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NOTE-BOOK

PRACTICAL SOLID

DESCRIPTIVE GEOMETRY

NOTE-BOOK

PRACTICAL SOLID

DESCRIPTIVE GEOMETRY

CONTAINING

PROBLEMS WITH HELP FOR SOLUTIONS.

J. H. EDGAR, M.A,

LECTURER ON MECHANICAL DRAWING AT THE ROYAL SCHOOL OF MINES, LONDON ;
/ ND FORMERLY LECTURER ON GEOMETRICAL DRAWING, KING's COLLEGE, LONDON ;

AND

G. S. PRITCHARD,

1 ATE PROFESSOR OF DESCRIPTIVE GEOMETRY, ROYAL MILITARY ACADEMY, WOOLWICH.

FOURTH EDITION,
BY

ARTHUR G. MEEZE,

." 3SISTANT LECTURER ON MECHANICAL DRAWING, ROYAL SCHOOL OF MINES, LONDON.

Hontron :
MACMILLAN AND CO.

1880.

\_The Right of Translation is reserved,']

Camftrftge :

PRINTED BY C. J. CLAY, M.A.
AT THE UNIVERSITY PRESS.

PREFACE.

In teaching a large class, if the method of lecturing and
demonstrating from the black-board only is pursued, the I
more intelligent students have generally to be kept back,
from the necessity of frequent repetition, for the sake of 1/
the less promising ; if the plan of setting problems to each
pupil is adopted, the teacher finds a difficulty in giving
to each sufficient attention. A judicious combination of
both methods is doubtless the best, though this is not
always easy of attainment in working with numbers — the
use of this book may help in accomplishing it.

It is suggested that at the beginning of a chapter, and in
some cases with each problem, the teacher should give a
black-board explanation, carefully pointing out any fresh
steps, before sending his pupils to their work. The number ''"
of examples in each chapter to be worked out by the
student is, of course, left to the teacher's judgment of the
progress and requirements of his pupil.

The student will here be assisted, it is beheved, because
he will find help givenVhen necessary, and thus will often
be able to go on with his work by himself, with the satisfac-
tion of also feeling that he is progressing systematically.
Moreover he will become practised in dealing with written
questions — a point not to be lost sight of by those who are
preparing for Examinations,

The Second Edition has been enlarged by the addition
of chapters on the straight line and plane with explanatory
diagrams and exercises, on tangent-planes, and on the cases
of the spherical triangle.

E. G. b

vi PREFACE.

It is hoped that the work, thus rendered more complete,
i^may prove still more useful as a class-book and means of
V self -instruction to the various and constantly increasing
classes of students for whom it is designed. It was originally
intended as an aid in teaching the Mechanical Drawing
Class at the Royal School of Mines from Professor Bradley's
Elements of Practical Geometry. The authors of this work
were associated with him in his duties at King's College,
London, and the Royal Military Academy, and learnt
practically the value of his treatise; but the cost of that
work has rendered it inaccessible to many for whom the
present book may be available.

A greater number of diagrams have not been added, in
order that students may be thrown upon their own resources,/
and encouraged to consider the principles upon which their
work proceeds, more than they would probably do if there
were figures always at hand for reference.

October, 1871.

ADVERTISEMENT TO THE FOURTH EDmON.

A STEADY demand for the "Note-Book" having shown
that the principles on which it was designed met with the
recognition of teachers and students, the successive editions
have been enlarged to satisfy the requirements of those
competing in the various public examinations, as well as to
adapt it to the gradual development of the class at the
Royal School of Mines.

In preparing the Fourth Edition the original scope of the
book has been considerably extended. The aim being to
make it, without prejudice to the elementary character of
its earlier portions, an exhaustive vade-mecum for the work- ^
ing student. Attention may, therefore, be called to the
following points in the treatment of the present edition
which have consequently become more prominent : —

1. While retaining its original form of a classified
collection of "Problems with Help for Solutions," the book
has been placed upon a more independent and scientific

'• footing ; by adding the necessary definitions and theorems,
and introducing a large amount of matter of a general
character, not elsewhere accessible, in a cheap and compendi-
ous form, to the majority of English students.

2. A large number of carefully selected problems have
been added to illustrate the leading principles and afford

^ specific exercises for self-examination, while in writing the
solutions the constructions have been studiously varied so
as to bring together as many as possible of the methods
likely to be of practical utility to the scientific draughts- V
man.

viii AD VER TI SEME NT TO FO UR TH EDITION.

3. Diagrams have been purposely eschewed and a
systematic attempt made to reduce the whole subject, as
far as possible, to a purely verbal form — sufficient description
of the various constructions being given to enable a student
who works steadily through the earlier chapters to build up,

h under his own hand and eye, illustrative drawings for
himself. The verbal formulae, besides being much less
fatiguing to peruse than the plates usually accompan3dng
books on this subject, soon become immeasurably more
powerful and suggestive helps, and leave a wider field for
the exercise of the learner's ingenuity. Moreover, the style

\J of treatment adopted' has rendered it possible to compress
into a very small compass a quantity of matter which would
otherwise have augmented the bulk and cost of the book so
as to have placed it out of the reach of many whose wants
it is intended to supply.

4. Some of the more advanced problems have been
dealt with in general terms and very briefly, experience

1/ having shown that when a student's own thought or theoreti-
cal studies have prepared him for the necessary geometrical
conceptions, these higher developments are really the most
easy, and a hint is almost all that is required as a starting-
point for fresh knowledge — which is not so much learned as

V self-evolved when once the mind is set upon the proper
track.

June, 1880.

CONTENTS.

PAGE

Introductory Explanations . . . . • • ^

Definitions and Theorems 3

Definitions Illustrated [Notation) 12

Orthographic Illustrations [Points, Lines, and Planes) . 16

Memoranda for Working the Problems .... 20

CHAPTER I.
Solids in Simple Positions 2t

CHAPTER II.
Combinations and Groups of Solids 38

CHAPTER III.

Problems on the Straight Line and Plane ... 42
Exercises ...... ... . .86

CHAPTER IV.

Solids with the Inclinations of the Plane of One

Face and of One Edge or Line in that Face given . 92

CHAPTER V.

Solids with the Inclinations of Two Adjacent Edges

or Lines given 97

CHAPTER VI.

Solids with the Inclinations of Two Adjacent Faces

GIVEN . . 99

X CONTENTS.

CHAPTER VII.

PAGE

Properties of Curved Surfaces and Problems Based

THEREON 103

Problems on the Projection of Curved Surfaces: — 105

The Sphere . . . . . . . . . • • 05

The Cylinder . . . . . . . . .112

The Cone . . . . . . . . . .119

Tangent-Planes and Normals to Curved Surfaces : — 125

General Remarks . . 125

Tangent- Planes to Cones . . . . . . -127

Tangent- Planes to Cylinders . . . . . • '3^

Tangent- Planes to Spheres . . . . . . • ' 37

Problems on Surfaces of Revolution .... 144

Tangent-Planes and Normals to Surfaces of Revolu-
tion . . 149

The Hyperboloid of Revolution of One Sheet . .150

Undevelopable Ruled Surfaces generally . . -154

CHAPTER VIII.

Sections by Oblique Planes 158

Preliminary . .158

General Pemarhs 160

Problems 161

CONTENTS.

CHAPTER IX.

PAGE

Intersections 173

General Remarks , . . . . . . . • ^73

Forms Bounded by Plane Surfaces . . . . -174

Forms Bounded by Curved Surfaces 182

Combination of Curved Forms zvith Forms Bounded by

Plane Surfaces 198

Shadows .......... 201

CHAPTER X.
Figured Projections and Scales of Slope . . . 208

CHAPTER XI.

Trimetric Methods of Projection 215

Trimelric Projection . . ■ . . . . .216

Isometric Projection .220

Axial Projection . . . . . . . . .221

Trihedral Projection . . . . . . . .224

CHAPTER XII.
Miscellaneous Problems . 228

CHAPTER XIII.

Solutions of the Trihedral Angle {^'Spherical Triangles''^) 250

ELEMENTARY
SOLID OR DESCRIPTIVE GEOMETRY.

INTRODUCTORY EXPLANATIONS.

"The object of Descriptive Geometry is the invention of methods
by which w^e may represent upon a plane having only two dimensions,
namely length and breadth, the form and position of a body which pos-
sesses three dimensions, namely length, breadth, and height."

Hall's Elements of Descriptive Geometry.

For this purpose two planes, called the co-ordinate planes,'/
are conceived at right angles to one another, intersecting in
a line called the ground line, xy, and named from their usual
positions the horizontal zxid. vertical planes of projection. See
Fig. I.

Drawings or projections on the horizontal plane are
called plans ; on the vertical plane elevations.

The plan or horizontal projection of any point A in space,
is the foot of the perpendicular let fall from point A to the
E. G. -| t 1

2 SOLID OR DESCRIPTIVE GEOMETRY.

horizontal plane, and is marked a; the elevation of A is
marked a\ and is the point of intersection of the perpendi-
cular from A to the vertical plane. See Fig. i.

Fig. I.

The projection of a line may be defined to be the sum
of the projections of its points.

Forms possessing three dimensions, length, breadth, and
height, require to be represented or projected upon both
planes of projection, so that ^'height," as well as "length"
and "breadth," may be exhibited or determined by the
drawings.

But as we can have for drawing only one plane, that is,
one flat sheet of paper, not two at right angles to each other,
the vertical plane is supposed to rotate backwards upon xy,

DEFINITIONS AND THEOREMS. 3

its intersection with the horizontal plane, as upon a hinge,
• until it coincides and forms one plane with the horizontal.
The plan a and elevation c^ of any point A will then lie in
the same perpendiadar to the ground line.

This perpendicular, called the " projecting line " of the
point A, will thus form a locus of points a and a', that is to
say, must contain points a and a'.

The distance of a from xy shows the distance of the
point A from the vertical plane.

The distance of a' from xy shows the distance of the
point A from the horizontal plane, i.e. its height.

The object of all science is exact knowledge, and as
this is impossible of attainment unless language be made to
keep pace in precision with the advance of thought, it has
been considered desirable to rigidly define at the outset all
the important terms that have a meaning peculiar to the
subject. It is not intended however that the student at the
very commencement of his labours shall learn these by rote,
but he is earnestly requested to turn back from time to
time, and assure himself that he has at least mastered the
significance of such technical terms as form part of the
description of the problem he is working. Many a difficulty
will be got over by a timely reference to the definitions,
-while full mastery of the language of the subject is the only
way to a thorough grasp of its principles.

DEFINITIONS AND THEOREMS.

The object of Solid Geometry is systematised knowledge
of the geometrical properties of forms having three dimen-
sions.

4 SOLID OR DESCRIPTIVE GEOMETRY.

Descriptive Geometry deals directly, with the relations
of these forms to two co-ordinate planes at right angles, and
indirectly, with their relations to one another. The object
of the science being to systematise our knowledge of the
methods by which, and the principles upon which, these forms
of three dimensions can be represented upon a plane or
surface of two dimensions, so that, from the representation
alone, the real shape, size, and relationship of the forms
may be unerringly inferred.

The method of Descriptive Geometry is almost wholly
a method of projections, as laid down in the definitions that
follow.

The principles are involved in the application of the
method to the solution of problems, many of the leading
results of which find expression in the subjoined theorems.

PRELIMINARY DEFINITIONS.

1. The projection of a point on a plane is the foot of
the perpendicular let fall from the point to the plane.

2. The projection of a line (straight, curved, or broken),
figure, or form is the sum of the projections of its points.

3. The plan of a point, line, figure, or form is its
projection on the horizontal plane; its elevation is its projec-
tion on the vertical plane.

4. The perpendicular which gives the projection of a
point on a plane, is called the projector of the point, and the
plane on which the point is projected, is called the plafie of
projection of the point. The vertical projector is the one
drawn to the vertical plane ; the horizontal projector the one
drawn to the horizontal plane of projection.

PRELIMINARY DEFINITIONS. 5

5. The surface containing all the projectors of a
straight or curved line, is called the projecting surface of the
line. The intersection of this surface with the plane of pro-
jection coincides with the projection of the line upon that
plane.

6. A trace is the intersection of a straight line or of a
surface (plane or curved) with, unless otherwise stated, one
of the planes of projection. The horizontal trace (H.T.) is
the intersection with the horizontal plane (H.P.); the
vertical trace (V.T.) is the intersection with the vertical
plane (V.P.).

Straight lines are represented either by their traces or
by the method of projections. By the latter when y?;zzV<?, and
by the former when the length is indefinite or disregarded
for the purposes of the problem. Similarly, figures are
represented by their projections, and planes, indefinitely
extended, by their traces. Curved surfaces are frequently
represented by a combination of both. The traces of an
indefinitely extended straight line, or of a plane, are always
sufficient to determine it. The traces of a curved surface
are not alone sufficient to determine it.

JVote. The projections made in Descriptive Geometry,
are called ^^orthographic" projections, to distinguish them
from radial ox perspective projections.

7. A section plane is a plane cutting through a form.
The ^^ section" is the trace of the form on this plane.

Planes may be vertical, horizontal, inclined, or oblique.

8. A vertical plane is one at right angles to the horizon-
tal plane. It may make with the vertical plane of projec-
tion any angle between 0° and 90°. In the former position

6 SOLTD OR DESCRIPTIVE GEOMETRY.

it will be called a parallel vertical plane, and in the latter, a
right vertical plane. See Theorems II. and III., and Fig. 5,
planes n^o' and hZv.

9. A horizontal plane is one at right angles to the
vertical plane and parallel to the horizontal plane of projec-
tion.

10. An inclined plane is one at right angles to the
vertical plane and not parallel to the horizontal plane. Ex.
qlm'. Fig. 5.

11. An oblique plane is one not at right angles to either
plane of projection, i.e. is a plane inclined to both. See
qlm,,, Fig. 2, and hav', Fig. 5. A plane incUned to both co-
ordinate planes and parallel to xy, is called a parallel oblique
plane. See plane rs, t'u, Fig. 5, An oblique plane equally-
inclined to both planes of projection is called a right
oblique.

12. A plane is said to be ^'constructed'' or ^' rabafted,''
when it is rotated, with all its contained points, lines, and
figures upon it, about one of its traces into the correspond-
ing plane of' projection.

13. A development is the figure produced by unrolling
or laying out in one plane the surface or surfaces of a form.
All surfaces are not developable, i.e. cannot be laid flat in one
plane without tearing or doubling. All forms bounded by
plane surfaces, and some bounded by curved surfaces, are
developable. See Chap. VII., Curved Surfaces.

THEOREMS.

Theorem I. The traces of a straight line ^& points. A
line parallel to either plane of projection has no trace in the
plane to which it is parallel, and therefore, a line parallel to

THEOREMS. 7

xy (the line in which the co-ordinate planes of projection
intersect), being parallel to both planes of projection, will
have no trace at all

Theorem II. The traces of a plane are straight lines,
and, unless the plane is parallel to xy, the traces will meet at
some point in that line. A plane parallel to xy, and not
perpendicular to either plane of projection, will have for its
traces two straight lines parallel to xy. A plane parallel to
xy, and at right angles to one of the planes of projection,
will be consequently parallel to the other plane of projec-
tion and will have no trace in that plane. Figs. 2 and 5.

Theorem III. If the horizontal trace of a plane is at
right angles to xy, the plane is perpendicular to the vertical
plane of projection. If the vertical trace of the plane is at
right angles to xy, the plane is perpendicular to the hori-
zontal plane of projection. Fig. 5 planes qhn' and n^o.

Theorem IV. The two projections of a point are in the
same projecting line or perpendicular to xy, after the vertical
plane has been turned into the horizontal. Fig. 3.

Theorem V. The perpendicular from the plan of a
point to xy is parallel to the vertical projector of the point,
and is equal to the distance of the point from the vertical
plane. Similarly, the perpendicular from the elevation of a
point to xy is parallel to the horizontal projector of the
point, and is equal to the distance of the point from the
horizontal plane.

Theorem VI. When two planes are at right angles,
every line drawn from a point in one of them perpendicular
to the other, lies wholly in the former, and meets the latter

8 SOLID OR DESCRIPTIVE GEOMETRY.

in, and is perpendicular to, the line of intersection of the
planes. Whence it follows :—

(i) A point in one plane of projection has its pro-
jection on the other in xy. That is, if the point is in the
vertical plane its plan is in xy. If in the horizontal plane
its elevation is in that line.

(2) If a given plane be perpendicular to one plane of
projection, it contains all the projectors drawn from any
points, hnes, or figures, in that plane, to the plane of pro-
jection to which the given plane is perpendicular. That is
to say, the given plane is the projecting plane of all figures
lying in it. The trace of this plane on the plane of pro-
jection to which it is at right angles, contains the projections
on this plane of all points, lines, or figures lying in the given
plane.

(a) Thus, if an inclined plane be perpendicular to the
vertical plane of projection, the elevations of all points in
the inclined plane will be in its vertical trace. E.g. A
circle of 3 inches diameter lying in a plane inclined at 30"
to the horizontal plane and perpendicular to the vertical
plane, has a segment of the vertical trace, 3 inches long, for
its elevation.

(/3) A plane inclined to the vertical plane of pro-
jection and perpendicular to the horizontal plane, has the
plans of all points, lines, and figures lying in it projected in
its horizontal trace. Thus, the plans of the vertical faces of
a right prism resting on its base coincide with the lines
forming the sides of that base.

Theorem VII. The projection of a straight line on a
plane is a straight line. The projections upon the same
plane of parallel straight lines are parallels.

THEOREMS. 9

The intersection of a plane with a series of parallel
planes will be a series of parallel lines, and the projections
of these intersections will, according to this theorem; be
parallels.

Theorem VIII. The orthographic projection of a finite
straight line is equal to the real line if the latter is parallel
to the plane of projection. The magnitude of the pro-
jection diminishes as the angle of inclination of the line
increases, and becomes a minimum, i.e. a point, when the
angle of inclination reaches 90", or when the line is perpen-
dicular to the plane.

The projection of a rectilineal angle on a plane parallel
to that of the angle, is an angle equal to the given one.

The projection, on the same plane, of the circular arc
subtending that angle will also be equal to that of the arc.
See Figs. 21 and 22.

Generally, the projection of any plane figure on a plane
parallel to it is a figure equal in all respects to the figure of
which it is the projection. And the area of the projection
of a given figure decreases as the angle between the plane
of projection and the plane of the figure increases, —
becoming a minimum, i.e. a straight line, when this angle is
a right angle.

Note. This is a principle often taken advantage of in
finding the true shape of sections. See Prob. i, Chap. I.

Theorem IX. The projections of a perpendicular to a
plane are perpendiculars to the traces of the plane. The
plan is at right angles to the horizo?ital trace ; the elevation
is at right angles to the vertical trace.

lO SOLID OR DESCRIPTIVE GEOMETRY.

Theorem X. If a straight line lies in a plane and has
traces (Theorem L), they will be in those of the plane.
Thus a plane can be drawn to contain a given line by-
taking the traces of the plane through those of the line,
when the latter are determinable.

Theorem XL The angle which the tangent plane to a
right cone makes with the plane of the base of the cone, is
measured by the plane angle which a generatrix of the cone
makes with the base, or is the complement of the angle
between a genera.trix and the axis.

Theorem XII. If a plane touches a conic surface and
a second plane intersects them, the trace of the tangent
plane on the intersecting one is a tangent to the trace of the
conic surface on the same plane. Generally, any plane
cutting a curved surface and its tangent plane through a
point of contact of the tangent plane with the surface, will
intersect the tangent plane in a Hne which is a tangent to
the curve in which the cutting plane intersects the surface.

Theorem XIII. The tangent plane at any point of a
curved surface contains the tangent lines drawn at the
same point to all the lines traced on the surface at that
point.

Theorem XIV. The tangent plane to a cone or a
cylinder at a given point is the tangent plane to the surface
at every other point in the generatrix passing through the
given point, and all the normals to the surface along the
generatrix are in a plane containing the generatrix and per-
pendicular to the tangent plane.

Theorem XV. Parallel sections of a conic surface are
similar but unequal curves.

THEOREMS. II

Theorem XVI. Points in the generatrix of a surface of
revolution describe circles in planes perpendicular to the
axis of revolution whose centres are the points of inter-
section of the axis with their respective planes.

Theorem XVII. All sections of a surface of revolution
made by planes containing the axis are similar and equal
curves.

Theorem XVIII. A plane which bisects a chord of a
sphere at right angles passes through the centre of the
sphere.

Theorem XIX. The section of a sphere by a plane
passing through its centre, is a circle whose radius is equal
to the radius of the sphere — hence called a great circle.
Any other plane section cuts a small circle.

Theorem XX. When a plane is '■^constructed" or
^' rabatted" (Def. 12) about one of its traces every point in
it will have for its locus a straight line drawn at right angles
to the trace used through the projection of that point on
the plane of projection containing the trace.

Thus if the horizontal trace be used as the axis of
"rabatment" or "construction," the loci of all points in the
plane will be the perpendiculars to the trace through their
respective plans.

The reason for this relation or "locus" will be obvious by
considering that every point in a "constructed" plane must
describe a circle about the trace forming the axis, and that
the projecting plane of the path of the point is a plane
perpendicular to that trace.

These loci are made use of, for example, in determining
the true shape of a section, and the projection of a frustum

12 SOLID OR DESCRIPTIVE GEOMETRY.

of a solid on a plane which passes through it. The solution
of many problems whose data lie in one plane is often
greatly facilitated by this method of rabatments.

DEFINITIONS ILLUSTRATED.

The co-ordinate planes, upon which the projections are
made, intersect at right angles in a line called the ground
line {xy), and are named from their usual positions the
horizontal and vertical planes of projection. Fig. i.

The four dihedral angles formed by the intersecting
planes of projection are known as the ist, 2nd, 3rd,
and 4th.

The ist dihedral angle is that contained between the
upper face of the H.P. and the front face of V.P.

The 2nd is that between the upper face of the H.P. and
the back face of V.P.

The 3rd is between the lower face of H.P. and the back
.face of V.P.

The 4th between the lower face of H.P. and the front
face of V.P.

Notation. . - - .

A point in space is indicated by a capital letter A, its
projections by itaHcs a and a': the accented letter denoting
the vertical projection or elevation of the point, and the
unaccented italic the plan. Figs, i and 3.

Similarly, a line in space is denoted by capital letters, as ;
AB: and its projections by italics, as; ab, a'b'. Figs, i and 4.

DEFINITIONS ILLUSTRATED.

A plane is denoted by three letters, one on each trace
and the third on the point of intersection in xy. Figs. 2
and 5.

Fig. I.

Note. Figs. I and 2 are merely pseudoperspective
sketches illustrative of the elementary principles and defini-
tions, and must on no account be mistaken for the kind
of drawing to be made in the solution of the problems.
Figs. 3, 4, and 5 are orthographic illustrations of the mode
of determining points, lines, and planes, and may be taken
as elementary types of the kind of drawing to be made.

14 SOLID OR DESCRIPTIVE GEOMETRY.

Fig. I. The points aa' are the projections of the point
A. Def. X.

The lines ah, a'b' are the projections of the line AB.
Def. 2. The line ^^ is the plan, and «y the elevation.
Def. 3.

The perpendiculars Aa, Aa', and j5/^, Bb', are the
projectors of the points ^ and ^; Aa, Bb, and ^«', ^i!'',
being the horizotital and vertical projectors of the points ^
and B respectively, and the planes H. and V. the horizontal
and vertical pla?ies of projection. Def. 4.

The planes AB, ab, AB, a'b', containing the given
straight line AB and the projectors Aa, Aa',. and Bb, Bb',
are the projecting planes of the line AB, and the intersection
of these planes with the planes of projection give the lines
ab, ab' which are the projections of the lines AB. Def. 5.

As our drawings can only be made conveniently on one
plane, we assume that the vertical plane, with the vertical
projection on it, is rotated backwards in the direction shown
, by the arrows, about xy as an axis, until it coincides with the
horizontal plane. In this position the vertical is oftentimes
more or less superposed upon the horizontal projection, and
as in all our- reasonings upon this subject it is necessary to
be clear as to what points, lines, &c. are in the vertical, and
what in the horizontal planes, some conventional system of
notation, such as that described above, has to be adopted.
When the planes are thus rotated the plan a and the elevation
(i! of any point A will lie in the same perpendicular to the
ground line, xy. Theorem IV.

This perpendicular is therefore a locus of points a and a',
that is, will contain a and a'.

DEFINITIONS ILLUSTRATED.

The distance of a from xy shows the distance of the
point A from the vertical plane; and the distance of a' from
xy shows the distance of the point A from the horizontal
plane, i.e. its height. Thus aa is parallel and equal to Aa\
and a'o. is parallel and equal to Aa. Theorem V.

Fig. 2.

hor. trace

Fig. 2 shows the traces (Def. 6) of an "oblique" plane
(Def 1 1) meeting in xy (Theorem II.). The traces of every
plane which is not a parallel vertical, a horizontal or a
parallel oblique (Defs. 8, 9, and 11), will if produced meet
in xy.

SOLID OR DESCRIPTIVE GEOMETRY.

ORTHOGRAPHIC ILLUSTRATIONS.

Fig- 3-
Points.

Examples of points determined in various positions by
their projections.

Point A or aa! is in the ist dihedral angle of the planes of

projection.
„ B or bb' ,, 2nd „ „ „ „ „
,, C or « ,, 3^^ 11 11 11 " "

„ D or dd! ,, 4th „ „ „ „ „
„ E or e^ is in the horizontal and behind the vertical

plane.
„ Foxff is in both planes.
,, G ox gg is in the vertical and below the horizontal

plane.
„ H or hh' is equidistant from the horizontal and ver-
tical planes.
„ K or kk' is in the vertical and above the horizontal
plane.

ORTHOGRAPHIC ILLUSTRATIONS.

Observe, That to find the projections of points satis-
fying given conditions of position with regard to the planes
of projection the relations indicated in Theorems IV. and V.
will suffice.

Examples of lines in various positions determined by
;heir projections,

E. G.

1 8 SOLID OR DESCRIPTIVE GEOMETRY.

The segment AB or ab, a'b is in the ist dihedral angle of

planes.
„ CD or cd, c'd! „ and ,, „

„ EFoxef,e'f' „ 3rd „

„ GHoxgh.^H „ 4th ,,

„ AfJV or Pin, nin' „ ist „ „

and the line is perpendicular to the vertical plane.
„ OP or op, dp' is in the ist dihedral angle.

Fig. 5-

£*la.

ORTHOGRAPHIC ILLUSTRATIONS. 19

Examples of planes defined or expressed by their traces.

The portion included between the infinite branches ah
and av' of the given plane is in the first dihedral angle of
the planes of projection.

The portion between the branches a?/ and af is in the
second dihedral angle.

The portion between af and ag' is in the third angle,
and the portion between branches a^ and ah is in the
fourth angle.

The plane glm' is perpendicular to the vertical plane
md makes an angle of 6" with the horizontal.

The plane ;z/?^' is at right angles to the horizontal and
nakes an angle of i^" with the vertical plane.

The plane rs, t'u' is parallel to xy and meets both
)lanes.

The plane hlv' is at right angles to xy and therefore to
both planes of projection.

MEMORANDA

FOR WORKING THE PROBLEMS IN ENSUING CHAPTERS.

All the solids given are assumed to be ^ right,' unless
Otherwise expressed.

All dimensions and arrangements, not expressly men-
tioned and limited, may be assumed at pleasure.

The inclinations of all lines and planes must be under-
stood, unless otherwise mentioned, to be to the horizontal
plane.

Invisible edges are those which are hidden by the solid
from the eye when looking in the direction in which the
projectors are drawn.

Sections. The teacher should give directions for sections
and developments, as far as possible, with each problem.

Also several plans and elevations may advantageously be
completed of the solids given in many of the problems.

SOLIDS IN SIMPLE POSITIONS.

I. Draw plan and elevation of a pyratnid, 3-5 inches
highy with square base 0/2'^ inches side, when resting ivith
its base on the horizontal plane, and with one side of the base
?naki?ig an a?tgle of 30" with the vertical plane.

Commence with the plan. This will be the square
ABCD of the base with the opposite corners joined for the
plans of the slant edges of the solid. The point in which
the diagonals of the square cross will be the plan v of the
vertex V.

For the elevation take the ground line xy inclined 30"
with one side of the square and draw perpendiculars to xy
(Theorem IV.) from the four corners abed and the centre v of
the square. The elevations cib'c'd' of the four corners of
the square will be in the ground line, because the base of
the solid rests on the horizontal plane. The height of the
pyramid, 3*5 inches, must be set up for its vertex v' from
the point where the perpendicular from the centre of the
square meets the ground line, and at right angles to that
line (Theorem v.). The whole elevation will be completed
by joining the elevation v' of the vertex to the four points
determined cib'c'd' on the ground line for the elevations of
the four corners of the base.

Invisible edges. In this plan all the edges of the solid

22 SOLID OR DESCRIPTIVE GEOMETRY.

are visible. In this elevatmi the slant edge of the pyramid
nearest xy is invisible. The two sides of the base nearest
xy are also invisible, but as they coincide with the two other
sides in the xy, they cannot be shown in dotted lines.

Note. It must be understood that from any one plan
and elevation any number of elevations can be drawn by
assuming vertical planes, that is, by taking new ground lines,
in different positions round the solid, and working from the
plan to them. For example, if a solid were placed on the
floor of a room, four elevations of it might be shown on the
four walls of the room, which are so many vertical planes of
elevation. And so from any one elevation aftd plan any
nnmber of plans may be determined by assuming horizontal
planes in the desired position about the elevation, and
working from the elevation to them.

E.g. For a new elevation. The new ground line, xy,
being taken, draw perpendiculars to it from the plans of the
various points of the solid, and on these perpendiculars from
xy mark the various heights of the points, to be taken from
the elevation already drawn, and complete by joining the
points thus found as already joined in plan.

For a new plan. The new xy being taken, draw perpen-
diculars to it from the elevations of the various points of the
solid, and on these perpendiculars mark from xy the various
distances which the points are from the first plane of eleva-
tion to the pla7i first drawn, and complete by joining the
points as already joined in elevation.

To work a section of the solid. Draw a line through
the middle point of the elevation of the axis making an
angle of 40" with xy. Assume this to be the vertical trace

SOLIDS IN SIMPLE POSITIONS. 23

(Def. 6 and Theorem II.) of an inclined section plane (Defs.
7 and 10). Produce the vertical trace to meet xy and from
the point of intersection of these hnes draw the horizojital
trace perpendicular to xy. (Theorem III.).

The points i', 2', 3', 4', in which the given line or
vertical trace cuts respectively the lines a'v' , b'v', c'v', and
d'v' are the elevations of the points I., II., III., IV. in which
the section plane cuts the edges AV, BV, CV and DF of
the pyramid. (Theorem VI. (2) and (a)). The plans of
these points will be found by drawing lines from their
elevations at right angles to xy to cut the corresponding
lines in plan. (Theorem IV.). £.g. A line at right angles to
xy, drawn from the point i' in a'v' to cut the plan av, will
give the point i which is the plan of the real point I., in the
section, of which i' is the elevation. Similarly, the points
2, 3, 4 in the plan may be found. The figure got by
joining i to 2, 2 to 3, 3 to 4, 4 to i is the plan of the
section. This for clearness should be conventionally
crossed with a series of parallel lines. The elevation of the
section is shown by the points i', 2', 3', 4' in the vertical
trace. (Theorem VI. (2) and (a)).

To find the true shape of the section.

This is done by the inethod of rabatments or '■^con-
struction'^ (Def 12 and Theorem XX.). That is, the
plane of section is rotated about one of its traces as an axis
into the plane of projection containing the trace which is
chosen for the operation. See Theorem XX.

a. To find the true shape of the section by rabatting the
plane about its vertical trace.

The vertical projectors (Def. 4) of the points I., II., III.,

24 SOLID OR DESCRIPTIVE GEOMETRY.

IV. of the section, i.e. the perpendiculars from these points
to their elevations i', 2', 3', 4', lie in the plane of section
(Theorem 'VI. (2)), are at right angles to the vertical
trace (Theorem VI.), and are equal in length to the per-
pendiculars drawn from the plans of the points i, 2, 3, 4
to xy, (Theorem V.) — i.e. \he plane of section is \h& projecting
plane of the section. If, therefore, lines be drawn from the
points i', 2', 3', 4', of the elevation, at right angles to the
vertical trace of the section plane, equal in length to the
respective distances of the points i, 2, 3, 4 of the plan from
xy, and the points I., II., III., IV., so found, at the extremi-
ties of the perpendiculars, be joined in order, the figure
produced will be that resulting from the rabatment of the
plane of section about its vertical trace, i.e. will be the true
shape of the section.

p. To find the true shape by rabatfing the plane about
its horizontal trace. If the plane be rotated about the H.T.
as an axis into the horizontal plane, the points I., II., III.,
IV. of the section will describe segments of circles about
the H.T. in planes perpendicular to it, and the loci of these
points in plan will be straight lines drawn from the plans of
the points i, 2, 3, 4 of the section at right angles to the
H.T. of the section plane. If on these lines are measured
off from H.T. the distances of the corresponding points
i', 2', 3', 4' up the vertical trace from xy (i.e. the radii of the
segments of the circles described by the points of the
section when the plane is being rabatted), the figure pro-
duced by joining these points will be the true shape of the
section.

Note. To avoid the confusion that sometimes results
from rabatting the section plane directly, the section may

SOLIDS IN SIMPLE POSITIONS. 25

be projected on a plane taken parallel to the plane of
section, and will then be shown in true shape. (Theorem
VIIL). This, which is merely the method of rabatments
under a disguise, is of little use unless the section plane be
perpendicular to one of the planes of projection. The xy
for the parallel plane must obviously be taken parallel to
the trace of the section plane on that plane of projection
to which the section plane is perpendicular. Finish by
drawing the projection of the frustum of the pyramid on
this plane.

To find the true length and inclination of a slant edge of
the pyrainid.

a. Make a projection on a plane parallel to the edge
whose length is required, i.e. take xy parallel to the plan of
the slant edge. The elevation will show the inclination
and the edge in true length. Theorem VIIL or.

By a special construction,

13. Make a right-angled triangle having its two sides

respectively equal to the plan of the edge and the length of

the axis — the hypotenuse will give the true length of the

edge; and the angle subtended by the axis will be the

angle which the edge makes with the horizontal plane.

•
To work the development of the frusitcm. See Def. 13.

Describe a circle with radius equal to the slant edge of
the pyramid and cut off four consecutive chords of this
circle equal to AB, BC, CD, DA, the sides of the base of
the pyramid. Draw AB, BC, CD, DA, and join the points
A, B, C, D, with the centre V of the circle. Describe a
square on one of the bases of the four triangles thus
:)roduced. The four triangles and the square together

26 SOLID OR DESCRIPTIVE GEOMETRY.

forhi the development of the pyramid. Find now the points
I., II., III., IV. in the edges VA, VB, VQ VZ>, by dividing
each of these edges in the same proportion as the corre-
sponding elevations are divided by the vertical trace of the
section plane. Join the points in order (noting that in the
development one edge and its point of division necessarily
occurs twice) and cut the part next the vertex away. This
is the development of the frustum. Build up into a model
and fit on the true shape of the section.

2. P/an and elevation of a pyramid with base a regular
hexagon 0/1')$ in. side, when resting on its base, and with
one side of base inclined 40° to the vertical plane. Axis 3*5
inches.

Work also a section by a vertical plane [Fig. 5 n^o'^^, the
angle <^" being 35", and the horizontal trace '25 in. from the
plan of the axis. Show new plan and new elevation of the
frustum.

Find true shape of section by rabatting or "constructing"
the plane of section about one of its traces, and develop the
frustum.

3. Plan and elevation of a square prism 3*5 in. long,
bases 2*5 in. side.

(a) Wheti standing on one of its bases and a rectangular
face making an angle of ^0° with the vertical plane of
projection.

()8) When resting on a rectangular face and its bases
making a?igles of ^o^ with the vertical plane.

To work the prism in (a) with section, true shape, pro-
jections of the frustu?n, aJid development.

SOLIDS IN SIMPLE POSITIONS. 2/

The plan will be a square with one side inclined at 30"
to xy.

Begin by drawing xy. Set off from xy a line at 30° in
the H.P. and describe on any segment ab, 2-5 in. long, of
this line a square abed. This will be the plan of the prism,
i.e. of its two bases — one on H.P. and the other 3*5 in.
above — and its four vertical faces. The elevations db'c'd'
of the four corners A BCD of the lower base will be in xy,
and the elevations a^^^^c^d^ of the four corners A^B^CJ)^
of the upper base will be in a line parallel to xy and 3*5 in.
above it. The segments da^, b'b^, c'c^, d'd^ of the lines
drawn from the four corners of the plan abed, at right angles
to xy for the elevation, will be the elevations of the edges
AA^, BB^, CQ, DD^, respectively. Make the edge AA^,
which is nearest the V.P., dotted in the elevation; and the
plan and elevation will be complete.

Section. Take a line cutting the elevations of the
upper base a^b^e^d^, and three of the edges a'a^, b'b^, e'c^,
in the points i', 2', 3', respectively, for the vertical trace
of the section plane. The elevation of the section will be
in this line. For the plan, draw a line from the point in
,vhich the vertical trace cuts the elevation of the upper base,
it right angles to xy, and produce it to cut the sides ad and
ie of the plan abed in the points j" and z. The figure sabez
vill be the plan of the section. Find the true shape by
"abattiftg the section plane about one of its traces as in
?rob. I.

New elevation of frustum. Take a new xy at right

ingles to the xy first drawn, and project as for a complete

' ilevation of the prism. Show the elevations of S and Zin the

lew elevation of the upper base, and find the points i', 2', 3',

2S SOLID OR DESCRIPTIVE GEOMETRY.

by measuring the heights of these points from xy in the
elevation first drawn and setting them off on the cor-
responding edges in the new elevation. Join up the points
of the section for the complete elevation of the frustum.

A new plan of the frustum can be made in a similar
manner by taking a new xy at, say 30", with the elevations of
the long edges of the prism — using either of the above
elevations of the prism.

Development of frustum. Draw two parallel lines 3*5
in. apart and measure off on the lower one the segments
AB, BC, CD, DA, equal to the sides of the square base of
the prism. From the points A, B, C, D, A, draw lines at
right angles to cut the second parallel line in the points
A^., Bi, Ci, D.2, A2. On DA and D-^A^ describe the
squares DABC and D^A^B^Q. This completes the de-
velopment of the prism.

For the frustum find the points I., II., III. in the edges
AA2, BB.^, CC2, and the points S and Zin AiD^ and D^C^.
The lengths AL, BIl., CIIL, can be got from the eleva-
tions and the lengths D^S and D^Z from the plan. Fit on
the true shape of the section and build up the model.

The prism in (/8) the student should now be able to
work for himself.

4. Blan and elevation of a right cylinder; axis 3*5 ///.
horizofital and inclined 40° to the vertical plane. Diameter of
base 2*5 in.

