B 448526

ARTES
1817
SCIENTIA
VERITAS
LIBRARY
OF THE
UNIVERSITY OF MICHIGAN
TUEBOR
SUQUÆRIS-PENINSULAM AMINAM
CIRCUMSPICE

From the Library of
C. S. Denison
PROBLEMS, THEOREMS
AND EXAMPLES
IN
DESCRIPTIVE GEOMETRY.
WITH THE DISTINCTIVE FEATURES OF
1. ESPECIAL FULLNESS AND SYSTEM ON THE POINT, LINE AND PLANE.
II. MANY FIGURED AND OTHER EXAMPLES FOR PRACTICE.
III. VALUABLE NEW AND ORIGINAL PLATES, WITH USEFUL SPECIAL
CONSTRUCTIONS IN SURFACES OF REVOLUTION.
IV. SPECIMENS OF DEMONSTRATIONS BY THE METHOD OF PROJECTIONS,
FOR
COLLEGES AND MATHEMATICAL STUDENTS, AND ENGINEERING
AND ARCHITECTURAL SCHOOLS.
BY
S. EDWARD WARREN, C.E..
FORMERLY PROFESSOR IN THE RENSSELAER POLYTECHNIC INSTITUTE AND ELSEWHERE,
AND AUTHOR OF A SERIES OF TEXT-POOKS ON DESCRIPTIVE
GEOMETRY AND ITS APPLICATIONS.
NEW YORK :
JOHN WILEY & SONS.
1883.
F
Entered according to act of Congress, ir the year 1874, by
S. EDWARD WARREN, J.E.,
in the office of the Librarian of Congress, at Washington.
Library of C. S. Daviser
3.4-82
A
NOTE.
THE title page of this edition is changed from this date in order to distin-
guish this work more plainly from the one entitled Elements of Descriptive
Geometry, Shadows, Perspective, etc., which was published later, and is an
entirely independent work; but which may hitherto have been sometimes con-
founded with this one, or mistaken for an abridgment of it. Each, however,
is intended to have its permanently distinctive character, use, and value for
students of mathematics and mathematical graphics.
At the same time, misprints, and a few not serious errors, incident to a
first edition, necessarily hastened by circumstances, are indicated in the list.
of corrections, which will facilitate the use of this volume, and which will
appear in their proper places in future editions.
NEWTON, March, 1883.
PAGE LINE
20. 11 from bottom.
25.
4
6 C
38. (8-14) from top.
CORRECTIONS.
For plane, read line.
a"",
P',
rands,
p.
d'.
the corresponding point of rs.
"to that plane, read to the other plane.
་
45.
7
51.
8
bottom.
67.
Bottom line.
68.
Top
75.
15 from bottom.
81. 12
.
83. 9
top.
97. 20
100. 8
bottom.
V, read Y.
12,
?
C
areas,
0.
ares.
these points, read aa', bb', ce', dd'.
After circle, insert, perpendicular to each
other at O₁.
For, second, etc., read, principle of Prob.
XVI. (In space).
For VB'E', read V'D'E'.
107. (Last paragraph).
127.
13 from top.
134.
4 and 5 from top.
11 and 12
140. Top line.
bottom.
153.
In EX. 5°.
Rottom line.
165. 9 from top.
N',
acb,
E,
E,
N'',
afe.
H.
D.
Read: The arcs drawn from the centre F.
For C'D', read jl — j'l'.
TIT',
TII'.
axis,
axes.
CORRECTIONS.
PAGE
LINE
174.
14 from bottom.
179.
1 and 2 from bottom.
180.
13 from bottom.
184.
4
188. 14
top.
197. 9
207. 15
bottom.
5
209. 17 66
215. 15 66
top.
217.
Last line.
225.
9 from bottom.
6
226.
14
233.
12
237. 5
240.
10
46
250. 18
251. 5
top.
bottom.
top.
For cE=H', read cE - p'H'.
After "other," read: the tangent plane on
either element cut from that other, is
osculatory to the curve at the point on
that element.
For XXVIII., read XXXII.
plane,
After 145,
Forgiven point,"
"projection,
when,
radius,
Part II.,
prism.
and 146.
given exterior point.
projections.
66
where.
•
radii.
subsequent volumes on
Shadows, Machine Drawing, and Stone
Cutting.
For Ph, read Rb.
"P and P' read p and p'.
"at a given point, read: at some point.
"3° and 4°, read 1° and 5°.
SS', read ss'.
NQ (denominator), read NQ..
U', read Ut.
After OB, insert = Mq1.
1
1
PREFACE...
CONTENTS.
PAGE
xvii
INTRODUCTION.
CHAPTER I.
THE SUBJECT MATTER OF DESCRIPTIVE GEOMETRY.-SURFACES AND LINES.
Preliminary..
Magnitudes as considered in descriptive geometry
Points and Lines.-Definitions
Classification of Surfaces.
1 Q2 00 4
CHAPTER II.
MEANS OF REPRESENTATION IN DESCRIPTIVE GEOMETRY.
Systems of projection..
CO
6
THEOREM I.-A point is determined by its projections on two planes.
7
77
Planes of projection...
Perpendicular projections in space.
8
9
Projections on paper……
Ciasses of problems.
11
Notation.-Literal, graphical, verbal.
12
BOOK I.-RULED SURFACES OF REVOLUTION.
CHAPTER I.
THE POINT, LINE, AND PLANE.
SECTION I.
FUNDAMENTAL PROBLEMS; OR, THE ALPHABET OF THE POINT, LINE,
AND PLANE.
A-PROJECTIONS.
a-Projections of Forms.
PROBLEM I.-To mark the projections of points, in any of the four
diedral angles.
16
iv
CONTENTS.
PAGE
PROBLEM II. -To draw the projections of various positions of lines.... 18
b-Projections of Tangencies.
PROBLEM III.-To draw the projections of lines lying in the planes of
projection.
19
PROBLEM IV. To make a new plane of projection contain a given line. 19
c-Projections of Intersections.
PROBLEM V.—To mark the traces of lines, having given kinds of posi-
tions.
PROBLEM VI.-To draw lines whose traces shall be on given parts of the
planes of projection..
PROBLEM VII. -To draw the traces of every position of a plane.
B-DEVELOPMENTS.
20
•
21
21
PROBLEM VIII. To show any plane figure, immediately, in its true size 23
SECTION II.
GENERAL PROBLEMS OF THE POINT, LINE, AND PLANE.
A-PROJECTIONS.
a-Projections of Forms.
PROBLEM IX.-Having given two projections of a point, to find its pro-
jections upon any other planes of projection, taken perpendicular to
either of the given planes..
24
General methods of operation in descriptive geometry….
26
PROBLEM X.-Having given the traces of lines, to construct their pro-
jections.
27
b-Projections of Tangencies.
PROBLEM XI.-To draw one or more straight lines, tangent to a plane;
and a plane tangent to a straight line.
30
c--Projections of Intersections.
THEOREM II.— If at least one of the sides of a right angle be parallel to
a plane of projection, the projection of the angle will be a right
angle.
32
PROBLEM XII.—Having given a line by its projections, to find its traces
PROBLEM XIII.—Having given one trace, and one point of a plane, to
find its other trace....
32
35
•
PROBLEM XIV. -To find the traces of a plane which shall contain three
given points
36
•
PROBLEM XV.-To pass a plane through one given line, and parallel to
another given line..
38
י

CONTENTS.
V
PROBLEM XVI.—To find the traces of a given plane upon new planes of
projection..
PROBLEM XVII.-Having given two planes by their traces, to find their
intersection..
•
PROBLEM XVIII.-To find the intersection of two planes, whose traces
are nearly or quite parallel to the ground line.
PROBLEM XIX.-To find the intersection of two planes which are nearly
perpendicular to the ground line.
First case.
Second case.
PAGE
39
40
42
43
PROBLEM XX.-To find where a line, perpendicular to a plane of projec-
tion, pierces a given plane; or, having one projection of a point in
an oblique plane, to find the other projection of the same point.... 45
THEOREM III.—If a line be perpendicular to a plane, the projections of
the line will be perpendicular to the traces of the plane..
PROBLEM XXI.-To find where a line, given by its projections, pierces
any plane given by its traces.
PROBLEM XXII.-To find where a line, given by its projections, pierces
a plane, given, not by its traces, but by any other two lines.
PROBLEM XXIII. To pass a plane through a given point, and perpen-
dicular to a given line...
46
47

48
•
50
B-DEVELOPMENTS.
a-Direct, or required Development, found from given projections.
PROBLEM XXIV.-To find the true length of the line between two given
points, by revolution about an axis perpendicular to a plane of pro-
jection...
THEOREM IV.-The distance from a point to a line parallel to a plane
of projection, is the hypothenuse of a right triangle, whose plane
is perpendicular to the line, and whose sides are parallel and per-
pendicular to the plane of projection...
52
52
•
PROBLEM XXV.—To find the true length of a line, by revolving it about
an axis, in, or parallel to a plane of projection.
PROBLEM XXVI.-To find the shortest distance from a given point to a
given line..
53
54
PROBLEM XXVII.—To find the perpendicular distance between two lines
which are not in the same plane.
55
PROBLEM XXVIII.—To find the true size of the angle between two
intersecting lines....
56
•
PROBLEM XXIX.-To find the angle between the traces of a given plane;
or, to revolve a given plane into either plane of projection. . . .
First method.
58
Second method.
PROBLEM XXX. —To find the angle between a given plane, and either
plane of projection....
60
PROBLEM XXXI.-To find the true size of the angle between two given
oblique planes..
61
First method.
Second method.
vi
CONTENTS.
PROBLEM XXXII.-To find the true size of the angle which a line makes
with a plane.
•
PAGE
61
b-Inverse or counter-developments; determining required projections
from auxiliary developments.
PROBLEM XXXIII.-Having given one projection of a point on a line
whose projections are nearly perpendicular to the ground line, to find
its other projection.
•
PROBLEM XXXIV.-To find the shortest distance from a point to a line,
and the projections of the same....
PROBLEM XXXV.-To find the shortest distance, or common perpen-
dicular between two lines, not in the same plane, by the method of
rotation.
·
PROBLEM XXXVI.-To find the shortest distance between two lines, not
in the same plane, by the method of new planes of projection..
PROBLEM XXXVII.—To draw a plane through a given point, and per-
pendicular to a given line, by the method of developments...
PROBLEM XXXVIII.-To find the true size of the angle between two
planes by the method of development.
First method.
Second method.
• •
PROBLEM XXXIX.-Having given the angles made by an oblique plane
with the planes of projection, and its perpendicular distance from a
point on the ground line, it is required to construct the traces of
that plane...
63
64
65
65
66
66
69
PROBLEM XL.-To bisect an angle in space hy counter-development.... 70
CHAPTER II.
DEVELOPABLE SURFACES OF REVOLUTION-CYLINDERS AND CONES.
SECTION I.
DEFINITIONS AND GENERAL PRINCIPLES, SURFACES AND CURVES.
THEOREM V.-A surface composed of straight lines meeting at a com-
mon point, is developable...
SECTION II.
PROBLEMS ON CYLINDERS AND CONES.
A-PROJECTIONS.
71
b-Projections of Forms.
Definitions and principles..
PROBLEM XLI.—Having given one projection of a point, on the circun-
ference of a circle contained in a profile plane, it is required to find
the other projection of the same point...
First solution.
Second solution.
3
78
CONTENTS.
vii
PROBLEM XLII.-To construct the projections of a circle of given centre
and radius, lying in a given oblique plane.
1°.
The centre of the circle..
2. The highest and lowest points, and tangents.
3.-The foremost and hindmost points, and tangents..
4'.-The right and left points, and tangents.
5˚.—The points on tangent lines of declivity.
•
PAGE
80
80
SO
81
81
81
THEOREM VI.—The projection of a circle, seen obliquely, is an ellipse. 82
PROBLEM XLIII.—To construct an ellipse, having given its axes, or other
conjugate diameters.
First solution.
Second solution.
PROBLEM XLIV.-To construct the projections of a cylinder of revolu-
tion, limited by parallel circular bases, when its axis is oblique to
both planes of projection..
PROBLEM XLV. To construct the projections of a cone of revolution,
84
86
whose axis is oblique to both planes of projection; and of any point,
or element, of its surface...
1°. The projections of the axis.
2.-The axes of the projections of the base, and the traces of its
planes.
•
3.-Other points of the base.
33 23
87
87
88
83
b-Projections of Tangencies.
Tangent lines to curves..
89
PROBLEM XLVI.-To construct a tangent to an ellipse, through a given
point on the curve..
91
First method.
Second method.
PROBLEM XLVII.-To construct a tangent to an ellipse from a given
exterior point.
92
PROBLEM XLVIII.-To construct a tangent to an ellipse, and parallel to
a given line.
92
95
THEOREM VII.-The tangent plane to a surface of revolution, is per-
pendicular to the meridian plane containing the point of contact.
PROBLEM XLIX.-To construct the extreme elements of the projections
of a cone, the projection of whose vertex falls without that of its
base...
PROBLEM L.-To construct a plane, tangent to a cone of revolution, and
through a given element of the surface, the axis of the cone being
a bi-parallel....
95
96
1˚.—Preliminary construction.
96
2.-Construction of the tangent plane..
97
PROBLEM LI.-To draw a plane, tangent to a cylinder of revolution, and
through a given point in space.
98
PROBLEM LII.-To draw a plane, tangent to a cylinder, and parallel to
a given line, when the axis of the cylinder is a bi-parallel.
PROBLEM LIII.-To construct a plane, tangent to a cone of revolution,
and parallel to a given line....
99
100
1.-The axis parallel to only one plane of projection..
100
viii
CONTENTS.
2.-The axis oblique to both planes of projection...
PROBLEM LIV.-To construct a plane, tangent at once to two cones,
whose axes are the projecting lines of their common vertex..
PACE
101
102
c-Projections of Intersections.
Intersections of planes with cylinders and cones..
PROBLEM LV.—Having given any pyramid by its projections, and any
plane by its traces, to find the intersection of the plane with the
pyramid.
First solution.
Second solution.
PROBLEM LVI.-To find the intersection of a vertical cylinder of revo-
lution, with a plane, perpendicular to the vertical plane of projec-
tion.
The conic sections.
•
•
104
105
107
108
109
The special, or elementary conic sections..
THEOREM VIII. -The conic section, in all its forms, is a symmetrical
curve; the parabola, relative to one axis; and the ellipse and
hyperbola, relative to two axes, at right angles to cach other.... 110
PROBLEM LVII.—To construct the projections of the intersection of a
perpendicular plane, with a cone of revolution, whose axis is verti-
cal; and to construct a tangent at any point of the curve...
1°.-Without auxiliary planes.
2.-With auxiliary planes.
•
3. To construct the tangent.
•
•
PROBLEM LVIII.-Having given a cylinder and a cone of revolution,
whose axes are oblique to both planes of projection, by their axes
and diameters; to find their horizontal traces..
1. A cylinder.
2. A cone.
•
THEOREM IX.-The plane section of a cone of revolution, when all its
element are cut, is a curve, such that the sum of the distances
from any point on its circumference to two fixed points within, is
constant, and equal to the greatest chord of the curve.
TпEOREM X.-Each point of the parabolic section of a cone of revolu-
tion, is equally distant from a fixel poiut, the focus, and a fixed
line, the directrix.
111
111
112
112
113
113
114
115
116
•
PROBLEM LIX.-To construct a cone of revolution, from which a given
conic section can be cut.
117
THEOREM XI.-The foci of parallel sections of a cone of revolution, are
in a straight line through the vertex.
118
Elementary properties of the ellipse.
118
Constructions on the ellipse..
119
Elementary properties of the parabola.
121
•
Constructions on the parabola...
122
Elementary properties of the hyperbola..
I. Properties corresponding to those of the ellipse.
II. Propertics peculiar to the hyperbola..
Constructions on the hyperbola...
123
•
123
•
124
124
·
CONTENTS.
ix
PAGE
The Ellipse.
PROBLEM LX.-To place a given ellipse upon a cone of revolution, whose
base shall be in the horizontal plane, and shall have a common tan-
gent with the given ellipse..
THEOREM XII.-The horizontal projection of the transverse axis of an
ellipse, cut from a vertical cone of revolution, is the transverse
axis of the horizontal projection of that ellipse; which therefore
cannot be a circle.
THEOREM XIII.-The fixed points, F and F₁, called foci, exist only for
the transverse axis of the ellipse.
PROBLEM LXI.-Having given the axes of an ellipse, to find its foci...
THEOREM XIV. -In the ellipse, the sum of the distances from any point
of the curve to the foci, is constant, and equal to the transverse
axis of the curve.
PROBLEM LXII.-To construct points of an ellipse; having given its
transverse axis and foci.
125
127

