92. Given a line in the fourth angle and parallel to the ground lino,
pass an oblique plane through it. Use a profile plane as a part of the construction.
93. Given a point in the second angle, pass an oblique plane through
it. Show only second angle traces.
94. Given a point in the third angle, pass an oblique plane through it.
Show only third angle traces.
95. Given a point in the fourth angle, pass an oblique plane through
it. Show only fourth angle traces.
96. Given the second angle traces of two oblique planes, find their
intersection in the second angle.
97. Given the third angle traces of two oblique planes, find their inter-
section in the third angle.
98. Given the fourth angle traces of two oblique planes, find their
intersection in the fourth angle.
99. Given the first angle traces of an oblique plane and one projection
of a second angle point, find the corresponding projection.
100. Given the first angle traces of an oblique plane, and one projection
of a third angle point, find the corresponding projection.
101. Given the first angle traces of an oblique plane, and one projection
of a fourth angle point, find the corresponding projection.
102. Given the second angle traces of an oblique plane, and one projec-
tion of a third angle point, find the corresponding projection.
138 GEOMETRICAL PROBLEMS IN PROJECTION
103. Given the third angle traces of an oblique plane, and one pro-
jection of a fourth angle point, find the corresponding pro- jection.
104. Given the first angle traces of an oblique plane, and a point in the
second angle, pass a plane through the given point and parallel to the given plane.
105. Given the first angle traces of an oblique plane, and a point in the
third angle, pass a plane through the given point and parallel to the given plane.
106. Given the first angle traces of an oblique plane, and a point in the
fourth angle, pass a plane through the given point and parallel to the given plane.
107. Given the second angle traces of an oblique plane, and a point in
the third angle, pass a plane through the given point and parallel to the given plane.
108. Given the third angle traces of an oblique plane, and a point in
the fourth angle, pass a plane through the given point and parallel to the given plane.
109. Given the first angle traces of an oblique plane and a point in the
second angle, draw a line through the given point perpendicular to the given plane.
110. Given the first angle traces of an oblique plane and a point in the
third angle, draw a line through the given point perpendicular to the given plane.
111. Given the first angle traces of an oblique plane and a point in the
fourth angle, draw a line through the given point perpendicular to the given plane.
112. Given the second angle traces of an 'oblique plane and a point in
the fourth angle, draw a line through the given point perpendicular to the given plane.
113. Given the first angle traces of a plane perpendicular to the hori-
zontal plane but inclined to the vertical plane, and a point in the third angle, draw a line through the given point perpendicular to the given plane.
114. Given the third angle traces of a plane perpendicular to the vertical
plane but inclined to the horizontal plane, and a point in the fourth angle, draw a line through the given point, perpendicular to the given plane.
115. Given the first angle traces of a plane parallel to the ground line
and a point in the third angle, draw a line through the given point perpendicular to the given plane.
116. Draw a profile plane of the diagram in Question 115.
117. Make an oblique projection of the diagram in Question 115.
118. Given the fourth angle traces of a plane parallel to the ground line
and a point in the second angle, draw a line through the given point perpendicular to the given plane.
119. Draw a profile plane of the diagram in Question 118.
120. Make an oblique projection of the diagram in Question 118.
THE POINT, THE LINE, AND THE PLANE 139
121. Given a line in the first angle and a point in the second angle, pass
a plane through the given point perpendicular to the given line.
122. Given a line in the first angle and a point in the third angle, pass a
plane through the given point perpendicular to the given line.
123. Given a line in the first angle and a point in the fourth angle, pass
a plane through the given point perpendicular to the given line.
124. Given a line in the second angle and a point in the third angle,
pass a plane through the given point perpendicular to the given line.
125. Given a line in the third angle and a point in the fourth angle, pass
a plane through the given point perpendicular to the given line.
126. Given three points in the second angle, pass a plane through them.
127. Given three points in the third angle, pass a plane through them.
128. Given three points in the fourth angle, pass a plane through them.
129. Given two points in the first angle and one point in the second
angle, pass a plane through them.
130. Given a point in the first angle, one in the second angle, and one
in the third angle, pass a plane through them.
131. Given a point in the second angle, one in the third angle, and one
in the fourth angle, pass a plane through them.
132. Given a line in the horizontal plane and a point in the second angle,
revolve the point about the line until it coincides with the hori- zontal plane.
133. Given a line in the horizontal plane and a point in the third angle,
revolve the point about the line until it coincides with the hori- zontal plane.
134. Given a line in the horizontal plane and a point in the fourth angle
revolve the point about the line until it coincides with the hori- zontal plane.
135. Given a line in the vertical plane and a point in the second angle,
revolve the point about the line until it coincides with the vertical plane.
136. Given a line in the vertical plane and a point in the fourth angle,
revolve the point about the line until it coincides with the vertical plane.
137. Given two points in the second angle, find the true distance between
them.
138. Given two points in the third angle, find the true distance between
them.
139. Given two points in the fourth angle, find the true distance between
them.
140. Given one point in the first angle and one point in the second angle,
find the true distance between them.
141. Given one point in the first angle and one point in the third angle,
find the true distance between them.
142. Given one point in the second angle and one point in the third
angle, find the true distance between them.
140 GEOMETRICAL PROBLEMS IN PROJECTION
143. Given one point in the second angle and one point in the fourth
angle, find the true distance between them.
144. Given the first angle traces of a plane and a line in the second angle,
find where the given line pierces the given plane.
145. Given the first angle traces of a plane and a line in the third angle,
find where the given line pierces the given plane.
146. Given the first angle traces of a plane and a line in the fourth angle,
find where the given line pierces the given plane.
147. Given the second angle traces of a plane and a line in the third
angle, find where the given line pierces the given plane.
148. Given the second angle traces of a plane and a line in the fourth
angle, find where the given line pierces the given plane.
149. Given the third angle traces of a plane and a line in the fourth
angle, find where the given line pierces the given plane.
150. Given the first angle traces of a plane and a point in the second
angle, find the distance from the point to the plane.
151. Given the first angle traces of a plane and a point in the third angle,
find the distance from the point to the plane.
152. Given the first angle traces of a plane and a point in the fourth
angle, find the distance from the point to the plane.
153. Given the second angle traces of a plane and a point in the third
angle, find the distance from the point to the plane.
154. Given the second angle traces of a plane and a point in the fourth * angle, find the distance form the point to the plane.
155. Given the third angle traces of a plane and a point in the fourth
angle, find the distance from the point to the plane.
156. Given a line in the first angle and a point in the second angle, find
the distance from the point to the line.
157. Given a line in the first angle and a point in the third angle, find
the distance from the point to the line.
158. Given a line in the first angle and a point in the fourth angle, find
the distance from the point to the line.
159. Given a line in the second angle and a point in the third angle,
find the distance from the point to the line.
160. Given a line in the second angle and a point in the fourth angle,
find the distance from the point to the line.
161. Given a line in the third angle and a point in the fourth angle, find
" the distance from the point to the line.
162. Given two intersecting lines in the second angle, find the angle
between them.
163. Given two intersecting lines hi the third angle, find the angle between
them.
164. Given two intersecting lines in the fourth angle, find the angle
between them.
165. Given two intersecting lines, one in the first and one in the second
angle, find the angle between them.
166. Given two intersecting lines, one in the first and one in the third
angle, find the angle between them.
THE POINT, THE LINE, AND THE PLANE 141
167. Given two intersecting lines, one in the first and one in the fourth
angle, find the angle between them.
168. Given two intersecting lines, one in the second and one in the third
angle, find the angle between them.
169. Given two intersecting lines, one in the second and one in the fourth
angle, find the angle between them.
170. Given two intersecting lines, one in the third and one in the fourth
angle, find the angle between them.
171. Given the second angle traces of two intersecting planes, find the
angle between them.
172. Given the third angle traces of two intersecting planes, find the
angle between them.
173. Given the fourth angle traces of two intersecting planes, find the
angle between them.
174. Given the first angle traces of one plane and the second angle traces
of another plane, find the angle between them.
175. Given the first angle traces of one plane and the third angle traces
of another plane, find the angle between them.
176. Given the first angle traces of one plane and the fourth angle traces
of another plane, find the angle between them.
177. Given the second angle traces of one plane and the third angle
traces of another plane, find the angle between them.
178. Given the second angle traces of one plane and the fourth angle
traces of another plane, find the angle between them.
179. Given the third angle traces of one plane and the fourth angle traces
of another plane, find the angle between them.
180. Given a plane hi the second angle, find the angle between the given
plane and the horizontal plane.
181. Given a plane in the second angle, find the angle between the given
plane and the vertical plane.
182. Given a plane in the third angle, find the angle between the given
plane and the horizontal plane.
183. Given a plane in the third angle, find the angle between the given
plane and the vertical plane.
184. Given a plane in the fourth angle, find the angle between the given
plane and the horizontal plane.
185. Given a plane in the fourth angle, find the angle between the given
plane and the vertical plane.
186. Given the second angle traces of a plane, draw another parallel
plane at a given distance from it.
187. Given the third angle traces of a plane, draw another parallel plane
at a given distance from it.
188. Given the fourth angle traces of a plane, draw another parallel
plane at a given distance from it.
189. Given the first angle traces of a plane and a line in the second angle,
project the given line on the given plane.
190. Given the first angle traces of a plane and a line in the third angle,
project the given line on the given plane.
142 GEOMETRICAL PROBLEMS IN PROJECTION
191. Given the first angle traces of a plane and a line in the fourth angle,
project the given line on the given plane.
192. Given the second angle traces of a plane and a line in the third angle,
project the given line on the given plane.
193. Given the second angle traces of a plane and a line in the fourth
angle, project the given line on the given plane.
194. Given the third angle traces of a plane and a line in the fourth angle,
project the given line on the given plane.
195. Given the fourth angle traces of a plane and a line in the second
angle, project the given line on the given plane.
196. Given the first angle traces of a plane and a line in the second angle,
find the angle between the given line and the given plane.
197. Given the first angle traces of a plane and a line in the third angle,
find the angle between the given line and the given plane.
198. Given the first angle traces of a plane and a line in the fourth angle,
find the angle between the given line and the given plane.
199. Given the second angle traces of a plane and a line in the third angle,
find the angle between the given line and the given plane.
200. Given the second angle traces of a plane and a line in the fourth
angle, find the angle between the given line and the given plane
201. Given the third angle traces of a plane and a line in the fourth angle,
find the angle between the given line and the given plane.
202. Given the fourth angle traces of a plane and a line in the second
angle, find the angle between the given line and the given plane.
203. Given the fourth angle traces of a plane and a line in the third angle,
find the angle between the given line and the given plane.
204. Given a line in the first angle and another line in the second angle
(skew lines), find the shortest distance between them.
205. Given a line in the first angle and another line in the third angle
(skew lines), find the shortest distance between them.
206. Given a line in the first angle and another line in the fourth angle
(skew lines), find the shortest distance between them.
207. Given a line in the second angle and another line in the third angle
(skew lines), find the shortest ditstance between them.
208. Given a line in the second angle and another line in the fourth
angle (skew lines) ; find the shortest distance between them.
209. Given a line in the third angle and another line in the fourth angle
(skew lines), find the shortest distance between them.
210. Through a given point in the second angle, draw a line of a given
length, making given angles with the planes of projection.
211. Through a given point in the third angle, draw a line of a given
length, making given angles with the planes of projection.
212. Through a given point in the fourth angle, draw a line of a given
length, making given angles with the planes of projection.
213. Through a given point in the second angle, draw a plane making
given angles with the principal planes.
214. Through a given point in the third angle, draw a plane making
given angles with the principal planes.
THE POINT, THE LINE, AND THE PLANE 143
215. Through a given point in the fourth angle, draw a plane making
given angles with the principal planes.
216. Given the second angle traces of a plane, a line, and a point in the
plane, draw another line through the given point, making a given angle with the given line.
217. Given the third angle traces of a plane, a line, and a point in the
plane, draw another line through the given point, making a given angle with the given line.
218. Given the fourth angle traces of a plane, a line and a point in the
plane, draw another line through the given point, making a given angle with the given line.
219. Given the second angle traces of a plane, and a line in the plane,
draw through the line another plane making a given angle with the given plane.
220. Given the third angle traces of a plane, and a line in the plane,
draw through the line another plane making a given angle with the given plane.
221. Given the fourth angle traces of a plane, and a line in the plane,
draw through the line another plane making a given angle with the given plane.
222. Given the second angle traces of a plane, the diameter and the
centre of a circle, construct the projections of the circle.
223. Given the third angle traces of a plane, the diameter and the centre
of a circle, construct the projections of the circle.
224. Given the fourth angle traces of a plane, the diameter, and the
centre of a circle, construct the projections of the circle,
CHAPTER IX CLASSIFICATION OF LINES
901. Introductory. Lines are of an infinite variety of forms. The frequent occurrence in engineering of certain varieties makes it desirable to know their properties as well as their method of construction. It must be remembered that lines and points are mathematical concepts and that they have no material existence. That is to say, a line may have so many feet of length but as it has no width or thickness, its volume, therefore, is zero. Hence, it cannot exist except in the imagination. Like- wise, a point is a still further reduction and has position only; it has no dimensions at all. Of course, the rraterial representa- tion of lines and points requires finite dimensions, but when speaking of them, or representing them, it is the associated idea, rather than the representation, which is desired.
902. Straight line. A straight line may be defined as the shortest distance between two points.* It may also be described as the locus (or path) of a generating point which moves in the same direction. Hence, a straight line is fixed in space by two points, or, by a point and a direction.
903. Singly curved line. A singly curved line is the locus
(plural :-loci) of a generating point which moves in a varying direction but remains in a single plane. Sometimes the singly curved line is called a plane curve because all points on the curve must lie in the same plane.
904. Representation of straight and singly curved lines.
Straight and singly curved lines are represented by their pro- jections. When singly curved lines are parallel to the plane
* Frequently, this is called a right line. There seems no reason, however, why this new nomenclature should be used; hence, it is here avoided.
144
CLASSIFICATION OF LINES 145
of projection, they are projected in their true form and require only one plane of projection for their complete representation. If the plane of the curve is perpendicular to the principal planes, a profile will suffice. If the plane of the curve is inclined to the planes of projection, both horizontal and vertical projections may be necessary, unless a supplementary plane be used, which is parallel to the plane of the curve.* This latter condition is then similar to that obtained wyhen the plane of the curve is parallel to one of the principal planes.
905. Circle. The circle is a plane curve, every point of which is equidistant from a fixed point called the centre. The path described -by the moving point (or locus) is frequently designated as the " circumference of the circle."! To define such curves consistently, it is necessary to limit the definition to the nature of the line forming the curve. Therefore, what is usually known as the circumference of the circle should simply be known as the circle and then subsequent definitions of other curves become consistently alike.
In Fig. 144 a circle is shown; abedc is the curve or circle proper. The fixed point o is called the centre and every point on the circle is equidistant from it. This fixed distance is called the radius ; oa, oc, and ob are all radii of the circle; be is a diameter and is a straight line through the centre and equal to two radii in length. The straight line, de, limited by the circle is a chord and when it passes through the centre it becomes a diameter; when the same line is extended like gh it is a secant. A limited portion of the circle like ac is an arc; when equal to one-quarter of the whole circle it is a quadrant; when equal to one-half of the whole circle it is a semicircle as cab or bfc. The area included between two radii and the circle is a sector as aoc; that between any line like de and the circle is
* In all these cases, the plane of the curve may be made to coincide with the plane of projection. This, of course, is a special case.
t The introduction of this term makes it necessary to define the circle as the enclosed area. This condition is unfortunate, as in the case of the parabola, hyperbola, and numerous other curves, the curves are open and do not enclose an area.
146
GEOMETRICAL PROBLEMS IN PROJECTION
a segment. If any other circle is drawn with the same centre of the circles are concentric, otherwise eccentric; the distance between the centres in the latter case is the eccentricity.
906. Ellipse. The ellipse is a plane curve, in which the sum of the distances of any point on the curve from two fixed points is constant. Fig. 145 shows this curve as adbc; The major axis is ab, and its length is the constant distance of the definition; cd is the minor axis and is perpendicular to ab through its middle point o. The fixed points or foci (either one is a focus) are e and f and are located on the major axis.
If ab be assumed as the constant distance, and e and f be assumed as the foci, the minor axis is determined by drawing arcs from e and f with oa, equal to one-half of the major axis,
FIG. 145.
FIG. 146.
as a radius. Thus ec+cf = ab and also ed+df = ab. To find any other point on the curve, assume any distance as bg, and with bg as a radius and f as a centre, draw an arc fh. With the balance of the major axis ag as a radius, draw an arc eh from the focus e as a centre. These intersecting arcs locate h, a point on the curve. Thus eh-hhf = ab and therefore satisfies the definition of the curve. With the same radii just used, the three other points i, j and k are located. In general, four points are determined for any assumption except for the points a, c, b or d. It can be shown that a and b are points on the curve, because oa = ob and oe = of. Hence ae=fb, therefore, fa+ea = ab and also eb+fb = ab.
In the construction of this, or any other curve, the student should avoid trying to save time by locating only a few points. This is a mistaken idea, as, within reasonable limits, time is saved by drawing numerous points on the curve, particularly
CLASSIFICATION OF LINES
147
where the curve changes its direction rapidly. The direction of the curve at any point should be known with a reasonable degree of accuracy.
Another method of drawing an ellipse is shown in Fig. 146. It is known as the trammel method. Take any straight ruler and make oc = a and od=b. By locating c on the minor axis and d on the major axis, a point is located at o, as shown. This method is a very rapid one, and is the one generally used when a true ellipse is to be plotted, on account of the very few lines required in the construction. Both methods mentioned are theoretically accurate, but the latter method is perhaps used oftener than the former.
In practice, ellipses are usually approximated by employing four circular arcs, of two different radii as indicated in Art. 405. The major and minor axes are laid off and a smooth looking curve drawn between these limits. Of course, the circular arcs do not produce a true ellipse, but as a rule, this method is a rapid one and answers the purpose in conveying the idea.
907. Parabola. The parabola is a plane curve, which is the locus of a point, moving so that the distance from a fixed point is always equal to the distance
from a fixed line. The fixed point is the focus and the fixed line is the directrix. A line through the focus and perpen- dicular to the directrix as eg, Fig. 147, is the axis with respect to which the curve is symmetrical. The intersection of the axis and the curve is the vertex shown at f .
The point a on the curve is so situated that ac = ad; also b is so situated that be = be. Points beyond b become more and
more remote, from both the axis and directrix. Hence, it is' an open curve, extending to infinity. Discussions are usually limited to some finite portion of the curve.
908. Hyperbola. The hyperbola is a plane curve, traced by a point, which moves so that the difference of its distances from two fixed points is constant. The fixed points a and b,
FIG. 147.
148
GEOMETRICAL PROBLEMS IN PROJECTION
FIG. 148.
Fig. 148, are the foci. The line ab, passing through them, is the transverse * axis ; the point at which either curve crosses the axis as e or f is the vertex (plural: — vertices), and the line kl, perpendicular to ab at its middle point, is the conjugate axis. To draw the curve, lay off the foci a and b; also lay off ef,
the constant distance, so that eo = of and o is the middle of ab. It must be observed that ef is always smaller than ab otherwise the curve cannot be constructed. Take any radius be, greater than bf, and draw an indefinite arc; from a, draw ac, so that ac — bc = ef, hence, ac=bc-f-ef. The point c is thus on the curve. Similarly, draw an arc ad, from a, and another arc bd = ad + ef . This locates d, which, in this case, is on
another curve. In spection will show that there are two branches of this curve. The point c has been selected so as to be on one branch, and d, on the other. To construct the curve, accurately, many more points must be located than shown. Both branches are open and symmetrical with respect to both axes.
A tangent to the curve through o is called an asymptote when it touches the hyperbola in two points, each at an infinite distance from o. As will be observed, there are two asymptotes.
909. Cycloid. The cycloid is a plane curve, traced by a point on a circle which rolls over a straight line. The straight line over which the circle rolls is the directrix; the point on the circle may be considered as the generating point.
The curve is shown as abc . . . i in Fig. 149. To construct it, lay out the directrix ai, and to one side, draw an auxiliary circle equal in diameter to the rolling circle as shown at 1' 2', etc. On a line through the centre of the auxiliary circle, draw the line 3'-9 parallel to the directrix ai. Lay off the distance
* Sometimes known as the "principal axis."
CLASSIFICATION OF LINES
149
1-9 on this line, equal to the length of the circle (3. 14 X diameter). Assume that when the centre of the rolling circle is at 1, a is the position of the generating point. After one-eighth of a revolution the centre has moved to 2, where the distance 1-2 is one-eighth of the distance 1-9. The corresponding position of the generating point is shown at 2' in the auxiliary circle at the left. Hence, for one-eighth of a revolution the centre of the rolling circle has moved from 1 to 2 and the generating point has moved a distance vertically upward equal to the distance of 2' above the line ai. Therefore, draw a line through 2', parallel to ai; and then from 2 as a centre, draw an arc with the radius of the rolling circle intersecting this line in the point b, a point on the required curve. After a quarter of a revolution, the gen- erating point is above the directrix, at a height equal to the
FIG. 149.
distance of 3' above ai. It is also on an arc from 3, with a radius equal again to the radius of the rolling. circle; hence, c is the point. This process is continued until the generating point reaches its maximum height e after one-half of a revolution, when it begins to descend, to i, as shown, after completing one revolution. Further rolling of the circle causes the generating point to dupli- cate its former steps, as it continues along the directrix to infinity if desired.
The cycloid is the same curve that is produced by a mark on the rim of a car-wheel, while rolling along the track, only, here, no slipping is permitted. The cycloid is, thus, continuous, each arch being an exact duplicate of the preceding.
910. Epicycloid. The epicyloid is a plane curve, which is generated by a point on a circle which rolls on the outside of another circle. The directrix is here an arc of a circle, instead of a straight line.
150
GEOMETRICAL PROBLEMS IN PROJECTION
The construction is indicated in Fig. 150. The centre of the rolling circle assumes successive positions as 1, 2, 3, etc. The length of the arc ai is equal to the length of the rolling circle. The points are located much the same as in the cycloid, and, as
the necessary construction lines are shown, the student should have no difficulty in following the construction.
911. Hypocycloid. The hypocycloid is a plane curve/ generated by a point on a circle which rolls on the inside o
FIG. 151.
another circle. The directrix of the epicycloid and the hypocy- cloid may be the same, the epicycloid is described on the outside of the arc while the hypocycloid is described on the inside.
All the necessary construction lines required to draw the hypocycloid have been added in Fig. 151 and the method is
CLASSIFICATION OF LINES 151
perhaps evident. No confusion should arise even though the curves are drawn in positions other than those shown. For instance, the cycloids (this includes the cycloids, epicycloid and hypocycloid) could be shown upside-down; the curves are gener- ated in much the same way and their properties, therefore, do not change.
912. Spiral. The spiral is a plane curve, generated by a point moving along a given line while the given line is revolving about some point on the line. An infinite number of spirals may exist, because the point may have a variable velocity along the line, while again the line may have a variable angular velocity about the point. The one having uniform motion of the point along the line, and uniform angular velocity about the point is called the Spiral of Archimedes.
Fig. 152 shows the Archimedian spiral which is perhaps the simplest type of spiral. If ox be assumed as the primitive position of a line revolving at o, and the point also starts at o, then o is the starting point of the curve. x_ Suppose that after one-eighth of a revo- lution the generating point has moved a distance oa, then, after one-quarter of a revolution, the point will be at b, where ob = 2Xoa, and so on. The spiral becomes j?IG 152.
larger and larger as the revolution con- tinues. Had the line revolved in the other direction, the curve would have been the reverse of the one shown.
913. Doubly curved line. A doubly curved line is one whose direction is continually changing and whose points do not lie in one plane. A piece of wire may be twisted so as to furnish a good example of a doubly curved line.
914. Representation of doubly curved lines. As no
single plane can contain a doubly curved line, it becomes neces- sary to use two planes and project, orthographically, the line on them. Familiarity is had with the representation of points in space on the principal planes. The line may be conceived as being made up of an infinite number of points and each point can be located in space by its projections.
152 GEOMETRICAL PROBLEMS IN PROJECTION
Fig. 153 represents a curve ABCD, shown by its horizontal projection abed and its vertical projection a'b'c'd'. Any point
on the curve, such as B, is found by erecting perpendiculars from b and b' and extending them to their intersection; this will be the point sought. The principal planes must be at right angles to each other if it is desired to locate a point by FlG 153 erecting these perpendiculars; other-
wise, the curve must be imagined
from its projections when the planes are revolved into coinci- dence, as is customary in orthographic projection.
915. Helix. The helix is a doubly curved line described by a point having motion around a line called the axis, and in addition, a motion along it. Unless it is noted otherwise, the helix will be considered as having a uniform circular motion around the axis and also a uniform motion along it. The curve finds its most extensive application on that type of screw known as a machine screw. This is then a uniform cylindrical helix. Wood screws furnish examples of conical helices.* The helix is also frequently used in making springs of a type known as helical springs.!
Fig. 154 shows the construction of the helix. Assume that the drawing is made in the third angle of projection; the plan is therefore on top. The ground line is omitted because the dis- tance of the points from the principal planes is not required, but only their relative location to each other. To return, 1', 2', ... B' is the plan (horizontal projection) of the helix, showing the circular motion of the point around the axis ab. In the elevation (vertical projection), the starting point of the curve is shown at 1. If the point revolves in the direction of the arrow, it will, on making one-eighth of a revolution, be at 2' in plan
* The distinction might be made between cylindrical and conical helices by considering the curve as being drawn on the surface of a cylinder or a cone as the case may be.
t Helical springs are frequently although incorrectly called spiral springs. The spiral has been previously denned; and a spring of that shape is a spiral spring.
CLASSIFICATION OF LINES
153
and at 2 in elevation. After a quarter of a revolution, the point is at 3' in plan, and at 3 in elevation. The position 3 is its extreme movement to the right, for at 4, the point has moved to the left, although continually upward as shown in the elevation. On completion of one revolution, the point is at 9, ready to proceed with an identical curve beyond it.
The distance between any point and the position of the point after one complete revolution is known as the pitch. The dis- tance p is that pitch, and may be given from any point on the curve to the succeeding position of that point after one
FIG. 154.
FIG. 155.
revolution. The distance 01', 02', etc. is the radius of the helix.
Fig. 155 is a double helix and consists simply of two distinct helices, generated so that the starting point of one helix is just one-half of a revolution ahead or behind the other helix. The portion of the helix that is in front of the axis is shown in full; that behind the axis is shown dotted to give the effect of a helix drawn on a cylinder. The pitch again is measured on any one curve and requires a complete revolution. It must not be measured from one curve to a similar position on the next curve as the two helices are entirely distinct. Fig. 155 will show this correctly. Double, and even triple and quadruple helices are used on screws; they are simply interwoven so as to be equal distances apart.
154
GEOMETRICAL PROBLEMS IN PROJECTION
916. Classification of lines.
Straight Lines: Unidirectional.
Lines
Curved Lines;
Singly Curved Lines: Di- rection continually chang- ing, but always in a single plane.
Doubly Curved Lines: Di- rection continually chang- ing, but not in any one plane.
circle
ellipse
parabola
hyperbola
cycloid
epicycloid
hypocycloid
spiral
etc.
helix etc.
FIG. 156.
917. Tangent. A tangent is the limiting position of a secant as the points of secancy approach and ultimately reach coincidence. Suppose, in Fig. 156, pq is a secant to the curve
and that it revolves about p as a centre. At some time, the secant pq will assume some such position as pr; the point q has then moved to r and if it continues, it will pass through p and reach s. The secant then intersects on the opposite side of p. If the rotation be such that q passes through r and ultimately coincides with p, then this limiting
position of the secant, shown as pt, is the tangent to the curve and p is the point of tangency. It will be observed that the tangent is fixed by the point p, and the direction of the limiting position of the secant.
The condition of tangency is a mutual relation. That is, the curve is tangent to the line or the line is tangent to the curve. Also, two or more curves may be tangent to each other because the tangent line may be considered (at the point of tangency) as the direction of the curve (see Art. 920).
918. Construction of a tangent. If a tangent is to be drawn to a curve from an outside point, the drafting room method is to use a ruler of some sort and place it a slight distance away from the point and then revolve it until it nearly touches the
CLASSIFICATION OF LINES 155
curve; a straight line then drawn through the point and touching the curve will be the required tangent.
If the problem is to draw a tangent at a given point on a curve, however, the quick method would be to estimate the direction so that the tangent appears to coincide with as much of the curve on one side as it does on the other.
A more accurate method of constructing the tangent at a given point on a curve is shown in Fig. 157. Let a be the desired point of tangency. Draw through a, the secants ab, ac, ad, the number depending upon the degree of accuracy, but always more than here shown. With a as a centre, draw any indefinite arc eh, cutting the prolongations of the secants. Lay off the chord ab = ei, ac = fj, and ad=hk, but this etter chord is laid
FIG. 157.
off to the left of the arc, because its secant cuts the curve to the left of the desired point of tangency. A smooth curve may now be drawn through i, j, and k, and where this curve inter- sects the indefinite arc at g, draw ga, the required tangent. The proof of this is quite simple. If ab, ac and ad be considered as displacements of the points b, c and d from the positions that they should occupy when the secant becomes a tangent, then, when these points approach a, so as ultimately to coincide with it, the displacement is zero. The secant then becomes a tangent. The curve ijk is a curve of displacements from efh, the indefinite arc. Hence, the desired tangent must pass through g, as it lies on the curve efh and its displacement from that curve is therefore zero.
919. To find the point of tangency. On the other hand, if the tangent is drawn, and it is desired to find the point of
156
GEOMETRICAL PROBLEMS IN PROJECTION
v
V
Oik
FIG. 158.
tangency, the problem becomes slightly different. Suppose TT, Fig. 158, is the given tangent, and atb is the given curve. It is desired to find the point of tangency t. Draw ab, cd and ef,
any chords parallel to TT. Lay off hg, through a, perpendicular to TT, and br, through b, perpendicular also to TT. Make gh = qr = ab. In the other cases, make ij=op = cd, and kl=mn = ef. Now draw a — T smooth curve through hjlnpr. Where this curve intersects TT at t, the desired point of tangency is found. Again, the proof is quite simple. The lengths of the vertical lines above and below TT are equal to the lengths of the corresponding chords. As the chords approach
the tangent, they diminish in length and ultimately become zero. Where the curve crosses the tangent, the chord length is zero, and, hence, must be the point of tangency.
920. Direction of a curve. A curve continually changes its direction, but at any given point its direction is along the tangent to the curve by definition. It is proved in Mechanics, that if forces act on a particle so as to give it a curved motion, the particle will fly off along a tangent when the impressed forces cease to act.*
921. Angle between curves. The angle between two inter- secting curves is the same as the angles, made by the tangents at the point of intersection, because the tangents determine the direction of the curve at that point. Hence, to draw smooth curves, it is necessary that the tangent at the end of one curve should coincide with the tangent at the beginning of the next curve.
922. Intersection of lines. Two lines intersect when they have a point in common. When the term intersection is used in connection with a pair of straight lines, it necessarily implies that the lines make an angle with each other greater than zero;
* See Newton's laws of motion in a text-book on Physics.
CLASSIFICATION OF LINES 157
since, otherwise, if there is a zero angular relation, the lines become coincident and have all points in common.
When a line meets a curve, the angle between the line and the curve at the point of intersection is the same as the angle between the line and the tangent to the curve at that point. When this angle of intersection becomes zero, however, by having the line coincide with the tangent, then it is the special case of intersection known as tangency.
Similarly when two curves intersect their angle of intersection is the same as the angle between their tangents at the point of intersection. Also, when this angle becomes zero then the curves are tangent to each other.
Thus, in order to differentiate between the two types of intersection, angular intersection means intersection at an angle greater than zero; and, as a consequence, the intersection at a zero angle is known as tangential intersection. When inter- section is unmodified, then angular intersection is implied.
923. Order of contact of tangents. A tangent has been previously defined as the limiting position of a secant, as the points of secancy approach and ultimately coincide with each other. This contact, for simple tangency, is of the first order. Two curves may also be tangent to each other so as to have con- tact of the first order, as for example, two circles, internally or externally tangent.