Conceive the cylinder to be inscribed in the prism (/S)
Prob. 3. The plan of the prism, which is a rectangle 3-5 in.
long and 2-5 wide, is also the plan of the cylinder. The
elevation of the cylinder is made by inscribing ellipses in the

SOLIDS IN SIMPLE POSITIONS. 29

rectangles which are the elevations of the square ends of the
prism.

To work a section of the cylinder take an auxiliary xy at
right angles to the plan of the axis and draw the circle
which is, in this position, the elevation of the cylinder.
Divide this circle into, say, sixteen equal parts and draw, from
each of the points, lines in plan parallel to the axis. These
will be the plans of lines lying on the surface of the cylinder.
Show these lines on the first elevation of the cylinder by
measuring their heights in the auxiliary elevation and
transferring them. A vertical or inclined section plane may
now be taken and the section determined by finding the
projections of the points in which the given plane cuts these
lines. The points found must be joined by a curved line
put in by hand. The section of a right cylinder by a plane
which is neither parallel nor at right angles to its axis, as in
this case, is an ellipse. Find the true shape by rabatting the
plane.

To work the development. Measure a straight line equal
to the circumference (found by multiplying the diameter of
the cyHnder by 3-1416) and divide this line into sixteen
equal parts. Draw lines i, 2, 3, 4, &c. from these points and
make each equal to the length from the base to the section
plane of the corresponding line on the cylinder.

Join the points so found by a curve for the development
of the cylindrical surface.

5. Right cotie with axis 3*5 in. and diameter of base
2*5 in.

a. Plan and elevation when standing on its base, and a
ucond plan when resting o?i its conical surface.

30 SOLID OR DESCRIPTIVE GEOMETRY.

p. Plan and elevation when the axis is horizontal, 2 in.
above the H.P., and itidined 35" /^ the V.P.

a. The plan of the cone when standing on its base will
be simply a circle 2*5 in. diameter, and its elevation an
isosceles triangle, base 2*5 in., and vertex 3*5 in. high. For
the second plan xy must be taken parallel to one of the
slant sides of the isosceles triangle. To find the new plan
of the base, circumscribe the base first drawn by a square
with one side parallel to xy. The new plan of this square
will be a rectangle in which an ellipse can be drawn for
the new plan of the base of the cone, and the projection
completed by drawing tangents from the new plan of the
vertex to this ellipse.

(3. The projection of the cone in this position presents
no special difficulty.

Sections of the cone by inclined and vertical planes may
here be worked as in the case of the cylinder by dividing
the base of the cone into any number of parts and drawing
the generatrices of the cone from these points in the base
to the vertex. Treating these generatrices as the edges of a
polyhedral pyramid, the projections of their intersections
with the section plane may be found, as in Probs. i and 2,
and joined by a curved line.

Note on the true shape of the sections of a right conic
surface. When the section plane cuts all the generatrices,
the section is a closed curve and is generally an ellipse, the
exception being when the plane is at right angles to the
axis, in which case the section is a circle. A plane parallel
to two generatrices cuts an hyperbola, the limit of which is
the pair of generatrices to which the section plane is parallel.

SOLIDS IN SIMPLE POSITIONS. 3 1

When the section plane is parallel to only one generatrix
the section is 2^ parabola^ and its limit is the parallel genera-
trix, in which case the section plane becomes a tangent plane
to the cone.

The development of the conic surface will be a segment

of a circle with radius equal to a generatrix of the cone, the

vxdo
circular arc of which segment subtends an angle of ^^ —

g
degrees at the centre; where r = radius of base and g= length
of generatrix.

6. T'/an and elevation of a square prism 3 -5 in. long,
'.dge of base 2*5 i7i., when its axis is horizontal, inclined
it 45" to the vertical plane, and its lowest face inclined 30"
'0 H.P.

Assume xy. Since the horizontal axis is at right angles
: o the base and inclined 45", the base will be inclined also at
45" to the vertical plane, i.e. at the complement of the
inclination of the axis. Take, therefore, a line in the H.P.
;.t 45" with xy and assume this line as the xy of an auxiliary
-^ ertical plane at right angles to the axis of the prism. The
] projections of the two bases of the prism on this plane will
( oincide and will be shown in true shape. (Theorem VIIL).
Draw the square base on this auxiliary vertical plane with
< ne side at 30° with the xy and deduce the plan. This
1 Ian will be the plan of the square prism with its axis
1 orizontal, inclined 45" to the vertical plane first assumed,
i nd its lowest face inclined at 30". Complete the elevation
c n first vertical plane as in other problems.

For the section, assume three points — one in each of
t iree edges of the solid — and apply the principle enunciated

32 SOLID OR DESCRIPTIVE GEOMETRY.

in Theorem VII. to the completion of the plan and eleva-
tion of the section made by a plane passing through these
points.

7. Plan and elevation of a prism, 3-5 in. long, with bases
regtdar hexagons of i'^ in. side, when its lowest face is inclined
20**, its axis horizontal, and inclined 40" to the V.P. Section
as i?i the last problem.

The work is similar to that of the last problem, the aux-
iliary xy being taken at 50" to the xy first drawn.

8. Plan and elevation of your instrument case when the
lid is open 60" ; the long edges horizontal and inclined 30° to
the V.P.

This is worked in a similar manner to Probs. 6 and 7.

9. Plan and elevation of a tetrahedron with one face
horizontal and i inch above the horizontal plane. Edge 3
inches. One horizontal edge to be inclined 1$" to the vertical
plane.

HYiQplan will be an equilateral triangle of 3 inches side,
with the corners joined to the centre for the plans of the
slant edges of the solid.

The base being horizontal, its elevation will be a
straight line and at the required height, i inch, above the
ground line.

The edge of the tetrahedron being known, the height of
the vertex above the horizontal base can be determined by
making a vertical section containing the vertex and one slant
edge. To find this, make a right-angled triangle with the
length from a comer of the base to its centre for the base of
the triangle, and the true length of the edge of the tetrahe-

SOLIDS IN SIMPLE POSITIONS. 33

dron for the hypotenuse. The perpendicular will be the
leight required.

10. The surface of a sphere, diameter 3 feet, is divided
'nto eqtcal portions by 4 great circles, one horizontal and three
\ vertical. Draw an elevation of the sphere and circles on a plane
parallel to one of them.

Work also a section by a plane, perpendicular to V. P.,
'inclined 40", and '5 in. from centre of sphere. Scale -jV-

11. A street-lamp is formed of a right pyramid, and a
J yustum of one ; the slant edges of former inclined 40°. The
cnnmon base of the forms is a square of 20 inches side, and the
I'twer base or end of frustmn is a square of 12 inches side and
2 o inches below the other.

(a) Determine an elevation on a plane parallel to a
diagonal of the plan.

(b) Dcteri7iine an elevation on a plane which makes an
angle of 2,0'^ with a horizontal edge.

Scale i.

IhQplati is a square of 2*5 in. with the opposite corners
j( ined for the slant edges, and with a second square of i "5 in.
w ;thin and parallel to the first.

In the elevation (a) the slant edges are drawn at 40" with
tl e elevation of the base of pyramid.

12. A ladder, ^o feet long and of tJu uniform width of
T) feet, rests against a vertical wall — 37^<?/' thick, $ofcet high,
2( feet wide — and makes an angle of ']o'^ with the horizontal
pi ine on which it stands.

(a) Draw apian, with an elevation on a plane parallel

to that of the wall.
E. G. 3

34 SOLID OR DESCRIPTIVE GEOMETRY.

(b) Aiso an elevation on a plane making an angle of 50"
with that of the wall.

Scale TO feet to i inch.

N,B. — Rungs and sides to be represented each by a single
line. Three rungs only to be shown, dividing the length of the
ladder equally.

Commence with an auxiliary elevation on a vertical
plane parallel to the ladder.

For this, draw xy and lay off the angle 70", and on this
line measure the length of the ladder : thence determine
plan,

13. A right prism, 4 inches lo?ig, with square bases of
2*5 inches side, is to be drawn

(a) with one diagonal of the solid vertical,
(f) with the same diagonal horizontal.

Draw a plan and elevation of the solid resting on one
of the square bases, with one diagonal of the square at right
angles to the ground line.

If two opposite corners of the rectangle which is the
elevation of the solid be joined, this Une will be the eleva-
tion of the diagonal of the solid, shown from its position
in its true length. A plan may then be determined with
this diagonal vertical or horizontal by taking (for a) xy
at right angles to, or (for b) parallel with, the elevation of this
diagonal.

14. Flan and elevation of a pyramid with hexagonal
base.

a. With one face ABV horizontal.

b. When one face is vertical.

SOLIDS IN SIMPLE POSITIONS. 35

€. When one edge is horizontal.
d. When one edge is vertical.
Determine plan of the pyramid when resting on its base.

For a. Determine elevation on a vertical plane taken
£t right angles with the plane of face ABV. Then the
raquired plan may be deduced by taking xy parallel to
t le line which represents the elevation of face AB V, and
V orking to it.

For b. Take xy perpendicular to the line representing
the elevation of face ABV.

For c and d. Determine an elevation in which the edge
is shown in its true length. Then xy taken parallel for c,
p irpendicular for d, to the elevation of this edge will place
tl e p)n-amid in proper position for the plans required.

15. Plan and elevation of an odahedrofi, edge 3 inches.

(a) When one axis of the solid is vertical and another is
inclined at TS° to the vertical plane.

{b) When resting with one face on the horizontal plane.

(a) Describe a square of 3 inches side and join the
di Lgonals, which will be the plans of the horizontal axes.
Tl.is square will be a plan of the solid in the required
pciition. For the elevation it is only necessary to know
th. .t the length of the vertical axis is equal to a diagonal of
th( : square, and that the heights of the horizontal edges

!ar( half that diagonal. Take xy at an angle of 75° with the
Ipk n of the diagonal.

(b) Work this first in the same way as Prob. 14, (a), and
Ithe n by the following special construction : —

3—2

36 SOLID OR DESCRIPTIVE GEOMETRY.

Describe an equilateral triangle of 3 inches side,
and from its centre with radius to one of the corners de-
scribe a circle. In this circle complete the hexagon, of
which the three corners of the triangle are alternate points.
This hexagon with the alternate points duly joined is the
required plan.

For the elevation it is necessary to find the distance
between the parallel faces. This is done by a vertical sec-
tion through one axis of the solid. This section is a rhom-
bus, the shorter diagonal of which is an edge of the solid,
and the side of which is the altitude of the equilateral tri-
angle forming the horizontal or any other face of the solid.
The distance between the opposite sides of the rhombus
gives the distance between the parallel faces of the solid.

Note. — It should be observed that this distance is equal
to the height of the tetrahedron whose side is equal to that
of the given octahedron.

{c) Develop the solid.

16. A solid is formed from a cube of 2 in. edge by planes
which pass through the middle points of its edges and cut off
the eight corners.

(a) Plan when resting on one of the cubic faces ; and
elevation on a plane taken at 30° to one of the vertical faces of
the primary cube.

{b) Flan when a diagonal of the primary cube is vertical.

17. A solid is formed from a cube i"5 in. edge by
producing a diagonal of the cube to equal distances of 2' 5 in. on
opposite sides, and then taking planes through the extremities of
the produced diagonal and the six lateral edges. Show plan and
elevation when resting on one face.

SOLIDS IN SIMPLE POSITIONS. 3/

1 8. A square, 2 in. side, in the H.P., is the base of an
o'^liqtie prism ; the edges, 4 in. long, are inclined at 55" and in
f'an parallel to a diagonal of the square. Elevation on a
vertical plane inclined 20° with the plans of the long edges.

Work also a section and the development of the prism.

1 9. An equilateral triangle abc, of 3 in. edge, centre v, is
ti e plan of a solid, bounded by plane faces, of which the points
A, B, C, andN, are respectively '25 in., '5 in., i in., and 3 '5 in.
h'gh. Determine the elevatio?i on a plane taken at an angle
of \^ with the line ab in plan, a?id work a section and
development of the solid.

II.

COMBINATIONS AND GROUPS OF SOLIDS.

I. A prism, with bases equilateral triangles, is placed
7vith one face on the horizontal plane. On the centre of it
rests a brick, g bj> $ bj> 4'$ inches, with one of the short
edges of a large face touching the ground, with t/ie centre of that
face touching the highest edge of the prism and with the face
itself inclined 45". Horizontal edge of prism 9 inches. The
side of the equilateral base to be determined from the question.
Scale \. Draw plan and elevation, and a second elevation
taking the vertical plane at 30° with any horizotital edge of the
prism.

Commence with an elevation so placed that both bases
of the prism coincide. To do this, draw first the elevation
of the brick, which will be simply a rectangular parallelo-
gram. Through the lowest corner take xy making the re-
(juired angle with the line that represents the face inclined
45°, and make the centre of this line the vertex of the tri-
angle, recollecting that the elevation of the base of that
triangle must according to the conditions be in xy, the
ground line. The work for \htplan is simple. The second
elevation will be determined by taking the new ground line
30" with any horizontal edge of the prism, and working to it.

COMBINATIONS AND GROCTPS OF SOLIDS. 39

2. Instead of the prism take a cylinder with brick touch-
ing it and otherwise disposed as above. The base of the cylin-
der to be determined from the question; i. e. so t/iat the horizon-
lal bisecting line of the face of the brick may be in contact with
I he cylinder.

3. A block, 3 in. square a?td i in. thick, is pierctd by a
rexagonal prism i in. side and 4 in. long. Axis of prism
2' as sing through centres of square faces of block and projecting
ipially on each side. Two faces of the prism to be parallel to
two narrow faces of the block.

(a) Plan when the block stands on one of its narrow faces
7nth two faces of the prism vertical: elevation on a vertical plane
inclined at 40" to the short edges of the horizontal faces of the
I lock.

Section and tru£ shape by art inclined plane.

[b) Plan when two of the faces of the prism not parallel to
those of the block are horizontal : elevation on a plane at 30" with
the axis of the prism.

Section and true shape by a vertical plane.

Commence with an auxiliary elevation so placed that
the projections of the bases coincide. In this position the
e evation of the block will be a square, and ■ of the prism a
haxagon, with its .centre in the centre of the square and two
c f its sides parallel to two sides of the square. From this
aixiliary elevation both the plans in {a) and ip) can be
c 2termined, and thence the required elevations.

4. The faces of a tetrahedron 3" edge are the bases of four
p nsms whose axes are equal in length to half the height of the
ti trahedron. Plan and elevation when an axis of the tetrahe-
a 'on is vertical.

5. Draw a line 5 in. long in the H.P. parallel to xy.

40 SOLID OR DESCRIPTIVE GEOMETRY,

From each end of this line cut off segmcjits of i'^ in., and
describe on them two regular pentagons. Consider these as the
bases of two prisms 2*5 in. long, and make a plan and
elevatio7i when spanned by a pentagonal Gothic arch — the
cetitres of the circles as shown in elevation being the middle
points of the elevation of the upper bases.

6. A solid is forjned of a slab i in. thick, having regular
hexagons of 2 in. side for its faces, and aii equilateral triangular
prism 4 i7i. long passing through the slab and projecting
equally from each of its faces — the long edges of the prism
containing the alternate short edges of the slab. Plan and
elevation when resting on one corner of the slab atid one edge of
the base of the prism.

7. Plan and elevation of a cylinder, 4 inches long,
diameter of base 2 inches, passing through the centre of a
circular slab, i ifich thick and 3 inches in diameter; the axis
of the cylifider at right angles to and extending equally from
both faces of the slab. The whole to be placed resting on the
rim of the slab and the rim of the cylinder.

8. Describe three circles with radii i'5 in., each touching
the other two, and join their centres. Co?isider the circles as the
plans of three spheres resting on the horizontal plane, and the
triangle as the plan of the base of a tetrahedron resting upon
them. Co?nplete plan and elevation of the whole, and draw
a second plan when ojie of the slant edges of the tetrahedron is
horizontal.

For the second plan take xy parallel to the elevation of
any slant edge of the tetrahedron.

9. A tetrahedron, edge 2*5 inches, stands on a square
block, with an edge parallel to a side of the sqjiare. The block

COMBINATIONS AND GROUPS OF SOLIDS. 4 1

is I inch thick, a?id its square faces have sides of 2,'S i^<^hes.
An axis of the tetrahedron, if produced, passes through the
centres of the squares.

Draw plan and elevation —
{a) When the square faces of block are horizontal and
the plane of elevation makes 30" with a side of a
square,
{b) When only one slant edge of tetrahedron is horizontal.

10. A circular slab, i inch thick, diameter 3 inches, rests
on ofie face. On it is placed a tetrahedron of 2 inches edge,
axis vertical and over the centre of the slab. Draw plan and
elevation, and a second plan when the rim of the slab and the
vertex of the tetrahedron tojich the horizontal plane.

For the second plan take xy touching the elevations of
the vertex and the rim of the slab.

11. A solid is formed from a cube of 7, in. edge by placing
a square pyramid on each of its faces; axis of pyramid equal
to half the diagonal of the face. Plan and elevation when one
face of the solid is horizontal.

12. A regular hexagonal slab 75 in. thick, side of
hexagon 1*5 in., stands on one of its rectangular faces, the long
edges of which itiake angles of 40° with xy. A sphere of
2*25 in. diameter is embedded in the slab, the centre of the
sphere coinciding with the centre of one of the hexagonal faces.
Show plan and elevation.

13. An equilateral triangle of ^ in. side is the outline of
the plan of four tetrahedrons, 2 in. edge each, three of which
stand upon the H. P. ; the fourth rests upon the others with
the three corners of its horizontal face coinciding with their
summits. Plan and elevation.

III.

PROBLEMS ON THE STRAIGHT LINE
AND PLANE.

Obs. To express points, lines and planes, see Problems
I to 7.

Problem I.

Given the projectmis ab, a'b' of a finite straight line to
determine:

1. The length of the segment AB of the given line, or
the distance between points A and B.

2. The angle of inclination of the line to each plane of
projection.

3. The traces of the given line and the distance between
them.

I. The line AB lies in each of its projecting planes:
'construct' or rabat either of these into the corresponding
plane of projection and the segment AB is shown.

Let the projecting plane for the plan ab revolve about
this line as an axis, into the horizontal plane.

STRAIGHT LINE AND PLANE.
Fig. 6.

From points a and b set off the projectors aA and bB,
equal to aa' and )8^' respectively, and at right angles to ab,
on the same side of it.

Join A and B ; AB is the required distance or length
of segment.

Observe. The length AB (being the hypotenuse of the
right-angled triangle of which the sides are ab and the
difference of the projectors Bb and Aa) may be found by
setting this difference .from b along bB to B' and joining
B' to a.

2. The angle a straight line makes with a given
plane is the angle between the line and its orthographic
projection on that plane, or the angle which AB makes
with ab, and is therefore the angle 6.

44 SOLID OR _DESCRIPTIVE GEOMETRY.

For the same reason ^ is the angle which the given line
makes with the vertical plane of projection.

3. The 'traces' of the line being points in the
planes of projection have each of them a projection in xy,
viz. h' and v (Theorem VI.) ; but the elevation db' pro-
duced contains h' , therefore the intersection of these lines,
namely, xy and the elevation, gives point //'; and Theorem
IV. and the plan of the line give point h. The correspond-
ing loci of V and v' determine those points.

The distance between the traces H and V of the given
line is obviously the length h F, as shown in figure.

Corollaries given below.

As an exercise on this problem determine the inclina-
tions of the given lines, AB, CD, EF, &c., and their traces,
and the lengths of the segments indicated. See Fig. 4.

Converse of Problem i :

(i) Given the traces of a straight line; or

(2) Its inclination and plan ; or

(3) Its length and inclination ; or

(4) Length and the position or loci of the extremities
of a finite portion ; to draw or determine the projections of
the line.

(i) If the traces hh' and vv' be given, —
Join h and v and h' and v'.
Then hv, h'v' are the projections required.

(2) If the plan hv be given making a given angle hva.
with xy, and the line inclined at an angle of 0 degrees.

At any point h in the given plan draw h V, making the
angle 6" with hv.

STRAIGHT LINE AND PLANE. 45

Assume points a and b in the plan, draw the perpen-
diculars aA and bB, and find points a' and b' by Theorems
IV. and V. and join a , b'.

Then ab, a'b' are the projections.

(3) If the length AB be given, its inclination 6, and
the position aa' of one extremity A.

Through a draw the indefinite plan hv.
Set a A at right angles to ab and equal to ad.
Through A draw h V, making the angle 6 with kv.
From A set off AB along this line and determine b and b'.
ab, a'b' are the required projections.

(4) If the length AB be given and the positions of
its extremities A and B with regard to the two planes.

Draw ad the projections of one extremity A, from the
conditions. Theorems IV. and V.

Also any point //' from the conditions for the other
extremity B and through / draw a parallel to xy. this is
a locus of b the other extremity of the plan ab required.
Determine the length ab"^ of the plan ; measure this length
from a to a point b in the parallel, and the position of ab
is fixed ; b' may be found of the height 'of /', and ab, a'b'
are determined.

The student will observe that the problems admit of
one, two, or more, or an infinite number of solutions. These
cases should be studied. See Problem 2.

* The base of a right-angled triangle, hypotenuse AB and altitude
equal the difference of the heights of A and B%

4^ SOLID OR DESCRIPTIVE GEOMETRY.

I. (3), Cor. I. To determine the ^ traces^ of a given plane
when the data are / , -

(i) Two straight lines {parallel or meeting.

(2) A straight line and a point,

(3) Three points.

(1) The horizontal and vertical traces of the given
lines must be found by Problem i, and the corresponding
traces of the plane drawn through the points thus found.
Theorem X.

(2) Find the traces of the given line.

Draw the projections of a line parallel to it, through
those of the given point (Theorem VII.), and find its traces.

Or, by Theorem IV., determine the projections of any
point qq' in the given straight line ; join the projections of
this point with pp' those of the given one; then the line
FQ lies in the given plane.

The traces of the two lines give those of the plane.

(3) Join the corresponding projections of the given
points; and determine the traces of the three lines of
which those joining lines are the projections.

The traces of the three lines give those of the plane.

Obs. If the given plane meets xy only three traces of
tie lines lying in it need be found ; e.g. two in the horizon^
tal trace of the plane and one in its vertical trace : the
latter joined to the point of intersection of the horizontal
trace of the plane with xy gives the vertical trace of the
plane.

I. Cor. 2 (see figure). Given the traces of a plane and
one projection zh of a straight line which lies in that plane;
to determine the second projection a'b' of the line. ■

STRAIGHT LINE AND PLANE.

Since the traces M' and vv' of the line must be m those
of the given plane (Theorem X.), determine these points,
then the line drawn through points /i', z/ is a If' the required
elevation.

Fig. 7.

Observe. Point H lies in the given straight line AB,
hence the projections M' of the point are in those of the
line. Thus ab is a locus of h^ and of v; the horizontal trace
Df the plane is a second locus of h, and xy of v, which
ooints are thus determined : perpendiculars to xy through
hem are loci of >^' and v\ Theorem IV., which points are
hus also determined.

SOLID OR DESCRIPTIVE GEOMETRY.

I. Cor. 3. To determine a point pp' of given altitude
a inches {or a") in a given plane hav'.

Fig. 8.

In the given vertical trace arj , take a point/', «" high,
Theorem V. ; find/ in x}\ Theorem IV. ; then/?)' is the re-
quired point.

I. Cor. 4. To determine a horizontal line a" high^ in a
given plane hav'. Fig. 8.

Determine/^" as before.

Through/ draw/<^ parallel to ho, Theorem VII.

Through/' draw/'/ parallel to xy.

Then/^, /'/ is the required line.

STRAIGHT LINE AND PLANE.

Problem II. ; ,

Through a given point aa' to draw a straight line in-
dined 6° to the horizontal plane.

Fig. 9.

If a right cone be determined having its vertex in
tlie given point A, axis equal to a' a and generatrix in-
< h'ned dy every generatrix of the cone fulfils the given
E. G. 4

50 SOLID OR DESCRIPTIVE GEOMETRY.

conditions. The conic surface is therefore a locus of the
required line.

To work the problem : "

Through point d draw a! s' making the angle d s' a equal
to b.

Through a draw as parallel to xy, then as, a's' is parallel
to the vertical plane and inclined 6".

With centre a and radius equal to as' describe the
circular base klm of the cone which gives the complete
solution.

Observe. The four lines AH, AK, AL, AM are in-
clined $°, and make equal angles with the vertical plane.

Problem III.

To determine a straight line which shall make a given
angle 6 with the horizontal plane and an angle <}> with the
vertical plane of projection.

A line inclined 6° will lie on the surface of a right
cone with axis vertical and generatrix inclined B" to the
base.

Determine such a cone having its vertex in point oo'
in the plane of elevation : Rps will be its circular base.
That generatrix of the cone which makes an angle ^ with
the vertical plane will be the line required. To find this.
Knowing the length dR of this generatrix, determine the
length dr of its elevation as in the figure ; and since the
-position of o' one extremity of it is fixed, that of the other,
viz. p', in xy, may be found as indicated. The perpendi-
cular to xy from /' is a locus of/, and intersects the cir-

STRAIGHT LINE AND PLANE.

cular arc Rps, which is another locus of /, which point
is thus determined. Join op^ o'p' \ it is the required line.
Fi-j. 10.

It is obvious that there are four such lines on the
uirface of the cone, and therefore there are four solutions
Df the problem, when 00' is a given point in the line.

It will also appear from the above, that the second
ingle <^ cannot exceed the complement of 6 or (90-^)
legrees.

4—2

52 SOLID OR DESCRIPTIVE GEOMETRY.

Problem IV.

To draw or determine the traces of a platie inclined 6 de-
grees and perpejidicular to the vertical plane of projection.
See Fig. 5, 'Planes.'

Draw the horizontal trace ql of the required plane at
right angles to xy, meeting it in point /; and through /
draw hn! making the angle ylm' equal 0 : qh/i' is the re-
quired plane.

It is obvious that Im and ly are the two perpendiculars
to ql, the horizontal trace, or common section of the
planes, at the point / in it, which form the measuring-angle
of the inclination of these planes to each other.

The plane of Im' and ly having been turned
into the horizontal plane, Im' in Fig. is no
longer perpendicular to ql.

Similarly (see Fig. 5, 'Planes') a plane n^o' may be
drawn at right angles to the horizontal and making an
angle <l> with the vertical plane of projection.

Problem V.

To determine the angles 6 and ^ which a given plane
hav' or am, mV makes with the planes of projection ; and
to determine the angle betweeti its traces.

Since a dihedral angle is measured by the plane angle
formed by two perpendiculars drawn one in each plane
to the same point in their common section, the plane of
this rectilineal angle is perpendicular to the 'trace' of the
given plane. ' Rabat ' the plane of this angle, revolving
it about that line of it which is in the plane of pro-
jection in question, and the required angle of inclination
is obtained.

STRAIGHl^ LINE AND PLANE.

Thus at point a in the horizontal trace ha. (Fig. ii)
two such perpendiculars av and aV were found, and at

Fig. ir.

p )int d' in the vertical trace perpendiculars dli and d'H,
a id the required angles shown.

To determine the magnitude of the angle ha.v' between
tl e traces of the plane.

Let the plane be 'constructed' or rabatted about its
h )rizontal trace ha until it coincides with the horizontal
p ane.

54 SOLID OR DESCRIPTIVE GEOMETRY.

Then the length a V measured from a along av pro
duced to V^ gives a second point in the second line of
the required angle. Join aFg, and ha.V^ is the angle be-^
tween the traces of the plane.

Problem VI.

Given the horizontal trace ah. of a plane inclined 6 to
determine its vertical trace av'.

At any point a in the horizontal trace draw av at right
angles to it and meeting xy in v. Fig. ii. Draw aV
making the angle z'aF equal to the given angle 0. From v
draw vV 2i\. right angles to aVy and from v set oflf the length
vv' at right angles to xy and equal to vK Then the
straight line through a and v is the vertical trace required.

Problem VII.

To determine the traces of a plane making an angle of
B degrees with the horizontal and a?i angle of ^ degrees with
the vertical plane.

When a plane touches a right cone it is inclined at the
angle which the generatrix of the cone makes with its base.

Determine two right cones (Fig. 1 2) with vertices q' and
v and bases tuw and sr'ni in the horizontal and vertical
planes respectively; the generatrix // making an angle 6
with its base tuw, and generatrix vs an angle <^ with its base
sr'm'. The common tangent plane to these cones is the
plane required.

The surfaces of the cones have a common point //
which is also on the surface of a common sphere enveloped

STRAIGHT LINE AND PLANE.

by them and having its centre o in xy. See Figure 13.
Hence the plane of points P, Q, and Fis that required.

Fig. 12. . '

/^l

\aI

: ?/

/\v

\)^ //

^^^^

/ y^

To work the problem.

Commence with one of the vertices, assuming point Q

or / in the vertical plane. Fig. 12. Draw^V indefinitely at

ight angles to xy^ meeting it ino; and draw ^V as above

lescribed, faking angle d with xy, on which let fall the

)erpendicular otf from 0; the circle described with centre 0

. .nd radius oc gives the common inscribed sphere of the

SOLID OR DESCRIPTIVE GEOMETRY.

two cones, vs drawn as above mentioned, making the angle
^ with xy, and touching this circle in d^ meets <^o produced
in the point v^ which is the second vertex. Tangents to
the bases of these cones through v and / meet xy in point
a and give the traces of the required plane qav.

Fig. 13.

Limits. The sum of the given inclinations or 5 + ^ lies
between 90" and 180**.

STRAIGHT LINE AND PLANE.

Obs. To express given combinations of points, lines,
and planes, see Problems 8 to 28.

Problem VIII.
To determine a straight line inclined &^ to lie in a given
plane qb'm'. ,

Fig. 14.

5^ SOLID OR DESCRIPTIVE GEOMETRY.

Limits. The angle ^ is the angle of inclination of the
given plane. It is therefore obvious that the angle of incli-
nation 0 of the given line must be not greater than <f>.
For no straight line can lie in a plane and have a greater
inclination than that of the plane.

Construction. Assume point aa' in the given plane.

Make the angle a'fia equal 0.

With centre a and radius a^ describe the arc ^ed meet-
ing the trace ^d' in d or d; then alf', db' or ad^ a'd' is the
required line.

Observe, that each of the lines AB and AD satisfies
the conditions of the problem.

The cone, vertex A and base deb, is the locus of all
lines through A inclined 6.

The given plane is a second locus of the required line.
And these loci intersect in the lines AB and AD only.

Problem IX {converse of Problem VIII).

Through the given straight line ab, a'b' {which is inclined
6) to draw a plane inclined ^.

Since the given line is to lie in the required plane, its
traces must lie in those of the plane, Theorem X. De-
termine these traces bb', vv'. Problem i.

Determine a right cone, with vertex in point aa' of the
given line, and base tuk in the horizontal plane, its gene-
ratrix making the given angle <^ with the base. Every
tangent plane to this cone is inclined <f>^, Theorem XL ;
and the straight line through b touching the base of the

STRAIGHT LINE AND PLANE.
. Fig. 15.

\«'

cone in / is the horizontal trace of such plane, Theorem
XII. By producing this line to meet xy in s and joining
. v', we obtain the vertical trace of tht required plane.

For this plane tsv, passing through points A and B in
the given line, contains that line; and touching the cone
l.a5 the required inclination.

Note : That the radius at of the base of the cone is taken
( qual to a^ in figure.

Also, that a second plane bwv', having ub for its hori-
2 ontal trace, likewise satisfies the conditions of the problem.

Generally, two planes are possible as long as </>" exceeds
l^ ; when «^'' is equal to 6° only one plane can be drawn,
i nd when <^'* is less than 6" the problem is impossible.

SOLID OR DESCRIPTIVE GEOMETRY.

Problem X.
To determine the magnitude of the angle which the two
given straight lines OH, OF make with each other.

Fig. I 6.

<r\

Determine the horizontal traces ff' and hH of the given
lines, then (Theorem X.) the horizontal trace //3 of the

STRAIGHT LINE AND PLANE, 6 1

plane of the lines passes through points / and h, and its
vertical trace ^v through ?/, vv being the vertical trace
oi OF.

'Rabat' the plane j'^Z'', revolving it about its tmce/ji
into the horizontal plane; then fOA the required angle is
exhibited.

The indefinite perpendicular on the horizontal trace
through the point <? is a locus of O (Theorem XX.), and the
length Oa is made equal to the hypotenuse of the right-
angled triangle, of which oa is the base and o'e the altitude.

Cor. To bisect the angle formed by two given straight
lines ^ as OY and OH.

'Rabat' the plane of the angle as above; and bisect
the angle fOh thus found by a straight line Os meeting the
horizontal trace used in point s; obtain s' in xy^ and join os,
ds' ', these are the projections of the required bisecting line.

Problem XI. (See Fig. 16.)

To draw a straight line oh, o'h' through a given point
pp' to make a given angle 9 with a given straight line of, o'f .

Determine the horizontal trace f^ of the plane of the
Tiven straight line OF and point P (Prob. i. Cor. i)
•Rabat' this plane, and through P so 'rabatted,' draw
;he line OP making the given angle 0 with OF in the
lorizontal plane. Determine the projections 00' of point O,
oin op, o'p', they are the projections of the required line.

Note: That the construction of this problem is made
nore simple by using a second plane of elevation perpen-
iicular to that of the angle. Theorem VI. (2).

SOLID OR DESCRIPTIVE GEOMETRY.

Problem XII.

To determine the projections of the intersection of two given
planes hav' and h(3v'.

Fig. 17. " . • ■

/'Sf

The planes intersect in a straight line, which lying in
both planes has its traces in those of the planes. Hence
z', in which the vertical traces meet, is the vertical trace of
the line of intersection, the plan v of this point is in xy^
and is found by a perpendicular to it (Theorem IV.). Simi-

STRAIGHT LINE AND PLANE. 63

larly, h is the horizontal trace of the line required, and //',
its elevation, is found in xy by a perpendicular to that line.

Join hVy h'v', these are the projections of the required
line.

Cor. I. To determine the intersection of three given planes.

Determine the intersection ZTF as in the problem, of
planes (i) and (2) : then the intersection AB of planes (2)
md (3). AB will meet HV in a point P which is the
ntersection required and is the vertex of the solid angle
"ormed by the three given planes.

Cor. 2. To determine a straight line which shall pass
^hrough a given point 00' and meet two given straight lines
■lb, a'b', and de, d'e'.

Determine the traces of the plane of the point O and
he line AB ; likewise those of the plane of point O and
' he straight line DE. Then the intersection of these two
)lanes is the line required.

Or, find the intersection of one of the given lines with
he plane containing the other and the given point O. The
ine drawn from this point of intersection to O will be the
ine required.

' ■ Problem XIII. ,

To determine a straight line at right angles to a given
j'lane qlm' which shall pass through a given point pp'.
:<ig. 18.

Since the projections of a, perpendicular to a plane are
] erpendiculars to the 'traces' of the plane, Theorem IX. ;
.^ s drawn through a point / at right angles to ql, and /'/ at
1 ght angles to lm\ are the projections of the required line.

SOLID OR DESCRIPTIVE GEOMETRY.

Conversely. To determine a plane at right angles to a
given straight line ps, p's' which shall pass through a given
point ii' in the same.

Fis. 1 8.

Draw the traces ^a, ea. of any plane da£\ by the pro-
blem, at right angles to the given line.

STRAIGHT LINE AND PLANE. 6$

Determine the projections of any line lying in this plane
and through the given point draw a parallel ih, i'h', to the
line determined, and find its trace hh'.

Through h draw gl at right angles to ps meeting xy in /,
and through / draw Im' at right angles to p's' : qlm' is the
required plane.

Cor. I. To determine the distance of a given point pp'
from a given plane qlm'.

Before reading this corollary study the following Pro-
)lem No. 14, and by it determine point //' in which the
])erpendicular PS to the plane intersects it.

Then by Problem 1 determine the length IP of the
])erpendicular of which ip, i'p' are the projections. IP is
the required distance of the given point /'from the plane.

Cor. 2. To determine two parallel planes at a given dis-
iince (n) inches from each other.

Determine a straight line BC, and in it mark off the
Imgth PQ^ points P and Q_ being («) inches apart; the
planes through these points at right angles to the line BC
are those required.

Problem XIV.

To determine the point of intersection of the given straight
li'ie ps, p's', with the given plane. Fig. 18.

Every plane which contains the given line meets the
g ven plane in a straight line which passes through the
r( quired point of intersection of the given straight line and
p ane. Assume any such plane pmm\ of which pm and
m 'n' are the traces.

Determine the line of intersection hm^ Jim! of this as-
si med plane with the given one.

E. G. 5

SOLID OR DESCRIPTIVE GEOMETRY.

The projections ii' of the Intersection of these two lines
HM and PS are those of the required point.

Another mode of solving this problem is to make an
elevation of the given line and plane on a plane at right
angles to the latter : point /' in this elevation will be the
intersection of the elevation of the line with the vertical
trace of the plane : whence the plan / may be obtained.

Problem XV.
To determine the angle which the two given planes qlm',
qq'm' make with each other, i. e. to determine their ^profile'
or '■measuring angle.''

Fig.

STRAIGHT LINE AND PLANE. 67

This plane-angle is formed by two straight lines drawn
one in each plane from a point in their common section
QM at right angles to that line. (Euc. Bk. xi. def. 6.)
The plane of these perpendiculars is therefore (Euc. Bk. xi.
\) at right angles to the line QM: hence if this plane be
determined and 'rabatted' about its horizontal trace,
he required angle wTr is shown.

To do this, assume the horizontal trace rw of the plane
of required angle, at right angles to qnt^ meeting it in j, and
1 he horizontal traces of the given planes in r and w.

Fold the plane about rw bringing the point T in QM
down into the horizontal plane in qm : join Tr, Tw.

The distance Ts is found by making it equal to Ts on
the vertical plane (as shown), this line being the common
section of the 'projecting plane' of the plan qjn with the
1 lane of the measuring angle.

Conversely. Given Q the profile or measuring angle of
t le inclination of two planes to each other, to determine the
t aces of the planes.

Draw the plane qq'm and determine the straight line
q m, q'ni in it.

Draw rs at right angles to qm and determine j 7" from
^T as before and join rT: then make the angle rTw — 6^
a id point w is determined, through which draw ql the
h Drizontal trace of the second plane : Im' its vertical trace
i^ drawn through the vertical trace of the line QM
("heorem X.).

SOLID OR DESCRIPTIVE GEOMETRY.

Problem XVI.

Through a given point aa' to draw a plane parallel to a
given plane qlm'.

Fig. 20.

Since if two parallel planes be cut by a third plane,
their sections with the third are parallel straight lines (Euc
XI. 1 6), therefore the traces of the given and required planes
will be parallel straight lines.

STRAIGHT LINE AND PLANE. 69

^ If therefore the given plane be not parallel to xy, one
ioint in either of the required traces being found both those
ines can be obtained.

Through the given point A let a straight line be drawn
] parallel to either trace of the given plane; the trace of this
line supplies the required point.