127
128
128
139
THEOREM XV.-Each half, or branch, of the ellipse has a directrix,
such that the ratio of the distances of any point of the curve
from the focus and directrix of either branch is constant, and less
than unity.
130
First method.
Second method.
PROBLEM LXIII.-To construct an ellipse, having given the focus and
directrix of either branch...
131
PROBLEM LXIV.-Having given an ellipse, to find its directrix.
PROBLEM LXV.-To construct a tangent to an ellipse, at a given point
on the curve.
131
132
THEOREM XVI.-The tangent to an ellipse at a given point bisects the
supplement of the angle included between lines from that point to
the foci..
133
PROBLEM LXVI.-To draw a tangent to an ellipse from a given point
without the curve...
133
PROBLEM LXVII.-To draw a tangent to an ellipse, and parallel to a
given line..
134
2. From a point without the curve.
•
PROBLEM LXVIII.-To draw normals to an ellipse.
1.—At a given point on the curve.
3.-Parallel to a given line.
—
The Parabola.
PROBLEM LXIX. To construct the cone of revolution which will contain
a given parabola, which has a common tangent with the cone's base 135
THEOREM XVII.-The parabola has a focus and directrix, from both of
which each point of the curve is equally distant.
PROBLEM LXX.-To construct a parabola, having given its focus and
·
136
134
134
•
134
135
directrix..
137
A
PROBLEM LXXI.-Having given a parabola, and its axis, to find its focus
and directrix.
137
X
CONTENTS.
PROBLEM LXXII.-To construct a tangent at a given point on a para-
bola
•
First method.
PAGE
137
Second method.
Third method.
• •
The Hyperbola.
PROBLEM LXXIII.-To construct the cone of revolution containing a
given hyperbola, and whose base shall have a common tangent with
the hyperbola.
THEOREM XVIII.-In the hyperbola, the difference of the distances from
any point of the curve to the foci, is constant, and equal to the
transverse axis...
PROBLEM LXXIV. -To construct a hyperbola, having given its trans-
verse axis and foci...
THEOREM XIX.-The hyperbola has directrices, such that the ratio of
the distances of any point of the curve from the like focus and
directrix is constant, and greater than unity...
PROBLEM LXXV.-Having given the transverse axis and foci of a hyper-
bola, to find its asymptotes..
• •
The focal curves of the conic sections.
THEOREM XX.-An infinite number of cones of revolution may be made
to envelope the same ellipse...
THEOREM XXI.-The foci of all the different projections of an ellipse,
E', upon planes containing the tangent at the extremity of the
transverse axis, lie on a circle perpendicular to the plane of the
ellipse, and whose diameter is the distance between the foci of
the ellipse...
THEOREM XXII.-The focal of the ellipse is a hyperbola, whose vertices
are the foci of the ellipse, and whose foci are the vertices of the
ellipse
138
139
139
140
141
141
112
143
144
THEOREM XXIII.-If we take any two points whatever, on the focal,
one on each branch, the sum of the lines from these points to any
point of the ellipse, is constant, wherever that point is taken.... 145
Determination of the conic sections by points..
THEOREM XXIV.-If a hexagon, inscribed in an ellipse, be such that it
has two pairs of opposite parallel sides, the two remaining sides.
will be parallel....
146
146
THEOREM XXV.-If any hexagon be inscribed in a conic section, the
three points of intersection of its opposite pairs of sides, produced,
will be in the same straight line..
THEOREM XXVI.-A conic section is determined by five given points,
or their equivalent.
147
117
•
The unity of the conic section...
143
PROBLEM LXXVI.—To draw a tangent to any one of the conic sections,
from a given exterior point...
149
PROBLEM LXXVII.—To draw a line, tangent at once to two conic sections 150
PROBLEM LXXVIII-Having given a cylinder of revolution, by its
CONTENTS.
xi
diameter and axis, to find its intersection by a plane perpendicular
to the axis, and a tangent to the curve..
1º.—The four initial points..
2°.-Other points.
3'.-The tangent line.
to find its intersection by any plane...
PROBLEM LXXIX.-Having given a cylinder of revolution by its trace,
PROBLEM LXXX.-Having given a cylinder of revolution by its axis and
diameter, to find its intersection with any plane, by means of aux-
iliary planes, and to draw a tangent to the curve..
1°.-Projections of points of the right section or base.
2.-Points of the required curve.
3°.-Visibility..
4'.-The tangent line.
PROBLEM LXXXI.-Having given any cone of revolution, by its axis and
diameter, to find its intersection by a plane, and a tangent to the
curve.
1°. The projections of the base.
2.—The common point of the traces on the given plane.
3°.-Points of the required curve.
4'.-Tangents to the intersection.
5.-Particular tangents...
•
•
•
Intersections of pairs of developable surfaces..
THEOREM XXVII.-The construction of the intersection of two develop-
able surfaces having a vertex, depends on the direction of their
elements.
THEOREM XXVIII.-The curve of intersection of two cylinders, etc., is
tangent to those elements of each which are cut from it by an aux-
iliary plane tangent to the other....
PROBLEM LXXXII.-To construct the intersection of two right cylin-
ders, one of which is vertical, and the other parallel to the ground
line...
PAGE
151
152
152
152
153
154
154
155
155
155
156
157
157
. 158
158
158
159
160
162
162
•
THEOREM XXIX.-Two equal cylinders of revolution, whose axes inter-
sect, will intersect in two ellipses..
PROBLEM LXXXIII.-To construct the intersection of two cones of
revolution, whose axes are horizontal and vertical, and in a plane
parallel to the vertical plane..
164
164
THEOREM XXX.-When two cones of revolution, with different vertices,
have two common tangent planes, they will intersect in two plane
curves..
166
PROBLEM LXXXIV.-To find points of the intersection of two cylinders
of revolution, given by their horizontal traces; their axes being
oblique to both planes of projection....
1°.--The construction of the cylinders and of the directing plane 167
2.-The determination of the number of curves, and the con-
166
struction of their points.
168
PROBLEM LXXXV.—To connect the points, found in a required inter-
section of two cylinders, etc., and to determine its principal features 168
1. The connection of the points..
169
xii
CONTENTS.
2°.-The visibility of the curve.
3. The visibility of the extreme clements.
•
PAGE
169
170
4.-The predetermination of the character of the intersection... 170
PROBLEM LXXXVI.-To construct the intersection of a prism and a
pyramid, each having any position in space.
172
PROBLEM LXXXVII.-To draw a tangent to the intersection of two
developable surfaces of revolution, at a given point of the curve.... 174
1.-To the intersection of a vertical and a bi-parallel cylinder... 174
2.-To the intersection of two right cones.
174
3.-To the intersection of two oblique cylinders..
4.-To the intersection at the multiple point..
5.-To the intersection parallel to a given plane.
THEOREM XXXI.-There is the same kind of variety in the intersection
of two cones, as in that of a plane and cone .
Case 1.-Case 2.-Case 3.
175
175
•
176
•
177
THEOREM XXXII.-When an element of a cylinder, or cone, is tangent
to a curve of double curvature upon the surface, the osculatory
plane of the curve, at the point of contact, is a tangent plane to
the surface...
THEOREM XXXIII.-The projections of a curve of double curvature
may present a point of inflection, or a cusp.
B-DEVELOPMENTS.
•
1-Development of Plane Intersections.
PROBLEM LXXXVIII.-To find the axes of the clliptical intersection of
a plane with a cylinder, or a cone; both being oblique to both planes
of projection.
PROBLEM LXXXIX.-To show the true form and size of the intersec-
tion of a developable surface by a plane, together with its tangent,
having given their projections..
1.-The plane and pyramid..
•
2.-The plane and vertical right cylinder.
2. The plane and vertical right cone..
•
177
179
180
182
183
183
184
184
2-Development of Cylinders and Cones, and of their Intersections.
THEOREM XXXIV.-If a line be tangent to any curve, traced on a de-
velopable surface, it will continue tangent to that curve, as found
on the development of the surface.
PROBLEM XC.—To develop the surface of a cylinder of revolution,
together with one of its oblique sections and a tangent to that sec-
tion
2°. The tangent..
•
185
186
1'.-The general development.
186
186
3. The points of inflection.
187
·
THEOREM XXXV.-The development of the oblique section of a cylin-
der of revolution is a sinusoid...
PROBLEM XCI.-To develop the surface of a cone of revolution, also
187
•
an oblique section of it, and its tangent.
187
1.-The general development.
188
2. The points of inflection
188
CONTENTS.
xiii
PAGE
CHAPTER III.
WARPED SURFACES OF REVOLUTION.
THEOREM XXXVI.-The surface, generated by the revolution of a
straight line around an axis, not in the same plane with it, is
warped....
THEOREM XXXVII.-The warped surface of revolution has two sets of
elements, each element of either of which intersects all those of
the others.
•
190
191
A-PROJECTIONS.
a-Projections of Forms.
PROBLEM XCII.-To construct the projections of the warped surface of
revolution.
•
192
. 193
PROBLEM XCIII.-To represent the projections of the warped surface
of revolution, by means of its apparent contour.
THEOREM XXXVIII.-The meridian curve of the warped surface of
revolution in a hyperbola..
193
PROBLEM XCIV.-Having given one projection of a point on a warped
surface of revolution, to find the other projection of the same point 194
b-Projections of Tangencies.
THEOREM XXXIX.-Every tangent plane to a warped hyperboloid of
revolution, cuts the surface in two elements, one of each genera-
tion, and is tangent only at their point of intersection...
PROBLEM XCV.-To construct the tangent plane to a warped hyperbo-
loid of revolution, at a given point..
PROBLEM XCVI.-To construct a plane, tangent to a warped hyperbo-
loid of revolution, and through a given line.....
PROBLEM XCVII.—To construct a plane, tangent to a warped hyperbo-
loid of revolution, and parallel to a given plane...
First method.
Second method.
THEOREM XL.-Every plane tangent to the conic asymptote of a warped
hyperboloid of revolution, contains two elements of the hyperbo-
loid; one of each generation, and both parallel to the element of
contact with the cone.
c-Projections of Intersections.
THEOREM XLI.-When two concentric surfaces are cut by a plane in
two curves, one surrounding the other, the inner one being a conic
section and so that any chord of the outer one, which is tangent
to the inner one, is bisected at its point of contact, the two curves
are similar and similarly placed conic sections..
•
196
197
. 198
200
201
203
THEOREM XLII.-Every plane section of the warped hyperboloid of
revolution, is a conic section, of the same kind as that cut from
its asymptote cone by the same plane.
203
xiv.
CONTENTS.
PAGE
204
PROBLEM XCVIII.-To find the intersection of a warped hyperboloid of
revolution, by a meridian plane, and by any other plane, parallel
to it..
THEOREM XLIII.-Every hyperbolic plane section of the warped hyper-
boloid of revolution, has asymptotes, parallel to the elements con-
tained in the parallel tangent plane.
PROBLEM XCIX.-To find the intersection of a warped hyperboloid of
revolution by a plane; and a tangent to the curve at a given point. 206
I.—Regarding the hyperboloid as a ruled surface...
207
205
1.-The auxiliary surfaces, the projecting planes of elements.... 207
2.-The auxiliary surfaces, tangent planes to the hyperboloid... 207
3°. The auxiliary surfaces, tangent planes to the asymptote cone 207
II.—Regarding the hyperboloid as a surface of revolution.
4.-The auxiliary surfaces, horizontal planes..
5.-The auxiliary surfaces, meridian planes.
C'.-The auxiliary surfaces, concentric cones.
III.-Regarding the hyperboloid as homologous with its asymptote cone
director.
•
7.-The auxiliary surface, the cone director, to whose intersec-
tion with the given plane, the required curve is made
similar..
8.-Construction of the tangent line.
•
•
PROBLEM C.-To find the intersection of any ruled surface, not plane,
with a warped hyperboloid of revolution..
Case 1.-Case 2'.-Case 3°....
208
208
208
208
209
209
210
210
210
B-DEVELOPMENTS.
Raccordment.
THEOREM XLIV.-When two warped hyperboloids have a common tan-
gent plane at each of three points of the same clement, they will
raccord along that element....
213
PROBLEM CI.—To construct two warped hyperboloids in raccordment.. 214
Other warped surfaces...
. 215
BOOK II.-DOUBLE CURVED SURFACES OF REVOLU-
TION.
A-PROJECTIONS.
a-Projections of Forms.
PROBLEM CII.-To construct the projections of double curved surfaces
of revolution...
1.-The sphere.
2°.-The ellipsoid.
·
3°. -The paraboloid and double curved hyperboloid..
42. The annular torus.
220
220
220
220
220
PROBLEM CIII.-To construct the projections of a regular dodecaedron 221
1°.
The horizontal projection..
2°.-The vertical projection.
221
222
CONTENTS.
XV
PAGE
b-Projections of Tangencies.
225
PROBLEM CIV.-To construct a tangent plane to a double curved sur-
face of revolution, at a given point of contact..
PROBLEM CV.-To construct a plane, through a given point in space,
and tangent to an ellipsoid at a point on one of its plane sections.. 226
First. The given section a parallel..
1'.-Auxiliary cones..
2'.-Auxiliary spheres..
227
227
227
Second. The given section a meridian.
228
•
PROBLEM CVI.-To draw a plane, tangent to a sphere, and containing a
given line.....
230
PROBLEM CVII.-To draw a plane, containing a given line, and tangent
to a sphere, by means of one auxiliary tangent cone..
PROBLEM CVIII.-To draw a plane, through a given line, and tangent
to a given double-curved surface of revolution, by means of two
auxiliary tangent cones.
232
233
c-Projections of Intersections.
PROBLEM CIX.-To find the intersection of a double curved surface of
revolution by a plane: and the tangent line at any given point of
the curve....
235
1.-The highest and lowest points.
236
2.-The points on the apparent contours of the ellipsoid.
237
3-Other points.
237
4'.-The tangent..
237
238
PROBLEM CX.-To circumscribe a sphere about a given triangular
pyramid..
.
239
241
THEOREM XLV.-Every plane section of an ellipsoid is an ellipse.
THEOREM XLVI.-If a cone be circumscribed, tangentially, about an
ellipsoid, its curve of contact will be an ellipse, whose centre will
be on the line joining the vertex of the cone with the centre of
the ellipsoid..
241
THEOREM XLVII.-Every plane which is parallel to the axis of a para-
boloid of revolution, cuts the surface in a parabola, identical in
form with the meridian parabola of the surface..
THEOREM XLVIII.—Two curves are similar, when all the tangents to
one are parallel to the homologous tangents to the other、...
THEOREM XLIX.-Any two similar and concentric parabolas are also
identical, and may be superposed upon each other...
THEOREM L.-Every plane section of a paraboloid of revolution, oblique
to the axis of the surface, is an ellipse....
THEOREM LI.-If two surfaces, S and S₁, be cut by a plane in two
curves, E and E., such that E, is a conic section, and that the
point of contact of every tangent to E bisects that portion of this
tangent which is a chord of E,, then the curve E is a conic section,
similar and concentric to E、………….
242
242
243
241
THEOREM LII.-The plane sections of a hyperboloid of revolution of
two separate nappes, and of its conic asymptote, made by the
same plane, are concentric and similar conic sections.
·
245
xvi
CONTENTS.
PROBLEM CXI.—To find the intersection of any two surfaces of revolu-
tion, whose axes are in the same plane...
I.—Axes parallel..
1°.—A cylinder and another surface..
2º. —A cone and any non-developable surface.
-A
3°. --Two non-developable surfaces..
· ·
II.--Axes intersecting...
1.-Particular cases; a-b—c.
2°.-The general case..
PAGE
246
246
246
247
247
247
•
247
247
PROBLEM CXII.—To construct the tangent line at a given point of any
of the intersections described in Prob. CXI.
249
B-DEVELOPMENTS.
PROBLEM CXIII.-Having given a pair of conjugate diameters of an
ellipse, to find its axes.
• •
250
PROBLEM CXIV. To develop the intersection of any plane with any
double curved surface; also the tangent to the given intersection.. 251
PROBLEM CXV.-To develop any double curved intersection, contained
in a double curved surface.
•
252
PREFACE.
In preparing a new work on Descriptive Geometry, I have
sought, in behalf of the subject, and of the schools in which it
is taught, to approach the completeness of the larger foreign
treatises on the same subject; and at the same time to suit the
present, and, more and more, the future wants of our scientific
schools.
In aiming at this result, while desiring to give the student
due occasion to think for himself, I have studied brevity; with-
out forgetting that if a book is to be used at all, its explanations
should be full enough to be a real relief from the peculiar per-
plexities of oral instruction, only, in exact subjects.
The following are the principal points of difference between
this, and my former work:
First. The mode of generating surfaces, rather than their
algebraic "order," has, appropriately to the subject, been made
the ground of their classification; thus securing an easy and
natural succession of subjects.
Second. I have paid regard to the observation of Gournerie,
that descriptive geometry has too much been treated in the inte-
rest of abstract, rather than of applied geometry; especially by
admitting into this first part only cylinders and cones of revolu-
tion. Certain useful constructions, of the kind required in
practice, have thus alone been secured.
Third. The improved re-arrangement, with the addition of
xviii
PREFACE.
the interesting group under the title of counter-development, of
the problems of the point, line, and plane; with a view to the
removal of all unnecessary difficulties from that fundamental
portion of the subject.
Fourth. The reduction of the main divisions of the problems
to two, viz., projections, of forms, tangencies, and intersections;
and developments; placing tangencies before intersections, since
they are applied under the latter head.
66
Fifth. The important distinction of "solution in space," and
graphical construction," is better exhibited, as well as better
named, in the spirit of descriptive geometry, "in space," "in
projection."
Sixth. Varied examples for practice have been freely given
under most of the problems. On this point I would add a
special remark.
In arithmetical and algebraic examples, a change in the num-
bers or the letters denoting the given quantities, will merely
cause a change in those expressing the answer, without chang-
ing the operations of the solution.
In like manner, a change in the position of the given parts,
in a problem of descriptive geometry, will cause a change in the
position of the required parts, but likewise, without changing
ut all the operations for finding them. IIence I would call
carnest attention to this most important feature, hitherto gen-
erally overlooked, as a necessary means for enabling the student
to thoroughly master the subject.
Seventh. The demonstration of properties of form, by the
method of projections, has been given, whenever necessary to
make the work more complete in itself; and in conformity to
the doctrine established by Olivier and others, that the method
of projections is a means of research, relative especially to pro-
perties of form and position, as well as a means of solving pro-
blems in space, upon planes.
PREFACE.
xix
The works of LEROY, OLIVIER, and GOURNERIE, whose pages
I have often consulted, have left little opportunity for origi
nality of matter, in subsequent works of more limited scope.
Most of whatever is peculiar to the present work, consists in
the features just indicated.
There are advantages and disadvantages, in binding the plates
separately, or with the text, or not at all. Students who prefer
the first method, can easily remove the plates and fasten them
together by their longer edges, and so that two plates at once
shall be exposed to view.
Finally, the interest and value of Descriptive Geometry as a
branch of mathematical study, and the desirableness of a more
general acquaintance with its elements as being the foundation
of much of the work of Technical Schools, suggests that it
should be included at least as an elective or alternative mathe-
matical study in all colleges. This volume is accordingly
complete in itself for Colleges, as well as sufficient for all
Technical Schools, in connection with the subsequent volumes
of application to Shades and Shadows, Perspective, Machine
Drawing and Stone Cutting; where those warped surfaces
which are here omitted, will be found, as many may prefer,
in close connection with their applications.
NEWTON, MASS, OCT., 1875.
DESCRIPTIVE GEOMETRY.
INTRODUCTION.
CHAPTER I.
THE SUBJECT MATTER OF DESCRIPTIVE GEOMETRY-SURFACES
AND LINES.
Preliminary.
1. SCIENCE is the whole body of truth, considered as a system
derived from some initial, or central truth, which thus gives
unity to the whole.
A science is a partial body of truth, similarly considered,
and embracing some one conveniently limited department of
knowledge.
2. DESCRIPTIVE GEOMETRY, thus regarded, is derived from
the primary truth, that if straight lines be drawn from each of
two fixed points, S and S,, Pl. I., Fig. 1, to all the points of
any fixed geometrical form, G, they will form two conical
figures, whose intersections with any two planes, P and P,,
respectively, will be two plane figures, F and F,, of definite
form, and having definite relations to each other, and to the
given form, G.
3. The points S and S, are supposed positions of the cye.
The lines from them to G are called projecting lines. The
planes, P and P,, are called planes of projection; and the
plane figures, F and F, are called the projections of G upon
the planes P and P₁.
1
1
2
DESCRIPTIVE GEOMETRY.
The usual case is that P and P, are perpendicular to each
other, and to the lines from S and S,, respectively, these points
being therefore infinitely distant from P and P₁.
4. Again, viewing the subject from the practical side:-Every
structure of a geometrical character, whether large or small, in
nature or art, from a crystal to the most intricate works in
masonry and mechanism, consists, as to its form, of a combina-
tion of surfaces and lines of various kinds, in various relations.
of tangency or of intersection, on which its stability, beauty,
and adaptation to use, all depend.
But as descriptive geometry is the geometry of forms and
positions in space, treated by means of their projections ( 3 ),
its utility is thus as evident from the outset, as in its many in-
dustrial applications, where nothing can be properly done with-
out it.
5. Descriptive geometry, as just defined, is to be distinguished
from the geometry of ratio and measure, that is, of metrical re-
lations, whether treated by the ordinary assumed diagrams, or
by algebra; also from the geometry of two dimensions, or
plane geometry.
The proposition: A plane which is perpendicular to each of
two planes, is perpendicular to their intersection, is one of form
and position.
The proposition: Two prisms, having equal bases and alti-
tudes, are equivalent, is one of ratio and measure, or a metrical
problem.
6. Descriptive geometry is properly so called, because the
projections of any regular body, having an exact relation to it
(2), form a description of that body, written on the planes of
projection, in what may be called the language of projections,
or the graphic language.
Magnitudes as considered in Descriptive Geometry.
7. Analyzing article ( 4 ), every form is ultimately a definite
assemblage of points. The path of a moving point is a line of
some kind. The line is said to be generated by the point,
DESCRIPTIVE GEOMETRY.
3
which is called the generatrix of the line. Two successive
positions of the generatrix are called consecutive points.
In like manner, a moving line generates a surface. Hence
the line, or the surface, is the sum of all the positions of its
generatrix. Any position of the generatrix is called an ele-
ment of the surface; and any element, and the one next to it,
are called consecutive elements.
8. Any fixed point, or line, which guides the motion of the
generatrix, is called a directrix. A similarly guiding surface
is a director.
9. Since Descriptive Geometry is concerned with the forms
and positions, rather than with the solid contents of objects, it
treats only of Points, Lines, and Surfaces.
Points and Lines.-Definitions.
10. For the purposes of Descriptive Geometry, lines are
divided into the two grand divisions of Straight Lines and
Curved Lines.
Likewise, Curves are divided into
Curves of Single Curvature.
Curves of Double Curvature.
A straight line is generated by a point as a generatrix (7)
moving without change of direction towards a fixed point taken
as a directrix (8). A curved line, by a point whose motion
continually changes its direction.
11. A curve of single curvature, or plane curve, is generated
by a point which moves so that all its positions lie in the same
plane; as a circle, oval, letter S, figure 8, or any curve capable
of being traced upon a plane.
12. A curve of double curvature is generated by a point
which moves so that all its positions do not lie in the same
plane; as the rail of spiral stairs; and, generally, any curve
traced at random on a curved surface.
Detailed explanations and examples will be given from time
to time as occasion requires.
4
DESCRIPTIVE GEOMETRY.
Classification of Surfaces.
13. Surfaces, as considered in Descriptive Geometry, are best
classified according to the way in which they are generated.
In the generation of a surface we note first, the motion, and
second, the form, of the generatrix.
The motion may be either one of revolution around a fixed
axis, or some less simple successive change of position, called
transposition.
The form of the generatrix may be either straight or curved
14. Hence, regarding the motion of the generatrix, surfaces
are first divided into
Surfaces of Revolution.
Surfaces of Transposition.
In surfaces of revolution, the sections made by planes, per-
pendicular to the axis, are called right sections, or parallels.
Those made by planes containing it, are called meridians.
15. Again, regarding the form of the generatrix: Any sur-
face which may be generated by a straight line, is called a
ruled surface, since through any point upon it at least one line
can be ruled with a straight-edge.
When a surface, as a sphere, or egg-form, can only be gen-
erated by a curved line, no straight line can be drawn upon it,
and it is called a double curved surface, because at any point of
it, two planes, cutting it perpendicular to each other, will inter-
sect it in curves.
16. Ruled surfaces are thus divided.-When the generatrix
moves parallel to itself, and upon another straight line as a
directrix (8); or revolves around an axis to which it is perpen-
dicular; or moves upon two intersecting straight lines, it gene-
rates a plane. IIence, through any point of a plane, any num-
ber of straight lines can be drawn in that plane.
17. When the generatrix moves upon a curved directrix,
each of its points describes some curve, and the resulting sur-
face is called a single curved surface, since it is thus straight in
the direction of the generatrix at any point, but curved in a
plane perpendicular to it, as in case of a cylinder.
DESCRIPTIVE GEOMETRY.
5
18. Again, when the generatrix moves so that its consecu-
tive positions intersect, as on a cone, or are parallel, as on a
cylinder, the surface is called developable, since its consecutive
clements are thus in the same plane, and hence can all be
brought into one plane without destroying their relative posi-
tion.
19. But when the generatrix moves so that its consecutive
positions are not in the same plane, the result is a warped sur-
face; so-called because, like the surface of a screw thread, it
cannot be rolled out flat without rending it; that is, destroying
the relative position of the elements. Thus, a warped surface is
non-developable.
20. Summing up, we have
Plane (16).
Ruled (15).
SURFACES.
J Developable (18).
Single Curved. Warped (19).
Double Curved, also non-developable.
This general table now being understood, as an outline map
of the field to be surveyed, we will proceed to a general descrip-
tion of the methods of operation.
•
6
DESCRIPTIVE GEOMETRY.
CHAPTER II.
MEANS OF REPRESENTATION IN DESCRIPTIVE GEOMETRY.
21. Continuing the analysis of (4), if, from a point, A, in
space, Pl. I., Fig. 2, a line be drawn, meeting the plane of pro-
jection, P, at some point, as a, then a will be the projection of
A upon P, as seen in the direction Aa.
The line Aa is the projecting line (3) of the point A.
Systems of Projection.
22. In Descriptive Geometry, the eye, E, is supposed to be a
point of vision, seeing equally well in all directions, and at all
distances from the object seen.
Thus, in Pl. I., Figs. 3, 4, and 5, let the curve, NQR, repre-
sent any given magnitude in space, and P, a plane of projec-
tion. If the eye, E, Pl. I., Fig. 3, be at any finite distance from
the plane of projection, P, then, first, only one perpendicular,
Er, can be drawn from it to this plane; second, the projecting
lines, from E, to all points of the given object indicated by
NQR, will form a conical surface, E-NQR, of which E will
be the vertex, and whose intersection, ngr, with the plane, P,
will be the projection of NQR upon the plane P.
23. If, now, first, the point E be transferred on the perpen-
dicular, ER, to an infinite distance from the plane of projec-
tion, P, all the projecting lines will be sensibly parallel to ER,
and hence to each other, Pl. I., Fig. 4, and hence will be per-
pendicular to the plane, P. Second, if the eye, E, be trans-
ferred on any of the oblique lines, as EO, and to an infinite
distance from P, the projecting lines will all become parallel to
EO, and hence parallel to each other, as before, but will all be
oblique to the plane P, as in Pl. I., Fig. 5.
24. Thus we have two great systems of projections, parallel
DESCRIPTIVE GEOMETRY.
7
or cylindrical projections, Figs. 4 and 5, in which the project-
ing lines are parallel to each other; and radial or conical pro-
jections, Fig. 3, in which the projecting lines radiate from a
given point.
And in the first system there are two species, perpendicular
or orthogonal projections, Fig. 4, in which the projecting lines
are perpendicular to the plane of projection (3), and oblique
projections, Fig. 5, in which these lines are oblique to that
plane.
25. Summing up the preceding principles, we have the fol-
lowing table:
Perpendicular, also called
Paralld or Cylindrical.
PROJECTIONS.
Orthogonal, or, Ortho-
graphic.
Oblique.
Radial, Conical, or Scenographic.
The latter, or scenographic projection, is the same as Natural
Perspective.
THEOREM I.
A point is determined by its projections on two planes.
26. In Pl. I., Fig. 2, a would equally be the projection of
any point on the line Aa. Hence, a second plane of projec-
tion, Q, is taken, on which another projection, a', is made as
before. Then, if from given projections, as a and a', project-
ing lines, aA and a'A, be drawn in the given directions, they
will intersect at the point A in space, and at no other point.
Hence, the point A is determined by its projections a and a'.
Planes of Projection.
27. Horizontal and vertical directions, being most familiar
and simple, the planes of projection are usually taken, the one
horizontal, and thence called the horizontal plane of projec-
tion; and the other vertical, and thence called the vertical
plane of projection.
8
DESCRIPTIVE GEOMETRY.
These planes are briefly distinguished as the plane H, and
the plane V. Their intersection is called the ground line.
28. Parts of the planes Hand V.-Each plane of projection
extends indefinitely in both directions from the ground line.
The two parts, F and B, Pl. I., Fig. 6, of the horizontal plane,
are called its front and back parts; and the two parts, U and K,
of the vertical plane, are called its upper and lower parts.
29. The four diedral angles.—Thus the two planes include
four angular spaces or diedral angles. These are called the
first, second, third, and fourth angles, as indicated in Pl. I.,
Fig. 6; the first angle being the one in which the eye is situa-
ted, and included between the front part of the horizontal
plane and the upper part of the vertical plane. The bounda-
ries of the other angles may be similarly defined.
Perpendicular Projection in Space.
30. Projections.-Let FB, Pl. I., Fig. 6, be the horizontal
plane, and KU the vertical plane. If now a line, Aa, be
drawn from any point, A, in space, perpendicular to the plane,
H, the point, a, where it meets that plane, will be the horizon-
tal projection of A. Likewise, if AA' be drawn from A, per-
pendicular to the plane V, the point A' will be the vertical
projection of A. And the like is true for any other point.
31. The distance of a point in space above the horizontal
plane is equal to the distance of its vertical projection above
the ground line. Thus, Aa, Pl. I., Fig. 6, =\'».
Likewise, AA', the distance of a point A, in space, in front
of the vertical plane, is equal to the distance, an, of its hori-
zontal projection in front of the ground line.
Thus, again, a point is determined by its two projections
(Th. I).
32. The projecting lines of all points of a straight line, ab',
Pl. I., Fig. 7, form the projecting plane of that line. Its inter-
section with that plane of projection, to which it is perpendic-
ular, will be the projection of the line on that plane.
Thus, a line has two projecting planes, as a point has two
DESCRIPTIVE GEOMETRY.
9
projecting lines; and two given projections of a line fully de-
termine the line, because the projecting planes erected on those
projections intersect each other only in the given line.
Finally, as two points determine a straight line, and as two
parallel lines determine a plane, such a line is practically pro-
jected by projecting any two of its points.
33. Traces. Since two points determine a straight line, it
may be determined by its intersections, as a and b, Pl. I., Fig.
7, with the planes of projection. Likewise, as a plane is deter-
mined by any two lines in it, it is determined by its intersec-
tions with these planes; and the like is true of other surfaces.
These intersections are called traces. Those of any line are
points. Those of a plane are straight lines; and those of
any other surfaces are generally curves, sometimes called the
bases, of those surfaces.
34. Traces, like projections (30), take their names from the
plane of projection in which they are found. Thus, a and b',
Pl. I., Fig. 7, are the horizontal and vertical traces of the line
ab'; and PQR and P'QS', Pl. I., Fig. 8, are the horizontal and
vertical traces of the plane PQP'.
35. The two traces of the same plane necessarily meet the
ground line at the same point. For, by the definition of the
traces, that point is at once in the given plane, and in each of
the planes of projection. Hence it is on the ground line, and
is the point where the given plane cuts that line.
Projections on Paper.
36. Hitherto the planes of projection have been considered
only in their real position, at right angles to each other, as re-
presented in Pl. I., Figs. 6 and 9.
But the inconvenience of operating on planes actually so
situated, leads to the representation of them upon one flat sur-
face of paper, as will now be explained.
37. The projecting lines, Aa and AA', Pl. I., Fig. 6, being
perpendicular to the two planes H and V, from the same point
A, their plane, AanA', is perpendicular to both planes of pro-
10
DESCRIPTIVE GEOMETRY.
}
jection, H and V, and hence to the ground line, GL. Hence,
the traces, na and nA', of this plane, are perpendicular to the
ground line at the same point, n. Indeed, they are simply the
projections of the projecting lines AA' and Aa, respectively, in
space.
38. Hence, to make the two planes of projection coincide,
we simply revolve the vertical plane backward, about the
ground line as an axis, till it coincides with the horizontal
plane, when we have the following results:
39. 1°. All that part of the draughtsman's paper, above, or
beyond the ground line, GL, Pl. I., Figs. 10, 11, and 13, re-
presents both the upper part of the vertical plane, and the back
part of the horizontal plane. All that part, below, or in front
of the ground line, represents both the front part of the hori-
zontal plane, and the lower part of the vertical plane.
2º. The vertical projection, A', Pl. I., Fig. 6, of A in space,
revolves in the arc A'a', about n as a centre, and in the plane,
A'na', to a'; whence a'n A'n Aa, and ana' is perpendicu-
lar to GL.
=
Thus, like points and distances are indicated by the same
letters in Figs. 6 and 10, and in Figs. 9 and 11.
40. From all that has now been said, the following princi-
ples, which are conveniently grouped together, are evident.
1º. If a point be on a line, its projections will be on the
projections of that line (32).
2º. If a point be in either plane of projection, it coincides
with its projection on that plane, and its other projection is on
the ground line.
3°. The two projections of the same point are on the same
perpendicular to the ground line (37).
4°. If a line, MN, or P'T (Pl. I., Fig. 8), be in a plane, its
traces, as M and N, are in the traces of that plane, and are
where its corresponding projections meet those traces.
5º. If a plane be perpendicular to a plane of projection, its
trace on that plane will be, or will contain, the projection of
everything in the given plane.
41. Also, note the following elementary principles of con-
struction:
*
DESCRIPTIVE GEOMETRY.
11
1º. Lines and planes are understood to be of indefinite ex-
tent. Hence, the projections of lines, and the traces of planes,
may always be produced indefinitely in either direction.
2º. Points may be imagined as the intersections of lines with
surfaces. They must be constructed as the intersections of lines
with lines.
3°. No constructions are generally considered geometrical,
unless effected with straight or circular lines of construction,
made with ruler and compasses, since no others can generally be
made with instruments by a continuous movement, conformed,
at every point, to the definition of the line.
Classes of Problems.
42. Problems may be distinguished in respect to the forms
represented, as indicated in the classification shown in the gene-
ral table (20).
But, as to position, two surfaces may be, first, exterior to
each other, in which case we may consider only the form of
each. Next, in approaching each other, they may be tangent;
and next, they may intersect. Further, each may, if possible,
be shown in its true size, or by an equivalent plane area.
43. Hence, we naturally have, in respect to the constructions
made, problems of PROJECTION, and those of DEVELOPMENT.
Problems of projection represent forms; separately, and as
combined in relations of tangency and intersection. Problems
of development show the real sizes of lines, of angles, or of
plane figures, by transposition till parallel to a plane of projec-
tion; and the true areas of curved surfaces, by transformation
into equivalent plane areas.
Thus there are:
Of Projection.
GRAPHIC PROBLEMS.
Of Development.
Of Forms.
Of Tangencies.
Of Intersections.
(By transposition.
By transformation.
12
DESCRIPTIVE GEOMETRY.
Notation.-Literal, Graphical, and Verbal.
44. Literal notation, Points.-All the projections of a point
are denoted by the same letter; unaccented in horizontal pro-
jection, with one accent in vertical projection; and with two
or more accents in other projections. Thus, P, P', p", etc.
Lines are similarly lettered, as ab, for the horizontal projec-
tion, and a'b' for the vertical projection.
Planes are denoted by lettering their traces, as PQP', where
Q is on the ground line, PQ is the horizontal trace, and P'Q is
the vertical trace.
45. Graphical notation.- Visible given and required lines
and traces, are made in full lines. Hidden lines are dotted.
Hidden traces are broken and dotted. Auxiliary lines are
also dotted, and traces of auxiliary planes are also broken and
dotted.
It is somewhat confusing to complicate the graphical notation
beyond this, but hidden lines may be distinguished from aur-
iliary lines by point-dotted lines, instead of by dotted lines
composed of short dashes. Also hidden traces may be distin-
guished from auxiliary ones, by simple broken lines without
dots.
The planes of projection are considered as opaque; but other
given or required planes are supposed to be transparent.
46. Verbal notation.-Points and lincs being determined by
their projections, and lines and planes by their traces, and the
like being true of bodies composed of, or bounded by points,
lines, and planes, magnitudes themselves are considered as
named by naming their projections, or traces.
Thus, the point, aa, means the point in space, whose projec-
tions are a and a'. Or, if, as sometimes happens, the vertical
projection were found first, we should say the point a'a.
The line ab-a'b', or a-b', means the one whose projections
are ab and a'b'; or, whose traces are a and b'. Thus, also, the
projections (37) of projecting lines are called the projecting
lines.
The plane PQP', means the one whose horizontal trace is
PQ, and vertical trace, P'Q.
DESCRIPTIVE GEOMETRY.
13
47. Notation of elementary positions.
1º. Descriptive Geometry is constantly concerned with lines.
in two radical elementary positions, relative to the planes of
projection, namely, parallel to the plane H (27), and parallel
to the plane V.
2º. Of these two general positions, there are three particular
cases, also of constant or very frequent use, namely, that par-
allel to the horizontal plane which is also perpendicular to the
vertical plane; that parallel to the vertical plane which is also
perpendicular to the horizontal plane; and that parallel to each
plane which is also parallel to the other, and hence parallel to
both planes, or to the ground line.
3°. We have then this summary of general and special ele-
mentary positions.
1. Parallel to the horizontal plane, = horizontal.
9.
<<
vertical
(C
3. Perpendicular to the vertical plane.
4.
66
66
horizontal"
5. Parallel to the ground line.
= vertical.
As these positions are of constant occurrence, it is highly
desirable, for the sake of urevity, to describe each by a single
word, as we do the first and fourth only.
48. In this work the following terms are used:
2—Parallel to the vertical plane = co-parallel.
3-Perpendicular
63
5-Parallel to the ground line
perpendicular.
bi-parallel.
CO-PARALLEL is chosen as the name of any line in the second
position, from its analogy with similar usage in Trigonometry,
and because the first and second positions are complementary;
the second being the same thing relative to the vertical plane,
that the first is, relative to the horizontal plane.
PERPENDICULAR is chosen as the name of any line in the third
position, because the same term is already universally used for
the same thing in Perspective; where the vertical plane of pro-
jection is the perspective plane.
BI-PARALLEL is then briefly expressive of the position of that
one of the co-parallels which is parallel to both of the planes at
once; that is, parallel to the ground line.
14
DESCRIPTIVE GEOMETRY.
49. Likewise for PLANES: parallel to the horizontal plane, a
horizontal plane.
Perpendicular to the same, a vertical plane.
Parallel to the vertical plane, a co-parallel plane.
Perpendicular to the vertical plane, a perpendicular plane.
Parallel to the ground line, a bi-parallel plane; not, however,
meaning parallel to both planes; but, in the same kind of
position, relative to the ground line, as a line having the same
name.
Perpendicular to the ground line, a profile plane.
DESCRIPTIVE GEOMETRY.
BOOK I.-RULED SURFACES OF REVOLUTION.
CHAPTER I.
THE POINT, LINE, AND PLANE.
SECTION I.
FUNDAMENTAL PROBLEMS; OR, THE ALPHABET OF THE POINT,
LINE, AND PLANE.
50. THE PLANE is classed among surfaces of revolution, be-
cause it can be generated by the revolution of a straight line
about another straight line, as an axis, to which it is perpendi-
cular or parallel; and at an infinite distance from it.
51. In either case, the sections made by planes perpendicular
to the axis, are circles, but of infinite radius, and therefore
straight. In the first case, these planes are parallel to the given
plane; but parallel planes meet at infinity in every direction,
and hence are said to intersect in a circle of infinite radius, or
a straight line at infinity.
In the second case, any plane, perpendicular to the axis,
would be perpendicular to the given plane, and hence would
intersect it in a straight line, that is, in a circle of infinite
radius.
52. Fundamental problems, or the alphabet of the "graphic
16
DESCRIPTIVE GEOMETRY.
language” of projections, relate to the immediate representa-
tion of things given by description; that is, without successive
steps of construction, for finding things required, from things
given by projection, as in the general problems which follow
after.
A-PROJECTIONS.
a-Projections of Forms.
PROBLEM I.
To mark the projections of points in any of the four diedral
angles.
In Space.-A point may have nine radically different posi-
tions relative to the planes of projection; four positions, taken
at pleasure, one in each diedral angle; one lying in each part
of each plane of projection, and one in the ground line.
Finally, these points may be given in space, either numeri
cally, by given distances from the planes of projection, or
graphically, Pl. I., Fig. 12, by an end view of the planes of pro-
jection, showing the points themselves as well as their projec-
tions.
In Projection.-Pl. I., Figs. 12 and 13.-In Fig. 12, the plane
of the paper is supposed to be perpendicular to both planes of
projection, so that HH is an edgewise view of the horizontal
plane, and VV a similar view of the vertical plane.
The planes of projection being thus represented, q will be
the ground line; BB and B'B' the traces-on the paper-of
the planes which bisect the diedral angles; and points, P, etc.,
will evidently be shown at their real distances, as Pp and P',
from their projections, p and p'.
Fig. 13 represents the planes of projection in the usual con-
ventional way (39). It also shows the projections of the points
whose position is indicated on Fig. 12, as will now be explained
in detail.
P, Fig. 12, is a point on the plane BB which bisects the first
and third diedral angles; it is, therefore, equidistant from the
planes of projection, and its projections, p and p', are equidis-
tant from the ground line, q. Then in Fig. 13, make p and p
1
DESCRIPTIVE GEOMETRY
17
equidistant from the ground line, and on the same perpendicu-
lar to it, and pq" = p'q″ = pq, Fig. 12.
Qis a point in the upper part of the vertical plane, hence
it coincides with its vertical projection, q', while its hori-
zontal projection, 7, is in the ground line. In Fig. 13, Qq′
and q are the two projections of the same point, and q'q = I'L
Fig. 12.
R is a point taken at pleasure in the second angle. and
are its projections, as shown on Fig. 12. In Fig. 13, make r'
and as far from the ground line as are the same points from
the ground line, s', in Fig. 12, and on the same perpendicular,
7'r; then will », in the back part of the horizontal plane, and
7', in the upper part of the vertical plane, be the two projec
tions of a point, R, in the second angle. A brief notice of the
remaining cases will now be sufficient. See both figures.
S is a point in the back part of the horizontal plane; there-
fore, S-s is its horizontal, and s' its vertical projection.
T is in the third diedral angle; therefore, t is its horizontal,
and tits vertical projection. Thus only the notation (4) dis-
tinguishes it from pp' in the first angle.
U is a point in the lower part of the vertical plane; its pro-
jections, therefore, are u, on the ground line, and u', coinciding
with U, in the lower part of the vertical plane.
W is in the fourth angle; therefore, its horizontal projection
is w, and its vertical projection, w'.
Y is a point in the front part of the horizontal plane; there- •
fore, y, coinciding with Y, is its horizontal projection, and y' is
its vertical projection.
Finally, Xa, Fig. 13, are the projections of a point in the
ground line; i. e., the point coincides with both of its projec-
tions.
EXAMPLES.-1°. Letter this point as one in the 3d angle, and tell to which
plane it is nearest.
Ex. 1.
#
Ex. 2.
Ex. 2'. This one as in the 2d angle, and as nearest to the horizontal plane.
2
18
DESCRIPTIVE GEOMETRY.
Ex. 3°. Define the position of this point, and its relative distance from the
planes.
Ex. 3.
Ex. 4.
P
'p'
Ex. 4°. Letter these points, one as in the back horizontal, the other in the
upper vertical plane.
PROBLEM II.
To draw the projections of various positions of lines.
In Space. The various elementary facts concerning the pro-
jections of straight lines in particular positions, can now easily
be found by analysis of the two projections of some simple
solid.
See Pl. I., Fig. 9, where ABDF represents a rectangular
prism, placed so that its edges are all parallel or perpendicular
to the planes of projection.
a-A vertical line, as BE, is its own projecting line relative
to the horizontal plane, and the point, b, is its horizontal pro-
jection. Its vertical projection is l'e', equal and parallel to the
line BE, in space, and perpendicular to the ground line.
Briefly for the remaining cases, as elementary projections are
here supposed to have been previously studied:
b-BA is a perpendicular (48). Its horizontal projection is
ba, equal and parallel to BA, and its vertical projection is the
point l'.
c-BC represents a bi-parallel (48), and its projections, be
and l'e', are equal and parallel to each other, and to BC, and
are parallel to the ground line.
d-AC represents a horizontal. Its horizontal projection,
ac, is equal and parallel to AC, and thus makes the same angle
with GL that AC does with the vertical plane. Its vertical
projection 'e is parallel to the ground line and less than AC.
e-Conversely EC represents a co-parullel, and be and e'c'
are its two projections, whose positions, analogous to the fore-
going, are evident on inspection.
DESCRIPTIVE GEOMETRY.
19
f-AE represents a line oblique to both planes of projec-
tion, but in a plane, ABE, which is perpendicular to both
planes of projection. Its projections ab and b'e' are both per-
pendicular to the ground line and shorter than AE.
g-Finally, the line AF, projected in ac and b'f', represents
the general case of a line oblique to both planes of projection.
In Projection.-See Pl. I., Fig. 11, which, being lettered to
correspond with Fig. 9, explains itself.
b-Projections of Tangencies.
53. Planes and lines can touch, or be in contact, without in-
tersection, only as the line lies in the plane, or as the plane
contains the line.
PROBLEM III.
To draw the projections of lines lying in the planes of
projection.
In Space.-If a line lies in either plane of projection, it will
coincide with its projection on that plane, and its other projec-
tion will be in the ground line. A line thus situated may be
parallel to the ground line, perpendicular to it, or oblique to it.
In Projection.-Pl. II., Fig. 14, ab-a'l' is on the plane H,
and parallel to the ground line, GL; ed-c'd' is in the same
plane, and perpendicular to the ground line; and ef—e'ƒ' is
also in that plane, and oblique to the ground line.
Similarly, gh-g'h', kl-k'l', and mn-m'n' are all in the
plane V.
EXAMPLES.-12. Construct lines, similar to the foregoing, and lying in the
back part of the horizontal plane, and with the obliquc one slanting either
way.
Ex. 2. Construct similar lines in the lower part of the vertical plane.
PROBLEM IV.
To make a new plane of projection contain a given line.
In Space.-Since one projection of a line which lies in a
plane of projection is in the ground line, a plane of projection
20
DESCRIPTIVE GEOMETRY.
.
is made to contain a given line by taking a projection of that
line as a ground line.
In Projection.-In Pl. II., Fig. 15, G'L' coinciding with ab,
would be the ground line of a new vertical plane of projection.
through the line ab-a'b'; or G"L", coinciding with a'b',
would be the ground line of a new plane of projection, per-
pendicular to the vertical of projection. By turning the paper
to bring G"L" into a directly right and left position like GL,.
the new plane may be considered as a new horizontal plane of
projection.
c-Projections of Intersections.
PROBLEM V.
·To mark the traces of lines having given kinds of positions.
In Space.—By (33) one projection of the trace of a line
coincides with the trace itself. Its other projection must be in
the ground line ( 44 ).
In Projection.-Pl. II., Fig. 16. Here a, alone, is the hori-
zontal trace of any line parallel to the vertical plane; V', as an
only trace, is the vertical trace of any line parallel to the hori-
zontal plane.
c and d are the traces of a line which crosses the first angle
in a plane perpendicular to the ground line.
e is the trace of any line which intersects the ground line.
g and f' are the traces of a line which crosses the first angle
obliquely forward, downward to the right; from the second
to the fourth angle.
k and h' are the traces of a plane which crosses the second
angle, backward, downward and to the right, from the first to
the third angle.
7 and m' are the traces of a line which crosses the third angle
forward, downward and to the right, from the second to the
fourth angle.
n and o' are the traces of a line which crosses the fourth
angle, backward, downward and to the left, from the first to
the third angle.
EXAMPLES. -1°. Mark the trace of any line parallel to the vertical plane and
behind it; also that of any line parallel to the horizontal plane but below it.
DESCRIPTIVE GEOMETRY. '
21
Ex. 2°. Mark the traces of lines crossing each of the four angles, and, each,
in a plane perpendicular to the ground line.
Ex. 3°. Mark the traces of lines crossing each of the first three angles, down-
ward and to the left, instead of as in Pl. II., Fig. 16, and crossing the fourth
angle downward to the right.
Ex. 4. Mark the traces of a line nearly horizontal and making a small angle
with the plane V. Do., and making a large angle with the plane V; also, the
traces of a line which is nearly a co-parallel, but which makes a large angle
with the plane H.
PROBLEM VI.
To draw lines whose traces shall be on given parts of the planes
of projection.
In Space. Having seen from the last problem, how lines
having given traces are inclined, it may be seen, conversely,
from Pl. I., Figs. 7, 9, and 11, how lines in space must be situ-
ated in order that their traces should fall on given parts of the
planes of projection.
In Projection.-Pl. II., Fig. 17.
qr→q'r' inclines forward and downward to the right.
st-s't' inclines forward and downward to the left.
uv-u'v' inclines backward and downward to the right.
wx-w'x' inclines backward and downward to the left.
EXAMPLES.-1°. The limited portions of all the lines just described are with-
in the first angle. Apply Prob. I. to the construction of similar limited por-
tions in cach of the three remaining angles. Thus ij-ij is in the second angle,
state its direction; yz-yz' is in the third angle, state its direction.
Ex. 2. Make a separate figure of each position shown in Pl. I., Fig. 11, and
add the projections of a line which shall intersect the ground line.
PROBLEM VII.
To draw the traces of every position of a plane.
In Space. The simplest classification of the positions of a
plane relative to the planes of projection, is as follows:
PLANES.
(In the same direction as the (Containing it.
ground line.
Parallel to it.
In a diferent direction from the (Perpendicular to it.
ground line.
Oblique to it.
22
DESCRIPTIVE GEOMETRY.
In Projection.-Pl. II., Fig. 18, GH, coinciding with the
ground line, is the trace of a plane containing that line. Such
a plane is not determined by its trace alone; but it may be, by
that and a given point which it must contain; or by this trace,
together with one upon any auxiliary plane, or by making a
given angle with either plane of projection.
IJ, parallel to the ground line, is the horizontal trace of a
co-parallel plane, whose vertical trace is in the same direction,
but at an infinite distance from the ground line.
K'L' is in like manner the vertical trace, parallel to the ground
line, of a horizontal plane.
MN-M'N', both parallel to the ground line, are the traces
of a bi-parallel plane, which is oblique to both planes of pro-
jection (49).
Proceeding to planes which meet the ground line, OPO' is a
profile plane. Its traces, OP and PO', are then both perpen-
dicular to that line. See also ana', Pl. I., Fig. 6.
QRQ' is a plane which is perpendicular to the vertical plane,
and its horizontal trace, QR, is perpendicular. to the ground
line.
STS' is a plane, perpendicular to the horizontal plane, and
its vertical trace, S'T, is perpendicular to the ground line.
UVU' is a plane which is oblique to both planes of projection
in such a way as to make acute angles with both of them, be-
hind, or on the left side of itself; and accordingly, its traces, UV
and U'V, both make acute angles with the ground line, reckoned
in the same direction from V.
YZY' represents the most general position of a plane, that
in which it makes an obtuse angle with one plane of projec-
tion, and an acute angle with the other, both on the same side
of itself.
Every plane which meets the ground line enters all of the
four angles, and thus, Pl. I., Fig. 8, shows that the kinds of
position, as now described, which such a plane has, relative to
the planes of projection, arc alike in the alternate angles, as in
the 1st and 3d, where the plane makes acute angles on the same
side of itself with both planes of projection; and in the 2d and
4th angles, where the position is of the same nature as that
of YZY'.
Finally, WXW' is the bounding position between all posi-
tions like UVU' and YZY'. That is, if YZ remain fixed as an
DESCRIPTIVE GEOMETRY.
23
axis around which YZY' revolves, ZY' may have any position
either to the right or left of YZ produced. Hence, there will
be one position where, as in WXW', the two traces will coincide,
as shown upon paper, but not in space.
EXAMPLES.—1°. Mark the traces of planes parallel to the vertical plane and
behind it, and parallel to the horizontal plane and below it.
Ex. 2°. Of oblique planes parallel to the ground line, and one of the planes
more nearly vertical than horizontal, and one more nearly horizontal than ver-
tical; and both crossing the first angle.
Ex. 3°. Do the same for like planes crossing the second angle, observing the
notation (46).
Ex. 4°. The same for like planes crossing the third angle.
Ex. 5. The same for like planes crossing the fourth angle.
Ex. 6. Construct planes like QRQ' and STS', but both inclined to the right,
about their perpendicular traces as axes.
Ex. 7°. Construct planes like UVU', 1st, more nearly horizontal than vertical;
2d, more nearly vertical than horizontal; and 3d, in both these positions, but
inclined to the right of V instead of to the left. The 2d position, moreover,
has two varieties, one nearly parallel and the other nearly perpendicular to the
vertical plane.
Ex. 8°. Make a like set of positions of YZY'.
Ex. 9°. Make a pictorial diagram like Pl. I., Fig. 8, for each of the foregoing
examples.
B-DEVELOPMENT.
54. Fundamental problems being defined as those in which
things required are shown without construction from things
given, those of development here consist of problems of plane
figures, so situated that they may be immediately represented
in their real form and size. This will be when these figures
are parallel to a plane of projection.
PROBLEM VIII.
To show any plane figure, immediately, in its true size.
In Space.-This problem simply reduces to that portion of
practical plane geometry which embraces the construction of
plane figures (5). We, therefore, only take here, as an exam-
ple of it, an operation often required in future problems of
development, viz., the representation of the path of a revolving
point in its real form.
24
DESCRIPTIVE GEOMETRY.
When a point revolves about an axis, it describes a circle, or
arc, whose plane is perpendicular to that axis. Therefore, in
order that this plane should be parallel to a plane of projection,
the axis must be perpendicular to the same plane.
In Projection.-The point, mm', Pl. II., Fig. 19, by making
an entire revolution about the vertical axis, a-a'b', will describe
the horizontal circle ABCD-A'B'; or by revolving to any
point as nn', will describe the arc, mn-m'n'.
By simply reversing the figure, similar operations relative to
an axis perpendicular to the plane, V, may be represented.
EXAMPLES.-1°. Make constructions similar to the above in the second angle.
Ex. 2. Make constructions similar to the above in the third angle.
Ex. 3. Make constructions similar to the above in the fourth angle.
Ex. 4. In each case let the revolution be less than 90°.
Ex. 5°. In each case let the revolution be greater than 90'.
55. In revolving a system of points, about an axis, four
things are to be noted.
1º. The axis is a fixed straight line.
2º. Therefore, every point in the axis remains fixed during
the revolution.
3º. Every other point describes a circle whose centre is in
the axis, and whose plane is perpendicular to it.
4°. Points and lines in a revolving plane do not change their
relative position.
SECTION II.
GENERAL PROBLEMS OF THE POINT, LINE, AND PLANE.
A-PROJECTIONS.
a-Projections of Forms.
PROBLEM IX.
Having given two projections of a point, to find its projections
upon any other planes of projection, taken perpendicular
to either of the given planes.
In Space.-In Pl. II., Fig. 20, let A be a point in space, GH
a horizontal plane of projection, and GV and G,V, any two
DESCRIPTIVE GEOMETRY.
25
vertical planes. Then let Aa, Aa' and Aa" be the projecting
lines of the point, A, and hence perpendicular to these planes,
respectively. Hence a'n a'm. That is, the heights of the
vertical projections, a' and a', above their ground lines GL and
GL, are equal.
Likewise, Pl. II., Fig. 21, let the plane of projection G, H,
be taken perpendicular to the vertical plane, GV, and let A be
a point in space. Aa, Aa and Aa" are the lines which project
it upon the planes H,V, and II,; and the projections, a and a"",
are equidistant from their ground lines, GL and G,L,; that is,
a'''r
= an.
56. And, in general, whenever two or more planes are each
perpendicular to a given plane of projection, then by (31),
first, the projections of a given point upon the former planes
are at equal perpendicular distances from their ground lines in
the latter plane; and, second, all of these perpendiculars pass
through the projection of the point on this plane.
In Projection.-First. Let aa', Pl. II., Fig. 22, be the given
projections of a point, and let G, L, be the ground line of a new
plane, V. Then draw am, indefinitely, and perpendicular to
G,L,, and make ma" na', and a and a" will be the projec
tions of aa' upon the planes which are perpendicular to each
other on GL,, after revolving the new plane, V, backward
about G,L,, into the plane, H.
Second. Given aa', as before, let G,L, be the ground line of
a new plane of projection, which, being perpendicular to the
plane, V, we will call a new plane, H, and will suppose it to
be revolved downward, about G,L, as an axis, into the plane of
the paper, taken now as the vertical plane of projection. Then,
making a'a"" perpendicular to G‚L„, and a'"'r = an, we shall
have a'"'a' for the new projections of the point aa'.
Third. To show that we may proceed indefinitely in this
way, let G,L, be the ground line of still another new vertical
plane, relatively, that is, to a"" GL, as a horizontal plane, then
draw a""a""" perpendicular to G,L,, and make pa'""' = ra', and
a'"' and a'''' will be the projections of aa' upon the two planes
of projection, whose ground line is G,L. That is, a""a" is
where a perpendicular from ad', to the plane, G.L,, pierces
that plane.
EXAMPLES.—1°. Let aa', Pl. II., Fig. 22, be in the second angle. Find a"
26
DESCRIPTIVE GEOMETRY.
in the vertical plane whose ground line is G, L₁, when this plane is viewed and
revolved as shown by the arrow.
Ex. 2°. Let aa' be in the third angle.
Ex. 3°. Let G₁L₁ be so taken as to make aa" in the second angle.
Ex. 4°. Let G₂L₂ be placed as G₁L, now is.
2
Ex. 5°. Let aa' be in the fourth angle.
General Methods of Operation in Descriptive Geometry.
57. From the previous pages, it is clear that we have—
1º. Planes of projection.
2º. Given magnitudes upon which to operate.
3°. Auxiliary magnitudes with which to effect a solution.
58. Corresponding to these there are three general modes of
procedure:
1º. The planes of projection and the given magnitudes may
both be fixed, and the auxiliary magnitudes may be so chosen,
and transposed when necessary, as to effect the solution with
the least labor.
2º. The planes of projection still being fixed, the given mag-
nitudes may be brought, by rotation, around one or more axes
successively, into such simple positions relative to the planes of
projections, that the desired result can be immediately found in
the new position, and then by counter-revolution, in the primi-
tive one.
3°. The given magnitudes remaining fixed, one or more new
planes of projection, each perpendicular to some one of the
preceding ones, may be taken till a pair is found having the
same simplest relative position to the given magnitudes that
was obtained in the second method by rotation of those magni-
tudes. The desired results can then be immediately noted on
the projections of the given magnitudes on the new planes,
whence they can be found on the primitive planes.
59. Of these three methods, the first is most simple to the
imagination, and the last is perhaps the most perplexing, espe-
cially when more than one new plane is used. The first is,
therefore, generally employed, though both the others are occa-
sionally useful, and will, therefore, be employed whenever con-
venient.
The operations in the second and third methods are quite
analogous to those of elimination in algebra.
DESCRIPTIVE GEOMETRY.
27
•
PROBLEM X.
Having given the traces of lines, to construct their projections.
In Space.-Four general varieties of position may be made
from (Prob. II., g) thus: A line which is oblique to both planes
of projection, may cross any one of the four diedral angles,
piercing the parts of the planes of projection which bound that
angle, and entering the adjacent diedral angles. To construct
the projections of the four positions just named, remember that
the projections of two points of a line determine the projec-
tions of that line, and assume the traces of the lines as such points.
In Projection.-Pl. III., Figs. 23-26: First. Let the line
cross the first angle. Then points as a and b will be respec-
tively its vertical and horizontal traces. The other projection
of each trace is on the ground line, whence ab-a'b' is the line
whose given traces are a' and b.
Such a line evidently enters the second angle at aa', and the
fourth angle at bʊ', as indicated in the figure, and the parts in
those angles, being hidden, are dotted.
Second. Let a line cross the second angle. It will pierce the
upper part of the vertical plane, and enter the first angle, and
it will also pierce the back part of the horizontal plane and
enter the third angle. Then points as a', Fig. 29, and b, in the
back part of the horizontal plane,will be its traces. Then, pro-
ceeding as before, ab and a'b' will be the two projections of a
line crossing the second angle. The part to the right of aa', is
seen, being in the first angle, and is made full; the parts in the
second and third angles are dotted.
Third. Let a line cross the third angle. It will enter the
fourth and second angles, and its traces will be points as a', Fig.
30, and b; and ab and a'b', found as before, will be its projec-
tions. Since no part of this line enters the first angle, its pro-
jections will be wholly dotted.
Fourth. ab-a'b', Fig. 31, represents a line which, coming
from the third angle, at a', taken in the lower part of the ver-
tical plane, crosses the fourth angle and enters the first at b,
taken in the front part of the horizontal plane. The part in
the first angle is seen, and its projections are full.
tions of the other parts are dotted.
The projec-
28
DESCRIPTIVE GEOMETRY.
60. If two lines are parallel, their projections on any given
plane will be parallel, for their projecting planes will be paral-
lel, and therefore the like traces of these planes will be parallel.
But these traces are the projections of the parallel lines in space.
Hence, the like projections of parallels are parallel.
EXAMPLES.—Construct the projections; define the positions of the given
traces, and mention the manner of crossing the diedral angles, of the lines
whose traces are given in each of the following figures:
Ꮳ
.b
Ex. 1°.
Ex. 3°.
.α
Ex. 5°.
.a
م.
Ex. 2°.
'q'
· α
Ex. 4°.
.b
• a
Ex. 0°.
Ex. 7°.
.α
Ex. 8°.
.6
Ex. 9°.
..m
..m
Ex. 10.
n
.m
ぴ
​DESCRIPTIVE GEOMETRY.
29
Ex. 11°. Letter these traces as those of a line cross-
ing the first angle.
Ex. 12°. And these as of a line crossing
the third angle.
Ex. 13°. And these as of lines cross-
ing the second angle, and the fourth
angle.