Suppose two curves acb and abe, Fig. 159, have first order contact at a, and cut each other at the point b. If the curve abe be made to revolve about the point a as a centre, so as to maintain simple tangency * and also to have the point b approach a, then at some stage of the revolution the point b can be made to coincide with a. Under this condition there is tangency FIG. 159.
of a higher order because three points
were made to ultimately coincide with each other; and it is called second order of contact. It is possible to have third order of contact with four coincident points; and so on. The
* In order to make this a rigid demonstration, the centre of curvature •of the two curves, at the point a must not be the same. See Art. 924.
158 GEOMETRICAL PROBLEMS IN PROJECTION
order of contact is always one less than the number of points that approach coincidence.
924. Osculating circle. Centre of curvature. Let abc,
Fig. 160, be a curve and b, a point through which a circle gdbe passes, cutting the curve abc in three points d,b and e. If the diameter of the circle is properly chosen, it may be revolved about b as a centre so that the points d and e will both approach and ultimately coincide with b at the same instant. This position of the circle is shown as bf . Hence, the circle bf is tangent to the curve abc, and is of the second order of contact. This circle is the osculating circle. As this osculating circle must more nearly approach the curvature of the curve abc than any other circle, its
radius at the point b is the radius of curvature. At every other point on the curve, there is a new osculating circle, of a new centre, and of a new radius. Thus, the osculating circle is the second order tangent circle at the point; and the radius of curvature may be denned as the radius of the osculating circle through the point, the centre of curvature being the centre of the osculating circle.
925. Osculating plane. If a tangent be drawn to any doubly curved line, an infinite number of planes may be passed so as to contain the tangent. If some one position of the plane be selected so as to contain the tangent and a piercing point of the doubly curved line on it, then by proper revolution of the plane the piercing point can be made to approach and ultimately coincide with the point of tangency; in this position, the plane is an osculating plane.
To put the matter differently, suppose it is desired to find the osculating plane at some point on a doubly curved line. In this case, draw a pair of secants to the doubly curved line which intersect at the given point. A plane passed through these secants will cut the doubly curved line in three points. As the secants approach tangency, the plane will approach osculation, and this osculating plane is identical with that of the former discussion for the same point on the curve. If the curve under
CLASSIFICATION OF LINES 159
consideration be a plane curve, then the secants will lie in the plane of the curve, and, hence, the osculating plane of the curve will be the plane of the curve.
926. Point of inflexion. Inflexional tangent. Assume a curve dae, Fig. 161, and through some one point a draw the secant be. If the secant be revolved about the point a so that the points of intersection b and c approach a, at some stage of the revolution they will coincide with
it at the same instant if the point a
is properly chosen. The point a must
be such that three points on the curve
ultimately coincide at the same instant,
Further than this, the radius of curva- FIG. 161.
ture (centre of the osculating circle)
must make an abrupt change from one side of the curve to the
other at this point. The point of inflexion therefore is a point
at which the radius of curvature changes from one side of the
curve to the other. The inflexional tangent is the tangent at the
point of inflexion. It may also be noted that the inflexional
tangent has a second order of contact (three coincident points)
and therefore is the osculating line to the curve at the point of
inflexion.
927. Normal. A normal to a curve is a perpendicular to the tangent at the point of tangency. The principal normal lies in the osculating plane. As an infinite number of normals may be drawn to the tangent at the point of tangency, the nor- mal may revolve about the tangent so as to generate a plane which will be perpendicular to the tangent and thus establishes a normal plane. In the case of a circle, the radius at the point of tangency is normal to the tangent; in such cases, the tangent is easily drawn, as it must be perpendicular to the radius at the point of contact.
928. Rectification. When a curve is made to roll on a straight line, so that no slip occurs between the curve and the line, the distance measured on the line is equal to the corresponding length of the curve. This process of finding the length of the curve is called rectification. Commercially applied, the curve is measured by taking a divider and stepping off very small distances; the
160
GEOMETRICAL PROBLEMS IN PROJECTION
number of steps multiplied by the distance between points will give approximately the length of the curye. It must be noticed that the divider measures the chord distance, instead of the arc distance and is therefore always less than the actual length of the curve, but when the distance is taken small enough, the accuracy of the final result is proportional to the care taken in making the measurement.
929. Involute and evolute. When a tangent rolls about a fixed curve, any point, on the tangent describes a second curve
which is the involute of the first curve. Fig. 162 shows this in construction. Let aceg be the fixed curve, and ab, be the posi- tion of a taut string that is wound on the curve aceg. If a pencil point be attached to the string and unwound, the pencil point will describe the curve bdfh which is the involute of the curve aceg. The process is the same as though the tangent revolved about the curve aecg and some point on the
tangent acted as the generating point. At a, the radius is ab; at c, the radius is cd, which is equal to ab plus the rectified arc ac (length of string). It may be observed that the curve aceg is the curve of centres for the curve bdfh.
If the string be lengthened so that ai is the starting position, it will describe the curve ijkl which again is an involute. This second involute is parallel to the first, because the distance between the two curves measured along the rolling tangent (or radius of curvature) is constant between the two curves.
The primitive curve aceg is the evolute. The tangent rolls on the evolute and any point on it describes an involute. As any point on the tangent will answer as the tracing point, it follows that every evolute has an infinite number of involutes, all of which are parallel curves.
Reversing the process of the construction of the involute, the method of drawing the primitive curve or evolute is obtained. If normals are drawn to the involute, consecutive positions of
FIG. 162.
CLASSIFICATION OF LINES 161
the normals will intersect. The locus of these successive inter- sections will regenerate the primitive curve from which they have been evolved.*
Again, the length of the tangent to the e volute is the radius of curvature for the involute. As the radius of any circle is perpendicular to the tangent at that point, it follows that the involute is always normal, point for point, to the evolute, since the rolling tangent is the direction of the evolute at the point of contact and that again is normal to the involute.
930. Involute of the circle. The involute of the circle is a plane curve, described by a point on a tangent, while the tangent revolves about the circle.
Let o, Fig. 163, be the centre of a circle whose radius is oa. Let, also, a be the starting point of the involute. Divide the circle in any number of parts, always, however, more than are shown in the illustra- tion. Draw tangents to the various radii. On them, lay off the rectified arc of the circle between the point of tangency and the starting point. For FIG 163
instance, eb equals the rectified length
of the arc ea; also fc equals the rectified semicircle fea; gd equals the rectified length of the arc gfea. The involute may be continued indefinitely for an infinite number of revolutions, but, discussion is usually centred on some limited portion.
The curve here shown is approximately the same as that described by the end of the thread, when a spool is unwound.
QUESTIONS ON CHAPTER IX
1. Why are lines and points considered as mathematical concepts?
2. How is a straight line denned?
3. By what two means may a straight line be fixed in space?
4. What is a generating point?
5. What is a locus?
6. What is a singly curved line?
7. What is a plane curve?
8. How are plane curves represented?
* The evolute of a circle, therefore, is its centre.
162 GEOMETRICAL PROBLEMS IN PROJECTION
9. Show the mode of representing curves when their planes are parallel to the plane of projection? When perpendicular? When inclined?
10. Is it desirable to use the plane of the curve as the plane of projection?
11. Define the circle.
12. What is the radius? Diameter? Sector? Segment?
13. When the chord passes through the centre of the circle, what does
it become?
14. What is a secant? Quadrant? Semicircle?
15. Define concentric circles; eccentric circles; eccentricity.
16. What is an ellipse?
17. Define major axis of an ellipse; n:inor axis; foci.
18. Describe the accurate method of c rawing an ellipse by the intersection
of circular arcs.
19. Describe the trammel method of drawing an ellipse.
20. Draw an ellipse whose major axis is 3" long and whose minor axis
is 2" long. Use the accurate method of intersecting circular arcs.
21. Construct an ellipse whose major axis is 3" long and whose minor
axis is 1|" long. Use the trammel method.
22. What is a parabola?
23. Define focus of a parabola; directrix; axis; vertex.
24. Is the parabola symmetrical about the axis?
25. Is the parabola an open or a closed curve?
26. Construct the parabola whose focus is 2" from the directrix.
27. What is a hyperbola?
28. Define foci of hyperbola; transverse axis; conjugate axis; vertex;
asymptote.
29. How many branches has a hyperbola? Are they symmetrical about
the transverse and conjugate axes?
30. How many asymptotes may be drawn to a hyperbola?
31. Is the hyperbola an open or a closed curve?
32. Construct a hyperbola whose distance between foci is 2" and whose
constant difference is \" .
33. What is a cycloid?
34. Define rolling circle of a cycloid; directrix.
35. Construct a cycloid whose diameter of rolling circle is \\" . Draw
the curve for one revolution only.
36. What is an epicycloid?
37. What form of directrix has the epycycloid?
38. Construct the epicycloid, whose diameter of rolling circle is \\"
and whose diameter of directrix is 8". Draw the curve for one revolution only.
39. What is a hypocycloid?
40. What is the form of the directrix of the hypocycloid?
41. Construct a hypocycloid whose diameter of rolling circle is 2" and
whose diameter of directrix is 9". Draw the curve for one revolu- tion only.
42. What is a spiral?
CLASSIFICATION OF LINES 163
43. Construct an Archimedian spiral which expands 1|" in a complete
revolution. Draw the spiral for two revolutions.
44. What is a doubly curved line?
45. Draw a doubly curved line with the principal planes in oblique
projection.
46. Construct the orthographic projection from Question 45.
47. What is a helix?
48. Define uniform cylindrical helix; conical helix; diameter of helix;
pitch.
49. Construct a helix whose diameter is 2", and whose pitch is I".
Draw for two revolutions.
50. Construct a triple helix whose diameter is 2" and whose pitch is 3".
The helices are spaced equally and are to be drawn for 1| revolu- tions.
51. Make a classification of lines.
52. What is a tangent?
53. Is the tangent fixed in space by a point and a direction?
54. Show that the tangent is the limiting position of a secant.
55. How is a tangent drawn to a curve from a point outside?
56. Given a curve and a point on it, draw a tangent by the accurate
method. Prove that the chord length is zero for the tangent position.
57. Given a curve and a tangent, determine the point of tangency. Prove.
58. What is the direction of a curve?
59. What is the angle between two intersecting curves?
60. When several curves are to be joined, show what must be done to
make them smooth curves.
61. Define intersection of lines.
62. Show that the tangent intersects the curve at a zero angle.
63. Define order of contact of tangents.
64. If three points become coincident on tangency, what order of contact
does the tangent have?
65. Define osculating circle.
66. Define centre of curvature.
67. Is the radius of the osculating circle to a curve the radius of curvature
at that point?
68. Show how an osculating circle may have second order contact with
a plane curve.
69. What is an osculating plane?
70. When is the osculating plane the plane of the curve?
71. Define point of inflexion.
72. What is an inflexional tangent?
73. Does the radius of curvature change from one side of the curve to
the other at a point of inflexion?
74. Define principal normal.
75. Show that the centres of curvature at a point of inflexion lie on
opposite sides of the normal to the curve through the point of inflexion.
164 GEOMETRICAL PROBLEMS IN PROJECTION
76. When a normal is drawn to a curve is the one in the osculating plane
the one generally understood?
77. How many normals may be drawn to a doubly curved line at a
given point?
78. What is meant by rectification?
79. Rectify a 2" diameter circle. Compute the length of the rectified
circle (=2x3.1416) and express the ratio of rectified to computed length as a percentage.
80. Define involute and evolute.
81. Show that all involutes to a curve are parallel curves.
82. Show that the involute is always normal to the evolute at the point
for which it corresponds.
83. Show that the drawing of the evolute is the reverse process of drawing
the involute.
84. Draw the involute of a circle.
85. Draw the involute to an ellipse.
86. Draw the evolute to an ellipse.
87. Draw the involute to a parabola.
88. Draw the evolute to a parabola.
89. Draw the involute to one branch of a hyperbola.
90. Draw the evolute to one branch of a hyperbola.
91. Draw the involute to a cycloid.
92. Draw the evolute to a cycloid.
93. Draw the involute to an epicycloid.
94. Draw the evolute to an epicycloid.
95. Draw the involute to a hypocycloid.
96. Draw the evolute to a hypocycloid.
97. Draw the involute to an Archimedian spiral.
98. Draw the evolute to an Archimedian spiral.
CHAPTER X
CLASSIFICATION OF SURFACES
1001. Introductory. A surface may be generated by the successive positions of a line which moves so as to generate an area. As there are infinite varieties of lines and as their motion may again be in an infinite variety of ways, therefore, an infinite variety of possible surfaces result. In engineering it is usual to limit the choice of surfaces to such as may be easily reproduced and easily represented. Surfaces, like lines and points, are mathematical concepts because they have no material exist- ence.
When curved surfaces, of a more or less complex nature, are to be represented, they may be shown to advantage, by the effects of light on them. Examples of this kind are treated in Chapters XIV and XV.
1002. Plane Surface. A plane surface is a surface such that when any two points in it are joined by a straight line, the line lies wholly within the surface. Thus, three points may be selected in a plane and two intersecting lines may be drawn through the three points; the intersecting lines lie in the plane and, therefore, may be used 'to determine it. Also, a line and an external point may determine a plane.
The plane surface may also be conceived as being generated by a straight line, moving so as to touch another line, and con- tinually remaining parallel to its original position. Hence, also, two parallel lines determine a plane.
In the latter case, the moving straight line may be consid- ered as a rectilinear generatrix, touching a rectilinear directrix, and occupying consecutive positions in its motion. Any one position of the generatrix may be used as an element of the surface.
165
166
GEOMETRICAL PROBLEMS IX PROJECTION
Fig. 164 shows a plane surface ABCD on which straight lines ab, cd, ef and gh are drawn, all of which must lie wholly within the plane, irrespective of the direction in which they are drawn. Any curve drawn on this surface is a plane curve.
1003. Conical surface. If a straight line passes through a given point in space and moves so as to touch a given fixed curve, the surface so generated is a conical surface. The straight line is the rectilinear generatrix, the fixed point is the vertex and the given fixed curve is the directrix, which need not be a closed curve. The generatrix in any one position is an element of the surface.
Fig. 165 shows a conical surface, generated in the manner
FIG. 164.
FIG. 165.
indicated. Either the upper or the lower curve may be consid- ered as the directrix. In fact, any number of lines may be drawn on the resulting surface, whether the lines be singly or doubly curved, and any of which will fill the office of directrix. A plane curve is generally used as the directrix.
With a generatrix line of indefinite extent, the conical sur- face generated is a single surface (not too surfaces as might appear) ; the. vertex o is a point of union and not of separation. The portion of the surface from the vertex to either side is called a nappe;* hence, there are two nappes to a conical surface.
1004. Cone. The cone is a solid, bounded by a closed conical surface of one nappe and a plane cutting all the elements. The * Pronounced "nap."
CLASSIFICATION OF SURFACES
167
curve of intersection of the rectilinear elements and the plane cutting all the elements is the base. A circular cone has a circle for its base and the line joining the vertex with the centre of the base is the axis of the cone. If the axis is perpendicular to the plane of the base the cone is a right cone. When the base is a circle, and the axis is perpendicular to the plane of the base, the cone is a right circular cone or a cone of revolution.* A cone of revolution may be generated by revolving a right tri- angle about one of its legs as an axis. The hypothenuse is then the slant height of the cone. The perpendicular distance from the vertex to the plane of the base is the altitude of the cone. The foot of the perpendicular may fall outside of the centre of the base and in such a case, the cone is an oblique cone.
The frustum (plural : — frusta) of a cone is the limited portion of the solid bounded by a closed conical surface and two parallel planes, each cutting all the elements, and giving rise to the upper base and the lower base of the frustum of a cone. The terms upper and lower base are relative; it is usual to consider the larger as the lower base and to represent the figure as resting on it. When the cutting planes are not parallel, then the solid is a truncated cone.
1005. Representation of the cone. A cone, like any other object, is represented by its projec- tions on the principal planes. For con- venience in illustrating a cone, the plane of the base is assumed perpendicular to one of the principal planes, as then its projection on that plane is a line. Fig. 166 shows a cone in orthographic projec- tion. The vertex O is shown by its projections o and o'; dV is the vertical projection of the base since the plane of the base is assumed perpendicular to the vertical plane. The extreme limiting ele- ments oV and o'd' are also shown, thus
completing the vertical projection. In the horizontal projection, any curve acbd is assumed as the projection of the base so
* This distinction is made because a cone with an elliptical base may also be a right cone when the vertex is chosen so that it is on a perpendicular to the plane of the base, at the intersection of the major and minor axes.
FIG. 166.
168
GEOMETRICAL PROBLEMS IN PROJECTION
that d and c are corresponding projections of d' and c'. From o, the lines ob and oa are drawn, tangent to acbd, thus completing the horizontal projection.
It must here be emphasized, that acbd is not the actual base of the cone, but only its projection. It is impossible to assume two curves, one in each plane of projection, and call them corre- sponding projections of the same base. The corresponding points must be selected, so that they will lie in one plane, and that plane must be the plane of the base. If it be desired to show the base in both projections, when the plane of the base is in- clined to the principal planes, it is necessary to assume one pro- jection of the base. Lines are then drawn in that plane, through the projection of the base and the corresponding 'projections
FIG. 167.
FIG. 168.
of the lines are found. The points can then be determined as they must be situated on these lines. (Arts. 704 and 811.)
1006. To assume an element on the surface of a cone.
To assume an element of a cone, assume the horizontal projec- tion oa in Fig. 167. There are two elements on the cone which have the same horizontal projection and they are shown as o'a' and o'b' in the vertical projection. If oa be considered visible, while viewing the horizontal plane, then o'a' is its corre- sponding projection. If o'b' be the one assumed projection then oa is on the far side and should in this case be drawn dotted.
1007. To assume a point on the surface of a cone. To
assume a point on the surface of a cone, assume c' in Fig. 168 as the vertical projection, somewhere within the projected area.
CLASSIFICATION OF SURFACES
169
Draw the element o'b' through c' and find the corresponding, projection of the element. If o'b' is visible to the observer, then ob is the corresponding projection, and c on it is the required projection. If o'b' is on the far side, then d is the desired pro- jection.
A slightly different case is shown in Fig. 169. If c is assumed on the visible element oa then c' is the corresponding projection on o'a'. Otherwise, if ob is dotted (invisible) then o'b' is the corresponding element and d and d' are corresponding projections.
1008. Cylindrical surface. When a straight line moves so that it remains continually parallel to itself and touches a given fixed curve, the surface generated is a cylindrical surface.
FIG. 169.
FIG. 170.
The straight line is the rectilinear generatrix, the fixed curve is the directrix and need not be a closed curve. The generatrix in any one position is an element of the surface.
Fig. 170 shows a cylindrical surface, generated in the manner indicated. Any curve, drawn on the resultant surface, whether singly or doubly curved, may be considered as the directrix. The limiting curves that are shown in the figure may also be used as directrices. A plane curve is generally used as a directrix
1009. Cylinder. A cylinder is a solid bounded by a closed cylindrical surface and two parallel planes cutting all the elements. The planes cut curves from cylindrical surface which form the bases of the cylinder and may be termed upper and lower bases if the cylinder is so situated that the nomenclature fits. When the planes of the bases are not parallel then it is called a truncated cylinder.
170 GEOMETRICAL PROBLEMS IN PROJECTION
Should the bases have a centre, a figure such as a circle for instance, then a line joining these centres is the axis of the cylinder. The axis must be parallel to the elements of the cylinder. If the axis is inclined to the base, the cylinder is an oblique cylinder. On the other hand, of the axis is perpendicular to the plane of the base, it is a right cylinder and when the base is a circle, it is a right circular cylinder, or, a cylinder of revolution. The cylinder of revolution may be generated by revolving a rectangle about one of its sides as an axis. A right cylinder need not have a circular base, but the elements must be perpendicular to the plane of the base.
1010. Representation of the cylinder. A cylinder, represented orthographically is shown in Fig. 171. Suppose the base is assumed in the horizontal plane, then e'g' may be
taken as the vertical projection of the base. Also e'f and g'h', parallel to each other, may be taken as the projections of the extreme limiting elements. Any curve, as aecg, may be drawn for the horizontal projection so long as e and g are corre- sponding projections of e' and g'. In addi- tion, draw ab and cd parallel to each other and tangent to the curve aecg. The hori- zontal projection is thus completed. It is to be noted that aecg is the true base FIG. 171. because it lies in the horizontal plane. If
the plane of the base does not coincide
with the horizontal plane, then, as in Art. 1005, what applies to the selection of the projection of the base of the cone applies here.
1011. To assume an element on the surface of a cylinder.
To assume an element on the surface of a cylinder, select any line, ab, Fig. 172, as the horizontal projection. As all parallel lines have parallel projections, then ab must be parallel to the extreme elements of the cylinder. If ab is assumed as a visible element, then a'b' is its corresponding projection, and is shown dotted, because hidden from view on the vertical projection. If c'd' is a visible element, then cd should be dotted in the horizontal projection.
CLASSIFICATION OF SURFACES
171
1012. To assume a point on the surface of a cylinder.
To assume a point on the surface of a cylinder, select any point c, Fig. 173, in the horizontal projection, and draw the element ab through it. Find the corresponding projection a'b' and on it, locate c', the required projection. What has been said before (Art. 1005) about the two possible cases of an assumed projection, applies equally well here and should require no further mention.
1013. Convolute surface. A convolute surface is a surface generated by a line which moves so as to be continually tangent * to a line of double curvature. For purposes of illustration, the uniform cylindrical helix will be assumed as the line of double
11 13
FIG. 172.
FIG. 173.
FIG. 174.
curvature, remembering, in all cases, that the helix may be variable in radius and in motion along the axis, so that its char- acteristics may be imparted to the resulting convolute.
The manner in which this surface is generated may be gained from what follows. Let abed, Fig. 174, be the horizontal pro- jection of a half portion octagonal prism on which is a piece of paper in the form of a right triangle is wound. The base of the triangle is therefore the perimeter of the prism and the hypoth- enuse will appear as a broken line on the sides of the prism. If the triangle be unwound from the prism and the starting point of the curve described by the hypothenuse on the horizontal
* It may be observed that the tangent to a line of double curvature must ile in the osculating plane (Art. 925).
172
GEOMETRICAL PROBLEMS IN PROJECTION
plane be at o, then the portion of the triangle whose base is oa will revolve about the edge a so as to describe the arc ol. At the point 1 the triangle is free along the face ab and now swings about b as a centre and describes the arc 1-2. As the process goes on to the point 2, the triangle is free on the face be and then swings about c as a centre and describes the arc 2-3; and so on. In the vertical projection, the successive positions of the hypothenuse are shown by a'l', b'2', c'3' etc. It will be noted that a'l' intersects b'2' at b', and b'2' intersects c'3' at c', etc.; but, a'l' does not intersect c'3' nor does b'2' intersect d'4'. Hence, the elements of the surface generated by the hypothenuse intersect two and two. The first element intersects the second; the second the third; the third the fourth, etc.;
but the first does not intersect the third, or any beyond, nor does the second intersect the fourth or any elements beyond the fourth. When the prism approaches a cylinder as a limit by increas- ing the number of sides indefinitely, the hypothenuse wound
FIG. 174.
FIG. 175.
FIG. 176.
around the cylinder approaches a helix as a limit; the unwind- ing hypothenuse will become the generatrix, tangent to the helix, and will approach the desired convolute surface. The ultimate operation is a continuous one and may be seen in Fig. 175. The curve abed described by the hypothenuse fd on the plane MM
CLASSIFICATION OF SURFACES
173
FIG. 177.
is the involute of a circle, if the cylinder is a right circular cylinder.
Fig. 176 may indicate the nature of the surface more clearly. Examples of this surface may be obtained in the machine shop on observing the spring-like chips, that issue on taking a heavy cut from steel or brass. The surfaces are perhaps not exact convolutes, but they resemble them enough to give the idea.
It is not necessary to have the tangent stop abruptly at the helix, the tangent may be a line of indefinite extent, and, hence, the convolute surface extends both sides of the helical directrix. No portion of this surface intersects any other portion of the surface, but all the convolutions are dis- tinct from each other. Fig. 177 will perhaps convey the final idea.
1014. Oblique helicoidal screw surface.* When the helical directrix of a convolute surface decreases in diameter, it will
ultimately coincide with the axis and the helix will become a line. The oblique helicoidal screw sur- face, therefore, resolves itself into FIG. 178. tne surface generated by a recti-
linear generatrix revolving about
another line which it intersects, at a constant angle, the inter- section moving along the axis at a uniform rate. The application of this surface is shown in the construction of the V thread screw which in order to become the United States Standard screw, must make an angle of 60° at the V as shown in Fig. 178.
* The helicoid proper is a warped surface (1016). If a straight line touches two concentric heb'ces of different diameters and lies in a plane tangent to the inner helix's cylinder, the line will generate a warped surface. When the diameter of the cylinder becomes zero, the helix becomes a line and the helicoidal surface is the same as that here given.
174
GEOMETRICAL PROBLEMS IN PROJECTION
FIG. 179.
1015. Right helicoidal screw surface. If the diameter of the helical directrix still remains zero, and the rectilinear
generatrix becomes perpendicular to the axis and revolves so that the in- tersection of the axis and the genera- trix moves along the axis at a constant rate,* a special case of the convolute is obtained. This special case of the convolute is called a right helicoidal screw surface, and when applied gives the surface of a square threaded screw as shown in Fig. 179.
In both cases of the oblique and right helicoidal screw sur- faces the helices at the outside and root (bottom) of the thread, are formed by the intersection of the screw surface and the outer and inner concentric cylinders. The pitch of both must be the same, as every point on the generatrix advances at a uniform rate. Hence, the angle of the tangent to the helix must vary on the inner and outer cylinders. For this reason, the helices have a different shape notwithstanding their equal pitch.
1016. Warped surface. A warped surface is a curved surface, generated by a rectillinear generatrix, moving so that no two successive elements lie in the same plane. Thus, the consecutive elements can neither be parallel nor intersect, hence, they are skew lines. An example of this surface may be obtained by taking a series of straight sticks and drilling a small hole through each end of every stick. If a string be passed through each end and secured so as to keep them together, the series of sticks may be laid on a flat surface and thus represent successive elements of a plane. It may also be curved so as to represent a cylinder. Lastly, it may be given a twist so that no single plane can be passed through the axis of successive sticks; this latter case would then represent a warped surface.
Warped surfaces find comparatively little application in engineering because they are difficult to construct or to duplicate. At times, however, they are met with in the construction of
* If the pitch becomes zero when the diameter of the helix becomes zero, it is the case of a line revolving about another line, through a fixed point; the surface is therefore a cone of revolution, if the generatrix is inclined to the axis. If the generatrix is normal to the axis, the surface is a plane.
CLASSIFICATION OF SURFACES 175
" forms " for reinforced concrete work, where changes of shape occur as in tunnels, and similar constructions; in propeller screws for ships; in locomotive " cow-catchers/' etc.
1017. Tangent plane. If any plane be passed through the vertex of a cone, it may cut the surface in two rectilinear elements under which condition it is a secant plane. If this secant plane be revolved about one of the rectilinear elements as an axis, the elements of secancy can be made to approach so as to coincide ultimately. This, then, is a limiting position of the secant plane, in which case it becomes a tangent plane, having contact with the cone all along one element.
If through some point on the element of contact, two inter- secting curved lines be drawn on the surface of the cone, then, also, two secants may be drawn to these curved lines and inter- secting each other at the intersection of the curved lines. The limiting positions of these secants will be tangent lines to the cone, and as these tangents intersect, they determine a tangent plane. The tangent plane thus determined is identical with that obtained from the limiting position of a secant plane.
Instead of drawing two intersecting curves on the surface of the cone, it is possible to select the element of contact and any curve on the cone intersecting it. The tangent plane in this case is determined by the element of contact and the limiting position of one secant to the curve through the intersection of the element and the curve.
As another example, take a spherical surface and on it draw two intersecting lines (necessarily curved). Through the point of intersection, draw two secants, one to each curve and deter- mine the tangent positions. Again, the plane of the two inter- secting tangents is the tangent to the sphere. Hence, as a general definition, a tangent plane is the plane established by the limiting position of two intersecting secants as the points of secancy reach coincidence.
1018. Normal plane. The normal plane is any plane that is perpendicular to the tangent plane. If a normal plane is to be drawn to a sphere at a given point, for instance, then construct the tangent plane and draw through the point of tangency any plane perpendicular to the tangent plane. An infinite number of normal planes may be drawn, all passing through the given
176 GEOMETRICAL PROBLEMS IN PROJECTION
point. The various normal planes will intersect in a common line, which is normal to the tangent plane at the point of tangency.
1019. Singly curved surface. A singly curved surface is a surface whose successive rectilinear elements may be made to coincide with a plane. Hence, a tangent plane must be in contact all along some one rectilinear element. As examples, the conical, cylindrical, convolute and the helicoidal screw surfaces may be mentioned.
1020. Doubly curved surface. A doubly curved surface is a surface whose tangent plane touches its surface at a point. Evidently, any surface which is not plane or singly curved must be doubly curved. The sphere is a familiar example of a doubly curved surface.
1021. Singly curved surface of revolution. A singly curved surface of revolution is a surface generated by a straight line revolving about another straight line in its own plane as an axis, so that every point on the revolving line describes a circle whose plane is perpendicular to the axis, and whose centre is in the axis. Thus, only two cases of singly curved surfaces "can obtain, the conical and the cylindrical surfaces of revolution.
1022. Doubly curved surface of revolution. A doubly curved surface of revolution is a surface generated by a plane curve revolving about a straight line in its own plane as an axis so that every point on the revolving curve describes a circle whose plane is perpendicular to the axis, and whose centre lies in the axis. Hence, there are infinite varieties of doubly curved sur- faces of revolution as the sphere, ellipsoid, hyperboloid, etc., generated by revolving the circle, ellipse, hyperbola, etc., about their axes. In the case of the parabola, the curve may revolve about the axis or the directrix in which cases two distinct types of surfaces are generated. Similarly with the hyperbola, the curve may generate the hyperboloid of one or two nappes depend- ing upon whether the conjugate or transverse axis is the axis of revolution, respectively.
Sometimes, a distinction is made between the outside and inside surfaces of a doubly curved surface. For example, the outside surface of a sphere is called a doubly convex surface, whereas, the inside is an illustration of a doubly concave surface.
CLASSIFICATION OF SURFACES
177
A circular ring made of round wire, and known as a torus, is an example of a doubly concavo-convex surface.
1023. Revolution of a skew line. An interesting surface is the one generated by a pair of skew
lines when one is made to revolve about the other as an axis. Fig. 180 gives such a case, and as no plane can be passed through successive elements, it is a warped surface. While revolving about the axis, the line generates the same type of surface as would be gen- erated by a hyperbola when revolved about its conjugate axis. The surface is known as the hyperboloid of revo- lution of one nappe, and, incidentally, is the only warped surface of revolu- tion.
1024. Meridian plane and me- ridian line. If a plane be passed through the axis of a doubly curved surface of revolution, it will cut from the surface a line which is the me- ridian line. The plane cutting the -pic 180 meridian line is called the meridian
plane. Any meridian line can be used as the generatrix of the sur- face of revolution, because all meridian lines are the same. The circle is the meridian line of a sphere, and for this particular surface, every section is a circle. In general, every plane perpendicular to the axis will cut a circle from any surface of revolution, whether singly or doubly curved.
1025. Surfaces of revolution having a common axis. If two surfaces of revolution have a common axis and the surfaces intersect, tangentially or angularly, they do so all around in a circle which is common to the two surfaces of revolution. Thus, the two surfaces shown in Fig. 181 intersect, angularly, in a circle having ab as a diameter, and intersect, tangentially, in a circle having cd as a diameter.
FIG. 181.
178
GEOMETRICAL PROBLEMS IN PROJECTION
1026. Representation of the doubly curved surface of revolution. Fig. 182 shows a doubly curved surface of revolution shown in two views. One view shows the same as that produced by a meridian plane cutting a meridian line and the other shows the same as concentric circles. When considered as a solid, no ground line is necessary as the distance from the principal planes is unimportant. Centre lines, ab, cd and ef
FIG. 182.
should be shown, ab and ef being represented as two lines, be- cause both views are distinct from each other. The lines indi- cate that the object is symmetrical about the centre line as an
axis.