Therefore through A or ad draw the straight line A V
(>r av, a'v' parallel to ql which is a horizontal line; the
trace -mf of this line gives the point v' in the vertical trace
/?z/ of the required plane : and ph drawn parallel to ql
t irough p gives the horizontal trace.

Problem XVII.

Through a given point slr to draiv a straight tine which
s, mil be parallel to a given plane qlm' and have a given in-
clination 6. Fig. 20.

Obs. The angle 6 cannot exceed the angle of inclinar
ti m of the given plane.

Through the given point ad let a plane hfiv' be drawn
as in Problem 16, parallel to the given plane. And
through point A draw the straight line ar, dr', by Problem
8, to lie in the plane h^v\ and have an inclination of 0
d( grees.

Another mode of solving the problem is, to determine
in the given plane a straight line having the required incli-
ne tion, and to draw through the given point A a straight
lit e parallel to the line so formed.

SOLID OR DESCRIPTIVE GEOMETRY.

Problem XVIII.

In a given plane bb'd' to determine a straight line which
shall be perpendicular to a given straight line ba, b'a' lying
in that plane.

Fig. 11.

STRAIGHT LINE AND PLANE. 7 1

'Rabat' the given plane, revolving it about its hori-
zontal trace bb' into the horizontal plane. Then AB will
lie in the horizontal plane in the position Ab. Fig. 21.

The point b (or B) does not move during the rotation
Df the plane, being in the axis or trace bb' of the rotating
plane. The perpendicular to bb' through point a; is a locus
Df A, and the position of point A in it is found by taking
;he length b'd along the vertical trace b'd' of the plane
ind setting it off from the line bb'. A and b joined gives
he line Ab or AB 'rabatted.'

Through point A draw the indefinite line AD perpen-
dicular to Ab^ and determine the plan d of any point D
in this perpendicular. To do this : — Through D draw a
Jine at right angles to bb' for one locus of ^ (Theorem XX.).
vieasure the distance of point D from the line bb' and
; et this length from point b', along the vertical trace of the
])lane, to determine d'. A perpendicular from d to xy is
;, second locus of d. The intersection of these loci of d
determine that point. Join a and d, then ad, a'd' are the
jirojections of the required perpendicular.

Observe. In the same manner a straight line AD may
1 e drawn in a given plane to make any given angle Q with
'<. given straight line AB in that plane.

N.B. Theorems VI. and VIII. should be studied in
c onnexion with this problem and figure.

72 SOLID OR DESCRIPTIVE GEOMETRY.

-Problem XIX.

To determme the projections of a given plane figure, when
the inclination of its plane is given and that of a straight line
lying in the plane of the figure. Fig. 21.

I St. When the given line forms part of the perimeter
of the figure.

Let the given figure be a square, its plane inclined
0 degrees and one side AB of the figure incUned ^ de-
grees.

Determine the plane bb'd' : in it place the line ab,
a'b'. 'Rabat' the plane as in Problem 18, and on
AB describe the square ABCD : revolve the plane back
into its former position and obtain abed, the plan of the
figure.

2ndly. When the given straight line is a diagonal of
the given figure or its bisecting line : a tangent, a chord, or
a diameter of a given ellipse or circle.

Determine the plane bb'd' of the figure and the projec-
tion of the given line in it. ' Rabat ' the plane, and the
line as lying in it ; draw the given figure in the horizontal
plane having the required position with regard to the 'ra-
batted ' line : and from thence proceed as before to obtain
the projections of the entire figure when turned back into
the inclined position.

STRAIGHT LINE AND PLANE.

Problem XX.

Given two straight lines AB, AC making an angle BAG
of 0 degrees with each other; to determine the projections of
the lines when AB is inclined at an angle of a degrees and
AC inclined /3.

Fig. 22.

3C.

Observe the conditions of the problem are impossible if
a+^ + ^exceed 180° :

74 SOLID OR DESCRIPTIVE GEOMETRY.

if a + ^ + e equal 180"
the plans of AB and A C will be in one straight line, /. e.
the plane of the angle BAC will be a perpendicular one.

The case in which a+^ + d is less than 180" will be
that here considered.

Draw the angle BA C on the horizontal plane. It will
be necessary to revolve the plane of the angle BAC into
the inclined position it will occupy when its containing
lines have the required inclinations. For this purpose a
horizontal line or trace BM of the plane must be disco-
vered, about which the plane is to revolve. Assume a
point B in AB and

make the atigle ABr = a" and ,
ArB = go\

Then Br will be the length of the plan of BA when BA
is in the required position.

Now for the same height Ar{ = A/), by the aid of a
tangent fM to the arc rsf determine
the angle AMt= ^ and
... ... AtM= 90" : and point M is found.

The straight line BM is the required horizontal trace of
the given angle. A plane of elevation having its xy at right
angles to this line will be always perpendicular to the
plane of that angle ; and the elevation of the angle upon it
will be in the vertical trace, Theorem VI. 2.

The elevation of the path of point A is the circular
arc a"a', in which a' is determined by the parallel to xy
at the distance Ar.

From a the plan a is found in AB (Theorem IV.).

STRAIGHT LINE AND PLANE. 75

Then BaM is the required plan, and 8'a' the elevation
and vertical trace of the plane of the given angle. «"SV is
the angle of the inclination of the plane of the lines.

Cor. To determine three straight lines meeting in a point,
each line perpendicular to the other two.

Proceed, as in this problem, to obtain the projections
of the two lines AB and AC; and through point aa', by
Problem 13, determine the projections ae, ae' of the third
line, at right angles to the plane MM. AB, A C, AE are
the required lines.

Problem XXI.

To determine the projection of a given plane figure when
the heights of three points in the plane of the figure are given.

Limits. When the given heights are equal, the projec-
tion of the figure is an equal and similar figure to the
given one.

When the given heights are unequal, the problem is a
possible one only when the difference between the heights
of any two points does not exceed the distance between
these points.

If this difference is equal to that distance the plane of
:he figure is vertical; and the projection required will be a
straight line.

Let the three given points A, B and C be at the angles
)f an equilateral triangle of m inches side, and the heights
)f A, B and C be /, ^ and r inches respectively : to deter-
nine the plan abc of the triangle. Fig. 23.

(i) Draw .^^C in the horizontal plane.

7^ SOLID OR DESCRIPTIVE GEOMETRY.

Fig- 23.

le^eZ

^ ^^iU-

(2) Determine point h in ^C which will have the same
height as B, viz. q inches. Then hB is a horizontal line
when the triangle is in its inclined position.

(3) By the aid of an elevation at right angles to hB as
shown, the plan. a^^ is determined.

STRAIGHT LINE AND PLANE. . 77

Problem XXII.

To determine two parallel planes which shall contain
the given straight lines AB or ab, aV and CD or cd, c'd'.
ist. If the given lines A£ and CZ) are parallel.

Determine the horizontal traces of the lines; parallel
lines through these will be the horizontal traces of the
parallel planes required, whence the vertical traces can be
obtained.

2ndly. If the given lines AB and CD be neither
parallel nor in the same plane.

Through any point A in AB draw a straight line AQ
parallel to CD.

Determine the horizontal traces of the lines AB and
AQ; the straight line drawn through these points is the
horizontal trace of one of the required planes.

Determine the horizontal trace of CD; through this
ooint a parallel line to the trace just found is the horizon-
al trace of the second plane required : and the vertical
races of the two planes can be obtained.

3rdly. If the given lines be not parallel but in the
; ame plane, the problem is impossible.

Problem XXIII.

To determine a straight line at right angles to each of two
I iven straight lines AB and CD.

I St. If the given lines be parallel.

Any Straight line drawn in the plane of the parallels per-
1 endicular to one of the lines is also perpendicular to the
c ther.

7^ SOLID OR DESCRIPTIVE GEOMETRY.

2ndly. When the given lines are neither parallel nor
in the same plane.

From any point P in AB draw a line PQ parallel to
CD : then the plane of AB and PQ is parallel to CD.

From any point C in CD draw CS perpendicular to
the plane of APQ, meeting it in .S*. A parallel to PQ or
CD through point S meets AB in H. Then HT, parallel
to CS from IT, meets CD in point 7^ and is the required
perpendicular to the given lines.

Problem XXIV.

To determine the angle of inclination B of a given straight
line ab, a'b' to a given plane qlm'. Fig. 24.

In this problem it is convenient to determine the com-
plement of the angle required.

This complementary angle is the angle between the
given straight line and a line meeting it in point A and per-
pendicular to the given plane.

Determine this perpendicular ah, a'h\ by Problem XIII.,
and find its horizontal trace ha'; likewise bb' that of the
given line ; the line hb through these points is the horizon-
tal trace of the plane of the lines. 'Rabat ' their plane
about hb bringing A into the horizontal plane. Ah, Ab,
when drawn, give bAh the complementary angle or 90 — 6,
and 6 is the angle sought.

Conversely :

To determine a straight line through a given point aa', to
make a given angle 6 with a given plane qlm'.

Through ad draw ah, a'h' at right angles to the given
plane, and determine its horizontal trace hh'. Through

STRAIGHT LINE AND PLANE.
Fig. 24.

p Dint h draw in any direction the line hh as the horizontal
t; ace of the plane of the perpendicular and the required
li le. Fold this plane about hb into the horizontal one and
tlirough point A thus 'rabatted' let Ab making an angle

So SOLID OR DESCRIPTIVE GEOMETRY.

go-0 with A/i meet the trace Ad in point d. Then d' being
found in xy and joined to a, the required line ad, a'b is
determined.

Observe. An infinite number of such lines ab, a'b' can be
determined to satisfy the given conditions; and their 'locus'
is the right-conical surface having its circular base in the
given plane and its generatrix inclined 6 degrees to the
same.

Problem XXV.

To determine two planes which shall be perpendicular to
each other and inclined at angles of 0 and <f> degrees respectively
to the horizontal plane.

It will be obvious that the angle of inclination of the
second plane cannot be less than the complement of that of
the first ; or ^ not less than 90 - ^. -

Therefore

if ^ is greater than 90 - B, there are two solutions :
...<^ is equal to 90 - Q, there is one solution ;
...^ is less th^n 90 - 0, there is no solution.
This will appear by what follows.

Draw the plane qltrl inclined B degrees; also the straight
]ine/w,/V;/ at right angles to it (Problem 13), and find the
horizontal trace pp' of PM.

Then every plane which contains that line is at right
angles to the plane B (Eucl. xi. 18). Wherefore if such a
plane revolve about PM, as an axis, until it makes an angle
^ with the horizontal plane, it is the plane required.

STRAIGHT LINE AND PLANE.
Fig. 25.

Let it therefore be made to touch a right cone, vertex
ir any point mm in PM and generatrix m'^ making an
a: igle of ^ degrees with its base gko. The horizontal trace
o this plane must touch ^/^^ (Theorem XL).

E. G. 6

S2 SOLID OR DESCRIPTIVE GEOMETRY.

And since the plane is to contain PM, the horizontal
trace rp of the plane must pass through p (Theorem X.).

Through the point mm' draw the horizontal line mn^
m'n parallel to the trace rp^ and determine its vertical
trace nn.

Through ii and s draw the vertical trace of the plane.

Observe that the plane rsn' by containing PM is per-
pendicular to plane B ; and, by touching the cone de-
scribed, has the required inclination <^ ; it thus satisfies the
conditions of the problem.

Note also that the inclination of PM is 90 — ^ ; since if
a straight line and an inclined plane be perpendicular to
each other their inclinations are complementary.

Cor. To determine three planes each perpendicular to the
other two.

For the first and second, determine planes qhn and rsfi
as in the above problem.

Find their intersection mq^ m'^ by Problem 1 2.

The plane which is perpendicular to MQ is perpendi-
cular to the planes which meet in this line (Eucl. xi. 19).
Therefore such plane, determined by Problem 13, is the
third plane required.

Problem XXVI.

To determine a plane inclined at an angle of 6 degrees and
making a given angle a with a given inclined plane.

Let qlm' be the given inclined plane.

To fulfil both conditions, the required plane must touch
two right cones having a common vertex in any point vtf

STRAIGHT LINE AND PLANE.
Fig 25.

^i \ / 1 \\

a,/

^ /

! t/

>^tr V

V ith circular bases op and ^V in the horizontal and inclined

f lanes respectively* ; the generatrix of the former making an

* b' is the point of intersection of the lines z/V and lm\ and is not
li ttered in the fig. : also the plan of the circular base b's' is not shewn
ii the fig.

6—2

§4 SOLID OR DESCRIPTIVE GEOMETRY.

angle Q with its base, and that of the latter an angle of
a degrees with the inclined plane.

The horizontal trace pi of the common tangent plane to
the cones is (Theorem XI T.) a tangent to the horizontal
traces of the two conic surfaces, /. e. the trace // touches the
circle op and the ellipse rtu. Since four such lines can be
drawn, as appears by the figure, it is obvious that in the case
here indicated four planes can be determined which satisfy
the given conditions of the problem.

Should the circle op fall within the ellipse, without
touching it, the given conditions are impossible.

Problem XXVII.

Given three lines ag, ak, ar meeting in a point a ,• as the
plans of the straight lines AG, AK, AR, each of which is at
right angles to the plane of the other two ; to determine an
elevation of the lines.

Since the plane of lines AK, AR is perpendicular to
the line AG (Eucl. xi, 4), the horizontal trace, cd, of the
plane of AK, AR is perpendicular to ag, the plan of AG
(Theorem IX.).

Therefore draw cd at right angles to ga produced, and
meeting ah, ar in points c and d.

ce drawn through c at right angles to ar, and de drawn
through d at right angles to ak, will meet ag in e and be the
horizontal traces of the other two planes (Theorem IX.).

Take the projecting plane of the plan oi AG for a plane
of elevation, and let it revolve about the line ag as an xy.

This plane cuts the plane of AK and AR in AP or a'p,
which is therefore perpendicular to AG or a'g in the plane
of elevation (Eucl. xi. def. 3).

STRAIGHT LINE AND PLANE.
Fig. 37.

^ — r

Therefore on ep describe the semicircle ec^p, through a
d aw aa! perpendicular to xy, and ea^ a'p are the required
elevations of the three lines (Theorem VI. 2).

Problem XXVIII.

Given two points B and C at unequal distances from a given
pi ine, to determine a point Q in the plane at which the straight
Ih es QB and QC shall make equal angles with the plane.

86 SOLID OR DESCRIPTIVE GEOMETRY.

From either of the points, as B^ draw the straight line
BR at right angles to the given plane (Problem 13), meeting
it in S, determined by Problem 14.

Make SR - SB, point R being on the opposite side of
the plane to B ; join R, C, cutting the given plane in point
Q '• Q.B, QC are the required lines.

Exercises on Chapter III.

1. Determine the projections of four points, each 2
inches from the planes of projection, viz. one in each
dihedral angle.

2. Determine four points each in a plane of projection
and 3 inches from xy ; no two of them to be on the same
side of that line.

Supposing the loci or projecting lines of these points
(Theorem IV.) to be 2 inches apart, determine the distance
between any two points.

3. A point is 2*5 inches from each plane of projection,
draw two lines through it inclined 50°; their plans making
angles of 35" with xy. Problem i, converse (2).

Determine the traces of the lines : also the angle which
the lines make with each other. Problem 10.

4. Through a point 2 inches from each plane of pro-
jection draw lines inclined 30", making angles of 40° with
the vertical plane. Problem 3 and Theorem IX.

STRAIGHT LINE AND PLANE. 8/

5. A straight line in the horizontal plane makes an
angle of 45" with xy and is a trace of planes which are
inclined 50"; find their vertical traces. Problem 3, "tan-
gent planes to cones," Chap. viii.

6. A regular hexagon, side 2 inches, has its diameter
and adjacent side inclined 30" and 45° respectively : draw
its plan and determine the incHnation of its plane. Pro-
blem 20.

7. Draw two parallel planes 2 inches apart and inclined
5o°. Problems i and 13, and Problem 25, note.

8. A plane is inclined 40", and is at right angles to
mother which is inclined 70°: draw the traces of the planes.
Problem 25.

9. The horizontal traces of two planes, inclined 45°
md 30° respectively, make an angle of 60": determine a
hird plane at right angles to the given ones. Problems 1 2

and 13.

10. Two planes are inclined 60" and 70° respectively
;.nd make an angle of 50" with each other: determine
1 heir vertical traces on a plane not perpendicular to them,
^'roblem 26. • .

11. A regular pentagon, side 2*5 inches, has three of
i;s angular points respectively i, 2, and 3 inches high.
Draw its plan. Problem 21.

12. A square, side 3 inches, lies in a plane inclined
^5", one side of the figure makes an angle of 40" with
t le horizontal trace of its plane. Draw its plan. Pro-
l lem 18.

88 SOLID OR DESCRIPTIVE GEOMETRY.

13. A plane is inclined 80" and makes an angle of
60** with a plane which is inclined 50". Draw the traces of ;
the planes. Problem 26. 1

14. The two traces of a plane make an angle of 50"
with each other and make equal angles with xy : determine
these latter angles and the inclination of the plane.

15. Three straight lines are perpendicular to one
another and two of them are inclined 25" and 40**: deter-
mine the inclination of the third line. Problems 20 and 13.

16. The horizontal and vertical traces of a plane
make angles of 30" and 55° respectively with xy; a straight
line parallel to xy is 2 inches from the vertical and 3 inches
from the horizontal plane : find the intersection of the given
line and plane. Problem 14.

17. Two planes contain a right angle; one of them
is inclined 60" and their intersection is inclined 50". Draw
the traces of the planes and find the inclination of the
second. Problems 8 and 13.

18. The hypotenuse of a right-angled triangle is 4
inches long, and horizontal; the plans of the sides make
an angle of 115° with each other: determine the lengths
and inclinations of the sides. Can more than one set of
answers be given ? Eucl. iii. t,},, and Problem 20.

19. A straight line inclined 40" lies in a plane in-
clined 60" ; determine a plane containing the given line
and perpendicular to the given plane. Problems 8 and 13.

20. Through a point 3 inches high draw three planes,
inclinations 35°, 45", and 60" respectively; forming a tri-

STRAIGHT LINE AND PLANE. 89

hedral angle at the given point. Show the real mag-
nitude of the angle of each face of the solid angle. Pro-
blems 12 and 10.

21. The plans of two lines contain an angle of no".
The lines themselves are at right angles, and one is in-
clined 27". Find the inclination of the other. Science
Exam. Hon. 1872.

(i) Draw plans of the lines and make an elevation of
3ne, assumed as that inclined 27°, on a vertical plane
;aken parallel to it.

(2) Determine a plane perpendicular to this line andcon-
;aining the other. The point in which the horizontal trace of
his plane meets the plan of the second line is the horizontal
race of the line whose angle of inclination is required.

(3) Determine this angle by Prob. i.

22. Two lines are inclined at 30" and 40". They are
; inches apart where they are nearest together, and this
ine of 2 inches is inclined 28". Show plan and elevation.
Science Exam. Hon. 1873.

(i) Determine a line inclined 28°, and mark off a
•egment of this line 2 inches in length. Prob. i.

(2) Draw two parallel planes at right angles to line of
;8" passing through the extremities of the segment of 2
nches.

(3) From the point where line of 28" meets one plane
flraw a line lying in that plane and inclined at 40", and
irom the point where the former line meets the parallel
])lane draw another line lying in that plane and inclined

90 SOLID OR DESCRIPTIVE GEOMETRY.

The two lines so drawn are those required.

Note. Four pairs of Hnes fulfilling the conditions are
possible.

Another solution may be based upon Problems 20,
r (converse), 13 (converse) and 8.

23. Given a plane by its traces, and two lines, not
passing through the same point, by their projections. The
lines to be neither in the given plane nor parallel to it.
Find the projections of a line which shall meet the two
given lines, be parallel to the given plane, and at a given
distance from it.

Determine a plane parallel to the given plane at a
distance from it equal to the given distance of the parallel
line from the same plane. Problem 13, Cor. 2.

Find the points of intersection of the two given lines
with this plane. Prob. 14. The line joining the two points
thus found is the line required.

24. Two horizontal lines AB, AC, contain an angle
of 56°, a plane inclined at 30" contains AB, another in-
clined at 65" contains AC. Draw two lines passing through
A, each inclined at 20" and lying one in each plane. Deter-
mine the angle between these two lines. Science Exam.
Hon. 1872. Problems 4, 8 and 10.

25. A hne TV 3*3 inches long has its extremities T
and Fin the horizontal and vertical planes respectively, it
is inclined to the horizontal plane at 40°, and has its trace
T 2X 75 inch from the ground line. Draw its plan and
elevation. Science JSxam. 1S6S.

STRAIGHT LINE AND PLANE. 9 1

26. An indefinite line is inclined to the horizontal
plane at 40" and makes an angle of 30' with the vertical
plane of projection, the distance between its traces being
3 "5 inches. Science Exam. Hon. 1869.

27. AB is a line parallel to the horizontal plane, AC
s a line parallel to the vertical plane. The angles bac be-
ween the plans and b'a'c' between the elevations are 120".
Find the real angle between the lines. Science Exam. 187 1.

28. Determine a line inclined at 33" lying in a plane
inclined at 50". This is the orthographic projection on
1 hat plane of a line which makes an angle of 40" with it.
Determine the projections of this latter line on the co-
ordinate planes and its inclination. Science Exam. Hon.
:87i.

29. Find the traces of a plane which contains three
assumed points — one in each plane of projection and
ttie third in space.

30. Bisect the dihedral angles formed by the three
J lanes in the exercise 20 above.

31. Find the centre of a sphere of given radius that
s lall touch three assumed planes.

IV.

SOLIDS WITH THE INCLINATIONS OF THE
PLANE OF ONE FACE AND OF ONE EDGE
OR LINE IN THAT FACE GIVEN.

Before commencing this chapter the student should
refer to Chapter iii. Problems i8 and 19.

I. Plan and elevation of a pyramid, 4 inches high, with
square base 0/2'$ inches side, when its base is inclined e^o'^,
and no side horizontal.

(i) Determine by its traces a plane 50" at right angles
to the vertical plane ; /. e. draw its horizontal trace at right
angles to, and its vertical trace at 50" with, the ground line.

(2) 'Rabat,' that is, fold this plane on its horizon-
tal trace into the horizontal plane, and draw the square as
then in the horizontal plane; i.e. simply draw the square,
taking care that no side is parallel to xy.

(3) Lift the plane back to the required position and
determine required plan and elevation of the square base.
Remember that this plane being at right angles to the verti-
cal plane will have all points in it shown in this elevation ///
the vertical trace (Theorem VI. 2). The square base will
therefore have its elevation wholly in the vertical trace.

SOLIDS WITH ONE FACE GIVEN. 93

To draw this elevation, take the distances from the
:orners of the square to the horizontal trace when the plane
s folded down, and set them along the vertical trace from
ts junction with xy. This will give these points in eleva-
'ion. Their plans will lie in the perpendiculars to xy from
\\&\r elevations (Theorem IV.) and also in the perpendiculars

0 the horizontal trace from the points as 'rabatted'
Theorem XX.). Thus let the point of which a' is the eleva-

lion when folded down or 'rabatted' be marked A. A
] )erpendicular from a to xy contains the plan a, as also
does the perpendicular from A to the horizontal trace.
The plan a therefore lies in the intersection of these per-
] )endiculars ; in other words, these perpendiculars are loci
( if the point a, and their intersection gives the point itself.

(4) The axis will be drawn from the centre of the base
ii plan and elevation at right angles to the traces, and
being parallel to the vertical plane will be set in true length
tn elevation. It only remains to join the proper points, and
tie projections of the solid are completed.

2. Flan and elevation of a cube of i-e^ inches edge, when
one face is inclined 60°, and no side horizontal.

3. Flan and elevation of a pyramid, 4 inches high, with
t'-gular hexagonal base of i'^ in. side inclined 70", 7io side
h orizontal, and its lowest corner i inch above the ground.

4. Flan and elevation of the pyrajnid in the last problem

1 'hen the base is inclined 70" and one side of the base 45".

In this problem there is given not only the inclina-
t on of the plane of the base, but also of one edge in that
tase.

94 SOLID OR DESCRIPTIVE GEOMETRY.

Limits. The problem is therefore to draw a line of
given inclination in a plane of given inclination, and this is
always possible if the inclination of the line is not greater than
that of the plane.

(i) Determine by its traces the required plane in-
clined 70" and at right angles to the vertical plane. Assume
any point A in the plane; its elevation a' is by the position
of the plane in the vertical trace; its plan « on a perpendi-
cular to xy from a', and for convenience it, i.e. the plan a,
may be assumed at the point where this perpendicular
meets xy; and from a' draw am to meet the ground line
at the given angle 45°. Rotate the triangle thus formed
on the vertical line a'a, the projector of the point A, until
the point m meets the horizontal trace. As A and fn are
now both in the given plane, the line joining a and m
will be the plan of Am and will fulfil the conditions by
being the plan of a line wholly in plane 70° and itself
incHned 45".

(2) * Rabat' plane 70° on its horizontal trace into
the horizontal plane. On the line Am, thus brought into
the horizontal plane, describe the required hexagon, and
complete plan and elevation of the whole as in former
problems.

5. Plan and elevation of a cube of 2'^ inches edge when
the plane of one face is inclitied 65" and one diagonal of that
face 25*'.

6. Plan and elevation of a tetrahedron of 3 inches edge
with one face 70" and one edge in that face 40°. Determine
also the inclinations of the other edges of the solid.

7. Plan and elevation of hexagonal right prism when the
plane of one of its faces ABCD is inclined 50° and the edge
AB of the base inclined 35".

SOLIDS WITH ONE FACE GIVEN. 95

(i) Determine the plan of rectangular face ABCD as
in preceding problems.

(2) It will next be necessary to determine the plane of
the base, which, by the definition of the solid, is perpendi-
cular to the plane of the face. The line AB already deter-
mined is in the base, being common to both planes, and its
horizontal trace will therefore be a point in the horizontal
trace of the required plane of base. And as the projections
of a perpendicular to a plane are perpendicular to its traces,
the horizontal trace of the plane of the base may be at
once drawn through this point at right angles to the long
edges of the prism, that is, in this case to be or ad.

(3) Take a ground line at right angles to the horizon-
tal trace of plane of base, and set up the height of ^ or ^
(both known points) in the new elevation. By the position
Df the new plane of elevation, the vertical trace will contain
:he elevations of both A and B^ and therefore can at once
DC drawn.

Fold the plane of the base about its horizontal trace to
jring AB down into the horizontal plane. On the AB
has rabatted describe hexagon and complete as in former
)roblems.

8. The plane containing one edge of a tetrahedron and
(isecting aiiother is inclined at 50", and the former edge is
inclined at 30". Draw plati and elevation of the solid, the
idge beingT, inches. Science Exam. Hon. 1870.

(i) Determine plane 50° containing line inclined 30",
; nd ' rabat ' it.

(2) Measure off on the rabatted line 3 inches, and on
1 lis segment as a base draw an isosceles triangle having its

96 SOLID OR DESCRIPTIVE GEOMETRY.

equal sides equal to the altitude of an equilateral triangle of 3
inches side. The figure thus drawn is the true form of the
section of the tetrahedron made by the given plane.

(3) Rerabat plane, and from vertex of triangle draw
a perpendicular to the plane and produce it on both
sides to equal distances of 1*5 in. On joining the
extremities of this perpendicular with the extremities of the
line of 3 inches measured off on line of 30°, we obtain the
projections of tetrahedron required. ^

9. Plan and elevation of octahedron 3 in. edge.

a. Plane of one face inclined at 65" and one edge of that
face at 30°.

/8. Plane containing two axes inclined at 70" and one edge
of the solid in that plane at 2^^^.

y. When two of its axes are inclined at 60° and 20"
respectively.

10. Pentagonal pyramid, side of base i"5 in., axis 3*5 in.,
7vhen the plane containing the axis and one of the slant edges
is inclined at 50" and the axis at 2^.

1 1. Octagonal right prism, side of base 75 in., axis 3 in.,
when the latter is inclined at 30" and one side of the base at 2 7".

SOLIDS WITH THE INCLINATIONS OF TWO
ADJACENT EDGES OR LINES GIVEN.

Limits. It must be remembered that the sum of these
inclinations with the angle contained by the t7vo edges must not
exceed i8o°.

For these problems refer to Chapter in., Problem 20.

I. Plan and elevation of a cube when the inclinations of
two adjacent edges as BA, BC, 20" and 48", are given.

(i) Draw the square A BCD to represent the face
rabatted into the horizontal plane. Then to find a
horizontal in the plane of this face : — At any point E in
BA, or BA produced, make the angle BEF 20". Then
if a perpendicular be drawn from B to EF, it will be the
length of the line expressing the height to which B, when
in required position, will be raised above horizontal plane
:ontaining point E. From ^ as a centre with this perpendi-
;ular for radius, describe a circle. The tangent to this
:ircle to make an angle of 48° {i.e. the inclination of the
)ther edge) with BC will give a point H where it meets
BC in the horizontal required. E being another point in it,
he straight line through H and E will be the horizontal
lought in plane of face ABCD.

(2) Assume a plane of elevation at right angles to this
E. G. - 7

98 SOLID OR DESCRIPTIVE GEOMETRY.

inclined plane of face ABCD. The height to which point
B is to be raised being known, a line of level that distance
above the xy taken will contain the elevation of B. The
rest of the work is similar to that in preceding problems.

Note. To understand the principles of this construction,
the student had better cut a card-board model, by which to
illustrate the folding up and down of the plane of the
angles and of the lines.

2. Plan and elevation of a tetrahedron when two of the
adjacent edges are inclined 30" and 45®.

3. Flan and elevation of hexagonal right prism when the
two edges AB, AD, in one face are inclined respectively 45°
and 2 7", AB being the edge of the base.

Determine the plan of the rectangular face ABCD as in
Problem i, and refer to explanations of Problem 7, Chapter
IV., to find the plane of base and to complete the projections
of the solid.

4. Octagonal prism, 3 in. long, side of base i in., when
its axis is inclined at 40" and an edge of the base «/ 27".

5. Cube, 2*5 in. edge, when two diagonals of the solid are
inclined at 40" and 35" respectively.

6. Octahedron 3 in. edge. \
a. When two adjacent edges in one face are inclined at

32" and 40°.

/8. When two adjacent edges not in one face are at 20°
and 48".

7. A regular pentagonal pyramid, 2 -5 in. side, axis 3 '5
in., has three of the angular points of its base taken in order,
r /;/., 2 in., and 3 /;/. high respectively. Plan and elevation.

VI.

SOLIDS WITH THE INCLINATIONS OF TWO
ADJACENT FACES GIVEN.

For these problems refer to Chapter in., Problem 25.

Limits. Case i. If the given planes are at right
angles to each other, their inclination being 6" and ^'*,
6" + <f>^ must not be less than 90".

Case 2. If the given planes are not at right angles
to each other, refer to Chapter iii.. Problem 26.

I. Plan and elevation of a cube of 3 inches edge, when
the inclinations, 45° and 70°, of the planes of two adjacent
faces are given.

(i) Determine the traces of plane 45° assumed at right
angles to the vertical plane.

(2) Determine the projections of a perpendicular to
plane 45", in order to determine plane 70", since every
plane containing this perpendicular is at right angles to
the former plane.

' (3) Determine horizontal trace of plane 70". To do
this, make any point in the perpendicular — e.g. that in which
it pierces the plane of 45° — the vertex of a right cone, genera-
trix inclined 70°, base in horizontal plane. Then, since

7—2

100 SOLID OR DESCRIPTIVE GEOMETRY.

every plane touching this cone is inclined 70", a tangent to
the circular base of the cone will be the horizontal trace of
such a plane, and if it be drawn as well through the horizon-
tal trace of the perpendicular to plane 45**, it will be the
horizontal trace required, i.e. of a plane perpendicular to
plane 45** and inclined itself 70°.

(4) Determine the plan of the intersection of the two
planes. One point in this plan will be at the intersection
of the horizontal traces of the two planes, and a second
point will be the plan of the point in which the perpendicular
pierces the plane. The line drawn through these two points
will be the plan of the intersection required. Rabat one of
the planes (say that at 45") containing this line into the hori-
zontal plane, draw the square face of the cube on the line so
rabatted, and determine plan of the face therefrom. Complete
plan and elevation of the cube as in former problems.

2. A prism 3 inches long has an equilateral triangle of
2 "3 inches side for its base ; draw its plan and an elevation of it
7vhen the planes of its base and one face are equally inclined at
60" to the horizontal plane. Science Exam. Hon. 1869.

3. Plan and elevation of a prism with base a regular
hexagon. Plane of one face inclined 48". Base 68".

(i) Determine the plan of the rectangular face of
prism as for the face of the cube first determined in
problem i.

(2) On a vertical plane at right angles to plane 68",
i.e. with its ground line taken at right angles to the
horizontal trace of plane 68", determine the vertical trace
of plane 68", and rabat this plane into the horizontal
plane. On AB, the edge common to plane of face and
plane of base so rabatted, describe the regular hexagon.

SOLIDS WITH TWO ADJACENT FACES GIVEN. 10 1

Determine plan therefrom and complete problem as in former
cases.

4. Plan and elevation of a tetrahedron. Planes of adja-
cent faces 6 and <^.

For example, let 6 - 45'' and ^ = 75".
(i) Determine the angle a between the two adjacent
faces of the solid.

(2) Draw the traces of plane 6, taking the plane at
right angles to the vertical plane of elevation.

(3) Determine a second plane inclined «^ and making
an angle of a with plane 6, Chapter in. Problem 26. The
intersection of these two planes B and ^ is the indefinite
edge of the solid. Rabat plane 0 or ^, and on the
intersection so rabatted mark off the length of the given
edge of the tetrahedron, describe the face upon it, and
proceed as in former problems.

5. Plan and elevation of a tetrahedron with one face
vertical and an adjacent face inclined 50".

Determine the angle Q between adjacent faces. •

Determine a plane inclined 50° to the horizontal plane
and 0 to the vertical plane by Chapter in. Problem 7.

On the vertical trace place c^b' the real length of an edge
of the tetrahedron, and on it describe the equilateral triangle
db'd. Find the centre v'. a'b'c'v' is the required eleva-
tion.

In plan the vertical face c^b'c' will be a straight line abc
parallel to xy, and the plan v of the vertex will be a point
in the projector drawn from v' perpendicular to xy at a
distance in front of abc equal in length to the axis of the
tetrahedron. Lines joining v with the points abc of the
vertical face complete the plan of the solid.

102 SOLID OR DESCRIPTIVE GEOMETRY.

Note. This problem should also be worked by the
general method as given in Problem 4.

6. Substitute the octahedron for the tetrahedron in Probs.
4 and 5.

7. Pentagonal pyramid in Prob. 10, Chap. iv.

a. When the planes of two adjacent faces are inclined at
6° and f^\

/8. When the plane of the base is inclined at 6° and of one
face at ^.

8. Plan and elevation of right prism, regular hexagonal
base, when the plane of ofie face is inclined at 50" and its axis
a/ 35". To be worked by the method given for this Chapter.

VII.

PROPERTIES OF CURVED SURFACES AND
PROBLEMS BASED THEREON.

The investigation of the properties of curved surfaces
can be greatly simplified by making certain assumptions as
to the mode in which these surfaces have been generated by
the motion of a line, straight or curved, of constant or
variable magnitude, according to a definite law.

The line whose motion generates the surface is called
l;he generatrix. Any lines which limit or direct the motion
jf the generatrix are called directrices.

All surfaces having straight lines for their generatrices
ire called Ruled Surfaces.

■Ruled Surfaces are divisible into two classes; Develop-
ible Surfaces and Twisted or Skew Surfaces.

Any two consecutive generatrices of a Developable Surface
ie in one plane. This is not true of Twisted or Skew
Surfaces. The former can therefore be laid flat, or de-
veloped, without extension or rupture of any part, by turning
iach element of the surface successively, about the consecu-
ive generatrices, into the plane of development. The latter
ire not developable.

104 SOLID OR DESCRIPTIVE GEOMETRY.

The simplest of all Ruled Surfaces is the plane, which is
generated by a straight line moving parallel to itself along
another straight line fixed in position.

To understand clearly what is meant by a Developable
Surface, observe that since the consecutive generatrices lie
in one plane they must either meet or be parallel. Let A,
B, C, D, &c. be a series of consecutive generatrices of a
Ruled Surface, and let B meet A in the point i, C meet B
in 2, D meet C in 3 and so on. The limit of the points i,
2, 3,...??, will be a curve, generally of double curvature,
called the Edge of Regression, to which every generatrix of
the surface is a tangent. The parts of the surface generated
by the tangent line to the edge of regression on each side of
the point of contact, form what are called the two sheets of
the surface.

In the cone the points i, 2, 3, &c. coincide, and the
edge of regression becomes, therefore, a point — i.e. the vertex
of the cone.

In the cylinder, the generatrices being parallel, the edge
of regression is at an infinite distance.

From the mode of generation, it is evident that through
every point on a Ruled Surface one straight line, at least,
can be drawn coinciding throughout its whole length with
the surface.

Some surfaces may be conceived as generated in a
variety of ways, but there is one class which it is most
convenient to regard as generated by the motion of a line
about a fixed axis in such a manner that every point in the
generating line is always at the same distance from two
fixed points in the axis. These are called Surfaces of
Revolution: The generating line may be straight, or of

CURVED SURFACES— THE SPHERE. 10$

single or double curvature. The treatment of these surfaces
is generally facilitated by assuming the generatrix to be the
line in which a plane containing the axis would cut the
surface. Such a section is called a meridian section.

The limits of this book do not admit of an exhaustive
treatment of the properties of curved surfaces, but the
student who wishes to pursue the subject, by the aid of
Descriptive Geometry, into its higher developments, is
strongly advised to do so on the lines indicated in the
foregoing general remarks.

PROBLEMS ON THE PROJECTION OF CURVED
SURFACES.

The construction of special generatrices of curved
surfaces is always a problem of importance, and, generally,
the projection of a curved surface is determined by the
projection of a sufficient number of its generatrices.

The Sphere.
Def. a sphere is the surface generated by the revolu-
tion of a semicircle about its diameter, which remains fixed
in position during the motion.

A sphere may, therefore, be regarded as the locus of all
points in space which are equidistant from a fixed point.

Theorem I. Every orthographic projection of a sphere
is a circle equal in radius to the radius of the sphere.

Theorem II. Every plane section of a sphere is a
circle. That cut by a plane passing through the centre is a
circle equal in radius to the radius of the sphere, hence called
a ^^great circle.'" Other plane sections are small circles.

I . To determine the centre and radius of a sphere which shall

I06 SOLID OR DESCRIPTIVE GEOMETRY.

pass through four given points, A, B, C, D, not in the same
plane, and no three whatever of which are in the satne straight
lifie.

Theorem. One sphere and one only is possible.

First method. If the line joining any two of the
points be bisected by a plane perpendicular to it, since every
point in that plane is equidistant from the extremities of the
line which it bisects, the plane must contain the centre
of the sphere (Theorem XVIII. page ii). For the same
reason if two other planes be assumed similarly bisecting
any other two of the lines joining the given points, three
planes will have been determined, each containing the
required centre. This will therefore be the point which is
the common intersection of the three planes.