•
Ex. 14°. Letter this line as one which crosses the
first angle.
Ex. 15. And this, as one which crosses the second
angle.

Ex. 16°. And this, as one which crosses the third
angle.
1
Ex. 17°. And this, as one crossing the fourth angle.
b-Projections of Tangencies.
61. A line in a plane, or a plane through, that is, containing
a line, is an example of tangency. For, conceive any number
of lines tangent to a sphere, at one point. Then if the radius
of the sphere becomes infinite, its surface will become a plane
containing all these tangent lines. Any line lying in a plane,
is thus a tangent to that plane considered as a sphere of infinite
radius.
30
DESCRIPTIVE GEOMETRY.
PROBLEM XI.
To draw one or more straight lines tangent to a given plane,
and a plane, tangent to a straight line, whose traces are
given.
62. In Space. It is evident from the rudest model, that if a
line lies in a plane, its traces will be in the traces of that plane.
Conversely, if the traces of a line are in those of a plane, the
line joining those points will lie in the plane. See Pl. I., Fig. 8.
Hence, having the traces of a plane, we have only to assume
points in each of its traces for the traces of the required lines,
and then to construct the projections of the lines by Prob. X.
If the assumed points be so taken that the lines joining them
shall not be parallel, it will be found that the intersections of
the respective projections will be in a perpendicular to the
ground line (40, 3°).
63. If a plane contains a line, its traces will evidently con-
tain the traces of the line. But an infinite number of planes.
can be passed through a single line, hence, lines drawn from any
point on the ground line to the traces of a given line, will be
the traces of a plane which will contain that line.
In Projection.- First. Pl. III., Fig. 27. Let PQP' be the
given plane. The line, ab-a'b', joining the points aa' and bb',
taken at pleasure in the traces of the given plane PQP', lies in,
that is, is tangent to, that plane. The same is true of c-c'd'.
Second. Also, Let ab-a'b' be a line, given by its traces, an'
and ', and to which a plane is to be drawn tangent. Join
any point, Q, of the ground line, with these traces, and this
will give the traces, QP and QP', of the required plane.
EXAMPLES. -1°. Let the lines be in any other augle than the first, and in the
given plane.
Ex. 2. Let the position of the given plane be that of any of the variations
on UVU' or YZY', Pl. LI., Fig. 18.
Ex. 3. Let the given line, also, cross any one of the four angles.
64. IIaving two lines in a certain plane, any line which inter-
sects both of them will also lie in the same plane. If the third
line intersects either of the given ones at an infinite distance, it
DESCRIPTIVE GEOMETRY.
31
would be said to intersect one of the lines and be parallel to
the other.
This principle, with the last, and the elementary ones ( 62,
63), are of frequent use in finding the traces of planes which
are determined by given lines, not conveniently situated in
those planes.
65. If a line in a given plane, P, is parallel to either plane of
projection, it will be parallel to the trace of P upon that plane
of projection.
Conversely, if a line in a plane, P, is parallel to one of the
traces of that plane, it is parallel to the plane of projection con-
taining that trace.
c-Projections of Intersections.
66. If two lines intersect, their projections will intersect, on
the same perpendicular to the ground line. For the lines can
only intersect at one point, and the projections of any one
point are always situated as described. See Pl. III., Fig. 27,
where the two lines, being in the same plane, necessarily inter-
sect, as at pp'.
But if the projections do not intersect as described, the lines
themselves do not intersect.
67. The following conclusions are true for all angles:
(a.) If the sides of any angle, abe, Pl. III., Fig. 28, are on
opposite sides of the projecting line, bd, of the vertex, and if it
be revolved about an axis, mn, through its vertex, in its own
plane, and parallel to a plane of projection, PQ, its least projec-
tion will be when it is parallel to the same plane of projection
--when the projection will be equal to the angle in space; and
its greatest projection will be when its plane is perpendicular to
the plane of projection-when it will be = mbn = 180°.
(b). If the sides of an angle, efy, are on the same side of
the projecting line of the vertex, and if it be revolved about
pq, as before, the greatest value of its projection will be when
it is parallel to the plane of projection-the projection then
being equal to the angle in space; and the least value of its
projection will be when its plane is perpendicular to the plane
of projection, for then its projection = pfp = 0°
32
DESCRIPTIVE GEOMETRY.
THEOREM II.
If at least one of the sides of a right angle be parallel to a
plane of projection, the projection of the angle will be a
right angle.
Let ABC and DBC, Pl. III., Fig. 29, be two planes, perpen-
dicular to each other, and to the plane, Q.
Their traces, PC, and RC, upon Q, will then be perpendicu-
lar to each other. But by (32) these traces are the projections
of all the lines lying in these planes, respectively. Hence, they
are the projections of all the right angles, formed by any line
as BD, parallel to Q, and hence perpendicular to ABC, in one
of the planes, with all the perpendiculars to it, as AB, FB,
GB, etc., in the other plane.
But if BD be revolved about any of those perpendiculars to
it, as AB or GB, which are not parallel to the plane, Q, it will
immediately leave the projecting plane, DBC, and hence the
right angle thus transposed will no longer be projected on Q as
a right angle, which proves the theorem.
PROBLEM XII.
Having given a line by its projections, to find its traces.
In Space. See Pl. I., Fig. 7: First. Each trace, as B, of a
line being a point of that line, its projections, b and b', will be
on the projections, ab and a'b', of the line.
Second. As each trace, as B, is in a plane of projection, its
projection, as b', on that plane is the trace itself, and its other
projection, as b, is on the ground line. Whence arises the rule,
to be carefully memorized, it being of constant use. To find
either trace of a given line. 1st. Note where the projection of
the line, on the plane not containing the required tracé, meets
the ground line (40; 1º, 2º). 2d. Project the latter point into
the other projection of the given line.
In Projection.-Comparing Pl. III., Figs. 23-26, with
the above rule, in Fig. 23 the given indefinite line, ab—a'b',
pierces the horizontal plane at b, whose vertical projection, b',
DESCRIPTIVE GEOMETRY.
33
is found first, and projected into ab at b. Likewise, it pierces
the vertical plane at a', whose horizontal projection, a, is found
first, and projected into a'b' at a'. This line crosses the first
angle.
In like manner, Fig. 24, ab—a'b', crossing the second angle,
pierces the horizontal plane at b, and the vertical plane at a';
and by a like lettering of like points, in Figs. 25 and 26, the
same letters, b and a', denote the horizontal and vertical traces,
respectively, of lines crossing the third and the fourth angles
respectively.
68. Another theory of the same figure depends on the prin-
ciple of (32). For a line is common to both of its projecting
planes, and therefore pierces the plane H, for example, at the
intersection of the horizontal traces of its two projecting planes.
Thus, the horizontal trace, a, of the line ab-a'b', Pl. I., Fig. 7,
is the intersection of the horizontal traces, ab and aa' of its two
projecting planes; where bb' is perpendicular to the ground line,
and ab coincides with the horizontal projection of the line. In
like manner may be found.
69. To indicate, in the figures, this theory of the construc-
tion, the lines, aa' and bb', would merely be made broken and
dotted, as traces of the auxiliary projecting planes; instead of
dotted, as the projections of projecting lines.
In every one of the following examples, tell which angle
contains the given position of the given line, and which angle
the line crosses.
EXAMPLES.-Find the traces of the line given in each of the following
figures:
Ex. 1°.
712
772
Ex. 2°.
772.
m
Ex. 3°.
Ex. 4°.

m"
772
272
3
34
DESCRIPTIVE GEOMETRY.
Ex. 5°.
m
m
m
Ex. 7°.
m
Ex. 9°.
Ex. 11°.
m
m
m
m
Ex. 6°.
Ex. 8°.
Ex. 10°
m
M
m²
Ex. 12°.
m
m
Ex. 13. Find the traces of lines in a plane perpendicular to the ground
line. (Begin by revolving this plane into a plane of projection.)
Ex. 14°. Take the given portion of the line in each of the four angles suc-
cessively, and parallel to each plane of projection in succession, and find its
traces in these eight cases.
*
DESCRIPTIVE GEOMETRY.
35
70. With the perfect familiarity with Prob. XII., which the
solution of the examples under it will give, the following will
be at once seen to be easy applications of it:
PROBLEM XIII.
Having given one trace, and one point, of a plane, to find its
other trace.
In Space.-Any line from the given point to any point on
the given trace, will be in the required plane. Hence, draw
one or more such lines, and their traces will by ( 62 ), fall on the
traces of the plane. It is generally convenient to make the
auxiliary line, either parallel to the given trace, or parallel to
that plane of projection which contains the required trace (65).
In Projection.-Let Q'P', Pl. III., Fig. 30, be the given
trace, and pp' the given point. Then by the general solution,
draw the line, pm-p'm', from pp' to any assumed point mm'
on the given trace. n is the horizontal trace of this line, and,
therefore, one point of the now required horizontal trace of the
plane. The given trace is so nearly parallel to the ground line.
as to make it inconvenient to use its intersection with that line
in finding the horizontal trace. Therefore, and to illustrate
the special solution; draw the line, p'q'-pQ, parallel to the
given trace, Q'P'. The horizontal trace of pQ-p'q' is Q,
hence, Qn is the required horizontal trace.
EXAMPLES.—1°. Explain the order of construction of the same figure, sup-
posing QP to be the given trace.
Ex. 2°. Let the second line through pp' be parallel to the horizontal trace.
(It may be drawn of course from any point of_mn—m'n'.)
Ex. 3°. Either trace being given, let it also be in either part of either plane
of projection.
Ex. 4'. Let the given point be in either of the other angles than the first. By
combining the variations, indicated in this example and the last, particular
examples, too numerous for separate mention, may be found.
Ex. 5°. In examples 3 and 4, let the auxiliary line be parallel either to the
given, or to the required trace.
Ex. 6°. In example 3, let the given trace be parallel to the ground line.
36
DESCRIPTIVE GEOMETRY.
PROBLEM XIV.
To find the traces of the plane which shall contain three given
points.
In Space. Since an infinite number of planes may be passed
through one straight line, it is necessary, in order to make the
problem determinate, that the given points should not be in
such a line. Join any one of the points with the other two;
then, since these lines intersect, the plane which contains them
will be fixed, and will contain the three points. For, conceive
the plane to revolve about either of the lines described, until it
contains the third point. It will then evidently be fixed, and
thus a plane is determined by three given points, and thence by
any two lines joining those points.
The traces of the required plane will contain the traces of the
auxiliary lines.
In Projection.-Pl. III., Fig. 31. Let GL be the ground
line, and let aa', bb', and ce', be the projections of the given
points.
The vertical trace of the line ab-a'b' (62), is P', which is
one point of the vertical trace of the required plane.
On ac-
count of the position of this line, its horizontal trace cannot be
conveniently found. The traces of the line ac-a'c', are n' and
s, giving another point, n', of the vertical trace, and a point, s,
of the horizontal trace of the required plane. The horizontal
trace of the line bc-b'c', is e, which is, therefore, a second point
of the horizontal trace. Hence, PQs is the horizontal, and
P'Q the vertical trace of the required plane. If the construc-
tion be accurately made, these two traces will intersect the
ground line at the same point, Q (35).
EXAMPLES.-1°. General example. Let the points be in any one, two, or
three of the diedral angles; as 2 in the first, and 1 in the second; 1 in the
first, and 2 in the third; 1 in the second, 1 in the third, and 1 in the fourth,
etc.
Ex. 2°. Let two of the points be in a line parallel to the ground line. (The
directions of the traces thus being known, only one point of each need be
found.)
Ex 3°. Let one of the points be in the ground line.
DESCRIPTIVE GEOMETRY.
37
Ex. 4°. Let one of the points be in some part of the horizontal plane, and
one in some part of the vertical plane.
Ex. 5°. Let two of the points be in the horizontal plane.
Ex. 6'. Let two of the points be in the vertical plane.
Ex. 7°. Let two of the points be in a vertical line.
Ex. 8°. Let the vertical projections of the three points be in the same straight
line.
Ex. 9°. Let two of the points be in a parallel to the horizontal plane; and
either of them, with the remaining one, be in a parallel to the vertical plane.
Ex. 10°. Pass a plane through a given point, and parallel to a given plane.
71. Since three points determine a plane, it may be revolved
about an axis by revolving, equally, any three points of it.
Its new traces will then be found, as in the last Problem, by
joining the traces of the lines which connect the revolved posi-
tions of these points.

EXAMPLES.-1°. Revolve this plane about
the axis a-a'.
C

Ex. 2°. Revolve this plane 45° about the
'xis b-b'.

Ex. 3°. Revolve this plane 60° about
the axis, c-c'd', in the second angle.
ď
38
DESCRIPTIVE GEOMETRY.
PROBLEM XV.
To pass a plane through one given line, and parallel to another
given line.
In Space.-The required plane will evidently be determined
by the first line, with any line intersecting it and parallel to the
second given line. Its traces will therefore be determined by
those of its determining lines.
In Projection.-Pl. III., Fig. 32. Let cd-c'P' be the line
through which the plane is to be passed, and mn-m'n' the line
to which it is to be parallel. Pe-e'f' is a line drawn through
any point, ɑɑ', of the line cd-c'P, and parallel to the line
mn-m'n'.
The intersecting lines, cd-c'P' and eP-e'f', which deter-
mine the required plane, pierce the plane, V, at e' and P', and
the plane, H, at e and P.
Hence, PQP' is the required plane, represented by its traces
PcQ and Qe'P'.
n'
n
mn
Ex. 5. Find the traces of the plane
through m,m', and parallel to n‚n'.
Ex. 6. Proceeding likewise with this
set of examples, let the parallel line n,n'
cross the third angle.
Ex. 7. Let the parallel line n,n' cross
the fourth angle.
Ex. 8°. Let both m,m' and n,n' cross
the second angle.
EXAMPLES.-1°. Find the traces of
a plane through m,m' and parallel to
n,n'.
Ex. 2°. Let the contained line m,m'
cross the second angle.
Ex. 3. Let the contained line m,m'
cross the third angle.
Ex. 4. Let the contained line m,m'
cross the fourth angle.
n
m
m
The student can multiply examples of this nature till all
such combinations are exhausted.
Then-
Ex. 9°. Let m,m' be in any angle, and parallel to the ground line.
DESCRIPTIVE GEOMETRY.
39
Ex. 10°. Let n,n' be in the last named position.
Ex. 11°. Let the line m,m' intersect the ground line.
Ex. 12°. Let the line m,m' be in any horizontal position, not parallel or per-
pendicular to the vertical plane.
Ex. 13. Let either of the given lines be perpendicular to either plane of
projection.
Ex. 14°. Pass a plane through a given point and parallel to a given plane.
PROBLEM XVI.
To find the traces of a given plane upon new planes of pro-
jection.
In Space. The required traces will contain the traces of any
lines lying in the given plane, upon the new planes of projec-
tion.
We may also often employ the special principles, that where.
a given trace meets a new ground line, will be a point of a new
trace; and that wherever the intersection of two planes of pro-
jection meets a given trace, will be a point common to the traces
on those planes.
In Projection.-Let PQP', Pl. III., Fig. 33, be the given
plane, and GL₁, the ground line of a new vertical plane. Then
by the special principles just given, Q,, where the trace PQ
meets GL,, is a point of the new vertical trace. Also, the two
vertical planes intersect each other in a vertical line at a, which
appears at ab', when the vertical plane on GL is revolved back
into the paper, and at ab", perpendicular to G,L,, and equal to
ab', when the vertical plane on G,L, is revolved back into the
paper. This vertical line contains the point b' of the vertical
trace P'Q, which appears as above, at b', giving b'Q, for the
required new vertical trace.
If ' and Q, are inconveniently situated, we may resort to
the general method, first described, and find one or more points,
as nn', where any line cd-c'd', newly projected in cd-c''d",
meets the new plane of projection.
Similar constructions will serve to find the traces of the given
plane, upon any plane, perpendicular to the given vertical plane,
and taken as a new horizontal plane of projection.
EXAMPLES.—1°. Let the ground line, G₁L₁, be parallel to PQ.
Ex. 2°. Find the trace upon a new horizontal plane.
40
DESCRIPTIVE GEOMETRY.
Ex. 3°. Let the auxiliary line, cd-c'd', be parallel to a trace of the given
plane.
Ex. 4°. Take the new ground line, GL1, so as to meet PQ in front of GL.
PROBLEM XVII.
Having given two planes by their traces, to find their inter-
section.
In Space. The intersection of two planes is necessarily a
straight line, any point of which is by (41, 2°) the intersection
of two lines, one cut from each plane by any auxiliary plane
which intersects them both. The planes of projection can
usually be taken as such auxiliary planes, and hence, the inter-
section of two planes joins the intersection of their horizontal
traces with that of their vertical traces.
In Projection.-Pl. III., Fig. 34. Let QS be the ground
line, and PQP' and RSR' the given planes. Their intersection
joins a', the intersection of their vertical traces, to b, the inter-
section of their horizontal traces. Now a is the horizontal pro-
jection of a', and ' is the vertical projection of b. IIence, ab
and a'b' are the projections of the required intersection.
This is also a problem of frequent and varied application,
which should therefore be solved in a wide variety of positions
of the given planes.
EXAMPLES.-1°. Find the intersection of these planes, and note which angle
that intersection crosses, in this, and in all the following examples. (See 41, 1˚.)
Ex. 1°.
P'
es
Ex. 2°.


T
S
י
DESCRIPTIVE GEOMETRY.
41
9
Ex. 3°.
Ex. 4°.


?
S
P
Ex. 5°. Find the intersection of these planes, where the horizontal traces are
parallel.
Ex. 5°.
Ex. 6°.


S
r
p
Ex. 7. Find the intersection of these planes, where one of the planes is par-
allel to the ground line.
Ex. 8°. Find the projections of the intersection of these planes. See (40,5°.)
S
Ex. 7°.
Ex. 9°.
r
200
Ex. 8°.


Ex. 10°.