1027. To assume a point on a doubly curved surface of revolution. Let c, Fig. 183, be assumed as one point on the
surface.* With oc as a radius, draw the arc ca, a is, there- fore, the revolved position of c when the meridian plane through co has been revolved to ao. Hence, a is at a' or a" in the corresponding view. On counter-revolution, a' de- scribes a circle, the plane of which is perpendicular to the axis, and the plane is shown by its trace a'c'; in the other view, a returns to c and, hence, c' is the final position. It may also be at c" for the same reason, but then c is hidden in that view. If, on the other hand, d is chosen as one projection, its corre-
FIG. 183.
* These views bear third angle relation to each other.
CLASSIFICATION OF SURFACES 179
spending projection is d', if d is visible; or, it is d", if d is
hidden.
1028. Developable surface. When a curved surface can be rolled over a plane surface so that successive elements come in contact with the plane and that the area of the curved surface can be made to equal the plane surface by rectification, the surface is a developable one. Hence, any singly curved surface, like a cylinder, can be rolled out flat or developed. A sphere cannot be rolled out as a flat surface because it has point con- tact with a plane, and is, therefore, incapable of development. If a sphere is to be constructed from flat sheets, it may be ap- proximated by cutting it into the type of slices called lunes, resembling very much the slices made by passing meridian planes through the axis of a sphere. To approach more nearly the sphere, it would be necessary to take these lunes and hammer them so as to stretch the material to the proper curvature. Simi- larly, in the making of stove-pipe elbows, the elbows are made of limited portions of cylinders and cut to a wedge shape so as to approximate the doubly curved surface known as the torus.
1029. Ruled surface. Every surface on which a straight line may be drawn is called a ruled surface. A ruled surface may be plane, singly curved or doubly curved. Among the singly curved examples may be found the conical, cylindrical, convolute and helicoidal screw surfaces. The hyperboloid of revolution of one nappe furnishes a case of a doubly curved ruled surface (1023).
1030. Asymptotic surface. If a hyperbola and its asymp- totes move so that their plane continually remains parallel to itself, and any point on the curve or on the asymptotes touches a straight line as a directrix, the hyperbola will generate a hy- perbolic cylindrical surface and the asymptotes will generate a pair of asymptotic planes. Also, if the hyperbola revolves about the transverse or conjugate axis, the hyperbola will generate a hyperboloid of revolution and the asymptote will generate a conical surface which is asymptotic to the hyperboloid. In all cases, the asymptotic surface is tangent at two lines, straight or curved, at an infinite distance apart and the surface passes within finite distance of the intersection of the axes of the curve.
180
GEOMETRICAL PROBLEMS IN PROJECTION
1031. Classification of surfaces.
Surfaces.
Ruled surfaces. Straight lines may be drawn on the resulting sur- face.
Singly curved sur- faces. May be developed into a flat surface by rec- tification.
Doubly curved surfaces. Tan- gent plane touches surface at a point.
Planes. Any two
points when joined
by a straight line lie
wholly within the
surface.
Conical surfaces. Rectilinear ele- ments pass through a given point in space and touch a curved directrix.
Cylindrical surfaces. Rectilinear ele- ments are parallel to each other and touch a curved di- rectrix.
Convolute surfaces. Rectilinear ele- ments tangent to a line of double curva- ture. Consecutive elements intersect two and two; no three intersect in a common point. Warped surfaces. No two consecutive elements lie in the same plane; hence, they are non-devel- opable.
Doubly curved surfaces of revolution.
Generated by plane curves revolving about an axis in the plane of the curve. All meridian lines equal and all sections per- pendicular to axis are circles.
Unclassified doubly curved surfaces. All others which do not fall within the fore- going classification.
QUESTIONS ON CHAPTER X
1. How are surfaces generated?
2. What is a plane surface?
3. What is a rectilinear generatrix?
4. WHat is a directrix?
5. What is an element of a surface?
6. What is a conical surface?
CLASSIFICATION OF SURFACES 181
7. What is the directrix of a conical surface?
8. What is the vertex of a conical surface?
9. Is it necessary for the directrix of a conical surface to be closed?
10. What is a nappe of a cone? How many nappes are generated in a
conical surface?
11. What is a cone?
12. What is the base of a cone?
13. Must all elements be cut by the plane of the base for a cone?
14. What is a circular cone?
15. What is the axis of a cone?
16. What is a right circular cone? Is it a cone of revolution?
17. What is the altitude of a cone?
18. What is the slant height of a cone?
19. What is the an oblique cone?
20. What is a frustum of a cone?
21. How are the two bases of a frustum of a cone usually designated?
22. What is a truncated cone?
23. In the representation of a cone, why is the plane of the base usually
assumed perpendicular to the plane of projection?
24. Is it necessary that the base of a cone should be circular?
25. Draw a cone in orthographic projection and assume the plane of
the base perpendicular to the vertical plane.
26. Draw a cone in projection and show how an element of the surface
is assumed in both projections. State exactly where the element is chosen.
27. Draw a cone in projection and show how a point is assumed on its
surface. Locate point in both projections.
28. What is a cylindrical surface?
29. Define generatrix of a cylindrical surface; directrix; element.
30. Is it necessary that the directrix of a cylindrical surface be a closed
curve?
31. How is a cylinder differentiated from a cylindrical surface?
32. How many bases must a cylinder have?
33. What is the axis of a cylinder?
34. Must the axis of the cylinder be parallel to the elements? Why?
35. What is an oblique cylinder?
36. What is a right circular cylinder? Is this cylinder a cylinder of
revolution?
37. Is it necessary that a right cylinder have a circle for the base? Why?
38. Draw an oblique cylinder whose base lies in the horizontal plane.
39. In Question 38, assume an element of the surface and state which
element is chosen.
40. Which elements are the limiting elements in Question 38?
41. Draw a cylinder in projection and then assume a point on the surface
of it. Show it in both projections.
42. What is a convolute surface?
43. What is an oblique helicoidal screw surface? Give a prominent
example of it.
182 GEOMETRICAL PROBLEMS IN PROJECTION
44. What is a right helicoidal screw surface? Give a prominent example
of it.
45. Show that the helicoidal screw surfaces are limiting helicoids as the
inner helix becomes of zero diameter.
46. What is a warped surface? Illustrate by sticks.
47. What is a secant plane?
48. Show how a tangent plane is the limiting position of a secant plane
to a conical surface.
49. Show how a tangent plane is the limiting position of a secant plane
to a cylindrical surface.
50. Show that a tangent line to the surface is the limiting position of a
secant drawn to a curve on the surface.
51. Show how two intersecting lines may be drawn on a curved surface
and how the limiting positions of two secants drawn through this point of intersection determine a tangent plane to the surface.
52. Show how one element of a conical surface and one limiting posi-
tion of a secant determine the tangent plane to the conical surface.
53. Show how one element of a cylindrical surface and one limiting
position of a secant determine the tangent plane to the cylindrical surface.
54. Show how two intersecting lines may be drawn on the surface of a
sphere and how the limiting positions of two secants drawn through the intersection determine a tangent plane to the sphere.
55. Define tangent plane in terms of the two intersecting tangent lines
at a point on a surface.
56. Define normal plane.
57. How many normal planes may be drawn through a given point
on a surface?
58. What is a normal (line) to a surface?
59. Define singly curved surface. Give examples.
60. Define doubly curved surface. Give examples.
61. Define singly curved surface of revolution. Give examples.
62. Define doubly curved surface of revolution. Give examples.
63. What is a doubly convex surface? Give examples.
64. What is a doubly concave surface? Give examples.
65. What is a doubly concavo-convex surface. Give examples.
66. Describe the surface of a torus. Is this a concavo-convex surface?
67. What surface is obtained when a pair of skew lines are revolved
about one of them as an axis?
68. Construct the surface of Question 67.
69. What is a meridian plane?
70. What is a meridian line?
71. Why can any meridian line be assumed as a generatrix for its partic-
ular surface of revolution?
72. What curves are obtained by passing planes perpendicular to the
axis of revolution?
CLASSIFICATION OF SURFACES 183
73. When two surfaces of revolution have the same axis, show that the
intersection is a circle whether the surfaces intersect tangentially or angularly.
74. Show how a doubly curved surface of revolution may be represented
without the ground line. Draw the proper centre lines.
75. Assume a point on the surface of a doubly curved surface of revolu-
tion.
76. What is a developable surface?
77. Is development the rectification of a surface?
78. Are singly curved surfaces developable?
79. Are doubly curved surfaces developable?
80. Are warped surfaces developable?
81. What is a ruled surface? Give examples.
82. What is an asymptotic plane? Give an example.
83. If a hyperbola and its asymptotes revolve about the transverse
axis, show why the asymptotic lines generate asymptotic cones to the resulting hyperboloids.
84. Show what changes occur in Question 83 when the conjugate axis
is used.
85. Make a classification of surfaces.
CHAPTER XI
INTERSECTIONS OF SURFACES BY PLANES, AND THEIR DEVELOPMENT
1101. Introductory. When a line is inclined to a plane, it will if sufficiently produced, pierce the plane in a point. The general method involved has been shown (Art. 823), for straight lines, and consists of passing an auxiliary plane through the given line, so that it cuts the given plane in a line of intersection. The piercing point must be somewhere on this line of intersection and also on the given line; hence it is at their intersection.
In the case of doubly curved lines, the passing of auxiliary planes through them is evidently impossible. Curved surfaces, instead of planes, are therefore used as the auxiliary surfaces. Let, for example, Fig. 184 show a doubly curved line, and let the
object be to find the piercing
point of the doubly curved line on the principal planes. If a cylindrical surface be -passed through the given line, the elements of which are perpen- dicular to the horizontal plane, it will have the curve ab as FIG. 184. its trace, which will also be the
horizontal projection of the
curve AB in space. Similarly, the vertical projecting cylindrical surface will cut the vertical plane in the line a'b', and will be the vertical projection of the curve AB in space. If a perpendicular be erected at a, where ab crosses the ground line, it will intersect the vertical projection of the curve at a', the vertical piercing point of the curve. The entire process in substance consists of this: the surface of the horizontal projecting cylinder cuts the vertical plane in the line aa'; the piercing point of the curve AB must
184
INTERSECTIONS OF SURFACES BY PLANES 185
lie on aa' and also on AB, hence it is at their intersection a'. Likewise, b, the horizontal piercing point is found by a process identical with that immediately preceding.
1102. Lines of intersection of solids by planes. The
extension of the foregoing is the entire scheme of finding the line of intersection of any surface with the cutting plane. Ele- ments of the surface pierce the cutting plane in points; the locus of the points so obtained, determine the line of intersection.
A distinction must be made between a plane cutting a sur- face, and a plane cutting a solid. In the former case, the surface', alone, gives rise to the line of intersection, whether it be an open surface, or a closed surface; the cutting plane intersects the surface in a line which is the line of intersection. In the latter case, the area of the solid, exposed by the cutting plane, is a section of the solid. This is the scheme of using section planes for the elucidation of certain views in drawing (313).
1103. Development of surfaces. The development of a surface consists of the rolling out or rectification of the surface on a plane, so that the area OH. the plane is equal to the area of the surf ace* before development. If this flat surface be rolled up, it will regenerate the original surface from which it has been evolved (1028).
For instance, if a flat rectangular sheet of paper be rolled in a circular form, it will produce the surface of a cylinder of revolution. Similarly, a sector of a circle may be wound up so as to make a right circular cone. In both cases, the flat sur- face is the development of the surface of the cylinder or cone as the case may be.
1104. Developable surfaces. A prism may be rolled over a flat surface and each face successively comes in intimate con- tact with the flat surface; hence, its surface is developable. A cylinder may likewise be rolled over a flat surface and the con- secutive elements will successively coincide with the flat sur- face, this, again, is therefore a developable surface. The sur- faces of the pyramid and the cone are also developable for similar reasons.
A sphere, when rolled over a flat surface, touches at only one point and not along any one element; hence, its surface is
186 GEOMETRICAL PROBLEMS IN PROJECTION
not developable. In general, any surface of revolution which has a curvilinear* generatrix is non-developable.
A warped surface is also a type of surface which is non-develop- able, because consecutive elements even though rectilinear, are so situated that no plane can contain them.
Hence, to review, only singly curved surfaces, and such surfaces as are made up of intersecting planes, are developable. Doubly curved surfaces of any kind are only developable in an approximate way, by dividing the actual surface into a series of developable surfaces; the larger the series, the nearer the approximation.!
1105. Problem 1. To find the line of intersection of the surfaces of a right octagonal prism with a plane inclined to its axis.
Construction. Let abed, etc., Fig. 185, be the plan of the prism, resting, for convenience, on the horizontal plane. The elevation is shown by a'b'c', etc. The plane T is perpendicular to the vertical plane and makes an angle a with the horizontal plane. In the supplementary view S, only half of the intersec- tion is shown, because the other half is symmetrical about the line a"V". Draw from e'd'c'b' and a', lines perpendicular to Tt' and draw a'"e"", anywhere, but always parallel to TV. The axis of symmetry ea, shown in plan, is the projection of e'a' in elevation, and e'a' is equal to e'"a'"; the points e"' and a'" are therefore established on the axis. To find d'", draw dm, perpendicular to ea; dm is shown in its true length as it is parallel to the horizontal plane. Accordingly, set off m"'d'" = md, on a line from d', perpendicular to Tt', from e'"a"' as a base line. As nb = md, then n'"b'"=m'"d'", and is set off from e"'a'", on a line from V, perpendicular to Tt'. The final point c'" is located so that o'"c"' = oc and is set off from e'"a'", on a line from c', perpendicular to Tt. The "half section" is
* If a curvilinear generatrix moves so that its plane remains continually parallel to itself and touches a rectilinear directrix, the surface generated is a cylindrical surface. Hence, this surface cannot be included in this connection.
t Maps are developments of the earth's surface, made in various ways. This branch of the subject falls under Spherical Projections. The student who desires to pursue this branch is referred to the treatises on Topographical Drawing and Surveying.
INTERSECTIONS OF SURFACES BY PLANES
187
then shown completely and is sectioned as is customary. The line of intersection is shown by *'"$"s'"\>'"9!" ; the half section is the area included by the lines e'''d'''c'"b'"a'"n'"o'"m'"e'".
1106. Problem 2. To find the developed surfaces in the preceding problem.*
Construction. As the prism is a right prism, all the vertical edges are perpendicular to the base; the base will develop into a straight line and the vertical edges will be perpendicular to it,
FIG. 185.
spaced an equal distance apart, because all faces are equal. Hence, on the base line AA, Fig. 185, lay off the perimeter of the prism and divide into eight equal parts. The distance of a" above the base line AA is equal to the distance of a' above the horizontal plane; b" is ablove AA, a distance equal to b' above the horizontal plane. In this way, all points are located. It will be observed that the development is symmetrical about the vertical through e".
To prove the accuracy of the construction, the developed surface may be laid out on paper, creased along all the verticals
* In the problems to follow the bases are not included in the development as they are evident from the drawing.
188
GEOMETRICAL PROBLEMS IN PROJECTION
and wound up in the form of the prism. A flat card will admit of being placed along the cut, proving that the section is a plane surface.
The upper and lower portions are both developed; the dis- tance between them is arbitrary, only the cut on one must exactly match with the cut on the other.
1107. Problem 3. To find the line of intersection of the surface of a right circular cylinder with a plane inclined to its axis.
Construction. Let Fig. 186 show the cylinder in plan and
FIG. 186.
elevation. The plane cutting it is an angle a with the horizontal plane and is perpendicular to the vertical plane. It is customary, in the application, to select the position of the plane and the object so that subsequent operations become most convenient, so long as the given conditions are satisfied. Pass a series of auxiliary planes through the axis as oe, od, etc., spaced, for con- venience, an equal angular distance apart. These planes cut rectilinear elements from the cylinder, shown by the verticals through e'd'c', etc. Lay off an axis e'"a'", parallel to TV, so that
e'" e' and a'"a' are perpendicular to Tt'. To find d'
it is
INTERSECTIONS OF SURFACES BY PLANES 189
known that md is shown in its true length in the plan; hence, lay offm'"d'"=md from the axis e^a'", on a line from d', perpen- dicular to Tt'. Also, make o'"c'" = oc and n'"b"' = nb in the same way. Draw a smooth curve through e^d"^'"^"^"^"^"^"^" which will be found to be an ellipse. The ellipse may be awn by plotting points as shown, or, the major axis e'"a'" and the minor axis c'"c"' may be laid off and the ellipse drawn by any method.* Both methods should produce identical results. The area of the ellipse is the section of the cylinder made by an oblique plane and the ellipse is the curve of intersection. As an example, a cylindrical glass of water may be tilted to the given angle, and the boundary of the surface of the water will be elliptical.
1108. Problem 4. To find the developed surface in the preceding problem.
Construction. As the cylinder is a right cylinder, the ele- ments are perpendicular to the base, and the base will develop into a straight line, of a length, equal to the rectified length of the circular base. Divide the base AA Fig. 186, into the same number of parts as are cut by the auxiliary planes. Every element in the elevation is shown in its true size because it is perpendicular to the horizontal plane; hence, make a" A equal to the distance of a? above the horizontal plane; b"A equal to the distance of b' above the horizontal plane, etc. Draw a smooth curve through the points so found and the development will appear as shown at D in Fig. 186. Both upper and lower portions are shown developed, either one may be wound up like the original cylinder and a flat card placed across the intersection, showing, that the sur- face is plane.
1109. Application of cylindrical sur- faces. Fig. 187 shows an elbow, approximat- ing a torus, made of sheet metal, by the use of short sections of cylinders. The ellipse is symmetrical about both axes, and, hence, the FIG. 187.
upper portion of the cylinder may be added to the lower portion so as to give an offset. Indeed in this way elbows
* See Art. 906.
190
GEOMETRICAL PROBLEMS IN PROJECTION
FIG. 188.
are made in practice. The torus is a doubly curved surface, and, hence, is not developable, except by the approximation
shown. To develop these individual sheets, pass a plane ab, perpendicular to the axis. The circle cut thereby will then develop into a straight line. Add corresponding distances above and below the base line so established and the development may be completed. Fig. 188 shows the sheets as they appear in the development.
A sufficient number of points must always be found on any curve,
so that no doubt occurs as to its form. In the illustration, the number of points is always less than actually required so as to avoid the confusion incident to a large number of construction lines.
1110. Problem 5. To find the line of intersection of the surfaces of a right octagonal pyramid with a plane inclined to its axis.
Construction. Let Fig. 189 represent the pyramid in plan and elevation, and let T be the cutting plane. The plane T cuts the extreme edge o'e' at e', horizontally projected at e, one point of the required intersection. It also cuts the edge o'd' at d', horizontally projected at d, thus locating two points on the required intersection. In the same way b and a are located. The point c cannot be found in just this way. If the pyramid be turned a quarter of a revolution, the point c' will be at q' and c'q' will be the distance from the axis where the edge oV pierces the plane T. Hence, lay off oc = c'q' and complete the horizontal projection of the intersection.
To find the true shape of the intersection, draw the supple- mentary view S. Lay off e"'a'" as the axis, parallel to Tt'; the length of the axis is such that e'a' = e'"a"'. From the horizontal projection, dm is made equal to d'"m'" in the supple- mentary view, the latter being set off from the axis e'"a'", and on a line from d', perpendicular to Tt'. Also, co = c'"o'" and bn = b'"n'". Thus, the true intersection is shown in the supple- mentary view S.
INTERSECTIONS OF SURFACES BY PLANES
191
1111. Problem 6. To find the developed surfaces in the preceding problem.
Construction. The extreme element o's', Fig. 189, is shown in its true length on the vertical plane because it is parallel to that plane. Accordingly, with any point o" as a centre, draw an indefinite arc through AA, so that o"A = o's'. Every edge of the pyramid meeting at the vertex has the same length and all are therefore equal to o"A. The base of each triangular face is shown in its true length in the plan, and, hence, with any one of them as a length, set the distance off as a chord on AA
. FIG. 189.
by the aid of a divider, so that the number of steps is equal to the number of faces. Draw these chords, indicate the edges, and it then remains to show where the cutting plane intersects the edges. The extreme edge oV is shown in its true length, hence, lay off o"e"=o'e'. The edge o'd' is not shown in its true length, but if the pyramid is revolved so that this edge reaches the position o's', then d' will reach p' and o'p' is the desired true length as it is now parallel to the vertical plane; therefore, set off o'p' = o"d". In this way, o'q' = o"c", and oY = o"b". The edge o'a' is shown in its true length, therefore, o'a' = o"a"; and a" A on one side must equal a' 'A on the other.
192 GEOMETRICAL PROBLEMS IN PROJECTION
as, on rolling up, the edges must correspond, being the same initially. Join a"b", b"c", etc., to complete the development. The proof, as before, lies in the actual construction of the model and in showing that the cut is a plane surface.
1112. Problem 7. To find the line of intersection of the surface of a right circular cone with a plane inclined to its axis.
Construction. Let Fig. 190 show plan and elevation of the cone, and let T be the cutting plane. Through the axis, pass a
FIG. 190.
series of planes as ao, bo, co, etc. These planes cut rectilinear elements from the cone, shown as o'a', o'b', oV, etc., in the eleva- tion. At first, it is a good plan to draw the horizontal projection of the line of intersection, as many points are readily located thereon. For instance, e' is the point where the extreme ele- ment oV pierces the plane T, horizontally projected at e. Sim- ilarly, the actual element OD pierces the plane so that d and d' are corresponding projections. The points b and b' and the points a and a' are found in an identical manner, but, as in the case of the pyramid, the point c cannot be located in the same way. If the cone be revolved so that the element oV occupies
INTERSECTIONS OF SURFACES BY PLANES 193
the extreme position o's', then c' will be found at q' and c'q' will be the radius of the circle which the point c' describes; hence, make oc = c'q' and the resultant curve, which is an ellipse, may be drawn.
The true line of intersection is shown in the supplementary view S. As in previous instances, first the axis e'"a"' is drawn parallel to TV, and equal in length to e'a' so that e"'e' and a'"a' are both perpendicular to TV. From the axis, on a line from d', lay off d'"m'" = dm; also o'" c'"=oc and b'"n'"=bn. A smooth curve then results in an ellipse.
If accuracy is desired, it is better to lay off the major and minor axes of the ellipse and then draw the ellipse by other methods,* since great care must be used in this construction. The major axis is shown as e'"a'". To find the minor axis, bisect e'a' at u'and draw uV, the trace of a plane perpendicular to the axis of the cone. This plane cuts the conical surface in a circle, of which wV is the radius. The minor axis is equal to the length of the chord of a circle whose radius is wV and whose distance from the centre of the circle is u'w'.
1113. Problem 8. To find the developed surface in the preceding problem.
Construction. The extreme visible element o's', Fig. 190, is shown in its true length and is the slant height of the cone. All elements are of the same length and hence, any indefinite arc AA, drawn so that o"A = o's' will be the first step in the development. The length of the arc AA is equal to the rectified length of the base, and, as such, is laid off and divided into any convenient number of parts. Eight equal parts are shown here because the auxiliary planes were chosen so as to cut the cone into eight equal parts. If the conical surface is cut along the element OA, then o'a' .may be laid off on both sides equal to o"a" in the development. The elements o'd', o'c' and o'b' are not shown in their true length, therefore, it is necessary to revolve the cone about the axis so as to make them parallel, in turn, to the vertical plane. The points will ultimately reach the positions indicated by p,' q' and r', and, hence, o"b" = o'r', o"c" = o'q' and o"d" = o'p'. The element o'e' is shown in its true length and is laid off equal to o"e". A smooth curve through a"b"c"d", etc., completes the required development.
'".etc., co
* See Art. 906 in this connection.
194
GEOMETRICAL PROBLEMS IN PROJECTION
When frustra of conical surfaces are to be developed the elements of the surface may be produced until they meet at the vertex. The development may then proceed along the usual lines.
1114. Application of conical surfaces.
As the ellipse is a symmetrical curve, the upper and lower portions of the cone may be turned end for end so that the axes intersect. The resultant shape is indicated in Fig. 191, used at times in various sheet metal designs, such as
oil-cans, tea-kettles, etc. FIG. 191.
1115. Problem 9. To find the line of inter- section of a doubly curved surface of revolution with a plane inclined to its axis.
Construction. Let Fig. 192 show the elevation of the given surface of revolution.* Atten- tion will be directed to the construction of the section shown in the plan. The high- est point on the curve is shown at a, which is found from its corresponding projection a', at the point where the plane T cuts the meridian curve that is parallel to the plane of the paper. The point b is directly under b' and the length bb is equal to the chord of a circle whose radius is m'n' and whose distance from the cen- tre is b'n'. Similarly, cc is found by drawing an indefi- nite line under c' intersected by op = o'p' as a radius. As
many points as are necessary are located, so as to get the true shape. One thing, however, must be observed: The plane XT
* Many of these constructions can be carried to completion without actually showing the principal planes. The operations on them are per- formed intuitively.
FIG. 192.
INTERSECTIONS OF SURFACES BY PLANES
195
cuts the base of the surface of revolution at h' in the elevation and therefore hh is a straight line, which is the chord of a circle, whose radius is q'r' and which is located as shown in^the plan.
The construction of the supplementary view resembles, in many respects, the construction in plan. The similar letters indicate the lengths that are equal to each other.
1116. Problem 10. To find the line of intersection of a bell- surface with a plane.
Construction. Fig. 193 shows the stub end of a connecting rod as used on a steam engine. The end is formed in the lathe by turning a bell-shaped surface of revolution on a bar of a
rectangular section, as shown in the end view, the shank being circular. The radius of curvature for the bell surface is located on the line ab as shown. The plane TT is tangent to the shank of the rod and begins to cut the bell-surface at points to the left of ab; hence, the starting point of the curve is at c. To find any point such as d, for instance, pass a plane through d perpendicular to the axis. It cuts the bell-surface in a circle (1024) whose radius is od" = o'd'". Where the circle intersects the plane TT at d', project back to d, which is the required point on the curve. The scheme is merely this: By passing auxiliary planes perpendicular to the axis, circles are cut from the bell- surface which pierce the bounding planes in the required points of intersection. Attention might be directed to the piont e which is on both plan and elevation. This is true because the planes TT and SS intersect in a line which can pierce the bell-
196
GEOMETRICAL PROBLEMS IN PROJECTION
surface at only one point. The manner in which nn = n'n' is located in the plan will be seen by the construction lines which are included in the illustration. To follow the description is more confusing than to follow the drawing.
1117. Development by Triangulation. In the developments so far considered, only the cases of extreme simplicity were selected. To develop the surface an oblique cone or an oblique cylinder requires a slightly different mode of procedure than was used
FIG. 194.
heretofore. If the surface of the cone, for instance, be divided into a number of triangles of which the rectilinear elements form the sides and the rectified base forms the base, it is possible to plot these triangles, one by one, so as to make the total area of the triangles equal to the area of the conical surface to be developed. The method is simple and is readily applied in practice. Few illustrations will make matters clearer.
1118. Problem 11. To develop the surfaces of an oblique hexagonal pyramid.
Construction. Fig. 194 shows the given oblique pyramid. It is first necessary to find the true lengths of all the elements of the surface. In the case of the pyramid, the edges alone need
INTERSECTIONS OF SURFACES BY PLANES
197
be considered. Suppose the pyramid is rotated about a perpen- dicular to the horizontal plane, through o, so that the base of the pyramid continually remains in its plane, then, when oa is parallel to the vertical plane, it is projected in its true length and is, hence, shown as the line o'a" '. All the sides are thus brought para lei, in turn, to the plane and the true lengths are found. In most cases it will be found more convenient and less confusing to make a separate diagram to obtain the true lengths. To one with some experience, the actual lengths need not be drawn at all, but simply the distances a", b", c", etc., laid off. Then from any point o'", draw arcs o'a" = o"'a'", o'b" = o'"b'", o'f" =
FIG. 195.
o'"f", etc. From a'",
lay off distances equal to the respective in this case, ab = bc = cd, etc., and hence any one side will answer the purpose. Therefore, take that length on a divider and step off ab = a''V"=b'/'c'" = c'"d"' = a''/f''', etc. The development is then completed by drawing the proper lines. The case selected, shows the triangulation method applied to a surface whose sides are triangles. It is extremely simple for that reason, but its simplicity is still evident in the case of the cone as will be now shown.
1119. Problem 12. To develop the surface of an oblique cone.
Construction. Fig. 195 shows the given oblique cone. The
horizontal projection of the axis is op and the base is a circle,
198
GEOMETRICAL PROBLEMS IN PROJECTION
hence, the cone is not a cone of revolution because the axis is inclined to the base. Revolve op to oq, parallel to the vertical plane. Divide the base into any number of parts ab, be, cd, etc. preferably equal, to save time in subsequent operations. The element oa is parallel to the plane, hence, o'a' is its true length. The element ob is not parallel to the plane, but can be made so by additional revolution as shown; hence o'b' is its true length. From any point o", draw o"a" = o'a', o"b" = o'b', o"c" = o'c', etc. On these indefinite arcs step off distances a"b" = b"c" = c"d", etc., equal to the rectified distances ab = bc = cd, etc. A smooth curve through the points a"b"c", etc., will give
FIG. 195.
the development. As in previous problems, the cut is always made so as to make the shortest seam unless other requirements prevail.
1120. Problem 13. To develop the surface of an oblique cylinder.
Construction. In the case of the oblique cone, it was pos- sible to divide the surface into a series of triangles which were plotted, one by one, and thus the development followed by the addition of these individual triangles. In the case of the cylin- der, however, the application is somewhat different, although two cones may be used each of which has one base of the cylinder
INTERSECTIONS OF SURFACES BY PLANES
199
as its base. The application of this is cumbersome, an i the better plan will be shown.
Let Fig. 196 show the oblique cylinder, chosen, for convenience, with circular bases. Revolve the cylinder as shown, until it is parallel to the vertical plane. In the revolved position, ass a plane T through it, perpendicular to the axis. The curve so cut when rectified, will develop into a straight line and the elements of the surface will be perpendicular to it. The true shape of the section of the cylinder is shown by the curve a"b"c"d", etc., and is an ellipse. Its construction is indicated in the diagram. Draw any base line AA and lay off on this the rectified portions
FIG. 196.
of the ellipse a"b" = a"'b'", b"c"=b'"c'", etc. The revolved positions of the elements shown in the vertical plane are all given in their true sizes because they are parallel to that plane. Hence, lay off a"'o'" = a'o', b'"p'" = b'p', etc., below the base line. Do the same for the elements above the base line and the curve determined by the points so found will be the development of the given oblique cylinder.
1121. Transition pieces. When an opening of one cross- section is to be connected with an opening of a different cross- section, the connecting piece is called a transition piece. The case of transforming a circular cross-section to a square cross- section is quite common in heating and ventilating flues, boiler flues and the like. In all such cases, two possible methods offer
200
GEOMETRICAL PROBLEMS IN PROJECTION
themselves as solutions. The two surfaces may be connected by a warped surface, the rectilinear elements of which are chosen so that the corners of the square are joined with the quarter points on the circle while the intermediate elements are distributed between them. A warped surface, however, is not developable, it is also difficult of representation and therefore commercially unsuited for application. The better method is to select some singly curved surface, or, a combination of planes and singly curved
FIG. 197.
surfaces since these are developable. The application, which is quite a common one, is shown in the following problem.
1122. Problem 14. To develop the surface of a transition piece connecting a circular opening with a square opening.
Construction. Let the two upper views of Fig. 197 represent the transition piece desired. The square opening is indicated by abed and the circular opening by efgh. The surface may be made up of four triangular faces aeb, bfc, cgd, dha, and four conical surfaces hae, ebf, fcg, gdh, whose vertices are at a, b, c and d.