Second method. Determine the centre F of the circle
which passes through any three of the points, as A, B, C,
and through the point F draw a perpendicular to the plane
of these three points. This perpendicular will contain the
centre of the sphere. Similarly, determine the centre Q of
the circle which shall pass through the fourth point F) and
two of the others, as A, B. The perpendicular through Q
to the plane of A, B, F>, will also contain the centre of the
sphere. The point O in which the perpendiculars meet will
therefore be the centre of the sphere required.

The radius of the sphere is the true length of a line
from O to any one of the given points. Prob. i. Chap. iii.

Note. To simpHfy the problem, three of the points may
be taken in one of the planes of projection.

2. Three spheres, A, B, and C, of i'6 in., '8 in., and -4
in. radii respectively, lie on the horizontal plane, each touching
the other two. Plan and elevation.

CURVED SURFACES— THE SPHERE. 10/

Begin with an elevation of the spheres A and ^ on a
vertical plane assumed parallel to a vertical plane containing
their centres. In this position the elevations will be two
circles resting upon xy and touching each other. Draw two
similar elevations of C, one touching A and the other
touching B. Take the horizontal distances between the
elevations of the centres of A and C, and of B and C, and
from the plans of the centres of A and B with these
distances for radii respectively, describe arcs. The inter-
section of these arcs will determine the plan of the centre of C,
the third sphere. Its elevation can then easily be determined.

3. To determine the projections, pp', of a point, P, on the
surface of a given sphere.

Let 00' be the projections of the centre O of the sphere,
and / the plan of the given point. The plan / must of
course be assumed somewhere within the circle which is the
plan of the sphere. The elevation /' of P will be in the
projecting line drawn from / at right angles to xy, and
its height may be found by taking an auxiliary vertical
section plane through the point P and the centre O, and
rabatting it into one of the planes of projection.

4. A sphere of i in. radius revolves uniformly about its
fixed vertical diameter. A tracing point is carried wiiformly
over its surface in a fixed vertical plane from the highest to the
^owest point of the sphere, during the time in which the latter
makes one complete revolution. Show the projections of the
■urve. Centre of sphe^-e 2 in. high.

(i) Beginning at the highest point, divide one half of a
/ertical great circle of the sphere into any number of equal
)arts. Through these points take a series of auxiliary

I08 SOLID OR DESCRIPTIVE GEOMETRY.

horizontal planes cutting the sphere. The sections in plan
will be circles, and in elevation, straight lines parallel to xy.

(2) Divide the complete circle, that is the plan of the
sphere, into the same number of equal parts as the half of
the great circle above is divided into, and draw the radii of
the circle from each of these points. Starting from the
highest point of the surface, construct the plan of the curve
through the points where the first circle is cut by a radius,
the second circle by the next radius, and so on to the
circumference of the plan, and then back again to the lowest
point.

(3) For the elevation, project the points through which
the plan of the curve is drawn to the elevations of the
horizontal sections in which they lie.

The case should also be worked in which the velocity of
the revolving sphere is doubled and the vertical velocity of
the tracing point constant.

5. The plane of a circle 2*5 in. diameter is inclined
at 25". Its centre is 2*8 in. above the horizontal platie.
Determine the sphere which resting on the horizontal plane
has this circle on its surface. Science Exam. Hon. 1873.

(i) Determine plane of 25° and the projections of the
circle lying therein.

■ A straight line drawn from the centre of the given circle
at right angles to its plane will pass through the centre of
the required sphere.

(2) Draw projections of a line at right angles to given
plane from centre of given circle and find the horizontal
trace of this line.

(3) Draw a vertical plane containing the line deter-
mined in (2). This plane will be perpendicular to plane 25"

CURVED SURFACES— THE SPHERE. I09

and its horizontal trace will pass through the horizontal trace
of the perpendicular to this plane (2) and will also be at
right angles to the horizontal trace of the given plane.

The vertical plane will cut the given circle at two points
which are the extremities of a diameter, shown in plan as the
minor axis of the ellipse which is the plan of the circle.
These points are, moreover, two points in a great circle of
the sphere which has the given circle on its circumference
and which rests on the horizontal plane at a point in that
great circle.

The plane of this great circle is vertical and coincides
therefore with the vertical plane just drawn. The horizontal
trace of this plane is a tangent at the point of the great
circle mentioned above as that in which the sphere touches
the horizontal plane.

(4) Rabat this vertical plane and with it the two
points into the H.P.

(5) Determine a circle passing through these points
and touching the horizontal trace of the vertical plane (3).
The centre of this circle is that of the sphere constructed,
and its radius that of the sphere required.

Find plan and elevation of the centre and describe
circles therefrom with the radius found.

The circles so drawn are the required projections of the
sphere.

6. Determine three points^ A, B, C, at unequal distances
from an assumed oblique pla7ie, and construct the sphere which
shall touch the oblique plane and pass through the three points.

Note. The three points must be on the same side of
the plane and must not lie in the same straight line.

no SOLID OR DESCRIPTIVE GEOMETRY.

(i) Find the projections a, )8, 7, of the three points
A, B, C, upon the oblique plane.

(2) Determine a plane passing through A, B, C, and
find its intersection with the same plane.

(3) Rabat the oblique plane with its contained points,
a, /3, y, and the intersection found in (2), into the horizontal
plane.

Then, since the lengths of the projectors, aA, (3B, yC,
are known, the points A, B, C, relatively to the rabatted
oblique plane, can be set up, the inclination of their plane
determined, and a circle described through them.

(4) Find the centre and radius of the sphere that
touches the rabatted oblique plane and has this circle on its
surface, as in the preceding problem, and lift the oblique
plane, and the centre of the sphere, back into position.

For another construction see Bradley's Ektnents of
Geometrical Drawings Part 11. Plate xxxii.

7. Three spheres^ A, B, C, of 2-3, 1*9, and 1*3 itiches
diameter^ rest upon the horizontal plane, each touching the other
tivo. Determijie plan and elevation of a fourth sphere, 75
inch radius, touching all three.

(i) Draw plan and elevation of the three spheres, A, B,
C, as in Problem 2 above.

(2) Determine an elevation of two of the spheres, as
A, B, on a vertical plane parallel to the line joining their
centres. The elevation will be two circles touching at a
point on their circumferences which is a projection of the
point in which the spheres themselves touch.

(3) Draw a third circle of 75 inch radius touching
each of the two circles determined above. Consider the

CURVED SURFACES— THE SPHERE. HI

circle last drawn as a projection of the sphere of 75 inch
radius which is required to touch all three given spheres.
From the drawing it will be apparent that the elevation
drawn is that of a sphere 75 inch radius touching two only
of these spheres, viz. A and B. Conceive the sphere of 75
inch radius to roll round the two spheres A, B, and preserve
its contact with each during the revolution. The centre of
the revolving sphere will describe a circle in a plane at right
angles to the line AB, which circle, in elevation, will be a
straight line drawn from the centre of the circle of 75 inch
radius perpendicular to a'b' and produced to an equal
distance on the opposite side of this line.

Note. The circle described by the centre of the re-
volving sphere is the locus of the centres of all spheres of
75 inch radii which touch each of the spheres A, B.

(4) Determine plan of this circle. The plan will be
an ellipse with its major axis at right angles to ab.

(5) Proceed similarly with one of the two spheres
taken above and the third, C, and determine the projections
of the locus of the centre of a sphere of 75 inch radius
in simultaneous rolHng contact with both. The plan of this
locus will likewise be an elHpse, and the two points where

. the loci intersect will be shown in plan by the intersection
of the ellipses.

The points thus determined are the centres of spheres
of 75 inch radii which satisfy the required conditions.

For practice another example may be taken in which
the three spheres, A, B, C, are at different heights above the
horizontal plane and not in contact, taking care not to
remove the centres so far apart as to render the problem
impossible.

112 SOLID OR DESCRIPTIVE GEOMETRY.

Note. After working Chapter x. the student will see
that these points are also the two intersections common to
three spheres described from the centres of the three
given spheres A^ B, C, with radii equal to that of .(4 + 75,
B+"}$, ^+75, from the centres oi A, B, C, respectively.

8. Detertnine two points on the surface of a given sphere^
and find the projections of the shortest line that can be traced
on the sphere between them.

Note. It will be a segment of the great circle passing
through the points.

The Cylinder.
Def. a cylinder is the surface generated by a straight
line which moves parallel to itself and always passes through
a given curved line.

A right cylindrical surface is one whose plane section
at right angles to its generatrices is a circle.

The traces of a cylinder are the curves in which it inter-
sects the horizontal and vertical planes of projection.

The directrix of a cylindrical surface may be a line of
single or double curvature, but it can in eveiy case be
reduced to a plane curve — its trace^ for example. Generally,
any curved line whatever can be regarded as the intersection
of two cylindrical surfaces whose generatrices are respec-
tively perpendicular to the horizontal and vertical planes
of projection, and directrices plane curves lying in those
planes. Conversely, the projection of any curved line
whatever on a plane is the trace, on that plane, of a
cylindrical surface whose generatrices are perpendicular
to the plane of projection, and directrix the given curved
line.

CURVED SURFACES— THE CYLINDER. 113

" A cylinder is given when one trace and a generatrix
are given.

1. Given ab, the major axis of the ellipse forming the
section of a right cylindrical surface by the horizontal plane
when the axis of the surface is inclined 6'^ ; determine the
base and the plan of the frustum.

Note. The minor axis of the ellipse is equal to the
diameter of the required base.

The vertical plane through ab contains the axis of
the cylinder and is at right angles to its base. Rabat
this plane into the horizontal. Bisect ab in o, and draw op'
making the angle bop' equal to ff'. Through p' draw I'p'm'
at right angles to op' , and meeting parallels to that line
through a and b in / and m'. Then I'm' gives the length
of the required minor axis cd. From these data the curve
may be readily drawn. ' "

Through p' a perpendicular to ab produced meets that
line in /, which is the centre of the plan of the base of the
cylinder. The plan of this base is likewise an ellipse,
having its major axis nq equal and parallel to the minor of
the former, its own minor Im being determined by perpen-
diculars on ab drawn from /' and m' . Tangents to these
:urves parallel to the plan of the axis complete the plan
)f the frustum of the cylinder.

2. Given an ellipse as the horizontal trace of a right
cylindrical surface. Find its plan, and an elevation on a
■ 'ertical plane making an angle of -^d^ with the plan of the
i 'xis.

(i) Describe a semicircle on the major axis ab, and
Irom one extremity, as a, set off the length of the minor

E. G. '8

114 SOLID OR DESCRIPTIVE GEOMETRY.

axis, cd, as a chord of the semicircle to m. The line drawn
from 0, the centre of ab, at right angles to am, is an
auxiliary elevation of the axis of the required cyUnder on
a vertical plane passing through ab.

Parallels to the elevation of this axis from a and b
complete the auxiliary elevation of the indefinite cylinder.
Tangents to the ellipse, parallel to its major axis, complete
the indefinite plan.

{2) Draw xy making an angle of 30° with the plan
of the axis, for the new elevation. Assume any point P
in the auxiliary elevation of the axis, and find its plan p.
From plan p draw a projecting line at right angles to xy,
arid measure off on this line above xy the height of P, —
the point thus found will be the elevation, /', on the
new vertical plane, of the point P in the axis. A line
from /' to the point 0' where the perpendicular from 0
meets xy, will be the elevation of the axis on the new
vertical plane.

The elevation of the cylinder may be now completed
by drawing two lines parallel to the elevation of the axis,
one on each side, at distances equal to the semi-minor axis
of the ellipse.

3. To determine the projections^ pp', of a point , P, on the
surface of a cylinder.

Let / be the plan of a given point, P, on the surface
of the given cylinder. Through / draw the plan of a
generatrix of the surface, and find its elevation. The point
/', in which the projecting line from the plan / cuts the
elevation of the generatrix, will be the elevation of P.

CURVED SURFACES— THE CYLINDER. 11$

4. Draw the cylinder which would envelope a sphere
of I'i, in. radius and centre 2 "5 iti, above the ground. Axis
of cylifider inclined 50", and making in plan an angle ^30°
with xy. Show the circle of contact and the horizontal trace
of the cylinder.

(i) Draw projections of the sphere and plan of the
axis of the cylinder according to the conditions.

(2) Use the plan of the axis as the ground line for
an auxiliary V.P., in which draw a second elevation of the
sphere. A line drawn through the centre of this elevation
of the sphere, to meet the new ground line at the given
inclination, will be the elevation of the axis of the required
cylinder, and parallels to it, tangential to the circle which
represents the sphere, will determine the major axis of
the ellipse in which the cylinder cuts the horizontal plane.
The minor axis of the curve will be marked off by parallels
to the axis of the cylinder in plan, tangential to the circle
which is the plan of the sphere. From the new or auxiliary
elevation, an elevation on the original vertical plane can
now be easily determined.

(3) The circle of contact is shown in the auxiliary
elevation as a straight line, and from this can readily be
determined in plan, and from thence on the original vertical
plane.

Note i. The horizontal trace of the cylinder might
also be found by means of generatrices drawn through a
number of assumed points in the circle of contact.

Note 2. The trace of this cylinder is the boundary
of the shadow of the sphere, cast upon the horizontal
plane by rays of light parallel to the generatrices of tha

Il6 SOLID OR DESCRIPTIVE GEOMETRY.

cylinder; and the "circle of contact" marks the boundary
of light and shade on the sphere. See Shadows, Chap, ix,

5. The axes of two equal cylindrical surfaces are inclined
in opposite directions at 25°, 50°, the line perpendicular to both
these axes is 3 inches long and inclined at 30", the two surfaces
touch each other i?i one point only. Draw plan of cylinders
showing point of contact. Elevation at pleasure.

Science Exam. Hon. 1869.
(i) Determine projections of the axes and their
common perpendicular by Problem 22, Exercises on
Chap. III.

(2) Bisect the plan and elevation of the line of 3 inches
which is perpendicular to the axes. The points of bisection
are the projections of the point in which the cylinders
touch.

(3) The plans of the cylinders can be completed by
making auxiliary elevations of the axes on vertical planes
assumed containing these axes. Lines parallel to the
elevations of the axes and in the same vertical planes
with them, at distances of i'5 in. on each side, meet
the horizontal plane in points, on the plans of the axes
produced, which are the extremities of the major axes
of the elliptic traces of the cylinders on the horizontal plane.
The semi-minor axes of the same sections are equal to
1*5 in., the radius of the cylinders, and hence the ellipses
can be drawn and the plans and elevations completed.

6. The centre, O, of a sphere of i in. radius is i "5 in.
above the horizontal plane and 2*5 in. ifi front of the vertical
plane. A straight line in the horizontal plane 2*25 in. from
the plan, o, of the centre of the sphere and making an angle of
30' with xy, is the plan of the indefinite axis, inclined at 50*,

CURVED SURFACES— THE CYLINDER. H/

of a right cylindrical surface touching the sphere. Determine
the traces of the surface and the point of contact with the sphere,
wJun tfie latter is without the cylinder.

(i) Determine a plane containing the centre O of
the sphere and the given axis. This plane cuts the sphere
in a great circle, and the touching cylinder in a generatrix
which is a tangent to this circle at a distance from the
axis equal to the semi-diameter of the cylinder required.

(2) Rabat this plane, with the great circle and given
axis, into the horizontal plane. A line touching the rabatted
great circle parallel to the rabatted axis will give the point
F of contact, from which the projections, //', may be readily
obtained. The distance between these parallels will give
the semi-diameter of the cylinder, whence the traces and the
projections of the surface may be determined.

Note. The student will understand the methods by
which touching surfaces are determined better after he
has worked systematically through the problems on tangent-
planes to curved surfaces given later on.

Two surfaces are said to touch at a point when they
have a common tangent plane at that point. A given
straight line may, in general, be the axis of two right cylinders
touching a given sphere. When the given axis passes
through the centre of the sphere one cylinder only is
possible, and it in this case envelopes the former surface.
When the axis is a tangent to the sphere one cylinder
only is possible, and it contains the sphere. If the axis
be entirely free from the sphere two cylinders — one includ-
ing and the other excluding the sphere — can be drawn.
When the axis meets the sphere in two points and does
not pass through its centre, two tangent cylinders — one con-
taining and one intersecting the sphere — are determinable.

Il8 SOLID OR DESCRIPTIVE GEOMETRY.

7. A right cylindrical surface of 2 in. diameter rawlves
uniformly about its axis fixed in position, and a line is traced
upon it by a point which moves over the surface parallel
to its axis at a uniform velocity. Draw the projections
of the curve formed when the tracing point moves i ///.
forward during the time of each complete revolution of the
cylinder.

Work two cases, (i) when the axis is vertical, (2) when
the axis is oblique, and a developme7it.

Note. This curve is the well-known helix or screw-
thread. One of its properties is that the tangent line
to the curve at any point is inclined to the generatrix
of the surface through that point at an angle constant for
every part of the curve. Hence when the cylinder is
developed the curve becomes a straight line inclined at a
determinable angle, a", to the line which is the developed
circular base of the cylinder. This angle is called the
angle of the screw, and is equal to that which the tangent
line at any part of the curve makes with the plane of
the base of the cylinder. The distance, P, through which
the tracing point moves while the cylinder makes one
revolution, is called the pitch of the helix. Obviously
P= 2Trr tana.

To work the problem, divide a circular base of the
cylinder into any number of equal parts and draw genera-
trices through the points of division. Let iV"be the number
of parts into which the circular base is divided. Starting
from one of the points of division, measure off from the
base, on the successive generatrices, distances successively

P P P P

equal to ■^, 2-^, 3-^ A^— . P in this case being

I in<

CURVED SURFACES— THE CONE. HQ

Join the projections of these points for the projections of
the helical curve.

The development presents no difiSculty.

8. Assume any two points on the surface of the cylinder
in the last problem not in the same generatrix, and show
the projections of the shortest line that can be traced on its
surface betiveen them.

Develop the surface with the two contained points,
and join them by a straight line. The points in the
development in which this line meets certain of the genera-
trices can then be easily transferred to the corresponding
projections of these generatrices, and the points so found
be joined for the projections of the required line, which,
in this case, is clearly a portion of a helix.

Note. This problem may be taken as a type of the
method by which the shortest line between any two points
on any developable surface whatever may be found, or any
given line laid upon the surface. See Prob. 7 on "The
Cone."

The Cone.

Def. a cone is the surface generated by the motion
of a straight line which always passes through a fixed point
and a given curved line.

The fixed point is called the vertex of the cone.

The directrix may be of single or double curvature, but
every conic surface may be represented by a plane directrix,
a fixed point, and a rectilineal generatrix.

If the generatrices extend beyond the vertex, a second
similar cone is generated. These simultaneously generated
cones are called the two sheets of the surface.

I20

SOLID OR DESCRIPTIVE GEOMETRY.

The curves in which the cones intersect the planes of
projection are called its traces.

A Right Conic surface is one which can be represented
by a circle as its plane directrix, and a point for its vertex
in the perpendicular to the plane of the circle at its
centre.

A Cone is given when its vertex and a trace are given.
. I. Given ab, the major axis of the ellipse which forms the
oblique section of a right cone by the horizontal plane, and vv',
the projections of the vertex, to draw the plan of the frustum.
Fig. 28.

Fig. 28.

Since the plan v in ab produced is given, it will be
obvious that when the ellipse is obtained, the required plan
will be completed by drawing tangents to the ellipse through
V. Hence the problem consists chiefly in finding cd^ the

CURVED SURFACES—THE CONE. 121

minor axis of the curve. This line is the horizontal trace of
the plane of a circular section of the cone made through
point o\ the bisection of ab^ and forms a chord of that circle.

This circle being perpendicular to the axis, VP, of the
cone, is inclined to the horizontal plane at the complement
of the angle of inclination of that axis. The circle must
therefore be rabatted and its chord shown.

Take a vertical plane through Fand ab; and on it draw
v'a, v'b, the traces of the conic surface. Through o' draw
k'l' making equal angles with these lines ; this is the vertical
trace, and elevation of the circular base.

Rabat this circle into the vertical plane as shown in
Fig. 28, about k'l\ its vertical trace and diameter. Draw
through 0' the chord CD at right angles to k'T. CD gives
the length cd of the required minor axis.

2. To determine the projections, pp', of a point, P, on the
surface of a cone.

Assume /, the plan of the point, and through p draw a
plan of the generatrix of the cone and show its elevation.
The point /', in which the projecting line from the plan /
cuts the elevation of the generatrix, will be the elevation oiP.

3. A cone with vertex 5 iji. high, axis inclined 50", and
in plan making an angle ^30" with the ground line, envelops
a sphere of i inch radius and centre 2 in. high. Determine
horizontal trace of cone and circle of contact.

As in Problem 4 on " The CyHnder," use the plan of the
axis as the ground Hne for an auxiliary elevation. In this
auxiliary plane set up the elevation of the axis by drawing a
line inclined 50° with xy through the point which is the
elevation of the centre of the sphere, and determine vv\ the

122 SOLID OR DESCRIPTIVE GEOMETRY.

vertex of the cone, 5 in. high. Tangents from if to the circle
representing the elevation of the sphere will mark off on xy
the line which is the elevation of the horizontal trace of the
cone. This same line is also the length of the major axis of
the ellipse which is the horizontal trace of the cone.

To find the minor axis of the same ellipse, the point
must be determined, on the circle of contact, through which
a line from the extremity of the minor axis to the vertex
would pass. In the auxiliary elevation this line will be
shown by joining the vertex and the centre of the major
axis. The intersection of this line with the straight line
representing the elevation of the circle of contact gives the
elevation of the point required, through the plan of which
point a line drawn from the plan of the vertex marks off the
extremity of the minor axis sought.

See also "Note" i, Prob. 4, on "The Cylinder."
Note. The straight hne which, in the auxiliary eleva-
tion, will be the elevation of the circle of contact, will not
pass, as in the case of the enveloping cylinder, through the
elevation of the centre of the sphere. It is the line joining
the tangent points before determined.

The plan and new elevation of the circle of contact can
now be readily determined. The student should show the
traces of the plane of the circle of contact.

The horizontal trace of the cone would represent the
shadow of the sphere, cast upon the H.P. by rays of light
proceeding from the fixed point V; and the circles of contact
would mark the boundary of the light and shade.

4. The centre, O, of a sphere of i in. radius, is i '5 in.
above the horizontal plane, and 2*5 ;«. in front of the vertical
plane. A straight line in the /wrizontal plane, 2-2^ in. from

CURVED SURFACES— THE CONE. 123

the plan, o, of the centre of the sphere, mid making an angle of
30" with xy, is the plan of the axis, inclined at 50°, of a right
cone whose convex surface touches the sphere. The vertex of
the cone is 5 in. high, and its circular base touches the hori-
zontal plane. Determine its plan and elevation and the point
of contact of the surfaces.

(i) Find the projections vi/ of the vertex V.

(2) Determine a plane containing the centre O of the
sphere and the axis, as in Prob. 6 on " The Cylinder."

(3) Rabat this plane with its contained great circle and
axis. Let V be the rabatted vertex. From V draw a Hne
touching the rabatted great circle in P'. The angle between
the rabatted axis and this tangent line is the semi-vertical
angle of the cone. When this angle is found, the projections
of the cone are easily determined by the aid of an auxiliary
vertical plane of elevation containing the axis.

(4) The projections//' of the point of contact P are at
once determinable from the rabatted point P'.

The "Note" to Prob. 6 on "The Cylinder" applies,
mutatis mutandis, to this.

5. Refer to Prob. 4 on ^^The Sphere.^ ^ A point I'sin.
dista?it in plan from the centre of the sphere and 4*5 in. high, is
the vertex, Y,ofa conic surf ace of which the traced curve is the
directrix; find its horizontal trace.

It is only necessary to draw a number of generatrices
from V through points in the curve, the line joining the
horizontal traces of these generatrices will be the required
trace of the conic surface.

6. A right cone revolves uniformly about its axis fixed in
position., and a curve is traced upon its surface by a point

1-24 SOLID OR DESCRIPTIVE GEOMETRY.

moving with uniform velocity in a straight line from the vertex
to the base of the cone in a fixed plane containing the axis.
Determine the projection of the curve when the rotation of the
cone and the motion of the point are so timed, that the point
moves from the vertex to the base while the cone is makifig one
revolution. Radius of base i'5 in.; axis 3 "5.

The student after working Prob. 4 on "The Sphere," and
Prob. 7 on " The CyHnder," will find this problem and its
development an easy exercise. It may be interesting to
note that this curve is Hogarth's famous "Hne of beauty."

The gradient of this spiral depends entirely upon the
relative velocity of the tracing point and the cone.

7. A semicircle is described upon a generatrix of the cone
in the last problejti as a diameter, and the plane cofitaining it is
wrapped round the conic surface. Draw the projections of the
resulting curve.

Develop the cone and describe the semicircle on a
generatrix in the development. If the segment of the circle
which is the development of the base of the cone be divided
into any number of equal parts, and the circular base of the
solid be divided into the same number of equal parts, a set
of generatrices in the development may be drawn of which
the corresponding projections are easily obtainable. The
points in which certain of the generatrices in the develop-
ment cut the semicircle may now have their corresponding
projections found, in the projections of these generatrices, by
proportionate division, and the curve be drawn through them
in plan and elevation. Similarly any plane curve may be
wrapped round any developable surface whatever, or the
shortest line between any two given points be traced on the
surface. For an example of the latter problem, see No. 8
on "The Cylinder."

TANGENT-PLANES AND NORMALS. 12$

Tangent-planes and Normals to Curved Surfaces.
General Remarks.
Def. The tangent line to a curve at any point P is the
limiting position a straight line passing through P, and
cutting the curve in another point (2, assumes when the
said straight line is turned about P in such a manner that <2
continually approaches P and ultimately coincides with it.

The tangent line to a curve at any point is also a
tangent, at the same point, to any surface on which the
curve is traced.

Def. The tangent-plane to a curved surface at any
point P is the plane containing two tangent lines, at the
same point P, to any two curves traced upon the surface
through that point.

Theorem, The tangent-plane to a curved surface at any
point P contains all the tangent lines drawn at the same
point to all the lines that can be traced upon the surface at
that point.

Cor. If one, or more, straight lines can be traced on a
curved surface at a point /*, the tangent-plane to the surface,
at that point, meets the surface in the said line or lines.

Thus, the tangent-plane to a Ruled Surface at any point
meets the surface along a generatrix drawn through that
point. When the surface is Developable, e.g. the cone and
cylinder — the tangent-plane at a point is also a tangent-
plane along a generatrix.

Def. The normal to a curved surface at any point P is
the perpendicular drawn through P to the tangent-plane to
the surface at that point.

126 SOLID OR DESCRIPTIVE GEOMETRY.

Only one normal can be drawn to a surface at the same
point.

All the normals to a cone or a cylinder (or other Develop-
able surface) along a generatrix lie in one plane perpendicular
to the tangent-plane which touches the surface along the
same generatrix. This is not true of Twisted or Skew surfaces.

Two curved surfaces touch at a point when the tangent-
plane to one of the surfaces at that point is also a tangent-
plane to the other.

Two lines that meet completely determine the plane
containing them. From this and the definition of a tangent-
plane given above, it follows that, in general, the problem of
constructing a tangent-plane to a surface is reducible to
that of drawing two tangent lines to any two lines that can
be conveniently traced on the surface at the point of contact
of the tangent-plane. When the surface is a Rtded one the
problem is further simplified by one at least of the lines
traceable on the surface through the point of contact being
straight. Sometimes, as for example in the hyperboloid of
one sheet and the hyperbolic paraboloid, two such straight
lines can be drawn.

When the normal to the required tangent-plane is easily
determinable, it may be first drawn and the tangent-plane
subsequently taken at right angles to it. Occasionally,
auxiliary surfaces, having the same tangent-plane as the
given surface, may be advantageously assumed. Thus the
following theorem is frequently required: —

Theorem. The tangent-plane to a cone or cylindrical
surface is a tangent-plane to the spheres or surfaces of
revolution enveloped by it.

TANGENT-PLANES TO CONES. • 12?

It was pointed out when treating of the cylinder, that
any curved line in space might be regarded as the inter-
section of two surfaces, consequently the line of intersection
of two tangent-planes, one to each surface, at the common
point, P, would be a tangent line to the curve at the same
point. This tangent line is obviously, therefore, perpen-
dicular to the plane containing the two normals to the
surfaces. Thus, if P were a point in the intersection of
two spheres of centres X and Y, a line through P perpen-
dicular to the plane of XYP would be a tangent line to
the curve of intersection at the point P.

PROBLEMS ON TANGENT-PLANES TO CURVED
SURFACES.
In the foregoing problems on curved surfaces it was
found convenient to treat the sphere first, and to follow with
the cylinder and cone. In the treatment of Tangent-Planes
to these surfaces this order can be inverted with consider-
able advantage.

Tangent-Planes to Cones.

I. A right cone, base 2 in, diameter, and axis 4 in. long, is
so placed that the circular base touches the horizontal plane ;
the axis is inclined at 50", and its plan makes an angle 0/^0"
with xy. Determine a point P on the conic swface and draiv
a tangent-plane to the cone at that point.

Find the horizontal trace of the cone. Draw the genera-
trix VP, and find its horizontal trace. This will be a point
in the ellipse which is the horizontal trace of the surface.
Through this point draw a tangent line to the ellipse for the
H.T. of the tangent-plane.

The point in which the H.T. meets xy will be one
point in the V.T. ; another will be the vertical trace of a

128 SOLID OR DESCRIPTIVE GEOMETRY.

line drawn from V parallel to H.T. Through these two
points the vertical trace (V.T.) of the tangent-plane can
be drawn.

Note. The plane containing the generatrix VP and the
axis of the cone, contains all the normals to the tangent -
plane, and is at right angles to it. Therefore, if a line be
drawn from any point in the generatrix (as V) at right
angles to the plane of the normals its horizontal trace will
be a point in the H.T, of the tangent-plane.

This enables us to find the H.T. of the tangent-plane
without depending on the plane geometrical problem for
drawing a tahgent line to a conic section.

2. Take the same cone as in the last problem, and assume
a point P outside the conic surface. Draw tangent-plane as
before, and the line of contact.

a. Find the horizontal trace of the conic surface, and
the horizontal trace of the line KP. The H.T. of the
tangent-plane will be the line drawn from the horizontal
trace of the line VP to touch the horizontal trace of the
conic surface. Two such lines may be drawn, i.e. two
tangent-planes, each satisfying the conditions, are possible.
The V.T.s may be found as in the last problem,

(3. When the horizontal trace of VP is inaccessible,
the point P, or any convenient point in the line VP, may be
made the vertex of a similar cone, and the two H.T.s drawn
touching the two horizontal traces of the conic surfaces. In
the particular case in which the line VP is parallel to the
horizontal plane, the H.T.s will be tangents drawn to the
horizontal trace of the conic surface parallel to the plan of
the line VP.

TANGENT-PLANES TO CONES. 1^9

The lines of contact will be the generatrices of the cone
drawn from the points in which the H.T.s of the tangent-
planes touch the horizontal trace of the surface.

3. Take the same cone as before a fid any line inclined at
6" and <^° to the horizotital and vertical planes respectively.
Draw a tangentplane to the cone parallel to this line.

Draw a line from F parallel to the given line and find its
horizontal trace. The H.T.s of the tangent-planes which
satisfy the conditions can then be drawn from this point
tangents to the horizontal trace of the surface.

Note the limits. When the line from the vertex parallel
to the given line falls inside the conic surface the problem is
impossible.

The above solution enables us to draw a tangent-plane
to a cone perpendicular to a given plane : such a tangent-
plane being of course parallel to any line taken at right
angles to the given plane.

4. To the same cone as before draxv a tangent-plane which
shall be inclined at 70" to the horizontal plane.

Theorem. A common tangent-plane can always be
drawn to two cones having the same vertex, unless one of
the sheets of one of the conic surfaces lies wholly within the
surface of the other without touching it.

The number of tangent-planes possible depends upon
the relation between the two cones, and may be one, two,
three, or four.

To work the problem, make V, the vertex of the given
cone, the vertex also of a right cone standing on the
horizontal plane, with generatrices inclined at 70".

E. G. - Q

130 SOLID OR DESCRIPTIVE GEOMETRY,

Any line that can be drawn a tangent to both the
horizontal traces of the cones will be a H.T. of a tangent-
plane satisfying the requirements of the question ; its V.T.
can be found as before.

Limits. The given angle of inclination of the tangent-
plane must lie between 90" and a", where a" is the angle
which the generatrix of minimum inclination makes with
the horizontal plane.

5. Given an oblique plane and a straight line VO meeting
it in the point V, determine a right cone that shall touch the
given plane and have the straight line NO for its axis.

(i) Construct the plane of the base at right angles to
the line VO through the point O, and find the intersection
of this plane with the given plane.

(2) Rabat the plane of the base, with the point O and
the line of intersection, into one of the planes of projection,
and determine the radius of the circle having O for its
centre and the intersection line for its tangent. This will
be the radius of the base of the cone.

(3) Take an auxiliary vertical plane containing VO,
and thence complete the projections of the cone.

6. Given a right cone, axis oblique to both planes of
projection, and a point P in space. Determine a second cone
having the point P for its vertex and a given straight line
passing through V for its axis that shall touch tlu given cojie.

The two cones will have for their common tangent-plane,
a plane passing through their vertices.

Draw, therefore, a tangent-plane to the given cone
through the given point F, and determine by the last

TANGENT-PLANES TO CYLINDERS. I3I

problem a right cone touching this plane and having the
given straight line for its axis.

Note. Two tangent-planes can be drawn to the given
cone through the given point, hence two cones can be
found satisfying the conditions of the problem.

7. General Case. To draw a normal to a conic
surface through a given point not in the surface.

Find the curve which is the locus of the feet of all the
perpendiculars let fall from the given point upon all the
tangent-planes to the cone. The point, or points, in which
this curve meets the surface will determine the normal or
normals.

To find the intersection of this curve with the given
surface a second cone may be taken, having the same vertex
as the first and the curve for its directrix. Lines from the
point, or points, in which the horizontal traces of these cones
meet, to the common vertex, will determine the intersection
of the curve and the given surface.

8. General Case. To draw a normal to a given cotie
parallel to a given plane.

Draw a tangent-plane to the given cone perpendicular to
the given plane (see Prob. 3). Any perpendicular to this
tangent-plane from a point in its line of contact Avith the
surface will be normal to the given cone, and parallel to the
given plane.

Tangent-Planes to Cylinders.

I . A right cylinder y base two inches diameter and axis four
inches long, is so placed that the circular base touches the hori-
zontal plane; the axis is inclined at 50° and its plan makes an

132 SOLID OR DESCRIPTIVE GEOMETRY.

angle of 30" with xy. Determine a point, P, on its surface,
and draw a tangent-plane to the cylinder at that point.

Draw a generatrix of the cylinder through P, and find its
horizontal trace in that of the cylinder. The tangent to
the horizontal trace of the cylinder at the point which is the
horizontal trace of the generatrix will be the H.T, of the
required tangent-plane, the V.T. of which can then be
found.

The generatrix through P is the line of contact of the
tangent-plane with the surface. The plane through this and
the axis contains all the normals to the tangent-plane.
Hence the latter might be determined at right angles to the
plane of the normals as in the parallel case of the right cone.

2. Same cylinder as in the last probletn, and a point,
P, without the surface. Draw tangent-plane and show line
of contact.

Through P draw a line parallel to the generatrices of
the cylinder. The H.T. of the tangent-plane can be drawn
through the horizontal trace of this line to touch the
horizontal trace of the surface. Find V.T. and the line
of contact as in other problems and compare with the
parallel problem on the cone.

3. Sajne cylinder as before^ and a line inclined 6° and ^°
to the horizontal and vertical planes respectively. Draw
tangent-plane parallel to the given line.

Determine a plane containing the given line and pa-
rallel to a generatrix of the cylinder. A straight line,
touching the horizontal trace of the cylinder, and parallel
to the horizontal trace of the plane determined above, will
be the H.T. of the required tangent-plane, whence the line
of contact and the V.T. may be found.

TANGENT-PLANES TO CYLINDERS. 133

Note the limits. The angle 6" must lie between 90"
and the angle of inclination of the generatrices of the
cylinder.

This problem enables us to determine a tangent-plane
to a cylinder perpendicular to a given plane.

4. Draw a plane inclined 70" to the horizontal plane and
touching the cylinder used in the foregoing problenis.

Determine any right cone standing on its circular base
having its generatrices inclined at 70".

Find the horizontal trace of a tangent-plane to this cone
parallel to a generatrix of the given cylinder. A line pa-
rallel to this horizontal trace touching the horizontal trace of
the cylinder will be the H.T. of the required tangent-plane.
Complete by drawing line of contact and V.T.

Limits. The given angle must lie between 90" and
the angle of inclination of the generatrices of the given
cylinder.

5. Given two poi?its, B and C, at equal distances from
a given plane, as the foci of an elliptic section of a right
cylinder which touches the plane: determine the cylinder atid its
line of contact with the plane.

The plane of the ellipse is perpendicular to the given
plane, and the curve touches the given plane at the ex-
tremity, (2, of its minor axis. This point may be determined
by a perpendicular to the plane through (9, the point of
bisection of BC. The line BC being parallel to the given
plane, the foot of this perpendicular from O to the plane will
be (2-

The major axis will be equal to BQ-¥ CQ, whence the
ellipse is completely determined.

134 SOLID OR DESCRIPTIVE GEOMETRY.

The position of a circular section of the cylinder per-
pendicular to the given plane, with radius equal to OQ, can
be determined, with one extremity of a diameter coinciding
with an extremity of the major axis of the ellipse. See the
construction given in Prob. i on "The Cylinder."

The straight line through Q and the point in which the
circle touches the plane, is the generatrix in which the
cylinder touches the plane.

Work the problem :

(i) When the given plane is horizontal.
(2) „ „ „ oblique.

6, To determine a right cylinder which shall have a given
straight line for its axis and shall touch a given right cylinder.

It is obvious that the two cylindrical surfaces will have
a common tangent-plane at their point of contact. This
point, which lies in a straight line at right angles to both
axes, must therefore be determined.

Let AB be the axis of the given cylinder, and CD that
of the required one : AB not lying in the same plane with
CD. Determine FQ the common perpendicular to these
axes by Problem 23, Chap, iii., meeting AB in point Pj
and CD in Q.

Find T'the point in which FQ pierces the surface of the
given cylinder. T is the point of contact, and QT the
radius of the circular section of the required cylinder,
which may hence be constructed. See Problem 5 on "The
Cylinder " for special exercise.

Another Method.

Assume any point, F, and draw through it parallels to
the two given axes. Determine the horizontal traces of

TANGENT-PLANES TO CYLINDERS. 135

these lines and join them for the horizontal trace of the
plane containing them. This plane will be parallel to the
generatrices of both cylinders.