90
42
DESCRIPTIVE GEOMETRY.
PROBLEM XVIII.
To find the intersection of two planes, whose traces are nearly
or quite parallel to the ground line.
In Space. The general problem of intersection of nearly
parallel planes, has two main divisions: first, where both of the
planes are nearly or quite parallel to the ground line; second,
where they are nearly perpendicular to it. In either case, pro-
ceed according to Prob. XVII, and take one or more auxiliary
planes, each conveniently disposed for cutting each of the given
planes in a line, so that these lines will meet within the limits
of the figure, and thus give a point of the required intersection.
In Projection.-Pl. IV., Fig. 35. Let PQ and P'Q', parallel
to the ground line, be the traces of one plane, and RS and R'S'
those of the other, nearly in the same position.
Take any auxiliary plane, as RTP', and by the last problem
find its intersections, Pa-TP' and R-TR', with the given
planes. The intersection, cc', of these two lines, being thus in
both of the given planes, is a point of their required intersec-
tion. Again, take any auxiliary plane, SOQ', whose intersec-
tions, Se-d'S', and Q'f' (whose horizontal projection is not
shown, it being not needed), give a second point, n'n, of the re-
quired intersection, whose projections are en-c'n'.
72. If both of the planes just given, had been parallel to the
ground line, their intersection would have been so also, and one
point of it, found as above by one auxiliary plane, would have
been sufficient.
If a profile plane were used, it would be revolved about
either of its traces into a plane of projection, in order to show,
separately, its auxiliary intersections with the given planes, and
thus give a point of their intersection.
EXAMPLES.-19. Let the given planes be bi-par-
allel ones, thus; and let the auxiliary plane, first,
have any oblique position; then, second, let it be
a profile plane.
Ex. 2. Let one of the given plancs be a bi-par-
allel, and cross the second angle.
Ex. 3. Let one of the given planes cross the
third angle.
Ex. 4°. Let one of the given planes cross the
fourth angle.
M'
M
N'
N
費
​43
DESCRIPTIVE GEOMETRY.
$
Ex. 5º. In Pl. IV., Fig. 35, let both of the planes cross the second angle.
Ex. 6°. Let both of the planes cross the third angle.
Ex. 7. Let both of the planes cross the fourth angle.
Ex. 8°. Let one of the given planes contain the ground line, and any given
point in space.
Ex. 9°. Let one of the given planes contain the ground line, and make a given
angle with a plane of projection.
PROBLEM XIX.
To find the intersection of two planes, which are nearly perpen-
dicular to the ground line.
In Space.-First case. The similar traces not very nearly
parallel. In this special case it is sufficient to take auxiliary
planes under the general solution, parallel to the original planes
of projection; knowing that the traces of a given plane upon
parallel planes will be parallel lines.
Second case. Let the planes, PQP' and RSR', be very nearly
perpendicular to the ground line. The difficulty in this case is,
that as such planes are very nearly parallel to each other, any
third plane will cut them in two nearly parallel lines, whose
intersection will, therefore, be both quite remote and too
acute.
But a plane, nearly horizontal, will cut from the given planes
two lines, whose vertical projections will intersect near the
ground line; and the like is true in horizontal projection, rela-
tive to a plane nearly parallel to the vertical plane.
In Projection.-First case. Let PQP' and RSR', Pl. IV.,
Fig. 36, be the given planes, and A'B' and C'D' the vertical
traces of two auxiliary horizontal planes. Cƒ-C'f' and dƒ—
D'f' are the lines cut from PQP' and RSR' by the plane C'D',
and they give the point, ff', of their required intersection.
Likewise ae-a'e' and be-b'e', cut by A'B' from the given
planes, meet at ce', another point of the required intersection,
which is ef—e'f'.
The auxiliary planes could equally well have been parallel to
the vertical plane; or one could have been parallel to each
plane, H and V.
Second Case. Let PR, Pl. IV., Fig. 37, be the horizontal
trace of a co-parallel plane, which cuts from the given planes
44
DESCRIPTIVE GEOMETRY.
the lines beginning at P and R, whose vertical projections, a'b'
and c'd', are parallel to the traces QP' and SR', and begin, as
shown, at the projections of P and Q, upon the ground line.
Now let a'd' be a horizontal line in the plane PR, and near
to the horizontal plane, and let a secondary auxiliary plane be
passed through a'd' and the ground line. Such a plane would
be nearly horizontal, as required; Q and a' would be points
common to it and to PQP'; S and c' would be common to it
and RSR', and hence Qa' and Sc' would be the vertical projec-
tions of the lines cut by the secondary auxiliary plane from the
two given planes. These lines, therefore, meet at a point, e',
of the vertical projection of the two planes. Another second-
ary plane passed through b'd', in the primary plane, PR, and
the ground line, would cut from the given planes the lines
whose vertical projections, QU' and Sd', give a second point, f',
of the vertical projection of the intersection of the two planes.
Similar operations, beginning with a herizontal plane, P'R',
and lines, qm and pk, lying in it and parallel to the ground
line, will give the points, r and s, in the horizontal projection
of the required intersection.
Qq
Finally, to avoid the inaccuracy in the direction of Qa', etc.,
and the resulting position of e', etc., arising from the nearness
of Q and a to each other, we have only to consider that if QP,
for example, were n times as great as it now is, Q' would also
be n times its present size. IIence, make Qy any convenient.
number of times Qt'; and Sh, the same number of times Su,
and gn,parallel to QP', and ho, parallel to SR', will be the lines
cut from PQP' and RSR' by some plane, K, far in front of PR;
and, assuming now that '' is in the plane, K, Qn and So will
be the lines cut from PQP' and RSR' by a plane containing
both b'd' and the ground line, and their intersection, y', will be
a point of the vertical projection, e'f'. Thus, y' is between the
determining points, Q and n, of Qn, and between S and o, which
is favorable to accuracy. Nevertheless, unless Qt', etc., can be
exactly expressed by numbers, so that Q, etc., can be laid down
at once, the errors in the distances, Qg, etc., as found by laying
off Qt over and over with dividers, may exceed those of the
former construction. Like constructions apply to rs'.
73. Parts of the solutions of problems in projections may
sometimes be neatly effected by regarding them as plane prob-
{
DESCRIPTIVE GEOMETRY.
45
1
lems. Thus, in the one just given, having found a single
point, as f', of either projection of the intersection of the two
planes, we can find another, as N, as follows. Knowing that
QP', SR', and e'f', all meet at one point; QN, for example,
will be a fourth proportional to C'D', C'f', and QS. Then N,
thus found, and joined with f' will give Nƒ". In like manner,
r and s may be found.
EXAMPLES. -1°. In Pl. IV., Fig. 36, let both of the auxiliary planes be co-
parallel.
Ex. 2°. In Pl. IV., Fig. 36, let both of them be oblique to both planes of pro-
jection.
Ex. 3'. In Pl. IV., Fig. 37, in using the parallel planes, PR, and the vertical
plane of projection, assume the vertical traces upon these planes, of the second-
ary auxiliary planes, as at b'd' on PR, and anywhere below QS on the vertical
plane, and produce the vertical traces, as P'Q, so as to make the determining
points of Qf', etc., further apart.
Ex. 4. Likewise, in using the parallel planes, P'R', and the horizontal plane
of projection, take the traces of the secondary auxiliary plane on P'R', and on
the back part of the horizontal plane, and intersecting the giren horizontal
traces produced.
Ex. 5. In Pl. IV., Fig. 36, let the required intersection cross any of the
other angles, making three more examples.
Ex. 6. The same in Pl. IV., Fig. 37, making three more examples.
PROBLEM XX.
To find where a line, perpendicular to a plane of projection,
pierces a given plane; or, having one projection of a point
in an oblique plane, to find its other projection.
In Space. The double enunciation of this problem is practi-
cally one, since, if a line be perpendicular to a plane of projec-
tion, its projection on that plane will be a point, which will be
the projection of every point of the line, as, for example, the
point where it pierces any given oblique plane. Taking the
problem, as expressed by the second enunciation, the point
itself, whose projection is given, is merely where a perpendicu-
lar, or projecting line, from the given projection, pierces the
given plane. Through the point itself, conceive any line to be
drawn in the given plane. Its traces will be in the traces of the
plane, and its projections will contain the projections of this
point.
46
DESCRIPTIVE GEOMETRY.
In Projection.-Pl. IV., Fig. 38. Let PQP' be the given
plane, and let p be the given projection of the point. The
simplest auxiliary line is a horizontal, or a co-parallel line of
the given plane. The horizontal line, pq-q'p', is such a line
(65). Hence, the required projection, p', is the intersection of
pp' with p'q', parallel to the ground line, and through the ver-
tical trace, g', of pq-p'q'.
EXAMPLES.-1°. Let p' be given and let the auxiliary line be horizontal.
Ex. 2°. Let p' be given, and let the auxiliary line be a co-parallel on the
given plane.
Ex. 3°. p being given, let the auxiliary line be a co-parallel.
Ex. 4°. Let p be given, and in the back part of the horizontal plane.
Ex. 5°. Let p' be given, and in the lower part of the vertical plane.

хр
хр
Ex. 6°. Let p be given in each of the
positions shown in this figure.
Ex. 7°. Let the auxiliary line have any direction in the given plane.

کر
Ex. 8. Let p' be given as in this figure.
Ex. 9°. Let the plane be parallel to the ground line.
THEOREM III.
If a line be perpendicular to a plane, the projections of the line
will be perpendicular to the traces of the plane.
At any point of the given line, L, imagine a perpendicular,
N, to either trace of the plane, PQP'; for example, to its hori-
zontal trace, PQ. Let N revolve about PQ as an axis, and it
will generate a plane, perpendicular to PQ; and hence, per-
DESCRIPTIVE GEOMETRY.
47
pendicular both to the given plane and to the horizontal plane.
For the former reason this plane will contain L, and thence for
the latter, it will be its projecting plane. But its horizontal
trace, being one position of N, will also be perpendicular to
PQ, while it is also the horizontal projection of L, which proves
the theorem.
74. The intersection of PQP' with the plane of revolution
of N, is also perpendicular to the trace, PQ, and is called the
line of greatest declivity, or, simply, the declivity of PQP'.
PROBLEM XXI.
To find where a line, given by its projections, pierces any plane,
given by its traces.
In Space. The required point is in the given line, and by
(62) in the intersection of the given plane with any auxiliary
plane containing the line. Hence, it is the intersection of these
lines. The auxiliary plane is most conveniently a projecting
plane of the line.
In Projection.-Let SQ, Pl. IV., Fig. 39, be the ground line,
ap-a'p' the given line, and PQP' the given plane. TUT' is
the plane which projects the given line on the horizontal plane,
and T't' and UT are the projections of its intersection with the
given plane, PQP'. The given line, ap-a'p', meets this inter-
section in space; therefore, p', where a'p' and T't' meet, is the
vertical projection of the required intersection of ap-a'p', with
the plane, PQP'. Then, p'p, perpendicular to the ground line,
determines p, the horizontal projection of the same point.
By using the plane, R'SP, which projects the given line on
the vertical plane, p, where ap meets Pr, the horizontal projec
tion of the intersection of RSP and PQP' will be found first,
and p' by drawing pp'.

EXAMPLES.1°. Find the intersection of
this line and plane-mm' being the line.
་་
m'
772
48
DESCRIPTIVE GEOMETRY.
*
Ex. 2°.
m'
m
m
m
Ex. 4°.
Ex. 6°.
Ex. 3°.


m'
m
m'
Ex. 5º.

a
m
N
P
"'
b
α
a
Q
Ex. 7°.


m
m
From the above examples, other variations can be made; of
the plane, as shown in Prob. VII., and of the line, as seen in
Probs. II. and VI.
PROBLEM XXII.
To find where a line, given by its projections, pierces a plane,
given, not by its traces, but by any other two lines.
In Space.-Construct the intersections of the lines which
determine the given plane, with either projecting plane of the
given line. The line joining these points will be the trace of
DESCRIPTIVE GEOMETRY.
49
the given plane upon the projecting plane, as in the last pro-
blem. Hence the intersection of this trace with the given line
will be the required point.
In Projection.--Pl. IV., Fig. 40. Let the given plane be
determined by the intersecting lines ab-a'b' and ac-a'c'; and
let de-d'e' be the given line.
The line ab-a'l' pierces the plane which projects de―d'e
upon the horizontal plane in the point FF'. The line ac-a'c'
pierces the same projecting plane in the point GG'. Hence
FG-F'G' is the trace of the given plane upon the projecting
plane, and HH', its intersection with de-d'e', is the required
intersection of that line with the given plane.

EXAMPLES.-Let aa' be the given line, and bb'
and cc' those of the given plane in each of the follow-
ing figured examples.
a
C
α
Ex. 2°.
Ex. 3°.
Q
Ex. 1°.
b
گا گئی

Ex. 4°. General Example.-aa' may cross each of the four angles in succes-
sion; dd', the point of intersection of the lines which determine the plane, may
be in each of the four angles successively; and, finally, dd' may be in the same
angle which aa' crosses, or in either of the other three. From this description
many particular combinations may be made.
4
50
}
DESCRIPTIVE GEOMETRY.
PROBLEM XXIII.
To pass a plane through a given point and perpendicular to a
given line.
In Space. Since (Th. III.) the projections of a line are per-
pendicular to the traces of a plane, when the line is perpendicular
to the plane, the traces of the required plane will be perpendi-
cular to the projections of the given line. Since a line is deter-
mined by a single point, if its direction is known, the traces of
the required plane will be determined by finding the traces of
any line containing the given point, and lying in the required
plane.
The most convenient auxiliary lines are those containing the
given point and parallel to the traces of the required plane,
since their direction in both projections will thus be known.
Pl. IV., Fig. 41. QG is the ground line, pp' is the given
point, and ab-a'b' is the given line. The auxiliary line,
Rpq-p'q', located as above mentioned, pierces the vertical
plane at q', which is a point of the vertical trace P'Q. This
trace is drawn through q' and perpendicular to a'b'. The hori-
zontal trace, QP, of the required plane, is drawn through Q
and perpendicular to ab.
If Q should not fall within the limits of the figure, a point
in the trace QP can be found by a process similar to that used
in finding the point q'. That is, a line parallel to QP' and con-
taining pp', would pierce the horizontal plane in a point of the
trace QP.
GENERAL EXAMPLES.-1º. Vary the position of the given point.
Ex. 2. Vary the position of the given line.
Ex. 3. Let the given point and line be in different angles.
B-DEVELOPMENTS.
75. In problems of the Point, Line, and Planc, development
consists in showing the true length of a line, or the true size of
a plane or diedral angle, or a plane figure; in either of two
ways. First, by bringing the given object into, or parallel to, a
DESCRIPTIVE GEOMETRY.
51
1st
}
plane of projection. Second, by taking a new plane of projec-
tion, through, or parallel to the given object. The change, in
either case, being one of position, and not of form, the only
kind of development here considered, is that by transposition
(43), either of the given objects, or of the plane of projection.
76. Conversely, having given the development, or true size
or relative position, of certain forms, we can find their projec-
tions in other positions; and we thus have a new method of
solving certain problems.
77. Hence, finally, problems under the general head of de-
velopment, may be arranged in two bodies: First, direct, or
required developments; being those in which we find the de-
velopment from given projections. Second, counter, or given
developments, being those in which we find the projections, or
some other development, from a given development.
78. The latter body of problems completes the full circle of
operations belonging to Descriptive Geometry, by enabling us
to find the projections of objects which are not given only by
description, but are given graphically, in definite size, form, and
position.
a-Direct, or required development, found from given
projections.
PROBLEM XXIV.
To find the true length of the line between two given points, by
revolution about an axis perpendicular to a plane of pro-
jection.
In Space.-By (55), revolve the line about an axis perpen-
dicular to either plane of projection, until it becomes parallel
to that plane, or lies in it, when its true size will appear on that
plane. The axis may contain a trace, or any other point of the
given line, or it may contain no point of the line.
Let ab a'b' be the line,
and bb', and let a-c'a' be
In Projection.-Pl. IV., Fig. 42.
connecting the two given points aa'
a vertical axis, taken through the point aa' of the line. Then,
if the line revolves about this axis till parallel to the vertical
plane, the point bb' will describe the horizontal are whose pro-
52
DESCRIPTIVE GEOMETRY.
jections are bB and b'B', where a is the centre of B; hence,
project B upon b'B, at B, and aB-a'B' will be the line when
parallel to the vertical plane, and a'B' will be its true length.
EXAMPLES.—1º. Let the axis be a vertical line through the vertical trace of
the given line.
Ex. 2°. Let it be any vertical line, as O-C'O', Pl. IV., Fig. 43.
Ex. 3. The axis, ef—ƒ', Pl. V., Fig. 44, perpendicular to the vertical plane,
contains the horizontal trace, e, of the line.
Ex. 4°. Let the axis be a perpendicular (48), without intersecting the given
line.
Ex. 5º. Let the vertical axis, Pl. IV., Fig. 42, intersect the given line any
where between aa' and bb'.
Ex. 6. Let the given points be in any two different angles, and let the axis
be vertical, in either of those angles.
Ex. 7. In examples five and six, let the axis be perpendicular to the vertical
plane.
THEOREM IV.
The distance from a point to a line parallel to a plane of pro-
jection is the hypothenuse of a right triangle, whose plane
is perpendicular to the line, and whose sides are parallel
and perpendicular to that plane of projection.
In Pl. V., Fig. 45, let A-aa' represent the point, with its
two projections, and MN any line, taken in the horizontal plane,
for example, since such a line is only one of a system of lines,
parallel to that planc, and above or below it. Then let AM be
the perpendicular distance of A from the axis MN; and it fol-
lows that AM is the hypothenuse of a triangle AɑM, whose base
is aM, perpendicular to MN from a, the projection of ▲ on the
plane containing the axis; and whose altitude is Aa, the perpen-
dicular distance of A from that plane. Aa is shown in its true
size at a'c (31), and thus AM can be constructed from aM and
a'c=Aa, both of which are known.
This method of revolution may be called that of parallel
axes.
a.-If MN had been in any other horizontal plane, a would
have been taken in that plane.
b. If MN had been in a vertical plane, aM would have
been replaced by a perpendicular to MN from the projection
DESCRIPTIVE GEOMETRY.
53
of A on that plane; and Aa by a perpendicular from A to that
plane.
c.-If ab, the horizontal trace of a projecting plane, were the
axis, the base, aM, would vanish; and the hypothenuse AM
would reduce to the altitude Aa.
PROBLEM XXV.
To find the true length of a line, by revolving it about an axis
in, or parallel to a plane of projection.
In Space. By the last theorem (IV.), the axis must intersect
the given line, in order that the latter, by lying in the same
plane with the axis, may be revolved into, or parallel to, that
plane of projection which contains, or is parallel to the axis.
In Projection.-Let al-a'l', Pl. V., Fig. 46, be the given
line, whose horizontal trace is k, found as in Prob. XII. Any
line through & will be in the same plane with ab-a'l'; then
assume M as the axis about which ab-a'b' shall revolve into
the horizontal plane. The point aa' will then revolve in an
arc perpendicular to M, whose horizontal projection is aA,
also perpendicular to Mk, and will be found at A; whose per-
pendicular distance from MZ is equal to the perpendicular dis-
tance of aa' from Mk, and is equal to the hypothenuse of a
right angled triangle, whose base is aM, and whose altitude is
a'd; B is similarly found, and AB is then the true length of
al-a'', shown in the horizontal plane. AB will also pass
through / (62).
EXAMPLES.-1°. Let the line be revolved into the vertical plane, about any
line, taken in the vertical plane, and through the vertical trace of the given
line.
Ex. 2°. Take an axis parallel to either plane of projection.
Ex. 3°. Let ab—a'b', Pl. II., Fig. 15, be the given line, and ab the axis, taken
as the ground line, G'L', of a new vertical plane of projection (Prob. IV.),
taken through ab-a'b'. Then by making aa"=ma' and bb"=nb', a'"b" will be
the line, ab-a'b', shown in its true length by revolving the vertical plane,
GL', back in the usual way into the horizontal plane.
Ex. 4. Let G"L", coinciding with a'b', taken as an axis, be called the ground
line of a new plane of projection, perpendicular to the vertical plane, and find
the true length of ab—a'b'.
54
DESCRIPTIVE GEOMETRY.
Ex. 5°. Find the true length of every line whose length is not shown in
Prob. II.
Ex. 6. Further, effect these constructions in each of the other three diedral
angles.
Revolutions of points, in all possible ways, being of very fre-
quent occurrence in descriptive geometry, the following is added
as a review, with variations, of the two preceding. The solu-
tion will be obvious from the figure only, Pl. V., Fig. 47.
Ex. 7°. To show the true length of a line in all possible positions, by re-
volving each of its projecting planes about its traces.
79. It is now evident, that the true length of a line in space
is the common hypothenuse of two right angled triangles, one
of which has for its base the horizontal projection of the line,
and for its altitude the difference of the perpendicular distances
of the two extremities of the line from the horizontal plane.
The other triangle has for its base the vertical projection of the
line, and for its altitude the difference of the distances of the
two extremities of the line from the vertical plane.
If the line be limited by its traces, its true length is the hypo-
thenuse of either of two right angled triangles, in which the
base is the projection of the line, on one of the planes of pro-
jection, and the altitude is the perpendicular distance, from its
trace on the other plane, to the ground line.
PROBLEM XXVI.
To find the shortest distance from a given point to a given line.
In Space.-Pass a plane through the given point, and per-
pendicular to the given line, by Prob. XXIII., and, by Prob.
XXI., find where the given line pierces this perpendicular
plane. Then a line joining the last-named point with the given
point would evidently be the required perpendicular. Its true
length would be found by Prob. XXIV. or XXV.
In Projection. The construction is left for the student,
EXAMPLES.-1°. Let the point and line be in the same angle.
Ex. 2. Let them be in different angles.
DESCRIPTIVE GEOMETRY.
55
PROBLEM XXVII.
To find the perpendicular distance between two lines which are
not in the same plane.
In Space.-1st. Suppose a plane to contain one of the lines,
M, and to be parallel to the other one, N. 2d. Project the line
N upon the parallel plane. 3d. This projection will evidently
intersect the line, M, lying in the plane, at the point from which
the required perpendicular may be drawn ; for this perpendicu-
lar is merely that one of the projecting lines of the line, N,
which meets the plane in a point of the line, M.
In Projection.-Pl. V., Fig. 48. Let ab—a'b' be one given
line, and cd—c'd' the other. The plane e Qb is drawn through
ab-a'b', and parallel to cd-c'd' by Prob. XV. The line hm
-h'm' is drawn through any point, h', of the line cd-c'd',
and perpendicular to the plane, e'Qb, by the method given in
Prob. XXI., and mm' is where it pierces the plane c'Qb. mn-
m'n', parallel to cd-c'd', is, therefore, the projection of cd-c'd'
on the plane e'Qb. The line mn-m'n' intersects ab-a'b', at
oo'. Therefore op-o'p', parallel to hm-'m', is a line perpen-
dicular to both of the given lines, ab-a'b' and cd-c'd'. Its
true length can be found by Probs. XXIV or XXV

EXAMPLES.-1°. Find the perpendicular
distance between these lines, a-a' and
b-b'.

Ex. 2°. Find the perpendicular
distance between these, a—a'
being parallel to the vertical
plane.
b
a

a
56
DESCRIPTIVE GEOMETRY.
Ex. 3. General Example. Take one or both of the given lines in any other
diedral angle.
b
α
b
Ex. 4°. Let one of the lines be perpendicular to a
plane of projection.
a
a
α
b
Ex. 5°. Let one of the lines be in a plane of projec-
tion.
Ex. 6°. Let one of the lines coincide with the
ground line. Here it will be convenient to use an
auxiliary plane of projection.
PROBLEM XXVIII.
To find the true size of the angle included between two inter-
secting lines.
In Space.-If a plane be passed through the given lines, its
traces will join the corresponding traces of the lines. If such
a plane be revolved about either of its traces till it coincides
with the plane of projection which contains that trace, the angle
between the given lines will appear in its real size.
In Projection.—Pl. V., Fig. 49. Let G be the ground line,
and ab-a'b' and cd-c'd' the given lines, making with each other
the angle whose projections are deb and d'e'b'.
These lines pierce the horizontal plane at d' and b', hence d
is the horizontal trace of the plane which contains them. If
now this plane be revolved about its horizontal trace bd, till it
coincides with its horizontal plane, the vertex,ee', of the angle
DESCRIPTIVE GEOMETRY.
57
deb-d'e'b', will revolve in an arc whose plane is perpendicular
to db, and whose horizontal projection is ee", perpendicular to
db. The point e" is at a distance, e'n, from the axis ɗb,equal
to the hypothenuse eu of a right angled triangle whose base
is en and whose altitude is e'k. (Th. IV.) Having found e",
draw de" and be", and de'b will be the true size of the angle
included between the given lines be-b'e' and de-d'e'.
EXAMPLES.-1°. Revolve the plane of the angle about its vertical trace,
into the vertical plane of projection.
Find the true size of the angle at aa' in the following figures:
and
a
4
+
Ex. 2°.
Ex. 4°.
a
Ex. 3°.
a
a

Ex. 5°. Use the true lengths of the sides of the angle, instead of the hypo-
thenuse above described, in finding the revolved vertex.
Ex. 6. Let one side of the angle be perpendicular to a plane of projection.
Ex. 7. General Example. Let the vertex of the given angle be in any one
of the four diedral angles, and revolve the plane of the angle about either of
its traces.
80. When two lines are not in the same plane, it is common
to speak of the angle made by their directions, since the lines
themselves do not meet. The direction of a line is shown by
any line parallel to it. Hence if A and B are two lines not in
the same planc, and if a line, b, be drawn through any point of
A and parallel to B, the angle between A and will be that
which the directions of A and B make with each
other.
58
DESCRIPTIVE GEOMETRY.
81. It is often desirable, as, for example, in making models,
to know the angular size of so much of a plane whose traces
are given, as is within any one of the diedral angles.
Hence we have
PROBLEM XXIX.
To find the angle between the traces of a given plane; or, to
revolve a given plane into either plane of projection.
First method. In Space.-Taking either trace as an axis, as-
sume any point on the other trace, and revolve it about the
axis, and into the plane of projection containing that axis. The
construction may be abridged by considering that the traces,
being in the planes of projection, show their true length, alike
before and after revolution.
In Projection.-Let PQP', Pl. V., Fig. 50, be the given
plane. Then, first, with the horizontal trace, PQ, as an axis,
take any point, aa', in the vertical trace. It will revolve about
the axis PQ in a vertical are, whose horizontal projection is
aa", and whose radius is the hypothenuse, constructed on ac
and aa' as sides, and laid off at ca". Then a'Q is the revolved
vertical trace, and a'QP the required angle between the traces
of the plane PQP. By reason of the equal length of a trace
before and after revolution, Qa" Qa' will intersect ca", per-
pendicular to PQ from a, at a' by describing an are with Q as
a centre, and Qa' as a radius.
Second. By taking P'Q as an axis, any point, as ce', of the
trace PQ, will revolve about P'Q and be found at c', by con-
structions similar to the foregoing.
Second method. In Space.-By perpendicular axes. In this
method two successive revolutions are necessary, the first to
bring the given plane perpendicular to a plane of projection;
the second, to bring it into a plane of projection.
In Projection.-Let the plane, PQP', Pl. V., Fig. 51, for ex-
ample, be revolved about the vertical axis aa' in the vertical
plane. Drawing aP", perpendicular to the horizontal trace,
PQ, revolve that trace, by the arc P"R', to the position, RR',
perpendicular to the vertical plane, which will bring the given
plane to the same position, and R'a' will be its new vertical
DESCRIPTIVE GEOMETRY.
59
trace. Q, on the former vertical trace, will still be on the hori-
zontal trace, hence at R, found by the arc, QR6Q, with a as
a centre, and having produced aP" to b. Then Ra and R'a'
are the projections of the revolved position of the former ver-
tical trace, P' Q. Now revolve the plane about RR' and into
the horizontal plane, and Ra-R'a' will appear at Ra", and
R'RN will be the angle between the traces PQ and P'Q. Or,
R'Ra" will be the true size of the angle between the traces
P"Q and P'Q, in the second angle.
EXAMPLES.—1°. General Example. Revolve each of the positions of a plane,
in Prob. VII., about each of its traces in succession.
Ex. 2°. General Example. Revolve each of the same positions about axes,
taken perpendicular to a plane of projection.
82. Let PQP' and PRP', Pl. V., Fig. 52, be any two planes,
intersecting each other in the line, PP'. If, at any point, A, in
the intersection, a line be drawn in each plane, and perpendi-
cular to PP, as at AM and AN, the angle contained by these
lines will be the measure of the diedral angle included between
the two planes. Since both of the lines are perpendicular to
PP' at the same point, they lie in a plane, AMN, perpendicular
to that line.
Thus, the diedral angle included between any two planes, is
contained by the lines cut from those planes by a plane perpen-
dicular to their intersection at any point.
83. If one of the planes including the required angle be a
plane of projection, the plane of the angle between them will,
by the last principle, be perpendicular to the trace of the other
plane upon that plane. Thus, the plane containing the angle
between PRP' and the horizontal plane, would be perpendicu-
lar to the trace, PR; and that of the angle between PRP' and
the vertical plane would be perpendicular to P'R.
60
DESCRIPTIVE GEOMETRY.
PROBLEM XXX.
To find the true size of the angle between a given plane and
either plane of projection.
In Space.-By (83) the auxiliary plane containing the re-
quired angle, is perpendicular to the trace of the given plane
on that plane of projection which bounds the required angle.
And the angle is included between the traces of this auxiliary
plane upon the planes which bound the angle.
Hence, by the usual method of movable auxiliary parts, take
such an auxiliary plane, and revolve it about either of its traces
into the plane of projection containing that trace; when it will
show the true size of the angle. Otherwise, by the method of
rotations of given parts (58, 2°), to find the angle bounded by
one of the planes of projection, revolve the given plane till per-
pendicular to the other plane of projection; when its trace on
that plane will make the required angle with the ground line.
In Projection.-Pl. V., Fig. 50. Let PQP' be the oblique
plane. To find the angle made by it with the plane H, caa' is
an auxiliary plane perpendicular to the trace, PQ. By revolv-
ing it about aa' as an axis, and into the vertical plane, c will
fall at c', c'a' will be the revolved intersection of the planes
PQP' and caa', and a'c'a will be the required angle; d'a being
the revolved position of the line ca cut from the horizontal
plane.
In like manner ceb is a plane perpendicular to the vertical
trace, P'Q, and containing the angle between PQP' and the
vertical plane.
In Pl. V., Fig. 51, RR'a' is the plane PQP', after revolution.
about the vertical axis aa', till perpendicular to the vertical
plane of projection (55). Then a'R'a is the true size of the
angle between PQP' and the horizontal plane.
EXAMPLES.-1°. Show the true size of the required angle in Pl. V., Fig. 50,
by revolving its plane caa' about its horizontal trace ca, into the horizontal
plane.
Ex. 2. Complete the construction of the angle made by the plane PQP' with
the vertical plane.
Ex. 3. General Example. Find the angles included between any of the
positions of a plane shown in Prob. VII., and either of the planes of projec-
tion.
DESCRIPTIVE GEOMETRY.
61
PROBLEM XXXI.
To find the true size of the angle between two given oblique
planes.
In Space.-Two methods, at least, may be used for finding
the required angle, without the previous development, com-
monly found, of parts used in the operation.
First Method.-Assuming any point on the intersection of
the two planes, pass a plane through it, by Prob. XXIII., per-
pendicular to the intersection. Joining the points where its
traces intersect those of the given planes, with the assumed
point, we shall have the two projections of the required angle;
whose true size can then be found by Prob. XXVIII.
Second Method.—Assuming either trace of the plane of the
angle, that trace must be perpendicular (Th. III.) to the corre-
sponding projection of the intersection of the two planes.
Since, as may be understood from Pl. V., Fig. 52, the perpen-
dicular distance, Ab, from the vertex, A, of the angle, to the trace,
as MN, of its plane, is perpendicular both to PP' and MN, we
may proceed by Prob. XXVII., to find the common perpendi-
cular to the assumed trace and to the intersection of the two
planes. Lines from the point, A, where this perpendicular bA,
Pl. V., Fig. 52, meets that intersection, to the points where the
assumed trace, as MN, crosses the traces of the given planes, will
again give the projections of the required angle; whose true
size can then be found as in Prob. XXVIII.
In Projection.-All the separate operations in this problem,
by both methods, having been given in previous problems, the
construction can readily be made by the student.
PROBLEM XXXII
To find the true size of the angle which a line makes with a
plane.
In Space.-A line makes an infinite number of angles with
a plane. Hence the angle made by a line with a plane means
the least of these angles, and this will be the lesser one of the