The development is shown at D, on the same figure. The
INTERSECTIONS OF SURFACES BY PLANES
201
triangle a"b"e" is first laid out so that the base a"b" = ab; and the true length of the sides are determined as found in the construction leading to the position a'e' = a"e"=b"e". The conical surfaces are developed by triangulation. For example, the arc he is divided into any number of parts em, mn, etc.; and the true lengths of the elements of the surface are found
FIG. 198.
by the construction leading up to the positions a'e', a'm', a'n'. Hence, a"e"=a'e', a"m" = a'm' ; and m"e"=me is the rectified arc of the base of the conical surface. When four of these com- binations of triangle and conical surface are laid out, in their proper order, the development is then complete.
If a rectangular opening is to be joined to an elliptical open- ing, the manner in which this may be accomplished is shown in Fig. 198.
1123. Problem 15. To develop the surface a transition piece connect- ing two elliptical openings whose major axes are at right angles to each other.
Construction. Fig. 199 shows three views of the elliptical transition piece. By reference to the diagram, it will be seen that the surface may be divided into
eight conical surfaces, turned end for end; that is, four vertices are situated on one ellipse and four vertices are situated on the
FIG. 199.
202
GEOMETRICAL PROBLEMS IN PROJECTION
other.
The shaded figure indicates the arrangement of the
conical surfaces and the a/ elements are shown as shade lines.
In all developments of this character, the ar- rangement of the surfaces has considerable effect on the appearance of the transition piece when completed. In the case chosen, the eight conical surfaces give a pleasing result. If the intersec- tion of the eight alter-
FIG. 199.
nate conical surfaces is objectionable then it is
possible to use a still larger number of divisions.
The development of the conical surfaces is performed by the
triangulation method. When finished, its appearance is that
/ \/\ /
b'" V" b'»
FIG. 200.
b"f
of Fig. 200. The similar letters indicate the order in which the surfaces are assembled.
1124. Development of doubly curved surfaces by approx=» imation. Doubly curved surfaces are non-developable because successive elements cannot be made to coincide with a plane. It is possible, however, to divide the surface into a series of singly curved surfaces and then to develop these. By dividing the doubly curved surface into a sufficiently large number of parts,
INTEKSECTIONS OF SURFACES BY PLANES
203
the approximation may be made to approach the surface as closely as desired.
For doubly curved surfaces of revolution, two general methods are used. One method is to pass a series of meridian planes through the axis and then to adopt a singly curved surface whose contour is that of the surface of revolution at the meridian planes. This method is known as the gore method.
The second method, known as the zone method, is to divide the surface into frusta of cones whose vertices lie on the axis of the doubly curved surface of revolution. The following examples will illustrate the application of the methods. .
1125. Problem 16. To develop the surface of a sphere by the gore method.
Construction. Let the plan P of Fig. 201 show the sphere.
FIG. 201.
Pass a series of meridian planes aa, bb, cc, dd, through the sphere and then join ab, be, cd, etc. From the plan P construct the elevation E and then pass planes through i' and g', perpendicular to the axis of the sphere, intercepting equal arcs on the meridian circle, for convenience. Find the corresponding circles in the plan and inscribe an octagon (for this case) and determine ef, the chord length for that position of the cutting planes through i' and g'. This distance may then be laid off as e'f in the eleva- tion E and the curve n'e'a'e'm' be drawn.
For the development D, it will be noted that ab = a'b' = a"b"
204
GEOMETRICAL PROBLEMS IN PROJECTION
m"g",
and ef = e'f' = e"f". Also, the rectified distance m'g' g'h'=g"h", h'i'=h"i", etc. A suitable number of points must be used in order to insure the proper degree of accuracy, the number chosen here is insufficient for practical purposes. By adding eight of these faces along the line a"b", the development is completed.
When this method is commercially applied, the gores may be stretched by hammering or pressing to their true shape. In this case they are singly curved surfaces no longer.
FIG. 201.
1126. Problem 17. To develop the surface of a sphere by the zone method.
Construction. Let P, Fig. 202, be the plan and E, the eleva- tion of a sphere. Pass a series of planes a'a', b'b', cV etc., perpendicular to the axis of the sphere. Join a'b', and pro- duce to h', the vertex of a conical surface giving rise to the frustum a'b'b'a'. Similarly, join bV and produce to g'. The last conical surface has its vertex e' in the circumscribing sphere.
The development D is s milar to the developnent of any frustum of a conical surface of revolution. That is, with h" as a centre, draw an arc with h"a"=h'a' as a radius. At the centre line make a"o" = ao, o"n" = on etc. The radius h"b" = h'b' so that the arc b"b"b" is tangent internally, to the similarly lettered arc. The radius for this case is g"b"=g'=b'. The development is completed by continuing this process until all the conical surfaces are developed.
INTERSECTIONS OF SURFACES BY PLANES
h"
205
FIG. 202.
1127. Problem 18. To develop a doubly curved surface of revolution by the gore method. Let Fig. 203 represent the doubly curved surface of revolution as finally approxi- mated. Through the axis at h pass a series of equally spaced meridian planes, and then draw the chords, one of which is lettered as dd. In the ele- vation E, pass a series of planes k'k', IT, m'm', perpendicular to the axis. The lines b'b', cV, d'd', etc., are drawn on the resulting singly curved surface and are shown in their true length. The corresponding projections of these lines in the plan P are also shown in their true length.
FIG. 203. To develop the surface of one face lay off the rectified dis-
206
GEOMETRICAL PROBLEMS IN PROJECTION
tance j'k' = j"
FlQ 203
, etc. At the points j", k", 1", etc., lay off a'a' = a"a", b'b' = b"b", etc., perpendicular to h"j" so that the resulting figure D is symmetrical about it. When eight of these faces (or gores) are joined together and secured along the seams, the resulting figure will be that shown in plan and elevation.
This surface may also be approxi- „ mated by the zone method. The ~// choice of method is usually governed by commercial considerations. The gore method is perhaps the more economical in material.
QUESTIONS ON CHAPTER XI
1. State the general method of finding the piercing point of a line on
a plane.
2. Show by an oblique projection how the piercing points of a doubly
curved line are found on the principal planes.
3. Why is the projecting surface of a doubly curved line a projecting
cylindrical surface?
4. What is a "line of intersection" of two surfaces?
5. Distinguish between "line of intersection" and "section" of a solid.
6. What is meant by the development of a surface? Explain fully.
7. What are the essential characteristics of a developable surface?
8. Why is a sphere non-developable?
9. Why is any doubly curved surface non-developable?
10. Why is a warped surface non-developable?
11. Prove the general case of finding the intersection of a right prism
and a plane inclined to its axis.
12. Show the general method of finding the development of the prism in
Question 11.
13. Prove the general case of finding the intersection of a right circular
cylinder and a plane inclined to its axis.
14. Show the general method of finding the development of the cylinder
in Question 13.
15. When a right circular cylinder is cut by a plane why must the ellipse
be reversible?
16. How is the torus approximated with cylindrical surfaces?
17. Prove the general case of finding the intersection of a right pyramid
and a plane inclined to its axis.
INTERSECTIONS OF SURFACES BY PLANES 207
18. Show the general method of finding the development of the right
pyramid in Question 17.
19. Prove the general case of finding the intersection of a right circular
cone and a plane inclined to its axis.
20. Show the general method of finding the development of the right
cone in Question 19.
21. Why is the ellipse, cut from a cone of revolution by an inclined plane,
reversible?
22. Prove the general case of finding the intersection of a doubly curved
surface of revolution and a plane inclined to the axis.
23. Is the surface in Question 22 developable? Why?
24. Prove the general case of finding the intersection of a bell-surface
with planes parallel to its axis.
25. Is the bell-surface developable? Why?
26. What is development by triangulation?
27. Prove the general case of the development of the surface of an oblique
pyramid.
28. Prove the general case of the development of the surface of an oblique
cone.
29. Prove the general case of the development of the surface of an oblique
cylinder.
30. What is a transition piece?
31. Why is it desirable to divide the surfaces of a transition piece into
developable surfaces?
32. Prove the general case of the development of a transition piece which
joins a circular opening with a square opening.
33. Prove the general case of the development of a transition piece
which joins an elliptical opening with a rectangular opening.
34. Prove the general case of the development of a transition piece
which joins two elliptical openings whose major axes are at right angles to each other.
35. What is the gore method of developing doubly curved surfaces?
36. What is the zone method of developing doubly curved surfaces?
37. Prove the general case of the development of a sphere by the gore
method.
38. Prove the general case of the development of a sphere by the zone
method.
39. Prove the general case of the development of a doubly curved sur-
face of revolution by the gore method.
40. A right octagonal prism has a circumscribing circle of 2" and
is 4" high. It is cut by a plane, inclined 30° to the axis, and passes through its axis, If" from the base. Find the section.
41. Develop the surface of the prism of Question 40.
42. A cylinder of revolution is 1\" in diameter and 3f " high. It is cut
by a plane, inclined 45° to the axis and passes through its axis 2" from the base. Find the section.
43. Develop the surface of the cylinder of Question 42.
208 GEOMETRICAL PROBLEMS IN PROJECTION
44. A right octagonal pyramid has a circumscribing base circle of 2\"
and is 4" high. It is cut by a plane, inclined 30° to the axis and passes through its axis 2" from the base. Find the section.
45. Develop the surface of the pyramid of Question 44.
46. A right circular cone has a base 2" in diameter and is 3f" high.
It is cut by plane inclined 45° to the axis and passes through its axis 21" from the base. Find the section.
47. Develop the cone of Question 46.
48. A 90° stove pipe elbow is to be made from cylinders 4" in diameter.
The radius of the -bend is to be 18". Divide elbow into 8 parts. The tangent distance beyond the quadrant is 4". Develop the surface to suitable scale.
49. Assume a design of a vase (a doubly curved surface of revolution)
and make a section of it.
50. The stub end of a connecting rod of a steam engine is 4"x6*'; the
rod diameter is 3£". The bell-surface has a radius of 12". Find the lines of intersection. Draw to suitable scale.
51. An oblique pyramid has a regular hexagon for a base. The circum-
scribing circle for the base has a diameter of 2"; the altitude is 3£" and the projection of the apex is 2|" from the centre of the base. Develop the surface of the pyramid.
52. An oblique cone has a circular base of 2|" in diameter and an altitude
of 3". The projection of the vertex on the plane of the base is li" from the centre of the base. Develop the conical surface.
53. An oblique cylinder has a circular base If" in diameter; it is 2£"
high and the inclination of the axis is 30° with the base. Develop the surface.
54. A transition piece is to be made joining a 3'-0"x5'-6" opening
with a 4'-0" diameter circle. The distance between openings 3'-6". Develop the surface. Use a suitable scale for the drawing.
55. A transition piece is to be made joining an opening having a 3'-0"
X4'-0" opening to a 2'-0"x6/-0" opening (both rectangular). The distance between openings is 4'-0". Develop the sheet and use suitable scale for the drawing.
56. A square opening 4"-0"x4'-0" is to be joined with an elliptical
opening whose major and minor axes are 5'-0" X 3' —0" respec- tively. The distance between openings is 3'-0". Develop the sheet. Use suitable scale in making the drawing.
57. A circular opening having a diameter of 3'-0" is to be connected
with an elliptical opening whose major and minor axes are 4'-0" x2'-6" respectively. The distance between openings is 6/-0". Develop the sheet. Use suitable scale in making the drawing.
58. An opening having two parallel sides and two semicircular ends has
overall dimensions of 3'-0" X5'-6". This opening is to be joined with a similar opening having dimensions of 3'-6"x4'-6". The distance between openings is 3'-6". Develop the sheet. Use suitable scale in making the drawing.
INTERSECTIONS OF SURFACES BY PLANES 209
59. A frustum of a cone has a base which is made up of two parallel
sides and two semicircular ends and whose overall dimensions are 3'-6" X6'-0". The height of the vertex above the base is 8'-6"; that of the upper base is 4'-0" above the lower base. Develop the sheet and use suitable scale in making the drawing.
60. An elliptical opening having a major and minor axis of 5'-0" and
4'-0", respectively, is to be joined with a similar opening but whose major axis is at a right angle. Develop the sheet and use suitable scale in making the drawing.
61. An elliptical opening having a major and minor axis of 4/-6// and
3'-6", respectively, is to be joined with a similar opening but whose major axis is at an angle of 30°. Develop the sheet and use suitable scale in making the drawing.
62. An elliptical opening having a major and minor axis of 5'-0" and
3'-6", respectively, is to be joined with another elliptical opening whose major and minor axes are 4/-6// and 3'-6", respectively. Develop the sheet and use suitable scale in making the drawing.
63. A sphere is 6" in diameter. Develop the surface by the gore
method and divide the entire surface into sixteen parts. Use a suitable scale in making the drawing.
64. A sphere is 6" in diameter. Develop the surface by the zone
method and divide the entire surface into twelve parts. Use a suitable scale in making the drawing.
65. Assume some doubly curved surface of revolution and develop the
surface by the gore method.
66. Assume some doubly curved surface of revolution and develop the
surface by the zone method.
CHAPTER XII
INTERSECTIONS OF SURFACES WITH EACH OTHER, AND THEIR
DEVELOPMENT
1201. Introductory. When two surfaces are so situated with respect to each other that they intersect, they do so in a line which is called the line of intersection. It is highly desirable in the conception of intersections to realize that certain elements of one surface intersect certain elements on the other surface, and that the locus of these intersecting elements is the desired line of intersection.
The line of intersection is found by passing auxiliary surfaces through the given surfaces. Lines will thus be cut from the given surfaces by the auxiliary surfaces, the intersections of \vhich yield points on the desired line of intersection. The simplest auxiliary surface is naturally the plane, but sometimes cylindrical,* conical and spherical surfaces may find application. It is not necessary only to use a simple type of auxiliary surface, but also to make the surface pass through the given surfaces so as to cut them in easily determinable lines — preferably, in the elements. As illustrations of the latter, the case of a concial surface cut by a plane passing through the vertex furnishes an example. It cuts the surface in straight lines or rectilinear elements. Likewise, a plane passed through a cylindrical sur- face parallel to any element will cut from it, one or more lines which are also rectiliner elements. For doubly curved surfaces of revolution, a plane passed through the axis (a meridian plane) cuts it in a meridian curve. When the plane is perpendicular to the axis, it cuts it in circles. An interesting and valuable
* A careful distinction must be made between solids and their bounding surfaces. The terms, cylinder and cylindrical surface are often used indis- criminately. The associated idea is usually obtained from the nature of the problem.
210
INTERSECTIONS OF SURFACES WITH EACH OTHER 211
property of spheres is that all plane sections in any direction, or in any place, are all circles, whose radii are readily obtain- able.
When prisms or pyramids intersect, auxiliary planes are not required to find their intersection, because the plane faces and the edges furnish sufficient material with which to accomplish the desired result. These types of surfaces are therefore exempt from the foregoing general method.
1202. Problem 1. To find the line of intersection o* the surfaces of two prisms.
Construction. Let Fig. 204 represent the two prisms, one of which, for convenience, is a right hexagonal prism. The face ABDC intersects the right prism in the line mn, one line of the required intersec- tion. The face CDFE is intersected by the edge of the right prism in the point o, found as shown in the con- struction. Hence, mo is the next line of the required intersection. Similarly, the edge EF pierces the faces of the right prism at p, and op is the con- tinuation of the intersection. In this way, all points are found. The curve must be closed because the inter- penetration is complete. Only the
one end where the prism enters is shown in construction. The curve where it again emerges is found in an identical manner.
1203. Problem 2. To find the developments in the preceding problem.
Construction. The development of the right prism is quite simple and should be understood from the preceding chapter. The points at which the oblique prism inter- sects the right prism are also shown. Thus, it is only necessary to make sure on what element the point pierces, and this may be obtained from the plan view of the right prism.
O,k c i eO
FIG. 204.
212
GEOMETRICAL PROBLEMS IN PROJECTION
In developing the oblique prism, D' of Fig. 205, the same general scheme is followed as was used in developing the oblique cylinder (1120). Revolve the prism of Fig. 204 so that the elements are parallel to the plane of projection; they are then shown in their true length. Pass a plane perpendicular to the edges and this line of intersection will develop into a straight line which is used as a base line. Find now the true section of the prism and lay off the sides perpendicular to the base line and at their proper distances apart. Lay off the lengths of the edges above and below the base line and join the points by lines to complete the development.
When actually constructing this as a model, it will be well to
D
D
FIG. 205.
carry out the work as shown, and make the oblique prism in one piece. It may then be inserted in the opening provided for in the right prism, as indicated.
1204. Problem 3. To find the line of intersection of two cylin- drical surfaces of revolution whose axes intersect at a right angle.
Construction. Let Fig. 206, represent the given cylindrical surfaces. Through o', pass a series of planes a' a', b'b7, cV, etc. The elements cut from the cylindrical surface by these planes interest the other cylindrical surface at the points a, e, b, d, c. The two upper views lead to the construct on of the lower view. It will be seen by this construction that the entire scheme of locating points on the curve lies in the finding of the successive intersections of the elements of the surfaces.
If the cylindrical surfaces have elliptical sections, instead of
INTERSECTIONS OF SURFACES WITH EACH OTHER 213
circular sections as shown, the method of procedure will be found to be much the same.
1205. Problem 4. To find the developments in the preceding problem.
Construction. As the elements of the surfaces depicted in Fig. 206 are parallel to the plane of the paper, they are therefore, shown in their true length. Hence, to develop the surfaces,
D'
FIG. 206.
FIG. 207.
rectify the bases and set off the points at proper distances from the base line and also their rectified distance apart. The appearance of the completed developments are shown in Fig. 207. This problem is similar to the development of the steam dome on a locomotive boiler.
1206. Problem 5. To find the line of intersection of two cylindrical surfaces of revolution whose axes intersect at any angle.
Construction. Fig. 208 shows the two cylindrical surfaces chosen. If a series of auxiliary planes be passed parallel to the plane of the intersecting axes they will cut the cylindrical surfaces in rectilinear ele- ments. The elements of one surface will intersect the elements of the other surface in the required points of the line of inter- section. The construction shown in the figure should be clear from the similar lettering for the similar points on the line of intersection.
FIG. 208.
214
GEOMETRICAL PROBLEMS IN PROJECTION
It may be desirable to note that this method of locating the required points on the curve is also applicable to the con- struction of Fig. 206, and vice versa.
If the plane of the intersecting axes is not parallel to the plane of the paper, the construction can be simplified by revolving the cylindrical surfaces until they become parallel. The method of procedure is then the same as that given here. If the axes cannot be made to lie in the same plane, then the construction is more difficult. The method in such cases is to pass a plane through one cylinder so as to cut in in elements; and the same plane will cut the other cylinder in some line of intersection.
The intersection or these will yield points on the desired line of intersection.
c"
FIG. 209.
1207. Problem 6. To find the developments in the pre- ceding problem.
Construction. Fig. 209 shows the development as it appears. If the plane of the base is perpendicular to the elements of the surface, the base will develop into a straight line. The elements will be at right angles to the base line and as they are shown in
their true length, they may be laid off directly from Fig. 208.
FIG. 210.
FIG. 211.
1208. Application of intersecting cylindrical surfaces to pipes. The construction of pipe fittings as commercially used furnish examples of intersecting cylindrical surfaces. Fig. 210
INTERSECTIONS OF SURFACES WITH EACH OTHER 215
shows two cylindrical surfaces whose diameters are the same and whose axes intersect at a right angle. The line of intersection for this case will be seen to consist of two straight lines at right angles to each other. In Fig. 211, the cylindrical surfaces have their axes intersecting at an angle of 45° and have different diameters.
1209. Problem 7. To find the line of intersection of two cylindrical surfaces whose axes do not intersect.
Construction. Fig. 212 shows two circular cylindrical sur- faces which are perpendicular to each other and whose axes are offset. Pass a series of planes ab, cd, ef, etc., through o. These planes cut the cylindrical surface in elements, which in turn, intersect elements of the other surface. As the construction lines are completely shown, the description is unnecessary.
FIG. 212.
FIG. 213.
1210. Problem 8. To find the developments in the pre- ceding problenio
Construction. First consider the larger cylindrical surface, D', Fig. 213, which is cut along any element. With a divider, space off the rectified distances between elements; and on the proper elements, lay off b'" to correspond with b" ; k"'d'" =k"d" i'"f'"=i"f", etc.
For the smaller cylindrical surface, rectify the entire circle ace ... la and divide into the number of parts shown. . As the elements are shown parallel to the plane of projection, in the side elevation, they may be directly plotted as indicated by the curve g'"i'"k'". . . e'"g"'. This completes the final development.
216 GEOMETRICAL PROBLEMS IN PROJECTION
1211. Intersection of conicaFsurfaces. If a plane be passed through the vertex of a conical surface it cuts it in rectilinear elements if at all. When two conical surfaces intersect it is possible to pass a plane through the vertices of both. Thus, the plane may be made to intersect both surfaces in rectilinear elements, the intersections of which determine points on the required line of intersection.
Fig. 214 shows this pictorially. If the vertices m and n be joined by a line and its piercing point o be found, then, any line through o will determine a plane. Also, if od be the line chosen, the plane of the lines mo and od will cut the conical surfaces in
FIG. 214.
the elements ma, mb, nc and nd. These elements determine the four points e, f, g, h. Any other line through o will, if prop- erly chosen, determine other elements which again yield new points on the line of intersection. The following problem will bring out the details more fully.
1212. Problem 9. To find the line of intersection of the surfaces of two cones, whose bases may be made to lie in the same plane, and whose altitudes differ.
Construction. Let A and B, Fig. 215 be the vertices of the two cones in question, whose bases are in the horizontal plane. If the bases do not lie in the principal planes, a new set of prin- cipal planes may be substituted to attain the result. Join A and B by a line and find where this line pierces the horizontal
INTERSECTIONS OF SURFACES WITH EACH OTHER 217
plane at c. Any line, through c, lying in the horizontal plane, in addition to the line AB will determine a plane, which may cut the conical surfaces in elements. Let, for instance, cd be such a line. This line cuts the bases at d, e and f, the points which will be considered. The elements in the horizontal pro- jection are shown as db, eb and fa; in the vertical projection, they are d'b', e'b' and fa'. The element fa' intersects the element d'b' at g', and the element e'b' at h', thus determining g' and h', two points on the vertical projection of the required line of intersection. The corresponding projections g and h may also be found, which determine points on the horizontal projection
FIG. 215.
of the same line of intersection. Consider, now, the line ci, tangent to the one cone. Its element is ib in the horizontal projection and i'b' in the vertical projection; the corresponding element cut from the other cone is ja in the horizontal projec- tion and j'a' in the vertical projection. The points k and k' are thus found, but they represent only one point K on the actual cones.
This process is continued until a sufficient number of points are determined, so that a smooth curve can be drawn through them. The cones chosen, have complete interpenetration, as one cone goes entirely through the other. Thus, there are two distinct curves of intersection. That, near the apex B, is iound in an identical way with the preceding.
218
GEOMETRICAL PROBLEMS IN PROJECTION
Attention might profitably be called to the manner in which points are located so as to miminize confusion as much as possible. To do this, draw only one element on each cone at a time, locat- ing one, or two points, as the case may be; then erase the con- struction* lines when satisfied of the accuracy. Several of the prominent points of the curve may be thus located and others estimated if considerable accuracy is not a prerequisite. If the surfaces are to be developed and constructed subsequently, however, more points will have to be established.
There is nothing new in the development of these cones, the case is similar to that of any oblique cone and is therefore omitted. One thing may be mentioned in passing however, and that is, while drawing the elements to determine the contour of the base, the same elements should be used for locating the line of inter- section as thereby considerable time is saved.
1213. Problem 10. To find the line of intersection of the surfaces of two cones, whose bases may be made to lie in the same plane and whose altitudes are equal.
Construction. Let A and B, Fig. 216, be the desired cones.
Join A and B by a line which is chosen, parallel to the planes of projection. Hence, this line cannot pierce the plane of the bases, and the preceding method of finding the line of intersection is thus inapplicable. It is possible, how- ever to draw a line cd parallel to ab. These lines therefore determine a plane which passes through both vertices and intersects the surface in rectilinear elements. The comple- tion of the construction becomes evident when the process which leads to the finding of F and G is understood.
The cases of intersecting cones so far considered have been so situated as to have bases in a common plane. This may not always be convenient. Therefore to complete the subject, and to cover emergencies, an additional construction will be studied.
FIG. 216.
INTERSECTIONS OF SURFACES WITH EACH OTHER 219
1214. Problem 11. To find the line of intersection of the surfaces of two cones whose bases lie in different planes.
Construction. Let A and B, Fig. 217, be the vertices of the two cones. The cone A has its base in the horizontal plane; B has its base in the plane T, which is perpendicular to the vertical plane. Join the vertices A and B by a line, which pierces the horizontal plane at c and the plane T in the point d' horizontally projected at d. Revolve the plane T into the vertical plane, and, hence, the trace Tt becomes Tt". The angle t'Tt" must
FIG. 217.
be a right angle, because it is perpendicular to the vertical plane, and, therefore cuts a right angle from the principal planes. The piercing point of the line AB is at d" in the revolved position, where d'd" is equal to the distance of d from the ground line. In the revolved position of the plane T, draw the base of the cone B and from it accurately construct the horizontal projec- tion of the base. Draw a line d"e" tangent to the revolved position of the base of the cone B at t" . From it, find f, the horizontal, and f, the vertical projection of this point, and draw the elements fb and f'b'. Lay off Te"=Te, and draw ce. The lines CD and DE determine a plane which cuts the horizontal
220
GEOMETRICAL PROBLEMS IN PROJECTION
projection of the base of the cone A in g and h, and thus, also, the two elements ag and ah which are horizontal projections, and a'g' and a'h' the corresponding vertical projections of the elements. The element BF intersects the elements AG and AH in the points M and N shown, as usual, by their projections. M and N are therefore two points of the required curve, being prominent points because the elements are tangent at these points. To obtain other points, draw a line d"i" and make Ti" = Ti. Where ci cuts the cone A, draw the elements as shown, also the correspond-
FIG. 217.
ing elements of the cone B. The points of the curve of intersec- tion are therefore as indicated.
One fact is to be observed in this and in similar constructions. Auxiliary planes are passed through the vertices of both cones, cutting, therefore, elements of the cones whose intersection determine points on the curve. All planes through the vertices of both cones must pass through c in the horizontal plane and through d" in the plane T. Also, such distances as Te" must equal Te because these auxiliary planes cut the 'horizontal plane and the plane T on their line of intersection Tt; the revolution of the plane T into the vertical plane does not disturb the location
INTERSECTIONS OF SURFACES WITH EACH 'OTHER 221
of any point on it. Hence such distance as Te will be revolved to Te" where Te = Te".
When applying this problem to a practical case, it would be better to select a profile plane for the plane T, as then It" would coincide with the ground line and all construction would be simplified. It has not been done in this instance, in order to show the generality of the method, and its adaptation to any condition.
1215. Types of lines of intersection for surfaces of cones.
When the surfaces of two cones are situated so that there is com-
FIG. 218.
FIG. 219.
plete interpenetration, the line of intersection will appear as two distinct closed curves. Fig. 215 is an example.
If the surfaces of the cones are such that the interpenetration is incomplete, only one closed curve will result. Figs. 216 and 217 are examples of this case.
When the surfaces of two cones have a common tangent plane then the curve is closed and crosses itself once. An example under this heading is given in Fig. 218.
It is possible to have two cones the surfaces of which have two common tangent planes. In this case there are two closed curves which cross each other twice. This illustration appears in Fig. 219.
222
GEOMETRICAL PROBLEMS IN PROJECTION
1216. Problem 12. To find the line of intersection of the surfaces of a cone and a cylinder of revolution when their axes intersect at a right angle.
Construction. Let Fig. 220 represent the cone and the cylinder. Through the cone, pass a series of planes perpendicular to its axis. If the planes are properly chosen, they will cut the conical surface in circles and the cylindrical surface in rectilinear elements, the intersection of which determine points on the line of intersection. A reference to Fig. 220 will show this in con- struction. As a check on the accuracy of the points on the curve, it is possible to draw elements of the conical surface through some point; the corresponding projections of the elements must con-
FIG. 220.
ta'n the corresponding projections of the points. In the illustration, the elements OA and OB are drawn through the points E and J respectively.
1217. Problem 13. To find the line of intersection of the surfaces of a cone and a cylinder of revolution when their axes intersect at any angle.
Construction. The given data is shown in Fig. 221. In this casexit is inconvenient to pass planes perpendicular to the axis of the cone, since ellipses will be cut from the cylinder. A better plan is to find m, the intersection of their axes, and use this as a centre for auxiliary spherical surfaces. The spherical surfaces intersect the surfaces of revolution in circles (1025).
INTERSECTIONS OF SURFACES WITH EACH OTHER 223
The "details of the construction are shown in Fig. 221. To check the accuracy of the construction it is possible to draw an element of one surface through some point and then find the corresponding projection of the element; the corresponding projection of the point must be located on the corresponding projection of the element. Two elements OF and OG are shown in the figure. The points determined thereby are shown at D.
1218. Problem 14. To find the line of intersection of the surfaces of an oblique cone and a right cylinder.
Construction. Let Fig. 222 illustrate the conditions assumed.
FIG. 222.
To avoid too many construction lines, the figures have been assumed in their simplest forms. Pass any plane through o, perpendicular to the horizontal plane; it cuts elements DO and CO from the cone and the element from the cylinder which is horizontally projected at f and 'vertically at f'g7. In the vertical projection, the elements c'o' and d'o' intersect the element f'g' at g' and h', two points of the required curve. The extreme element OE determines the point i by the same method. There are two distinct lines of intersection, in this case, due to complete interpenetration. The points on the line of intersection near the vertex are located in a manner similar to that shown.
224
GEOMETEICAL PROBLEMS IN PROJECTION
1219. Problem 15. To find the developments in the pre- ceding problem.
Construction. The oblique conical surface is developed by triangulation. In Figs. 222 and 223, the lines o'a' and o'b' are shown in their true length in the vertical projection; hence, lay off o'a' = o"a", and o'b' = o"b". If the element oe be revolved so that it is parallel to the vertical plane, the point i', in the vertical projection, will not change its distance above the horizontal plane during the revolution. Hence it moves from i' to k', the revolved position. Accordingly, lay off o"i" = o'k' and one point of the intersection on the development is obtained. Other points, of course, are found in absolutely the same manner.
* D
FIG. 223.
Extreme accuracy is required in most of these problems. Con- structions like this should be laid out to as large a scale as con- venient. The development of the cylinder is also shown and is perhaps clear without explanation.
1220. Problem 16. To find the line of intersection of the surfaces of an oblique cone and a sphere.
Construction. Let Fig. 224 represent the cone and sphere in question. The general scheme is to pass planes through the vertex of the cone perpendicular to the horizontal plane. These planes cut the cone in rectilinear elements and the sphere in circles; the intersection of the elements and the circles so cut will determine points on the curve.
Thus, through o, draw a plane which cuts the cone in a and b and the sphere in d and e. The elements cut are shown as o'a'
INTERSECTIONS OF SURFACES WITH EACH OTHER 225
and o'b' in the vertical projection. Attention will be confined to the determination of g', one point on the lower curve, situated on the element OA. The circle cut from the sphere by the plane through oa and ob has a diameter equal to de. If pc is a perpendicular to de from p the centre of the sphere, then c is the hori- zontal projection of the centre of the circle de, and cd and ce are equal. If the cutting plane is revolved about a perpendicular through o to the hori- zontal plane, until it is parallel to the vertical plane, a will move to a" and o'a'" will be the revolved position of this element. The centre of the circle _ will go to h in the horizontal projec- tion and h' in the vertical projection, because, in this latter case, the distance of h' above the horizontal plane does not change in the revolution. The
element and the centre of the circle in a plane parallel to the vertical plane are thus determined. Hence, with cd as a radius and h' as a centre, describe an arc, cutting o'a'" in f . On counter
revolution, f goes to g' on o'a', the original position of the element. The point g' is therefore one point on the curve. Every other point is found in the same way.