Another plane parallel to this, and touching the given
cylinder, will be the common tangent-plane of the two
cylinders, and its horizontal trace will, therefore, be a tangent
line to the horizontal traces of both surfaces.

The horizontal trace of the required cylinder may be
drawn in either of the following ways : —

(i) Find the horizontal trace of its given axis and the
real distance from any point M in this axis to the tangent
plane. This will be the radius of the contained sphere.
The cylinder enveloping this sphere and having the given
axis can be drawn by Problem 4 on " The Cylinder" and its
horizontal trace determined. Or^

(2) Draw from any point in the given axis a perpen-
dicular to the tangent-plane, and find the point in which
the perpendicular meets the plane. A line through this
point parallel to the given axis will be the line of contact of
the tangent-plane with the required cylinder, and the H.T.
of this line will be a point in the H.T. of the cylinder.
The true length of the perpendicular to the plane will be
the semi-minor axis of the elliptical trace. On drawing the
minor axis of the ellipse through the horizontal trace of the
given axis of the cylinder at right angles to the plan there-
of, we have one axis and a point in the curve, from which
data the ellipse can be drawn, by means of a strip of paper,
on the well-known principle of the trammel

The generatrices drawn from the two points in which
the ellipses are touched by the horizontal trace of the

13^ SOLID OR DESCRIPTIVE GEOMETRY.

tangent plane will meet in T^ the point of contact of the
cylinders.

7. Given a right cylinder, axis oblique, a point in space
and a straight line passi?ig through it; determine a right cone
that shall have the given line for its axis, the point for its
vertex and touch the given cylitider.

Refer to the parallel problem on "Tangent-Planes to
Cones," Problem 6, for method.

8. General Case. To draw a non?tal to a cylinder
through a point not in the surface.

Construct the curve which is the locus of the feet of all
the perpendiculars let fall from the given point upon all the
tangent-planes to the cylinder, as for the parallel case on
the cone. The intersection of this curve with the cylinder
will determine the normal or normals.

This intersection may be found by making the curve
the directrix of a new cylinder whose generatrices are paral-
lel to those of the given cylinder. The common genera-
trices of these surfaces, drawn from the points in which their
horizontal traces meet, will determine the intersections of
the curve with the given surface, and, therefore, the normals,
if there are more than one.

9. General Case. To draw a normal to a given
cylinder, parallel to a given plane.

Draw a tangent-plane to the given cylinder perpen-
dicular to the given plane (Prob. 3).

Any perpendicular to this tangent-plane from a point in
its line of contact with the cylinder will be normal to the
given surface, and parallel to the given plane.

1 0. Determine a helix of \'<^ in. pitch on the surface of
a right cylinder^ axis vertical, diameter of base 2 in., as in

TANGENT-PLANES TO SPHERES. 137

Problem 7 on '■^The Cylinder" and draw the curve in which
the surface generated by the motion of the tangent line to the
helix intersects the horizontal plane.

Let P be any point on the helix. Its elevation p' will
be a point in the elevation of that curve, and its plan p a
point in the circumference of the circle which is the plan of
the cylinder.

A tangent line to this circle at the point / will be the
plan of the tangent line to the helix at the point P in the
curve, and as this line is inclined at a" (the ajigle of the
screw) to the horizontal plane, its horizontal trace may be
found for one point in the required curve.

Similarly other points may be found, and the curve,
which is the involute of the circle forming the base of the
cylinder, drawn through them.

Note. This surface is a "Developable Helixoid" of
which the helix is the Edge of Regression.

Tangent-Planes to Spheres.

I. A point, O, 2 in., and i'^ in., from the horizontal
and vertical planes respectively, is the ccfitre of a sphere of i i?i.
radius. Draw the sphere, and a tangent-plane to it at a point
P on its surface, 2-3 in. from H.P., and i-8 in. from V.P.

To find P, take an auxiliary vertical plane parallel to the
vertical plane of projection, i*8in. from it, cutting the sphere.
The elevation of the section of the sphere by this plane will
be a circle shown in true shape. A point in the elevation
of the circle 2-3 in. above xy, will be/' the elevation of the
point P, whence the plan p can be readily found.

If any two planes be assumed passing through P and
the centre O of the sphere, they will cut the spherical

138 SOLID OR DESCRIPTIVE GEOMETRY.

surface in two great circles having a common radius OP^
and since the tangent lines at P to both of these circles are
at right angles to this radius, the tangent-plane to the sphere
at the same point is also perpendicular to OP ; i. e. OP is
the normal to the tangent-plane. Hence,

To draw a tangent-plane to the sphere at the point -P,
it is only necessary to draw a plane at right angles to OP
to pass through the point P, by Prob. 13, Chap. iii.
Converse.

Note. The normal to a sphere from any point whatever
is obviously the straight line drawn from the point to the
centre of the sphere.

2. Determine a tangent-plane to the sphere in the last
prohlem containing a given point, P, without the sphtrical
surface. Let P be in the vertical plane, 5 in. high and 375 in.
from the elevation o' of the centre of the sphere.

Make P the vertex of a cone enveloping the sphere
^Prob. 3 on " The Cone " ). Any tangent line to the hori-
zontal trace of the cone will be the horizontal trace of a
plane fulfilling the required conditions.

In cases in which the horizontal trace of the enveloping
cone is inaccessible the circle of contact may be determined,
and the tangent-plane to the sphere drawn at any assumed
point in this circle, by the last problem.

Note. There is no limit to the number of solutions
possible.

3. Take the sphere and point P in Prob. 2, and assume
another point, Q, also without the spherical surface. Deter-
mine a tangent-plane to the sphere that shall pass through
P and Q.

TANGENT-PLANES TO SPHERES. 139

Note. Since P and Q lie in the tangent-plane, the
straight line PQ indefinitely produced will also lie in the
tangent-plane. The tangent-plane to the sphere through P
and Q will therefore be a tangent-plane to the sphere
through every point in the straight line PQ. produced ; and
if this line do not meet the spherical surface two planes
touching the sphere can be drawn through it; if the line
touch the sphere one only can be drawn, and if it pass
through the spherical surface the problem is impossible.

First Method. Make P and (2, or any two other con-
venient points in the line PQ, produced, the vertices of two
cones enveloping the sphere. If the horizontal traces of
these cones are determinable, a line touching them may be
taken for the horizontal trace of the tangent-plane required ;
but if not, determine the circles of contact of the two cones
and the points R and 6" in which they intersect, and draw
the traces of the planes passing through the points PQR
and PQS, for those of the tangent-planes satisfying the
conditions. Or, tangent-planes to the sphere at the points
P and S may be drawn as in Prob. i.

To shorten the work, one cone may be assumed with its
vertex in PQ at the same level as the centre, (7, of the
sphere, the plan of the circle of contact being in this case
a straight line.

Second Method. Take a plane through the centre, O,
of the sphere at right angles to the straight line joining the
points P and Q. Let / be the intersection of this line and
plane. Straight lines, through the point 7, touching the
great circle in which this plane cuts the sphere, give the
points R and S, whence the plane may be determined
as before.

140 SOUD OR DESCRIPTIVE GEOMETRY.

Third Method. Find the horizontal trace of the line
PQ.t produced if necessary, and the horizontal trace of
a cone having its vertex in this line. The two tangent
lines to the conic trace, through the H.T. of the line PQ,
will be the horizontal traces of the two tangent-planes.
The vertical trace of the line will also be a point in the
vertical traces of the planes, which may be thus determined.

Fourth Method. Draw a cylinder enveloping the sphere
with its axis parallel to PQ. The two tangent-lines to the
horizontal trace of the enveloping cylinder, through the H.T.
of the line PQ, will be the horizontal traces of the tangent
planes.

4. Take the sphere, and poijit P, as in Prob. 2, and
determine a tangent-plane to the sphere that shall pass through
P and be inclined to the horizontal plane at an angle of 'j^^.

First Method. Make the given point P the vertex
of a cone enveloping the sphere, and draw a tangent-plane
to this cone at 75" to the horizontal plane, by Prob. 4 on
"Tangent-Planes to Cones."

Second Method. Envelop the sphere by a cone with its
axis vertical and generatrix inclined at 75",

A plane containing P and touching the cone will satisfy
the conditions. " Tangent- Planes to Cones," Prob. 2.

5. The centre, O,ofa sphere of 1 in. radius is i"5 in. from
eachpla?ie of projection. Draw the traces of a plane inclined
at 75" and 60° to the horizontal and vertical planes respectively,
and touching the sphere.

Determine a right cone, generatrix inclined 75" to the
horizontal plane, enveloping the sphere.

Similarly, find the vertex, V, of another right cone, genera-
trix inclined 60" to the vertical plane, enveloping the sphere.

TANGENT-PLANES TO SPHERES. I4I

The tangent-plane to the first Cone through the point F,
the vertex of the latter, satisfies the given conditions.

Note. Four planes — two parallel pairs — are possible.
Compare this problem with Prob. 7, Chap. in.

6. Determine the sphere, and the point P, as in Prob. 2.
Draw the traces of a tangent-plane to the sphere that shall
pass through P and be parallel to a line inclined at 60° to the
horizontal plane and 30" to the vertical.

Envelop the sphere by a cone having P iox its vertex,
and determine a tangent-plane to this cone parallel to the
given line. "Tangent-Planes to Cone," Prob. 3.

Note. Refer to Problem 3, "Tangent-Planes to Sphere,"
and compare.

7. Determine a tangent-plane to two spheres of miequal
sizes, unequal heights, and free from one another, which shall

(a) contain a given point P,

(fi) be parallel to a giveji straight line.

Envelop the two spheres by a cone.

Note. Two enveloping cones are possible. In one case
both spheres will be enveloped by the same sheet of the
conic surface, and in the other case, one sphere will be
enveloped by one sheet of the cone and the other sphere by
the other sheet.

For (a). Determine a tangent-plane to the enveloping
cone that shall contain P. Or, Make P the common vertex
of two cones — one enveloping one sphere and the other
the other sphere — and determine the tangent-plane to them.

For (^). Determine a tangent-plane to the enveloping
cone parallel to the given line.

142 SOLID OR DESCRIPTIVE GEOMETRY,

8. Three equal spheres, diameters 2 in. each, have their
centres at the angles of a vertical equilateral triangle of 4 in.
side, with one side horizontal. Draw a tangent-plane to the
three spheres : —

(a) To pass above the two lower and under the upper
sphere.

(fi) To pass above one of the lower and under the other
two spheres.

For (a). Draw an auxiliary elevation on a vertical plane
at right angles to the plane of the triangle. A tangent
to the two circles which are the elevations of the three
spheres can be drawn for the auxiliary vertical trace of
the required plane. The H.T. will of course be perpendi-
cular to the auxiliary xy.

For (/8). Make an elevation on a vertical plane parallel
to the plane of the triangle.

Determine a cylinder containing the upper and one
of the lower spheres, and a second cylinder parallel to the
first cylinder and containing the third sphere.

The horizontal trace of the required plane can now
be drawn as a tangent to the two ellipses which are the
horizontal traces of the two cylinders.

The tangent lines on the cylinders can be shown by
drawing generatrices from the tangent points of the ellipses.

Find the points of contact with the spheres by drawing
perpendiculars to the tangent-plane from their centres.
These points are fixed where the projections of the perpen-
diculars cut the generatrices which are the lines of contact
with the cylinders.

9. Draw the three spheres touching in Problem 2 on
**The Sphere" and determine a tangent-plane to the three.

TANGENT-PLANES TO SPHERES. 143

Note. The horizontal plane of projection is one
tangent-plane.

To find the other. Determine plan and elevation of the
axes of two cones, one enveloping A and B, and the other
enveloping A and C. Lines through the centres of the
spheres give these axes. The horizontal traces of these axes
will be the vertices of the enveloping cones, and will also be
points in the H.T. of the tangent-plane required. The
vertical trace will be most readily found by the aid of
an auxiliary vertical plane taken at right angles to the
H.T. of the tangent-plane.

10. Determine the tangent-planes to three spheres A, B, C,
of unequal sizes, unequal heights, and free from one another.

Determine two cones enveloping A and B, and A and C,
respectively. A tangent-plane to one of these cones con-
taining the vertex of the other will satisfy the conditions. Or,
the vertex of one of the cones enveloping two of the spheres
may be made the vertex of a second cone enveloping
the third sphere, and the tangent plane to these cones
determined for the tangent-plane required. The student
will here see that this affords a method of determining a
tangent-plane to a given cone and a given sphere.

Refer to "Note," Problem 7 above, and observe that
eight tangent-planes to the three spheres can be found.

The student can exercise himself by showing that the
vertices of the six possible enveloping cones are in four
straight lines — three vertices in each line.

11. Determine the cone, Prob. i. ^'^ Tangent- Planes to
Cones," and a sphere 75 in. radius touching the horizontal

plane in a point, P, assumed at pleasure. Show the traces of
a tangent-plane to the sphere and the cone.

144 SOLID OR DESCRIPTIVE GEOMETRY.

One solution of this problem was referred to in No. lo
above.

Another Method. Envelop the sphere by a cone
similar to the given one, and determine a tangent-plane
to one of these cones and containing the vertex of the
other. This will be the plane required.

1 2. To draw a normal to a sphere parallel to a given
straight line.

A line through the centre of the sphere parallel to
the given line will be the normal required.

13. To draw a tangent-plane to a sphere parallel to a
given pla7ie.

Draw a perpendicular from the centre of the sphere
to the given plane. This will be the normal to the required
tangent-plane, which may at once be drawn at right angles
to the normal through the point in which the latter meets
the surface.

Problems on Surfaces of Revolution.

Having already defined a surface of revolution in the
general remarks at the head of this chapter, it will be only
needful here to summarise the chief properties in the follow-
ing theorems : —

Theorem I. All sections of a surface of revolution by
planes at right angles to the axis, are circles having for their
centres the points in which the axis meets the section plane.

Theorem II. All meridian sections of a surface of
revolution are similar and equal curves.

Theorem III. All the tangent lines drawn from a
fixed point in the produced axis of a surface of revolution

SURFACES OF REVOLUTION. 145

to meridian sections, touch the surface in points which lie
on the circumference of a circle whose plane is at right
angles to the axis. That is, the locus of all tangent lines
to a surface of revolution drawn from a fixed point in the
produced axis, is a right cone.

Theorem IV. The tangent line to any circular section
of a surface of revolution at a point in it, is perpendicular
to the meridian section of the surface through that point.

Cor. I. The tangent-plane to the surface at any point
is perpendicular to the meridian plane through the same
point.

Cor. 2. The normal to a surface of revolution at any
point, lies wholly in the meridian plane passing through that
point, and is normal to the meridian section at the same
point.

Cor. 3. The normal from a given point to a surface of
revolution is the normal to the section of the surface made
by a meridian plane containing the given point.

Cor. 4. The tangent-planes to a surface of revolution
at different points in the same circular section pass through
the same point in the axis.

Cor. 5. The normals to a surface of revolution at
different points in the same circular section meet the axis at
the same point.

A surface of revolution is given when its axis and its
plane generatrix (meridian section) are given. But, what-
ever line be given as the generatrix, it can always be reduced
to an equivalent plane generator.

ll\^Q..right cone, and the right cylinder, may be regarded as
surfaces of revolution, but in the general treatment of conic

£. G. 10

146 SOLID OR DESCRIPTIVE GEOMETRY.

and cylindrical surfaces it is necessary to consider them in
another point of view. This has been already done as far
as the limits of this book allow. The sphere, which is
essentially a Surface of Revolution, has also for convenience
been already separately considered.

The surface generated by an ellipse revolving about
either of its axes is called a Spheroid. It is also a particular
case of an ellipsoid, namely an Ellipsoid of Revolution. It is
prolate or oblate according as it revolves about the major or
minor axis.

A few problems of a general character here follow: —

I. An ellipse, major axis 275 in.^ minor axis 2 in.,
generates a surface by revolving about its major axis. Show
plan and elevation of the surface when the axis of revolution
is vertical, and find the projections of a point, P, on it.

The plan of the surface will be simply a circle of 2 in.
diameter, and its elevation, an ellipse of ^he dimensions
given in the question.

To find pp', the projections of the point P. If the plan
p be given, assume a meridian section to be made through
F and to be turned round, with the point I* on it, till paral-
lel to the vertical plane. The height of F may then be
found by drawing a projecting line from the new position of
p in plan to cut the ellipse which is the elevation of the
surface. Either of the two points in which, in general, the
projecting line meets the ellipse, may be the height of /*,
according as we assume the point to be on the upper or
lower part of the surface.

If/' be given, p may be found by taking a horizontal
section through F, the plan of which section will be a circle

SURFACES OF REVOLUTION: HZ

passing through the plan /. This point may then be readily
determined by drawing a projecting line from /' to meet the
circle.

2, To determine a cone enveloping the spheroid in the last
problem f when the vertex is a point in the produced axis of the
surface.

Let v' be the elevation of the vertex. Draw from v' a
tangent line to the ellipse which is the elevation of the
spheroid and call the point of contact f. The line v't' is
the elevation of a generatrix of the enveloping cone, and its
plan will be a line drawn from v parallel to xy. The circle
of contact (the directrix) will be the circle in which a hori-
zontal section plane through T cuts the surface, and the
horizontal trace, or base of the cone, will be a circle passing
through the H.T. of the generatrix. VT, and concentric
with the circle which is the plan of the spheroid.

In a similar manner, any number of enveloping cones
can be drawn by taking other points in the axis for vertices.

3. Determine the spheroid as before, and assume a point,
P, in space, without the surface, and not in the axis. Find the
cone which has P for its vertex and erivelops the spheroid.

(i) Determine any cone, as in the last problem, envelop-
ing the spheroid, and having its vertex in the axis of
revolution produced.

(2) Draw a tangent-plane to this cone passing through
P. This will also be a tangent-plane to the spheroid, and
the point in which the circle of contact of the cone with
the spheroid meets the line of contact of the tangent-plane
with the same cone, will be the point of contact of the
tangent-plane with the spheroid. Two tangent-planes being

14^ SOLID OR DESCRIPTIVE GEOMETRY.

possible to the enveloping cone, there will, of course, be two
such points.

(3) Through these points of contact draw two lines to
the point P. Since these lines lie in tangent-planes to the
spheroid, and contain the points of contact of the tangent-
planes, they are tangents to its surface, and, inasmuch as they
pass through P, they are generatrices of the required en-
veloping cone.

Similarly, by taking a series of other points in the pro-
duced axis for the vertices of new cones, any required
number of pairs of generatrices may be found, and their
horizontal traces joined by a curve for the H.T. of the
enveloping cone. The curve joining the points of contact
of the tangent-planes will give the curve of contact of the.,
enveloping cone with the spheroid-

4. Determine a cylinder enveloping the spheroid in Pro-
blem I, when the generatrices are parallel to a given line inclined
at 60" to the horizojital plane, and itt plan making an angle
of 2)0° with xy.

(i) Envelop the spheroid by any cone, as in Problem
2, having its vertex in the produced axis,

(2) Draw a tangent-plane to this cone parallel to the
given line. This will also be a tangent-plane to the spheroid,
and its point of contact may be found as in the last
problem. Here too, since there are two tangent-planes,
there will be a pair of points of contact.

(3) Through this pair of points, two lines, parallel to
the given line, may be drawn for a pair of generatrices of
the enveloping cylinder.

SURFACES OF REVOLUTION: 149

Similarly, any number of pairs of generatrices can be
found, and the enveloping cylinder, with its curve of contact,
completely determined.

Note. The last two problems might be worked by the
aid of auxiUary sections of the surface by planes which pass,
in Prob. 3, through the given point F, and, in Prob. 4, by
planes taken parallel to the given line. Tangent lines to the
sections give pairs of generatrices of the enveloping surfaces.

5. A surface is generated by the revolution of a circle of
I in. diameter about a fixed straight line in the plane of the
circle and 1*5 in. from its centre. Detennine a poifit, P, on its
surface f and the enveloping cones and cylinder as in Probs. i,
2,3, and 4 above.

This surface, which is an anchor ring, or annulus of a
circular section, will afford the student an additional exercise
in the application of the methods used in the solution
of the problems mentioned, without presenting any special
features of difficulty.

Tangen'I*-Planes and Normals to Surfaces of
Revolution.

After what has been already given, it will be" only
necessary to speak of the following problems in the most
general terms.

1. To determine a tangent-plane to a surface of revolution
at a given point on it.

Draw the normal (Theorem IV. Cor. 2, above) and the
tangent-plane at right angles to it, through the given point.

2. To determine' a tangent-plane to a surface of revolution
that shall coiitain a given straight line or its equivalent^ tivo
points in spaa. .

ISO SOLID OR DESCRIPTIVE GEOMETRY.

The student on referring to the parallel problem, " Tan-
gent-Planes to Spheres," will not be at a loss for several
possible solutions, the easiest of which is to envelop the
surface by a cylinder with generatrices parallel to the given
line. A tangent-plane to this cyhnder through any point
in the line will satisfy the conditions.

3. Determine a tangent-plane to a surface of rez>olution
that shall make an angle 6° with H.P. and contain a given
point, P, in space.

Make P the vertex of a cone enveloping the surface of
revolution, and determine a tangent-plane to this cone
that shall make an angle of ff' with H.P. This will be
the plane required.

4. To draw a normal to a surface of revolution parallel
to a given straight line.

Determine a meridian section of the surface by a plane
parallel to the given straight line, and a tangent line to the
curve at right angles to the projection of this line upon the
plane. A perpendicular to the tangent line, through the
point of contact, will be the required normal.

5. To draw a tangent-plane to a given surface of revo-
lution parallel to a given plane.

Draw a perpendicular to the given plane and a normal
parallel to this line. The required tangent-plane may then
be determined at right angles to the normal through the
point in which it meets the surface.

The Hyperboloid of Revolution of one Sheet.

This surface is a particular case of a surface of greater
generality known as the Hyperboloid of one Sheet. It is
an undevelopable Ruled surface admitting of several modes

HYPERBOL 01 D OF RE VOL UTION. 1 5 I

of generation, one of which is, by the revolution of an
hyperbola about its conjugate axis, whence the name.

It will be treated here as a Twisted Surface of Revolution,
generated by a straight line revolving about an axis not
in the same plane with it; and the conjugate hyperbolas,
which form the meridian section of the surface, will be
determined as a type of the method by which any line
whatever, given as a generatrix, may be reduced to an
equivalent generator in a meridian plane.

I. A straight line inclined at 50" to the horizontal plane
is the generatrix of a surface of revolution, the axis of which is
vertical. The nearest point in the generatrix is '5 in. from the
axis. Draw the elevation of a meridian section on a parallel
vertical plane.

Since every point in the generatrix describes a circle
about the axis in a plane perpendicular to it, if a plane be
taken through any point, P, in the generatrix perpendicular to
the axis, meeting the latter in point A, and intersecting the
assumed meridian plane in a straight line through A, the
distance AP can be cut off on this line from A, both sides
of the axis, for two points, /" and P^, in the required
meridian section. Similarly other points may be found.

To draw the problem, begin by describing a circle of
•5 in. radius in the horizontal plane for the plan of the throat,
or collar, of the hyperboloid, and show the straight line,
which is its elevation, two or three inches above xy.

Draw the given generatrix parallel to the vertical plane.
In plan this will be a straight line touching the circle and
parallel to xy. In elevation it will be a straight line drawn
through the middle point of the elevation of the collar, and
making an angle of 50° with xy.

152 SOLID OR DESCRIPTIVE GEOMETRY.

Let 0 be the centre of the circle which is the plan of
the collar, then o is also the plan of the axis. Determine
the projections pp' of any point P in the generatrix. As-
sume a horizontal plane passing through P to cut the
surface ; the plan of the section will be a circle of centre o
and radius op ; the elevation, a straight line parallel to xy
through/'. The extremities of this elevation are two points
in the required elevation of the meridian section. Others
may be found in a similar manner, and the curve put in by
hand.

Observations. The student should note well the fol-
lowing facts :— ^

(i) The plan of the generatrix is, in every position,
a tangent to the collar.

(2) The surface has two systems of generatrices inclined
at equal angles in opposite directions. Hence two straight
lines, coinciding with the surface throughout their whole
length, may be drawn through any point on it, and, there-
fore, two generatrices of different systems are always in the
same plane.

(3) No two generatrices belonging to the same system
are in one plane, and no three can be parallel to the same
plane. Hence the surface cannot be developed.

(4) If any two generatrices of the same system be
determined, and a third belonging to the other system be
drawn from a point in one of them and produced, it will
meet the other. Similarly, if three such generatrices be
determined, and a fourth belonging to another system be
drawn through any point in one of them, it will, if produced,
meet the other two. Hence^ we see that if three generatrices

HYPERBOLOID OF REVOLUTION. 153

of the surface be assumed as directrices, it might be generated
by a fourth straight line which moves so as ahvays to pass
through them. That is, the Hyperboloid of Revolution is a
particular case of a surface called the Hyperboloid of One
Sheet, which may be defined as the surface generated by a
straight line which moves so as always to meet three straight
lines not in the same plane.

(5) The locus of all lines drawn through the centre of
the collar parallel to the generatrices is a right conic sur-
face called the asymptotic cone. A section plane passing
through the vertex of this cone and not at right angles
to the axis, cuts the Hyperboloid in an ellipse, parabola, or
hyperbola, according as it meets the cone in a point, a
straight line, or two straight lines. Parallel sections are
similar curves.

Note. If the hyperbolas which form the meridian
section of this surface were rotated about the transverse
axis, two distinct surfaces would be produced— the Hyper-
boloid of Revolution of Two Sheets. This however is not
a Ruled surface.

2. Given one projection (p or p') of a point, P, on the
surface of the hyperboloid of revolution in the last problem, to
find the other projection.

If p be given, draw a tangent to the plan of the collar
through/. This will be the plan of a generatrix. Find its
elevation and the point/' in it.

If /' be given, a horizontal section of the surface
through P, as in the parallel problem on the sphere, will
determine/.

3. To draw a tangent-plane to the surface at a point, P,
on it. -

154 SOLID OR DESCRIPTIVE GEOMETRY.

Draw the two generatrices through P. The plane con-
taining these will be the plane required.

Note. This is a tangent-plane at a point only, and not
along a generatrix.

Undevelopable Ruled Surfaces Generally.

The Hyperboloid of One Sheet has been already defined
as the surface produced by a straight Une which moves
so as always to meet three given straight lines not in the
same plane.

To determine the surface, then, it is only necessary to
draw a number of straight lines from different points in one
of the given rectilineal directrices to meet the other two.
(Prob. XII., Cor. 2, Chap, in.)

Theorem. If any three generatrices be found by this
construction, a straight line which moves so as always
to pass through these three lines will generate the same
surface.

That is, the Hyperboloid of One Sheet has two distinct
systems of rectilineal generatrices, a fact which was pointed
out when treating of the particular case of the Ruled Hyper-
boloid of Revolution.

Every plane section of the surface is a conic, and it is by
a comparison of the character of the sections made by planes
cutting the surface in particular directions that the student
will best get a mental picture of its form. Turning to the
Hyperboloid of Revolution of One Sheet, it is seen that the
sections by planes at right angles to the axis are circles.
Now, if these sections were ellipses, the surface would be the
general case of the Hyperboloid of One Sheet we are here
considering, and, from this point of view, might be called the

RULED SURFACES OF REVOLUTION. 155

Elliptic-Hyperboloid. Several other modes of generation are
obvious, such as by the motion of a variable ellipse, or a
variable hyperbola, under particular restrictions, and by the
motion of a straight line, always touching the elliptic collar,
in a plane perpendicular to the plane of the collar and in-
clined at a constant angle to the same plane.

A number of interesting problems might be discussed on
surfaces generated when the three directrices are not recti-
lineal, but the limits of this book exclude them.

Another group of surfaces, generated by a straight line
which moves parallel to a given plane and always passes
through two given lines not in one plane, may just be
mentioned.

When both the directrices are straight lines, a surface
called the Hyperbolic-Paraboloid is produced.

Like the Hyperboloid of One Sheet, this surface can be
generated by the motion of a straight line in two ways.
Thus, the surface may be generated by a straight line
moving along two generatrices ' for new directrices, and
always parallel to the plane to which the two original
directrices are parallel.

To picture the form of this surface in the mind, the
student should turn once more to the Hyperboloid of
Revolution of One Sheet. Let /* be a point on the collar,
and assume that the meridian section through /* is a
parabola, instead of an hyperbola, of which P is the vertex.
If now the section by the plane of the collar be also a
parabola, with P for its vertex, instead of a circle, and
the first parabola move with its vertex along the curve
of the second in a plane always parallel to itself, it will
generate the surface in question.

156 SOLID OR DESCRIPTIVE GEOMETRY.

Since through every point in these two surfaces two
rectilineal generatrices can be drawn a tangent-plane at
a point on either surface is easily determinable.

When one directrix is a straight line and the other a
curve, the surface generated is called a Conoid.

The only case that can be touched upon here is that
in which the axis of a right cylinder and a helix traced
on its surface are the directrices, and the plane of the
circular base that to which the generatrix is always parallel.
This is the well-known surface of the

Square- Threaded Screw.

I. Draw two helices, '5 in. apart and i '5 in. pitch, on the
surface of a right cylinder 2 in. diameter and vertical axis. A
rectangle, '^in. by 'T$in., moves between these helices in a
vertical plane which always passes through the axis, thus
generating a square-threaded screw. Show several turns of
the thread and a tangent-plane at a point on the screw
surface.

Noting that the two remote angles of the generating
rectangle trace helices of the same pitch, and same dis-
tance apart, as those given, on the surface of a right
cylinder of radius 175 in. having the same axis as the
given cylinder, these helices may be drawn as in Prob. 7
on "The Cylinder" and the drawing completed to show the
screw and a sectional elevation of a nut with the same
thread.

To work the tangent-plane : Let P be the point taken
on the helix, through P draw the rectilineal generatrix
and a tangent line to the helix at that point. The plane
containing the generatrix and tangent line is the required
tangent-plane to the surface .at P.

RULED SURFACES OF REVOLUTION. 157

Before dismissing the subject of curved surfaces, there is
one other Ruled Surface of Revolution of importance in the
arts which may be left to the ingenuity of the student to
work out for himself.

2. An angular-threaded screw is generated by an isosceles
triangle whose vertex is constrained to inove in a helix while
its base moves along the axis of the right cylinder on which the
helix is traced. Show the screw and nut, and determine a
tangent-plane to the surface at a given point on it.

Two examples of this surface will be found in Bradley's
Elements of Geometrical Drawing, Part II. Plate xxxvii.
Figs. I and 2.

The student who wishes for other exercises can deter-
mine tapering, square, and V-threaded screws on a cone.

VIIL

SECTIONS BY OBLIQUE PLANES.

Preliminary,

An Oblique Plane may be given in various ways of
which the following are some of the most important : —

1. By the angles a" and ^ which the vertical and
horizontal traces respectively make with xy. Fig. 29.

2. By the angles 6" and <^" which the plane makes with
the horizontal and vertical planes of projection respectively.
Prob. VIL Chap. iir.

3. By the angles a" and ^'.

4. By the angles a" and <^".

5. By the angles ^ and ^.

6. By the angles /S" and i^\

7. By three given points, or their equivalents, as; (i)
Two straight lines parallel or meeting, and (2) One straight
line and a point. Prob. I, (3), Cor. I. Chap. III., p. 46.

The student should solve each of these cases before
going on with the work of this Chapter.

As an example, 3 and $ (which are also types of 4 and 6)
are solved in Fig. 29.

SECTIONS BY OBLIQUE PLANES.
Fig. ag.

159

v^>

it*

(^*

A point ^ {aa') in the vertical plane is made the vertex
of a right cone standing on its base, cef, in H.P., with its
generatrices inclined at 6°. For 3, draw ct tangent to cefaX
c, meeting xy at t, and making the angle xtc equal to a"; join
a't. Then ct is the horizontal trace, and td the vertical trace,
of the required plane. For 5, draw a't making the angle
dtx equal to jS", and from t draw tc touching the circle cef.
This determines the plane as before.

Nos. 4 and 6 are worked similarly, the axis of the cone,
however, being in the H.P. instead of the V.P,

l60 SOLID OR DESCRIPTIVE GEOMETRY.

General Remarks on the Method of Working
Sections by Oblique JPlanes.

When the section is of a form bounded by plane faces,
the general solution is effected by finding the intersection of
the given section plane with the indefinite planes of each of
the faces. The parts of the indefinite intersections that are
common to the faces and the section plane, are lines of the
section required. Sometimes the work is simplified by
finding the points in which the edges of the form pierce the
section plane, by the aid of auxiliary planes containing these
edges.

When the section' is of a curved surface, the general
method is to assume some mode of generation of the
surface and construct the intersection of each generatrix
with the plane.

Here also recourse is frequently had to auxiliary planes
cutting the surface in generatrices and the section plane in
straight lines. Sometimes, however, the auxiliary planes are
made to cut the surface in some easily determinable curve
other than a generatrix.

When convenient, auxiliary planes should be taken
parallel or perpendicular to one of the planes of projection,
as in these cases one projection of the auxiliary section is
always a straight line.

For the most part the practical solution of the problems
in this chapter is greatly facilitated by assuming an auxiliary
plane of projection at right angles to one of the traces
of the section plane. The projection of the section on this
auxiliary plane being a straight line, the other projections,
true shape of section, and projection of frustum on the

. SECTIONS BY OBLIQUE PLANES. l6l

plane of section, may be easily deduced. The simplest
practical solution of a problem is, however, not necessarily
the one most meet for the purposes of the student, whose
aim should always be to develop a grasp of the method
and principles of the subject, and a knowledge of the
properties of Form on which they are based.

Problems.

I. A right prism standing on a square base of 2 in. side,
one side of the square making an angle of T^d^ with xy, is cut by
an oblique plane, inclined at 50", which passes through a point
in the axis of the pris7n 2 in. above the base, and has for its
horizontal trace a line inclined to xy at an angle of 40".
Show the projections of the frustum, the true shape of the
section, the development, and a new projection of the frustum
on the plane of section.

Determine plan and elevation of the prism and the
traces of the section plane.

To work the projection of the section.

First Method: — Assume auxiliary vertical planes con-
taining the edges of the prism and cutting the given plane.
The lines in which the assumed vertical planes cut the given
section plane, will meet the edges through which the auxiliary
vertical planes are taken in required points of the section.

If the horizontal traces of these auxiliary vertical planes
be taken parallel to the horizontal trace of the given oblique
section plane, the lines of intersection will be horizontal —
their plans coinciding with the horizontal traces of the
auxiliary planes, and their elevations being lines, parallel to
xy, drawn from the points in which the vertical traces of
these planes cut the V.T. of the given section plane.

E. G. II

1 62 SOLID OR DESCRIPTIVE GEOMETRY.

The auxiliary planes may however be taken so as to
contain two edges and thus find two points in the section
simultaneously, a particular case of which is the following : —

Second Method: — Determine the intersection of the plane
of each of the faces with the given plane of section. The
elevations of these intersections will cut the elevations of
the edges of the prism in the required points of section, and
the segments of the lines of intersection between the points
of section will be the sides of the section by the given plane.

Third Method: — Take an auxiliary vertical plane of
projection at right angles to the horizontal trace of the given
plane of section. The projection of the section on this plane
will lie in the vertical trace of the section plane on the same
plane. Hence, the heights may be determined iil this
auxiliary elevation and transferred.

To find the true shape of the section by rabatting the
plane of section into one of the planes of projection.

Let -f be a point in the section of which/,/' are the pro-
jections. Through the plan, /, draw/w perpendicular to the
horizontal trace and meeting it in m. Measure the hypo-
tenuse of the right-angled triangle which has pm for its
base, and//*, the height of P^ for its perpendicular, along
mp produced. This will give the point P rabatted into the
horizontal plane about the H.T. of the section plane.

If the other points of the section be similarly rabatted,
they may be joined up for the true shape of the section.

Note. If the auxiliary vertical plane of projection, given
in the third method above, be used, the hypotenuses of these
right-angled triangles may be measured at once from the
auxiliary vertical trace. For this, and the development of
the frustum, refer to similar work in Chap. i.

SECTIONS BY OBLIQUE PLANES. 163

To draw the projection of the frustum on the plane of
section.

Let F, Q, R, S, be the points in which the vertical
edges of the prism, drawn from the points A, B, C, B, of
the base respectively, meet the section plane, i.e. the points
of the section.

The feet T, U, V, W, of the perpendiculars to the
plane of section from the points A, B, C, D, will determine
the projection of the base of the frustum on the plane.
Lines joining the feet of the perpendiculars with the cor-
responding points of the section and with one another will
complete the projection of the frustum on the plane of
section, and the latter must now be rabatted about one of
its traces to show the projection.

For example, to rabat the plane about its vertical trace,
draw from /', and other points g, r, s', of the elevation of
the section, perpendiculars to that trace, and mark off from
it, on these perpendiculars, the actual distances of the points
F, Q, jR, S, from the vertical trace, i.e. the hypotenuses
of the right-angled triangles of which perpendiculars from
/', q', r, /, to the vertical trace, and from /, q, r, s, to xy,
are the sides.

Join these in order for the rabatted section.

Then proceed in the same manner for the rabatment of
the points ll, U, V, W, which determine the projection
of the base A, B, C, D, upon the plane. The rabatted
points T, U, V, W, may then be joined with F, Q, F, S,
respectively, for the complete projection of the frustum.

Note. When one point, as T, in the rabatted projection
of the base upon the plane of section is found, work may be
saved by joining TF, and drawing the projections of the
other edges from the points Q, F, S, parallel to TF to meet

II — 2

164 SOLID OR DESCRIPTIVE GEOMETRY.

the perpendiculars from b\ ^, d, to the vertical trace in the
points U, V, W.

Another Method : — If the auxiliary vertical trace, in the
third method of working the section, be treated as a new xy,
the plan of the frustum on the plane of section can be
deduced without any difficulty. The student should work
the problem both ways.

2. A solid is formed by truncating a right prism, which
stands upon a regular pentagonal base of I'e^ in. side with the
nearest face of the prism parallel to the vertical plane, by two
oblique planes inclined in opposite directions on either side ;
their horizontal traces equally inclined to xy at angles of
40", equally distant 175 in. from the centre of the plan of the
base, and the line in which they intersect inclined at 60° and in
the same plane with the axis of the prism.

Draw the plan of the prism and from the centre describe
a circle of 175 in. radius. Two tangents to this circle
making angles of 40" with xy will be the horizontal traces of
the oblique planes; and a line from the point in which
these horizontal traces intersect, through the axis and
inclined at the given angle 60°, will be the common section
of the two given planes, the vertical traces of which can be
drawn through the vertical trace of this line., ■

The rest of the work is the same as in the last problem.

Develop the form and determine its projection on one of
the planes of section. Work also a new plan, the xy for
which is to be taken at 30" with the elevation of the axis.