62
DESCRIPTIVE GEOMETRY.
two angles made by the line with its projection on the plane ;
that is, it will be the angle made by the line with the trace, upon
the given plane, of a plane passed through the line and perpen-
dicular to the given plane.
The latter plane will be determined by the given line, and
a perpendicular from any point of it to the given plane.
Having thus found the projections of the required angle, its
true size can be found by Prob. XXVIII.
In Projection.-Pl. V. Fig. 53. Let ap-a'p' be the given
line; PQP" the horizontal, and P'QP"" the vertical trace of the
required plane. By Prob. XVII., aB and BP" are the traces
of the plane which projects the given line upon the horizontal
plane, and P'''' is the vertical projection of its intersection with
the given plane. The given line therefore intersects the given
oblique plane at pp'. Next, aq-a'q' is a line drawn through
any point aa' of the given line, and perpendicular to the given
plane, and q' is where this line pierces the given plane. There-
fore Pq-P'q' is the projection of ap-a'p' on the given planc.
Now ap-a'p' pierces the horizontal plane at D, and aq-a'q'
pierces the same plane at C, hence CD is the horizontal trace
of the plane of the angle apq-a'p'q'. Revolving this plane
CD about its horizontal trace, into the horizontal plane, the
points aa', pp', and qq' will revolve in ares whose horizontal
projections will be perpendicular to the axis CD, and will be
found at a", p', and q', at distances from the axis equal to
their true distance in space from it; viz., at distances equal
to the hypothenuse of right angled triangles whose bases are an,
etc., and whose altitudes are d'o', etc.-the heights of the points
above the horizontal plane. a"p"q" is, therefore, the true size
of the required angle.
84. Note that if it be merely required to find a measure of
the required angle, and not the angle, a'p"q", itself, it will be
sufficient to find immediately the points C and D, without find-
ing pp' and 77'. Then by revolving the plane of the angle
CaD about CD, as before, the complement of the angle at a"
will measure the size of the angle required.
EXAMPLES.-1. Let the given line be the ground line.
Ex. 2. Let it be parallel to the horizontal plane.
Ex. 3. Let it be parallel to the horizontal plane, and its direction perpendi
cular to the horizontal trace of the given plane.
Ex. 4. Let the plane be parallel to the ground line.
DESCRIPTIVE GEOMETRY.
63
Ex. 5°. Let the plane be perpendicular to the ground line.
Ex. 6°. Let the plane be either plane of projection.
Ex. 7. General Example. Let the line cross any diedral angle, and let the
plane be in any of the positions shown in Prob. VII.
b-Inverse or counter development; determining re-
quired projections from auxiliary developments.
85. In the problems under this head, developments, as well
as projections, may be the things required; but, in either case,
they are found from a previous auxiliary development, which
therefore becomes a given one relative to the thing required.
PROBLEM XXXIII.
Having given one projection of a point, on a line whose other
projection is nearly perpendicular to the ground line, to
find its other projection.
In Space.-Let the horizontal projection be given. The ver-
tical projection will then be the intersection of a vertical line,
containing the given projection, with the line in space; but,
from the supposed positions of these lines, this intersection
would be too acute to be readily constructed with accuracy ;
hence we have simply to bring these lines into a position which
will form a well defined intersection, either by transposing the
lines, or the planes of projection.
=
In Projection.-Let G₁L,, Pl. V., Fig. 54, be the ground line
of a new horizontal plane ( 58, 3°), making a small angle with
the given line, ab-a'b', and an auxiliary line through e and in
the plane H. Then da, aa'; mb, mb; and me, = me;
giving a,b, and de, for the projections of ab-a'b', and a'e̱—a'm
on the new horizontal plane, whose ground line is GL,; and
c, the like new projection of e, whence c is accurately found by
drawing cc', perpendicular to G¸L¸.
EXAMPLES.-1°. Let the vertical projection of the point be the given one,
and still use the same figure, only with a difference in the order in which its
lines are constructed.
Ex. 2°. Let the projecting plane, bb'd', perpendicular to the vertical plane,
be used and revolved about any axis perpendicular to the vertical plane.
64
DESCRIPTIVE GEOMETRY.
Ex. 3°. By Prob. XXV., Ex. 3, let either of the projecting planes of the given
line, treated as a new plane of projection, be revolved back into the plane to
which it is perpendicular, in order thus to show together, and clearly, the re-
volved positions of the given line, and a projecting line of the required point.
Ex. 4. General Example. Let that portion of the given line on which the
point is situated be in any one of the four angles.
PROBLEM XXXIV.
To find the shortest distance from a given point to a given line,
and the projections of the same.
In Space. Suppose a plane to be passed through the given
point and the given line, and to be revolved into either plane
of projection, about its trace on that plane as an axis; the true
perpendicular distance from the point to the line can then be
shown at once, and from this its projections can be found by
counter revolution.
In Projection.—Pl. VI., Fig. 55. Let ab—a'b' be the given
line, and pp' the given point. Take the auxiliary line pr-p'r',
through pp' and parallel to ab-a'b', and rb joining the horizon-
tal traces, r and b, of these lines, is the horizontal trace of a
plane containing the given point and the given line. If now,
this plane be revolved about rb as an axis, and into the horizon-
tal plane, the line ab-a'b', and the point pp, will be found in
that plane; the latter at its true distance from rb. This dis-
tance is the hypothenuse of a right angled triangle, whose base
is pK, perpendicular to rb, and whose altitude is the vertical
height-equal to p'p'-of the point pp' above its horizontal
projection. Lay off this hypothenuse from K to p". Points
in the axis remain fixed, hence ba" and rp"" are the revolved
positions of the given line, and of its parallel through pp', and
p'''q" is the true length of the perpendicular from pp' to ab-
a'b'. In the counter revolution, p' returns to P ; I'' to I, I'″'I
being perpendicular to rb, the axis, and pq is the horizontal
projection of the required perpendicular: p'q' is its vertical
projection.
86. To find the true perpendicular distance from a point to a
plane. As is proved in (Th. III.), the projections of the required
perpendicular will be perpendicular to the traces of the plane.
DESCRIPTIVE GEOMETRY.
65
Find the point in which this perpendicular pierces the oblique
plane, and the true length of so much of it as is included be-
tween the given point and the oblique plane.
EXAMPLES.-1°. General Example. Let the point be in any one of the four
angles, and let the line cross any of them; either the one containing the point or
a different one.
Ex. 2°. Let the line be horizontal.
Ex. 3°. Let the line be parallel to the vertical plane.
Ex. 4°. Let the line be in a plane perpendicular to the ground line.
PROBLEM XXXV.
To find the shortest distance, or common perpendicular, between
two lines not in the same plane, by the method of rotation.
In Space. Having passed a plane through one of the lines
and parallel to the other, as in Prob. XXVII., pass a parallel
plane through the latter line. Then drop a perpendicular upon
these planes from any axis taken perpendicular to a plane of
projection. Revolve this perpendicular about this axis till it
becomes parallel to the other plane of projection; when the two
parallel planes will both be perpendicular to that plane, and the
required distance will be seen in its true size, from which, by
counter revolution, its projections can be found.
In Projection.-The construction can be made from the
above description.
PROBLEM XXXVI.
To find the shortest distance between two lines, not in the same
plane, by the method of new planes of projection.
In Space.-Take successive new planes of projection, each
perpendicular to one of the preceding (58, 3°), forming the
new projections of the given lines upon them by Prob. IX., till
we have one perpendicular to one of the lines, when the re-
quired distance, being perpendicular to this line, will be paral-
lel to the latter plane, and will therefore be projected upon it
in its true size.
In Projection.-The construction is left as an example.
5
66
DESCRIPTIVE GEOMETRY.
PROBLEM XXXVII.
To draw a plane through a given point, and perpendicular to
a given line, by the method of developments.
In Space. The line of declivity of the required plane (74),
relative to either plane of projection, and a line parallel to the
given line, both drawn through the given point, will be perpen-
dicular to each other, and in a plane perpendicular to that plane
of projection, relative to which the declivity is estimated. By
developing these lines, the traces of the former can be found,
and thence the traces of the required plane.
In Projection.—Let pp', Pl. VI., Fig. 56, be the given point,
and ab—a'b', the given line. Knowing, by (Theo. III.), that the
trace of the required plane will be perpendicular to ab and a'b',
we have dSP' for the plane, parallel to ab, which will contain
the line pc-p'c', parallel to ab-a'b', through pp', and the de-
clivity, which is perpendicular to pc-p'c'. Revolving dSP'
about dS, and into the horizontal plane of projection, pc-p'e
appears at p'c, by making p'p perpendicular to dS, and equal
to p'k. Then p'd, perpendicular to p'c, is the declivity of the
required plane, and d, its horizontal trace, is a point of the hori-
zontal trace, PQ, of the required plane. Projecting dat d',
gives d'p' for the vertical projection of the declivity, and P', its
vertical trace, a point of the vertical trace, P'Q, of the plane.
Should Q be inconveniently remote, consider that P'Q is per-
pendicular to a'l'; or, proceed as in Prob. XIII.
EXAMPLES.-1°. Let the given point be on the given line.
Ex. 2°. Let the projections of the given line be perpendicular to the ground
line.
Ex. 3°. Let the given line be parallel to a plane of projection.
PROBLEM XXXVIII.
To find the true size of the angle between two planes, by the
method of development.
In Space.-By (82), and Pl. V., Fig. 52, it is evident that,
in order to find the angle, as MAN, we must have either the
DESCRIPTIVE GEOMETRY.
67
true length of the perpendicular, Ab, from the vertex, A, of the
angle to the trace MN of its plane; or else the true length of
its sides MA and NA.
The true length of Ab may be found by developing the in-
tersection, PP', of the two planes, in any of the ways shown in
Probs. XXIV. and XXV.
That of MA and NA may be found by developing both of
the given planes into one of the planes of projection, by revo-
lution about their traces on that plane as axes.
In Projection.-First Method. For further variety, we shall
take the case where the traces of the given planes do not meet
near the ground line.
2 ༡
Pl. VI. Fig. 58. Let PQP' and RSR' be the given planes;
uq and u'q' are the projections of their intersection, found as in
Prob. XVIII., by the auxiliary horizontal planes R'P' and s't'.
Let the plane of the required angle be revolved about its trace on
the plane s't', and let VW, perpendicular to uq, be this trace.
The line perpendicular to VW from the vertex of the required
angle, is projected on all the horizontal planes in the indefinite
line xu. This line, and the intersection of the given planes, being
iu a vertical plane, whose trace on s't' is uY, let this plane be re-
volved about uY as an axis, till it coincides with the plane s'ť.
Then uu', being in the axis, remains fixed, and qq' falls at q″,
at a distance from 7 equal to its true vertical height above u—
equal to the distance q'' between the planes R'P' and s't'. This
operation gives uq", the revolved position of the intersection of
the given planes. From a, which, being in the axis, remains
fixed, draw xz', which is the revolved position, and true length,
of the line containing the vertex of the angle, and perpendicu-
lar to VW; therefore z' is the revolved position of the vertex
of the required angle. Now when the plane of the required
angle revolves about its trace VW into the horizontal plane s't',
the vertex of the required angle revolves in the vertical plane
Yu and will be found in the latter line. But æz' is the true
distance of the vertex of the angle from the line VW, hence,
with a as a centre, describe the arc z'z, draw zo and zm—o
and m being the fixed points in which the lines cut by the
plane VW from the given planes pierce the axis VW in the
plane s't'—and ozm will be the true size of the required
angle.
By counter revolving ' to its position on Vu, and joining the
68
DESCRIPTIVE GEOMETRY.
point so found with n and m, we should have the horizontal
projection of the required angle.
EXAMPLES.-1°. Make the constructions on the horizontal plane of projec-
tion, the traces being taken so as to meet near the ground line.
Ex. 2°. Show the true size of the angle on the vertical plane.
Ex. 3°. Find az' in vertical projection, by revolving the intersection uq—u'q',
till parallel to the vertical plane; or into it in case of the traces meeting near
the ground line.
Ex. 4°. Effect the solution in any other angle than the first.
Ex. 5. Effect the solution for any other position of the two planes.
•
In Projection.-Second Method. Pl. VI., Fig. 57. PQ and
QP' are the traces of one of the given planes, and PR and
RP' are the traces of the other given plane. Pb-a'P' are the
projections of the intersection of these planes (Prob. XVII.);
and ce' is any assumed point of this intersection. As cach plane
is revolved about its horizontal trace into the horizontal plane
of projection, Pb-u'P' and ce' will take the two revolved posi-
tions Pe" and Pe". To represent this revolution, consider that
c''n equals the hypothenuse of a right angled triangle of which
the base is en, and the altitude, dr. Likewise d'"g equals the
hypothenuse of a right angled triangle whose base is ge, and
whose altitude is c'r. The lines d'e and e" are the revolved
positions of the lines lying one in each given plane, and per-
pendicular to their intersection P-a'P' at ce'. The ares
c''''''' and ''c'"", represent the sides de" and ec", which in
space include the required angle between the given planes, as
revolving about the fixed points and e-and still keeping in
the horizontal plane-till the points " and ", which coincide
in space, coincide at e'""' in the horizontal plane and form the
vertex of the required angle ed", the true size of the diedral
angle between the given planes PQP' and PRP'.
C
A line from e to d would be the horizontal trace of the plane
of the angle, and eed would be the horizontal projection of the
angle. The vertical projections of e and d are in the ground
line, and by joining them with we should have its vertical
projection.
EXAMPLES. —1. Let both planes be parallel to the ground linc.
Ex. 2°. Let both be nearly perpendicular to the ground line.
Ex. 3. Let each be perpendicular to a plane of projection, but not to the
saine, for both.
Į
DESCRIPTIVE GEOMETRY.
69
PROBLEM XXXIX.
Having given the angles made by an oblique plane with the
planes of projection, and its perpendicular distance from
a point on the ground line, it is required to construct the
traces of that plane.
In Space. The plane of the diedral angle made by any plane,
P, with either plane of projection, may be perpendicular to the
trace of the plane P, on that plane of projection, at any point.
Therefore, the two diedral planes may intersect the ground line
at the same point. When thus located, they will intersect each
other in a line, y, containing the point in the ground line, and
to which the oblique plane, P, will be perpendicular.
In Projection.-Pl. VI., Fig. 59. Let QG be the ground line,
'the point at which both the diedral planes are supposed to
intersect the ground line, and a'G' the true length of the per-
pendicular from a to the required plane. Also let m be the
angle made by the required plane with the horizontal plane,
and the angle made by that plane with the vertical plane of
projection.
N.
With a radius a'G' and centre a', describe an indefinite are.
Draw P'd tangent to this are and forming with the ground line
the angle P'da' equal to the angle m. Draw also Pe tangent to
the same are, so as to make the angle Pea' equal to the angle
P'du̸' and Peu̸' are the positions of the diedral angles m
and n after revolution about P'a' and Pa' respectively; P'a'
and Pa' being those traces of the diedral planes which are per-
pendicular to the ground line at the point a'. P'd and Pe are
the revolved positions-in the same revolution-of the inter-
sections of the diedral planes with the given plane, and ab and
al" are the two revolved positions of the line in which the
diedral planes intersect each other in space. Hence P and P',
the points in which Pe and P'd intersect the planes of projec-
tion, are points in the traces of the required plane. As the
diedral planes counter revolve about their axes Pa' and P'a'
back to their primitive position in space, it is evident that the
points c and d will describe ares about a' as a centre, to which
the required traces will be tangent; hence, drawing these ares,
70
DESCRIPTIVE GEOMETRY.
י.
we have PQ and P'Q, tangent to them, for the traces of the
required plane.
87. A plane through a given point.-Observe that this prob-
lem, without the distance a'G', determines only the relative
position of the required plane, with respect to the planes of
projection, and not its absolute position. Hence any two lines,
drawn from the same point in the ground line, and parallel to
the traces PQ and P'Q would be the traces of a plane which
would make angles with the planes of projection equal to the
given diedral angles. To draw the plane through a given point,
t, we should construct, as above, a preliminary plane, as PQP',
and then, through t, construct the required plane by (Prob.
XXXIX).
88. This problem is one of a class of similar ones; another,
for example, being, to construct one trace of a plane, having
given its other trace and the diedral angle formed by the re-
quired plane with the plane of projection containing the re-
quired trace.
PROBLEM XL.
To bisect an angle, in space, by counter-development.
In Space. The bisecting line of the development of the
angle, will, by counter-revolution, give the projections of the
half angles.
In Projection. -Having found the true size of any planc
angle, as in Problem XXVIII., the developed angle deb, Pl.
V., Fig. 49, may be bisected. Note the intersection of the bi-
secting line with the axis bd. This point, which we will call
xx', will remain fixed in counter-revolution, while e" returns to
ee'. Then dex-d'e'x' and bex-b'e'x' will be the projections of
the halves of the bisected angle, a' being on the ground line.
EXAMPLES.—1°. Let one side of the angle be vertical.
Ex. 2°. Let it be a perpendicular.
Ex. 3. Bisect all of the angles in the examples under Problem XXVIII.
DESCRIPTIVE GEOMETRY.
71
CHAPTER II.
DEVELOPABLE SURFACES OF REVOLUTION-CYLINDERS AND
CONES.
SECTION I.
DEFINITIONS AND GENERAL PRINCIPLES, SURFACES AND CURVES.
89. A DEVELOPABLE SURFACE is one which can be made to
coincide with a plane without changing the relative position of
any two of the consecutive straight elements of which it is
composed. Such surfaces form a group, in the class called single
curved surfaces, of the division called ruled surfaces (15).
THEOREM V.
A surface, composed of straight lines meeting at a common
point, is developable.
Let V-ABQ, Pl. VII., Fig. 65, represent such a surface.
The elements VA and VB lie in the plane VAB. The ele-
ments VA and VB, lie in the plane VAB,; and these two
planes intersect in the element VA. Now, if the plane VAB,
be revolved about VA as an axis, till it coincides with the plane
VAB, the three clements, VA, VB, and VB, will all be in the
same plane. But similar reasoning will apply to any number
of elements and to consecutive ones, as well as to detached
And the proof is independent of the distance of the in-
tersection V of the elements. Hence the theorem is proved.
ones.
90. The surface, composed of a system of straight lines
which join a fixed point, V, with all the points of the curve, C,
is a cone. The fixed point V is called its verter; and the fixed
curve, C, its directrix (8). That is, a cone is generated by a
72
DESCRIPTIVE GEOMETRY.
straight line, L, which passes through a fixed point, while it
moves upon a given curve as a directrix.
91. The generatrix being of indefinite extent, the entire sur-
face will consist of two equal and opposite parts meeting at the
vertex. These are called the two nappes of the cone.
92. When the vertex is at an infinite distance, the elements
become parallel, and the cone becomes a cylinder, Pl. VII.,
Fig. 66, which is thus a particular case of the cone.
The cylin-
der may then be said to be generated by a straight line, ab,
moving parallel to itself, and upon a given curve as aec; that
is, by drawing lines in one given direction, and therefore paral-
lel to each other, through every point of the curve.
93. A cone of revolution, VAB, Pl. VII., Fig. 65, is the
simplest one, and is generated by the revolution of a straight
line, VA, around another straight line, VO, which it intersects.
This fixed line, VO, is the axis of the cone, and every point.
of the generatrix, VA, describes a circle, whose centre is in the
axis, and whose plane is perpendicular to it. That is, the
curved directrix of a cone of revolution is a circle, perpendicu-
lar to the axis of the cone.
94. In this case, likewise, when the vertex is at an infinite
distance, the cone becomes a cylinder of revolution, ucbd, Pl.
VII., Fig. 66, which may, therefore, be generated by the revo-
lution of a straight line about another straight line, parallel
to it, as an axis.
95. Any curved section of a cylinder or cone, made by a cut-
ting plane, and taken as the limit of the elements, may be called
the base of the cylinder or cone. If the plane is perpendicular
to the axis of the surface, the section is called a parallel of the
surface, or a right section, and such a section is usually called
the base, when the elements are limited by it.
96. It is now evident that a cone of revolution may vary be-
tween the opposite extremes of a plane and a cylinder; and
these extreme cases may be derived by considering extreme
values of the distance from the vertex to the base; and ex-
treme values of the radius of the base.
97. By conceiving of a cone as a pyramid of an infinite
number of sides; and of a cylinder as a prism of an infinite
number of sides, problems upon pyramids and prisms, with
their sections, which are polygons, are naturally associated with
those upon cones and cylinders.
DESCRIPTIVE GEOMETRY.
73
98. The intersections of cylinders and cones with planes, and
with each other, give rise to a variety of curves. Some general
principles concerning curves shall therefore be considered next.
99. Conceive of a point which we will call a. On any
straight line through a, place a point b, consecutive with a. As
no other straight line can contain both a and b, this line is de-
termined by those points, and may be called the line ab. Also,
the points a and b, alone, form an elementary straight line.
100. Again: On any straight line, through b, and making a
sınall angle with ab, place a third point, c, consecutive with b.
The line be has a different direction from that of ab; and thus
three points not on the same straight line, produce a single
change of direction. A curve changes its direction at every
point in passing to its next point. IIence, the three points, a,
b, and c, form a curve; which, since it has but a single change
of direction, may be called an elementary curve, or arc.
101. Since a plane can always be passed through three points
not in the same straight line, an elementary arc is a plane curve
(11); and, moreover, as a circle can always be drawn through
three such points, an elementary arc is a circular one.
102. Again, through e draw any straight line not quite coin-
ciding with be, and upon it place a fourth point, d, consecutive
with e, and so proceed, placing a point e, consecutive to d, on
any straight line through d, and not coinciding with ed, and we
shall form, point by point, a curve of some kind, each three
consecutive points of which will be one of the elementary cir-
cular arcs of which the curve may be conceived to be com-
posed.
103. As each of the successive straight lines, on which a new
component point is taken, may have an indefinite number of
different positions, an infinite variety of curves may evidently
be formed by the process just described.
104. The infinite variety of curves, here indicated, are taus
classified in Descriptive Geometry-FIRST. Curves of single
curvature, otherwise called plane curves, or curves of two dimen
sions; and-SECOND. Curves of double curvature, or curves
of three dimensions.
105. A curve of single curvature is one, all of whose points
74
DESCRIPTIVE GEOMETRY.
lie in one plane, as a circle. Being curved only in one plane,
it thence takes its name.
106. A curve of double curvature is one in which not more
than three consecutive points lie in the same plane. It is thus
curved in the direction of each of any two planes, perpendicu-
lar to each other, through any one of its points, as in the case
of the hand rail to circular stairs, and hence takes its name.
107. Though curves of single curvature may be thought to
belong only to plane geometry, yet they may be considered as
belonging to descriptive geometry: First. When treated, not,
as in plane geometry, as independent magnitudes; but as the
sections of surfaces. Second. By considering them as the pro-
jections of curves in space, which are referred to two planes of
projection. Third. By their occurrence in problems of devel-
opment, requiring their true size in their own plane.
108. A direct curve is everywhere convex towards space on
the same side of it.
A reversed curve is in some parts convex, and in others con-
cave, towards space on the same side of it.
A simple curve is a curve of one kind, as a circle, or ellipse,
or an arc of either.
A compound curve is composed of two or more arcs of the
same, or of different kinds of curves, tangent to each other, and
is thus artificial. When composed of circular arcs, it is also
called polycentral.
109. Every curve having definite properties, or a law of form,
is, in fact, geometrical; but so long as these properties are un-
known, it is only a graphical curve, being in the same condi-
tion as a curve drawn arbitrarily.
110. A closed curve is one in which, as in a circle, the gene-
ratrix returns to its original position without retracing its posi-
tions.
An open curve is one which does not thus return into itself.
111. A secant to a curve is a line which intersects it in two
or more points, as AB, Pl. VII., Fig. 65. When B, by ap-
proaching A, becomes consecutive with it, AB is in its last
secant position with B to the right of A, and may be called an
ultimate secant. Finally, when B coincides with A, the line
AB becomes a tangent, having A for its point of contact.
A
DESCRIPTIVE GEOMETRY.
75
A tangent to a curve is, therefore, the line which the secant
becomes when its points of intersection with the curve coin-
cide.
A chord is that portion of a secant included between two
successive points of its intersection with the curve.
A diameter of a curve is a line which bisects each of a sys-
tem of parallel chords.
An axis is a diameter which is perpendicular to its system of
parallel chords.
A centre is the intersection of two such diameters or axes.
112. Singular points are those at which the curve containing
them possesses some peculiar properties not belonging to its
other points. The following will appear in future problems:
A point of inflexion, Pl. VII., Fig. 69, is where the curve,
estimated each way from the point, lies on opposite sides of the
tangent at that point.
A multiple point, Pl. VII., Fig. 70, is where a curve crosses
itself, at least once, at the same point, forming a loop, so as to
have at least two tangents at the same point. Crossing itself
only once, the curve forms a double point.
A cusp, Pl. VII., Fig. 71, is where the loop, formed by a mul-
tiple point, disappears, and in such a manner that the two tan-
gents at that point coincide. A cusp is of two species; first,
when its two areas are, and second, when they are not on the
same side of its tangent.
A point of arrest, is a point where an open curve terminates,
as in the case described below.
113. The following curious distinction will also arise in actual
problems: The projection of a closed (110) curve in space,
may be a part of some known open plane curve.
That part of
the projection to which there is no corresponding reality in
space, is called a parasite, since its existence depends on that of
the other or useful portion.
Thus, the horizontal projection of a vertical circle is a limited
straight line, the extensions of which are parasites of the part
which represents the circle, and the extremities of which are
points of arrest.
114. The circle is the only curve which can be constructed
76
DESCRIPTIVE GEOMETRY.
by exact continuous movement of an instrument sufficiently
simple for general use. All other curves are nearly always
constructed by sketching them by hand, or with the curved
ruler, through a sufficient number of scattered points.
The following are three important practical rules relative to
the construction of curves:
115. First. A few points determine nearly every kind of
curve, as three points, for example, determine a circle. There-
fore, a few points carefully constructed, and about equally
distributed, or nearest together where the curvature is sharpest,
are better than many points; since the minute instrumental
errors which are inevitable, will induce an improper wavy
appearance in the curve sketched through a superfluity of points.
116. Second. The use of the irregular curve. First. It should
be made to join as many points at once as possible, and each
portion should be tangent to the preceding portion. Second. If
the curve to be drawn has four equal quarters, and two axes of
symmetry, a complete quarter should be drawn, if possible, with
some one portion of the instrument, and the same portion, care-
fully marked, should be used in drawing the other three quar-
ters. Third. Curves of familiar form, as the ellipse, and thus
known not to be pointed, should on no account be made pointed,
by an improper use of the curved ruler, as at the extremities of
their axes, where they cross their axes perpendicularly. Fourth.
When a curve, as in the case of a very narrow oval, is very
sharply curved at its extremities, those portions can be made by
hand; or by the compasses, as small circular ares, tangent to
the other portions.
117. Third. In sketching curves, a given tangent is nearly or
quite as useful as any two given points, since the tangent coin-
cides with the direction of the curve at its point of contact.
DESCRIPTIVE GEOMETRY.
11
SECTION II.
PROBLEMS ON CYLINDERS AND CONES OF REVOLUTION.
A-PROJECTIONS.
a-Projections of Forms.
Definitions and Principles.
118. The projections of a cylinder of revolution will consist of
the projections of at least one of its circular sections, taken as a
directrix, together with the projections of any convenient num-
ber of elements. The latter will be parallel, and drawn through
points on the directrix, or bases (95).
119. The projections of the mere unlimited convex surface
consist of the projections of the extreme visible elements seen
in each projection, and these will be at a distance apart equal
to the diameter of the cylinder.
120. If the axis of a cylinder be oblique to both planes of
projection, the projections of the cylinder will consist of the
projections of its bases, and extreme visible elements, as seen
in looking towards each plane of projection. Since every ele-
ment must contain some point of each base, these extreme ele-
ments can neither be secant to the bases nor exterior to them.
Hence they must be tangent to them.
121. The projections of the elements of a cone will pass
through the projections of the vertex.
Whenever the projection of the vertex falls outside of that
of the base, only a part of the surface will be seen, and the pro-
jections of the extreme visible elements will be tangents from
the projections of the vertex to that of the base; since, as there
must be elements from all the points of the base, there will be
two, between which all the others, and hence the entire base,
will be projected.
78
DESCRIPTIVE GEOMETRY.
Projections of Curves in Space.
122. A curve in space, is conceived to be projected by two
groups of projecting lines, from its various points. The group
which projects the curve upon the vertical plane of projection,
are the elements of a cylinder, perpendicular to that plane, and
whose intersection with it is the vertical projection of the curve.
The horizontal projection is similarly found, by means of a
vertical projecting cylinder. These two cylinders are the pro-
jecting cylinders of the curve. From this, it follows, that their
intersection is the given curve in space, and hence that a curve
in space is determined by its two projections.
123. When the curve is of single curvature, and in a plane
which is perpendicular to either plane of projection, it will be
projected on that plane of projection in the corresponding trace
of its own plane, since its plane will, in this case, be the pro-
jecting plane of the curve.
124. The circle, being the right section (14) of a surface of
revolution, will often be seen obliquely in projecting such sur-
faces. We shall, therefore, begin with the projections of a
circle, and its tangent, as seen obliquely, taking, first, a simple
particular case.
PROBLEM XLI.
Having given one projection of a point, on the circumference
of a circle contained in a profile plane, it is required to
find the other projection of the same point.
In Space. If the given circle be revolved 90 degrees about
any axis, perpendicular to either plane of projection, its pro-
jections will be a circle, and a straight line parallel to the ground
line. On these, both projections of any point of the circle can
be found; and then by counter revolution returned to their
required position.
In Projection.-Pl. VI.,Fig. 60. First solution. Let AB—
C'D' be the projections of the circle, and p the given projection
of the point on its circumference. AB-A'B' is the horizontal,
and CD-C'D' the vertical diameter of the given circle. Let
DESCRIPTIVE GEOMETRY.
79
this circle be revolved about its vertical diameter CD-C'D',
till it becomes parallel to the vertical plane. Its projections
will then be A"B"-A"C'D'. There are evidently two points
of the circumference, whose common horizontal projection is
pq. After the revolution, these points appear at p'q", in hori-
zontal projection, and at p''' and q'" in vertical projection. In
making the counter revolution, since the axis CD—C'D' is ver-
"tical, the point p"p"" moves in a horizontal arc, p''p—p'''p', and
p' is the required projection of the point whose given projec-
tion is p. The point q' is similarly found.
Second Solution.-This, which is valuable as an example,
showing different positions of the same point, consists in deter-
mining the positions of the same are of the circle in space, on
its different revolved positions. Thus: Cp is the horizontal
projection of the are included between the point p, and the
upper extremity CC' of the vertical diameter of the given cir-
cle. When the circle is revolved in the sense of the arrow, 2,
about its horizontal diameter, AB-A'B' as an axis, and till it
becomes parallel to the horizontal plane, Cp will appear at C"
p'''', in its true size. If now the given circle be revolved
about the vertical diameter CD-C'D', in the sense of the arrow,
1, and until it becomes parallel to the vertical plane, it will be
vertically projected in its true size at A'"'C'D'. Next, let the
are Cp be transferred to its place C'p" on the new re-
volved position, and p' may be found by drawing p'''p', which
represents the counter revolution of p"" about CD-C'D' as
an axis.
EXAMPLES. -1°. Revolve the circle around a vertical tangent.
Ex. 2. Around any vertical axis, taken in the plane of the circle.
Ex. 3. Around any vertical axis not in the plane of the circle.
Ex. 4°. Around any axis perpendicular to the vertical plane, whether in or
out of the plane of the circle.
Ex. 5°. Let the plane of the circle be vertical but oblique to the vertical
plane.
Ex. 6°. Let the plane be perpendicular to the vertical plane only.
Ex. 7°. Let the vertical projection of the point be the given one.
80
DESCRIPTIVE GEOMETRY.
PROBLEM XLII.
To construct the projections of a circle of given centre und
radius, lying in a given oblique plane.
In Space.-All those chords of the circle which in space are
parallel to either trace of the given plane, will be shown in
their real size on the plane of projection containing that trace.
All those chords which are also lines of declivity, of the given
plane, relative to either plane of projection, will appear least
on that plane of projection. Parallel tangents to the circle will
be parallel tangents to its projection, and such parallel tangents
being at opposite ends of diameters of the circle will be at
opposite ends of diameters of its projection. These principles
give the following construction:
In Projection.-Let PQP', Pl. VI., Fig. 61, be the given
plane.
1º. To find the centre of the circle.-Take a plane, SKP',
perpendicular to the trace, PQ, of the given plane at any point.
SP" is the revolved position of its intersection with the given
plane, hence the true distance of the centre of the circle, from
the trace PQ, will appear on SP". Let the distance be SO,.
In counter revolution O, returns to o,, and the centre will be
somewhere on 0,0,, parallel to PQ.
1
In like manner, take a plane, PKT', perpendicular to the
vertical trace, QP', of the given plane, and any given distance,
as T'O,, of the centre of the circle, from QP', will appear on
T'p, the revolved position of the intersection of these two planes.
Then by counter revolution, as before, the centre will be some-
where on Oo, parallel to P'Q, and whose horizontal projection
is, therefore, ¿O, parallel to the ground line. Hence, the centre
is 00', where O is the intersection of 20 with 0,0,.