1221. Problem 17. To find the line of intersection of the surfaces of a cylinder and a sphere.
Construction. Fig. 225 pic- tures the condition. Pass a series of planes, through the cylinder and the sphere, perpendicular to the plane of the base of the cylinder. One position of the cutting plane cuts the cylinder in a'b' and the sphere in c'd'. Construct a supplementary view S with the centre of the sphere at o" as shown. The elements appear as a"a" and b"b". The diameter of the circle cut from the sphere is c'd' and
FIG. 224.
FIG. 225.
226
GEOMETRICAL PROBLEMS IN PROJECTION
with c'e' as a radius, (equal to one-half of c'd'), draw an arc cutting the elements at f"g''h" and i". Lay off f"j" = fj; g"j" = gj, etc., and the four points f, g, h, i are determined on the required view. These points are on the required line of intersection.
1222. Problem 18. To find the line of intersection of two doubly curved surfaces of revolution whose axes intersect.
Construction. Let Fig. 226 represent the surfaces in ques- tion. With o, the intersection of the axes, as a centre, draw a series of auxiliary spherical surfaces. One of these spherical surfaces cuts the surface whose axis is on in a circle whose diameter is cd. This same auxiliary sphere cuts the surface whose axis is
FIG. 226.
om in a circle projected as ef. Hence cd and ef intersect at a, a point on the required line of intersection.
In the view on the right, only one of the surfaces is shown. The location of the corresponding projections of the line of intersection will be evident from Art. 1027.
When the axes do not intersect then the general method is to pass planes so as to cut circles from one surface of revolution and a curve from the other. The intersections determine points on the curve. It is desirable in this connection to chose an arrange- ment that gives the least trouble. No general method can be given for the mode of procedure.
1223. Commercial application of methods. In practice, it is always desirable as a matter of time to turn the objects so that the auxiliary surfaces may be passed through them with the least effort. Frequently, many constructions may be car- ried out without any special reference to the principal planes.
INTERSECTIONS OF SURFACES WITH EACH OTHER 227
There is no harm in omitting the principal planes, but the student should not go to the extreme in this ommission. With the principal planes at hand, the operations assume a familiar form, which will have a tendency to refresh the memory as to the basic principles involved. All the operations in the entire subject have a remarkable simplicity in the abstract; the confusion that sometimes arises is not due to the principles involved, but solely to the number of construction lines required. It, therefore, seems proper to use such methods as will lead to the least con- fusion, but it should always be borne in mind that accuracy is important at all hazards.
QUESTIONS ON CHAPTER XII
1. State the general method of finding the intersection of any two
surfaces.
2. When the surfaces are those of prisms or pyramids, is it necessary
to use auxiliary planes as cutting planes? Why?
3. Prove the general case of finding the line of intersection of the sur-
faces of two prisms.
4. Show the general method of finding the development of the prisms
in Question 3.
5. Prove the general case of finding the line of intersection of two
cylindrical surfaces of revolution whose axes intersect at a right angle.
6. Show the general method of finding the development of the surfaces
in Question 5.
7. Prove the general case of finding the line of intersection of two
cylindrical surfaces of revolution whose axes intersect at any angle.
8. Show the general method of finding the development of the surfaces
in Question 7.
9. Prove the general case of finding the line of intersection of two
cylindrical surfaces whose axes do not intersect.
10. Show the general method of finding the development of the cylin-
drical surfaces in Question 9.
11. State the general method of finding the line of intersection of two
conical surfaces.
12. Prove the general case of finding the line of intersection of the sur-
faces of two oblique cones whose bases may be made to be in the same plane and whose altitudes differ.
13. Show the general method of finding the developments of the surfaces
of the cones in Question 12.
14. Prove the general case of finding the line of intersection of the sur-
faces of two oblique cones whose bases may be made to lie in the same plane and whose altitudes are equal.
228 GEOMETRICAL PROBLEMS IN PROJECTION
15. Show the general method of finding the developments of the
surfaces of the cones in Question 14.
16. Prove the general case of finding the line of intersection of the sur-
faces of two oblique cones whose bases lie in different planes.
17. Show the general method of finding the developments of the cones
in Question 16.
18. When the surfaces of two cones have complete interpenetration,
discuss the nature of the line of intersection.
19. When the surfaces of two cones have incomplete penetration, discuss
the nature of the line of intersection.
20. When the surfaces of two cones have a common tangent plane, discuss
the nature of the line of intersection.
21. When the surfaces of two cones have two common tangent planes,
discuss the nature of the line of intersection.
22. Prove the general case of finding the line of intersection of the sur-
faces of a cone and a cylinder of revolution when their axes inter- sect at a right angle.
23. Show the general method of finding the developments of the surfaces
in Question 22.
24. Prove the general case of finding the line of intersection of the sur-
faces of a cone and a cylinder of revolution when their axes inter- sect at any angle.
25. Show the general method of finding the developments of the sur-
faces in Question 24.
26. Prove the general case of finding the line of intersection of the sur-
faces of an oblique cone and a right cylinder.
27. Show the general method of finding the developments in Question 26.
28. Prove the general case of finding the line of intersection of the sur-
faces of an oblique cone and a sphere.
29. Show the general method of finding the development of the cone in
Question 28.
30. Develop the surface of the sphere in Question 28 by the gore
method.
31. Prove the general case of finding the line of intersection of the sur-
faces of a sphere and a cylinder.
32. Show the general method of finding the development of the cylinder
in Question 31.
33. Develop the surface of the sphere in Question 31 by the zone method.
34. Prove the general case of finding the line of intersection of two
doubly curved surfaces of revolution whose axes intersect.
35. Develop the surfaces of Question 34 by the gore method.
36. Develop the surfaces of Question 34 by the zone method.
37. State the general method of finding the line of intersection of two
doubly curved surfaces of revolution whose axes do not intersect.
38. What items are to be considered when applying the principles of
intersections and developments to commercial problems?
39. Is it always necessary to use the principal planes when solving prob-
lems relating to intersections and developments?
INTERSECTIONS OF SURFACES WITH EACH OTHER 229
40. Find the intersection of the two prisms shown in Fig. 12-A. Assume
suitable dimensions.
41. Develop the prisms of Question 40.
42. Two cylinders of 2" diameter intersect at a right angle. Find the
line of intersection of the surfaces. Assume suitable dimensions for their lengths and position with respect to each other.
43. Develop the surfaces of Question 42.
44. Two cylinders of 2" diameter and If" diameter
intersect at a right angle. Find the line of intersection of the surfaces. Assume suitable dimensions for their lengths and position with respect to each other.
45. Develop the surfaces of Question 44.
46. Two cylinders of 2" diameter intersect at an angle FlG 12_A
of 60°. Find the line of intersection of the sur- faces. Assume suitable dimensions for their lengths and position with respect to each other.
47. Develop the surfaces of Question 46.
48. Two cylinders of 2" diameter intersect at an angle of 45°. Find
the line of intersection of the surfaces. Assume suitable dimen- sions for their lengths and position with respect to each other.
49. Develop the surfaces of Question 48.
50. Two cylinders of 2" diameter intersect at an angle of 30°. Find
the line of intersection of the surfaces. Assume suitable dimen- sions for their lengths and position with respect to each other.
51. Develop the surfaces of Question 50.
52. Two cylinders of 2" diameter and If" diameter intersect at an
angle of 60°. Find the line of intersection of the surfaces. As- sume suitable dimensions for their lengths and position with respect to each other.
53. Develop the surfaces of Question 52.
5?-. Two cylinders of 2" diameter and If" diameter intersect at an angle of 45°. Find the line of intersection of the surfaces. Assume suitable dimensions for their lengths and position with respect to each other.
55. Develop the surfaces of Question 54.
56. Two cylinders of 2" diameter and If" diameter intersect at an angle
of 30°. Find the line of intersection of the surfaces. Assume suitable dimensions for their lengths and position with respect to each other.
57. Develop the surfaces of Question 56.
58. Two cylinders of 2" diameter intersect at a right angle and have
their axes offset \" . Find the line of intersection of the surfaces. Assume the additional dimensions.
59. Develop the surface of one of the cylinders of Question 58.
60. A 2" cylinder intersects a If" cylinder so that their axes are offset
|" and make a right angle with each other. Find the line of intersection of the surfaces. Assume the additional dimensions.
230 GEOMETRICAL PROBLEMS IN PROJECTION
61. Develop the surfaces of Question 60.
62. Two 2" cylinders have their axes offset \" . The elements intersect
at an angle of 60°. Find the line of intersection of the surfaces. Assume the additional dimensions.
63. Develop the surfaces of Question 62.
64. Two 2" cylinders have their axes offset J". The elements intersect
at an angle of 45°. Find the line of intersection of the surfaces. Assume the additional dimensions.
65. Develop the surfaces of Question 64.
66. Two 2" cylinders have their axes offset \" . The elements intersect
at an angle of 30°. Find the line of intersection of the surfaces. Assume the additional dimensions.
67. Develop the surfaces of Question 66.
68. A 2" cylinder intersects a If" cylinder at an angle of 60°. Their
axes are offset \" . Find the line of intersection of the surfaces. Assume the additional dimensions. 99. Develop the surfaces of Question 68.
FIG. 12-B. FIG. 12-C.
70. A 2" cylinder intersects a If" cylinder at an angle of 45°. Their
axes are offset \". Find the line of intersection of. the surfaces. Assume the additional dimensions.
71. Develop the surfaces of Question 70.
72. A 2" cylinder intersects a If" cylinder at an angle of 30°. Their
axes are offset £". Find the line of intersection of the surfaces. Assume the additional dimensions.
73. Develop the surfaces of Question 72.
74. Find the line of intersection of the surfaces of the two cones shown
in Fig. 12-B. Assume suitable dimensions.
75. Develop the surfaces of Question 74.
76. Assume a pair of cones similar to those shown in Fig. 12-B, but,
with equal altitudes. Then, find the line of intersection of the surfaces.
77. Develop the surfaces of Question 76.
78. Assume a cone and cylinder similar to that shown in Fig. 12-C.
Then, find the line of intersection of the surfaces.
79. Develop the surfaces of Question 78.
INTERSECTIONS OF SURFACES WITH EACH OTHER 231
80. Assume a cone and cylinder similar to that shown in Fig. 12-C,
but, have the cylinder elliptical. Then, find the line of inter- section of the surfaces.
81. Develop the surfaces of Question 80.
82. Assume a cone and cylinder similar to that shown in Fig. 12-C,
but, have the cone elliptical. Then, find the line of intersection of the surfaces.
83. Develop the surfaces of Question 82.
84. Assume an arrangement of cone and cylinder somewhat similar
to that of Fig. 12-C, but, have both cone and cylinder elliptical. Then, find the line of intersection of the surfaces.
85. Develop the surfaces of Question 84.
86. Assume a cylinder and cone of revolution, whose general direction
of axes are at right angles, but, which are offset as shown in Fig. 12-D. Then, find the line of intersection of the sur- faces.
87. Develop the surfaces in Question 86,
FIG. 12-D.
FIG. 12-E.
88. Assume an elliptical cylinder and a cone of revolution, whose general
direction of axes are at right angles to each other, but, which are offset as shown in Fig. 12-D. Then, find the line of inter- section of the surfaces.
89. Develop the surfaces of Question 88.
90. Assume a circular cylinder of revolution and an elliptical cone, whose
general direction of axes are at right angles to each other, but, which are offset as shown in Fig. 12-D. Then, find the line of intersection of the surfaces.
91. Develop the surfaces of Question 90.
92. Assume an elliptical cylinder and an elliptical cone, whose general
direction of axes, are at right angles to each other, but, which are offset, as shown in Fig. 12-D. Then, find the line of inter- section of the surfaces.
93. Develop the surfaces of Question 92.
94. Assume a cylinder and cone of revolution, similar to that of Fig.
12-E, and make angle a =60°. Then, find the line of intersection of the surfaces.
95. Develop the surfaces of Question 94.
232
GEOMETRICAL PROBLEMS IN PROJECTION
96. Assume a clyinder and cone of revolution, similar to that of Fig.
12-E, and make angle a =45°. Then, find the line of inter- section of the surfaces.
97. Develop the surfaces of Question 96.
89. Assume a cylinder and a cone of revolution, similar to that of Fig.
12-E, and make angle « =30°. Then, find the line of intersection
of the surfaces. 99. Develop the surfaces of Question 98.
100. Assume an elliptical cylinder and a cone of revolution arranged
similar to that of Fig. 12-E, and make the angle a =60°. Then, find the line of intersection of the surfaces.
101. Develop the surfaces of Question 100.
102. Assume a cylinder of revolution and an elliptical cone, arranged
similar to that of Fig. 12-E, and make the angle a =45°. Then, find the line of intersection of the surfaces.
103. Develop the surfaces of Question 102.
FIG. 12-E.
FIG. 12-F.
FIG. 12-G.
104. Assume an elliptical cylinder and an elliptical cone arranged similar
to that of Fig. 12-E, and make the angle a =60°. Then find the line of intersection of the surfaces.
105. Develop the surfaces of Question 104.
106. Assume a cylinder and cone of revolution as shown in Fig. 12-F,
and make angle a =60°. Then, find the line of intersection of the surfaces.
107. Develop the surfaces of Question 106.
108. Assume a cylinder and a cone of revolution as shown in Fig. 12-F,
and make angle a =45°. Then, find the line of intersection of the surfaces.
109. Develop the surfaces of Question 108.
110. Assume a cylinder and a cone of revolution as shown in Fig. 12-F,
and make angle a = 30°. Then, find the line of intersection of the surfaces.
111. Develop the surfaces of Question 110.
112. Assume an elliptical cylinder and a cone of revolution, arranged
similar to that shown in Fig. 12-F, and make angle a =60°. Then find the line of intersection of the surfaces.
INTERSECTIONS OF SURFACES WITH EACH OTHER 233
113. Develop the surfaces of Question 112.
114. Assume a cylinder of revolution and an elliptical cone, arranged
similar to that shown in Fig. 12-F, and make angle a =45°. Then, find the line of intersection of the surfaces.
FIG. 12-H.
FIG. 12-1.
115. Develop the surfaces of Question 114.
116. Assume an elliptical cylinder and an elliptical cone, somewhat
similar to that shown in Fig. 12-F, and make angle «=60°. Then find the line of intersection of the surfaces.
117. Develop the surfaces of Question 116.
118. Assume a cone and cylinder as shown in Fig. 12-G. Then, find
the line of intersection of the surfaces.
119. Develop the surfaces of Question 118.
120. Assume a cylinder and a sphere as shown in Fig. 12-H. Then,
find the line of intersection of the sur- faces.
121. Develop the surface of the cylinder of Ques-
tion 120.
122. Develop the surface of the cylinder of Ques-
tion 120, and, also, the surface of the sphere by the gore method.
123. Assume a cylinder and sphere as shown in
Fig. 12-1. Then, find the line of intersec- tion of the surfaces.
124. Develop the surface of the cylinder in Ques-
tion 123.
125. Develop the surface of the cylinder in Ques-
tion 123, and, also, the surface of the sphere by the zone method,
126. Assume a cone and a sphere as shown in Fig. 12-J. Then, find
the line of intersection of the surfaces.
127. Develop the surface of the cone in Question 126.
128. Develop the surface of the cone in Question 126, and, also, the
sphere by the gore method.
FIG. 12-J.
234
GEOMETRICAL PROBLEMS IN PROJECTION
129. Assume two doubly curved surfaces of revolution, whose axes
intersect. Then, find the line of intersection of the surfaces.
130. Develop the surfaces of Question 129 by the gore method.
131. Develop the surfaces of Question 129 by the zone method.
132. Assume two doubly curved surfaces of revolution whose axes do
not intersect. Then, find the line of intersection of the surfaces.
FIG. 12-K.
133. Develop the surfaces of Question 132 by the gore method.
134. Develop the surfaces of Question 132 by the zone method.
135. Assume a frustum of a cone as shown in Fig. 12-K. A circular
cylinder is to be fitted to it, as shown. Develop the surfaces. HINT. — Find vertex of cone before proceeding with the devel- opment.
PART III
PRINCIPLES OF CONVERGENT PROJECTING-LINE
DRAWING
CHAPTER XIII , PERSPECTIVE PROJECTION
1301. Introductory. Observation shows that the apparent magnitude of objects is some function of the distance between the observer and the object. The drawings made according to the principles of parallel projecting-line drawing, and .treated in Part I of this book, make no allowance for the observer's posi- tion with respect to the object. In other words, the location of the object with respect to the plane of projection has no influence on the size of the resultant picture, provided that the inclina- tion of the various lines on the object, to the plane of projec- tion, remaiivrfre same. Hence, as the remoteness of the object influence^' its apparent size, then, as a consequence, parallel projecting-line drawings must have an unnatural appearance. The strained appearance of drawings of this type is quite notice- able in orthographic projection wherein two or more views must be interpreted simultaneously, and is less noticeable in the case of oblique or axonometric projection. On the other hand, the rapidity with which drawings of the parallel projecting-line type can be made, and their adaptability for construction purposes, are strong points in their favor.
1302. Scenographic projection. To overcome the foregoing objections, and to present a drawing to the reader which cor- rects for distance, scenographic projections are used. To make these, convergent projecting lines are used, and the observer is located at the point of convergency. The surface on which
235
236 CONVERGENT PROJECTING-LINE DRAWING
scenographic projections are made may be spherical, cylindrical, etc., and find their most extensive application in decorative painting.
1303. Linear perspective. In the decoration of an interior of a dome or of a cylindrical wall, the resulting picture should be of such order as to give a correct image for the assumed loca- tion of the eye. In engineering drawing, there is little or no use for projections on spherical, cylindrical, or other curved surfaces. When scenographic projections are made on a plane, this type of projection is called linear perspective. A linear perspective, therefore, may be defined as the drawing made on a plane surface by the aid of convergent projecting lines, the point of convergency being at a finite distance from the object and from the plane. The plane of projection is called the picture plane, and the posi- tion of the eye is referred to as the point of sight.
1304. Visual rays and visual angle. If an object — an arrow for instance — be placed a certain distance from the eye, say in
the position ab (Fig. 227) and c be the point of sight, the two extreme rays of light or visual rays, ac and ab, form an angle which is known as the visual angle. If, now, the arrow be FIG. 227. moved to a more remote position de,
the limiting visual rays dc and ec
give a smaller visual angle. The physiological effect of this variation in the visual angle is to alter the apparent size of the object. If one eye is closed during this experiment, the distance from the eye to the object cannot 'be easily estimated, but the apparent magnitude of the object will be some function of the visual angle. In binocular vision, the muscular effort required to focus both eyes on the object will, with some experience, enable an estimate of the distance from the eye to the object; the mind will automatically correct for the smaller visual angle and greater distance, and, thus, to an experienced observer, give a more or less correct impression of the actual size of the object. This experience is generally limited to horizontal distances only, as very few can estimate correctly the diameter of a clock on a church steeple unless they have been accustomed to making such observa- tions. As the use of two eyes in depicting objects in space causes
PERSPECTIVE PROJECTION 237
slightly different inpressions on different observers, the eye will be assumed hereafter as a single point.
1305. Vanishing point. When looking along a stretch of straight railroad tracks, it may be noted that the tracks appar- ently vanish in the distance. Likewise, in viewing a street of houses of about the same height, the roof line appears to meet the sidewalk line in the distance. It is known that the actual distances between the rails, and between the sidewalk and roof is always the same; yet the visual angle being less in the distant observation, gives the impression of a vanishing point, which may, therefore, be defined as the point where parallel lines seem to vanish. At an infinite distance the visual angle is zero, and, hence, the parallel lines appear to meet in a point. The lines themselves do not vanish, but their perspective projections vanish.
1306. Theory of perspective projection. The simplest notion of perspective drawing can be obtained by looking at a distant house through a window-pane. If the observer would trace on the window-pane exactly what he sees, and locate all points on the pane so that corresponding points on the house are directly behind them, a true linear perspective of the house would be the result. Manifestly, the location of each successive point is the same as locating the piercing point of the visual ray on the picture plane, the picture plane in our case being the window- pane.
1307. Aerial perspective. If this perspective were now colored to resemble the house beyond, proper attention being paid to light, shade and shadow, an aerial perspective would be the result. In brief, this aerial perspective would present to the •eye a picture that represents the natural condition as near as the skill of the artist will permit. In our work the linear perspective will alone be considered, and will be denoted as the " perspective," aerial perspective finding little application in engineering.
1308. Location of picture plane. It is customary in per- spective to assume that the picture plane is situated between the eye and the object. Under these conditions, the picture is smaller than the object, and usually this is necessary. The eye is assumed to be in the first angle; the object, however, is generally in the
238
CONVERGENT PROJECTING-LINE DRAWING
second angle * for reasons noted above. The vertical plane is thus the picture plane. A little reflection will show that this is in accord with our daily experience. Observers stand on the ground and look at some distant object; visual rays enter the eye at various angles; the mean of these rays is horizontal or nearly so, and, therefore, the projections naturally fall on a vertical plane interposed in the line of sight. The window- pane picture alluded to is an example.
1309. Perspective of a line. Let, in Fig. 228, AB be an arrow, standing vertically as shown in the second angle; the point of sight C is located in the first angle. The visual rays pierce the picture plane at a" and b", and a"b", in the picture plane, is the perspective of AB in space. The similar condition
f
FIG. 228.
of Fig. 228 in orthographic projection is represented in Fig. 229. The arrow and the point of sight are shown by their pro- jections on the horizontal and vertical planes. The arrow being perpendicular to the horizontal plane, both projections fall at the same point ab; the vertical projection is shown as a'b'; the point of sight is represented on the horizontal and vertical planes as c and c', respectively. Any visual ray can likewise be represented by its projections, and, therefore, the horizontal projection of the point of sight is joined with the horizontal pro-
* The object may be located in any angle; the principles are the same for all angles. It is perhaps more convenient to use the second angle, as the lines do not have to be extended to obtain the piercing points as would be the case for first-angle projections.
PERSPECTIVE PROJECTION
239
jections. Similarly, the same is true of the vertical projections of the arrow and point of sight.
It is now necessary to find the piercing points (506,805) of the visual rays on the picture plane (vertical plane); and these are seen at a" and b". Therefore, a"b" is the perspective of AB in space.
1310. Perspectives of lines perpendicular to the hori= zontal plane. In Fig. 230, several arrows are shown, in projec- tion, all of which are perpendicular to the horizontal plane. Their perspectives are a"b", d"e", g"h". On observation, it will be noted that all the perspectives of lines perpendicular to the horizontal plane are vertical. This is true, since, when a plane
FIG. 230.
FIG. 231,
is passed through any line and revolved until it contains the point of sight, the visual plane (as this plane is then called) is manifestly perpendicular to the horizontal plane, and its vertical trace (603) is perpendicular to the ground line.
1311. Perspectives of lines parallel to both principal planes. Fig. 231 shows two arrows parallel to both planes, the projections and perspectives being designated as before. This case shows that all perspectives are parallel. To prove this, pass a plane through the line in space and the point of sight; the vertical trace of the plane so obatined will be parallel to the ground line* because the lirfe is parallel to the ground line.
* If the line in space lies in the ground line, the vertical trace will coincide with the ground line. This is evidently a special case.
240
CONVERGENT PROJECTING-LINE DRAWING
1312. Perspectives of lines perpendicular to the picture plane. Suppose a series of lines perpendicular to the vertical plane is taken. This case is illustrated in Fig. 232. The per- spectives are drawn as before and designated as is customary. It may now be observed that the perspectives of all these lines
FIG. 232.
(or arrows) vanish in the vertical projection of the point of sight. This vanishing point for the perspectives of all perpendiculars to the picture plane is called the centre of the picture. The reason for this is as follows: It is known that the visual angle
FIG. 233.
FIG. 234.
becomes less as the distance from the point of sight increases; at an infinite distance the angle is zero, the perspectives of the lines converge, and, therefore, in our nomenclature vanish.
1313. Perspectives of parallel lines, inclined to the picture plane. Assume further another set of lines, Fig. 233,
PERSPECTIVE PROJECTION 241
parallel to each other, inclined to the vertical plane, but parallel to the horizontal plane. Their orthographic projections are parallel and their perspectives will be seen to converge to the point V. Like the lines in the preceding paragraph, they vanish at a point which is the vanishing point of any number of lines parallel only to those assumed. The truth of this is established by the fact that the visual angle becomes zero (1304) at an infinite distance from the point of sight and, hence, the perspectives vanish as shown. To determine this vanishing point, draw any line through the point of sight parallel to the system of parallel lines; and its piercing point on the vertical or picture plane will be the required vanishing point. It amounts to the same thing to say that the vanishing point is the perspective of any line of this system at an infinite dis- tance.
A slightly different condition is shown in Fig. 234. The lines are still parallel to each other and to the horizontal plane, but are inclined to the vertical plane in a different direction. Again, the perspectives of these lines will converge to a new van- ishing point, situated, however, much the same as the one just preceding.
1314. Horizon. On observation of the cases cited in Art. 1313, it can be seen that the vanishing point lies on a horizontal line through the vertical projection of the point of sight. This is so since a line through the point of sight parallel to either sys- tem will pierce the picture plane in a point somewhere on its vertical projection. As . the line is parallel to the horizontal plane, its vertical projection will be parallel to the ground line and, as shown, will contain all vanishing points of all systems of horizontal lines. Any one system of parallel lines will have but one vanishing point, but as the lines may slope in various directions, and still be parallel to the horizontal plane, every system will have its own vanishing point and the horizontal line drawn through the vertical projection of the point of sight will be the locus of all these vanishing points. This horizontal line through the vertical projection of the point of sight is called the horizon.
As a corollary to the above, it may be stated, that all planes parallel to the horizontal plane vanish in the horizon. As before,
242
CONVERGENT PROJECTING-LINE DRAWING
the visual angle becomes less as the distance increases, and, hence, becomes zero at infinity.
There are an unlimited number of systems of lines resulting in an unlimited number of vanishing points, whether they are on the horizon or not. In most drawings the vertical and hori- zontal lines are usually the more common. The lines perpendicular to the picture plane are horizontal lines, and, therefore, the centre of the picture (vanishing point for the perspectives of the per- pendiculars) must be on the horizon.
1315. Perspective of a point. The process of finding the perspective of a line of definite length is to find the perspectives
235.
FIG. 236.
of the two extremities of the line. When the perspectives of the extremities of the line are found, the line joining them is the perspective of the given line. In Fig. 235, two points A and B are chosen whose perspectives will be found to be a" and b".
When the point lies in picture plane it is its own perspective. This is shown in Fig. 236, where a' is the vertical projection and a is therefore in the ground line; hence, the perspective is a'.
In the construction of any perspective, the method is to locate the perspectives of certain points which are joined by their proper lines. The correct grouping of lines determine certain surfaces which, when closed, determine the required solid.
1316. Indefinite perspective of a line. Let AB, Fig. 237, be a limited portion of a line FE. Hence, by obtaining the
PERSPECTIVE PROJECTION
243
piercing points of the visual rays AC and BC, the perspective a"b" is determined. Likewise, DE is another limited portion of the line FE and its perspective is d"e". If a line be drawn through the point of sight C, parallel to the line FE, then, its piercing point g' on the picture plane will be the vanishing point of the perspectives of a system of lines parallel to FE (1313). Also, as FE pierces the picture plane at f, then f will be its own perspective (1315). Therefore, if a line be made to join f and g', it will be the perspective of a line that reaches from the picture plane out to an infinite distance. And, also, the perspective of any limited portion of this line must lie
FIG. 237.
on the perspective of the line. From the construction in Fig. 237, it will be observed that a"b" and d"e" lie on the line f g'-
Consequently, the line f'g' is the indefinite perspective of a line FE which reaches from the picture plane at f out to an infinite distance. Thus, the indefinite perspective of a line may be defined as the perspective of a line that reaches from the picture plane to infinity.
Before leaving Fig. 237, it is desirable to note that g' is the vanishing point of a system of lines parallel to FE. It is not located on the horizon because the line FE is not parallel to the horizontal plane.
244
CONVERGENT PROJECTING-LINE DRAWING
1317. Problem 1. To find the perspective of a cube by means of the piercing points of the visual rays on the picture plane.
Construction. Let ABCDEFGH, Fig. 238, be the eight corners of the cube, and let S be the point of sight. The cube is in the second angle and therefore both projections are above the ground line (315, 317, 514, 516). The horizontal projec- tion of the cube is indicated by abed for the upper side, and efgh for the base. The vertical projections are shown as a'b'c'd' for the upper side, and eTg'h' for the base. The cube is located in the second angle so that the two projections overlap. The point of sight is in the first angle and is shown by its horizontal projection s and its vertical projection s'. Join s with e and
FIG. 238.
obtain the horizontal projection of the visual ray SE in space; do likewise for the vertical projection with the result that s'e' will be the visual ray projected on the picture plane. The point where this pierces the picture plane is E, and this is one point of the required perspective. Consider next point C. This is found in a manner similar to point E just determined. The horizontal projection of the visual ray SC is sc and the vertical projection of the visual ray is s'c' and this pierces the vertical or picture plane in the point C as shown. In the foregoing man- ner, all other points are determined. By joining the correct points with each other, a linear perspective will be obtained.
It will be noticed that CG, BF, DH, and AE are vertical because the lines in space are vertical (1310), and hence their perspectives are vertical. This latter fact acts as a check after
PERSPECTIVE PROJECTION
245
locating the perspectives of the upper face and the base of the cube. It will be noticed, further, that the perspectives of the horizontal lines in space meet at the vanishing points V and V (1313, 1314).
1318. Perspectives of intersecting lines. Instead of locat- ing the piercing point of a visual ray by drawing the projections of this ray, it is sometimes found desirable to use another method. For this purpose another principle must be developed.
If two lines in space inter- sect, their perspectives inter- sect, because the perspective of a line can be considered as being made up of the perspec- tives of all the points on the line. As the intersection is a point common to the two lines, a visual ray to this point should pierce the picture plane in the intersection of the perspectives. Reference to Fig. 239 will show that such j?IG 239
is the case. CG is the visual
ray in space of the point of intersection of the two lines, and its perspective is g", which is the intersection of the perspectives a"b" and d"e".
1319. Perpendicular and diagonal. Obviously, if it is desired to find the perspective of a point in space, two lines can be drawn through the piont and the intersection of their perspectives found. The advantage of this will appear later. The two lines generally used are: first, a perpendicular, which is a line perpendicular to the picture plane and whose perspective therefore vanishes in the vertical projection of the point of sight (1312); and second, a diagonal which is a horizontal line making an angle of 45° with the picture plane, and whose perspective vanishes somewhere on the horizon (1313). As the perpendicular and diagonal drawn through a point are both parallel to the horizontal plane, these two intersecting lines determine a plane which, like all other horizontal planes, vanish in the horizon. Instead of using a diagonal making 45°, any other angle may be used, provided
246
CONVERGENT PROJECTING-LINE DRAWING
it is less than 90°. This latter line would be parallel to the picture plane and therefore would not pierce it.
It can also be observed that there are two possible diagonals through any point, one whose perspective vanishes to the left of the point of sight and the other whose perspective vanishes to the right of the point of sight.
It is further known that the perspectives of all parallel lines vanish in one point and therefore the perspectives of all parallel diagonals through any point must have a common vanishing point. With two possible diagonals through a given point in space, two vanishing points are obtained on the horizon.
1320. To find the perspective of a point by the method of perpendiculars and diagonals. Let a and a', Fig. 240,
Horizon
FIG. 240.
represent the projections of a point in the second angle, and let c and c' be the projections of the point of sight. By drawing the visual rays ac and a'c' the piercing point a" is determined by erecting a perpendicular at e as shown. Although the perspective a" is determined by this method, it may also be determined by finding the intersections of the perspectives of a perpendicular and a diagonal through the point A in space.
Thus, by drawing a line ab, making a 45° angle with the ground line, the horizontal projection of the diagonal is found. As the diagonal is a horizontal line, its vertical projection is a'b'. This diagonal pierces the picture plane at b', which point is its own perspective. Its perspective also vanishes at V, the vanishing point of all horizontal lines whose inclination to the picture plane is at the 45° angle shown and whose directions are parallel to each other. The point V is found by drawing a line
PERSPECTIVE PROJECTION
247
cv parallel to ab, c'V parallel to a'b', and then finding the piercing point V. The horizontal projection of a perpendicular through A is ad; its vertical projection is evidently a', which is also its piercing point on the picture plane and as it lies in the picture plane, it is hence its own perspective. The vanishing point of the perspective of the perpendicular is at c', the centre of the picture. Since the perspective of a given point must lie on the per- spectives of any two lines drawn through the point, then, as b'V is the indefinite perspective of a diagonal, and a'c' is the indefinite perspective of a perpendicular, their intersection a" is the required perspective of the point A. The fact that a" has already been determined by drawing the visual ray and its
v
FIG. 241.
piercing point found shows that the construction is correct, and that either method will give the same result.