3. A square, 2 in. side, in the horizontal plane, one side
inclined to xy at 30", is t/ie base of an oblique prism, the
long edges of which are inclined at 60" and in plan parallel td

SECTIONS BY OBLIQUE PLANES. 1 6$

a diagonal of the square. Distance between the planes of
the bases 3*5 in. Determine the section made by an oblique
plane inclined in the same direction as the edges of the prism
at an angle of ^o" and passing through a point assumed in
the upper base ; the horizontal trace to make an angle of ^f^
with xy.

Draw the projection of the prism and make the assumed
point in the upper base the vertex of a right cone with
generatrix incUned at 30". A tangent line to the circular
base of this cone and ma,king an angle of 50° with xy can be
drawn for the horizontal trace of the section plane. Find
the vertical trace, and work the section as in the other
problems.

Note. The method of working by means of the auxiliary
vertical plane of projection will be found most useful.

Show true shape of section and plan of frustum on plane
of section, and work the development.

4. A right cylinder 2 in. diameter, axis 4 in., stands on
its base and is cut by an oblique plane passing through the
middle point of the axis and inclined at angles of 60" and 50"
to the vertical and horizontal planes respectively. Show pro-
jections and true shape of section ; also the projection of the
frustum on the plane of section and its development.

Find the oblique plane by Probs. 7 and 16, Chap. in.

Take any convenient number of generatrices of the
cylinder, and find the points in which they meet the section
plane, as was done in the case of the edges of the prism,
Prob. I, and join them in the elevation by a curve, which in
this case will be an ellipse. In plan the base of the cyHnder
and the projection of the section will coincide.

l66 SOLID OR DESCRIPTIVE GEOMETRY.

For the projection of the frustum use the auxiliary
vertical plane of projection as before, and develop as in
Chap. I.

Note. The plan of the section of any right prism
standing on its base, or cylindrical surface with generatrices
perpendicular to the horizontal plane, by an inclined or
oblique plane, coincides with the trace of the form on the
horizontal plane of projection. Hence, any given figure
whatever may be the plan of a certain figure lying in a
given oblique plane : and the elevation and true shape of
this figure may be deduced in the same way as a section of
a vertical prism, or cylinder, having the given figure for its
horizontal trace, would be worked if the given plane were
the plane of section.

5. To work the section of an oblique cylindrical surface by
an oblique pla7te.

In the particular case in which the horizontal trace
of the cylinder is a circle, or part of a*x:ircle, a series
of auxiliary horizontal planes cutting the surface in similar
circles, and the section plane in lines parallel to its horizontal
trace, would be most convenient.

In the general case, in which the horizontal trace of the
cylinder is any curved line whatever, draw a number of
convenient generatrices and determine the points in which
these generatrices meet the section plane. An auxiliary
vertical plane of projection at right angles to the horizontal
trace of the section plane facilitates finding the required
points in the section.

6. Determine a regular hexagonal pyramid of ri^in.
side standing on its base, axis 4 in. long, and one side of the
hexagon making an angle of 20° with xy. Section by an

SECTIONS BY OBLIQUE PLANES. 167

oblique plane passing through a point in the axis i'5 in. above
the base ; t/ie horizontal and vertical traces making angles of
80" and 40" respectively with xy.

A line from the given point in which the section plane
meets the axis, parallel to H.P. and inclined at 80° to the
V.P., will determine a point in the vertical trace of the
section plane from which the traces of the plane may be at
once found.

Determine the section (i) as in Prob. i, by the aid of
auxiliary vertical planes containing the edges of the pyramid;
(2) by finding the intersections of the given oblique plane
with the indefinite planes of the faces of the solid ; or (3) by
means of an auxiliary elevation on a vertical plane taken at
right angles to the horizontal trace of the oblique plane of
section.

When the problem is worked by means of auxiliary
vertical planes containing the slant edges, the work may be
shortened if it be noted that the point in which the axis
of the pyramid meets the section is common to all the
intersections. It will be, therefore, only necessary to find a
second point in each intersection — such as that in which
the vertical or horizontal trace of any one of these vertical
planes cuts the corresponding trace of the section plane —
to determine the line of intersection and, therefore, another
point in the required section.

Find the true shape of the section by rabatting the section
plane as in other problems.

To draw the projection of the frustum on t/ie plane
of section.

(i) Work for the points in the section as in Problem i.

1 68 SOLID OR DESCRIPTIVE GEOMETRY.

(2) Treat the projection of the vertex on the plane of sec-
tion as the points T, U, &c. were treated in the same problem.

(3) The lines through the rabatted projection of the
vertex on the section plane and the points of section just
determined, will be the indefinite projections of the slant
edges of the pyramid. The points of the base may be
fixed by perpendiculars from their elevations to the vertical
trace to intersect these indefinite projections of the edges,
and when found may be joined consecutively to complete
the required projection of the frustum.

The development may be found by describing a circle
with radius equal to a slant edge of the pyramid, setting
along the circumference the edges of the base as chords, on
one of which the square for the base must be described,
and joining these points to the centre of the circle. The
points of section may be found by dividing the developed
edges in the same proportion as the corresponding plans or
elevations are divided by the projections of the points of
section. Fit on the true shape of the section.

7. Assume an irregular hexagonal pyramid and cut it as
in the last problem by an oblique plane. Show the true shape
of the section, projection of frustum on plane of section, and
development.

To work this problem, use an auxiliary vertical plane of
projection at right angles to the horizontal trace of the
given oblique plane.

8. A right cone, axis 3 "5 in., diameter of base 2 in., stands
upon its base and is cut by an oblique plane which passes
through three given points X, Y, Z, on its stiff ace. Show
true shape of section, projection of frustum on plane of section,
and development.

SECTIONS BY OBLIQUE PLANES. 1 69

Find the traces of the plane which contains the three
given points.

To work the section assume a series of horizontal auxili-
ary planes cutting the cone and the plane of section. The
sections of the cone will be circles, and of the plane straight
lines, parallel to the horizontal trace, drawn from the points
in which the vertical trace of the section plane meets the
vertical traces of the auxiliary planes. The points common
to the circles and the lines of section will be required
points in the section of the cone by the given plane.

Complete as in other problems. For the development
see Chap. i.

9. To work the section by a given oblique plane of a conical
surface whose directrix is any curve whatever and vertex a
t>oint anywhere in space.

Find a convenient number of generatrices and determine
their intersections with the given section plane. The line
drawn through these points will be the section required.

10. A right hexagonal prism, i'25 /;z. side, axis 4 in.
^ong, horizontal, and inclined at ^^'^ to V. P.; lowest edge 75
'n. above H. P., and a face containing it inclined at 20", is cut
'iy an oblique plane which passes through three points to be
xssumed at pleasure on any three edges of the solid. True
hape of section, projection of frustujn, ^'C. as in other probletns.

Use either vertical or horizontal auxiliary planes con-
aining the edges.

11. Take a right cylinder, axis horizontal, inclined at 35"
. 0 V. P., instead of the prism in the last problem, and work
i blique section, true shape, &'c.

I/O SOLID OR DESCRIPTIVE GEOMETRY.

Use a series of horizontal auxiliary planes cutting the
cylinder in generatrices.

12. Take a right pentagonal pyramid, side of base i"5 in.,
axis 3*5 in., when its highest face is horizontal and the plan
of its axis inclined at 40" to xy, and work a section of it by an
oblique plane.

Use auxiliary vertical planes containing the edges, except
for the edges of the upper face, in which case a horizontal
auxiliary plane will be more convenient.

Note. Since all the vertical planes containing the long
edges of the pyramid must also contain a perpendicular to
the horizontal plane from the vertex of the solid, all the
intersections of these planes with the plane of section must
pass through the point in which the perpendicular in
question meets this section plane. Hence this point may
be used to simplify the work, just as the intersection of
. the axis was used in Problem 6.

13. Work section by an oblique plane of a right cone with
its axis horizontal and inclined to the V. P. at 40°.

Use a series of auxiliary vertical planes cutting the cones
in generatrices, and observe that the "Note" to the last
problem applies also to this. Each plane will, in general,
give two points.

14. Work the section of a sphere by an oblique plane
•which passes through tivo given points on its surface and
makes an angle of 60" with If. P,

Determine the plane, and for the section use auxiliary
planes parallel to one of the planes of projection.

15. Section of the Hyperboloid of Ra^olution of one Sheets
given in the last Chapter, by an oblique plane.

SECTIONS BY OBLIQUE PLANES. I/^

Use horizontal auxiliary planes. These will cut the
surface of revolution in circles and the plane of section
n straight Hnes — the work being similar in every respect to
:hat of Problem 8 above.

Any surface of revolution with its axis at right angles to
Dne of the planes of projection might be worked in the
same way; e.g. the sphere in Prob. 14 above, the spheroid
Prob. I, "Surfaces of Revolution," Chap, vii., and the
"oUowing : —

16. Section of the annulus or anchor ring., Prob. 5,
^Surfaces of Revolution^' Chapter VII., by a tangentplane
'vhich passes through its centre and makes an angle of6Q° with
'he vertical plane of projection.

1 7. Section by an oblique plane of the screw surfaces given
it the end of Chap. VII.

Use auxiliary elevations at right angles to the H. T.
of the section plane, and deduce therefrom the required
projections of the points in which the generatrices meet
hat plane.

18. To find the section by an oblique plane of a surface
< 'f revolution with its axis oblique to both planes of projection

ind not at right angles to the oblique plane.

Use auxiliary planes of section perpendicular to the
ixis of the surface of revolution. These cut the oblique
)lane in straight lines and the surface in circles. The
)oints common to these lines and circles are points in the
;urve of intersection.

Note. The difficulty of projecting the surface of re-
■olution in the position given is generally sufficient to deter

1/2 SOLID OR DESCRIPTIVE GEOMETRY.

any but the most courageous of students from attempting
this problem. It may, however, be remarked that the
projection of a surface of revolution on a plane inclined to
its axis, is the section by that plane of a cylinder enveloping
the surface and perpendicular to the plane of projection.
For the construction of a cylinder enveloping a surface of
revolution see Prob. 4, "Problems on Surfaces of Revolu-
tion," Chap, VII.

IX.

INTERSECTIONS.

General Remarks.

The methods by which the intersection of two surfaces
i.re determined when one of them is a plane, have been
])retty fully exemplified in the preceding chapter, and it
remains now only to comment briefly on the general case.

If a section plane be taken through any two intersecting
surfaces whatever, and the lines of section with each of the
surfaces be drawn, the point, or points, in which the lines
c f section meet each other, will be in the intersection of the
sarfaces. Consequently, by taking a series of these auxi-
1 ary section planes, a sufficient number of points may be
f )und, and the line of intersection drawn through them.

Although the auxiliary planes may be chosen arbitrarily,
it is obviously convenient, in general, to take them parallel,
o • perpendicular, to one of the planes of projection, so that
o le projection of the section shall always be a straight line.
T his rule has, however, sometimes to give way to other
c )nsiderations, as when, for instance, we are dealing with
J uled Surfaces, in which case it often facilitates the work if
tl e auxiliary planes be so taken as to cut the surface along
it ; rectilineal generatrices.

1/4 SOLID OR DESCRIPTIVE GEOMETRY.

Note. For the application of the principle of parallel
sections to the intersection of forms given by their figured
plafis, see Prob. ii, Chapter x., "Figured Projections and
3cales of Slope."

FORMS BOUNDED BY PLANE SURFACES.

I. Two right prisms, with square bases of 2-^ in. side,
intersect. One is placed with its edges horizontal, and parallel
to the vertical plane ; its lowest face inclined 30°, and its
lowest edge i in. above the horizontal plane. The other stands
on its base with one face inclined 2^° to the vertical plane, and
its axis passing throtigh the middle point of the highest edge of
the horizontal prism. Axes of the prisms 4 in. long. Draw
plan and elojation showing the intersection, and develop the
vertical prism.

To get the prisms into position : —

Commence with an auxiliary elevation of the horizontal
prism on a vertical plane perpendicular to its axis, so that
the elevations ^of both bases coincide. The elevation of the
prism in this position will be a square, 2*5 in. side, the
lowest corner i in. above, and one side making an angle of
30" with xy. For the final elevation take a new xy parallel
to the plan of the axis.

For the plan of the vertical prism, describe a circle of 2-5
in. diameter from the middle point of the plan of the
highest edge of the horizontal prism as centre. A pair of
parallel tangents to this circle making angles of 25<* with
the new xy, and a second pair at right angles to the former,
will determine the square base of the vertical prism in the
required position. Thence determine the elevation on the
new vertical plane.

INTERSECTIONS.— PLANE SURFACES. 1/5

To fijid the intersection : —

The forms being bounded by plane faces, the lines of
intersection will be straight. And since but one straight
line can be drawn between two points, it is only necessary
to determine two points in each of the intersections of the
faces, and to draw the line of intersection through them.
The points most convenient for the purpose are those in
which each of the edges of the one prism pierces the faces
of the other.

Begin with the edges of the horizontal prism. The
plans of the points in which these meet the faces of the
vertical prism are shown where the plans of the horizontal
edges of the one prism cut the square which is the plan of the
other. Determine the elevations of these points by drawing
projecting-lines from their plans to meet the elevations of
:he edges in which they lie.

Observe : — Each edge should be consideied separately, and
xs much of it as is visible be put in as a dark continuous line ;
■ iny part that is invisible should be dotted, and that portion
which passes through the other prism rubbed entirely out.

Unless this rule be strictly attended to before any
ittempt is made to join up for the intersection, inextricable
:6nfusion will in general be the end of an abortive eftbrt
:he reverse of instructive.

For the edges of the vertical prism, Measure the heights
It which they pierce the faces of the horizontal prism from
he auxiliary elevation, and transfer them to the new vertical
)rojection.

Treat each edge of the vertical prism in the same way
is recommended for the horizontal edges in the observation
jrinted in italics above.

176 SOLID OR DESCRIPTIVE GEOMETRY.

The lines of intersection can, at this stage, be easily put 111 .
from inspection.

To work the development : —

Develop the entire prism as In Prob. 3, Chap, i, and for
the development of the intersection, find, first, the points on
the edges, by measuring from one of the elevations, and,
second, the points in the faces. These are determined by
their distances from the edges and their heights above the
base.

Prick off the development on a sheet of cardboard and
make the model of the intersection.

2. Using the same plan and auxiliary elevation as in the
preceding problem, make a new elevation on a vertical plane
inclined at 40" to the long edges of the horizontal prism, so as
to show the parts which were invisible in the elevation in
Problem i. Develop the horizontal prism and deduce a second
plan of the intersecting solids — xy at pleasure.

3. A regular pentagonal prism, i '5 in. side, 5 in. long,
stands on the horizontal plane with the vertical face passing
through the side of the base nearest to y^y parallel to V.P., and
is intersected by a similar and equal prism 7vith its axis
horizontal, 2'^ in. high, inclined at 32° to V.P. and 'ic^in.

from the axis of the other. The rectangular face of the
horizontal prism that is furtliest from V.P. to be vertical.
Plan, elevation, and development.

Begin with a plan of the vertical prism at a convenient
distance in front of xy.

Since the axis of the horizontal prism is Inclined at 30"
to V.P. the base must be inclined at 60°, the complement
of the angle of inclination of the axis, to the same plane.

INTERSECTIONS— PLANE SURFACES. lyj

Determine an auxiliary vertical plane of projection parallel
io the base of the horizontal prism, and show thereon an
elevation of the vertical prism. A point in this elevation,
2 "5 in. above xy and "25 in. from the elevation of the axis
of the vertical prism, will be the elevation of the axis of
:he horizontal one, and if a circle be described from this
:)oint, with the proper radius, the pentagon which is the
luxiliary elevation of the horizontal prism may be put
about it with one side vertical, and the plan and new
elevation be deduced without difficulty.

The intersection is worked as in Probs. i and 2.

4. A right pristn, 4 in. long, bases equilate?'al triangles
if 2 in. side; axis horizontal, inclined 30" to V.P. ; lowest
i-ige -z^in. above, and a face containing it inclined 15" to
/I. P., intersects a right pyramid, square base 2*5 in. side, axis
J in. long, vertical, and passing through the middle point of the
cxis of the prism. Side of square base inclined at 30" to xy
c nd not parallel to the long edges of the prism. Show intersec-
t 'on in plan and elevation and work development of the pyramid.

Determine plan and elevation of prism as in Problem 6,
Chap. I., page 31, by means of an auxiliary elevation on a
\ lane at right angles to the axis.

For the pyramid ; — From the middle of the plan of the
a ds of the prism as centre, describe a circle of 2 '5 in.
diameter and circumscribe it by a square, in the right
p osition, for the plan of the base of the pyramid. Thence
d2termine its elevation, and also an elevation on the auxi-
liiry vertical plane of projection which was used to find
t le plan of the prism.

To work the intersection : —

(i) Find the projections of the points in which the
e Iges of the pyramid pierce the faces of the prism.
E. G. 12

1/8 SOLID OR DESCRIPTIVE GEOMETRY.

The elevations of these points are shown in the auxi-
liary plane where the elevations of the edges of the pyramid
cut the equilateral triangle which is the projection of the
prism on that plane. From these the other required pro-
jections may be found by drawing projecting-lines from
them to meet the other projections of the lines that contain
the points.

When these points are found, put in the edges of the
pyramid in the way recommended for the edges of the
prism Prob. i.

(2) Find the projections of the points in which the
edges of the horizontal prism pierce the pyramid : —

Assume that the three corners of the equilateral triangle
in the auxiliary elevation are the elevations of the points in
which the horizontal edges of the prism meet the faces of
the pyramid, and, on this assumption, draw lines from the
vertex of the pyramid through these points in its faces and
produce them to the base. The projections of these lines
on the auxiliary plane will be drawn from the projection of
the vertex on that plane through the three corners of the
triangle. Each of these is the elevation of two lines — one
passing through the point in which the edge of the prism
enters, and the other through the point in which it leaves
the pyramid. Determine these lines in plan. The points
common to these lines and the plans of the horizontal
edges of the prism will be the plans of the required points,
from which the other elevations may be at once obtained.

Put in the projections of the edges in the proper manner
and then join up for the projections of the intersection.

Develop the pyramid as in Prob. i, Chap. i. The points in
the faces may be readily found by showing in the develop-

INTERSECTIONS— PLANE SURFACES. 179

ment the lines which pass through them, and determining
them in these Hnes in the same way as the points in the
edges are found.

5. For the prism in the preceding problem, substitute an
irregidar pentagonal prism with a re-entering angle. No face
to be either horizontal or vertical : axis of pyramid to pass
throjfgh the middle point of the highest edge of the prism, and
the latter form to have two long edges — one on each side of the
axis — entirely free from the pyramid. Other conditions as in
Problem 4. Plan, elevation, and development of pyramid
showing the lines of intersectioft.

Begin with a plan of the pyramid and an auxiliary
elevation on a vertical plane parallel to the plane of the
base of the prism. The end elevation of the prism may
then be put in so as to fulfil the conditions, and the problem
worked as in Prob. 4 above.

6. Intersection of a right square pyramid and a regular
i>entagonal prism.

Pyramid: — Axis 4 in. long, inclined at 35" to H.P., and
in plan making an angle of 2^ with xy : side of square base
2 in. and inclined at 30° to the horizontal trace of the plane of
'he base : lowest corner resting upon H.P.

Prison : — Standing on the H.P. tvith a regular pentagon
f \'^in. side for its base ; axis 3*5 in, long and vertical.

Inter sectio7i when the prism is so placed that the vertical

face furthest fro7n xy is parallel to the V.P. of projection, and

has one of its long edges free from the pyramid. All other

'dges of the prism to pass through the pyramid and one of

hem to meet it at a point in its base.

Work also a development of the pyramid.
The points in which the edges of the pyramid meet

12 — 2

l80 SOLID OR DESCRIPTIVE GEOMETRY.

the vertical faces of the prism are shown in plan where
the plans of these edges cross the sides of the penta-
gonal base, and can thence be determined in elevation.

Note : — It sometimes happens that an edge of the
pyramid pierces a base of the prism, in which case the
point of intersection is readily deduced from the elevation.

The intersections of the vertical edges of the prism
with the faces of the pyramid can be found either by the
general method for the intersection of a straight line and a
plane, or by the following equivalent special construction : —

Draw lines from the plan of the vertex of the pyramid
through the points which are the plans of the vertical edges
of the prism, and produce them to the base of the pyramid.
Assuming each of these lines to be the plan of two lines —
one drawn from the vertex through the point in which the
edge of the prism enters one face of the pyramid and the
other from the vertex through the point in which it leaves
another — show them in the elevation. The intersections
of the elevations of these lines with the elevations of the
edges through which they are drawn in plan, give the
elevations of the required points of intersection.

The intersection of the edge of the prism with the base
of the pyramid may be similarly found by drawing a line,
say, from one corner of the plan of the square base,
through the plan of the vertical edge meeting it, and
showing this line in the elevation of that base.

Complete the representation of the edges of the inter-
secting forms and then join up the points of intersection
from inspection.

7. A right hexagonal pyramid^ i"]^in. side, axis 51/7.

INTERSECTIONS— PLANE SURF A CES. 1 8 1

long, stajids on its base with one side making an angle of 40"
with xy, a7id is intersected by a right square prism, 2 in. side,
axis passing through that of the pyramid, inclined 30° to H.P.,
a?id in plan making an angle of ^0° with xy. JVo side of the
square to be horizontal. Show plan, elevation, and intersection.

Develop the prism.

Find the points in which the edges of the pyramid meet
the faces of the prism, by the aid of auxiliary vertical
section planes cutting the prism and containing the slant
edges of the pyramid. Similarly, find the intersections of
the edges of the prism with the faces of the pyramid by the
aid of vertical planes containing the former.

8. Work the following general cases : —

a. Two intersecting prisms of four and six sides respec-
tively, axes oblique to both planes of projection. Other data
^t pleasure.

p. Two intersecting pyramids of four and five sides
■espectively, axes oblique to both planes of projection. Other
lata as above,

y. General case of irregular prism and pyramid.

When the traces of the indefinite planes of the faces
] all within the limits of the paper the intersections of these
;)lanes may be found by Prob. 12, Chap. in. and that of
ihe forms put in therefrom. As, however, it seldom
] appens that all the traces of the planes intersect within the
1 mits of the drawing, recourse is had to auxiliary section
1 lanes. Two such planes — one parallel to one plane of
] rojection and the other to the other — will be found most
( onvenient for the purpose. From the points furnished by
tie sections, and such of the traces as meet on the paper,
1 le intersections can be put in without much difficulty.

1 82 SOLID OR DESCRIPTIVE GEOMETRY.

FORMS BOUNDED BY CURVED SURFACES.

As far as possible the following exercises have been
grouped according to the inost convenient methods of work-
ing them. Some of the problems admit of several con-
structions, but on the whole it will be found that the problems
worked in each group afford types of the most approved
method of working those of the same group to which no
solutions are appended.

^^ Critical lines" and ^^ Key points." In all intersections
that involve curved surfaces the student should direct his
attention to the lines in which the projecting surfaces touch
the curved forms and the points in which these lines on the
one form meet the surface, or surfaces, bounding the other.

The line of intersection always passes through these
points, and inasmuch as a habit of attending to them, and
to their corresponding lines, greatly simplifies matters by
developing a power of prospective insight into the characte-
ristic features of the intersection resulting from any given
combination, they have been specially designated the critical
lines and key points.

Group I.

I . Intersection of a right cotie and a sphere. Cone standing
on its circular base of 2'<^ in. radius, axis 3 in. long: sphere

INTERSECTIONS^CURVED SURFACES. 1 83

resting upon H.P.^ radius 1*25 in., centre i in. from the axis
of the cone ; line joining the plans of the centre of the sphere
and the vertex of the cone inclined at 2i^^ to xy. Development
of cone showing curve of intersection.

Take a series of sections of both surfaces by auxiliary
planes parallel to H.P. The elevation of each will be a
straight line parallel to xy, and the plan, two circles — one
described from the plan of the centre of the sphere and the
other from the plan of the vertex of the cone, as centres, with
radii that may be measured from the elevation. The points
in which each pair of circles intersect will be points in the
plan of the intersection of the forms, and from these the
2levations may be found by projecting, and the projections
Df the intersection put in.

Note. It will simplify the work if the auxiliary planes
ire taken at equal distances above and below the centre of
;he sphere — so that one circle in plan shall serve for two
jections of that surface.

For the development.

(i) Divide the base of the cone in plan into a number
)f equal parts and put in the plans of the generatrices from
;he points of division to the vertex.

(2) Develop the cone and show these generatrices
n the development.

(3) Mark the points where the curve of intersection, in
')lan, meets the plans of the generatrices, and divide the
corresponding generatrices in the development in the same
)roportion, for the points through which the curve of inter-
ection passes.

2. Arrange an example of two intersecting spheres of

184 SOLID OR DESCRIPTIVE GEOMETRY.

unequal sizes and unequal heights; the line joining their centres
oblique to both planes of projection.

3. Ititersection of a sphere and a surface of revolution
with vertical axis — e.g. the Spheroid, or the Hyperboloid, in
Chap. VIL

4. Intersection of a right cone and a right cylinder with
their axes vertical.

Note. The plan of the intersection in this problem, and
in the next, coincides with the circle that is the plan of the
cylinder.

5. Intersection of a sphere and a right cylinder standing
on its base.

Note. Any surface of revolution with its axis vertical
might be substituted for the sphere in this problem.

6.- Arrange an example of a sphere, or a spheroid, axis
vertical, intersecting an oblique cone or cylinder with a circular
horizontal trace.

7. Intersection of a sphere with the anchor ring, Prob. 5,
'■^Surfaces of Revolution^'' Chap. VII.

8. Intersection of an oblique cone, or a cylinder, having
a circle for its horizontal trace, with an anchor ring, axis
vertical.

9. Intersection of a right cone standing on its base and an
oblique cone, or cylinder, with circular horizontal trace.

I o. Work one of the following combinations : —

a. Two oblique cylinders with circular horizontal traces.

(3. Two oblique cones with circular horizontal traces.

INTERSECTIONS— CURVED SURFACES. 1 85

7. Oblique cone and oblique cylinder with circular horizon-
tal traces.

Group 1 1.

I. Intersection of two right cylinders. One 2 in. diameter ;
axis, horizontal, 2*5 in. above If. P., inclined 40° to V.P., and
5 in. long ; the other 2*5 in. diameter, axis vertical and 5 in.
long.

a. When the axis of the vertical cylinder passes through
the middle point of the horizontal one.

p. When the axis of one is "25 in. from that of the other
so that the cylinders have a common tangent-plane.

y. When t/u axis of one is '5 in. from that of the other.

Develop the horizontal cylinder in the last.

To find the intersection : — Divide one base of the horizontal
cylinder into any number of equal parts (say sixteen), and
through the points of division draw straight lines (i.e.
generatrices) on the cylindrical surface, and show them
in plan and elevation.

( 1 ) To show the lines in plan : — Rabat the base about its
horizontal diameter till parallel to the H.P. This will be
a circle described upon a short side of the rectangle which
is the plan of the horizontal cylinder. Draw half of the
circle only and divide it into eight equal parts. Lines from
the points of division, parallel to the plan of the axis, will
determine the plans of the generatrices— each line, with the
exception of the two outside ones, representing two gene-
ratrices— one above and one below the level of the axis.

(2) To show the lines in elevatioji: — Rabat the base
about its vertical diameter till parallel to the V.P. This

1 86 SOLID OR DESCRIPTIVE GEOMETRY.

will be a circle described upon the major axis of the ellipse
which is the elevation of the base of the horizontal cylinder.
Only half of this circle need be shown, and this, if divided
into eight equal parts, will furnish points through which lines
parallel to the elevation of the axis may be drawn for the
elevations of the generatrices. These lines in elevation, with
the exception of the highest and lowest, will each represent
two generatrices — one on the front and one on the back
part of the cylindrical surface.

Note. Any horizontal lines whatever may be taken on
the cylinder in plan, and their elevations determined in the
vertical projection by the aid of an auxiliary elevation of the
horizontal cylinder on a vertical plane at right angles to its
axis, whence the heights can be measured : but it is much
more convenient to take them at equal distances apart,
especially when a development has to be worked.

Having shown these generatrices in plan and elevation,
the points in which they meet the surface of the vertical
cylinder may be determined for points in the curve of inter-
section.

Iti plan., the curve of intersection coincides with the
circle that is the plan of the vertical cylinder, and the plans
of the points in which the lines on the horizontal cylinder
meet the surface of the vertical cylinder are shown where the
plans of these horizontal lines cross its circular base. Project-
ing-lines, drawn from these points to meet the corresponding
horizontal lines in elevation, determine the points through
which the elevation of the line of intersection is to be drawn.

Note. Pay attention to the critical lines and key points,
and number the generatrices in plan and elevation to avoid
confusion.

INTERSECTIONS— CURVED SURFACES. 1 8/

The development may be worked as in preceding
chapters.

2. Intersection of a right cylinder with a right cone.
Cylinder 2*5 in. diameter, axis 5 in. long and vertical ; cone
with base 2-5 in. diameter; axis, 5 in. long, horizontal, in-
clined 45" to V.F. and 2*5 in. above H.P.

a. When the axis of the cylinder passes through the middle
point of the axis of the co?ie.

p. When the two surfaces have a common iangentplane.
y. When part of the cone is without the cylinder.
Develop the cone in ft.

Divide the base of the cone into a number of equal parts,
as was done with the base of the cyhnder in the foregoing
problem, and show plan and elevation of the generatrices. —
The lines in this case instead of being parallel to the axis will
radiate from the vertex to the points on the base, and the
student must observe that although in the plan each line will,
as in the cylinder, represent two generatrices, in the elevation
this will not be the case, /. e. the front and back lines not
being parallel to the axis will not coincide.

When the plans and elevations of the generatrices are
determined, the intersection is found as before.

The development is worked as in other chapters.

3. Arrange a right cylinder standing on its base and find
its line of section with a right cylinder whose axis is inclined
to both planes ofprofection.

Take any number of convenient points in the horizontal
trace of the oblique cylinder and draw plans and elevations
of generatrices passing through them. The elevation of the

1 88 SOLID OR DESCRIPTIVE GEOMETRY.

intersection can then be found from the points where the
plans of the generatrices cross the horizontal trace of the
vertical cylinder, as in the other problems of this group.

Note. The oblique cylinder might be assumed with a
circle for its horizontal trace and worked as above, but in this
case it could be as conveniently worked by a series of hori-
zontal auxiliary sections, as in Group I.

4. Arrange a vertical cylinder and an oblique cone and
find their intersection.

The generatrices in this case are drawn to the vertex of
the cone ; with this exception the description of the con-
struction for the preceding problem, and the "Note," apply
to this.

5. Find the intersection of the anchor ring, Proh. 5,
'•'■Problems on Surfaces of Perolution" Chap. VIL, with a
vertical cylinder.

Divide a vertical circular section of the ring into a number
of equal parts, as for the base of the cylinder, Prob. i, of this
group, and through the points of division draw horizontal
circular lines on its surface. The ring when thus prepared
may be likened in some respects to a long cylinder bent
into a circular form, and the intersection will be found to
present no greater difficulty than that of the horizontal
cylinder in Prob. i above.

Note. This same construction, looked at in another
aspect, might be regarded as an illustration of the method of
working by a series of parallel sections as in Group I.

Group III.

I. Intersection of a right cone standing on a circular base
of I in. diameter, axis vertical and ^in. long, with a right

INTERSECTIONS— CURVED SURFACES. 1 89

cylinder resting upon H.P. ; base 2 '5 in. dia7neter, axis
horizofital, 5*5 in. long, inclined /Hf^" to xy, and in plari 75 in.
distant from the plan of the vertex of the cone. Develop the
cone with curve of intersection.

Determine horizontal lines (generatrices) on the surface
of the cylinder as in the horizontal cylinder, Prob. i, Group
II., and work sections of the cone by auxiliary horizontal
planes containing these lines.

In elevation the sections of the cone will be represented
by the segments of the elevations of the generatrices that lie
between the legs of the triangle forming the elevation of the
cone.

In plan these sections will be circles, described from the
plan of the vertex of the cone as centre, with radii measured
from their corresponding elevations.

The points in plan in which the generatrices of the
cylinder meet the corresponding circles of the same level,
will be points in the plan of the intersection, through which
the curve must be drawn by hand.

Projecting-lines, drawn from these points to the elevations
of the generatrices in which they lie, will determine points in
the elevation of the curve of intersection.

2. Liter section of a horizontal cylinder with a sphere.
Sphere 4 in. diameter, and centre 2 '5 in. high.

Cylinder 2 in. diameter^ axis horizotital, inclined 45° to
V.P., in plan '5 in, from, and in elevation 75 in. below, centre
f sphere.

Develop the cylinder.

3. If iter section of a horizontal cylinder with the Hyper-
'wloid of Revolution of one sheet given in Chapter VII.

1 90 SOLID OR DESCRIPTIVE GEOMETRY.

4. A surface generated by the revolution of a circle of 2, in.
diameter about a tangent-line as axis, rests upon H.P. with the
axis vertical ; a sphere ^2*5 in. diameter rolls along the H.P.
on a line inclined at 40" to V.P. and tangential to the circle in
which the surface of revolution touches the H.P. Show the
opening through which the sphere would pass.

5. Arrange an example of a horizontal right cylinder,
axis inclined to V.P., intersecting an oblique cone, or oblique
cylinder, which has a circle for its horizontal trace.

Group IV.

I . A r range an example of a horizontal right cylinder, axis
inclined to V.P., intersecting another right cylinder, axis
oblique to both planes of projection, and show the curve of
intersection.

Auxiliarysectionsof the cylinders determined by horizon-
tal planes as in Group III. would cut the horizontal cylinder
in generatrices and the oblique cylinder in ellipses. The
labour, however, of drawing these curves would be consider-
able, and it is, therefore, better to take auxiliary sections by
planes containing generatrices of the horizontal cyHnder and
parallel to those of the oblique cylinder.

( 1 ) Prepare the horizontal cylinder by drawing a number
of rectilineal generatrices on its surface, as in preceding
problems.

(2) Assume a point in a generatrix of the horizontal
cylinder and draw therefrom a line parallel to the generatrices
(or axis) of the oblique cylinder.

(3) Find the H.T. of this line and through it draw a
straight line in the H.P. parallel to the plan of the generatrices
of the horizontal cylinder.

INTERSECTIONS—CURVED SURFACES. IQI

The line last drawn will be the H.T. of an auxiliary
plane containing a generatrix of the horizontal cylinder
and parallel to the generatrices of the oblique cylinder.

(4) From each of the two points in which, in general,
the H.T. of this auxiliary plane meets the curve which is the
horizontal trace of the oblique cylinder, draw plans of the
generatrices of the latter surface. These will be the plans
of the lines in which the auxiliary section plane cuts the
oblique cylinder, and the points in which these lines meet
the generatirix of the horizontal cylinder will be points in
the plan of the curve of intersection of the surfaces. Simi-
larly, other points may be found, by taking auxiliary planes
through each of the generatrices of the horizontal cyHnder,
and the curve put in by hand.

The elevation may be readily found from the plan.

2. Arrange an example of a horizontal right cylinder,
axis inclined to V.F., intersecting an oblique right cone, and show
the curve of intersection.

Take the auxiliary section planes in this case containing
generatrices of the horizontal cylinder and the vertex of the
oblique cone.

Note. To find the H.T. of the section planes, the Hnes
drawn from the assumed points in the generatrices of the
horizontal cylinder, in the last problem parallel to the axis of
the oblique surface, must now be drawn from the vertex of
the cone. The oblique cylinder may in fact be likened to a
cone with its vertex at an infinite distance.

3. Arrange a right cone standing on its base intersecting
a cylindrical surface oblique to both platies of J>roJectiofi, and

192 SOLID OR DESCRIPTIVE GEOMETRY.

show curve of intersection. The H.T. of the cylinder not to he
a circle.

Take auxiliary section planes containing the vertex
of the cone and parallel to generatrices of the cylinder.

Note. The horizontal traces of these planes will all
pass through the H.T. of a line drawn from the vertex
of the cone parallel to the generatrices of the oblique
cylinder.

4. Arrange a right cone standing on its base intersecting
another right cone whose axis is inclined at, say, 40" to the V.P.,
and horizontal.

Take auxiliary planes containing generatrices of the cone
whose axis is horizontal and passing through the vertex of
the one whose axis is vertical.

Note. All the horizontal traces of these planes will
pass through the H.T. of the line joining the vertices of the
two cones.

(i) Divide the base of the horizontal cone into equal
parts and determine the lines on its surface as in Prob. 2,
Group II.

(2) Draw a straight line from the vertex of the vertical
cone through any convenient point in a generatrix of the
horizontal one and find its H.T. This will be a second point
in the H.T. of the auxiliary plane containing the generatrix.
Similarly, the horizontal traces of the other auxiliary planes
may be determined, and the intersection worked as before,

5. Arrange a right cotie standing on its base a?idfind its
intersection with another right cone, axis oblique to both planes
of projection.

INTERSECTIONS— CURVED SURFACES. 193

General Cases.

I. To find the intersection of two cylindrical surfaces
oblique to both planes of projection.

Take auxiliary section planes parallel to the generatrices
of both surfaces.

Note. Some little difficulty is usually experienced in
getting the cylinders into a position that will give a satis-
factory intersection, unless somewhat cumbrous and elaborate
data are supplied for their arrangement. One of the
simplest ways of arranging an intersection of these surfaces
is to begin by drawing two lines inclined at an angle of about
120°, and treat these as the plans of the axes of the inter-
secting cylinders. Let one, for example, be a right cylinder
of 2*5 in. diameter, axis inclined 55", and the other a right
cylinder of 2 in. diameter, axis inclined at 40", and, say, a
little below the other.

To find their intersection. Determine plan and elevation
of any assumed point in space, and the horizontal traces of
two lines drawn from the point, one parallel to the axis of
the one cylinder, and the other parallel to the axis of the
other. The line joining the horizontal traces of these Hnes
will be the H.T. of a plane containing them, and therefore
oi z plane parallel to the generatrices of the two surfaces.

A series of lines parallel to this horizontal trace can then
be taken, cutting the horizontal traces of the cylinders, for
the horizontal traces of the auxiliary planes.

Lines from the points in which the horizontal traces of
the auxiliary planes meet the horizontal traces of the surfaces,
parallel to the plans of their respective axes, determine points
in the plan of the curve of intersection, which may be drawn
n by hand and projected for the elevation.

E. G. 13

194 SOLID OR DESCRIPTIVE GEOMETRY.

Attention should be given to the critical lines and key
points.

2. To find the intersection of two conical surf aces oblique
to both planes of projection.

Take auxiliary section planes passing through the vertices
of both cones and, therefore, cutting both surfaces in
generatrices.

Note. The horizontal traces of all the auxiliary section
planes pass through the H.T. of the line joining the vertices
of the two surfaces, and the vertical traces of the same planes
pass through the V. T. of that line.

Sometimes, when the vertical trace of one of the cones
is more accessible than its horizontal trace, the vertical
traces of the auxiliary section planes are required.

3. To find the intersection of an oblique cylindrical
surface with an oblique cone.

Take auxiliary section planes passing through the
vertex of the cone and parallel to the generatrices of the
cylinder.