2°. The highest and lowest points, with their tangents. These
points are evidently on the line of declivity through the centre,
OO'. On the declivity, SP"" lay off O,II" and Ô,", each equal
to the given radius, and II", and " will be revolved positions
of points at the required height. Hence, II'II, parallel to PQ,
and '7, the parallel to it at l", will be the lines whose intersec-
tion with the centre line of declivity, II, will be the highest
point, and lowest point, II and I. The vertical projections of
DESCRIPTIVE GEOMETRY.
81
r
#
the horizontal tangents, H'He, and at l", are 'II, and the
parallel to it at l', and H' and l' are the vertical projections of
the required points.
3º. The foremost and hindmost points, with their tangents.—
These points being respectively the ones farthest from, and
nearest to the vertical plane, have the same relation to the ver-
tical trace P'Q that the highest and lowest have to PQ. Hence
lay off the given radius on T'p, and on each side of O,, as at
O,F". Then, by counter revolution, F"F'g' and the parallel
through will be the vertical projections of the required tan-
gents; Fg-F'g' and the parallel to it at hh', whose intersections
with the central line of declivity, 'O'F', give the required
points, FF' and hh'.
2
4°. The right and left points, with their tangents.—These
points are the ones whose tangents are parallel to any plane
which is perpendicular to the ground line. Let PKP' be such
a plane, and let it be considered as an auxiliary vertical plane
of projection. Making KP" KP', we have PP" as the trace
of the given plane PQP' upon this auxiliary plane. Then by
facing PK as a new ground line, QPP" represents the given
plane, and the required points will be found as just explained,
they being the hindmost and foremost, as we face the auxiliary
plane in the direction QK, mRs", etc., and by making the
heights R's", etc., R's, etc.
5º. The points on tangent lines of declivity in both projec-
tions.-Lines, parallel to the traces of a plane, appear in their
full size in the projections which are parallel to the same traces.
Thus, in the figure, ab and c'd' are each equal to the diameter
of the given circle. But the tangents at these points are by
(Theo. II.) perpendicular to these lines. But these tangents,
being lines of declivity of the given plane, their vertical pro-
jections ja', and the parallel at b', will be parallel to S'P', the
vertical projection of the declivity SK—S'P'.
Likewise, c'', and the parallel at d', are lines of declivity re-
lative to the vertical plane, and, in like manner, their horizon-
tal projections ke, and the parallel at d, are parallel to TP, the
horizontal projections of the declivity, PT—KT', relative to the
vertical plane.
125. Thus we have found five pairs of tangents and their
points of contact, which, since a straight line is determined by
6
82
DESCRIPTIVE GEOMETRY.
two points, are equivalent, as an aid in sketching the curve, to
twenty points, without any tangents. By finding the tangents
of declivity relative to the plane, PKP', two more points could
be found.
EXAMPLES. -1°. Find the two points just mentioned.
Ex. 2. Let the given plane be parallel to the ground line.
Ex. 3. Let the given plane be perpendicular to one of the planes of projec-
tion.
Ex. 4. Let its traces make obtuse and acute angles, on the same side, with
the ground line.
Ex. 5. Let the circle be in the second angle.
Ex. 6°. Let it intersect one of the planes of projection.
126. It is obvious from elementary geometry, that the for-
lowing principles are true for all systems of projection in which
the projecting lines are parallel to each other (24), whether
they are also perpendicular to the planes of projection or not.
1º. If one or more lines pass through a given point, their
projections will pass through the projections of that point.
2º. The projections of parallel lines will be parallel.
3º. If a line be divided equally, or in any proportion, its pro-
jections will be divided in the same proportion.
These principles enable us to demonstrate the following
theorem :
THEOREM VI.
The projection of a circle seen obliquely in an ellipse.
127. To prove this theorem, it is only necessary to show, that
the projection of a circle placed obliquely, possesses the charac-
teristic properties of an ellipse.
In doing this by the method of projections, it is sufficient to
refer to either projection separately, of the circle shown in Pl.
VI., Fig. 61.
Let us consider the horizontal projection.
128. Properties of Form.-First. On Ol produced, lay off,
from the trace PQ, a distance equal to SO,, and from the point
O, thus found, describe a circle with the given radius, and it
will be the revolved position and true size of the given circle,
shown in the horizontal plane. This done, the diameters, II,,,
and ab, bisect each other and are perpendicular to each other
DESCRIPTIVE GEOMETRY.
83
со
at the centre, and it is obvious from the principles of ( 55, 4°—65 )
that, after counter revolution about PQ as an axis, the projec-
tions, IIl and ab, will bisect each other perpendicularly.
Second. Again, every diameter of the circle is bisected at the
centre, O,, and it bisects the system of parallel chords to which
it is perpendicular. Hence by (126) the projection of every
such diameter will be bisected at O, and will bisect the projec-
tions of these parallel chords.
Third. If we take any two diameters of the circle, the tan-
gents at the two extremities of each will be parallel to the
other. Hence, by (126, 2d.) the projections of these tangents
will be parallel to that of this other diameter; and, moreover,
to the projections of the chords which this diameter bisects.
Such are called conjugate diameters.
Thus, the curve O (designated by its centre) has one centre
O, which bisects every chord drawn through it. Every such
chord is a diameter (111). Every diameter bisects a system of
parallel chords, and the tangents at its extremities are parallel
to those chords. These diameters occur in pairs, such that the
chords bisected by each are parallel to the other; whence, also,
the tangents at the extremities of each are parallel to the other.
Finally, in one of these pairs, each is perpendicular to the other,
and these are the ares of the curve. These properties of form,
with the fact that the curve is evidently a closed one (110),
show it to be an ellipse, of which ab is called the transverse, or
major axis, and IIl the conjugate, or minor axis.
129. A metrical property.-At any point as e of the curve,
draw cu perpendicular to ab and tep parallel to it. Then, in
the triangle SO,0,, these sides being divided proportionally by
parallels to the base 0,0,, we have
to, : pŨ, :: qo, : l''O,.
1
But to, cu which is called an ordinate to the transverse axis
of the ellipse; pO, being the true length of to,, is equal to the
corresponding ordinate CU of the circle; and go, and 7″O, are
respectively equal to the semi-axis 70, and ɑ0, of the ellipse.
That is, the ordinate to ab, in the curve O, is to the correspond-
ing ordinate CU in the circle, as the semi-minor axis of the curve
is to its semi-major axis.
But this is a property of the ellipse, and, with the previous
properties of form, sufficiently proves the curve O to be an
ellipse.
84
DESCRIPTIVE GEOMETRY.
130. From the preceding theorem we have the following re-
sults:
1°. Through all points of the circle in PQP', draw perpen-
diculars to that plane. The curve, O., formed by their hori-
zontal traces, will be the oblique projection of the given circle
upon the plane H. But, by (126) all the properties of O, can
be transferred to.O, in the same manner as they were to the
projection O. Hence, the projection O, is also an ellipse.
2°. Conversely, calling O, a given ellipse, and PQP' a plane
of projection, every ellipse may have its longer axis so inclined
to the plane of projection, that the projection of the ellipse
upon that plane will be a circle.
3°. Having established the properties of the ellipse O, from
those of the circle O,, let it replace this circle, in the plane
PQP'. Its properties can then obviously be transferred in like
manner to its projection. That is, the projection of any ellipse
will be an ellipse.
Some further useful constructions also follow from the last
theorem.
PROBLEM XLIII.
To construct an ellipse, having given its axes; or other conju-
gate diameters.
First Solution.-In Space. Pl. VII., Fig. 63, represents a
familiar construction of the circle, by points. Knowing (Th.
VI.) that the projection of a circle on a plane not parallel to it,
will be an ellipse, let Fig. 63 turn about AC as an axis, throngh
a certain angle. Then project it on a plane, parallel to AC,
but oblique to the plane of the circle. AC, being parallel to
the new plane of projection, will, with all its divisions, be pro-
jected in its real size. AE, being oblique to this plane, will
appear shorter than it really is; but its divisions, being equal in
space, will be equal in projection. In the same manner, CB
and CD will be equal in projection. By (126, 1°) the projec-
tions of the lines which meet at B and D, will meet at the pro-
jections of these points.
In Projection.-Pl. VII., Fig. 62. Let AX and BD be, re-
spectively, the transverse and conjugate axes of an ellipse. Make
DESCRIPTIVE GEOMETRY.
со
AE equal and parallel to CB, and divide AC and AE each into
the same number of equal parts, and number them from A, as
shown. Then lines from D, through the points on AC, will
meet those from B to the corresponding points on AE, in points
of the quadrant AbB of the required ellipse.
The similar construction of the three remaining quadrants
can be readily made. Or, points on BX may be found by mak-
ing na' etc.,=na, etc.; points on AD may be similarly laid off
on perpendiculars to AC through a, b, and e; and points on
DX may be found from those on BX or from those on AD,
either of the two latter operations serving as checks upon the
former one.
Second Solution.-In Space. This solution depends on the
principles that the oblique view of a circle is an ellipse, and
that the oblique view of an ellipse may be a circle (Th. VI.).
In Projection.-Let AB and CD, Pl. VII., Fig. 64, be the
given axes. Let the circles with radii OA and OD, be the hori-
zontal projections of two concentric vertical cylinders, and let
the circle OA be a right section of the larger cylinder. Also
consider the circle OĎ as the horizontal projection of that ellip-
tical section of the inner cylinder, whose axes are equal to AB
and CD. If, now, circle OA be revolved about AB as an axis,
any point, a, will describe an are, whose horizontal projection
is an, every point of which projection will be a point of some
ellipse having AB for its transverse axis.
Again, if ellipse OD be revolved about the axis CD, any
point, a', will describe an are, whose horizontal projection is
d'a", and every point of aa" will be a point of some ellipse
having CD for its conjugate axis.
Now, in order that a", the intersection of the horizontal pro-
jection of these ares, should belong to the ellipse having both AB
and CD for its axes, it must fulfil the condition (129).
That is,
na"
na::OD: OA.
na" : na:: Oa': Oa,
and a and a must be taken on the same radius Oa, which is
assumed at pleasure, in a sufficient number, gencrally two, of
different positions for each quadrant.
A simple construction of an ellipse, when two conjugate di-
ameters are given, is shown in Pl. IX., Fig. 83.
Draw a circle with the radius OC, one of the semi-diameters,
86
DESCRIPTIVE GEOMETRY.
and the radius OE, perpendicular to OC. Join D and E. Them
draw ordinates, as rA and sB, of the circle; and, Aa, Bb, etc.,
parallel to ED, will meet chords ra, sb, etc., parallel to OD, in
points a, b, etc., of the ellipse, which thus appears as an oblique
projection of the circle OC.
PROBLEM XLIV.
To construct the projections of a cylinder of revolution, limited
by parallel circular bases when its axis is oblique to both
planes of projection.
In Space.-Beginning with the projections of the cylinder
upon a pair of planes respectively parallel and perpendicular to
its axis, so as to show all of its dimensions, the required projec-
tions can be found upon the principle (Prob. IX.), that when.
two planes are both perpendicular to a third plane, the projec-
tions of any point on the two former will be at equal distances
from their respective ground lines upon the latter.
In Projection.-Let CD-C'D', Pl. VII., Fig. 72, be the axis
of the cylinder. Assume a new vertical plane of projection,
G,L,, parallel to this axis, where, by making the heights of C
and D" above G, L,, on the projecting lines CC" and DD", equal
to the heights of C' and D' above GL, we shall have C"D" the
projection of CD-C'D' on the new parallel plane. This plane
being revolved forward, about G,L,, into the plane of the paper,
may now be conceived of as a horizontal plane of projection,
and the part of the paper containing CD as a vertical plane.
Then, accordingly let G,L, be the ground line of a new vertical
plane perpendicular to the axis C"D". The projection of the
cylinder on this last plane will be a circle, whose centre, C'"',
is at a height, C'"'n'", above G₂L, equal to Cn.
Ilaving now two complete projections of the cylinder, show-
ing all its elements and diameters in their real size, any point of
the two required projections may be found as follows. Assume
any point, as a'', on the circle C"". Its horizontal projection,
on the right hand base, is a". Its vertical projection is a, at
the same height above G,L, that a"" is above GL. Turning
the whole system of planes back again, without changing their
relative position, till projection, CD, is the horizontal one, a' is
DESCRIPTIVE GEOMETRY.
87
found on aa', perpendicular to GL, and as high above GL as a"
is above G,L,.
In the same way all other points may be found.
2
GA, perpendicular to G,L, and Ga"", perpendicular to`
GL,, are the two different positions of the vertical intersectious
at G, of the two vertical planes which are perpendicular to the
plane, LG,L, of the paper, taken as a horizontal plane of pro-
jection. Hence e, for example, might have been found by pro-
jecting e' upon G₁a"" as at E"", and then revolving E" about G,
as a centre, to E, when e will be projected on Ee, drawn parallel
to the ground line, G,L.
1
EXAMPLES.-1°. Take the first auxiliary plane parallel to the axis CD-C'D'
and perpendicular to the vertical plane.
Ex. 2'. Instead of such an auxiliary plane, begin by revolving CD--C'D' till
parallel to one of the original planes of projection.
Ex. 3°. Let CD-C'D' be in a plane perpendicular to the ground line.
Ex. 4°. Substitute a cone for a cylinder in the above problem.
PROBLEM XLV
To construct the projections of a cone of revolution, whose axis
is oblique to both planes of projection; and of any point
or element of its surface.
In Space. This problem, like the last, requires an auxiliary
plane, parallel to the axis of the cone, on which the projection
of the cone shows its true height and diameter. This projection
is therefore the one first made.
1
In Projection. To vary this problem from the last as much
as possible, let the axis of the cone cross the first angle.
1º. The projections of the axis.-Let V"L", Pl. VIII., Fig.
79, be the ground line of an auxiliary vertical plane, parallel
to the axis of the cone, and let V"O" be the true length and
direction of this axis, shown on that plane, after revolving it
forward into the horizontal plane, in order to avoid confusion
with the plan.
VO, parallel to this ground line, and at any convenient dis-
tance from it, is then the horizontal projection of the axis. The
horizontal plane being taken through the vertex, the vertical
88
DESCRIPTIVE GEOMETRY.
projection, V', of the latter is on the ground line V'L; then,
making d'O'=d"O", V'O' is the vertical projection of the axis.
2º. The axes of the projections of the base, and the traces of
its plane.—C''D' is the auxiliary projection of that diameter of
the base which is parallel to the auxiliary planc. CD, parallel
to the ground line, V"L" is its horizontal projection, and the
minor axis of the horizontal projection of the base. O"-AB,
where AB C"D", is the transverse axis of the same projec-
tion.
G'II', the transverse axis of the vertical projection of the
base also equals C"D". Its minor axis, E'F', may be found
with precision, either from another auxiliary projection of the
cone, on a plane perpendicular to the vertical plane, and parallel
to the axis V'O'; or by Prob. XLII., after finding the traces of
the plane, X, of the base.
Since the plane X is perpendicular to the auxiliary plane V"
L', its auxiliary and horizontal traces are C"L"; and L'Q,
perpendicular to L". Then, being perpendicular to the axis,
its vertical trace, QP', will be (Th. III.) perpendicular to V'O'.
3°. Other points of the base.-These are found simply by re-
volving about its diameter CD-C"D", till parallel to the aux-
iliary plane, as at C"A"D", when J'J'", etc., show the real dis-
tances of points of its circunference from the diameter, C"D".
But these distances, being perpendicular, in space, to the same
plane, will also appear in their real size in horizontal projection,
as at aJ=aK=J"J"", giving the points J and K, whose verti-
cal projections, J' and K', are in the like projection, K'J, of
the horizontal chord, JK, whose height is e'K'=e"J".
Any additional points can be found in the same manner; or,
by Prob. XLIII.
131. The projections of the vertex and base really determine
the cone, since a line from the vertex to any point of the base
is an element.
The continuity of the convex surface can then be represented
by the projections of any number of sections parallel to the
base, and found as above explained, or by any number of ele-
ments. Of the latter, two must (121) be tangent to the base,
in projection, and they may be drawn with sufficient accuracy
by inspection.
Finally, V"I"-VI, and V'I', with p".pp', are the projections
of an element and of any point taken upon it.
DESCRIPTIVE GEOMETRY.
89
EXAMPLES.-12. Change the direction of the axis, in the first angle.
Ex. 2. Let the vertex and the base be in different angles.
Ex. 3°. Let the axis cross any one of the four angles.
Ex. 4º. Let the auxiliary plane, parallel to the axis, be perpendicular to the
vertical plane.
b-Projections of Tangencies.
Tangent Lines to Curves.
132. The tangent to a plane curve may be variously defined.
First, resuming (111) it is the line in which the secants
through any point unite, when, in turning in the plane of the
curve, and about that point, their intersections with the curve
fall together at that point.
Hence, second, the tangent to a plane curve (105) at a given
point, is a line which lies in the plane of the curve and touches
it only at that point.
The tangent is thus, third, a line having the same direction
that the motion of the generatrix of the curve has, when at the
point of contact.
133. But, fourth, it is sometimes convenient to regard a curve
as a polygon of an infinite number of sides, each consisting of
two consecutive points; which are, however, practically one.
The tangent at any point, is then the side which forms that
point produced. The apparent contradiction here is no more
than will generally be found in definitions involving an infinity,
large or small.
134. Since a circle may always be passed through three points
not in the same straight line, one may be traced through three
consecutive points of any other curve. Such a circle coincides
with that curve at those points, and hence its radius is called
the radius of curvature of the curve at that place. The circle
is called the osculatory circle of the curve.
135. Now, so long as the radius of curvature of any curve is
finite, the figure will be sensibly curved, and hence visibly dis-
tinct from its tangent. The larger the radius of the curve, the
more nearly will the latter approach to coincidence with its tan-
gent. IIence, when that radius is infinite, the curve becomes
sensibly straight, and then coincides with its tangent. That is,
the tangent to a straight line--considered as a curve of infinite
radius-is that line itself.
90
DESCRIPTIVE GEOMETRY.
136. The tangent to a curve of double curvature.―Three
consecutive points of a curve of double curvature (106) con-
stitute a small circular arc (101), which is, therefore, a plane
curve. Then if 1, 2, 3, 4, 5, n be any number of con-
secutive points of such a curve, it may be conceived as com-
posed of the elementary arcs, 1,2,3 : 2,3,4 : 3,4,5, etc., and
the tangent line to it at any point, as 3, may be defined as a
line lying in the plane 2,3,4, and containing only the point 3.
For the tangent, thus defined, is the limit between the ultimate
secants (111) 2,3 and 3,4, and lies in the same plane with them,
just as in the case of an extended plane curve (132).
The tangent at an infinitely remote point of an open curve
(110) is called an asymptote.
137. Every plane drawn through the tangent to a curve is
called a tangent plane to the curve. Such a plane is therefore
indefinite, there being an infinite number of them for any one
tangent line. That one of them which contains the three con-
secutive points, of which the point of contact is the middle one,
is called the osculatory plane at the given point, since it evi-
dently contains the osculatory circle through these three points.
All the osculatory planes of a plane curve evidently coincide
with the plane of the curve. A perpendicular to the tangent,
lying in the osculatory plane, is called a normal (134).
138. Any two successive osculatory planes to a curve of
double curvature intersect in an ultimate secant joining the two
points common to those planes, and make an infinitely small
angle with each other at that line, which is called the angle of
flexion. Thus, in (136) the planes 1, 2, 3, and 2, 3, 4, inter-
sect in the line 2, 3, and the plane of the angle of flexion is
perpendicular to that line.
139. Again, the successive secants, 1,2; 2,3; 3,4; etc., inter-
sect at successive points, 2, 3, 4, etc., forming angles differ-
ent for different curves, and for different parts of the same
curve, and called angles of contingence.
140. A straight line is determined by two points, either
given, or implied in other equivalent conditions.
Hence the tangent to a plane curve is determined by a given
point of contact (132, 133 ), or by a given exterior point and a
DESCRIPTIVE GEOMETRY.
91
required point of contact on the curve; or by being parallel to
a given line.
141. These are the three fundamental conditions because
others less elementary can be reduced to these. Thus, if a tan-
gent be required to make a given angle with another line in the
plane of the curve, it will simply be parallel to one of the sides
of the angle, by the third fundamental case; the given line being
the other side of the angle. Or if a tangent be required to two
curves at once, and if by any means the point of contact on one
can be found, this, by the second fundamental condition, be-
comes a given exterior point relative to the other curve.
PROBLEM XLVI.
To construct a tangent to an ellipse, through a given point on
the curve.
In Space. Two methods are here given; the first, on the
principles of revolution about an axis (55), the second, on the
same, together with that of conjugate diameters and their paral-
lel tangents.
In Projection.-Let the ellipse, ABCD, Pl. VIII., Fig. 77,
be given by its axes.
First method.-Let R be the given point of contact. ArBm
is the circle, which, when revolved about AB, and seen oblique-
ly, will be projected in the given ellipse. Then R revolves to
7,and Sr, perpendicular to the radius, rO, is the revolved posi-
tion of the required tangent. In counter revolution, S, being
in the axis, remains fixed, and therefore SR is the required
tangent.
Second method.-Let M be the point of contact. Its revolved
position in the paper is m, and mO is the radius perpendicular
to a tangent at m. Then O, perpendicular to mO, is the re-
volved position of the diameter conjugate to MO (Theo. VI.),
and hence parallel to the required tangent. NO is the pro-
jection of nO, hence MK, parallel to NO, is the required
tangent at M.
The first method, where S is near A, is thus better when the
point of contact R is near A or B. The second method is bet-
ter when M is near an extremity of the conjugate axis.
92
DESCRIPTIVE GEOMETRY.
EXAMPLES.-1°. Solve the same problem by making CD the axis, and revolv-
ing the ellipse, till its projection shall be a circle on CD as a diameter.
Ex. 2°. Let the axis of revolution be a tangent at C, or D.
PROBLEM XLVII.
To construct a tangent to an ellipse, from a given exterior point.
In Space. By using two planes of projection so as to show
two projections of the given ellipse and point, the construction
can readily be made by the principles of projections, that is, by
the methods of Descriptive Geometry, as follows:
In Projection.-Let the given ellipse, ABCD, Pl. VIII.,
Fig. 78, be the projection of a circle whose revolved position is
that on AB as a diameter; and let P be the given point.
Let O'', parallel to CD, be the trace of a plane of projec-
tion perpendicular to the paper. Then project dat d', and
describe d'D', the new projection, with O' for its centre, of the
are dD. Project D upon this arc at D', then D'O' will be the
trace of the plane of the ellipse, on which P will be projected
at P'.
By revolution PP' will appear at pp', in the plane of
the circle, and pm and pn will therefore be the revolved posi-
tions of the required tangents, there being evidently two.
By counter revolution, p returns to P, and m and n to M
and N, giving PM and PN for the required tangents.
If n were so near B, as to make nN quite obscure, its projec-
tion on the trace D'O' could be found, and then N could be
clearly found by projection.
EXAMPLES.-1°. Let P be nearer CD than to AB.
Ex. 2. Let P be in any other angle than DOB.
Ex. 3. Let P be so situated that both points of contact shall be on the same
quadrant of the ellipse.
PROBLEM XLVIII.
To construct a tangent to a given ellipse, and parallel to a
given line.
In Space.-This problem is here solved by the principles of
DESCRIPTIVE GEOMETRY.
93
conjugate diameters and their tangents, as proved by the method
of projections.
In Projection.-Let ABCD, Pl. VIII., Fig. 78, be again the
given ellipse, and E the given line.
Draw any two chords, as Aa and Db, parallel to E, then the
diameter GII which bisects them, will give the points of con-
tact, G and II, of the tangents, parallel to E.
The chords Aa and Db were taken, since as the ellipse was
traced with an irregular curve, only the points ABC and D are
exact, but a and b being very near them, are supposed to be
very nearly so; which, however, would not be the case, if the
curve were carelessly drawn, so as to be either pointed or
re-entrant at C and B.
EXAMPLE.-Solve this problem by two planes of projection, and the revolved
position of E, as in the last problem.
Tangent Planes to Cylinders and Cones.
142. The tangent plane to a cylinder, or cone, may be defined
by some sensible property, obvious on mere inspection; or by
some rational property, requiring demonstration.
143. The tangent plane to a cylinder or cone of revolution,
is seen by inspection to be a plane which merely touches that
surface along one element, as VD, Pl. VII., Fig. 73.
144. The tangent plane as thus defined, is entirely deter-
mined, since, if the element of tangency were taken as an axis,
the tangent plane evidently could not revolve about it in any
manner without instantly becoming a secant plane.
145. The tangent plane is thus more comprehensively defined
as being the limit of all the secant planes through the element
of contact, also as being the plane which contains the direc
tion of the motion of every point of the generatrix, at the in-
stant when the latter occupies the position of the clement of
contact.
There can, therefore, be but one tangent plane to a cylinder,
or cone, along a given element.
146. Moreover, as the directions of the motion of every point
of the same position of the generatrix are parallel, a plane, tan-
94
DESCRIPTIVE GEOMETRY.
gent at one point of a cylinder or cone, is tangent, at the same
time, at all points of the same element.
147. Every line of a tangent plane, not parallel to its element
of contact, intersects that element at a point which is its point
of contact with the given surface. That is, the tangent plane
contains all the tangent lines, whose points of contact are on its
element of contact.
There are an infinite number of such tangent lines at each
point of the element of contact of the tangent plane. Any two
of them will determine the tangent plane.
148. That is, if a plane, VDN, Pl. VII., Fig. 73, and a cylin-
der, or cone, VPQ, be tangent to each other, every secant plane,
S, which cuts both, will cut a line as TA, or TR, etc., from
the plane, and a curve, as TA, or TR, etc., from the surface,
which will be tangent to each other at the point, T, in which
the secant plane cuts the element of contact, VD, of the tangent
plane. Hence the traces of a tangent plane are tangent to
those of the given surface.
149. When the secant plane, S, taken through T, also con-
tains V, the line cut by it from the cone becomes straight and
no other than the element, VD, which therefore coincides with
its tangent. That is, VD is its own tangent (135), and is
therefore to be reckoned as one of the tangent lines contained
in the tangent plane on VD (147).
150. It follows from (148), that if a line and a curve in
space be tangent at a point, T, their projections will be tangent.
For the projecting plane of the line will be tangent to the pro-
jecting cylinder of the curve along the element which is the
projecting line of T. Hence the trace of the plane, which is
the projection of the line, will be tangent to that of the cylin-
der, which is the projection of the curve, at t, the projection of
T. For these traces are in the same plane and have only the
point t in common.
151. The tangent trace, just described, is the projection, not
only of the supposed tangent to the curve in space, but of every
line in the projecting plane described.
That is, if the tangent plane to a curve be perpendicular to
a plane of projection, the projection of any line in that plane
will be tangent to the projection of the curve.
DESCRIPTIVE GEOMETRY.
95
THEOREM VII.
The tangent plane to a surface of revolution is perpendicular
to the meridian plane containing the point of contact.
Let the cone, V, OAB, Pl. VII., Fig. 65, represent the sur-
face of revolution, the demonstration being the same for all
such surfaces.
The tangent plane at t is determined by the element Vt, and
the tangent bt to the parallel through t. But bt is perpendicular,
both to the radius, ta, of the parallel, and to the line Nto,
which is perpendicular to the given surface at t, and intersects
the axis. Hence, as the tangent plane contains bt, it is perpen-
dicular to the plane of at and ot. But as at and ot both inter-
sect the axis, VO, they determine the meridian plane through t;
which is thus perpendicular to the tangent plane at t. See also
Fig. 67, where any plane through Tt is perpendicular to the
meridian plane, atc. IIence the tangent plane at t, and contain-
ing the tangent Tt, is perpendicular to the plane, atc.
152. Besides problems of tangent planes to cylinders and
cones, singly, there may be problems of planes tangent to two
such surfaces at once; and of cylinders or cones tangent to
each other. But these can generally be solved by means of a
knowledge of the problems of tangent planes.
153. Two cylinders or cones tangent to each other, will have
a common tangent plane. There will then be two cases: first,
when their elements of contact with this plane coincide, in
which case the two bodies will have a common element of con-
tact: second, when these elements of contact intersect, in which
case the bodies will have only a common point of contact.
PROBLEM XLIX.
To construct the extreme elements of the projections of a cone,
the projection of whose vertex falls without that of its
base.
In Space. In this case, the projections of the extreme ele-
ments are tangents to the elliptical projections of the base from
96
DESCRIPTIVE GEOMETRY.
those of the vertex, considered as a given exterior point. They
are also the traces of the projecting planes of those elements;
and these planes are tangent to the cone and parallel to a
given line, viz., a perpendicular to that plane of projection on
which the extreme elements, supposed, are shown (151).
In Projection.-In Pl. VIII., Fig. 79, V, and the ellipse
ABCD, may be considered as the horizontal projection of the
circle, C"D", and of a point in its plane. Then the intersec-
tion of VV", produced, with D"C" produced, would be the
auxiliary projection of the latter point. This projection falls
without the paper, hence make Ov=OV, and project into the
plane, C"L", at X, and X will be symmetrical with the point
sought on C"L", and on the other side of O". That is, XT"
is the revolved position, about CD-C"D" as an axis, of the
previous revolved position about O"A" as an axis, of that tan-
gent to the base, C"D", whose horizontal projection is a tan-
gent from V to ABCD.
Hence make A"U""A"T"", revolve U"" back to U", and
project it at U, and M,where Mƒ=Uƒ=U"U". Then VU and
VM are the extreme elements of the horizontal projection.
Those of the vertical projection were drawn by inspection only.
U and M, and the points symmetrical with them, relative to
AB, will give four more points in each projection of the base.
PROBLEM L.
To construct a plane, tangent to a cone of revolution, and
through a given element of the surface, the axis of the
cone being a bi-parallel.
In Space. By (147, 149) the plane will be determined by
the given element, and by a tangent line at any point of the
element, to any section of the surface. For this tangent line
we can take a line parallel to it, through the vertex.
In Projection.-Preliminary construction. Pl. X., Fig. 87.
Let ABV-C'D'V' be the projections of the cone, cV—c'V' be-
ing the projections of the axis. Va-V'a' are the projections
of the given element of tangency. Both projections of this
element cannot, in this case, be assumed at pleasure. Either of
them, as Va, being assumed, the other, being the intersection
DESCRIPTIVE GEOMETRY.
97
of its vertical projecting-plane with the surface of the cone,
must be constructed. This plane cuts the base of the cone in
two points whose horizontal projection is a. Revolve D'QA,
the plane of the cone's base, about its vertical trace DQ, and
into the vertical plane of projection. The centre, cc', of the
base will describe the horizontal arc cc'-c'c', and the circle
with e' as a centre is the revolved position of the base. The
points whose horizontal projection is a, appear at a"a", and
after the counter revolution, at a'a'. IIence Va is the hori-
ontal projection of the two elements whose vertical projections
re V'a' and V'a". Of these, let Va-V'a' be taken as the
iven element of contact.
2°. Construction of the Tangent Plane.-Proceeding by the
solution in space, the required plane may be determined by the
element, Va-V'a', and any tangent; as the tangent to the base
at da. But the element does not here pierce the planes of pro-
jection conveniently, hence we take two such tangents. And
as these tangents are really traces of the required tangent plane
upon the planes AQD' and MNM', we finally proceed by the
second method of the solution in space. Thus, the tangent
plane has Va-V'a' for its element of contact; and its trace on
the plane of the base is therefore tangent to the base at a""
hence, a''t's is this trace in its revolved position. The point
t' is the revolved position of the point in which it pierces the
horizontal plane; and s is where it pierces the vertical plane.
Ilence, after the counter revolution, in which s, being in the
axis, remains fixed, t' will appear at t, as one point of the hori-
zontal trace of the tangent plane, and s as a point of its vertical
trace, in the lower part of the vertical plane.
Take any other auxiliary profile plane, as M'NM, construct
the revolved position e" of the circle ce' cut by it from the
cone; the revolved position, d''h'n, of the trace upon it of the
tangent plane; and, by counter revolution, n will be another
point of the vertical, and h another point of the horizontal
trace of the tangent plane, whose traces, sn and th, are thus
determined.
EXAMPLES.—1°. Let the axis of the cone be oblique to both planes of pro-
jection. (This simplifies the problem.)
Ex. 2°. In place of a cone take a cylinder whose axis shall be a bi-parallel.
Ex. 3. Let the cone be vertical, and let the plane be tangent to it along the
foremost element.
í
Um
98
DESCRIPTIVE GEOMETRY.
Ex. 4°. Let the plane be tangent to a cylinder, both projections of whose
axis are oblique to the ground line.
Ex. 5°. In the last case let both projections of the axis of the cylinder be
perpendicular to the ground line.
MU
PROBLEM LI.
To draw a plane, tangent to a cylinder of revolution, ana
through a given point in space.
In Space.-Every plane, tangent to a cylinder, is parallel to
its axis. Hence, if through the given point a line be drawn
parallel to the axis of the cylinder, the plane will be determined
by this line, together with a tangent to the cylinder from any
point of the line (147). If the base of the cylinder be in the
horizontal plane, the tangent to it from the horizontal trace of
the line (148) will also be the horizontal trace of the plane,
whose vertical trace can then be found in any of the ways
already explained; remembering that the element of contact of
the plane and cylinder will be drawn from the point of contact
of the trace of the plane with that of the cylinder.
In Projection.-Let AB-A'B', Pl. VIII., Fig. 80, be the
axis of the cylinder, oblique to both planes of projection.
The projections, abcdk—a'b'c'd'k', of the cylinder can be
found by Prob. XLIV.
Let pp' be the given point through which the tangent plane
is to be drawn. Then q, the horizontal trace of pq—p'q', paral-
lel to the axis, is one point of the horizontal trace of the tan-
gent plane. By Prob. XXI., the same line pierces the plane
PQP' of the base of the cylinder at FF'. The traces of this
plane are parallel to the axes, ab and c'd', of the projections of
the base, and contain the traces, as e', of any diameter, as ab-
a'b', of the base.
Having thus found FF', the tangent Fg-F'g' to the base is
a line of the tangent plane, and is its trace upon the plane of
the base. Hence P' and S (S not shown), its intersections with