In Fig. 241, a similar construction is shown. The points V and V7 are the vanishing points of the left and right diagonals respectively. The perspective a" may be determined by the use of the two diagonals without the aid of the perpendicular. For instance ab and a'b' are corresponding projections of one diagonal; ad and a'd' are corresponding projections of the other diagonal. Hence W is the indefinite perspective of the right diagonal (since cv' is drawn to the right) and d'V is the indefinite perspective of the left diagonal. Their intersection determines a", the required perspective of the point A. The indefinite per- spective of the perpendicular is shown as a'c' and passes through the point a" as it should.
248
CONVERGENT PROJECTING-LINE DRAWING
Any two lines may be used to determine the perspective and should be chosen so as to intersect as at nearly a right angle as possible to insure accuracy of the location of the point. The visual ray may also be drawn on this diagram and its piercing point will again determine a".
1321. To find the perspective of a line by the method of perpendiculars and diagonals. Suppose it is desired to construct the perspective of an arrow by the method of perpendic- ulars and diagonals. Fig. 242 shows a case of this kind. AB
is an arrow situated in the second angle; C is the point of sight. A perpendicular through the point of sight pierces the picture plane in the vertical projection c'. The two possible diagonals whose horizontal projections are given by cm and en pierce the picture plane in V and V respectively. FIG. 242.- The perspective of any diagonal
drawn through any point in
space will vanish in either of these vanishing points depending upon whether the diagonal is drawn to the right or to the left of the point in question. Likewise, the perspective of a per- pendicular drawn through any point in space will vanish in the centre of the picture.
In the case at issue, the horizontal projection of a diagonal through b is bo and its vertical projection is b'o'; the piercing points is at o'. As the perspectives all lines parallel to the diagonal have a common vanishing point, then the perspective of the diagonal through B must vanish at V and the perspective of the point B must be somewhere on the line joining o' and V. Turning attention to the perpendicular through B, it is found that its vertical projection corresponds with the vertical pro- jection of the point itself and is therefore b'. The perspectives of all perpendiculars to the vertical or picture plane vanish in the centre of the picture; and, hence, the perspective of B in space must be somewhere on the line b'c'. It must also be somewhere on the perspective of the diagonal and, hence, it is at their intersection b". This can also be shown by drawing
PERSPECTIVE PROJECTION
249
the visual ray through the point of sight C and the point B. The horizontal and vertical projections of the visual ray are cb and c'b' and they pierce the picture plane at b", which is the same point obtained by finding the indefinite perspectives of the perpendicular and the diagonal.
By similar reasoning the perspective B!' is obtained and, therefore, a"b" is the perspective of AB.
1322. Revolution of the horizontal plane. The fact that the second and fourth angles are not used in drawing on account of the conflict between the separate views has already been con- sidered (315, 317). This is again shown in Fig. 238. The two views of the cube overlap and make deciphering more difficult. To overcome this difficulty the horizontal plane is revolved
-f
|
V |
A Horizon |
|
|
\—~ -^_^ |
4\0' |
|
|
\ \ i |
b< |
|
|
1 |
y ! \\ |
FIG. 243.
180° from its present position. This brings the horizontal projections below the ground line and leaves the vertical pro- jections the same as before. Now the diagonals through any point slope in the reverse direction, and care must therefore be used in selecting the proper vanishing point while drawing the indefinite perspective of the diagonal.
1323. To find the perspective of a point when the hori= zontal plane is revolved. Let Fig. 243 represent the conditions of the problem. The vertical projection of the given point is a' and its corresponding horizontal projection is at a, below the ground line due to the 180° revolution of the horizontal plane. The vertical projection of the point of sight is at c' and its cor- responding horizontal projection is at c, now above the ground line. The conditions are such that the given point seems like a first angle projection and the point of sight seems like a second
250
CONVERGENT PROJECTING-LINE DRAWING
angle projection, whereas the actual conditions are just the reverse of this.
Through a, draw ab the horizontal projection of the diagonal; the corresponding vertical projection is a'b' with V as the piercing point. As ab is drawn to the right, the indefinite perspective of the diagonal must vanish in the left vanishing point at V, Hence, draw Vb', the indefinite perspective of the diagonal.
|
y |
|||
|
1 |
y |
||
FIG. 243.
The indefinite perspective of the perpendicular remains unchanged and is shown in the diagram as a'c'. Therefore the intersection of Vb' and a'c' determine a", the required perspective of the point A. The accuracy of the construction is checked by drawing the visual ray ca and then erecting a perpendicular at o, as shown, which passes through a" as it should.
1324. To find the perspective of a line when the hori- zontal plane is revolved. The point B in space will again be
located in Fig. 244 just as was done in Fig. 242. The horizontal projection of the diagonal is bo and its vertical projection is b'o'; the piercing point, therefore, is o'. The horizontal projection of the point of sight c is now above the ground line instead of below it, due to the 180° revolution of FlG 244 the horizontal plane; the orig-
inal position of this point, is marked c". A diagonal through the point of sight, parallel to
PERSPECTIVE PROJECTION 251
the diagonal through the point B, is shown as en and pierces the vertical plane at V and, therefore, o'V is the indefinite per- spective of the diagonal. The indefinite perspective of the perpendicular, as before, is b'c', vanishing in the centre of the picture. The intersection of these two indefinite perspectives determines b", the perspective of the point B in space.
Instead of using the left diagonal bo, through b, it is possible to use the right diagonal bq, and this pierces the picture plane at q', and vanishes at V which (as must always be the case) again determines the perspective b". It must be observed that it matters little which diagonal is used with a perpendicular, or, whether only the diagonals without the perpendicular are used. In practice such lines are selected as will intersect as nearly as possible at right angles since the point of intersec- tion is thereby more accurately determined than if the two lines intersected acutely. Whatever is done the principal points can always be located by any two lines, and the other may be used as a check. Still another check can be had by drawing the hori- zontal projection of the visual ray cb whereby b", the perspective, is once more determined. The point A in space is located in an identical manner.
1325. Location of diagonal vanishing points. On observa- tion of Figs. 242 and 244, it is seen that the distance V to c' and V to c' is the same as the distance of the point of sight is from the vertical plane. This is true because a 45° diagonal is used and these distances are the equal sides of a triangle so formed. The use of any other angle would not give the same result, although the distance of the vanishing points from the centre of the pic- ture would always be equal.
All the principles that are necessary for the drawing of any kind of a linear perspective have now been developed. The subsequent problems will illustrate their uses in a variety of cases. Certain adaptations required for commercial application will appear subsequently.
1326. Problem 2. To find the perspective of a cube by the method of perpendiculars and diagonals.
Construction. Let ABCDEFGH in Fig. 245 represent the cube. A case has been selected that is identical to the one shown in Fig. 238, in order that the difference between the two
252
CONVERGENT PROJECTING-LINE DRAWING
methods may be clearly illustrated. Suppose it is desired to locate the perspective of the point E. Draw from e, in the hori- zontal projection, the diagonal eo, the vertical projection is e'o' with o' as a piercing point. Join o' with v, and the indefinite perspective of the diagonal is obtained. The perpendicular from e pierces the vertical plane at e' and e's' is the indefinite perspective. The intersection E of these indefinite perspectives is the required perspective of the point. This point can also be checked by drawing the other diagonal which pierces the picture plane at p', thereby making pV the indefinite perspective of the latter diagonal, which passes through E, as it should.
The point C is located by drawing the perpendicular and
FIG. 245.
diagonal in the same way as was done for the point E. The rest of the construction has been omitted for the sake of clearness. As in Fig. 242, the vanishing points V and V are laid out by drawing a line through the point of sight parallel to the system of lines. Fig. 245 does not show the construction in this way because the horizontal projection of the point of sight is now above the ground line, due to the 1 80 °-re volution of the horizon- tal plane. Instead, the lines have been drawn from the original horizontal projection shown as s" and care has been taken to see that the lines are parallel to the reversed position of the horizon- tal projection. A reference to both Figs. 238 and 245 will make this all clear.
In Fig. 245 two vanishing points v' and v will be noted which
PERSPECTIVE PROJECTION
253
are (as has already been shown to be) the vanishing points of the perspectives of all diagonals. It is also known that the centre of the picture s' is the vanishing point of the perspectives of all perpendiculars. Altogether, there are five vanishing points; the points V and V are the vanishing points of the perspectives of the horizontal lines on the cube.
1327. Problem 3. To find the perspective of a hexagonal prism.
Construction. Reference to Fig. 246 shows that one edge of the prism lies in the picture plane while the base is in the hor- izontal plane.* The diagonals used to determine the vanishing points are here chosen to be 30°, as in this way they are also parallel to some of the lines of the object and, therefore, have
FIG. 246.
common vanishing points. V' and V are the vanishing points of the diagonals, and of those sides which make an angle of 30° with the picture plane. The centre of the picture is at s' and, therefore, the vanishing point of the perspectives of all perpendic- ulars. The edge AG lies in the picture plane, and, therefore, is shown in its actual size. The indefinite perspectives of the diagonals from the extremities A and G will give the direction of the sides FA and AB for the top face and LG and GH for the base. The points F and L are located by drawing a diagonal and perpendicular through f and 1 in the horizontal projection. The distance mm' and nn' is equal to AG, the height of the prism.
* This picture has a strained appearance due to the selection of the point of sight. See Art. 1331 in this connection.
254
CONVERGENT PROJECTING-LINE DRAWING
1328. Problem 4. To find the perspective of a pyramid superimposed on a square base.
Construction. In Fig. 247, the edge GK is shown in the picture plane, a fact which enables the immediate determination of the direction of JK, KL, FG and GH by joining G and K with the vanishing points V' and V. It is only necessary to show the points J and L to determine the verticals FJ and HL. The method for this has already been shown. One other important point to locate is the apex A. This is shown by drawing a diagonal and perpendicular through a. The line AO, in space, pierces the
picture plane at o', and AP at p' at a height oo' equal to the height of the apex above the horizontal plane. Joining p' with s', the indefinite perspective of the perpendicular is obtained. At its inter- section with the indefinite perspective of the diagonal oa, locate A, the perspective of the apex. The horizontal projection of the point of sight is not shown as the location of v' from s' and v from s' also gives the distance of the point of sight from the pic- ture plane (1325).
1329. Problem 5. To find the perspective of an arch.
Construction. Fig. 248 shows plan and elevation of the arch. The corner nearest to the observer is in the picture plane, and therefore the lines in the plane are shown in their true length.
PERSPECTIVE PROJECTION
255
256 CONVERGENT PROJECTING-LINE DRAWING
This length need not be in actual size but may be drawn to any scale desired. From the extremities of these lines, others are drawn to the vanishing points V and V, as these are the van- ishing points of the perspectives of the parallels to the sides of the arch.
It is to be remembered that through s, a line sm is drawn parallel to the reversed direction of op, because the plan would ordinarily be above the ground line. This convention may or may not be adopted. In Fig. 249 it is shown in another way and perhaps this may seem preferable. The matter is one of personal choice, however, and to the experienced, the liability to error is negligible.
The most important feature of this problem is the location of points on the arch.* The advantage of the use of perpendiculars and diagonals, to locate the points A, B and C, may be shown here. Were these points located by finding the piercing points of the visual ray, the reader would soon find the number of lines most confusing, due to the lack of symmetry of direction; the use of the 45° and 60° triangles for the diagonals is much more convenient.
In commercial perspective drawing, the draftsman estimates such small curves as are shown at the base. Only such points are located which are necessary for guides. In this case, the verticals and their limiting position are all that are required.
1330. Problem 6. To find the perspective of a building.
Construction. The plan and elevations of the building are shown on Fig. 249 drawn to scale. In drawing the perspective, the plan only is necessary to locate the various lines. Continual reference must be made to the elevations for various points in the height of the building. Considerable work can be saved by having one corner of the building in the picture plane. Thus, that corner is shown in its true length even though it may not be the actual length. In the case shown, it is drawn to scale.
The vanishing points of the horizontal lines on the building have been located from the horizontal projection of the point
* This arch may also be drawn by craticulation. See Art. 1331.
PERSPECTIVE PROJECTION
257
258 CONVERGENT PROJECTING-LINE DRAWING
of sight, above the ground line. The construction lines are therefore drawn parallel to the sides of the building. The dis- tances of the window lines may be laid out to scale on the corner of the building which lies in the picture plane, as shown. Strictly speaking, the term " scale " cannot be applied to perspective drawing, as a line of a given length is projected as a shorter line as the distance from the observer increases.
1331. Commercial application of perspective. The artistic requirement of perspective involves some choice in the selection of the point of sight. In general, the average observer's point of sight is about five feet above the ground, unless there is some reason to change it. The center of the picture should be chosen so that it is as nearly as possible in the centre of gravity of area of the picture and this may modify the selection of the observer's point of sight. For instance, if a perspective of a sphere is made and the centre of the picture is chosen in the centre of gravity of the area, then the perspective is a circle. Otherwise, if the centre of the picture is to one side, the perspective is an ellipse,* and, to properly view the picture, it should be held to one side. This is contrary to the usual custom; an observer holds the picture before him so that the average of the visual rays is normal to the plane of the paper. On building or similar work, the roof and base lines should not converge to an angle greater than about 50°. This can be overcome by increasing the distance between the vanishing points, or, what amounts to the same thing, increasing the distance between observer and object.
It is also desirable to choose the position of the observer so as to show the most attractive view of the object most prominently. Where a building has two equally prominent adjacent sides, it is a good plan to adjust the observer's position to present the sides approximately in proportion to their lengths. That is, when one side of a building is 200 feet long and 50 feet wide, the length of the building should appear on the drawing about four times as long as the width (these recommendations have been purposely ignored in Fig. 249).
* The limiting rays from the sphere are elements of the surface of a cone of revolution; hence, its intersection with a plane inclined to its axis will be an ellipse.
PERSPECTIVE PROJECTION
259
When impressions of magnitudes of objects are to be con- veyed, it can be done by placing men at various places on the drawing. A mental estimate of the height of a man will roughly give an estimate of the magnitude of the object.
When curves are present in a drawing, the use of bounding figures of simple shape again finds application (210, 408). Architects know this under the name of craticulation. Fig, 250
shows an example as applied to the drawing of a gas-engine fly-wheel. A slightly better picture could have been obtained by giving the wheel a tilt so as to show the " section " more prominently. This would also bring the centre of the picture nearer to the centre of gravity of the area. If the reader will study a few photographs, keeping these suggestions in mind he will note the conditions which make certain pictures better than others.
260 CONVERGENT PROJECTING-LINE DRAWING
To make large perspectives is sometimes inconvenient due to the remoteness of the vanishing points. This requires a large drawing board and frequently the accuracy is impaired by the unwieldiness of the necessary drawing instruments. To over- come these objections, it is possible to make a small drawing of the object and then to redraw to a larger one by means of a pro- portional divider or by a pantograph. Only the more important lines need be located on the drawing. Such details as windows, doors, etc., can usually be estimated well enough so as to avoid suspicion as to their accuracy. The artist, with a little practice, soon accustoms himself to filling in detail.
Freehand perspective sketches can be made with a little practice. To make these, lay out the horizon and two vanish- ing points for building work and perhaps one or more for machine details. Accuracy is usually not a prerequisite, and, therefore, sketches of this kind can be made in almost the same time as oblique or axonometric sketches. The experience gained from this chapter should have furnished sufficient principles to be applied directly.
The chief function of perspective is to present pictures to those unfamiliar with drawing. It is therefore desirable to use every effort possible to present the best possible picture from the artistic viewpoint. The increased time required to make per- spectives is largely offset by the ease with which the uninitiated are able to read them. The largest application of perspective is found in architect's drawings to clients, and in the making of artistic catalogue cuts. With perspective is usually associated the pictorial effect of illumination so as to call on the imagination to the minimum extent.
1332. Classification of projections.* When drawings are made on a plane surface then there are two general systems employed: those in which the projecting lines are parallel to each other and those in which the projecting lines converge to a point.
* See Art. 413 in this connection.
PERSPECTIVE PROJECTION
261
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Commonly |
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called ortho- |
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Showing two |
graphic pro- |
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dimensions |
jections or |
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Mechanical |
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drawings. |
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Orthographic • |
Showing three dimensions |
Isometric |
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' Parallel projecting |
and known as axonome- |
Dimetric |
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lines |
tric projec- |
Trimetric |
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Projections on plane " |
Oblique |
tions. |
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surfaces |
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Convergent projecting . |
\ Outline alone is shown perspective I |
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lines |
Aerial ( Color and illumination added |
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perspective I to linear perspective |
QUESTIONS ON CHAPTER XIII
1. What is a scenographic projection?
2. What is a linear perspective?
3. Define visual rays.
4. Define visual angle.
5. What is the physiological effect of a variation in the visual angle?
6. How does binocular vision afford a means of estimating distance?
7. Why is the eye assumed as a single point in perspective?
8. What is a vanishing point? Give example.
9. Is the vanishing point the vanishing point of the line or of its per-
spective?
10. Show the theory of perspective by a window-pane illustration.
11. What is an aerial perspective?
12. What is the picture plane?
13. How is the picture plane usually located with respect to the object
and the observer?
14. Show by an oblique projection how the perspective of a line is
constructed.
15. Show how the perspective of a line is constructed in orthographic
projection.
16. Prove that all perspectives of lines perpendicular to the horizontal
plane have vertical perspectives.
17. Prove that all perspectives of lines parallel to both principal planes
have perspectives parallel to the ground line.
18. What is the centre of the picture?
19. Prove that the perspectives of all lines perpendicular to the picture
plane vanish in the centre of the picture.
262 CONVERGENT PROJECTING-LINE DRAWING
20. Is the centre of the picture the vertical projection of the point of sight?
21. What is a system of lines?
22. Prove that the perspectives of all systems of lines have a common
vanishing point.
23. How is the vanishing point of a system of lines found?
24. What is the horizon?
25. Prove that the perspectives of all horizontal systems of lines have
a vanishing point on the horizon.
26. Why is the centre of the picture on the horizon?
27. Find the perspective of a point which is located in the second angle,
by means of the piercing point of the visual ray on the picture plane.
28. Find the perspective of a point which is located in the first angle,
by means of the piercing point of the visual ray on the picture plane.
29. Show that the vertical projection of a point which is situated in
the picture plane is its own perspective.
30. Show that the vertical projection of a line which is situated in the
picture plane is its own perspective.
31. What is the indefinite perspective of a line? Give proof.
32. Find the perspective of a cube by means of the piercing points'of
the visual rays on the picture plane.
33. Prove that if two lines in space intersect, their perspectives intersect
in a point which is the perspective of the point of intersection on the lines.
34. What is a perpendicular when applied to perspective?
35. Where does the perspective of a perpendicular vanish?
36. What is a diagonal when applied to perspective?
37. How many diagonals may be drawn through a given point?
38. Where do the diagonals vanish? Why?
39. What angle is generally used for the diagonals? What other angles
may be used?
40. Find the perspective of a second angle point, by the method of
perpendiculars and diagonals.
41. Find the perspective of a second angle point, by drawing two diagonals
through it. Check by drawing the perpendicular.
42. Find the perspective of a first angle point, by the method of per-
pendiculars and diagonals.
43. Find the perspective of a first angle point, by drawing two diagonals
through it. Check by drawing the perpendicular.
44. Find the perspective of a second angle line, by the method of per-
pendiculars and diagonals.
45. Find the perspective of a second angle line, by drawing the diagonals
only. Check by drawing the perpendicular and visual rays.
46. Find the perspective of a first angle line, by the method of perpen-
diculars and diagonals.
47. Find the perspective of a first angle line, by drawing the diagonals
only. Check by drawing the perpendiculars and visual rays.
PERSPECTIVE PROJECTION 263
48. What is the object of revolving the horizontal plane in perspective?
49. What precaution must be used in perspective with reference to the
diagonals, when the horizontal plane is revolved?
50. Find the perspective of a second angle point, by the method of
perpendiculars and diagonals, when the horizontal plane is revolved.
51. Find the perspective of a first angle point, by the method of per-
pendiculars and diagonals, when the horizontal plane is revolved.
52. Find the perspective of a second angle line, by the method of per-
pendiculars and diagonals, when the horizontal plane is revolved.
53. Find the perspective of a first angle line, by the method of perpen-
diculars and diagonals, when the horizontal plane is revolved.
54. When the diagonals make an angle of 45° with the picture plane,
how far are the vanishing points from the centre of the picture?
55. Find the perspective of a cube by means of perpendiculars and
diagonals.
56. Why should the centre of the picture be chosen as near as possible
to the centre of gravity of the perspective?
57. What is the perspective of a sphere, when the centre of the picture
is in the centre of gravity of the perspective?
58. What is the perspective of a sphere, when the centre of the picture
does not coincide with the centre of gravity of the perspective?
59. What should be the approximate angle of convergence of the roof
and base lines of a building on a perspective?
60. If the roof and base lines of a building converge to an angle con-
sidered too large, what remedy is there for this condition?
61. When adjacent sides of a building are equally attractive, what should
be their general relation on the perspective?
62. How may impressions of magnitude be conveyed on a perspective?
63. What is craticulation?
64. How may large perspectives be made when the drawing board is
small?
65. When perspective is commercially applied, is it necessary to locate
every point on the details or can it be drawn with sufficient accuracy by estimation?
66. Make a complete classification of projections having parallel and
convergent projecting lines.
Note. In the following drawings, keep the centre of the picture as near as possible to the centre of gravity of the drawing.
67. Make a perspective of Fig. 1 in text.
68. Make a perspective of Fig. 10 in text.
69. Make a perspective of Fig. 17 in text.
70. Make a perspective of Fig. 2A. (Question in Chap. II.)
71. Make a perspective of Fig. 2B. (Question in Chap. II.)
72. Make a perspective of Fig. 21 in text.
73. Make a perspective of Fig. 3 A. (Question in Chap. III.)
74. Make a perspective of Fig. 3B. " "
75. Make a perspective of Fig. 3C. " " "
264 CONVERGENT PROJECTING-LINE DRAWING
76. Make a perspective of Fig. 3D. (Question in Chap. III.)
77. Make a perspective of Fig. 3E.
78. Make a perspective of Fig. 3F.
79. Make a perspective of Fig. 3G.
80. Make a perspective of Fig. 3H.
81. Make a perspective of Fig. 37.
82. Make a perspective of Fig. 3-7.
83. Make a perspective of Fig. 3K.
84. Make a perspective of Fig. 3L.
85. Make a perspective of Fig. 40 in text.
86. Make a perspective of Fig. 41 in text.
87. Make a perspective of Fig. 42 in text.
88. Make a perspective of a pyramid on a square base.
89. Make a perspective of an arch.
90. Make a perspective of a building.
PART IV PICTORIAL EFFECTS OF ILLUMINATION
CHAPTER XIV
PICTORIAL EFFECTS OF ILLUMINATION IN ORTHOGRAPHIC
PROJECTION
1401. Introductory. The phenomena of illumination on objects in space is a branch of the science of engineering drawing,, which aims to give the observer a correct imitation of the effect of light on the appearance of an object. In the main, it is desired to picture reality. The underlying principles of Rumination are taken from that branch of physics known as Optics or Light. The application of these principles to graphical presentation properly forms a part of drawing.
It is needless to say that the subject borders on the artistic; yet it is sometimes desirable to present pictures to those who are unfamiliar with the reading of commercial orthographic pro- jections. The architect takes advantage of perspective in con- nection with the effects of illumination, and in this way brings out striking contrasts and forcibly attracts attention to the idea expressed in the drawing.
It is not always essential, however, to bring out striking effects, as occasion often arises to picture the surfaces of an object. This is done by pic- torially representing the effects of illumination.
1402. Line shading applied to straight lines. The simplest application of line shading is shown in Fig. 251; the object
is a rectangular cover for a box. In the illumination, it is assumed
265
FIG. 251.
266
PICTORIAL EFFECTS OF ILLUMINATION
FIG. 251.
that the light comes in parallel lines, from the upper left-hand corner of the drawing. The direction is downward, and to the
right, at an angle of 45°. The illuminated surfaces are drawn with any thickness (or, as is commercially termed, " weight ") of line; the surfaces not in the light are made heavier as though the line or surface actually cast a shadow. Irrespective of the
location on the sheet, the drawing would always be represented in the same way since the light is assumed to come in parallel lines.
Evidently there is (strictly speaking) no underlying theory in this mode of shading, the process being merely a convention adopted by draftsmen, thus its extended use merely is the neces- sary recommendation for its value.
1403. Line shading applied to curved lines. In Fig. 252
FIG. 252.
is shown one part of a flange coupling, the illustration being chosen to show the application of line shading to the drawing of curves. In this latter application all is given that is necessary to make drawings using this mode of representat'on.
As before, the light comes from the upper left-hand corner
IN ORTHOGRAPHIC PROJECTION 267
of the drawing. By the aid of the two views (one-half of one being shown in section) such surfaces as will cast a shadow are easily distinguished. The shade lines of the circles are shown to taper gradually off to a diagonal line ab. In drawing the shade line of a circle, for instance, draw the circle with the weight of line adopted in making the drawing; with the same radius and a centre located slightly eccentric (as shown much exaggerated at c in Fig. 252), draw another semicircle, adding thickness to one side of the circle or the other, depending upon whether it is illum'nated or not.
It will be seen that this second circle will intersect the first somewhere near the diagonal ab. For instance, the extreme outside circle casts a shadow on the lower right-hand side, whereas the next circle is shaded on the upper left-hand side. These two shaded circles indicate that there is a projection on the surface. In the same manner the drawing is completed. It makes no difference whether the surface projects much or little, the lines are all shaded in the same way and with the same weight of shade line.
Notice should be taken that for all concentric circles, the eccentric centre is always the same. The six bolt holes are also shown shaded, and each of which has its own eccentric centre.
Many cases arise in practice where there are projec- tions or depressions in the surface. This convention helps to interpret, rapidly and correctly, such drawings. The time taken to use this method is more than offset by the advantages thereby derived; only in extremely simple drawings does it become unnecessary.
1404. Line shading applied to sections. It makes no dif- ference whether the outside view is shown, or whether the object is shown in section (Fig. 252) the shade lines are drawn as though the supplementary planes actually cut the object so as to expose that portion. This is done for uniformity only.
1405. Line shading applied to convex surfaces. Occa- sionally, curved surfaces must be contrasted with flat surfaces, or, perhaps, the effect of curvature brought out without an attempt to
268 PICTORIAL EFFECTS OF ILLUMINATION
contrast it with flat surfaces. Fig. 253 shows a cylinder shaded. The heavier shade is shown to the right as the light is supposed to come from the left; the reason for this will be shown later. The effect of shading and its graduation is produced by gradually altering the space between the lines, however, keeping the weight of line the same. A somewhat better effect is produced by also increasing the weight of the line in connection with the decrease in the spacing, but the custom is not general.
1406. Line shading applied to concave surfaces. A hollow cylinder is shown shaded in Fig. 254. Again, the light comes from the left and the heavier shade falls on the left, as shown.
FIG. 253. FIG. 254.
A comparison between Figs. 253 and 254 shows how concave and convex surfaces can be indicated.
1407. Line shading applied to plane surfaces. Flat surfaces are sometimes shaded by spacing lines an equal distance apart. The effect produced is that of a flat shade. The method is used when several surfaces are shown in one view whose planes make different angles with the planes of projection, like an octagonal prism, for instance.
PHYSICAL PRINCIPLES OF LIGHT .
1408. Physiological effect of light. Objects are made evident to us by the reflected light they send to our eyes. The brain becomes conscious of the form of the object due to the physiological effect of light on the retina of the eye. The locality from which the light emanates is called the source. Light travels in straight lines, called rays, unless obstructed by an opaque body. If the source of light is at a considerable distance from the object, the light can be assumed to travel in parallel lines.
IN ORTHOGRAPHIC PROJECTION 269
The chief source of light is the sun and its distance is so great (about 93,000,000 miles) that all rays are practically parallel.
1409. Conventional direction of light rays. The source of light may be located anywhere, but it is usual to assume that it comes from over the left shoulder of the observer who is viewing an object before him. The projections of the rays on the hor- izontal and vertical plane make an angle of 45° with the ground line. The problems to follow will be worked out on this assumed basis.
1410. Shade and shadow. The part of the object that is not illuminated by direct rays is termed the shade. The area from which light is excluded, whether on the object itself or on any other surface, is called the shadow. As an example, take a building with the sun on one side; the opposite side is evidently in the shade. A cornice on this building may cast a shadow on the walls of the building, while the entire structure casts a shadow on the ground, and sometimes on neighboring buildings.
1411. Umbra and penumbra. When the source of light is chosen near the object, two distinct shadows are observable, one within the other. Fig. 255 shows a plan view of a flat gas flame ab and a card cd. Rays emanate from every point of the flame in all directions. Consider the point a, in the flame. The two rays ac and ad will determine a
shadow cast by the card the area of which is located away from the flame and limited to that behind fcdh. Similarly, from the point b, the two rays be and bd cast a a shadow, limited by the area behind ecdg. These two areas overlap. From the posi- tion of fcdg, no part of the flame can be seen by an observer standing there; but from any one of the areas behind eof or gdh, a portion of the flame can be seen. The effect of this is that the shadow in fcdg is much more pronounced and therefore is the darker.
That portion of the shadow from which the light is totally excluded (fcdg) is called the umbra; that portion from which the light is partly excluded (ecf or gdh) is called the penumbra.
270
PICTORIAL EFFECTS OF ILLUMINATION
This effect is easily shown by experiment, and is sometimes seen in photographs taken by artificial illumination when the source of light is quite near the object. In passing, it may also be mentioned that if there are two sources of light, there may be two distinct shadows, each having its own umbra and penumbra, but such cases are not usually considered in drawing.
When applying these principles to drawing, the shadow is limited to the umbra, and the penumbra is entirely neglected. The light is supposed to come from an infinite distance, and, therefore, in parallel rays. So little in the total effect of a draw- ing is lost by making these assumptions, that the extra labor involved in making theoretically correct shadows is uncalled for.
GRAPHICAL REPRESENTATION
1412. Application of the physical principles of light to drawing. In the application of the physical principles enumerated above, it is merely necessary to draw the rays of light to the limiting lines of the object, and find their piercing points with the surface on which the shadow is cast. This is the entire theory.
1413. Shadows of lines.* Fig. 256"exhibits the shadow cast by an arrow AB shown by its horizontal projection ab and its
FIG. 256.
FIG. 257.
vertical projection a'b'. The rays of light come, as assumed,
* Lines are mathematical concepts and have no width or thickness. They can, therefore, fetrictly speaking, cast no shadow.
IN ORTHOGRAPHIC PROJECTION 271
in lines whose horizontal and vertical projections make an angle of 45° with the ground line. A ray of light to A is shown as ca in the horizontal projection and c'a' in the vertical projection. The ray of light to B is shown by the similar projection db and d'b'. These two rays pierce the horizontal plane at a?' and b" and a"b" is therefore the shadow of the arrow AB in space.
If the arrow is as shown in Fig. 257, then the shadow is on the vertical plane. A similar method is used for drawing the pro- jections of the rays on the horizontal and vertical planes. Instead, however, the piercing points on the vertical plane are found, and these are shown as a"b", which again is the shadow of AB in space.
In Fig. 256 the rays pierce the horizontal plane before they pierce the vertical plane and the shadow is therefore on the horizontal plane. Since the plane is assumed opaque there is no vertical shadow. Fig. 257 is the reverse of this. A vertical shadow and no horizontal shadow is obtained. Cases may occur where the shadow is partly on both planes, similar to a shadow cast partly on the floor and partly on the wall of a room. — Fig. 258 shows this in construction. It differs from the preceding in so far as the two horizontal and the two ver- tical piercing points of the rays are FIG. 258. obtained, and thus the direction of the
shadow is determined on both planes. If the construction is carried out accurately, the projections will intersect at the ground line, and the shadow will appear shown as a"b".
