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I THE THEORY

ENGINEERING DEAWING

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

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

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

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

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

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

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

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.

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

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

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

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

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

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

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

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).

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

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?

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

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.

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

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.

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

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

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

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

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.

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-

i i
^N

1
1
1
1

— 1

I i

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.

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

-1 —

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

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?

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

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?

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.

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

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.

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

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.

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

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

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.

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

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.

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.

Inclination of

projecting

lines 45°

Oblique. Pro-

with plane of

•

jecting lines

projection.

inclined to

plane of pro-

Inclination of

jection but

projecting

parallel to

lines greater

each other.

than 45°

with plane of

projection.

Mechanical

' On Plane

Drawing.

Surfaces:

showing two

principal di-

mensions on .

a single view.

Hence, at

least two

views needed

Orthographic.

'•

Projecting

lines perpen-

dicularto

the plane of

projection.

PROJEC-
TIONS

Axonometric.

projection,

showing

three dimen-

sions in a

single view.

On r

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

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

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.

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

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.

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

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

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.

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

(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

2 ^«

X c 1 n 4

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

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

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.

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.

1«

b',

d'l

2 3

,, ..

„'

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?

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.

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.

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

REPRESENTATION OF PLANES

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

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

IT'

FIG. 72.

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

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

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

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~ ^^ \.* * / *

/ 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

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',

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

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

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

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.

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

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

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

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

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

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

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.

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

1 <4 a',

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

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

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

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,

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.

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

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

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

A Horizon

\—~ -^_^

4\0'

\ \ i

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.

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

Commonly

called ortho-

Showing two

graphic pro-

dimensions

jections or

Mechanical

drawings.

Orthographic •

Showing three
dimensions

Isometric

' Parallel
projecting

and known
as axonome-

Dimetric

lines

tric projec-

Trimetric

Projections
on plane "

Oblique

tions.

surfaces

Convergent
projecting .

\ Outline alone is shown
perspective I

lines

Aerial ( Color and illumination added

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

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

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

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

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

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

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

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

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

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

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

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

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

Quadrant of a circle 905

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

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

Umbra, defined ................................

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

Zone method of developing a sphere ............................... 1125

Zone method of developing surfaces ........................... 1124

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SHORT=TITLE CATALOG

|f ttkltottimts

SCIENTIFIC AND ENGINEERING

BOOKS

This list includes the technical publications of the following
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AUGUST, 1912

SHORT-TITLE CATALOG

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ABC Code. (See Clausen-Thue.)
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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

Adams, H. Theory and Practice in Designing 8vo, *2 50

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Adams, J. W. Sewers and Drains for Populous Districts 8vo, 2 50

Addyman, F. T. Practical X-Ray Work 8vo, *4 oo

Adler, A. A. Principles of Projecting-line Drawing 8vo, *i oo

Theory of Engineering Drawing 8vo, (In Press.}

Aikman, C. M. Manures and the Principles of Manuring 8vo, 2 50

Aitken, W. Manual of the Telephone 8vo, *8 oo

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Alexander, J. H. Elementary Electrical Engineering. . i2mo, 2 oo

Universal Dictionary of Weights and Measures 8vo, 3 50

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Allen, H. Modern Power Gas Producer Practice and Applications.. i2mo, *2 50

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Anderson, Capt. G. L. Handbook for the Use of Electricians 8vo, 3 oo

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