the traces of the plane PQP', are its traces, and therefore points
in the traces of the tangent plane. The vertical trace of the
line pq-p'q' is not here within the limits of the figure, hence
we take the parallel, Ff-F'f, to the horizontal trace Sy of the
DESCRIPTIVE GEOMETRY.
99
tangent plane, to find a second point, f', of its vertical trace
P'f'. Then HqS and P'f' are the required traces of the tan-
gent plane through pp'.
The two tangent lines from FF', of which but one is shown,
may be drawn with exactness by Prob. XLVII., and the traces
of a second tangent plane, not shown, can be found.
The elements of contact, as gk-g'k, can be used as a check
in finding these traces.
EXAMPLES.—1°. Let the tangent plane be tangent to a cone, instead of a
cylinder. (pq-p'q' wil then join pp' with the vertex of the cone.)
Ex. 2°. Let the cylinder be parallel to the ground line. (Then use an aux-
iliary plane perpendicular to the ground line.)
Ex. 3°. Let its axis be parallel to either plane of projection and oblique to
the other.
PROBLEM LII.
To draw a plane, tangent to a cylinder, and parallel to a given
line, when the axis of the cylinder is a bi-parallel.
In Space.-A plane either through the axis, and parallel to
the given line, or through the given line and parallel to the
axis, will be parallel to the required plane. The trace of the
latter, on any auxiliary plane, P, will be tangent to the traces
of the cylinder (148) on the same plane, and parallel to the
trace of the auxiliary parallel plane upon the plane P.
In Projection.-Pl. X., Fig. 88. Let ABG-C'D'G' be
the projections of the cylinder, and HK-H'K', those of
the given line. P'K'-PH are the traces of an auxiliary
parallel plane containing the given line. PQP' is a pro-
file plane, perpendicular to the axis of the cylinder. Let
this plane be revolved about its horizontal trace PQ into
the horizontal plane of projection, then the trace of the
parallel plane upon it will appear at PP"; and the circle, or
trace, ur-pq, will take the position of the circle whose centre
is c'. nN, drawn parallel to P''P, and tangent to this circle
e', is the revolved position of the trace of the tangent plane
upon PQP'. By making the counter revolution of the plane
PQP', the trace nN will take the position NN', and will pierce
Um
100
DESCRIPTIVE GEOMETRY.
the planes of projection at the points N and N'. Hence MN
and M'N' are the traces of the required tangent plane.
EXAMPLES.-1°. Let the axis of the cylinder be oblique to both planes of
projection.
Ex. 2°. Let it be in any other than the first angle.
Ex. 3°. Let the given line cross the fourth angle.
PROBLEM LIII.
To construct a plane, tangent to a cone of revolution, and
parallel to a given line.
In Space. Two intersecting lines determine a plane. The
lines to be chosen in this case, according to the conditions of
the problem, are, a line, L, through the vertex of the cone, and
parallel to the given line, and a tangent line to the cone from
any point of the line L. The former will make the required
plane parallel to the given line, and both together will make it
tangent to the cone, since every tangent plane to a cone con-
tains the cone's vertex.
In Projection. There are four principal cases to this im-
portant problem, viz.:
First. When the axis of the cone is perpendicular to a plane
of projection.
Second. When it is parallel to the ground line.
Third. When it is parallel to one only of the planes of pro-
jection.
Fourth. When it is oblique to both of these planes.
We shall illustrate the last two cases, when it will clearly
appear that the solution given above will be alike for all, and
that the first two can be constructed without further aid.
1º. The axis parallel to only one plane of projection.—Let
VBC—V'B'E', Pl. IX., Fig. 81, be the given cone, constructed
like the plan and auxiliary projection in Prob. XLV., and AA”
the given line.
Va-V'a' is the auxiliary parallel line through the vertex,
VV', and its traces, b and a', are points respectively of the hori-
zontal and vertical traces of the required tangent plane. This
line, Va-V'a', pierces the cone's base at c', whose horizontal
projection is c, and whose revolved position, around E'D' as an
DESCRIPTIVE GEOMETRY.
101
axis and into the vertical plane DV, is c', found by making
c'e'=cd. Then c'e', tangent to the base revolved in like man-
ner, is the revolved position of the auxiliary tangent line pro-
posed.
Projecting e' at e, ce-c'e' is, by counter revolution, the same
tangent, or the trace of the required plane upon the plane of
the cone's base. Its traces, f and ', are points of the traces,
bf and a'h', of the required tangent plane PQP'.
If a', or h', had been without the diagram, a'h' might have
been made parallel to V'e', the trace of the tangent plane on
the vertical plane DV. Or, T'T'T is the point of contact of the
trace c'e'-c'e'-ce of the tangent plane upon the plane fGE' of
the cone's base. Hence TV-T'V' (not drawn), would be the
element of contact of the tangent plane, and its traces would
therefore be points of the required traces of the tangent plane.
2°. The axis oblique to both planes of projection.—Let
V"C"D"-VAB-V'C'D', Pl. VIII., Fig. 79, be the given
cone, and SS' the given line. The vertex being already in the
horizontal plane, is one point of the horizontal trace of the tan-
gent plane. Vg-V'g', parallel to SS', and through VV', gives
its own vertical trace, l', for a point in the vertical trace of the
plane.
Projecting gg' at g″, gives V"g", and thence 'k for the inter-
section of V"g"-Vg-V'g' with the plane C"L"QP' of the
cone's base. Revolve the latter plane about its auxiliary trace
C"L" into the auxiliary plane V"L", and the cone's base
appears as the circle O""', and ''k revolves to k", by making
O"O"" = Od", and "k"=ik. Then "Nm"-k"m" -kno
are the three projections of the trace of the tangent plane upon
the plane of the cone's base; this trace is thus tangent to that
base (148) at N, a', n, n', and V"a"-Vn-V'n' is the cle-
ment of contact of the tangent plane.
The latter line thence gives o', its vertical trace, as another
point of the required vertical trace h'o'R, which can now be
drawn. The horizontal trace, RV, contains r, the horizontal
trace of the tangent mk-m'k", to the cone's base, from the
point, kk". VRh' is thus the required plane.
154. It may add interest to this and the last problem to indi-
cate their uses. Besides possible applications to architecture,
masonry, and machinery, if the given line in this problem were
a ray of light, the tangent planes, parallel to it, would therefore
102
DESCRIPTIVE GEOMETRY.
be planes of rays, and their elements of contact would there-
fore be those which bounded the illuminated portion of the
convex surface of the cone. Also, if the given point in the last
problem were the position of the eye, the tangent planes through
it would be planes of vision, and their elements of contact would
be the visible boundaries of the surface as seen from the given
point, in forming a perspective of the surface.
EXAMPLES. -1°. Let the axis of the cone be vertical.
Ex. 2°. Let the axis of the cone be a perpendicular ( 48 ).
Ex. 3. Let the axis of the cone be a bi-parallel. The plane of the base,
used as above, being then a profile plane, the construction will be very simple.
Ex. 4'. In both cases of the above problem, find the traces of the second
tangent plane which can be drawn, beginning, in Pl. IX., Fig. 81, with the
other tangent from c" to the circle O'E'; and in Pl. VIII., Fig. 79, with the
other tangent from "" to the circle O".
Ex. 5. In Pl. IX., Fig. 81, let the axis of the cone be parallel to the hori-
zontal plane only.
Ex. 6. Substitute a cylinder for a cone in each of the above cases.
PROBLEM LIV.
To construct a plane, tangent at once to two cones, whose axes
are the projecting lines of their common verter.
In Space.-One axis will thus be vertical, and the other a
perpendicular (48). The common tangent plane will thus
make the same angles with the planes of projection that are
made by the elements of the cones with the same planes, and
will be determined, in thought, by its elements of contact with
the cones.
There can be but one perpendicular from a point to a plane.
Hence if we first find a plane at any given distance from some
point on the ground line and making the required angles with
the planes of projection, the required plane will be parallel
to it.
In Projection.-Pl. IX., Fig. 82. Let ABC-A'B'C' and
ADE-A'D'E', be the two cones, the former, and the tangent
plane, making the angle m with the plane, H, the latter, and
the same tangent plane, making the angle n with the plane V.
Let BC, Fig. 83, represent any plane parallel to the tangent
DESCRIPTIVE GEOMETRY.
103
plane, and placed perpendicular to the paper. Let OP repre-
sent a perpendicular to this plane, from some point on the
ground line. Now OP in its real position, Fig. 82, will be
the common perpendicular of two right triangles, in each of
which the hypothenuse will be the perpendicular from O to a
trace of the plane, and the base, the perpendicular from the
foot of OP to the same trace. Fig. S3 represents these tri-
83
angles, when revolved about OP till their planes coincide.
Then, in Fig. 82, make Ohp"" equal to the triangle OBP, Fig.
83, and Ofp" equal to OCP. Counter revolution about Oh
and Of, gives Op' and Op as the length of the projections of
the perpendicular, Op"" Op", to the parallel plane. Then
the traces of this plane will be tangent to the arcs having Oh
and Of as radii; and the projections of p and p' will be on ares
having Op and Op′ as radii.
Produce fp" to meet OR", perpendicular to Of, and R" will
be the revolved position of the vertical trace of the intersec-
tion, ƒR", of the vertical plane, Of, with the vertical plane of
projection. Hence, make OR' = OR", and R'S tangent to the
are H, will be the vertical trace of the parallel plane, and SR,
tangent to the arc of radius Of, is its horizontal trace.
The traces UT and UT' of the tangent plane are parallel to
SR and SR', and tangent to the bases of the cone.
A check on the construction is afforded by revolving Oh to
OH, perpendicular to R'S, and Op to OP'; also Of with p to
OE, perpendicular to SR. Then PP' will be the true position
of the foot of the perpendicular from 0 to the plane RSR',
hence PP' will be perpendicular to the ground line.
155. Figure 82 would evidently serve the enunciation: To
draw a plane through a given point, AA', and making given
angles, m and n, with the planes of projection.
156. After the full solutions now given of the last five prob-
lems, the following remarks will serve to guide the student in
constructing planes, tangent at once to two cones (152).
The two determining lines of the proposed plane may be
considered as its elements of contact. Now:
157. First. If the two cones have a common vertex, and
bases exterior to each other, the supposed elements of contact,
meeting at the common vertex, will be in one plane.
158. Second. If the cones have a common base, elements of
Upp
104
DESCRIPTIVE GEOMETRY.
contact may be conceived, joining the separate vertices with
some point of that base.
159. Third. And generally, if the line joining the two ver-
tices, and a common tangent to the two bases, are in the same
plane, that common tangent will be the trace of a common tan-
gent plane on the plane of the bases.
The preceding solutions also lead to that of the following:
EXAMPLES.—1°. Construct a plane, through a given line and making a given
angle with the horizontal plane.
Ex. 2. Construct a plane making a given angle with any given plane.
Ex. 3°. Construct a plane tangent to the cone in Pl. IX., Fig. 81, and making
a given angle with either of the planes of projection.
Ex. 4. The same for Pl. VIII., Fig. 79.
c-Projections of Intersections.
160. There are here two groups of problems:
FIRST. Intersections of cylinders and comes with planes.
SECOND. Their intersections with each other.
Cylinders and cones, being merely prisms and pyramids of
an infinite number of sides, the latter forms are here included.
Their plane intersections are polygons.
I.-Intersections of planes with cylinders and cones.
161. Since the convex surface of every cylinder or cone is
composed of rectilinear elements, its intersection with any given
plane may always be found as follows:
General Solution, stated simply as an application of Prob.
XXI.-Find, by Prob. XXI., where a sufficient number of ele-
ments, taken at pleasure, pierce the given cutting plane, and
join these points.
162. Auxiliary planes, in order to intersect a conic surface in
straight lines, must pass through its vertex. To cut such lines
from a cylindrical surface, they must be parallel to its axis.
Circular elements are next simplest to rectilinear ones, and
accordingly, the auxiliary planes may be parallel to the base of
the cylinder, or cone,
DESCRIPTIVE GEOMETRY.
105
163. If the intersection of a prism or a pyramid by a plane
be required, the auxiliary planes would be passed through its
edges; but they may be passed in any direction-not parallel
to the given secant plane-since they will always intersect the
faces of the plane-sided body in straight lines.
164. The tangent line at a given point of the intersection of
any developable surface by a plane, must by (132) be in the
plane of the curve, and by (147) in the tangent plane to the
surface along the element containing the given point. Hence
it must be the intersection of these planes; the traces of which
on one plane of projection will be sufficient, since the given
point is one point of their intersection.
PROBLEM LV.
Having given any pyramid, by its projections, and any plane,
by its traces; to find the intersection of the plane with the
pyramid.
In Space.-The figure of intersection will be a polygon of
some kind. Now a polygon is determined when all its vertices,
or when all its sides, are determined. These two methods of
determining a polygon lead to the two following solutions:
First Solution.-The vertices of the required polygon are
the points where the given plane cuts the edges of the given
pyramid; hence, by repeated applications of Prob. XXI., find
the point in which each edge of the pyramid pierces the given
plane, and join the points thus found by straight lines. The
figures so formed, will be the projections of the intersection of
the plane and the pyramid.
Second Solution.-The sides of the required polygon are so
much of the intersections of the planes of the faces of the
pyramid with the cutting plane, as lie on the face of the
pyramid.
In Projection.-If the traces of the cutting plane, PQP',
Pl. X., Fig. 89, make a large angle with the ground line, its
horizontal trace will meet the horizontal trace of an auxiliary
plane, as V'A'A, at an inconvenient distance from the ground
line. This case affords a good opportunity for changing the
planes of projection. (58, 3°.) Thus:
106
DESCRIPTIVE GEOMETRY.
Let GL be the original ground line; ABC-V the horizon-
tal, and A'B'C'-V' the vertical projection of a pyramid, which,
for the sake of simplicity, is a triangular one; and let PQP' be
the cutting plane. Next take G' L', a new ground line, making
a less angle than GQP with PQ the horizontal trace of the cut-
ting plane. Now the ground line is always the horizontal trace
of a vertical plane of projection; and two intersecting vertical
planes will intersect in a vertical line, one point of which is the
intersection of their respective ground lines. That is, in the
figure, the angle GTG' indicates a revolution of the primitive
vertical plane of projection, about a vertical axis at T, through
the diedral angle GTG'.
Remembering, now, that the different vertical projections of
the same point are, in space, at equal heights above their several
ground lines, we have A"B"C"-V" as the new vertical projec-
tion of the given pyramid, V" being at the same height above
G'L' that V' is above GL. Q', where the trace PQ meets the
new ground line, is one point of the trace of the cutting plane on
the new vertical plane. Likewise, the cutting plane meets the in-
tersection of the two vertical planes, in a point, TT'—T", which,
in space, is common to its traces on the two vertical planes.
Therefore, TT', when the vertical plane is revolved backward
to coincide with the horizontal plane, about G'L' as an axis,
takes the position TT", and T" is another point of the new
vertical trace T" Q', which can next be drawn.
The figure of intersection in space cannot change, so long as
the relative position of the pyramid and cutting plane with
respect to each other remains unchanged, hence in this case, the
horizontal projection of any point of the figure of intersection
should be the same, whichever vertical projection is referred to
in finding it. Accordingly, a, the horizontal projection of the
point in which the edge VC-V'C' pierces the plane PQP'
coincides with a, the horizontal projection of the same point in
space, in which the same edge, represented at VC-V"C",
pierces the same cutting plane, represented at PQ'P".
EXAMPLES.-1°. Let the cutting plane make acute and obtuso angles with
the planes of projection; and let the pyramid be a regular square, hexagonal,
or octagonal one,
Ex. 2. Let the axis of the pyramid be horizontal.
Ex. 3'. Let the horizontal trace of the cutting plane intersect the base of the
pyramid.
DESCRIPTIVE GEOMETRY.
107
Ex. 4°. Take any plane, perpendicular to the vertical plane, for a new hori-
zontal plane of projection.
Ex. 5. Let the pyramid be either partly or wholly in any other angle than
the first.
Ex. 6. Substitute any prism for the pyramid.
Ex. 7. Let the pyramid be inverted.
PROBLEM LVI.
To find the intersection of a vertical right cylinder of revolu-
tion with a plane, perpendicular to the vertical plane of
projection; together with a tangent to the curve.
In Space. The given magnitudes being placed in the sim-
plest position in relation to the planes of projection in this
problem, the general solution (161) may be abridged as follows.
Since the cutting plane is perpendicular to the vertical plane,
and since the required intersection is also on the surface of the
cylinder, its vertical projection will be that portion of the ver-
tical trace of the cutting plane which lies between the extreme
elements of the vertical projection of the cylinder. Again,
the cylinder being vertical, its convex surface, and therefore
the required intersection, being on that surface, will be hori-
zontally projected in the circle which is the horizontal projec-
tion of the cylinder.
In Projection.-Pl. X., Fig. 90. AB-A'B'A"B" is the
given cylinder; PQP' is the given cutting plane; a'b'-Am Bn
are the projections of the required curve of intersection, and
any points,as and ?', taken on these projections of the curve,
are the two projections of one of its points.
165. Though the foregoing is sufficient for this case, it will
be a useful exercise to state all the steps of the general solution,
thus (161, 162):
TN'm' is an auxiliary profile plane. It cuts two vertical ele-
.ments from the cylinder, whose horizontal projections are m
and T, and which are vertically projected at m'N'. It also
cuts a horizontal line from the given plane, whose vertical pro-
jection is T' and whose horizontal projection is N'T. This
line, being in the same plane with the elements m-m'N' and
T-N'm', intersects them at TT' and mT', which are therefore
two points of the required intersection.
108
DESCRIPTIVE GEOMETRY.
By proceeding similarly with other auxiliary planes, as n''n'''n,
we shall find other points, as nn' and on', of the required inter-
section.
166. To construct a Tangent Line at the point TT', of the
curve of intersection of a plane PQP' and a vertical cylinder-
Pl. X., Fig. 90 (132) and (147).
The cylinder being vertical, the tangent plane will be verti-
cal, and TL, tangent to the horizontal projection of the cylin-
der, is its horizontal trace and also the horizontal projection of
the tangent line. The horizontal traces of the tangent and cut-
ting planes intersect at L. The vertical projection of the tan-
gent line LT, is QT', which coincides with the vertical trace of
the cutting plane, since it lies in that plane, and that plane is
perpendicular to the vertical plane.
EXAMPLES. —1°. Let the cylinder be parallel to the ground line.
Ex. 2°. Let the cylinder be perpendicular to the vertical plane, and the given
cutting plane parallel to the ground line.
Ex. 3. State the solution (165), when the auxiliary planes are taken parallel
to the vertical plane.
The Conic Sections.
167. A plane through the vertex of a cone can have three,
and only three, essentially different positions; which, taking
the simplest first, are these:
1º. It may contain only the vertex.
2º. It may be tangent to the cone, and thus contain one ele-
ment of the cone.
3º. It may be a secant plane, containing therefore two ele-
ments of the cone.
168. Now any plane parallel to the first position, MN, Pl.
VII., Fig. 75, cuts the cone in the curve called an ellipse.
Any plane, hhp, parallel to the second position, cuts the cone
in the curve called a parabola.
Any plane, CDE, parallel to the third position, abV, cuts the
cone in the curve called a hyperbola.
169. From (168), and Pl. VII., Fig. 75, the following table
of conic sections may be readily understood:
DESCRIPTIVE GEOMETRY.
109
The conic section.
Its plane may be
1º. Parallel to a plane containing only the vertex:
ELLIPSE.
Therefore, 1. Cutting but one nappe.
2. Cutting all the elements.
3. Making a greater angle with the axis
than the elements do.
2º. Parallel to a tangent plane: PARABOLA.
Therefore, 1. Cutting both nappes; one actually, and
the other in one point at infinity, on
the element of contact of the tangent
plane.
2. Cutting all the elements but one; the
parallel element of contact of the
tangent plane.
3. Making the same angle with the axis,
or the base, that the elements do.
3º. Parallel to a secant plane containing two elements :
HYPERBOLA.
Therefore, 1. Cutting both nappes.
2. Cutting all the elements but two; those
in the parallel plane through the
vertex.
3. Making a less angle with the axis than
the elements do.
The Special, or Elementary Conic Sections.
170. These are: 1°. The point, cut from a cone by a plane,
MN, Pl. VII., Fig. 75, containing only its vertex.
2º. The straight line, cut from any plane, considered as a
cone (96) by another plane; or from any cylinder or cone, by
a tangent plane.
3°. The circle, cut from any cylinder or cone of revolution,
by a plane perpendicular to its axis.
4°. Parallel straight lines, cut from a cylinder of revolution
by any secant plane parallel to its axis.
5°. Intersecting straight lines, cut from a cone of revolution
by any plane cutting the surface through the vertex.
171. These five sections, with polygons, or the plane sections
of prisms and pyramids, are the whole subject matter of ele-
mentary plane geometry, and may therefore be called the ele-
mentary conic sections.
110
DESCRIPTIVE GEOMETRY.
It is evident from (169) that the five sections just described,
together with the limited straight line from any point of an
element to the vertex, as an extreme case of the ellipse, are all
particular cases of the general conic sections.
THEOREM VIII.
The conic section in all its forms is a symmetrical curve; the
parabola relative to one axis, and the ellipse and hyperbola
relative to two axes, at right angles to each other.
Let VAB, Pl. XI., Fig. 93, be a cone of revolution, whose
axis is in the paper, and from which the three forms of conic
section shall be cut by planes perpendicular to the paper, only
one of which, PQ, is shown, containing the ellipse cc'. It is
evident that the plane of the paper is a plane of symmetry of
the cone and of each of these three planes. Hence its inter-
section, as cc', with each of them is a line or axis of symmetry
of the curve contained in each plane respectively.
The parabola has but one such axis of symmetry, since it is
an open or infinite curve of but one branch. The theorem,
therefore, mainly relates to the proof of a second axis of sym-
metry perpendicular to the first. The demonstration for the
ellipse will show the method of proof, both for it and the
hyperbola.
If the ellipse, cc', has a second axis of symmetry, it must be
perpendicular to the axis cc', at its middle point, and we shall
have the thickness of the cone, or width of the ellipse, equal,
when taken perpendicular to the paper at a and a', points equi-
distant from c and c'. That is, take the horizontal planes
through a and a', containing the circular sections, half shown
at bde and b'd'e', and it is to be shown that ad a'd'. In the
semicircles,
ad² = аб хас
2
a'd'² = a'V'×a'e'.
The triangles abc and a'b'e', having equal altitudes, give
abc
ab
a'b' c
a 'V'
(1)
aec'
ае
Likewise,
a'e'c a'e'
DESCRIPTIVE GEOMETRY.
111
Hence
abc aec'
аттех
ab
a'e' c = ab'Xae
ае
de
or by decomposing,
ae
when by cancelling (1)
ae
ae + a'b'
a'e'
a'e' + ab
(2)
abc (aea'b'+ a'b'c') ab
a'b'c' (a'e'ab+ abc ) = a'b'Xa'e'
aea'b'
a'e'ab
since trapezoids of equal altitudes are as the sums of their
parallel bases.
From the second and third members of (2)
Hence
and
ae (a'e' + ab) = a'e' (ae + a'b')
ae × ab = a'e' × a'b'
ad = a'd'.
By making the corresponding figure for the hyperbola in
which the trapezoid aea'b' will become the sum instead of the
difference of aec' and a'b'e', the same demonstration will apply.
PROBLEM LVII.
To construct the projections of the intersection of a perpendicu-
lar plane, with a cone of revolution, whose axis is vertical ;
and to construct the tangent at any point of the curve.
In Space. The vertical projection of the required curve,
according to the conditions of the problem, will be that por-
tion of the vertical trace of the cutting plane which is included
between the extreme elements of the cone, as seen in vertical
projection. To find the horizontal projection of any point of
the curve, merely project its vertical projection, either upon the
horizontal projection of the element drawn through it, or upon
the horizontal projection of the horizontal circle containing it,
as may be most convenient.
In Projection.-1°. Without auxiliary planes.-Pl. X., Fig.
91. Let V-AeBE and V'-a'b'A'B' be the projections of the
cone of revolution, inverted, to give additional variety to the
problems, and let UQP' be the given cutting plane.
We know at once that p'' is the vertical projection of the
required intersection. In finding definite points, as tť' and yt',
of this intersection, assume t' at pleasure, then V't'u' is the ver-
tical projection of the two elements, one on the front, and one
112
DESCRIPTIVE GEOMETRY.
Jo
on back of the cone, which contain those two points of the
curve, of which t' is the vertical projection. Projecting u' at u
and u”, Vu and Vu" are the horizontal projections of the same
elements, on which therefore t' must be projected, giving t and
g for the required horizontal projection of the points projected
at t'. Other points may be similarly found.
The second method applies most usefully to points like
and n', each of which may be projected down upon the hori-
zontal projection of the horizontal circle drawn through it.
2º. With auxiliary planes.-As in Prob. LVI., while the
foregoing is enough, it will be a useful exercise to make the
application of (161, 162) to this case in full form, thus:
To find points of the required intersection, we may pass an
auxiliary plane through the vertex of the cone and perpendicu
lar to the vertical plane of projection. Such a plane will cut
two rectilinear elements from the cone and a straight line, L,
perpendicular to the vertical plane, from the cutting plane.
This line will intersect each of the elements, cut from the cone
by the same auxiliary plane, in a point of the curve of intersec-
tion. Both of these points will be vertically projected in the
point which is the vertical projection of the line, L, containing
them; viz., the point of intersection of the vertical traces of
the auxiliary and cutting planes.
Thus, EE'u' is an auxiliary plane, perpendicular to the ver-
tical plane of projection. This plane cuts from the cone the
elements, Vu-V'u' and Vu"-V'u', whose horizontal projec-
tions are determined by joining V with the points E" and T,
where the horizontal trace E'E of the auxiliary plane cuts the
horizontal trace TE"a of the cone; or with u and u”, projected
from u'. This plane also intersects the given plane UQP' in
the line tỳ-t' which intersects the elements just named in the
points tt' and gt' of the required intersection.
172. 3.º To construct a tangent line at the point tt', Pl. X.,
Fig. 91, of the intersection of the plane P'QU with the cone.
The horizontal trace of the tangent plane along the element
containing t, must be tangent to the horizontal trace alT
of the cone at T. Hence UT is the horizontal trace of
the tangent plane. The horizontal traces of the tangent and
cutting planes intersect at U, which is therefore one point of
the tangent line. But t is another point, hence Ut is the hori-
zontal projection of the required tangent line. Its vertical pro-
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DESCRIPTIVE GEOMETRY.
113
jection, Q, coincides with the vertical trace of the cutting plane,
because that plane is perpendicular to the vertical plane.
EXAMPLES.-1°. Let the cone be the lower nappe of a complete cone.
Ex. 2°. Let the auxiliary planes be radial planes through the axis.
Ex. 3°. Let the axis of the cone be horizontal.
Ex. 4. Let the cone be placed with both projections of its axis perpendicu-
lar to the ground line, and then first with only its vertex in the vertical plane;
second, tangent to that plane; third, with any two of its elements in that
plane. Then find the intersection of each of these positions of the cone with
a plane parallel to the vertical plane. [This construction will evidently show
each of the three conic sections in its true form.]
Ex. 5'. In 4', let the axis of the three positions be horizontal.
PROBLEM LVIII.
Ilaving given a cylinder and a cone of revolution, whose axes
are oblique to both planes of projection, by their axes and
a diameter, to find their horizontal traces.
In Space. The given data being the diameter of the base
and the projection of the axis, in either case we can imme-
diately represent the two diameters which are parallel to the
planes of projection. Thence the four elements through their
extremities can be drawn, and the traces of these can be found.
Finally, by finding the axes of the traces, any number of points
in them can be found.
In Projection.—1°. A Cylinder: Four Points.-Let Oo—
O'o', Pl. IX., Fig. 84, be the indefinite axis of a cylinder, OO'
the centre of its base. Then AB—A'B', and jl-j'l', are those
diameters which are parallel to the planes of projection, and
therefore show their true length; each on the plane to which it
is parallel. Then parallels to the axis, through AA', BB', jj',
and W', evidently are the four extreme elements, whose horizon-
tal traces, as K and H, found by Prob. XII., are four points of
the required trace of the cylinder.
The transverse axis of this trace.-This is here found by re-
volving the axis about its horizontal projection Oo, and into
the plane H, by making 00″ = O'ɑ. Then D'M, parallel to
the revolved axis, O'o, and at a distance from it equal to the
radius, OA, of the cylinder, is the revolved position of the
S
114
DESCRIPTIVE GEOMETRY.
highest element of the cylinder, whose horizontal projection is
DM, and horizontal trace is M. Then MN 2Mo is the re-
quired transverse axis. IIK is the shorter axis of the trace
MIINK of the cylinder.
2°. A Cone: Four Points.-Let Vo-V'o', Pl. XVI., Fig.
128, be the axis of a cone of revolution; cd-c'd', and ab-a'b',
the diameters of its base, parallel to the planes of projection.
The horizontal traces, as CC', of the elements, as Ve'-V'e,
through the extremities of these diameters, will be four points
of the like trace of the cone.
The axes of the trace.-These might be found as in the case
of the cylinder. But, for variety, let the cone be projected
upon a new vertical plane, G,L,, parallel to its axis, Vo-V'o',
whose new vertical projection is therefore Vo". The new
vertical projection of the base is "y"; ed or a''; and giving
the extreme elements V"E" and V"F". Hence E"F" is the
new vertical projection of the required horizontal trace, and
true size of its transverse axis, EF, and nn" is its centre. The
conjugate axis at n" is found by revolving the section parallel
to the base, at n'", till parallel to the new vertical plane, as at
hn", when n"n"" is one half of this axis; whose horizontal
projection MN, appears by making MnN = "n"".
>
The horizontal trace of the given cone is therefore the ellipse
whose axes are EF and MN.
The extreme elements of a cone, not being parallel to the axis,
do not contain the extremities of the diameters, cd and a'b', but
are, in horizontal projection, tangents from V to the horizontal
trace of the cone; and, in vertical projection, those whose hori-
zontal traces, as T, are the points of contact of projecting lines,
as TT'. These and the elements through ad' and bb', so nearly
coincide in vertical projection in this figure, that they are shown
separately only at V'T' and V'B'.
EXAMPLES. -1°. Let the axis of the cylinder cross any one of the four angles.
Ex. 2. Let the axis of the cone cross any one of the angles.
Ex. 3. Find the vertical trace of a cylinder.
Ex. 4. Find the vertical trace of a cone.
Ex. 5. Find the vertical trace of either, when the projections of its axis are
perpendicular to the ground line.
Ex. 6. Find the transverse axis of the trace, by revolving the cylinder or
cone, till its axis is parallel to a plane of projection.
173. Having gained from Probs. LVII. and LVIII., with
DESCRIPTIVE GEOMETRY.
115
the solution of examples 4° and 5° of Prob. LVII., a clear idea of
the forms of all the curved sections of a cylinder and a cone, it
will be convenient here to demonstrate some of their properties,
before proceeding to the more general problems, in which any
of these forms may occur.
And, first, having indicated by Theo. VIII., the identity of
the oblique closed (110) section of the cylinder and the cone,
in showing its symmetry relative to two axes in both cases, this
identity will now be completely proved by establishing the usual
definitive property of the ellipse, viz., that concerning the rela-
tion of its points to its foci.
THEOREM IX.*
The plane section of a cone of revolution, when all its elements
are cut, is a curve, such that the sum of the distances from
any point on its circumference to two fixed points within,
is constant, and equal to the greatest chord of the curve.
Make S the
VB, and ab;
Let VAB, Pl. XI., Fig. 94, be a cone of revolution, and PQ
a secant plane perpendicular to the meridian plane, VAB, of
the cone, and containing the ellipse (168) abp.
centre of a circle, tangent to the three lines VA,
then let it revolve along with VA, about the vertical axis, VO,
of the cone. It will then generate a sphere having a horizontal
circle of contact, ce, with the cone generated by VA, and only
the point of contact, f, with the plane PQ; which point ƒ will
be on the axis ab of the curve.
Likewise, make s the centre of a sphere similarly tangent to
the cone and plane, but above the plane, PQ.
Now all tangents to the same sphere from the same point, as
Ve, Vm, Ve, etc., are equal; and all the segments of elements
of a cone of revolution, included between parallel right sec-
tions, as ce and dy, of that cone are equal.
Hence, if p be any point on the ellipse ab, we have
pf (in space) = pm, and pf, = pn;
whence pf+pf₁ = pm + pn = mn = cd = eg. (1)
But c = ca + ad = aƒ + af₁ = 2aƒ±ƒƒr
af
(2)
*From Quetelet and Dandelin, Belgian gcometers, as given by OLIVIER.
116
DESCRIPTIVE GEOMETRY.
and
eg
hence as cd eg; af =
and substituting in (2)
or
pf +pf
lf + bf₁ = 2bƒ₁ + ff.
bf,
bf + af ab,
ci
pf + pf₁ = ab.
The fixed points, f and f, are called the foci of the ellipse ab.
174. Thus the cllipse, abp, is a curve, such that the sum of
the distances of any one of its points, from two fixed points
within, called foci, is equal to the longer axis, ab, of the curve.
175. It immediately follows that the points of the ellipse exist
in sets of four, symmetrically placed with respect to ab, and a
perpendicular to it at its middle point. That is, the ellipse has
two axes of symmetry, at right angles to each other, called the
transverse or major axis, ab, and the conjugate or minor axis
(Theo. VIII).
176. The separate proof by the method of Theo. IX., that
the oblique section of a cylinder of revolution is an ellipse, is
very similar to the above, as the two tangent spheres are drawn
in the same manner as there described, and is left as an exercise
for the student. But as no reference was made to the angle at
the vertex, we may infer that the proof is independent of that
angle, and therefore general, and true for the cylinder as well
as for any other form of the cone.
The similar proof of the fundamental property of the para-
bola is as follows:
That for the hyperbola is so similar to Theo. IX., that it is
left as an exercise for the student.
THEOREM X.
Each point of the parabolic section of a cone of revolution is
equally distant from a fixed point, the focus; and a fixed
line, the directrix.
Let VAB, Pl. XI., Fig. 95, be a cone of revolution cut by a
secant plane PDL, parallel to the tangent plane on VA, and
hence cutting the cone in a parabola. S is the centre of a
sphere, tangent to the cone in the circle cd, and to the plane
PDL at the point f. Let Vp be the element of the cone
DESCRIPTIVE GEOMETRY.
117
drawn through any point, p, of the parabola, and intersecting
the circle cd at m.
The planes cd and PDL intersect in a line, LD, parallel to
CF or pq. Now pf = pm (in space), these being tangents to
a sphere from one point, and dh = da + ah. But, by reason
of the isosceles triangles, dDa and qah
pf (in space)
= da + ah
=qa + aD = qD; or pf = qD.
LD is thus the directrix, and f the focus, from both which
any point, p, of the parabola is equidistant.
177. In Pl. XI., Fig. 94, the ellipse is determined by its axis,
ab, and foci, fand f. But for every new value of fs, new lines
can be drawn from a and b to the new sphere thus given, which
will be elements of a new cone of revolution containing the
ellipse abp, and having the new position of sV for its axis; for
every cone tangent to an inscribed sphere must be a cone of
revolution. That is, finally, as Theorem IX. required no par-
ticular value of the angle aVb, it follows that an infinite num-
ber of different cones of revolution may contain the same ellipse.