1414. Problem 1. To find the shadow cast by a cube which rests on a plane.
Let, in Fig. 259, abed be the horizontal projection of the top of the cube and a'b'c'd' be the vertical projection of the top of the cube. The three points in space A, B and C will be sufficient to determine the shadow and, as a consequence, the projections of these rays are drawn an angle of 45°, as shown. The hor- izontal projections of the rays are accordingly ae, bf and eg; the vertical projections are likewise a'e', b'f and c'g'. The
272
PICTORIAL EFFECTS OF ILLUMINATION
piercing points (506, 805) of these rays are e, f and g. The shadow ir therefore determined because the cube rests on the horizontal plane, and the base must necessarily be its own shadow.
1415. Problem 2. To find the shadow cast, by a pyramid, on the principal planes.
In order to show a case where the shadow is cast on both planes, the pyramid must be located close to the vertical plane. The pyramid, ABCDE, Fig. 260, is shown in the customary way
FIG. 259
as abcde or the horizontal projection, and a'b'c'd'e' for the vertical projection. The horizontal piercing point of what would be the shadow of the apex must be located in order to determine the shadow of the side BA, even though it forms no part of the actual shadow. Locate also the vertical piercing point g' which is the actual shadow of the apex. It will be noticed that the shadow of BA only continues until it meets the vertical plane and is therefore limited by the projection bh. Similarly, the shadow is limited by di, a portion of the line df. Where these two lines meet the ground line, join the points with the actual shadow of the apex and the shadow is then completed. It may be
IN ORTHOGRAPHIC PROJECTION
273
observed that the limiting lines of an object determine the shadow in all cases, and that the interior lines do not influence the figure unless they project above the rest of the object in any way.
1416. Problem 3. To find the shade and shadow cast by an octagonal prism having a superimposed octagonal cap.
Let Fig. 261 represent the object in question. Confine atten- tion at present to the shadow on the horizontal plane. Con-
c'b' d'a' e'fi fg
FIG. 261.
sider, first, the shadow cast by the line GH in space. It will be seen that the rays of light from G and H pierce the horizontal plane at points g" and h" and that these points are found by drawing the projections of the rays from the points G and H in space. If the construction is correct so far, then g"h" should be parallel to gh, as GH is parallel to the horizontal plane and as its shadow must be parallel both to the line itself and also to its horizontal projection. Similarly, locate a" and i" '. Thus, three lines of the shadow are determined.
It is now necessary to observe carefully that the shadow of
274
PICTORIAL EFFECTS OF ILLUMINATION
the superimposed cap is not only due to the top face, as the points m"n"o"p" are the piercing points of rays that come from the points MNOP, in space, and that they are on the lower face of the cap. The shadow of the points BCDE fall within the area a"h"g"f"p"o"m"n" and therefore need not be located. The shadow is completed by drawing r"s" and t"u" which are a part of the shadow cast by the octagonal prism below. It is unneces- sary to determine the Complete outline of the shadow of the
c'b' d'a! e'ti f'J
mf:
prism as it merges into that cast by the cap. The shadow on the horizontal plane is thus completely determined.
Consider, in addition, the shadow cast on the prism due to the superimposed cap. The shadow limited by the line v"w" is due to the limited portion of the lower face of the cap shown horizontally projected at vw and vertically as v'w'. The shadow of the point O in space is shown as x" and that of Y in space as y". All these latter points are joined by straight lines as they are shadows of straight lines cast upon a plane surface and may be considered as the intersection of a plane of rays through the line forming the outline of the object.
IN ORTHOGRAPHIC PROJECTION
275
The right of the face of the prism and of the cap is shaded entirely, because it is in the shade, and since it receives no direct illumination, it is represented as shown.
1417. Problem 4. To find the shade and shadow cast by a superimposed circular cap on a cylinder.
Fig. 262 shows the cylinder with its superimposed cap. The shadow cast by the cap will be some form of curve, and attention will be directed at present, to the location points on this curve. From a', in the vertical projection, draw a ray whose projection makes an angle of 45° with the ground line, and locate it so that it impinges on the surface of the cylinder at the extreme visible element to the left. The corre- sponding horizontal projection of a7 is a, and the point where this ray impinges is a," one point of the required shadow. Consider point b, vertically projected at b'. The projections of its ray will in- tersect the surface of the cylinder at the point b" along an element shown ver- tically projected at b'". Select another point c' in the vertical projection, so that the ray c'd' will be tangent to the cyl- inder. The element is horizontally pro- jected at c"d and the ray impinges at c" which is still another point of the shadow.
In practice, several more points are determined, but they are omitted here for the sake of clearness. The portion of the cylinder away from the source of light must be in the shade and d' shows the vertical projection of one element from which the illuminated portion is separated from the shade. Accord- ingly, the portion of the cylinder to the right of c"d is shaded entirely. The same effect occurs on the superimposed cap, and, hence, from the element e, to the right, the entire remaining area is shaded, being unilluminated from rays having projections which make an angle 45° with the ground line.
1418. High=light. When a body, such as a polished sphere, is subjected to a source of light, one spot on the sphere will appear
FIG. 262.
276 PICTORIAL EFFECTS OF ILLUMINATION
much brighter than the rest of the sphere. This spot is called the brilliant point, or, more commonly, the high-light. The effect is that due to the light being immediately reflected to the eye with practically undiminished intensity.
1419. Incident and reflected rays. It is a principle in optics * that the incident ray and the reflected ray make equal angles with the normal to the surface on which the light impinges. The tangent plane at the point where the light impinges must be perpendicular to the normal; the incident ray and the reflected ray also lie in the same plane with the normal and therefore this plane is perpendicular to the tangent plane. It may be observed that if a line be drawn through any point in space parallel to the ray of light or incident ray, and through the same point, another line be drawn parallel to the reflected ray, then the included angle will be the same irrespective of the loca- tion of the arbitrary point chosen. The normal through this point would therefore be parallel to a normal drawn through any other point in space, and a plane perpendicular to one normal would be perpendicular to all. By choosing the perpendicular plane so that it touches the surface a tangent plane to the sur- face is obtained; the point of contact is then the high-light. The foregoing can best be illustrated by an example.
1420. Problem 5. To find the high-light on a sphere.
To use a simple illustration, the high-light on the vertical
projection only will be deter- mined. Fig. 263 shows this con- struction. Assume that the eye is directed perpendicularly to the vertical plane. The light comes in a direction whose projections make an angle of 45° with the ground line. For convenience, take a ray through the centre of the sphere o'; the incident ray is therefore m'o'. The line FlG 263 through the point of sight is per-
pendicular to the plane of the
paper, and is therefore projected at o'. Revolve the plane of * See a text-book on Physics.
IN OKTHOGRAFHIC PROJECTION 277
these incident and reflected rays about mV until it coin- cides or is parallel with the vertical plane. The perpendic- ular to the plane of the paper through o' will fall to c" and the revovled angle will be c"o'm"; the angle m'o'm" being numerically * equal to 35° 16' for rays whose projections make an angle of 45° with the ground line. The point m" is graphically located as follows: Assume that a horizontal plane is passed through o', the centre of the sphere. The distance of the point m' from this new horizontal plane is m'q' and this distance does not change in the revolution. Hence, the point M in space falls at m", om" being the revolved position of the actual ray of light.
The bisector of the angle m"o'c" is the normal to the surface and a plane tangent to the revolved position of the sphere will be tangent at the point p". It is unnecessary to draw the plane in this case as the normal readily determines p". The counter revolved position of p" is p' ; therefore, p' is the required high- light.
1421. Multiple highlights. Before leaving this subject it must be noted that with several sources of light on an object, there may be several high-lights, although it is usual to assume only one on a surface like the sphere. A corrugated surface will have several high-lights for any one position of the eye and a single source of light.
1422. Highlights of cylindrical or conical surfaces. A
cylinder or a cone can have only one high-light for any one posi- tion and direction of vision of the eye. If the surface is highly polished, it will be the only point visible; the object itself cannot be distinguished as to the details of its form. As the eye is directed anywhere along the surface, a new high-light is observed for each direction of vision and the locus of these points forms approximately a straight line.
1423. Aerial effect of illumination. The foregoing princi- ples are true for the conditions assumed. In nature, however, it is impossible to receive light from only one source in directly parallel rays, to the total exclusion of any other light. Modifications must be introduced because the surfaces are not optically true
* Computed by trigonometry.
278 PICTORIAL EFFECTS OF ILLUMINATION
in the first instance. This means that a cylinder, although made round as carefully as possible, very slight deviations hardly discernible by measuring instruments are readily detected by their appearance in the light. The reflected light augments the imperfections, so that to the experienced eye the effect is noticeable. Other equally important facts are those due to the reflection from the walls of the room, and from other objects; the diffusion caused by the light being transmitted through the window-pane, and that reflected from dust particles suspended in the air. All these disturbing influences tend to illuminate certain other portions of the object and actually do so to such an extent that cognizance of this aerial effect of illumination must be taken.
1424. Graduation of shade. The effect on a highly polished sphere can be brought out by holding one in a room whose walls are a dull black, similar to a room used for photometric testing. Even this photometer room does not fulfil the requirements to the last degree, as there is no surface which does not reflect at least some light. Observing this sphere by aid of a ray of light coming through any opening in the room, the high-light will alone be visible. If there were no variation in the amount of light sent to the eye from the various points on the object, due to the aerial effect of illumination, it would be impossible to recognize its form. Hence no light can come to the eye except from the high-light unless the aerial effect is present.
With these difficulties to contend with, it may seem that the foregoing principles become invalid. Such is not the case, however, as the principles are true, in the main, but the proper correction is made for diffused light and a graduation in shade is introduced so as make the observer conscious of the desired idea to be conveyed.
1425. Shading rules. At times it may even be necessary to strain a point in order to bring out some detail of the object drawn. This latter is a characteristic of the skill of the artist. The rules for the representation of shades are here inserted, being taken from Mahan's Industrial Drawing:
" 1. Flat tints should be given to plane surfaces, when in the light, and parallel to the vertical plane; those nearest the eye being lightest.
FIG 264.
FIG. 265.
FIG. 266. FIG. 267.
Examples of Graduated Shades in Pictorial Effects of Illumination.
[To face page 278
IN ORTHOGRAPHIC PROJECTION 279
" 2. Flat tints should be given plane surfaces, when in the shade, and parallel to the vertical plane; those nearest the eye being darkest.
" 3. Graduated tints should be given to plane surfaces, when in the light and inclined to the vertical plane; increasing the shade as the surfaces recede from the eye; when two such surfaces incline unequally the one on which the light falls most directly should be lightest.
" 4. Graduated tints should be given to plane surfaces, when in the shade, and inclined to the vertical plane; decreasing the shade as the surfaces recede from the eye."
1426. Examples of graduated shades. In Figs. 264, 265, 266 and 267 are shown a prism, cylinder, cone and sphere shaded in accordance with the above rules. It will be seen that no mistake can occur as to the nature of the objects, thus certifying to the advisability of their adoption.
QUESTIONS ON CHAPTER XIV
1. What is the purpose of introducing the pictorial effects of illumina-
tion?
2. What is meant by line shading?
3. How are the straight lines of an object line shaded? Give example.
4. How are the curved lines of an object line shaded? Give example.
5. How are the "sections" line shaded? Give example.
6. How are convex surfaces line shaded? Give example.
7. How are concave surfaces line shaded? Give example.
8. How are plane surfaces line shaded? Give example.
9. How are objects made evident to us?
10. What is the source of light?
11. What are light rays?
12. What conventional direction of ray is adopted in shading drawings?
13. What is the shade on an object?
14. What is the shadow of an object?
15. What is the umbra? Give example.
16. What is the penumbra? Give example.
17. Why is only the umbra used on drawings?
18. What is the fundamental operation of finding the shadow of an
object?
19. Construct the shadow of a line that is situated so as to have a
shadow on the horizontal plane.
20. Construct the shadow of a line that is situated so as to have a
shadow on the vertical plane.
280 PICTORIAL EFFECTS OF ILLUMINATION
21. Construct the shadow of a line that is situated so as to have a
shadow on both principal planes.
22. What is a high-light? Explain fully.
23. What is the incident ray?
24. What is the reflected ray?
25. What angular relation is there between the normal and the incident
and reflected rays?
26. Do the normal, the incident ray and the reflected ray lie in one
plane?
27. Find the high-light on a sphere.
28. How are multiple high-lights produced?
29. What is the aerial effect of illumination?
30. What is meant by graduation of shade?
31. To what is the graduation of shade due?
32. Shade a sphere with lead pencil so as to show the high-light and
the graduation of the shade.
33. Shade a cylinder with lead pencil so as to show the high-light and
the graduation of the shade.
34. Shade a cone with lead pencil so as to show the high-light and the
graduation of the shade.
35. Shade an octagonal prism with lead pencil so as to show the high-
light and the graduation of the shade.
36. Construct the horizontal shadow of a cube resting on the horizontal
plane.
37. Construct the horizontal shadow of a cube which is some distance
above the horizontal plane.
38. Construct the horizontal shadow of a triangular prism which rests
on the horizontal plane.
39. Construct the horizontal shadow of a hexagonal prism which rests
on the horizontal plane.
40. Construct the horizontal shadow of a pyramid which rests on the
horizontal plane.
41. Construct the shade and horizontal shadow cast by a cylinder with
a superimposed circular cap.
42. Construct the shade and horizontal shadow cast by a cylinder with
a superimposed square cap.
43. Construct the shade and horizontal shadow cast by a*cylinder with
a superimposed octagonal cap.
44. Construct the shade and horizontal shadow cast by an octagonal
prism with a superimposed circular cap.
45. Construct the shade and horizontal shadow cast by an octagonal
prism with a superimposed square cap.
46. Construct the shade and horizontal shadow cast by an octagonal
prism with a superimposed octagonal cap.
47. Construct the shade and horizontal shadow cast by a cone which
rests on a square base.
48. Construct the shade and horizontal shadow cast by an octagonal
pyramid which rests on a square base.
IN ORTHOGRAPHIC PROJECTION 281
49. Construct the shadow cast by a cone which is situated so as to
cast a shadow on both principal planes.
50. Construct the shadow cast by an octagonal pyramid which is situated
so as to cast a shadow on both principal planes.
51. Construct the shadow cast by an octagonal prism which is situated
so as to cast a shadow on both principal planes.
52. Construct the shadow cast by a superimposed cap on the inside
of a hollow semi-cylinder.
CHAPTER XV
PICTORIAL EFFECTS OF ILLUMINATION IN PERSPECTIVE PROJECTION
1501. Introductory. The fundamental principles of the pic- torial effects of illumination are best studied in orthographic projection. Their ultimate use, however, is usually associated with perspective. Thus, in one color, an attempt is made to picture reality, whether it be used for engineering purposes, as catalogue illustrations, or whether it be used for general illustrating purposes. The principles, established here, hold equally well when color is added to the perspective (making it an aerial per- spective) and its illumination; but, as this is recognized as a distinct field, it will not be considered in this book.
1502. Problem 1. To draw the perspective of a rectangular prism and its shadow on the horizontal plane.
The first step in this case is to draw the object and its shadow orthographically. The object is shown in the second angle as
has been the custom in perspec- tive; the shadow is on the hori- zontal plane and its construc- tion is carried out in accord- ance with principles previously discussed (Chap. XIV). The shaded area in Fig. 268 shows the shadow so constructed. The advisability of the 180° revolu- tion of the horizontal plane has been given (1322), and carrying this into effect in the present FIG. 268. instance, the effect shown in
Fig. 269 is obtained. The hori- zontal projection of the prism and the shadow of the prism are then shown below the ground line. The perspective of the prism
282
a'd
dh
IN PERSPECTIVE PROJECTION
283
perhaps be clear from the illustration in Fig. 269 as all the necessary construction lines have been included. The steps necessary for the construction of the perspective of the shadow, however, will be considered.
The point of sight is at S, shown horizontally projected at s and vertically projected at s'. The vanishing points of the diagonals are shown as v and v'; the distance of v and v' from s' are equal to the distance s above the ground line (1325). A
FIG. 269.
perpendicular and a diagonal through the point d" (the shadow of D in space) will intersect at d'" which is the required perspective of the shadow of this point. The same procedure will locate c'" and b'". The corners F and G are the perspectives of those points in space, and as they rest on the horizontal plane, they are also the perspectives of their shadows. By joining Fb"'c'"d'" and H with lines the complete outline of the shadow is obtained, except in so far as the limited portion behind the prism from H which is hidden from the observer is concerned.
284
PICTORIAL EFFECTS OF ILLUMINATION
1503. General method of finding the perspective of a shadow.* The above method of constructing the perspectives of shadows is perfectly general, although lengthy. It is possible to economize time, however, by taking advantage of the method of locating the perspective of the shadow directly. The shadow of a point in space on any surface is the piercing point of a ray of light through the point on that surface; the perspective of that shadow must therefore lie somewhere on the perspective of the ray of light. It will also lie on the line of intersection of the plane receiving the shadow with a plane containing the ray of light. The perspective of this line of intersection will also
FIG. 270.
contain the perspective of the shadow. Hence, the perspective of the shadow of the point will lie on the intersection of these two perspectives (1318).
1504. Perspectives of parallel rays of light. The rays of light are assumed as coming in parallel lines; being parallels, they therefore have a common vanishing point. To find this vanishing point (1313) draw through the point of sight, a line parallel to these rays; and the piercing point of this line on the picture plane will be the required vanishing point. In Fig. 270, if a ray be drawn through the point of sight, then the vertical
* This method is similar, in general, to the finding of the piercing point of a given line on a given plane. See Art. 823.
IN PERSPECTIVE PROJECTION 285
projection of the ray will be sY; the horizontal projection of the ray will be sr (a careful note being made of the 180° revo- lution of this plane and hence the revolved direction of the ray) and the piercing point on the picture plane will therefore be atr7.
1505. Perspective of the intersection of the visual plane on the plane receiving the shadow. The line of intersection of the plane receiving the shadow and the plane containing the ray of light (or visual plane as this latter plane is called) is in our case a horizontal line, as the plane receiving the shadow is the horizontal plane. If the horizontal projecting plane of the ray be taken, it is known that the trace makes an angle of 45° with the ground line, and that this horizontal line must vanish in the horizon at v. It will be observed that this same point (v) is also the vanishing point of all diagonals drawn to the right of the point of sight. A further note may be taken of the fact that if the perspective of the horizontal projection of the point be joined with the right vanishing point of the diagonal, the perspective is identical with the perspective of the intersection of the horizontal plane and the visual plane, because the per- spectives of the horizontal projection of the point and the van- ishing point are common to the two.
1506. Application of the general method of finding the perspective of a shadow. The foregoing can be applied to the finding of the shadow of Problem 1. A reference to Fig. 270 in addition to what follows, will indicate the application. Suppose the perspective of the shadow of D is under consideration. The perspective of the visual ray is Dr'; the perspective of the hori- zontal projection of the ray is Hv; their intersection is d'", the perspective sought. Likewise, c'" is similarly located, and it is the perspective of the shadow of C, found by the inter- section of the Cr' (perspective of the ray) and Gv (perspec- tive of the horizontal projection of the ray). The point B has its shadow at b'", and its location is clearly shown in the figure.
The shadow is completed by joining the proper points with lines. In every way it is identical with the shadow determined in Problem 1.
286
PICTORIAL EFFECTS OF ILLUMINATION
1507. Problem 2. To draw the perspective of an obelisk with its shade and shadow.
Let AFGHK be the obelisk (Fig. 271) and S the point of sight. The perspective is drawn in the usual way. The problem is of interest in so far as the rays of light make an angle of 30° with the ground line, thus causing a longer shadow than when the 45° ray is used.
Through the horizontal and vertical projections of the point of sight, draw a ray parallel to the conventional ray adopted. This ray pierces the picture plane at r' the vanishing point of all the rays in space. The horizontal projections of all rays vanish
lie
FIG. 271.
at r" on the horizon, as all lines should that are parallel to the horizontal plane and at the same time belong to the system of lines parallel to the rays of light.
Inspection of Fig. 271 will show that the lines GC, CA, AE, and EK affect the shadow, and that the only points to be located for the shadow are C, A, and E. To locate c", the shadow of point C, draw the perspective of the ray through C; Cr' is this perspective. The perspective of the horizontal projection is c'"r"; c'" is the perspective of the horizontal projection of C, the necessary construction lines being shown in the figure. The intersection of these two perspectives is c", the required per- spective of the shadow of C in space.
The point A has the perspective of its shadow at a" which,
•
FIG. 273.— Commercial Application of the Pictorial Effects of Illumination in Perspective.
[To face page 287]
IN PERSPECTIVE PROJECTION 287
as before, is the intersection of the perspective Ar' (of the ray) and a'"r" (of the horizontal projection of the ray). The point E has its shadow e" located in an identical manner as the preceding points c" and a".
As the obelisk is resting on a plane (the horizontal in this case) the base is its own shadow, and it is only necessary to join the shadow of C with G and the line of the shadow c"G is deter- mined. Likewise, join c" with a", a" with e", and, finally, e" with K.
1508. Commercial application of the pictorial effects of illumination in perspective. A few general remarks, in cases where the shadow falls on itself or nearby objects, may not be amiss. The draftsman usually has some choice in the selection of the direction of the rays, and, sometimes, in the location of nearby objects.
Where the shadow is cast on the object itself or on neighboring objects, it will, in general, be found much easier to find the shadow, orthographically, and then to proceed with the making of a perspective from it. When the shadow is cast on a horizontal surface only, the general method outlined in Arts. 1503, 1504, and 1505 will find ready application.
The application of the pictorial effects of illumination in perspective in general, requires some consideration of the time required to make the drawings. The principles developed serve as a useful guide, so that the draftsman does not picture impos- sible shadows, even though the correct outline is not given. In fact, it is a difficult matter exactly to determine the assumed direction of the light from the picture itself. For artistic reasons, certain portions of an object are purposely subdued in order more strongly to emphasize some particular feature. The largest application of these principles lies in making illustrations for high-class catalogues, particularly for machinery catalogues. Figs. 272 and 273 give examples of this kind of work. It will be observed that the presented principles are ignored in many respects, yet, the effect is pleasing notwithstanding. After all, the theory indicates correct modes of procedure, but the time required to make such drawings is frequently prohibitive. Hence, common sense, based on mature judgment, must be used as a guide.
288 PICTORIAL EFFECTS OF ILLUMINATION
QUESTIONS ON CHAPTER XV
1. When rays of light are parallel, do their perspectives have a common
vanishing point on the picture plane? Why?
2. How is the vanishing point of the perspectives of a parallel system
of lines found?
3. What is a visual plane?
4. Show that the perspective of the intersection of a visual plane and
the horizontal plane vanishes on the horizon.
5. If the conventional direction of rays is used, show that the per-
spectives of their horizontal projections vanish in the right diagonal vanishing point.
6. State the general method of finding the perspective of a shadow
without first constructing it orthographically.
7. Show that the method of Question 6 is an application of finding
the perspectives of intersecting lines.
8. A rectangular prism rests on the horizontal plane. Find its shadow
on that plane by constructing it orthographically and then make a perspective of it.
9. Take the same prism of Question 8 and construct its horizontal
shadow directly.
10. An obelisk rests on the horizontal plane. Find its shadow on that
plane by constructing 'it orthographically and then make a per- spective of it.
11. Take the same obelisk of Question 10 and construct its horizontal
shadow directly.
Note. In the following problems construct the shade and shadow orthographically and then find its perspective.
12. A square-based pyramid rests on a square base. Construct the
shadow.
13. A square-based pyramid rests on a circular base. Construct the
shadow.
14. A square-based pyramid rests on a hexagonal base. Construct the
shadow.
15. A cone rests on a hexagonal base. Construct the shadow.
16. A rectangular prism rests on two square bases (stepped). Construct
the shadow.
17. A rectangular prism rests on two circular bases (stepped). Construct
the shadow.
18. A rectangular prism rests on two hexagonal bases (stepped). Con-
struct the shadow.
19. A cylinder has a superimposed square cap. Construct the shadow.
20. A cylinder has a superimposed circular cap. Construct the shadow.
21. A cylinder has a superimposed hexagonal cap. Construct the shadow.
22. A cylinder has a superimposed square cap and rests on a square
base. Construct the shadow.
IN PERSPECTIVE PROJECTION 289
23. A cylinder has a superimposed circular cap and rests on a circular
base. Construct the shadow.
24. A cylinder has a superimposed hexagonal cap and rests on a hexagonal
base. Construct the shadow.
25. A hollow semi-cylinder has a superimposed cap. Construct the
shadows on the inside of the cylinder and on the horizontal plane.
INDEX
A ART. NO
Aerial effect of illumination 1423
Aerial perspective 1307
Altitude of a cone " 1004
Altitude of a cylinder 1009
Angle between curves • 921
Angle between a given line and a given plane 831
Angle between a given plane and a principal plane 828
Angle between two given intersecting lines 826
Angle between two given intersecting planes 827
Angle, Line drawn through a given point and lying in a given plane
intersecting the first at a given 836
Angle of convergence of building lines (perspective) 1331
Angle of orthographic projection. Advantage of the third 315
Angle of orthographic projection. First 315
Angle, Through a given line in a given plane pass another plane intersect- ing it at a given 837
Angle, Visual 1304
Angles formed by the principal planes 502
Angles, Isometric projection of 406
Angles, Lines in all 514
Angles, Oblique projection of 207
Angles of orthographic projection 315
Angles, Points in aU 516
Angles projection to drawing. Application of 317
Angles, Traces of planes in all 609
Angles with the planes of projection. Through a given point to draw a
line of a given length and making given 834
Angles with the principal planes. Through a given point draw a plane
making given 835
Application of angles of projection to drawing . . . 317
Application of axonometric projection. Commercial 412
Application of drawing. Commercial 104
Application of oblique projection. Commercial 212
Application orthographic projection. Commercial 318
Application perspective projective. Commercial 1331
Approximate method of drawing an ellipse 906
Archimedian spiral 912
291
292 INDEX
ART. NO.
Arch, Perspective of 1329
Arc of a circle 905
Art of drawing 102
Asymptote to a hyperbola 908
Asymptotic surface 1030
Axes, Angular relation between dimetric 409
Axes, Angular relation between isometric 402
Axes, Angular relation between trimetric 410
Axes, Dimetric 409
Axes, Direction of dimetric 409
Axes, Direction of i&ometric 404
Axes, Direction of trimetric 410
Axes, Isometric 402
Axes, Trimetric 410
Axis, Conjugate, of a hyperbola 908
Axis, Major, of an ellipse 906
Axis, Minor, of an ellipse 906
Axis of a cone 1004
Axis of a cylinder 1009
Axis of a helix 915
Axis of a parabola 907
Axis, Principal, of a hyperbola (footnote) 908
Axis, Surfaces of revolution having a common 1025
Axis, Transverse, of a hyperbola 908
Axonometric drawing. Commercial application of 412
Axonometric drawing defined 411
Axonometric projection. Commercial application of 412
Axonometric projection defined 411
B
Base of a cone 1004
Base of a cylinder 1009
Bell-surface. Intersection of with a plane 1116
Bounding figures in isometric projection 408
Bounding figures in oblique projection 210
Bounding figures in perspective projection 1331
Branches of a hyperbola 908
Building, Perspective of 1330
C
Centre of a circle 905
Centre of a curvature 924
Chord of a circle 905
Circle defined 905
Circle, Graduation of the isometric 407
Circle, Involute of 930
INDEX 293
AKT.NO.
Circle, Isometric projection of 405
Circle, Oblique projection of 206
Circle, Osculating 924
Circle, Projection of, when it lie& in an oblique plane the diameter and
centre of which are known 838
Circular cone 1004
Circular cylinder 1009
Classification of lines 916
Classification of projections 413, 1332
Classification of surfaces 1031
Coincident projections, Lines with 515
Coincident projections, Points with 517
Commercial application of axonometric projection 412
Commercial application of drawing 104
Commercial application of oblique projection 212
Commercial application of orthographic projection 318
Commercial application of perspective projection 1331
Commercial application of pictorial effects of illumination 1508
Concave surfaces, Doubly 1022
Concave surfaces, Line shading applied to 1406
Concavo-convex surface 1022
Concentric circles 905
Concepts, Lines and points considered as mathematical 901, 1001
Concepts, Mathematical 501, 1001
Cone, Altitude of 1004
Cone and sphere, Intersection of 1220
Cone, Axis of 1004
Cone, Base of 1104
Cone, Circular 1004
Cone defined 1004
Cone, Development of an intersecting cylinder and 1219
Cone, Development of an oblique 1119
Cone, Development when intersected by a plane 1113
Cone, High-light on 1422
Cone, Intersection of a cylinder and 1216, 1217, 1218
Cone, Intersection of a right circular, and a plane 1112
Cone, Intersection of a sphere and 1220
Cone of revolution 1004
Cone, Representation of 1005
Cone, Right 1004
Cone, Right circular 1004
Cone, Slant height of 1004
Cone, To assume an element on the surface 1006
Cone, To assume a point on the surface 1007
Conical helix 915
Conical surfaces 1004
Conical surfaces, Application of 1113
•294 INDEX
ART. NO.
Conical surfaces, Line of intersection of 1211, 1212, 1213, 1214
Conical surfaces, Types of line of intersection of 1215
Conjugate axis of a hyperbola 908
Contact of tangents, Order of 923
Convergence of building lines. Angle of 1331
Convergent projecting lines 1302
Convex surface. Doubly 1022
Convex surface. Line shading applied to 1405
Convolute surface 1013
Craticulation 1331
Cube, Perspective of 1317t 1326
Cube, Shadow of 1414
Curvature, Centre of 924
Curvature, Radius of 924
Curve, Direction of 920
Curve,Plane 903
Curved lines, Doubly : 913
Curved lines, Line shading applied to 1403
Curved lines, Representation of doubly 914
Curved lines, Representation of singly 904
Curved lines, Singly 903
Curved surfaces, Development of doubly 1124
Curved surfaces, Development of doubly (gore method) 1125, 1127
Curved surfaces, Development of doubly (zone method) 1125
Curved surfaces, Doubly 1020
Curved surfaces, Intersection of doubly, with planes 1115
Curved surfaces of revolution, Doubly 1022
Curved surfaces of revolution, Intersection of two doubly 1222
Curved surfaces of revolution, Representation of doubly 1026
Curved surfaces of revolution, Singly 1021
Curved surfaces of revolution, To assume a point on a doubly 1027
Curved surfaces, Singly 1019
Curves, Angle between 921
Curves, Smooth 921
Cycloid, denned '. 909
Cylinder, defined 1009
Cylinder, Development of an intersecting cone and 1219
Cylinder, Development of an oblique 1120
Cylinder, Development when intersected by a plane 1108
Cylinder, High -light on 1422
Cylinder, Intersection of a cone and 1216, 1217, 1218
Cylinder, Intersection of a right circular, and a plane 1107
Cylinder, Intersection of a, with a sphere 1221
Cylinder, Representation of 1010
Cylinder, Shade and shadow on 1417
Cylinder, To assume an element on the surface 1011
Cylinder, To assume a point on the surface 1012
I INDEX 295
ART. NO.