Note, The horizontal traces of these auxiliary planes
all pass through the H.T. of a line drawn from the vertex of
the cone parallel to the generatrices of the cylinder.

Group V.

I. Arrange an oblique right cylinder intersecting a sphere^
and find the curve of intersection.

If an auxiliary section of the two surfaces were taken by
a horizontal plane the curve cut from the sphere would be
a circle, and that from the cylinder an ellipse, similar and

INTERSECTIONS— CURVED SURFACES. IQS

equal to that which is the horizontal trace of the latter
surface. The two points in which the ellipse and circle meet
would be points in the curve of intersection of the surfaces.
If, now, an auxiliary cylindrical suj-face were determined,
having the circle which is cut from the sphere for its
directrix and lines parallel to the generatrices of the oblique
cylinder for its generatrices, its horizontal trace would be a
circle similar and equal to that which is the directrix of the
auxiliary surface, and if generatrices of this cylinder were
drawn from the points in which the circular horizontal trace
of the auxiliary surface cuts the elliptic trace of the original
oblique cylinder, they would cut the directing circle in the
same points as the ellipse first mentioned. Hence the
following construction : —

(i) Determine a series of sections of the sphere by hori-
zontal planes.

(2) From the centre of one of these circles draw a line
parallel to the obHque cylinder.

(3) Find the H.T. of this line, and from that point as
;entre describe a circle in the H.P. with radius equal to that
)f the circle cut from the sphere.

(4) From the points in which this circle (which is the
'. I.T. of the auxiliary cylinder) cuts the ellipse, draw gene-
1 atrices meeting the directing circle, for points in the curve
( f intersection.

Similarly, by treating the other circles in the same way a
r umber of points in the curve of intersection will be found.

2. Arrange an oblique cone intersecting a sphere, and
J. >id the curve of intersection.

13—2

196 SOLID OR DESCRIPTIVE GEOMETRY.

The construction in this example is similar to that of the
preceding problem, except that the auxiliary surfaces are
cones, having the same vertex as the given oblique cone,
instead of cylinders; the directrices being, as before, the
circular sections of the sphere.

Note. The constructions given for the two problems of
this group apply to the intersection of an oblique cylinder or
cone with any surface whose sections by horizontal planes
are circles, e.g. the Anchor ring, or the Spheroid, with ver-
tical axis.

Group VI.

I. To find the intersection of two surfaces of revolution
when their axes lie in one plane.

There are two cases : —

a. When the axes are parallel.

ft. When the axes meet.

For a : — Use auxiliary section planes at right angles to
the axes.

For ft : — Take auxiliary sections of the surfaces by a
series of concentric spheres, described from the point in
which the axes meet as centre. Every such auxiliary sphere
will cut each of the surfaces of revolution in a circle whose
plane is perpendicular, to the axis of the surface, and the
points, in which the pair of circles cut by each sphere meet,
will be points in the curve of intersection. The following
special exercise is deserving of attention :—

A triangle ABC is the base of a pyramid whose vertex is
V. AB=sin.;BC=A in. ; CA = ^ in. The angle A VB
= 6o»; B VC= 45" / CVA^ 40".

INTERSECTIONS— CURVED SURFACES. I97

Find the plan of the point V afid its height above the base.

(i) Draw the triangle ABC, and on AB describe a
segment of a circle containing an angle of 60", on BC one
containing an angle of 45°, and on CA a third containing an
angle of 40". Euclid, Book III. Prop. 2,2»'

If these segments be rotated about the lines AB, BC,
CA as axes, surfaces of revolution will be generated whose
common intersection will determine V.

(2) Find the intersection of any two of these surfaces
thus : — AB and BC being the two axes of the surfaces
whose intersection we require, from B with any radius
describe a circle cutting the segment described on AB in /,
and the segment on ^C in ^.

If the segment on AB revolve about that line as an axis,
the point/ will trace a circle, in a plane perpendicular to
AB, on the surface of a sphere with centre B and radius Bp.
The plan of the locus of/ is, therefore, a straight line drawn
from / at right angles to AB and produced to an equal
distance on the other side of that axis.

Similarly, the plan of the locus of q — which also traces a
circle on the above-mentioned sphere — will lie in the perpen-
dicular from g io BC

The point of intersection of the plans of the loci is a
point in the plan of the intersection of the two surfaces of
revolution.

In like manner any number of points may be found
in the plan of the intersection, and the latter line drawn
through them.

(3) Find the intersection of the third surface with one
of the foregoing. The point v, where the plans of the two
intersections cross, is the plan of the vertex V.

IpS SOLID OR DESCRIPTIVE GEOMETRY.

It may be remarked that there are two points answering
to V, one above, and one below the plane of the triangle
ABC. The latter is neglected.

(4) Complete the plan by drawing lines from v to A, B,
and C, and find the height of Fthus : —

Draw a line from v perpendicular to one of the axes (say
AB) and produce it to meet the segment of the circle described
on this line in the point x. Then Bx will be the real length
of the edge of which Bv is the plan. From these data the
height of V can be at once determined.

2. To find the intersection of two surfaces of revolution
when their axes are not in the same plane.

Take a series of sections of both surfaces by auxiliary
planes and construct the points in which the curves of section
meet for points in the intersection of the surfaces.

Note. This problem presents no difficulty except the
mechanical one of drawing the curves of section, and to
simpUfy this as much as possible assume one of the planes of
projection at right angles to one of the axes, and the other
plane of projection parallel to both. The auxiliary section
planes may then be taken parallel to the former plane of pro-
jection and the intersection deduced without much trouble.

COMBINATION OF CURVED FORMS WITH
FORMS BOUNDED BY PLANE SURFACES.

I. Intersection of a right hexagonal prism standing on
its base of 1 in. side with a sphere of 2'<, in. diameter. One
face of the prism making an angle ^20" with xy; centre of
sphere 2 in. above H.P. and in plan '5 in. from the plan of the
axis of the prism.

INTERSECTION'S— COMB IN A TIONS. 1 99

Take auxiliary sections of the sphere by a series of
horizontal planes in pairs at equal distances above and
below the centre of the sphere.

The plan of the intersection coincides with the base of
the prism. The elevation can be found by projection from
the points furnished in plan by the auxiliary sections.

Note. Any surface of revolution with vertical axis,
or oblique cylinder or cone with circular horizontal trace,
might be substituted for the sphere in this problem.

2. Arrange an example of a right square prism, axis
horizontal and inclined to V.F., intersecting a right cylinder,
axis vertical.

Divide each edge of the base in plan and in elevation
into a convenient number of similar parts, and draw through
the points of division a series of lines in each face parallel
to the long edges of the prism.

The plan of the intersection coincides with the base of
the cylinder. For the elevation, it will be necessary to draw
projecting-lines from the points where the parallel lines in
plan, and the edges of the prism, cross the circle, to the
corresponding lines in elevation. Special attention should
be given to the edges of the prism, which should be put in
dark or dotted, or partly rubbed out as they require, before
joining the points for the curves of intersection.

Note. The same construction would apply were the
prism oblique.

3. Arrafige an oblique pyramid intersecting a vertical
right cylinder.

Divide each edge of the base of the pyramid as before,
and draw the lines to the vertex. The rest of the work is
the same as in the preceding problem.

200 SOLID OR DESCRIPTIVE GEOMETRY.

4. Arrange a right pentagonal prism standing on its base
intersecting a horizontal right cylijider.

Use horizontal lines (generatrices) on the surface of the
cylinder as in previous cases of the intersection of the
horizontal cylinder.

Pay attention to the key points and critical lines.

Note, The cone might be substituted for the cylinder
in this problem, in which case the lines would be drawn
from the points of division of the base to the vertex.

5. Arrange a right pentagonal prism standing on its base
intersecting a right cone, axis oblique to both planes of projectioji.

6. Intersection of Cone and Prism.

Prism, axis horizontal, 4 in. long, and inclined -^d^ to V.P.,
bases equilateral triangles of 2 in. side, lowest face inclined 1 5°,
and lowest edge "25 in. above H.P.

Cone, axis vertical, 3 in. long, and passing through the
middle point of the axis of the prism, base 1*5 in. radius.

Draw a series of horizontal lines in each face of the
prism, as in the prism Prob. 2 above.

Horizontal planes, passing through these lines and
cutting the cone in circular sections parallel to its base, may
now be taken, and the points in which the plans of the
horizontal lines and circular sections of the same levels
meet, joined up for the plan of the intersection. The
elevation can be completed by projecting.

7. Arrange a square prism, axis horizontal, intersecting a
sphere.

Same method as in Prob. 6. Any surface of revolution
with vertical axis, or an oblique cone with a circular trace,
can be substituted for the sphere.

INTERSECTIONS—SHADOWS. 201

8. Arrange a square prism, axis horizontal or oblique,
intersecting a right cone, axis inclined to both planes of
projection.

Prepare the prism by drawing lines in its faces parallel
to the edges, as in other problems, and use section planes
containing these lines and the vertex of the cone.

Note. An oblique cylinder might be substituted for
the cone, but the vertex being at an infinite distance the
auxihary planes would be parallel.

9. Arrange a sphere intersecting an irregular pyramid.

Use a series of horizontal auxiliary sections of both
forms.

SHADOWS.

In a homogeneous medium light travels in straight lines,
called rays, which are parallel, divergent, or convergent,
according to the source whence the light proceeds and the
configuration of the media through which it has passed.

The sun's rays are sensibly parallel ; rays from the
electric light are divergent, and, for all practical purposes,
may be considered as diverging from a point ; while rays of
light that have passed .through a double convex lens, for
example, converge towards a point ox focus.

If an opaque body be placed between a source of light
and the object or objects illuminated, some of the rays will
be received upon the interposed body and will, consequently,
be cut off from illuminating certain parts of the objects
beyond. Popularly the opaque body would be said to
throw or cast a shadow. There is, however, an ambiguity
in the term shadow as commonly employed, which it is

202 SOLID OR DESCRIPTIVE GEOMETRY.

convenient, when treating of the projection of shadows, to
avoid by suitable nomenclature.

It will be evident after a little consideration that, when
the light from a given source is partly intercepted by an
opaque body, there is carved as it were out of the surround-
ing bundle of luminous rays, a central part or core of
darkness of a definite form, the surface of which derives its
configuration from the interposed body. To the surface
which separates this core from the surrounding rays of light
the term shadow-surface will be given. The line of inter-
section of the shadow-surface with the forms or surfaces it
meets will be called the shadow-trace. The figure bounded
by the shadow-trace is plainly the shadow cast by the
interposed body. The shade-line, or line of contact of the
shadow- surface with the form whose shadow is projected,
marks the separation of light and shade thereon. The shade-
line is oftentimes an edge or boundary of the form itself.

From these considerations it is apparent that the problem
for the student in any given case is to find the shadow-
surface, and its intersection with the surfaces on which the
shadow falls. Sometimes the shadow-surface is treated as
the boundary of a definite form, at others a convenient
number of lines (rays) are drawn in the shadow-surface
from points in the shade-line, and their intersections deter-
mined with the forms or surfaces on which the shadow is
cast; but in both cases the problem is one of "Sections by
Planes" (oblique or otherwise) or of "Intersections."

It may be remarked that the shadows we deal with are
purely geometrical, and their boundaries, or traces, sharp
distinct lines. Those actually seen in nature, called physi-
cal shadows, are more or less hazy about the region of the

INTERSECTIONS— SHADOWS. 203

shadow-trace, due partly to inflexion and partly to the
formation of a penumbra — a fringing semi-shadow which
makes its appearance when the source of light has an
appreciable angular magnitude.

I. Shadows cast by Parallel Rays.

The rays of light are in all cases represented by straight
lines given in direction.

One of the parallel rays being determined by its plan
and elevation, the shadow-surface is supposed everywhere
parallel to the given ray and enveloping the form whose
shadow is to be projected.

For example, if the form is a sphere the shadow-surface
is the enveloping cylinder whose axis is the line drawn
from the centre of the sphere parallel to the given ray.
The shadoiv-trace is the intersection of this cylinder with
the forms or surfaces on which the shadow falls. The
"circle of contact" is the shade-line which separates the
part of the sphere in light from the part in shade.

Taking in lieu of the sphere a right cone in space, axis
oblique to the co-ordinate planes, its shadow-surface would
be partly determined by a pair of tangent planes parallel to
the ray, thus : — Draw a line from the vertex of the cone
parallel to the given ray and determine two tangent-planes to
the conic surface containing this line. The planes so drawn
form a portion of the shadow-surface required, and if the
plane of the base of the cone were parallel to the given ray,
that plane produced would intersect the two tangent-planes
in parallel lines, giving rise to a triangular prism as the
complete shadow-surface. The intersections of this prism
with the surfaces or forms on which the shadow falls will

204 SOLID OR DESCRIPTIVE GEOMETRY.

give the shadow, and shadow-trace, as in other cases. The
"lines of contact" of the tangent-planes, and the arc of the
base that lies between them on the portion of the conic
surface turned towards the light, form the shade-line. Were
the plane of the base not parallel to the given ray, this part
of the shade-line would be treated as the directrix of a
cylindrical surface ; that is, points would be assumed in it,-
and lines drawn from them parallel to the given ray, and the
intersection of these lines with the objects they meet deter-
mined for points in the shadow-trace. Generally, lines drawn
parallel to the given ray from assumed points in the shade-
line will intersect the plane or curved surfaces on which the
shadow is cast in points which determine the shadow-trace.
In many cases a few points suffice. For example, in finding
the shadow cast on the horizontal plane by a cube, it will
be seen that the shadow-surface is bounded by planes
parallel to the given ray containing certain edges (those
forming the shade-line) of the solid, and that the shadow-
trace is made up of parts of the horizontal traces of these
planes. If, therefore, two lines be drawn from the extremi-
ties of one of these edges parallel to the given ray, and their
horizontal traces found and joined by a straight line, this
line will be one side of the shadow-trace.

Exercises.

I. The shadow of a cone on a cylinder and of both on the
H.P. Cylinder resting on If. P., axis horizontal and inclined
at 30° to V.F., base 1*5 in. diameter. Cone resting on H.P.^
base 2 in. diameter, axis 4*5 in. long, vertical and 1*5 in. from
nearest generatrix of cylinder. Rays inclined 45° and in plan
at right angles to the plan of the axis of the cylinder. Show
the shade-lines on both surfaces.

INTERSECTIONS— SHADOWS. 205

2. A circular slab 3 m, diameter^ '4 in. thick, rests symme-
trically over the upper base of a right hexagonal prism, axis
2*5 in., side of base v^. Vertical face of prism inclined at
20° to V.P. Shadow of the slab on the prism and of both on
the H.P. by rays of light making angles of 45" with xy in
plan and elevation.

■ 3. A right cone axis 4 in. long, base 2 in. diameter, resting
upon H.P. with the rim of its base touching the V.P. and the
plan of its axis making an angle of 60° with xy, is truncated
by a plane at right angles to the axis and 1 in. from the vertex.
Shadow of frustum by rays of light inclined 45" to xy in
elevation and 50** in plan.

4. Assume an oblique prism and a sphere, and show the
shadow cast by the sphere on the prism.

5. A niche is formed, in a vertical wall parallel to the
V.P, by a semi-cylindrical recess of 2'^ in. diameter capped by
a semi-hemispherical dome. Show the shadow in the niche
as cast when the rays of light are inclined in plan and elevation
at equal angles of 4^^^ to xy.

II. Shadows cast by Rays that meet in a Point.

This division includes the projection of shadows cast
by rays divergent and convergent. The former is the only
case that will receive attention, inasmuch as whatever is said
concerning the one, is, with very obvious modifications,
equally applicable to the other.

The source of light is a point, given by its projections,
from which the rays diverge in every direction.

A line passing through the given point and moving

206 SOLID OR DESCRIPTIVE GEOMETRY.

round the body whose shadow is to be projected, generates
the shadow- surface and determines the shade-line.

Thus, the shadow- surface of a sphere is an enveloping
cone having its vertex in the given luminous point. The
"circle of contact" will be the line of shade, and the
shadow-trace will be determined in any given case by the
methods specially applicable to the intersection of the cone
and the particular form or surface on which the shadow is
cast.

If the cone selected as an illustration under parallel
rays be taken, the work is substantially the same as there
given. The two tangent-planes to the cone that pass
through the given luminous point constitute a portion of the
shadow-surface, the rest of which will be a cone having the
luminous point as vertex and a segment of the base as
directrix. Any required points in the shadow-trace can
always be found by lines from the vertex, through points in
the shade-line, produced to meet the objects on which the
shadow falls.

The shadow-surface of a solid bounded wholly by plane
figures will be a more or less irregular pyramid having its
vertex in the luminous point. The plane faces of this
pyramid will contain certain of the edges of the form whose
shadow has to be projected, and in this case, as in the case
of parallel rays, the work may sometimes necessitate that
two lines only be drawn from the point — one through each
extremity of an edge— to determine one side of the
shadow-trace.

Exercises.
I. Arrange a right cone standing on its base and a
cylindrical slab resting on H.P. with its axis vertical, and

INTERSE CTIONS—SHAD 0 WS. 207

show the shadow cast by the cone upon the cylinder, and of
both upon H.P., by rays diverging from an assumed point in
space.

2. Assume an oblique plane, a vertical circular disc, and
a point in space, and shotv the shadow of the disc upon the
oblique plane as cast by rays of light diverging from the point.

3. A hemispherical bowl rests upon the H.P. with the
cavity upwards, shoiv the shado7V cast inside, the shadow of
the hemisphere on the H.P. and the line of shade, when the
rays radiate from an assumed point in space.

4. Arrange an example showing the shadow of a sphere
cast upon an oblique pentagonal pyramid by rays diverging
from a point.

III. Shadows cast bv Luminous Bodies of appreciable
Angular Magnitude.

For example, suppose the luminous body to be the sun,
and the body whose shadow is cast the moon. There are
in this case two shadow-surfaces, y'lt. the two enveloping
cones j the trace of the inner surface bounding the true
shadow, and that of the outer surface bounding the pen-
umbra.

The arrangement of a few examples to illustrate the
projection of the penumbra must be left to the ingenuity of
the student. One that may be suggested, is the shadow of
a disc by a white-hot platinum wire of a given length.
The source of light in this case being treated as a line of
ight, show the section of the shadow-surfaces by an
oblique plane.

FIGURED PROJECTIONS AND SCALES OF
SLOPE.

Def. The index of a point is a number affixed to its
projection on a plane to denote the distance of the point
from that plane.

K point in space is completely determined when its plan
and the affixed index denoting the length of its projector are
given. Similarly, a straight line is determined by its plan
and the indices denoting the length of the projectors of any
two fixed points in it. The height of any other point in
such a line, is, of course, readily determinable by a rabat-
ment of the projecting plane of the line into the plane of
projection.

A minus sign is prefixed to indices that denote points
below the plane. Points above the plane are indexed by
numbers in the usual way, the positive sign being under-
stood. The index of a point in the plane is of course always
zero.

Def. Points, lines, figures, and forms, determined by
one projection and its affixed indices, are said to be given
by XhQix figured plans.

A plane is determined when three contained points, or

FIGURED PLANS AND SCALES OF SLOPE. 209

two straight lines (meeting or parallel) in it, are given by
their figured plans. But for convenience in representing in-
definite planes by the method of figured plans, recourse is had
to a device called the Scale of Slope, the nature of >vhich will
be readily understood from the following considerations : —

If a pair of horizontal lines be drawn in a given plane,
they will both be parallel to its horizontal trace, and, there-
fore, to one another. Any two such horizontals (one of which
might be the H.T. itself), if properly indexed, would deter-
mine the plane. A line drawn at right angles to the figured
plans of the pair of horizontals might be the plan of a line
measuring the inclination of the plane, and from the known
indices of the two points where the line of inclination meets
the pair of parallels, a number of other indexed points
(denoting a series of horizontals) might be at once deter-
mined in the line of inclination, and a scale showing heights
in the plane constructed. This is what is done in the Scale
of Slope, which is always drawn as a double line with an
interspace of about ^g th of an inch. The line on the left-
hand side of the scale in the ascending direction should be
made thicker than the other, as a convention to assist the
eye in reaching the drawings in which Scales of Slope are
employed.

A circle is completely determined when its plan and the
index of its centre are given. Also, a sphere is similarly
determined when its plan and the index of its centre are
given.

A few problems to be worked by the method of figured
plans follow. The student who wishes for others will have
no difficulty in selecting good examples from among the
exercises already given for solution in the usual way.

E. G. 14

2IO SOLID OR DESCRIPTIVE GEOMETRY.

Problems.

1. An equilateral triangle abc of t, in. side, is the plan
of another triangle whose corners A,B,C, are i in., 1*5 in., and
2*5 in. respectively above H.P,; determine the Scale of Slope of
the plane of the triangle ABC, and the index of a point, P, in
this plane of which the pla?i, p, is the centre of the equilateral
triangle.

Find in ac the plan of a point having the same index as
B {i.e. i'5). A line from b through this point will be the
plan of a line of level in the plane of the triangle. Draw
the pair of lines for the Scale of Slope at right angles to this
line. A second horizontal from ^ or C will determine
another point of known height, whence the Scale of Slope
may be completed by divisions showing, for example,
differences of level of (say) ^th of an inch.

A horizontal from p to the Scale of Slope will ascertain
the index of P.

2. To find the index ofV without recourse to the Scale of
Slope.

Join one of the points, as a, with p, and produce the line
ap to meet be in 0. Find the index of 0 in JBC, and then
of P in A O, by rabatting the projecting planes of these lines.

3. To find the true shape of the triangle ABC.

Rabat the plane about one of the horizontals till parallel
to H.P.

4. Assume two inclined planes by their Scales of Slope,
and deter?nine the figured plan of the intersection, and its
inclination.

Draw in the one plane a pair of horizontals through any

FIGURED PLANS AND SCALES OF SLOPE. 2ll

two points of division in its Scale of Slope. A similar pair
in the second plane, having the same indices as those first
drawn, will meet them in points in the plan of the inter-
section, whence the line may be indexed, and its inclination
determined by rabatting its projecting plane.

5. Given a plane by its Scale of Slope and a straight line
by its figured plan, find the point of intersection of the plane
and line.

Assume any coriprenient plane containing the line, and
determine its intersection with the given plane. The point
in which the plan of the line meets the plan of the inter-
section of the planes will be the point required.

6. Given a regular hexagonal pyramid by its figured
plan, and an inclined section plane at 60" by its Scale of Slope,
determine the figured plan and true shape of the section.

7. Assutne two lines, neither parallel nor meeting, by
their figured plans, and determine the surface generated by a
third straight line, which moves so as always to meet the two
given lines and be parallel to the plajie ofprojection.

This surface, which is the Hyperbolic-Paraboloid, is
determined by drawing a series of generatrices from points
in one directrix to points in the other having similar
indices.

8. Work a section of the above surface by an inclined
Mane assumed by its Scale of Slope.

9. Work Prob. 7 above when the directing plane is not
^he plane ofprojection, but is given by its Scale of Slope.

10. An ellipse, which is the figured plan of a circle, and
in inclined line, are the directrices of a Qonoidal surface, the.

14—2

212 SOLID OR DESCRIPTIVE GEOMETRY.

plane of projection being the directing-plane^ or plane to which
the generatrices are parallel; determine the surface, and the
section by an inclined plane givejt by its Scale of Slope.

Note, When working Probs. 7, 8, 9, and 10, the
student should refer to remarks on "Undevelopable Ruled
Surfaces," Chap. vii.

II. To determine by the method of parallel sections the
intersection of two or more forms with plane faces., when the
forms are given by their figured plans.

To work also a section by a given vertical plane.

For the first example take two irregular polygons with
several edges of the one cutting those of the other, and
assume them to be the bases of two pyramids resting upon
H.P. Assume two points for the plans of the vertices and
attach indices to denote their heights above the plane of
projection. Join the plans of the vertices with the corners
of the corresponding bases for the complete plans of the
intersecting forms.

To work the intersection: — A little attention bestowed
upon the plan and the heights of the vertices will make it
evident what faces intersect, and the general direction of the
intersections. Sections of any pair of intersecting faces by
two horizontal planes will determine two points in the inter-
section of the indefinite planes of those faces, through which
points the line of intersection can be drawn. The horizon-
tal plane of projection itself will serve as one of the auxiliary
section planes. Thus, the points in which the polygonal
bases meet will be points in the intersection of the faces
passing through them. To find another set of points, take
an auxiliary section of both forms by a horizontal plane at a
• convenient height ; that is, draw horizontal lines at the same

FIGURED PLANS AND SCALES OF SLOPE. 213

level in each pair of intersecting faces. The points in w?iich
the plans of these lines at the same level meet, will be the
plans of the points required.

When two points in each line of intersection are found,
the segment of the indefinite intersection that is common to,
and limited by the intersecting faces, may be put in, and so
on till the whole intersection is complete.

Note. It is sometimes necessary to produce the sides
of the bases, or the lines of level, beyond the limits of the
intersecting faces, in order to determine conveniently the
required segment of the indefinite intersection.

Work is sometimes saved by taking the horizontal lines
in the planes of the faces of the pyramid with the higher
vertex, at the same level as the vertex of the lower one.

For another exercise take two pyramids, one with its base
resting on H.P. and the other inclined.

The intersection is found, as before, by means of lines of
level in the intersecting faces, but in this case the inclined
base has to be taken into consideration. Generally it is
best to begin with the section of the one pyramid by the
inclined plane of the base of the other, and it sometimes
simplifies matters if the horizontal trace of the latter form is
found before working the rest of the intersection.

To work the section by the given vertical plane it is only
necessary to find the indices of the points where the line
which is the plan of the section plane cuts the edges and
lines of intersection of the forms. The section-plane, with
these points of section, can then be rabatted, and the true
shape of the section determined.

12. Find the Scale of Slope of a plane inclined at 40",

214 SOLID OR DESCRIPTIVE GEOMETRY.

and the plan and true shape of the section cut by it from the
intersecting pyramids in the preceding problem.

13. Determine the intersection of a right cone and an
irregular Pyramid, both resting upon H.P., by the 7nethod of
parallel sections, using only the figured plans of the forms.

14. A cube, diagonal of the form vertical, stands with its
lowest corner on the horizontal plane. Determine the Scales of
Slope of the planes of the three faces meeting in that corner.

TRIMETRIC METHODS OF PROJECTION.

In the preceding chapter a method is given of working
with figured plans, by which the necessity for a second
projection on a vertical plane is, in a measure, obviated. It
is intended to discuss here some developments of this
method, one of which is of considerable practical utility
from the ease with which it can be applied to the construc-
tion of very intelligible drawings.

All forms can be completely determined by referring
them to a system of three rectangular planes. Such a system
of reference planes intersect in, and are themselves deter-
mined by three rectangular axes meeting in a point.
Inasmuch as the forms most common in practical construc-
tions are chiefly developed in the three directions of length,
breadth, and thickness, it is obvious they might be readily
referred to such a system of rectilineal axes, and the lines
for their drawings laid off about the projections of these
axes on a plane oblique to them. The following consider-
ations will make this clear : —

If the three rectangular axes were projected on a plane
oblique to all and differently inclined to each, three scales
(one for each axis) might be constructed to show the
projected lengths of the unit when set off in the three several
directions of the axes. Drawings made with reference to

2l6 SOLID OR DESCRIPTIVE GEOMETRY.

these axes and their scales, would be true orthographic
projections, showing at a glance, in one projection, the three
principal directions in which the forms projected are deve-
loped, and affording a means of readily ascertaining their
dimensions. Such drawings might be conveniently called
"Trimetric Projections." If two of the axes were equally
inclined to the plane of projection at angles different from
the third, two scales only would be required, and the some-
what simplified projection might, in this sense, be called
Dimetric. In the particular case in which the three axes
are equally inclined, the projection is monometric, or, as it is
usually termed, "Isometric" — only one scale being required
for the three directions.

For some purposes {e.g. the projection of the forms of
crystals) it is convenient to regard the intersections of the
three planes as axes of symmetry. In the forms of the
cubic system, for example, since the faces can be disposed
symmetrically about three such axes of equal length, any
conventional sign denoting one face equally symbolises the
whole form, and if the data for fixing one face be given the
projection of the complete form can be readily determined.
This method is a particular case of Trimetric Projection, the
application of which to the projection of symmetrical
forms will be touched upon briefly under the heading of
"Axial Projection."

TRIMETRIC PROJECTION.

Before the exercises can be attempted the student must
.solve the following problem : —

Draw three straight lines, ox, oy, oz, forming three unequal
obtuse angles xoy, yoz, zox, at a pointy o, and assuming these

TRIMETRIC projection: 217

lines to be the plans of the three rectangular axes, OX, OY,
OZ, determine the set of trimetric scales corresponding to
them.

Read Prob. XXVII., Chap. iii.

It will be necessary first to find the inclination of each
axis to the plane of projection. This done, the scales can
be readily constructed.

To find the inclination of OX : — Since OX is perpendi-
cular to the plane of YOZ, the horizontal trace of this plane
will be at right angles to ox, the projection of OX. Draw,
therefore, an assumed horizontal trace of the plane YOZ at
right angles to ox and meeting oy, oz, in s and t respectively.
Similarly, from s and t draw the horizontal traces sr and tr
of the planes XO V, XOZ, meeting ox in r.

Assume the vertical plane containing ox to be rabatted
thus : — Produce xo to meet st in m, and on rm as diameter
describe a semicircle. A line at right angles to rm from o
will cut the semicircle in O' — the point O rabatted — and
the angle O'ro will be the inclination of OX required.

Similarly the inclinations of OY a.nd OZ may be found.

To construct the trimetric scale of OX : — Draw two lines
making an angle equal to the inclination of OX : on one of
them lay off the divisions of the scale of units employed
(inches in the following exercises), and from the points of
division draw a series of perpendiculars to the other line.
These will determine the divisions of the trimetric scale of
OX. Work similarly for the others, and finish the scales
neatly to show inches and tenths. The scales must be
carefully named to guard against the error which would
result from interchanging them.

2l8 SOLID OR DESCRIPTIVE GEOMETRY.

Exercises.

1. T?'imetric projection of a brick, 9 in. long, 4*5 wide,
and 3 in. thick, to half scale.

Measure the dimensions (taken from the trimetric
scales) along the corresponding axes, and complete by draw-
ing lines parallel.

2. Trimetric projection of a right cylinder, 3*5 in long
and 2 in. diameter.

Assuming the cylinder to be circumscribed by a square
prism, determine the projection of the latter. Two ellipses
inscribed in the trimetric projection of the square ends will
be the projections of the circular bases of the cylinder.
Two parallel tangents to the elHpses will complete the
required projection.

3. Show the trimetric projection of a circular hole of
2 in. diameter bored through the middle of the brick. Exercise i.,
at right angles to its large faces.

4. Trimetric projection of your T-square to a convenient
scale.

5. Trimetric projection of a mortise and tenon joint.

6. '' Trimetric projection of your instrument case with the
lid open at right angles.

7. A large central semicircular arch-way, "^oft. span and
40 ft. high to the springing of the arch, is flanked by two
smaller archways — one each side — of 10 //. span and 25 //.
to the springing of the arches. The ce?itral piers are ^ft.
and the outer ones 2 ft. 6 in. thick. Width at right-angles to

TRIMETRIC PROJECTION. 219

the plane of its face 2^ ft. Other dimensions and emhellish-
jnents at pleasure. Show its trimetric projection.
Scale lo ft. to i in.

8. Trimetric projection of a brick with two semicircular
grooves of I'^in. radius cut in a large face — one running
parallel to its long edges through the middle of the face, arid
the other bisecting the former at right angles. Show the
intersection of the grooves. To half scale.

9. Trimetric projection of hexagonal right pyramid ;
axis 3 "5 in. long, side of base i i?i.

Let abcdef be the hexagonal base. Produce two of its
parallel sides as ab, de, and through c and / draw a pair of
parallel lines at right angles to, and meeting the produced
sides ab, de. This forms a rectangle circumscribing the
hexagon.

Set off the reduced lengths of the sides of this rectangle
along two of the projected axes, as oy, oz, and complete the
parallelogram for the projection of the circumscribing
rectangle. The corners of the hexagonal base may now be
found, in the sides of the parallelogram, and the projection
of the base drawn in.

For the vertex, draw from the centre of the parallelogram
a straight line (the projection of the axis) parallel to ox, and
make it equal to 3 "5 in. on the scale of OX. Join the
vertex with the corners of the hexagonal base for the com-
plete projection of the pyramid.

Note. By the the device of an enveloping rectangle,
)r, more generally, by the aid of rectangular ordinates, any
)lane figure whatever, or any prismatic or pyramidal form,

220 SOLID OR DESCRIPTIVE GEOMETRY.

may be projected as above. The general cases will be,
however, treated under Trihedral Projection below.

10. A figure formed by describing two semicircles on two
parallel sides of a regular hexagon of i' 2^ in. side as diameter
— the concavity of the semicircles being turned in the same
direction — is the base of a prismatic form 4*5 in long. Show
its trimetric projection.

11. Show the Trimetric projection of your instrument
case when the lid is open at an angle of 60".

ISOMETRIC PROJECTION.

Isometric Projection has been already defined as the
particular case of Trimetric Projection that results from the
three axes being equally inclined to the plane of projection.
In this case it is clear that the obtuse angles formed by the
meeting of the plans of the three axes at the point, 0, are
all equal — /. e. are each= 120°.

We may therefore begin by drawing the plans of the
three axes at equal angles of 120", and proceed to deter-
mine the scale of one of the axes as for the trimetric scale
above. This scale, which also serves for the other two axes,
is called the Isometric Scale.

The ratio of a line, measured along an isometric axis,
to its isometric projection may be readily proved to be as
J 2, is to J 2. We may, therefore, avail ourselves of this
established ratio and construct as follows the

Isometric Scale: — Draw a right-angled triangle having
one side i in. long and the other equal in length to the
diagonal of a square of i in. side. The diagonal of a

AXIAL PROJECTION. ' 221

square whose side is unity is equal to Ji, and the
hypotenuse of a right-angled triangle with one side equal to
unity and the other to J 2, will be equal to J^ (Euclid
47, Bk. I.). That is, the hypotenuse of this triangle is to .
its longest side as J^ is to J 2. Therefore, real lengths may
be measured along line Jt, and the corresponding isometric
lengths found in line ^2 by drawing perpendiculars to that
line from the points set off in the hypotenuse ^3, or
a scale of isometric inches may be similarly constructed.

Isometric being but the most simple form of Trimetric
projection, it is unnecessary to give further exercises. All
those given under Trimetric projection, or any that the
student' has omitted, may, however, be worked here at his
own or the teacher's discretion.

AXIAL PROJECTION.

The term Axial Projection is used to denote a trimetric
method which lends itself with great facility to the pro-
jection of symmetric forms bounded by plane faces, such
as the forms of crystals.

The investigation of the geometrical properties of these
forms, and of the number and position of the axes of
symmetry to which their faces can be most conveniently
referred, is the special province of crystallography. The
principles on which they are projected fall, however, within
the scope of Descriptive Geometry, while many of the
forms have an interest attaching to them quite apart from
their crystalogical significance.

OX, OY, OZ, being the three rectangular axes as

222 SOLID OR DESCRIPTIVE GEOMETRY.

before, it is evident that an indefinite plane will in general
meet them in three points, which may be called the axial
traces of the plane. Let a, )3, y denote the distances of the
points of section or axial traces along OX, OY, OZ
respectively. A plane, relatively to the three axes, is
obviously completely determined by its axial traces. The
plane of the face of a form referred in this way to three
rectangular axes can therefore be expressed by the ratio of
the parameters of the face : — a : y8 : y.

The student will observe that it is in general convenient
to treat the axes as three lines drawn yr^;« the point, O, to
indicate three directions in which certain measurements are
to be set off, but for the present purpose the axes must be
produced through the origin, O, on both sides. The
system of intersecting planes that give rise to the indefinite
axes XX', YY', ZZ', in question, form eight similar trihe-
dral angles, whence it follows that for every plane determined
by its axial traces in one of these trihedral angles there will
be seven other planes similarly* related to the axes — one
in each of the other seven trihedral angles.

If the three fundamental lines, ox, oy, oz, used as the
projection of the axes in previous problems, be produced
and lettered xox', yoy', zoz', any required trihedral angle
can be specified by the extremities of the semi-axes, as xyz,
yzx', &c.

Exercises.

I. One face of a form symmetrical about three equal
rectangular axes is given by the formula a:/8:y=i : r :i;
determine its axial projection.

AXIAL projection; 222,

The projections of the three indefinite axes and the set of
scales may be determined once for all, as in the case of the
trimetric exercises above.

The greatest number of faces similar to the one given
that can be disposed about the axes is eight — one in each
trihedral angle. The faces are equilateral triangles, and the
form is, therefore, a regular octahedron.

To draw the projection it is only necessary to cut off, by
means of the scales, equal lengths on each axis from O, on
both sides of that point, and join up the points so found for
the edges of the octahedron. For example, let i, 2, 3, be
the plans of the points on the semi-axes ox, oy, oz, respect-
ively, then if i be joined with 2, 2 with 3, and 3 with i, the
triangle 123 will be the projection of the face situated in
the trihedral angle xyz. Similarly the other faces may be
drawn,

2. Determine the form produced by the intersection of the
indefinite planes of the alternate faces of the octahedron drawn
in the preceding problem.

Lines through the points i, 2, 3, parallel respectively to
lines 23, 31, 12, will give the intersections of the face 123
with the alternate faces of the trihedral angles xy'z', x'yz',
x'y'z. These intersections are three edges of one face of
the tetrahedron — which is the form produced. Similarly
the other edges can be found.

3. Axial projection of a syjnmetrical form one face of
which is given by the formula a:^:y: = oc:i:oc. The
sign oc = infinity.

The faces of this form are six in number — each meeting
one axis at unit distance from O and parallel to the other
two. The form produced is, therefore, a cube.

224 SOLID OR DESCRIPTIVE GEOMETRY.

The intersections of the planes being parallel to the
axes, no difficulty will be experienced in drawing the form.

4. Axial projection of the form produced by planes
parallel to one axis and meeting the other two at the unit
distance: a : /S : y = c< : i : i.

This form, which is the Rhombic Dodecahedron, will
serve as an additional exercise. There are several others,
the most general of which is a forty-eight-faced form having
the formula m : 1 \ n, but their discussion belongs more to
crystallography than to Descriptive Geometry.

TRIHEDRAL PROJECTION.

Hitherto all forms have been regarded as trimetrically
developed about rectangular axes in three directions from
their origin O. But the most general aspect in which a
problem can present itself is when the position of the form
is not referred directly to the axes but to the three intersect-
ing planes. A Trimetric projection of this kind would also
be Trihedral, and the latter term serves very conveniently to
denote this most general aspect of the trimetric method.

As was before observed, three indefinite rectangular
planes intersect in eight similar trihedral angles, but in order
to avoid the necessity for a cumbrous system of signs to
specify which of these trihedral angles is meant, it will be
best to represent all points, lines, figures, and forms in one
of these angles only, viz. that formed by the planes XOYy
YOZ, ZOX, and limited by the semi-axes OX, OY, OZ.
For the purposes of construction the planes may of course
be considered as indefinite.