Cylinders, Development of two intersecting 1205, 1207, 1210
Cylinders, Intersection of two 1204, 1206, 1209
Cylindrical helix 915
Cylindrical surfaces 1008
Cylindrical surfaces, Application of 1109, 1208
D
Developable surface 1028, 1104
Development by triangulation 1117
Development of a doubly curved surface by approximation 1124
Development of a doubly curved surface by the gore method 1127
Development of a cone intersected by a plane 1113
Development of a cylinder intersected by a plane 1108
Development of an intersecting cone and cylinder 1219
Development of an oblique cone 1119
Development of an oblique cylinder 1120
Development of an oblique pyramid 1118
Development of a prism intersected by a plane 1106
Development of a pyramid intersected by a plane . . 1112
Development of a sphere by the gore method 1121
Development of a sphere by the zone method 1 126
Development of a transition piece connecting a circle with a square.. . . 1125
Development of a transition piece connecting two ellipses. 1123
Development of surfaces 1103
Development of two intersecting cylinders 1205, 1207, 1210
Development of two intersecting prisms 1203
Diagonal vanishing points. Location of 1325
Diagonal when applied to perspective 1319
Diagrams, Transfer of, from orthographic to oblique projection 505, 608
Diameter of a circle 905
Dimensions on an orthographic projection 308
Dimetric axes, Angular relation between 409
Dimetric drawing 409
Dimetric projection « 409
Direction of a curve 920
Direction of light rays, Conventional 1409
Directrix of a cycloid 909
Directrix of an epycycloid 910
Directrix of an hypocycloid 911
Directrix of a parabola 907
Directrix of surface 1002, 1003, 1008
Directrix, Rectilinear 1003
Distance between a given plane and a plane parrallel to it 829
Distance between a given point and a given line 825
Distance between a given point and a given plane 824
Distance between two points in space 819, 820, 821, 822
296 INDEX
AKT. NO.
Distance between two skew lines 832
Distortion of oblique projection 211
Drawing an ellipse, Approximate method of 405
Drawing an ellipse, Exact method of 906
Drawing, Application of angles of projection to 317
Drawing, Application of the physical principles of light to 1412
Drawing, Art of '. 102
Drawing, Axonometric 411
Drawing, Commercial application of 104
Drawing, Commercial application of axonometric 412
Drawing, Dimetric 409
Drawing, Distinction between isometric projection and isometric 403
Drawing, Examples of isometric 408
Drawing, Nature of 101
Drawing of a line 502
Drawing, Scales used in making 209
Drawing, Science of 102
Drawing to scale 209
Drawing, Trimetric 410
Drawing, Use of bounding figures in isometric 408
Doubly concave surface 1022
Doubly convex surface 1022
Doubly curved line 913
Doubly curved line, Representation of 914
Doubly curved surface 1020
Doubly curved surface, Development by approximation 1124
Doubly curved surface, Development by the gore method 1127
Doubly curved surfaces of revolution 1022
Doubly curved surfaces of revolution, Intersection of 1222
Doubly curved surfaces of revolution, Intersection of by a plane 1115
Doubly curved surfaces of revolution, Representation of 1026
Doubly curved surfaces of revolution, To assume a point on a 1027
E
Eccentric circles 905
Eccentricity of circles 905
Element, Mathematical ». 501
Element of a surface 1002
Element on the surface of a cone, To assume 1006
Element on the surface of a cylinder, To assume 1011
Elevation, defined 301
Ellipse, Approximate method of drawing 405
Ellipse, as isometric projection of a circle 405
Ellipse as oblique projection of a circle 206
Ellipse denned 906
Ellipse, Extract method of drawing 906
INDEX 297
Epycycloid, defined 910
Evolute, defined 929
Examples of graduated shades 1426
F
Figures in isometric projection, Use of bounding 408
Figures in oblique projection, Use of bounding 210
Figures in perspective projection, Use of bounding 1331
Foci of an ellipse 906
Foci of an hyperbola 908
Focus of an ellipse 906
Focus of an hyperbola 908
Focus of a parabola 907
Frustum of a cone 1004
G
Generatrix of a surface 1002, 1003, 1008
Gore method, Development of a doubly curved surface 1127
Gore method, Development of a sphere 1125
Graduation shades, Examples of 1426
Graduation in shade 1424
Graphic representation of objects 101
Ground line, defined 302, 502
Ground line, Traces of a plane intersecting it 606
Ground line, Traces of a plane parallel to it 602
H
Helix, Conical 915
Helix defined 915
Helix, Uniform cylindrical 915
Height of a cone, Slant 1004
Helicoid (footnote) 1014
Helicoidal screw surface, Oblique 1014
Helicoidal screw surface, Right 1015
High-light 1418
High-light on a cylinder or a cone 1422
High-light on a sphere 1420
High-light, Multiple 1421
Horizon defined 1314
Horizontal lines inclined to the picture plane, Perspectives of parallel -. . . . 1313
Horizontal plane, Revolution of in orthographic projection 303
Horizontal plane, Revolution of in perspective projection 1322
Horizontal projection defined 302
Hyperbola, defined 908
Hypocycloid, defined 911
298 INDEX
Illumination, Aerial effect of 1423
Illumination, Commercial application of the pictorial effect of 1508
Incident rays 1419
Inclined lines, Isometric projection of 406
Inclined lines, Oblique projection of 207
Inclined planes, Traces of 605
Inflexion, Point of 926
Inflexional tangent 926
Interpenetration of solids 1215
Intersecting a given line at a given point, Line 804
Intersecting in space, Projection of lines 702
Intersecting lines, Angle between two given 826
Intersecting lines, Perspectives of 1318
Intersecting planes, Angle between two given 827
Intersecting the ground line, Traces of planes 606
Intersection of a bell-surface with a plane 1116
Intersection of a cone and cylinder 1216, 1217, 1218
Intersection of a cone and sphere 1 220
Intersection of a doubly curved surface of revolution and a plane 1115
Intersection of a given line at a given point and at a given angle 836
Intersection of a prism and a plane 1105
Intersection of a pyramid and a plane 1110
Intersection of a right circular cone and a plane 1112
Intersection of a right circular cylinder and a plane 1 107
Intersection of a sphere and a cylinder 1221
Intersection of conical surfaces 1211, 1212, 1213, 1214
Intersection of conical surfaces. Types of 1215
Intersection of lines 922
Intersection of two cylinders 1204, 1206, 1209
Intersection of two planes oblique to each other and to the principal
planes 809, 810
Intersection of two prisms 1202
Invisible lines, Representation of '. 208
Involute, denned 929
Involute of a circle 930
Isometric axes 402
Isometric axes, Angular relation 402
Isometric axes, Direction of 404
Isometric circle, graduation of 407
Isometric drawing. Distinction between isometric projection and 403
Isometric drawing, Examples of 408
Isometric drawing, Use of bounding figures in 408
Isometric projection and isometric drawing. Distinction between 403
Isometric projection considered as a special case of orthographic 402
Isometric projection. Examples of 408
INDEX 299
ART. NO.
Isometric projection. Nature of 401
Isometric axes of angles 406
Isometric projection circles 405
Isometric projection inclined lines 406
Isohietric projection, Theory of 402
Isometric projection. Use of bounding figures 408
L
Light, Application of the physical principles of, to drawing 1412
Light, High ' 1418
Light, High on a sphere 1420
Light, Multiple high 1421
Light, Perspective of parallel rays of 1504
Light, Physiological effect of 1408
Light rays, Conventional direction of 1409
Line, Angle between a given, and a given plane 831
Line, Convergent projecting 1302
Line, Distance between a given point and a given 825
Line, Doubly curved 913
Line, Drawing of 502
Line fixed in space by its projections 503
Line, Ground 302, 502
Line in a plane, pass another plane making a given angle 837
Line, Indefinite perspective of 1316
Line intersecting a given line at a given point 804
Line intersecting a given line at a given point and given angle 836
Line lying in the planes of projection 512
Line, Meridian 1024
Line, Oblique plane through a given oblique 806, 807
Line of a given length making given angles with planes of projection 834
Line on a given plane. Project a given 830
Line, Orthographic representation of 502, 504
Line perpendicular to a given plane through a given point 814, 815
Line perpendicular to planes of projection 513
Line, Perspective of 1309, 1321, 1324
Line piercing a given plane 823
Line, Piercing point of, on the principal planes 506
Line piercing the principal planes 805
Line, Plane through a given point, perpendicular to a given 816
Line, Plane through three given points 817
Line, Projecting plane of 502, 610
Line, Representation of doubly curved 914
Line, Revolution of a point about 707, 818
Line, Revolution of a skew 1023
Line shading applied to concave surfaces 1406
Line shading applied to convex surfaces 1405
300 INDEX
ART. NO.
Line shading applied to curved lines 1403
Line shading applied to plane surfaces 1407
Line shading applied to sections 1404
Line shading applied to straight lines % 1402
Line, Singly curved 903
Line, Singly curved, Representation of 904
Line, Straight 902
Lines, Straight, Representation of 904
Line through a given point parallel to a given line 803
Line, Traces of a plane intersecting the ground 606
Line, Traces of planes parallel to the ground 602
Lines, Angle between two given intersecting 826
Lines, Classification of 916
Lines considered as mathematical concepts 901
Lines, Distance between two skew 832
Lines in all angles, Projections of 514
Lines in oblique planes, Projection of 704
Lines of profile planes, Projection of 518
Lines intersecting in space, Projection of 702
Lines, Intersection of 922
Lines, Isometric projection of inclined 406
Lines, Line shading applied to curved 1403
Lines, Line shading applied to straight 1402
Lines, Non-intersecting in space, Projection of 703
Lines, Oblique projection of inclined 207
Lines, Oblique projection of parallel (footnote) 207
Lines, Oblique projecting 202
Lines parallel in space, Projection of 701
Lines parallel to both principal planes, Perspectives of 1311
Lines parallel to both principal planes, Projection of ....'. 511
Lines parallel to the plane of projection, Oblique projection of 202
Lines parallel to the principal planes and lying in an oblique plane .... 705
Lines perpendicular projecting 302
Lines perpendicular to given planes. Projection of 706
Lines perpendicular to the horizontal plane, Perspectives of 1310
Lines perpendicular to the plane of projection, Oblique projections of. . 204
Lines perpendicular to the vertical plane, Perspectives of 1312
Lines, Perspectives of intersecting 1318
Lines, Perspectives of systems of parallel 1313
Lines, Representation of invisible 208
Lines, Representation of visible 208
Lines, Shadows of 1413
Lines, Skew (footnote) 703
Lines, Systems of, in perspective 1313
Lines with coincident projections 515
Linear perspective 1303
Locus of a generating point 902
INDEX 301
M ART. NO.
Magnitude of objects 103
Mathematical concepts 501, 901, 1001
Mathematical elements 501
Mechanical representation of the principal planes 510
Meridian line 1024
Meridian plane 1024
Multiple high-light 1421
N
Nappes of a conical surface 1003
Nature of drawing 101
Nature of isometric projection 401
Nature of oblique projection 231
Nature of orthographic projection 301
Nomenclature of projections 507
Non-intersecting lines in space, Projection of 703
Normal defined 927
Normal plane 1018
O
Oblique and orthographic projections compared 309
Oblique cone 1004
Oblique cylinder 1009
Oblique helicoidal screw surface 1014
Oblique line, Oblique plane through a given 806, 807
Oblique plane of projection 202
Oblique plane, Projection of lines in an 704
Oblique plane, Projection of lines parallel to the principal planes and
lying in an 705
Oblique plane through a given oblique line 806, 807
Oblique plane through a given point 808
Oblique planes, Intersections of 809, 810
Oblique projection, Commercial application of 212
Oblique projection considered as a shadow 203
Oblique projection, Distortion of 211
Oblique projection, Examples of 210
Oblique projection, Location of eye in constructing 202, 211
Oblique projection, Location of object from plane of projection 202
Oblique projection, Nature of 201
Oblique projection of angles 207
Oblique projection of circles 206
Oblique projection of inclined lines 207
Oblique projection of lines parallel to the plane of projection 202
Oblique projection of lines perpendicular to the plane of projection .... 204 Oblique projection of parallel lines (footnote) 207
302 INDEX
ART. NO.
Oblique projection, Theory of 202, 203, 204, 205
Oblique projection, Transfer of diagrams from orthographic 505, 608
Oblique projection, Use of bounding figures 210
Oblique projecting lines 202
Objects, Graphic representation of 101
Objects, Magnitude of 103
Order of contact of tangents 923
Orthographic and oblique projections compared 309
Orthographic planes of projection 302
Orthographic projection, advantages of third angle 315
Orthographic projection, Angles of 315
Orthographic projection considered as a shadow 310
Orthographic projection, Commercial application of 318
Orthographic projection, Dimension on 308
Orthographic projection, First angle of 315
Orthographic projection, Location of eye in constructing 304, 316
Orthographic projection, Location of object with respect to the planes
of projections 306
Orthographic projection, Nature of . . 301
Orthographic projection, Size of object and its projection 305
Orthographic projection, Theory of 302
Orthographic projection, Third angle of 315
Orthographic projection, Transfer of diagrams from oblique 505, 608
Orthographic projections, Simultaneous interpretation of 308, 318
Orthographic projections, with respect to each other, Location of 307
Orthographic representation of lines 502, 504
Orthographic representation of points 508
Osculating circle 924
Osculating plane 925
P
Pantograph, Use of, in perspective 1331
Parabola, defined 907
Parallel lines, Oblique projection of (footnote) 207
Parallel lines, Orthographic projection of 701
Parallel lines to both principal planes, Perspectives of 1311
Parallel lines to planes of projection, Oblique projection of 202
Parallel lines to planes of projection, Orthographic projection of 511
Parallel plane at a given distance from a given plane 829
Parallel to a given plane, Plane which contains a given point and is. . . 813
Parallel rays of light, Perspectives of 1504
Parallel systems of lines, Perspectives of 1313
Parallel to a given line, Line through a given point 803
Parallel to the ground line. Traces of planes 602
Parallel to the principal planes. Traces of planes 601
Penumbra. . 1411
INDEX 303
/
ART. NO.
Perpendicular (when applied to perspective) 1319
Perpendicular line through a given point to a given plane 814, 815
Perpendicular lines to planes of projection, Projection of 513
Perpendicular lines to the horizontal plane, Perspective of 1310
Perpendicular plane to a given line through a given point 816
Perpendicular projecting lines 302
Perpendicular to both principal planes. Traces of planes 604
Perpendicular to given planes. Projection of lines 706
Perpendicular to one of the principal planes. Traces of planes 603
Perpendicular to the plane of projection. Oblique projection of lines. . 204
Perspective, Aerial 1307
Perspective and shadow of a prism 1502
Perspective, Commercial application of 1331
Perspective, Linear 1303
Perspective of a building 1330
Perspective of a cube 1317, 1326
Perspective of a hexagonal prism 1327
Perspective of a line 1309, 1321, 1324
Perspective of a line, Indefinite 1316
Perspective of a line parallel to both principal planes 1311
Perspective of a line perpendicular to the horizontal plane 1310
Perspective of a line perpendicular to the vertical plane 1312
Perspective of a point 1315, 1320, 1323
Perspective of a pyramid 1328
Perspective of a shadow, Application of the general method 1506
Perspective of a shadow, General method of finding 1503
Perspective of an arch 1329
Perspective of an obelisk with its shade and shadow 1507
Perspective of intersecting lines 1318
Perspective of parallel rays of light 1504
Perspective of parallel systems of lines 1313
Perspective of the horizontal intersection of the visual plane 1505
Perspective projection, Theory of 1306
Perspective sketches 1331
Picture, Center of 1312
Picture plane 1303, 1308
Picture plane, Location of 1308
Pictures 101
Pictorial effects of illumination, Commercial application of 1508
Physiological effect of light 1408
Plan, defined .' 301
Plane, Angle between a given line and a given 831
Plane, Angle between a given plane and a principal 828
Plane containing a circle of a known diameter and center, Projection
of 838
Plane'containing a line intersected by another at a given point and
angle 836
304 INDEX
ART. NO.
Plane, Corresponding projection of a given point when in a given. . . 811, 812
Plane curve 903
Plane, Distance between a given point and a given 824
Plane fixed in space by its traces 607
Plane, Location of picture 1308
Plane making given angles with the planes of projection 835
Plane, Meridian 1024
Plane, Normal 1018
Plane of projection, Horizontal 302, 502
Plane of projection, Oblique 202
Plane of projection, Oblique projection of lines parallel to 202
Plane of projection, Oblique projection of lines perpendicular to 204
Plane of projection, Vertical 302, 502
Plane, Osculating 925
Plane parallel to a given plane at a given distance from it 829
Plane passed through a line in a plane making a given angle 837
Plane, Perpendicular line through a given point to a given 814, 815
Plane, Perspective of the horizontal intersection of the visual 1505
Plane, Picture 1303, 1308
Plane, Piercing point of a line on the principal 506, 805
Plane, Projecting, a given line on a given 830
Plane, Projecting, of a line 502, 610
Plane, Revolution of the horizontal (in orthographic) 303
Plane, Revolution of the horizontal (in perspective) 1322
Plane surface 1002
Plane surfaces, Line shading applied to 1407
Plane, Tangent 1017
Plane through a given oblique line, Oblique 806, 807
Plane through a given point and perpendicular to a given line 816
Plane through a given point, Oblique 808
Plane through three given points , 817
Plane which contains a given point and is parallel to a given plane 813
Plane, Visual 1310
Planes, Angles between two given intersecting 827
Plants, Angles formed by principal 502
Planes in all angles, Traces of 609
Planes inclined to both principal planes, Traces of 605
Planes intersecting the ground line, Traces of 606
Planes, Intersection of two planes oblique to each other 809, 810
Planes, Lines in profile 518
Planes, Line piercing principal 506, 805
Planes, Mechanical representation of the principal 510
Planes of projection, Line drawn through a given point, length, and
angles with 834
Planes of projection, Lines lying in . 512
Planes of projection, Location of object with respect to 306
Planes of projection, Orthographic 302, 502
INDEX 305
ART. NO.
Planes of projection, Parallel lines to 511
Planes of projection, Plane drawn, making given angles with 835
Planes of projection, Principal 302, 502
Planes parallel to the principal planes, Traces of .• 601
Planes parallel to the ground line, Traces of 602
Planes perpendicular to both principal planes, Traces of 604
Planes perpendicular to one of the principal planes, Traces of 603
Planes, Points lying in the principal, 509
Planes, Principal 502
Planes, Profile 311
Planes, Projection of lines in oblique 704
Planes, Projection of lines perpendicular to given 706
Planes, Section 313
Planes, Supplementary 314
Point, considered as mathematical concept 901
Point, Corresponding projection when one is given 811, 812
Point, Distance between it and a given line 825
Point, Distance between it and a given plane 829
Point, Distance between two, in space 819, 820, 821, 822
Point, Generating 902
Point in all angles 516
Point, Line intersecting another at a given 804
Point, Line perpendicular to a given plane through a given 814, 815
Point, Line through a given, parallel to a given line 803
Point, Location of diagonal vanishing 1325
Point lying in the principal planes 509
Point, Oblique plane through a given 808
Point of inflexion 926
Point of sight, Choice of 1331
Point of tangency, To find 919
Point on a doubly curved surface of revolution. To assume 1027
Point on a surface of a cone, To assume 1007
Point on a surface of a cylinder, To assume 1012
Point, Orthographic representation of 508
Point, Perspective of 1315, 1320, 1323
Point, Piercing, of a given line on a given plane 823
Point, Piercing of a given line on the principal planes 506
Point, Plane through three given 817
Point, Plane through a given, perpendicular to a given line 816
Point, Plane which contains a given, and is parallel to a given plane . . . 813
Point, Revolution of, about a line 707, 818
Point, Through a given, draw a line of given length and angles 834
Point, Through a given, draw a plane making given angles 835
Point, Vanishing 1305
Point with coincident projections 517
Principal axis of a hyperbola (footnote) 908
Principal planes of projection 302, 502
306 INDEX
ART. NO.
Principal planes, Angles formed by 502
Principal planes, Angle between a given plane and 828
Principal planes, Intersection of two planes oblique to each other and to
the 809, 810
Principal planes, Line piercing 805
Principal planes. Mechanical representation of 510
Principal planes, Piercing point of line on 506
Principal planes, Points lying in 509
Principal planes, Traces of planes inclined to both 605
Principal planes, Traces of planes parallel to 601
Principal planes, Traces of planes perpendicular to both 604
Principal planes, Traces of planes perpendicular to one of the 603
Prism, Developments of two intersecting 1203
Prism, Developments when intersected by a plane: 1106
Prism, Intersection of two 1202
Prism, Intersection of with a plane 1105
Prism, Perspective of 1327
Prism, Perspective and shadow of • . . . 1502
Prism, Shadow of 1416
Profile planes 311
Profile planes, Lines in 518
Profile projections. Location of 312
Project a given line on a given plane 830
Projecting line, Convergent 1302
Projecting line, Oblique 202
Projecting line, Parallel 302, 304
Projecting plane of a line •„ 502, 610
Projection, axonometric 411
Projection, axonometric, Commercial application of 412
Projection, Classification of 413, 1332
Projection, Dimetric .x 409
Projection, Isometric, and isometric drawing compared 403
Projection, Isometric considered as special case of orthographic 402
Projection, Isometric, Examples of 408
Projection, Isometric, Nature of 401
Projection, Isometric of angles 406
Projection, Isometric of circles 405
Projection, Isometric of inclined lines 406
Projection, Isometric, Theory of 402
Projection, Isometric, Use of bounding figures 408
Projection, Oblique, Commercial application of 212
Projection, Oblique considered as shadow 203
Projection, Oblique, Distortion of 211
Projection, Oblique, Examples of 210
Projection, Oblique, Location of object from plane 202
Projection, Oblique, Nature of 201
Projection, Oblique of angles 207
INDEX 307
ART. NO.
Projection, Oblique of circles 206
Projection, Oblique of inclined lines 207
Projection, Oblique of lines parallel to plane 202
Projection, Oblique of lines perpendicular to plane 204
Projection, Oblique of parallel lines in space (footnote) 207
Projection, Oblique plane of 202
Projection, Oblique position of eye in constructing 202, 211
Projection, Oblique, Theory of 202, 203, 204, 205
Projection, Oblique, Use of bounding figures 210
Projection, Orthographic, angles of 315
Projection, Orthographic, Application of angles 317
Projection, Orthographic, Commercial application 318
Projection, Orthographic, Compared with oblique 309
Projection, Orthographic, Considered as shadow 310
Projection, Orthographic, Dimensions on 308
Projection, Orthographic, First angle 315
Projection, Orthographic, Horizontal plane of 302, 502
Projection, Orthographic, Line fixed in space by 503
Projection, Orthographic, Location of object to planes 306
Projection, Orthographic, Location of observer while constructing. . 304, 316
Projection, Orthographic, Location of profiles 312
Projection, Orthographic, Location of projections with respect to each
other 307
Projection, Orthographic, Nature of 301
Projection, Orthographic, Nomenclature of 507
Projection, Orthographic, of a circle lying in an oblique plane 838
Projection, Orthographic, of a point lying in a plane when one projec- tion is given 811, 812
Projection, Orthographic of lines in oblique planes 704
Projection, Orthographic of lines intersecting in space 702
Projection, Orthographic of lines lying in the planes of 512
Projection, Orthographic of lines non-intersecting in space 703
Projection, Orthographic of lines parallel in space 701
Projection, Orthographic of lines parallel to planes of 511
Projection, Orthographic of lines parallel to one plane and in an
oblique plane , 705
Projection, Orthographic of lines perpendicular to given planes 706
Projection, Orthographic of lines perpendicular to planes of 513
Projection, Orthographic of lines with coincident 515
Projection, Orthographic of points with coincident 517
Projection, Orthographic, Perpendicular projecting lines 302
Projection, Orthographic, Plane of 302
Projection, Orthographic, Position of eye in constructing 302, 334, 316
Projection, Orthographic, Principal planes of 302
Projection, Orthographic, Size of object and its projection 305
Projection, Orthographic, Simultaneous interpretation 308, 318
Projection, Orthographic, Theory of 302
308 INDEX
AHT. NO.
Projection, Orthographic, Third angle of projection 315
Projection, Orthographic, Transfer of diagrams from oblique to .... 505, 608
Projection, Orthographic, Vertical plane of 302, 502
Projection, Perspective, (see topics under perspective)
Projection, Scenographic 1302
Projection, Theory of perspective 1306
Projection, Trimetric 410
Pyramid, Development of an oblique 1118
Pyramid, Development when intersected by a plane 1111
Pyramid, Intersection of with a plane 1110
Pyramid, Perspective of 1328
Pyramid, Shadow of 1415
Q
Quadrant of a circle 905
R
Radius of a circle 905
Rays, Conventional direction of light 1409
Rays, Incident 1419
Rays of light, Perspective of parallel 1504
Rays, Reflected 1419
Rays, Visual . 1304
Rectification, defined 928
Reflected rays 1419
Representation of cones 1005
Representation of cylinders 1010
Representation of doubly curved surfaces of revolution 1026
Representation of invisible lines 208
Representation of lines, Orthographic 502, 504
Representation of objects, Graphic 101
Representation of points, Graphic 508
Representation of singly curved lines 904
Representation of straight lines 904
Representation of the principal planes, Mechanical 510
Representation of visible lines 208
Revolution, Doubly curved surface of 1022
Revolution, Cone of 1004
Revolution, Cylinder of 1009
Revolution, of a point about a line 707, 818
Revollution of a skew line 1023
Revo ution of the horizontal plane, (orthographic) 303
Revolution of the horizontal plane, (perspective) 1322
Revolution, Representation of doubly curved surfaces of 1026
Revolution, Singly curved surfaces of 1021
Revolution, Surfaces of, having a common axis 1025
INDEX 309
ART. NO.
Revolution, To assume a point on a doubly curved surface of 1027
Right cone 1004
Right cylinder 1009
Right helicoidal screw surface 1015
Ruled surface 1029
Rules for shading 1425
Scale, Choice of 209
Scale, Drawing to 209
Scales used in making drawings 209
Scenographic projection 1302
Science of drawing 102
Screw surface, Oblique helicoidal 1014
Screw surface, Right helicoidal 1015
Secant of a circle 905
Section plane 313
Sections, Line shading applied to 1404
Sector of a circle 905
Segment of a circle 905
Semicircle 905
Shade, defined 1410
Shade and shadow of an obelisk in perspective 1507
Shade and shadow on a cylinder 1417
Shade, Graduation in 1424
Shades, Examples of graduated 1426
Shading applied to concave surfaces, Line 1406
Shading applied to convex surfaces, Line 1405
Shading applied to curved lines, Line 1403
Shading applied to plane surfaces, Line 1407
Shading applied to sections, Line 1404
Shading applied to straight lines, Line 1402
Shading rules , 1425
Shadow defined 1410
Shadow, Application of general method of finding its perspective ...... 1506
Shadow and perspective of a prism 1502
Shadow and shade of an obelisk in perspective 1507
Shadow and shade on a cylinder 1417
Shadow, General method of finding its perspective 1503
Shadow, Oblique projection considered as 203
Shadow of a cube 1414
Shadow of a line 1413
Shadow of a prism 1416
Shadow of a pyramid «. 1415
Shadow, Orthographic projection considered as 310
Si^ht, Choice of point of 1331
Singly curved line, defined 903
310 INDEX
ART. NO.
Singly curved line, Representation of 904
Singly curved surface defined 1019
Singly curved surface of revolution 1021
Sketches, Perspective .' 1331
Skew lines, Distance between two 832
Skew lines (footnote) ' 703
Skew lines, Revolution of 1023
Slant height of a cone .' 1004
Solids, Interpenetration of 1202, 1215
Spiral defined 912
Spiral, Archimedian 912
Sphere and cone, Intersection of 1220
Sphere and cylinder, Intersection of 1221
Sphere developed by the gore method 1125
Sphere developed by the zone method 1126
Sphere, High-light on 1420
Straight lines defined 902
Straight lines, Representation of 904
Straight lines, Line shading applied to 1402
Supplementary plane 314
Surface, Asymptotic 1030
Surface, Bell, Intersection of, with a plane 1116
Surface, Concavo- convex 1022
Surface, Conical 1003
Surface, Convolute 1013
Surface, Cylindrical 1008
Surface, Developable 1028, 1103, 1104
Surface, Doubly concave 1022
Surface, Doubly convex 1022
Surface, Doubly curved 1020
Surface, Doubly curved, Developments by approximation 1124
Surface, Doubly curved, Developments by the gore method 1127
Surface, Doubly curved, Intersection of, with a plane 1115
Surface, Oblique helicoidal screw 1014
Surface, Plane, defined 1002
Surface, Right helicoidal screw 1015
Surface, Ruled 1029
Surface, Singly curved 1019
Surface, Warped 1016
Surface of a cone, To assume a point on the 1007
Surface of a cone, To assume an element on the 1006
Surface of a cylinder, To assume a point on the 1012
Surface of a cylinder, To assume an element on the 1011
Surface of revolution, Doubly curved / 1022
Surface of revolution, Singly curved 1021
Surface of revolution, To asusme a point on a doubly curved 1027
Surfaces, Application of conical 1114, 1211
INDEX 311
Surfaces, Application of cylindrical 1109, 1208
Surfaces, Classification of 1031
Surfaces, Development of doubly curved 1124
Surfaces, Gore method of developing 1124
Surfaces, Line of intersection of conical 1211, 1212, 1213, 1214
Surfaces, Line shading applied to concave 1406
Surfaces, Line shading applied to convex 1405
Surfaces, Line shading applied to plane 1407
Surfaces of revolution having a common axis 1025
Surfaces of revolution, Intersection of two doubly 1222
Surfaces of revolution, Representation of doubly curved 1026
Surfaces, Types of intersection of conical 1215
Surfaces, Zone method of developing ' 1124
System of lines (in perspective) 1313
T
Tangent, defined 917
Tangent, Construction of ' 918
Tangent, Inflexional 926
Tangent, Order of contact of 923
Tangent plane 1017
Tangency, To find point of 919
Theory of isometric projection 402
Theory of oblique projection 202, 203, 204, 205
Theory of orthographic projection 302
Theory of perspective projection 1306
Theory of shades and shadows 1412
Torus 1022
Torus, Development of 1109
Traces of planes in all angles 609
Traces of planes inclined to both principal planes 605
Traces of planes intersecting the ground line 606
Traces of planes parallel to the ground line t 602
Traces of planes parallel to the principal planes 601
Traces of planes perpendicular to both principal planes 604
Traces of planes perpendicular to one of the principal planes 603
Traces, Plane fixed in space by 607
Trammel method of drawing an ellipse 906
Transition piece, defined 1121
Transition piece connecting a circle with a square, Development of. ... 1122
Transition piece connecting two ellipses, Development of 1123
Transverse axis of a hyperbola 908
Triangulation, Development by 1117
Trimetric axes, angular relation 410
Trimetric drawing 410
Trimetric projection 410
312 INDEX
ART. NO.
Truncated cone
Truncated cylinder
U
Umbra, defined ................................
V
Vanishing points, defined .................
Vanishing points, Location of diagonal ............. ] 325
Vertex of a conical surface ......................
Vertex of a hyperbola ........................
Vertex of a parabola .....................
Vertical plane of projection ........................ 302 502
Vertical projection ............................... ' gQr>
Visible lines, Representation of ............................ 208
Visual angle ............................................. ....... 1304
Visual plane ............................. ^Q
Visual plane, Perspective of horizontal intersection of the ............ 1505
Visual ray ................ .' ................................. " 1304
W Warped surface ................................................ 1016
Z
Zone method of developing a sphere ............................... 1125
Zone method of developing surfaces ........................... 1124
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SHORT=TITLE CATALOG
OF
|f ttkltottimts
OF
SCIENTIFIC AND ENGINEERING
BOOKS
This list includes the technical publications of the following English publishers:
SCOTT, GREENWOOD & CO. CROSBY LOCKWOOD & SON
CONSTABLE & COMPANY, Ltd. TECHNICAL PUBLISHING CO.
ELECTRICIAN PRINTING & PUBLISHING CO.
for whom D. Van Nostrand Company are American agents.
AUGUST, 1912
SHORT-TITLE CATALOG
OF THE
Publications and Importations
OF
D. VAN NOSTRAND COMPANY
25 PARK PLAGE, N. Y.
Prices marked with an asterisk (*) are NET. All bindings are in cloth unless otherwise noted.
ABC Code. (See Clausen-Thue.) Ai Code. (See Clausen-Thue.)
Abbott, A. V. The Electrical Transmission of Energy 8vo, *$5 oo
A Treatise on Fuel. (Science Series No. 9.) i6mo, o 50
Testing Machines. (Science Series No. 74.) i6mo, o 50
Adam, P. Practical Bookbinding. Trans, by T. E. Maw i2mo, *2 50
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