TRIHEDRAL PROJECTION. 225

Problems.

1. To determine a point, P, in space trihedrally.

Let a, b, c, be the respective distances of P from the
planes YOZ, ZOX, XOY. Measure the trimetric length
of a from o along ox and from the point thus found
draw a parallel to oy equal to b as shown on the trimetric
scale oi OY. A parallel to oz from the point last deter-
mined, equal in length to c on the scale of OZ, will give p,
the trihedral projection of P.

2. Given p, the trihedral projection of P, and c, the
distance of P from the plane XOY, find a and b, its distances
frOm planes YOZ and ZOX,

Shotv also the trihedral projections of the orthographic
projection ofVon each of the reference planes.

Theorem. A point in space is completely determined
when its trihedral projection and its distance from one of
the three reference planes are given.

The particular trihedral angle in which P is situated
will depend on the position of / and the length of c. The
following construction although perfectly general, if the
student interpret from his drawing the direction in which
;he parallels are to be drawn from /, is intended to apply
:)nly to the one trihedral angle used before : —

Draw from / a parallel to zo equal to the trimetric
ength of c. The extremity of this parallel is the trihedral
)rojection of the orthographic projection of ^ on the plane
XOY. Two parallels to xo and yo from the extremity of
• he parallel first drawn, meeting oy and ox respectively,
( etermine a and b.

E. G. 15

226 SOLID OR DESCRIPTIVE GEOMETRY.

From a and b thus found the other required projections
of F can be readily determined.

Note. The student will observe that a line is deter-
mined trihedrally by two contained points and a plane by
three, but, more generally, by their trihedral traces.

Def. The trihedral traces of a line are the points in
which it pierces the planes of reference.

Def. The trihedral traces of a plane are the lines in
which it intersects the planes of reference.

It is obvious that the trihedral traces of a plane pass
through the axial traces of the same plane.

3. Given the data that determine the trihedral projections
of two points, P and Q, in space find the trihedral traces of the
line joining them.

Assuming that the three reference planes are limited by
the semi-axes OX, OV, OZ, it is plain a straight line can
have but two trihedral traces and these will always suffice to
determine it. Unless, however, the line be parallel to the
third plane the two will intersect if produced, and, for the
purposes of construction, this third trace might be found
useful.

It will be easily seen from inspection in which plane
there is no trace. This we will suppose to be the plane
ZOX. Through the projections of F and Q on this plane
draw a straight line (that is the projection of the Hne FQ
on plane ZOX), and produce it to meet oz and ox in / and
m. Parallels from / and m to oy will meet the trihedral
plan of the line FQ produced in the required traces of the
line.

TRIHEDRAL PROJECTION. 22/

Note. The trihedral traces of a plane containing three
given points will of course pass through the traces of the
lines joining them two and two. Hence the traces of such
a plane might be readily determined.

The foregoing problems are of a somewhat general
character and are offered merely as illustrations of a method
of projection that has never yet been sufficiently worked
out. To the student who has leisure, and the inclination,
to think out some of its applications for himself they will, it
is hoped, suggest a method that promises some interesting
results.

15—2

XII.

MISCELLANEOUS PROBLEMS.

I. Plan and elevation of a cube of 3 inches edge, when
three of its corners are i, 17, and 3 inches, respectively, above
the horizontal plane.

Refer to Chapter in. Problem 21.

Take one inch to equal 10 units.

(i) Draw the square ABCD, and index ^ as i inch or
10 units, -5 as 17 units, and C as 30 units, that is, the
respective heights to which these points are to be lifted.

(2) To find a 'horizontal,' in order that the square may
be folded about it into the required position \-t-

If A and 6", which are to be at heights of 10 and 30
units, be joined, there will be a point in the line ACi'j
units high, that is, at the same level as point B. Find this
point, M, by dividing the line in due proportion, and the line
drawn through B and J/ will be the horizontal required, at
level 17 units or 17 in. above the horizontal plane.

(3) A vertical plane at right angles to plane of face
ABCD may now be assumed by taking a line at right
angles to BM, for the intersection, xy, of this vertical plane

MISCELLANEOUS PROBLEMS. 22()

and the horizontal plane at level 17 units. The vertical
trace of plane of face ABCD may be easily determined
because a and c will, from the position of the assumed
vertical plane, be in the vertical trace, and their heights are
known.

(4) The square, ABCD, is one face of the cube rabatted
about the horizontal BM into a horizontal plane at level
1 7 units. The rest of the work is similar to that in former
problems.

Note. The heights of the three points determine the
inclination of two adjacent edges of the square face ; hence,
another construction based on Chap. v. (see Prob. 7) may
be used.

2. Plan and elevation of a tetrahedron, edge 3 inches,
when the three corners of its lowest face are '8, 1*5, and 2*2
inches above the horizontal plane.

3. A right pentagonal pyramid, side of base 1*5 in., axis
3*5 /;z., has its veriex 3 in. high and the extremities of one
edge of the base i in. and i'2 5 in. high respectively.. Flan and
elevation.

Prob. 26, Chapter iir. required.

4. Plan and elevation of a cube of 7, inches edge, when the
indefinite plans of three adjacent edges meeting in point A are
given.

Compare Chapter in. Problem 27, and Problem i, page
216, "Trimetric Projection."

(i) Draw the three given lines as ab, ac, and ^^ forming
three obtuse angles at a. Since the projections of a perpen-
dicular to a plane are at right angles to its traces, a line
drawn at right angles to any one of these edges, as to ab, will

230 SOLID OR DESCRIPTIVE GEOMETRY.

be a horizontal in the plane containing the other two.
Draw such a line to cut ac in ^ and ad in s. Then the
angle /^j is the plan of a right angle the plane of which may
be rabatted about the horizontal ps by bisecting ps and
describing a semicircle on it. The angle in a semicircle
being a right angle, point A when rabatted will be in the
intersection of the semicircle with fl;A If/ and A, and A
and s, be joined, the edge of the cube, 3 inches, can be
measured from A on these lines Ap and As^ or the same
produced, and the square can be completed.

(2) The plan and elevation of the whole may be finish-
ed by taking a vertical plane at right angles to the horizon-
tal ps and working from the square rabatted.

5. Given an oblique plane by its traces, equally inclined
at angles of j\o° to xy. Determine the plan and elevation of a
tetrahedron, 3*5 in. edge, resting upon it. No edge of tetra-
hedron to be horizontal.

Determine the inclination of the plane, Chapter iii.
Problem 5. Take a new ground line at right angles to the
horizontal trace, set up the vertical trace in this new vertical
plane, complete elevation and plan therein, and thence
derive the elevation upon the first or given vertical plane.

6. An oblique pyramid stands on its base, which is a
regular hexagon ABC...F ^ 1*5 inches side. V is the vertex,
the face Y KB is inclined at 75° to the base, VBC at 65°, VCD
at 60". Draw the plan of the solid. Science Examination,
1870.

( 1 ) Draw the hexagon.

(2) Determine the traces of plane 75", taking the xy at
right angles to the horizontal trace.

MISCELLANEOUS PROBLEMS. 23 1

(3) The same with plane 65*.

(4) In each of these planes determine a horizontal line
at the same height. The intersection of these lines gives a
point, /, in the plan of the intersection of the two planes.

Another point in the same intersection is b, and the line
bp being the plan of one edge of the pyramid contains v.

(5) Work similarly with planes 65° and 60°, and the
edge cv is determined, also containing v. The point v is
fixed by the intersection of the lines bpv and cv.

7. Given the plan of a sphere, required to determine the
plan of a cube inscribed in if, a, the plan of one of its corners^
and the direction, ab, in plan of one edge being also given.

(i) To find the true length of the edge of the inscribed
cube. On the diameter of the circle which is the plan of the
sphere, mark off one-third of its Length from one extremity,
and from the point so found erect a perpendicular to meet
the circumference. Join the extremities of the diameter to
this point in the circumference. The shorter line gives the
length of the edge of the inscribed cube, whilst the longer
gives that of the edge of the inscribed tetrahedron. Or, by
computation, if d equals the diameter of the sphere in
inches, and e the required edge of the cube, thei)

J 2^ •• J ^ •••• d : e.

(2) Make an elevation of AB on ab used .as a line of
level at the height of the centre of the sphere, and determine
b. To do this, describe a circle on the part of ab produced
which cuts the circle forming the plan of the sphere. Erect
a perpendicular to ab from point a to meet the cjrcumr
ferenca This gives a'. From a' measure AB as a chord.

232 SOLID OR DESCRIPTIVE GEOMETRY.

This gives b', and from b' a perpendicular to ab — which
being used as a line of level corresponds for this work to a
ground line — determines b.

(3) A line drawn from b through the centre o of the
sphere with the distance from b to 0 measured on it from 0,
will give the plan of^, another point in the face A...G.

(4) Determine by its traces the plane containing aa\
and perpendicular to ab, db'. This will be the plane of the
face A...G of the cube. To do this, through point a draw
the vertical trace at right angles to a'b', and through m,
where this vertical trace crosses ab, the line of level, draw
horizontal trace at right angles to ab.

(5) Rabat this plane and with it points aa!, gg. Join
AG which will be a diagonal of this face of the cube
rabatted. Complete the square, and therefrom determine
its plan by rotating the plane back to its original position.

(6) The edge ab being already drawn, parallels to ab,
and of the same length, through the points a...g will com-
plete the plan of the solid.

8. PlaH drid eteVation of a tetrahedron inscribed in a
sphere of 3 in^ radius. Two adjacent edges of the solid inclined
at angles of zo^ and 30° respectively. Centre of sphere 3 in.
above hotizontal plane.

(i) Find edge of tetrahedron inscribed in a sphere
of 2 inches radius. See preceding Problem.

Note. The edge of tetrahedron is a diagonal of face of
inscribed cube.

MISCELLANEOUS PROBLEMS. 233

(2) Draw an equilateral triangle with sides equal to
edge of inscribed tetrahedron, and determine the diameter
of a circle circumscribing this triangle. Set off the diameter
of this circle as a chord of a great circle of given sphere, and
find the length of a perpendicular from the centre of great
circle to the chord drawn. Call Ox the length of this per-
pendicular ; then Ox is the length of a perpendicular from
centre of given sphere to plane of face of inscribed tetra-
hedron.

(3) Determine on a convenient part of the paper the
plane of a face of any tetrahedron having two adjacent
edges of that face inclined as above. Chapter iii. Problem
20.

(4) Draw from Centre, O, of the given sphere, a line
perpendicular to the plane determined in (3) and cut off on
this line from the centre of given sphere a length equal to
Ox. A plane through the point x at right angles to the
perpendicular Ox, will be parallel to the plane of the face
of the auxiliary tetrahedron first drawn, and will also be the
olane of one face of the solid required.

(5) Rabat this plane with the circle in which it cuts
he given sphere.

(6) Inscribe in the rabatted circle the triangular face
of the required tetrahedron. To do this: — draw three
lingents to the circle parallel to the three sides of the
1 dangular face of the auxiliary tetrahedron first drawn for
the purpose of determining the plane of a face of that
J olid. The three tangents constitute an equilateral triangle
( ircumscribing the circle. Draw lines from the three corners
( f this figure to the centre of the circle, and join the points

234 SOLID OR DESCRIPTIVE GEOMETRY.

where these lines cut the circumference. This will be
another equilateral triangle with sides parallel to the one
inscribed in the rabatted circle, and will be, moreover,
the rabatted plan of the face of the required inscribed
tetrahedron, from which the projections can be at once
completed.

9. The centre, O, of a sphere of 2 inches radius is 3
inches above the plane of the paper. A point, a, on the surface,
4*25 inches high, is the plan of one corner, A, of an octahedron
inscribed in the sphere. The edge AB is inclined 45". Draw
the plan and an elevation on a verticcil plane inclined 30° to
the line ao.

(i) Determine the edge of the octahedron. This will
be equal to the side of a square inscribed in a great circle
of the given sphere.

If r = radius of sphere, x = side of inscribed octahedron,
then

r : X -.w : J 2.

(2) Make plan and elevation of sphere, and determine
the plan and elevation of A on its surfacp. To do this : —
draw, in the elevation, a line, at the height of 4*25 inches,
parallel to xy and cutting the elevation of the sphere. This
may be considered as the vertical trace of a horizontal plane
passing through sphere at the assumed level. The section
of the sphere by this plane will be a circle, shown in true
form in plan, and as a straight line in the trace of the plane
in elevation. The point A is on the circumference of the
circle constituting the plane section of sphere, and its pro-
jections a, a are on the corresponding projections of the
section. To find a, draw a line from 0, the plan of the

MISCELLANEOUS PROBLEMS. 235

centre of the sphere, making an angle of 30° with xy, and
produce it to meet the circle which is the plan of the
section. The point of intersection is a. d is at once
found by projecting a to the straight line drawn at 4*25
inches above xy.

(3) Make A the vertex of a right cone, axis vertical,
generatrix inclined 45" and equal in length to the side of
the octahedron (i).

The base of this cone in elevation is a straight line and
in plan a circle, with radius equal to that of the given
sphere, described from the point a as centre. The section
of the sphere by the plane of the base of the cone produced,
is another circle. Show the latter in plan. Either of the
two points where the plan of the circular base of the cone
meets the plan of the circle cut from the sphere by the
plane of the base of the cone produced, may be taken as
the plan b of the other extremity B of the edge AB.

(4) Determine // and draw ab, dV. These are the
projections of the edge AB.

(5) Determine the plane containing the straight line
4,B and the centre of the sphere 0. Rabat it and complete
he square A BCD, which will be inscribed in a great circle
jf the sphere,

(6) Determine the plan abed, and draw a perpendicular
o the plane of ABCD from O, meeting the surface of

• he sphere on both sides of the plane at P and R.

(7) Complete the plan of the solid by joining / with
( bed and r with the same points, and the elevation by joining
, ' and r with db'c'd'.

236 SOLID OR DESCRIPTIVE GEOMETRY.

10. A parallelogram, sides 3 inches and 2 inches, included
angle Go'', is Ihe plan of a square. Determine the side of the
square and the inclination of its plane. Science Exam. Hon.
1871.

(i) Draw parallelogram and inscribe in it an ellipse.

This ellipse will be the plan of a circle inscribed in
the square of which the parallelogram is the given plan,
and which circle will necessarily lie in the same plane with
the square; that is, in the plane whose inclination we have
to find.

(2) Determine the major and minor axes of this ellipse.

The former is a horizontal line lying in the plane of the
square, and is therefore the diameter of the circle of which
the ellipse is a plan. This latter property carries with it
two others, viz. that of being the diameter of the circle in-
scribed in the square whose side is required, and also that of
being eqiial to that side.

(3) To determine the inclination of the plane, it is
only necessary to make a projection of the minor axis on a
vertical plane assumed at right angles to the major axis.
The plane of the square and the assumed vertical plane are
perpendicular, hence, the elevation of the minor axis of the
ellipse will fall in the vertical trace of the plane of the square,
and its angle of inclination with xy will therefore be the
same as that of the plane.

The student will require no further assistance in deter-
mining this plane, or in rabatting it with its contained square
for the purpose of verifying the side found from the major
axis of the ellipse.

MISCELLANEOUS PROBLEMS. 237

11. A 7-ight hexagonal pyramid, 5 inches high, has its axis
inclined 50° and one dianieter AD, 4 inches long, of its hexago-
nal base inclined 25". Draw plan and elevation, and deter-
mine the circumscribing sphere.

(i) If the axis is inclined 50", the plane of the base
will be inclined 40°. Determine this plane by its traces, and
draw plan and elevation of the solid resting on it. Refer to
Chapter iv. Problem i.

(2) To determine the sphere. Rabat the plane con-
taining the vertex and a diameter of the base of the
pyramid. The horizontal traces of the two slant edges of
the pyramid springing from this diameter to the vertex will
be in the horizontal trace of plane required. Determine
the centre of the triangle formed by this diameter and the
slant edges of solid thus rabatted, and describe circle
:herefrom passing through the three corners of the triangle.
Phis gives the rabatted centre of the required sphere and its
"adius, whence the required projections of the circum-
scribing sphere can be readily determined.

12. Plan and elevation of a pentagonal pyramid when
me face is vertical and one long edge of that face inclined at 50".
Side of base i"25 in., axis 3*5 in. Science Exam. Hon. 1872.

(i) Determine angle 0 between base and one face of
)yramid.

(2) Determine a line in the vertical plane of projection
inclined 50", and on this line draw a triangular face of the
\ )yramid, one long edge of that face lying in the line.

If the vertical face of the pyramid be assumed parallel to
1 he vertical plane of projection, the triangular face of the
jiyramid drawn as above will be an elevation of a face of the

238 SOLID OR DESCRIPTIVE GEOMETRY.

form, and the plan of this face will be a straight line
parallel to xy at any chosen distance in front of the vertical
plane.

(3) Assume two points in the short side (i. e. the base),
or short side produced, of the triangular face of the pyramid,
and make them the vertices of two right cones having their
bases in the vertical plane and generatrices inclined to bases
at Q degrees.

(4) Draw the tangent plane to these cones. The
vertical trace will be a line tangential to the circular bases,
and the horizontal trace a line drawn through the horizontal
trace of the short side or base of the triangular face of the
pyramid and the point where xy is met by the vertical
trace.

(5) Rabat this plane and the short side of the triangular
face with it. On this side rabatted describe the pentagon
which is the base of the pyramid, and determine plan and
elevation therefrom.

The projections can be at once completed by joining the
proper points, all of which are thus determined.

13. A right pyramid, having a hexagon of 1*25 in. side
for its base, and an axis of 2,'S ^^-j ^^^^ with one edge on the
horizontal plane, a?td a face Containing that edge is inclined at
25". Draw plan and elevation. Science Exam. Hon. 1870.

(i) Determine the angle a" between the base and one
face of the pyramid.

(2) Determine plane of 25" by its traces.

(3) Rabat this plane and draw in it a triangular face
of the pyramid with one of its equal sides lying in the
horizontal trace, and thence determine plan.

MISCELLANEOUS PROBLEMS. 239

(4) Produce the plan of the base of the triangular face
and rabat it.

Note. The line thus rabatted will be the produced base
of the isosceles triangle first drawn.

(5) Describe a circle of any convenient radius touching
this line and mark, P, the point of contact. Assuming this
circle to be the base of a cone with generatrix inclined a",
find the vertex, V, and lift the plane, with the cone on it,
back to its former position^

(6) Find the projection s^//', vv, oi PV, and the hori-
zontal trace of a straight line drawn from V to P.

The line drawn through this tf-ace and the point where
tangent line to the rabatted base of the cone at point P
meets the horizontal trace of plafie 25", will be the hori-
zontal trace of a plafie inclined a" to this plane, and will
contain the base of the triangular face of pyramid deter-
mined in (3), and hence is the plane of the base of that
pyramid. This plane can now be rabatted and the
projections determined as in previous problems.

It may be remarked that the line PV is that in which
the plane of the base of the pyramid touches the cone.

14. T/ie hvo extremities A, B, of one edge of a cube are
at heights of'^ifi., i'6in. above the paper. The centre, O, of
the solid is at a height of \'\ in. Draw the plan of the solid.
Edge = 3 inches. Science Exam. Hon. 1870.

The section of the cube made by a plane passing through
the edge AB and centre O of the solid is a rectangle, two
sides of which are edges of the cube and two diagonals of
the faces.

240 SOLID OR DESCRIPTIVE GEOMETRY.

By the conditions, the heights of three points — two ex-
tremities of one short side, and the centre — of this rectangle
are given.

(i) Determine plan and elevation of this rectangle as
in Problem i.

(2) Determine a plane at right angles to one of the
short edges of the rectangle and containing one of the long
edges of the latter. This condition is complied with if the
plane be drawn through one extremity of a short side and
perpendicular to it. Converse of Chapter in. Prob. 13.

Or, the horizontal trace of the plane may be drawn at
once through the horizontal trace of one long side of the
rectangle at right angles to the plans of short sides of the
latter.

Note. This trace will be the point where the plan of
the long side produced meets the horizontal trace of the
plane of the rectangle.

(3) Rabat the plane thus drawn, which is the plane
of one face of the cube, and with it the long side of the rect-
angle.

The latter being a diagonal of a face of the cube, the
plan of the solid can be completed therefrom as in previous
problems.

15. An irregular pyramid has for its base a triangle
ABC- AB = 3 in. ; AC = 3'5 in. ; BC = 4 in. : the plan d of
the fourth corner projected on the plane of the base is 2 in. from
A, I "5 in. from ^; the true length of the remaining edge CD «•
37 in. Draw the plan of this pyramid, when standing on its
base, and an elevation on a plajie parallel to the edge AD.
Determine the lengths of the edges AD, BD, and the height of
D above the base.

MISCELLANEOUS PROBLEMS. 2\\

a. Determine the inclinations of the faces ABD, ACD, to
the plane of the base.

p. Draw a plan of this pyramid when the edge CD is
vertical. Science Exam. 1868,

(i) Draw the triangle ABC. The point d will be
found by describing a circle of 2 in, radius from A a.s a.
centre and another of i'5 in. radius from B ; the point where
the circles intersect on that side of AB nearest to C, is the
point required.

(2) Join dC, and from d draw a perpendicular to this
line. A circle of 37 in. radius described from C as a centre
will cut this perpendicular at D, and Dd will be the height
of the vertex of the pyramid.

(3) To determine the lengths of the edges AD, BD,
draw from d lines perpendicular respectively to Ad and Bd.
Make these perpendiculars equal to height of vertex = dD,
md complete the right-angled triangles AdD, BdD ; the
lypotenuses AD, BD, of these triangles are the edges

: equired.

(4) For the required elevation, xy must be taken
] 'arallel to the plan dA of the edge AD. The rest of the
A 'ork presents no difficulty.

Note. The elevation of AD on this vertical plane is in
t ue length.

(5) For the inclination of the faces ABD, ACD, eleva-
ti Dns may be made on vertical planes at right angles to the
li les AB and A C respectively.

(6) For the plan (^S) make an elevation on a plane
p irallel to CD, and from this elevation deduce the plan
n quired, by taking xy at right angles to the elevation c'd' of
CD.

E. G. 16

242 SOLID OR DESCRIPTIVE GEOMETRY.

1 6. A tetrahedron, of 3 //;, ^dge, stands with one corner on
the fwrizontal plane, so that the plans of tJie tiuo edges meeting
at that corner are I'g in. and i"5 /;/. respectively. Draw the
plan of the solid. Science Exam, Hon. 1871.

(i) Determine the plane of the face containing the
edges whose lengths in plan are given.

To do this : — draw a line and make it equal in length
to either of the plans of the given edges. Show an eleva-
tion of this edge on a parallel vertical plane. For this
purpose the xy must be taken parallel to the line drawn,
and should be assumed at a convenient distance therefrom.
An elevation on this plane will show the edge in true length,
i.e. 3 inches.

This determines one edge of the solid. The other is of
the same real length and inclined at a constant -angle of
60" to the first. It is plain, therefore, that the second will
lie somewhere on a right conical surface having the first for
an axis and the second for a generatrix ; that is, on a cone
whose vertex is at the point where the first edge meets the
horizontal plane, — vertical angle 120", generatrix 3 inches,
and axis coincident with the edge whose projections we
have found.-

Determine plan and elevation of this cone. The eleva-
tion of the base will be a straight line bisecting at right
angles the vertical projection of the edge first found, and
the plan, an ellipse, having its centre at the middle point of
the plan of that edge.

Describe a circle from the vertex of the cone in plan
with a radius equal to the length of the plan of the second
edge; and from either point common to circle and ellipse
draw lines to the two extremities of the first edge.

MISCELLANEOUS PROBLEMS. 243

The triangle thus drawn is the plan of the face of the
tetrahedron containing the two given edges. From this the
plane of the face can be at once drawn.

(2) Rabat this plane and complete plan of tetrahedron
as in previous examples.

17. An irregular triangular pyramid has its edges AB
= 2;AC = 2-5; BC = 3 5 AD^s'ss ; BD=3-45; €0 = 3-84
inches. Determine the plan of the sphere circumscribing this
solid when the plane of the face ABC is inclined 40" and one
edge AB of that face 30* to the horizontal plane.

(i) Determine the plane of 40** and the line inclined
30" lying in it.

(2) Rabat the plane with the line of 30", and on the
ine so rabatted describe the triangle ABC.

(3) On the side AB describe another triangle ABD.
[magine the triangle ABD, which is a face of the pyramid,
o revolve about AB as an axis ; the point D will describe

; , circle in a plane perpendicular to AB and the locus of D
;.bout the axis AB will be, in plan, a straight line drawn
1 "om the vertex D of the triangle ^ i5Z> at right angles to AB.
Draw the plan of this locus and produce it indefinitely.

(4) On the side BC describe a triangle BCD.

A line from D perpendicular to CB will be the plan of
t le locus of D about CB as axis.

Draw this line and produce it to cut the line determined
ii(3).

(5) The point d where the loci intersect, in plan, is
t e plan of the vertex D. The height of this point can
b i readily determined by an elevation on a vertical plane
p irallel to one of the slant edges.

16 — 2

244 SOLID OR DESCRIPTIVE GEOMETRY. .

The plan and elevation of the solid on the plane of 40"
present no difficulties. See Chapter iv.

The four points A, B, C, D, situated at the solid angles
of the pyramid, are those through which the spherical
surface has to pass, and the centre of this sphere is found
as in Problem i on "The Sphere," Chap. vii.

18. Draw a triangle ABC having KB = 2"j, BC = 3*2,
AC = 2-3 inches. The points A, B, C, are the stations of
three observers who at the same moment take the altitude of
a point P in space. The angle by A is ■T)'^, by B 37°, a?id
by C 46". Show the plan of the point and find its height
in inches above the plane of ABC. Science Exam. Hon.
1872.

The point P is the common intersection of three in-
verted cones with axes vertical, vertices at the points A, By
and C, and generatrices inclined to the planes of the bases
at angles of 2>-^y 37* and 46" respectively.

To work the problem : —

(i) Make elevations of the cones on a vertical plane,
and across the elevations draw a series of lines, at con-
venient distances apart, each parallel to xy. These lines
are to be regarded as the vertical traces of a series of hori-
zontal planes at the levels shown by the lines in elevation.

(2) The sections of the cones by the horizontal planes
are circles, which are shown in true form in plan. Deter-
mine the plans of the sections of any two of the cones.
The points where the circles which are the plans of the
sections of the two cones under consideration by the same
horizontal planes cut one another, are points in the plan of
the intersection of those cones, and a curve drawn through
these points will be the plan of that intersection.

MISCELLANEOUS PROBLEMS. 245

Similarly, the plan of the intersection of one of the above
cones with the third can be found.

The point where the two intersections in plan cross is
the plan, /, of the point P.

(3) The elevation /', and therefore the height of P,
can be easily determined thus -. From the plan of the vertex
Df one of the cones, as A^ describe a circle to pass through
J>. This circle is the plan of a section of the cone whose
vertex is ^ by a horizontal plane passing through P. A
line at right angles to xy and a tangent to the circle will cut
1 he slant side of the cone in elevation at a point the height
< )f P above xy.

19. ABCV is a pyramid of which ABC is the base.
iV.B = 3-5z>?., BC=3 in.^ CA-2"]^in. The vertex, V, is 4in.,
z '2^in., and 4."j^in. from A^ B, and C, respectively. Deter-
1. line the inscribed sphere.

Note. Find the vertex as the common intersection of
t iree spheres of 4 in., 4*25 in., and 475 in. radii, described
fiom the centres A, B, and C, respectively. (Compare
I rob. 17 of this Chap.)

The centre of the inscribed sphere will be the common
ir tersection of three planes — two bisecting any two of the
tl ree dihedral angles formed by the slant faces of the
p Tamid, and the third bisecting one of the dihedral angles
fcrmed by the plane of the base and a slant face. The
ra Jius will be determined by a perpendicular from the
ce itre to one of the faces.

20. Assume three intersecting planes and determine a
sp 'lere of i in. diameter touching them.

246 SOLID OR DESCRIPTIVE GEOMETRY.

The centre of the required sphere will be the common
intersection of three planes, each -5 in. distant from, and
parallel to, one of the three touching planes.

21, Plan and elevation of a right pentagonal pyramid
under the following conditions : — Plane of face VAB inclined
50" to horizontal and 75" to vertical planes, one long edge VA
of face inclined 25" and lowest corner. A, i inch above the
horizontal plane. Other di?nensions at pleasure.

(i) Determine angle a between the base and one of
the faces.

(2) Determine the face VAB lying in a plane inclined
50" to horizontal and 75* to vertical planes. Chapter iii.,t
Probs. 7 and 19.

(3) Determine a plane containing the edge AB of the
face VAB and inclined to the plane of that face at the
angle a. Chapter iii., Prob. 15, converse.

(4) Rabat the plane determined in (3), and on the
edge AB contained therein describe the pentagonal base
of the given pyramid. Determine the projections of the
base and join abcde, plans of the corners of the base, with
V, and a'b'c'd'e' with v', for the complete projections of the
solid.

22. A right cone, base J' 5 in. diameter, axis 7, in. long,
rolls over the convex surface of another right cone standing
upon a circtdar base of \-^ in. diameter. The two cones have
a common vertex and their generatrices are of equal length, so
that the circle which is the base of the one cone rolls round
that of the other in a plane inclined to it at a cojtstant angle.
Draiv the plan of the curve traced by a point in the circumfer-
ence of the rolling cone.

MISCELLANEOUS PROBLEMS. 247

It is obvious from the mode of generation, that every
point in the curve is upon the surface of a sphere having
the common vertex for its centre and a generatrix of one
of the cones for radius.

When a circle rolls over the circumference of another, so
that the two always have a tangent in common at the point
of contact, the curve generated by a point in the circumfer-
ence of the rolling circle is either an epicycloid or a
hypocycloid, according as it rolls along the convex or
concave side of the fixed circumference.

When the two circles are coplanar the curve generated
must of course be in the same plane with them. The
curve generated in the above problem is, however, not plane
but of double curvature — the Spherical Epicycloid.

To construct tJie plan of the curve : —

(i) Measure a segment of the circumference of the
larger fixed circular base equal in length to the whole
circumference of the rolling base.

Note. The circumferences of circles being directly
proportional to their diameters, the required segment will,
in this case, be exactly one third of the whole circumfer-
ence.

(2) Bisect the segment in 0, and rabat the base of the
-oiling cone that touches it in that point, about its horizontal
race {i. e. about the tangent to the segment at 0') into H.P.,
)y aid of an auxiliary vertical plane of elevation passing
hrough 0 and the plan, v, of the common vertex.

{3) Produce the line vo through the centre of the
abatted base to meet its circumference at P, and let P be
he rabatted point that traces the epicycloid. Find its plan,

248 SOLID OR DESCRIPTIVE GEOMETRY.

p, when the base is lifted back to its proper position. This
will be one point in the required curve.

(4) Divide half the rabatted base from 0 io F into a
number of equal parts i, 2, 3, &c., and the half of the seg-
ment of the fixed base on the same side into the same num-
ber of equal parts i', 2', 3', &c.

If, now, the rabatted circle roll so that i coincides with
i' the position of Pwill be one division nearer, and by aid
of a new auxiliary vertical plane of elevation through vi
the base may be rotated about the tangent to the segment
at the point 1', and a second point /' in the plan of the
epicycloid found.

Similarly, as the rabatted base rolls to the points 2', 3',
&c., other points in this half of the curve will be determined.
The other half may be found by a similar operation on that
part of the segment.

Note. The student will readily see from his drawing
that the vertical plane of elevation first drawn can be made
to serve for all the points (i', 2', 3', &c.), and thus somewhat
lessen the labour of construction.

23. To determine a tangent to the spherical epicycloid at
a point in the curve.

The tangent must evidently lie in the tangent-plane to
the sphere on which the curve is traced. Also, it can be
easily proved that the tangent lies in the tangent-plane to
the sphere whose centre is the point of contact of the
generating circle, and radius the line drawn from the point
of contact to the corresponding position of the tracing
point.

MISCELLANEOUS PROBLEMS. 249

Thus, if p' were the plan of the given point, the tangent
to the curve would be the common section of a tangent-
plane to the first-mentioned sphere at that point, with the
tangent-plane at the same point to a sphere whose centre is
1' and radius the true length of \'p'.

When the spheres are determined, the tangent-line can
be drawn through P at right angles to the plane of the
normals to the two spheres at that point. See General
Remarks, "Tangent-Planes and Normals to Curved Sur-
faces," Chapter vii.

XIII.
SOLUTIONS OF THE TRIHEDRAL ANGLE.

("SPHERICAL TRIANGLES.")

Case I.

Given the three faces a, b, and c ; to determine the three
dihedral angles A, B, and C.

Note, In 'spherical triangles' the dihedral angles A,
B, and C are those which are opposite to the faces a, b,
and c, each to each; namely A \.o a, B X.Q b, and C to c.

Let the three faces be developed on the horizontal plane
of one of them, viz., b.

Take a plane of elevation, xy, at right angles to 01, the
common edge of the faces c and h; and let it meet the
second edge of face c in point F^, and cut 01 in e and Om
in/

Make OF^ on the second edge of a equal to OP^ ; since
OP^ and OP„ are really the same line, viz., the common
edge of faces c and a.

Then eP^ ef, fP^ are the developed sections of the three
faces by the vertical plane through xy.

Rabat the section ep'f, in this plane ; taking the centre e
and radius eP„ and centre f and radius /Pj, and describing
arcs intersecting in /' : join p'e and /'/ Then p, the plan
of P, is determined in xy from /', and Op is the plan of
the edge OP on the plane of b. Angle p'ef is the angle
A. The plane through / at right angles to Otn gives the
angle C. And the third angle B is determined by a plane
through P, in space, at right angles to the third edge OP,
in space, by Problem 15, Chap. in. /;/; is the horizontal
trace of this plane, which is that of the profile angle of B.

'^SPHERICAL triangles:' 25 1

Case I.

Given tc. 5 ctnci c.

/\\y

\ /

V /\

'"■

N. /

'/-^ ^'>

/ ^

\ ^N

\ ^s

252 SOLID OR DESCRIPTIVE GEOMETRY,

Case IL

Given the faces a, b, and the angle C : to determine A, B,
and c — /. e. two faces and the included dihedral angle.

Let the face a be developed on the horizontal plane of
b : since the angle C between these faces is known, a plane,
xj, at right angles to their intersection Om, exhibits this
angle Jfp' ; fp' being the vertical trace of the face a, the
length^' being made equal to/P,, / in xy, determined from
/', gives Op, which produced is the indefinite plan of the
third edge OP on the plane of b. A plane through / at
right angles to the edge 01 gives the angle A, and enables
us to rabat the third face c about 01 — its horizontal trace.
A plane through P in space, and at right angles to the edge
OP in space, and having Im for its horizontal trace, will
determine the profile angle of B.

''SPHERICAL TRIANGLES."
Case II.

253

Giren a, d and C.

A\^

254 SOLID OR DESCRIPTIVE GEOMETRY.

Case III.

Giveji the dihedral angles B and C and the face a, that is,
two angles, and their adjacent face : to detennine the angle A
and tlie faces b and c.

On the horizontal plane take the face a, of which 01,
Ofn are the edges.

Take the plane xy at right angles to 01.

The vertical trace of the face c on this plane forms the
angle, B, with xy, which angle being known this trace can
be drawn.

On a plane xj\, at right angles to Oin, the vertical
trace of face b forming the known angle, C, with x^y^ can be
drawn.

By the aid of the elevation on xy determine the vertical
trace of face c on plane x^y^ ; the two vertical traces on the
latter plane intersect. in a point i', whence / is found in xj^
and //' is a point in the third edge 01, of which draw the
indefinite //<7« 01.

The faces b and c can now be readily rabatted into the
plane of a by rotation on Om and 01 respectively. Angle
A is found by Problem 15, Chapter iii., as before, by a
plane through point P ox I and perpendicular to 01.

" SPHERICAL triangles:'

255

Case III.

Given B , C izncC a.

<c/

256 SOLID OR DESCRIPTIVE GEOMETRY.

Case IV,

Given two faces a and b, and the dihedral angle, A, (fhat
is, two faces, and an angle opposite to one of them) : to deter-
mine the dihedral angles B and C, and face c.

Assume the given face b on the horizontal plane, edges
01 and Om, and face a rabatted into that plane about
Om.

On plane xy, at right angles to 01, make the given angle
A with xy, the second line of this angle is the vertical trace
of face c on this plane.

On a second plane, x.^y^, assumed at right angles to Om,
determine the vertical trace ep' of the face c by the aid of
the former elevation : cut this trace in /' by the circular arc
described on plane x^y^ by point P^ on the third edge :
join pf, this is the vertical trace of face a ; obtain/ in xj^,
then //' is a second point determined in the third edge OP
of the spherical triangle. Draw the indefinite plan Op and
obtain the third angle, B, as before, by Problem 15, Chapter
III.

'' SPHERICAL TRIANGLES.

257

Case IV.

Giren a, "b and A

E.G.

258 SOLID OR DESCRIPTIVE GEOMETRY.

Case V,

Given the dihedral angles B and C, and a face b opposite
to one of the7n : to determine faces a and c, and the third
dihedral angle A.

Let the face b be drawn on the horizontal plane, and
revolved about its edge Oni until it makes the angle C with
that plane.

The indefinite line^' will "then be the vertical trace of
face b on the plane xy of the profile angle of C ; which is,
of course, perpendicular to Om : and fp' on this trace,
taken equal to /P,, gives the elevation, /', of a point P in
the second edge of face b. p in xy, obtained from p\
determines the pointy' in this second edge : and Op is the
plan of that edge.

Determine a plane to contain the line OP, in space,
just found, and inclined at the known angle of B, by
Problem 9, Chapter iii. : 01, the horizontal trace of this
plane, is the third edge of the required spherical triangle.

The face c can now be rabatted on the plane of a
about 01. The angle A may be determined as before,
by a plane at right angles to OP through point P.

" SPHERICAL triangles:'

259

Case V.

Given B . Cf an^ t.

260 SOLID OR DESCRIPTIVE GEOMETRY.

Case VI.

Given the three dihedral angles A, B, and C .* to determine
the three faces a, b, and c.

If the dihedral angle C has one face in the horizontal
plane, the traces of a second face can be readily drawn ;
and of these the horizontal trace is an indefinite edge of the
required spherical triangle.

A third plane making the angle of A with its horizontal,
and of i8o-^ with the inclined plane, determined by
Problem 26, Chap, in., will complete the required spherical
triangle. The intersections 01, OP, of this third plane with
the planes of the angle C, form the second and third edges
of the spherical triangle. The face b will thus be deter-
mined in the horizontal plane; into which the remaining
faces a and c can be rabatted about their respective edges
or horizontal traces into the plane of the face b.

" SPHERICAL triangles:'

261

Case YI.

Given A. B cmd C.

Cambrftrge :

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