The like' is evidently true for the parabola, as also for the
hyperbola, as will appear on making the construction for the
hyperbola, analogous to Pl. XI., Fig. 94.
PROBLEM LIX.
To construct a cone of revolution from which a given conic
section can be cut.
In Space.-In the case of the ellipse and hyperbola, any cone
tangent to a sphere which is tangent to either of these figures
at a focus, will be a cone of the kind required.
The construction is made for the ellipse and parabola.
1º. The Ellipse.--Let E,E', Pl. XV., Fig. 125, be the given
ellipse. At either focus, as S,s', erect a perpendicular, S-—8′S′,
to the plane of the figure, and make any point, S', of this per-
pendicular, the centre of a sphere, tangent at Ss. Then tan-
gents to this sphere from the extremities of the transverse axis,
ab-a'b', will be elements of a cone of revolution, V,Eab-
V'a'b', containing the given ellipse. Tangents from V to E
DESCRIPTIVE GEOMETRY.
118
will also be found tangent to the horizontal projection of the
sphere.
2. The Parabola.-Let PQP', Pl. XI., Fig. 97, be the plane
of the given parabola, whose vertex is a, and focus, F. Then
take any point, S, or the perpendicular, FT, to PQP', at F, as-
the centre of a sphere, to which draw the tangent AV parallel
to P'Q (168), and aV. These will determine a cone of revo-
lution containing the given parabola.
To find that cone whose axis is vertical, as in the figure, take
S at the intersection of FT with aS, parallel to the ground line.
THEOREM XI.
The foci of parallel sections of a cone of revolution are in a
straight line through the vertex.
Let VAD, Pl. XI., Fig. 96, be the vertical projection, and
meridian section of a cone of revolution and AB and ab the
vertical traces of parallel planes, cutting the cone in parallel
sections, AB and ab, which will be ellipses, parabolas, or hyper-
bolas, according to the direction of the planes.
S and s are the centres of the inscribed tangent spheres whose
points of contact with the planes AB and ab are (Theos. IX., X.),
the foci of the given parallel sections.
Now SR and sr are parallel, being perpendicular to VD.
Also, SF and sf are parallel, being perpendicular to the parallel
lines AB and ab.
Hence,
But as
Vs: VS :: sr: SR.
=s,
s7, and SF = SR
sf: SF.
of
Vs: VS
Hence, VƒF is a straight line.
Elementary properties of the Ellipse, derived from its
circular projection, with consequent constructions.
178. In Pl. X., Fig. 90, every point and line of the circular
base of the vertical cylinder of revolution is the horizontal
projection of a corresponding point and line of the ellipse, a''.
From the principles of (126), we therefore have the proper
DESCRIPTIVE GEOMETRY.
119
ties found in Theo. VI., and some others; here assembled for
convenient reference.
1º. Every diameter of an ellipse, as of a circle, bisects a sys-
tem of parallel chords, to which that diameter is said to be con-
jugate.
2°. The centre is the intersection of any two such diameters;
all of which bisect each other, at that point, in the circle, and
therefore in the ellipse.
3°. The projecting cylinder of the ellipse being vertical, one
of the diameters of the curve will be horizontal, the other, the
most inclined. These are perpendicular to the chords which
they bisect (Theo. II.), are respectively the least and greatest
diameter, and are called axes.
4°. When a diameter is one of the set of chords bisected by
another diameter, the pair are called conjugate diameters.
5º. Conjugate diameters are, each, parallel to the tangents at
the extremities of the other.
6°. Diameters of the circle which are equally inclined to the
projection of an axis of the ellipse, are the projections of equal
diameters of the ellipse. There will evidently be one pair of
equal conjugate diameters.
7°. Lines from any point of an ellipse to the extremities of
any one diameter are called supplementary chords. They are
parallel to a pair of conjugate diameters, and, if perpendicular
to each other, to the axes.
8°. Tangents at the two extremities of any chord meet on
the diameter (produced) which bisects that chord.
9°. Equal segments of any chord parallel to the chord join-
ing the points of contact of two tangents, will be intercepted
between the curve and these tangents, since the like is true of
the circular projections of the ellipse.
Constructions upon the Ellipse.
179. 1°. To find the centre of a given ellipse. Take the mid-
dle point of any diameter (178, 1°).
2°. To find the axes of a given ellipse. Find any diameter,
D, and on it describe a semicircle, which will cut the ellipse at
a point p. The supplementary chords from p to the two ends
of D, will be at right angles, and hence parallel to the axes ;
which will also pass through the middle point of D. (178, 7°.)
120
DESCRIPTIVE GEOMETRY.
*
3º. To draw a tangent at a given point, t.-Joint and the
centre, giving the diameter through t. Then the tangent at t
will be parallel to the conjugate of this diameter.
Otherwise Find the diameter through t, and draw any chord
C parallel to it, and then the required tangent will be parallel
to the chord, supplementary to C.
4°. To draw a tangent, parallel to a given line.—Find a
diameter parallel to the line, and the required tangent will be
drawn through the extremity of its conjugate.
180. The principles of projections also yield the following
results:
The projection of a straight line is equal to the length of the
line, X, the cosine of the angle made by the line with its pro-
jection.
Now any plane figure may be conceived of as the sum of
consecutive parallel lines limited by its perimeter. Hence, the
area of its projection will be equal to its own area, X, the cosine
of the angle between its plane and the plane of projection.
Therefore, if the projections of figures in the same plane are
equal, the figures themselves will be equal. For example—
181. 1°. The circumscribing squares of a given circle are
equal; therefore, the circumscribing parallelograms of the
ellipse, projected in the circle, are equivalent.
2°. The diagonals of all rectangles inscribed in a circle are
diameters; hence, the diagonals of all parallelograms inscribed
in an ellipse are diameters.
3°. All the inscribed squares in a circle are equal, and their
diagonals being perpendicular to each other, are conjugate
(178, 1°). Hence, all the inscribed parallelograms in an ellipse,
whose diagonals are conjugate diameters, are equal.
4°. The area of a circle, C, whose radius is R, is R'. Let
A and B be the semi-axes of the ellipse E, whose projection is
C, and whose plane makes an angle a with that of C; and let
B be parallel to the circle C. Then RB = A cos a
π.В.A cos a
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DESCRIPTIVE GEOMETRY.
121
Elementary properties of the Parabola, with consequent
constructions.
182. As the parabola cannot be cut from any cylinder having
a circular or elliptic base, its properties are derived from the
corresponding ones of the ellipse, by treating both as sections
of a cone.
Then conceive of a cone of revolution, VCBD-V'C'B'D',
Pl. XI., Fig. 99, derived, for convenience, from the position
V"A"B", having one element vertical. At any point, gg', of
the vertical element, AV-A'V', conceive a tangent, TT', to
the cone. Also a tangent plane, parallel to TT', and deter-
mined by the parallel, VH-V'H', through VV', the vertex.
He is the trace of this plane on the cone's base, and Ve-V'é
is its element of contact. Finally, conceive two planes, P and
E, through TT'; one, P, parallel to the tangent plane Hc; the
other, E, perpendicular to the vertical plane, HA, tangent along
the element, AV-A'V'. The former will cut the cone in a
parabola, abƒ—a'b'f', whose vertex, ff', is the intersection of
ef—e'ƒ', parallel to Ve-V'e', with the opposite element, Vd—
V'd'; the latter is an ellipse, EFG-E'F'. And TT', the in-
tersection of the planes P and E, is also the common tangent
to the two curves at gg'.
This understood, we proceed as follows:
183. Any secant plane through the vertex VV', and parallel
to TT', the tangent at gg', will now cut a chord, both from the
ellipse and from the parabola, and parallel to that tangent.
Other such planes will cut, collectively, a series of chords of
each curve, all parallel to T,T', and hence to each other. But
all of these being parallel chords of a cone of revolution, and
the two chords of each pair, one in the ellipse and one in the
parabola, being in the same plane through the vertex, are in-
cluded between the same elements of the cone. Hence it
follows that a plane through the vertex, which will intersect the
ellipse in a diameter bisecting its parallel chords, will also inter-
sect the parabola in a line which bisects its corresponding chords.
That is, the diameters of a parabola are straight.
1º. The tangent plane He, and the plane of the parabola, being
parallel, are cut by the plane AH in two parallel lines, which are
TT' and HV-II'V'. But the plane, E, of the ellipse, being
122
DESCRIPTIVE GEOMETRY.
parallel to HV-H'V', cuts the tangent plane He in a line,
Gt—G't', parallel to HIV-H'V', and therefore to TT', and
which is therefore a tangent to the ellipse, since GG' is the
point of the ellipse on the element of contact, Ve-V'e', of the
plane He. Hence GA is a diameter of the ellipse conjugate
to the tangent TT', and to the chords and diameter parallel to
it. Now the plane of the parabola being parallel to the cle-
ment Ve-V'', the plane of the elements Ve and VA, and
containing the diameter, GA, of the ellipse, will cut the plane
of the parabola in the line, rg-r'g', parallel to ef—郔.
2º. That is, the chord-bisecting diameters, as rg, of the para-
bola, corresponding to the like diameters, as GA, of the ellipse,
are parallel to its axis.
3º. Hence, the parabola has no centre, or, in other words, its
centre is at an infinite distance from the vertex, on the axis.
4°. One of each pair of supplementary chords is parallel to
the axis, and is therefore also a diameter.
5º. The axis is merely that one of the diameters which is
perpendicular to the chords which it bisects.
Hence tangents
at the extremities of these chords will meet on the axis, they
being symmetrical relative to it. But the projections of the
axis and of its bisected chords, will not generally be perpen-
dicular to each other, and hence, as in the figure, will be only
a diameter, and its series of chords, of the parabolic projec-
tions. Yet, if the lines meet at a point, in space, they will do
so in projection. IIence the tangents at the extremities of
any chord will meet on the diameter conjugate to that chord.
Constructions on the Parabola.
184. 1°. Having given a parabola to find a diameter, bisect
any two parallel chords.
2°. To find the axis, draw a chord perpendicular to a diame-
ter; and a parallel to that diameter through the middle point
of the chord will be the axis.
3°. To construct the tangent line at a given point of the
curve, draw three equidistant parallel diameters, the middle one
of which shall pass through this point. Then a line through
the vertex of this diameter, and parallel to the chord connect-
ing the vertices of the others, will be the required tangent.
DESCRIPTIVE GEOMETRY.
123
Elementary properties of the Hyperbola, and resulting
constructions.
I.-Properties corresponding to those of the Ellipse.
185. Let VATB-V'A'B', Pl. XI., Fig. 98, be a vertical cone
of revolution, cut by the vertical plane CD, parallel to its axis,
in a hyperbola. This plane will thus cut all the elements of
the cone, except the two contained in the parallel plane, AB,
through the vertex, VV', of the cone.
The plane CD may be taken as described, since out of the
infinite number of cones of revolution from which a given
hyperbola can be cut (177), one will have its axis parallel to
the plane of the curve.
The properties of the ellipse derived from those of the circle
may now be elegantly transferred to the hyperbola, as follows:
186. 1°. The hyperbola, cabd-c'a'd g'b'h', evidently consists
of two equal, opposite and infinite branches, having two axes of
symmetry, XZ and Y'O'; for, as the figure is placed, the vertical
projection of the curve is identical in form with the curve itself.
2º. Let PQP' be a plane, perpendicular to the vertical plane,
and tangent to the hyperbola at tt', and cutting the given cone
in an ellipse.
The chord, ef, of this ellipse, is bisected by ot, its conjugate
diameter. Hence e't' = t'f'. Hence m'r' r'n'. That is, as
the like is true for other chords, parallel to m'n', any diameter,
t'V', bisects each of a system of parallel chords, and passes
through the centre, V', of the hyperbola.
3°. The tangents, as h'l', and at s'u', at the extremities t' and
s', of any diameter, as V't', are parallel to the chords, as m'n',
which that diameter bisects.
4°. By reason of the double symmetry of the curve, t'V'=
V's', and lines, m'p' and n'w', parallel to t's', or to any other
diameter, are also chords of the hyperbola, such that the line, as
O'I, bisecting any diameter, t's', and the chords parallel to it,
is straight, and therefore a diameter.
5°. Two diameters, such that each is parallel to the chords
which the other bisects, or to the tangents at the extremities of
the other, are called conjugate, and one, as t's', of any pair of
conjugates, cuts both branches of the curve, while the other,
IO', cuts neither branch.
124
DESCRIPTIVE GEOMETRY.
6°. Supplementary chords join any point of the curve with
the two extremities of the same diameter. They are parallel
to two conjugate diameters, as will appear by constructing the
chord p'w', equal and parallel to m'n', in the upper branch, and
the chord m'p', supplementary to m'n'.
7°. Tangents to the ellipse E, at e and f, will meet at a point,
x, on the conjugate axis, ot, produced. But these tangents lie
in tangent planes to the given cone along the elements Ve and
Vf; hence is a point of the intersection of these planes, and
therefore, t'V' is the vertical projection of their intersection.
But the tangents to the hyperbola at m and n, on the same ele-
ments, are in the same tangent planes, and therefore meet on
their line of intersection. That is, tangents at the extremities.
of a chord, m'n', meet on the diameter, t'V', which bisects that
chord.
II.—Properties peculiar to the Hyperbola.
187. 1°. A'V' and B'V' are, each, tangent to the curve c'a'd'
g'b'h', at an infinite distance each way from V'. Such tangents
are called asymptotes (136).
2°. The tangent, cd-kf", at any point, as tt', of the hyper-
bola, and limited by its asymptote, is bisected at its point of
contact. For this line is evidently also tangent, at tt', to the
ellipse, fl—'l', contained in the plane, PQP'; but, being
parallel to the vertical plane, it is tangent to this ellipse at the
extremity of its shorter axis, ot; and hence t' is the middle
point of 'l'.
3º. Each asymptote may be said to consist of two conjugate
diameters which coincide. For the further t' is removed from
a', the smaller will be the angle between t'V' and its conjugate
10', which are on opposite sides of A'V'; and when t' is re-
moved to infinity on the curve, this angle vanishes, and the two
conjugates fall together on the asymptote A'V'.
=
4°. We have 't't'l, therefore A'' ''; but m'r' r'n',
= r'q'
hence A'm' n'g'. That is, those segments of chords which
=
are included between the curve and its asymptotes are equal.
Constructions upon the Hyperbola.
188. The elementary constructions upon the given hyperbola
DESCRIPTIVE GEOMETRY.
125
are so much like the corresponding ones on the ellipse (179)
that they are left as a—
GENERAL EXAMPLE.-Find the centre, axes, tangent at a given point, or
parallel to a given line, for a given hyperbola.
189. By (187, 4°) any number of points of a hyperbola
can be found, when the asymptotes and one point are given.
For, draw through the given point, let it be m', Pl. XI., Fig.
98, any number of secants cutting both asymptotes. Then the
distance from B'V' to a point of the curve on each secant will
be equal to the known distance on the same secant, from m' to
A'V'.
190. From the preceding constructions, and Theorem VI.,
any parallel or cylindrical projection (24) of a conic section
will be a conic section of the same species.
191. For each new cone of revolution containing a given
section (177), as aph, Pl. XI., Fig. 94, there will be a different
circular section containing a or b. Hence we conclude that
any one of these circles and a given conic section will deter-
mine one, and the only one, cone of revolution that will con-
tain these two curves.
192. Thus Theo. IX. leads to the following problem, and
thence to a re-demonstration of the fundamental property of
Theo. IX., and of many others, in a novel and elegant manner,
more obviously framed in the spirit of descriptive geometry;
that is, according to the methods of ordinary, or perpendicular
projection.
The Ellipse.
PROBLEM LX.
To place a given ellipse upon a cone of revolution, whose base
shall be in the horizontal plane, and shall have a common
tangent with the given ellipse.
In Space. This problem is solved by the principles of (172),
applied in an inverse order, and by considering, 1st, that the
126
DESCRIPTIVE GEOMETRY.
element of contact of a tangent plane to a cone of revolution is
perpendicular to the trace of that plane upon the plane of the
cone's base; and 2d, that the projections of elements upon the
plane of the base are equal.
In Projection.—Let E,E', Pl. XII., Fig. 102, be the given
ellipse, E being constructed upon given axes, AB and CD, by
methods already given (Prob. XLIII).
Then AZA'B' will be the plane of the ellipse, and let A' be
the common tangent to it and to the base of the required cone.
Draw any other tangent to the ellipse; for convenience, Cl,
then, its intersection with the trace A' of the plane of the
curve, is a point in the horizontal trace of the tangent plane to
the required cone along the element through C.
But this element and this trace are perpendicular to each
other, and the elements limited by the base are equal, in reality
and in horizontal projection, and these conditions are fulfilled
together by drawing the element, VCo, and the trace, ko, of the
tangent plane, respectively a tangent and a radius of the circle
whose radius is A. For A'k is the horizontal trace of the tan-
gent plane along the element at AA', whose horizontal projec
tion is AV. But VA and Vo are equal, being tangents to the
circle, Ako, and as V and I are on lines, VA and A, at right
angles to each other, the circles A and VA intersect at right
angles, making the element Vo perpendicular to the trace ko,
as well as equal to VA.
Hence V is the horizontal projection of the vertex, and cir-
cle VA that of the base, of the required cone; and the like
construction and reasoning will apply if we take any point, n,
instead of C. V' is the intersection of II'B'V', vertical projec-
tion of the element HBV, with the projecting line VV'; and
V'A'II' is the vertical projection of the required cone.
Since two tangents can be drawn from C to the circle, ZA, the
complete construction gives a second cone whose vertex is vo',
and base, the circle vA.
193. Pl. XII., Fig. 103, shows that by applying the construc-
tion of Fig. 102 to the case where the horizontal projection, E,
of the ellipse EE' is a circle, the cone of Fig. 102 necessarily
becomes a cylinder, which further indicates the identity of the
ellipse, whether considered as cut from a cylinder or a cone (176)
of revolution.
DESCRIPTIVE GEOMETRY.
127
THEOREM XII.
The horizontal projection of the transverse axis of an ellipse,
cut from a vertical cone of revolution, is the transverse
axis of the horizontal projection of that ellipse; which
therefore cannot be a circle.
Let ab, Pl. IX., Fig. 85, be an ellipse, cut from the cone
VAB, by a plane ab, perpendicular to the meridian (14) plane
VAB. The middle point, c, of ab is the vertical projection of
the conjugate axis, whose half length, cd, is found by means of
the circular section fh, through c. Drawing ae and bg, perpen-
dicular to fh, gives eg, the length of the horizontal projection
of the transverse axis, ab.
Now by reason of the obviously equal triangles, acb and bhg,
eg = fh, = 2km. But the conjugate axis at e 2ed, which
will plainly always be less than 2km, which proves the theorem.
=
194. The ellipses, ab and pq, being symmetrical with respect to
the axis, Vk, of the cone, are equal; and ellipse, br, being paral-
lel to pq, is similar to it, and hence to ab. That is, ellipses as
ab and br, which make equal angles with the circular section
bq, are similar.
But as cd nearly equals mk, it thus appears that there will be
quite a wide angular range, abr, of positions within which
ellipses will be nearly circular in horizontal projection. And
the less the angle at V, the wider will be this range.
The horizontal projection of an ellipse as ub, would be dis-
tinctly elliptical, having nl and 2st for axes, where st is found
as cd was.
THEOREM XIII.
The fixed points, F and F₁, called foci, exist only for the trans-
verse axis of the ellipse.
All the tangents to any circular section of a vertical cone of
revolution are horizontal. Only two of the tangents to the
elliptical section of the same cone are horizontal, viz., those at
the extremities of its transverse axis. Hence only the two cir-
cular sections through these points have a common tangent with
128
DESCRIPTIVE GEOMETRY.
the ellipse, as at AA', Pl. XII., Fig. 102, giving their centres,
F and F₁, on AB, in horizontal projection.
Now, m being the intersection of the tangents at n and A,
we have in greater than mA wherever n is taken, so that when
the circle mA is tangent to AB, at A, tangents to it can be
drawn from n. And the like is true of the circle A. But if
the circle with centre, k, be tangent to CD at C; the point A
will be within it, so that no tangents to it from A being possi-
ble, no points analogous to F and F, can be found on CD.
Finally, for any other diameter, a circle, with centre m, and
drawn through n, would not be tangent at n to the diameter
nO; also, the centre of any circle, tangent to the ellipse at n,
would not be on the diameter O, as F is on the diameter AB.
Hence the points F and F, are found only on AB.
PROBLEM LXI.
Having given the axes of an ellipse, to find its foci.
Draw
Let E, Pl. XII., Fig. 104, be the ellipse. Draw the tangents
at the extremities, a and b, of its axes, and meeting at c.
the circle with centre c and radius ca, and the two tangents to
it from b, which, by the last theorem, will give the required
foci, F and F₁.
The usual metrical construction can be derived as follows
from the one just given, based on relations of position.
Draw ed to the contact d of bF. Then cdb and FOb are
equal triangles, being evidently similar, while cd = Ob, since
each equals ca. Then calling the semi-axes, Oa, (= be = Fb ;)
and Ob, respectively A and B,
OF = db = √ be²ed² = √ A²—B².
THEOREM XIV.
In the ellipse, the sum of the distances from any point of the
curve to the foci is constant, and equal to the transverse
axis of the curve.
Let the ellipse, E, Pl. XIII., Fig. 110, be the horizontal pro-
jection of the equal ellipses, E' and E", in vertical projection.
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DESCRIPTIVE GEOMETRY.
129
These ellipses are cut by the planes, AA'B' and Bl'a', from the
equal cones of revolution, whose vertices are VV' and vv; and
V and v are the foci of the ellipse E.
Let p be any point of E, whose vertical projections are p'
and p". Then draw the elements, p'o' and p'V', one on each
cone, and whose horizontal projections are pv and pV. The
true length of p'v' is P'v', and that of p''V' is P'V'. Now
E' and E", being equally inclined to the horizontal plane,
C'p'p" and C'a'A' are similar isosceles triangles, evidently
so related that a'm' A'n'. Hence a'P" A'P'. But a'V'
= v'd' =v'B′ = v′b″. Therefore, the line pv-p'v' + the
line pV— p'V' = P'v' + P''V'
That is,
= P'v' + A'P' + v'b'' — A'V".
(pv—p'v') + (pV—p'V') = A'V".
But as these lines are all equally inclined, their horizontal
projections will also be equal, and we have pv + pV = AB,
which expresses the theorem to be proved, and which is usually
taken as the fundamental definition of the ellipse when con-
sidered independently, or not as the section of a cone.
PROBLEM LXII.
To construct points of an ellipse, having given its transverse
axis and foci.
Let AB, Pl. XIII., Fig. 110, be the given transverse axis,
and F and f the given foci.
Assume any point as e, at pleasure, between F and f; and
with Ae and Be in succession as radii, describe arcs, both from
F and f as centres, and the ares, shown at q and 7, and drawn
with the different radii, as Ae and Be, will intersect so as to
give four points, as P, 2, r and s, of the ellipse. Thus each
point, as e, with its corresponding pair of radii, gives four points
of the curve.
The point e must be between F and f, because FA and FB
are evidently the least and greatest radii that can be used.
EXAMPLES.-12. Compare this construction with those of Prob. XLIII., for
the same axes, when the minor axis is less than half of the major one.
2º. Do., when the minor axis is greater than half of the major one.
9
DESCRIPTIVE GEOMETRY.
130
THEOREM XV.
Each half, or branch, of the ellipse has a directrix, such that
the ratio of the distances of any point of the curve from
the focus and directrix of either branch, is constant and
less than unity.
First Method.-Let EE', Pl. XII., Fig. 108, be an ellipse, a
section of the cones of revolution, Vag-V'a'g' and vah-v'a'h'
(Prob. LX.), made by the plane aa'p'. Its foci are F and f
(Prob. LXI.)
V'p' and 'N' are horizontal planes through the vertices VV'
and vv', which cut the plane, aa'p', of the ellipse, in the lines
p' and N', whose horizontal projections, pq and MN, are the
directrices of the ellipse E.
To prove this, take any point, xx', at pleasure, upon the ellipse,
and thence draw a perpendicular, ap-x'p', to the line pq-p'.
As xp-'p' is a co-parallel, it shows its true length; and all
parallels to it make the same angle, a, with the horizontal plane,
that the plane aa'p' does. IIence
xp x'p' cos a.
A horizontal plane, x'm, will cut the element V'g' of the
cone VV' at m. All the elements of both cones make the
same angle, 6, with the horizontal plane; hence V'm shows the
true length of V-V'a', and
√x = im
Ꮩ
V'm. cos ẞ.
Now the similar triangles, l'a'm and V'b'p', evidently give,
for every position of ex', the constant ratio
Hence
V'm
-R.
x'P'
Vx
XP
R.
cos B
which is also constant.
་
Cos a
But we have R < 1, because a < ß; and for the same reason
cos ẞ < cos a.
Hence
Vo
xp
= constant < 1.
The like can in the same manner be shown by considering
the cone vv'; and thus each branch of the ellipse E has its
focus and directrix; F and pq for the one, andƒ and MN for
the other.
Second Method. The same thing can be proved as follows,
from Pl. XI., Fig. 94.-Let ph represent a perpendicular from
DESCRIPTIVE GEOMETRY.
131
any point, p, of the curve to its axis ab, and ko a parallel to
ce, through h. Conceive perpendiculars to ab in the plane of
the ellipse, at P and D, then P and hD will be the perpen-
dicular distances of p from these lines.
=
=
Now ac af bf, bg, and cd eg. Hence, if a line,
=
br, be drawn parallel to VA, and meeting Pe in x, the triangles
Pbx and Dda will evidently be equal, and
Pb Da. Hence, taking ab from each, Pa = Db.
This being so, the triangles, ahk and acP,give
(ck = pm = Lf): hP :: ac: aP;
likewise, the triangles hbo and bgD give
(og = pn = pf₁): hD :: bg: bD
:: ac: aP.
Selecting the terms desired:
pf: hP: ac aP,
and
pf, hD :: ac: aP.
That is, the ellipse is such a curve that the distances from
any of its points to that focus, and to that perpendicular at P
or D, which are both on the same side of the centre of the
ellipse, have a constant ratio.
PROBLEM LXIII.
To construct an ellipse, having given the focus, vertex, and
directrix of either branch.
Let F, b and pq, Pl. XII., Fig. 108, be respectively a focus,
vertex, and directrix of an ellipse.
On any straight line, bn, Fig. 109, make by bq in Fig. 108,
and qn, any assumed distance. On any parallel to bn, as BX,
make BF = 6F in Fig. 108, draw 6B and qF to meet at S, and
draw Sn. Then, FX: qu:: BF: bq.
Hence draw ya in Fig. 108, parallel to the directrix pq, and
at the distance qnqn, in Fig. 109, from it, and ares with
centre F, and radius FX, will intersect ay at a and y, giving
x and y as points of the ellipse; for by the construction, Fa,
or Fy: ap:: Fb : bq.
PROBLEM LXIV.
Having given an ellipse, to find its directrix.
1º. As a Problem in Space.-Find the cone of revolution
132
DESCRIPTIVE GEOMETRY.
from which the ellipse, having the given ellipse E for its hori-
zontal projection, can be cut. Then the horizontal projection,
P7, of the intersection of the plane of the ellipse E' with the
horizontal plane through the vertex of the cone, will be the
required directrix.
2º. As a Plane Problem.-For the point d, Pl. XII., Fig.
108, we have
Fb: bq :: Fd: dn, or by (Prob. LXI.)
Fb : bq :: Ob: Oq, which, by putting
OF = c and Ob = α,
and substituting in the last proportion, becomes
a-c: Oq-a: a: Og; whence
or
whence
O2—a²
a × Oq-cx Oq = u × O9-a²
c × Oq = a²
c: a :: a : Oq.
That is, if the centre or both foci are known, the distance of
the directrix from the centre is a third proportional to the dis-
tance of the focus, and of the vertex, from the centre.
195. Again, the first proportion is,
Fb : bq :: a: Oq.
That is, if the focus, directrix, and vertex are given, the distance,
Og, of the centre from the directrix is a fourth proportional to
Fb, by, and the semi-transverse axis, a.
Hence, for the same focus and directrix there may be differ-
ent ellipses so long as Fb < 1.
bq
196. Having shown by the method of counter development
( Probs. XLVI.-XLVII.) how to draw tangents to an ellipse
considered independently of any curved surface of which it is a
section, the constructions of the general problem of the tangent
to an ellipse considered as the section of a cone, are here given.
PROBLEM LXV.
To construct a tangent to an ellipse, at a given point on the
curve.
Let p, Pl. XIII., Fig. 110, be the given point on the ellipse E.
The tangent at p is the intersection of the tangent plane
DESCRIPTIVE GEOMETRY.
133
along the element Vp-V'p' of the cone VV', containing the
ellipse EE' (148) with the plane BA'B' of the ellipse itself.
Ak and mk are respectively the horizontal traces of these two
planes. Hence kp is the horizontal projection of their inter-
section, and is the required tangent.
THEOREM XVI.
The tangent to an ellipse at a given point bisects the supple-
ment of the angle included between lines from that point
to the foci.
In Pl. XIII., Fig. 110, having found by (Prob. LX.) the two
cones of revolution, VV" and vo', which contain the ellipse
EE', the tangent at p is (Prob. LXV.) the intersection of the
tangent plane nk along the element fpn-v'p', with the plane
AA'B' of the ellipse. As before, & is the intersection of the
k
horizontal traces of their planes, and Ap the tangent. But note
that kA = km, being tangents to the same circle, FA; also
kA = kn = km, the two former being tangents to circle fA.
Ilence the circle A contains the points A m and n; and
thus fpn and Fmp are tangents from p to the same circle, A.
Hence the tangent, kp, to the ellipse, being drawn from the
centre, k, of this circle, bisects the supplement of the angle in-
cluded by the focal radius vectors from the point of contact.
This principle and resulting construction are the most fami-
liar ones used when the ellipse is treated only as a plane curve,
independently of the cone.
PROBLEM LXVI.
To draw a tangent to an ellipse from a given point without
the curve.
Let E, horizontal projection of the ellipse EE', Pl. XIII., Fig.
115, be taken as the given ellipse, and p (pp') the given point,
which is in the plane of the ellipse E (EE').
The point pp' may be considered as that in which the line
of intersection of two tangent planes to the cone VV' (Prob.
134
DESCRIPTIVE GEOMETRY.
LI., Ex. 1), containing EE', pierces the plane BB'A' of that
ellipse. Hence draw Vp-V'p' and find q'q, where it pierces
the plane of the cone's base; that is, here, the horizontal plane.
Then qE is the horizontal trace of one of these tangent planes,
EV its element of contact with the cone; and pe, its intersec-
tion with the plane of the ellipse, is therefore the tangent to the
ellipse, E, from the point p.
There will be another tangent from p, which the student can
construct.
PROBLEM LXVII.
To construct a tangent to an ellipse, and parallel to a given
line.
Let E, Pl. XIII., Fig. 115, horizontal projection of EE' in
the plane BB'A', and on the cone VV', be the given ellipse;
and let L-L', parallel to the plane of the ellipse EE', be the
line, whose horizontal projection, L, is the given line.
The tangent lines to EE', and parallel to LL', will be the in-
tersection of the plane, BB'A', of the curve, with a plane tan-
gent to the cone and parallel to LL'. Such a plane will contain
the line, VN-V'N', parallel to LL', and through the vertex
VV'. Hence NK is the horizontal trace of such a plane, KV
its clement of contact with the cone, and ET, drawn parallel to
L, through E, is its intersection with the plane BB'A'; and
is therefore the required tangent to E, parallel to L.
The student can construct the second tangent which the
problem allows.
PROBLEM LXVIII.
To draw normals to an ellipse, at a given point on the curve.
1º. The normal to an ellipse is at any point perpendicular
to its tangent (137) at the same point. It may, therefore, be
drawn on this principle, or as the bisecting line of the angle
between the radius vectors of the given point.
2°. The normal to an ellipse, from a point not on the curve,
DESCRIPTIVE GEOMETRY.
135
is most conveniently drawn by the method of trial curves, or
curves of error, on account of the absence of elementary geo-
metrical constructions. Thus, let E, Pl. XI., Fig. 100, be the
given ellipse, and P the given point, from which the normal is
to be drawn. Draw several secants through P, and note the
points, as G, H, and I, where they meet the conjugate axis.
Construct circles through the foci, F and F', and G; the foci
and H; and the foci and I; and note the points, g, h, and i,
where they meet the secants drawn from P. Then the line
through P, and N, where the trial curve ghi crosses E, will be
the required normal.
For ghi is, by construction, the locus of the intersection of
lines from F and F', and making equal angles with the lines
through P. Thus, FgG = F'gG, they being measured by half
of the equal arcs FG and F'G.
=
Hence FNQ F'NQ, or PQ bisects FNF', as it should do.
There are other ways of drawing a normal by trial curves,
but this is quite accurate, since the trial curve ghi is nearly
straight and well determined by a few points, far apart, afforded
by secants taken very near the normal, and hence crosses the
ellipse nearly at right angles.
3º. A normal, parallel to a given line, may be drawn perpen-
dicular to the tangent that is parallel to a perpendicular to that
line.
The Parabola.
PROBLEM LXIX.
To construct the cone of revolution which will contain a given
parabola, which has a common tangent with the cone's base.
Let P, Pl. XII., Fig. 105, be the given parabola, and MA
the common tangent to it and to the cone's base. Proceeding
as in Prob. LX., draw any circle tangent to AF, the axis of
the parabola P, and hence having its centre M on MA.
Draw by (184, 3°) Mt tangent to P, and tF tangent to the
circle MA. With centre F describe the circle C, to which MO
must be tangent at O. Then Mt being (148) the intersection
136
DESCRIPTIVE GEOMETRY.
of the plane MA of the curve, and the tangent plane MO, is
the horizontal projection of the tangent to the parabola in space,
tF is the element of contact of the tangent plane MO with
the required cone, and tF being perpendicular to MO, and the
element AF to MA, that cone is one of revolution, and the
circle C, with F as its centre, is its base.
The vertical projection of the cone will be found as shown in
Pl. XII., Fig. 106, where, if aa'm' be the plane of the parabola
P, the line d'v', parallel to x'a'm', will be the element parallel
to the plane of the curve, and will meet vv' at ', the vertical
projection of the vertex of the cone.
THEOREM XVII.
The parabola has a focus and directrix, from both which each
point of the curve is equally distant.
Pl. XII., Fig. 106. Let F (vv) be the vertex of the cone of
revolution, v ad-v' a'd', from which the parabola P is cut by
the plane, m'a'a, parallel to the element l-v'd', and perpen-
dicular to the vertical plane. Assume any point, a', on the
parabola, and va-v'a' will be the element through that point,
and vb-v'b' its true length, when revolved about the axis of
the cone, till parallel to the vertical plane. Draw v'm' parallel
to the ground line as the vertical trace of a horizontal plane
through the vertex vv', and it will intersect the plane aa'm' in
the perpendicular ( 48 ), mD—m'. Now, by reason of the equal
inclinations of the plane aa'm', and of the surface of the cone
to the plane H, we have the line
mx-m'x'′ = the line
Also the line
That is,
va--v'a'
b-v'b' — the line væ―v'x'.
=
the line ɑD-a'm'.
Fx = xm and Fa = aD;
and the like may be shown wherever x may be assumed, on
the curve. That is, F and mD are a fixed point and a fixed
line, both in the plane of the curve, and from both of which
any point of the curve is equidistant. F is called the focus,
and mD the directrix of the curve.
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DESCRIPTIVE GEOMETRY.
137
PROBLEM LXX.
To construct a parabola, having given its focus and directrix.
Let F, Pl. XII., Fig. 105, be the focus, and KD the directrix.
Assume any points, as B, on the axis, and at each of them draw
an indefinite line, as BP, perpendicular to the axis. Then, with
F as a centre, and BD as a radius, describe arcs which will cut
BP in points as P and Q, which will by construction be equally
distant from the given focus and directrix; and will therefore
be points of the required parabola (Theo. XVII.).
The vertex A bisects FD.
PROBLEM LXXI.
Having given a parabola, and its axis, to find its focus and
directrix.
Let P, Pl. XII., Fig. 105, be the given parabola. At any
point, M, on the tangent AS at the vertex A, describe a circle
with radius, MA. At S, where this circle cuts AS, draw the
tangent Kt parallel to AF. From t, where this tangent cuts.
the parabola, draw a tangent, tOF, to the circle MA. Its in-
tersection F is the required focus, and by making AD = AF,
or tK
tF, the directrix, KD, perpendicular to DF, will be
found.
197. The construction of a tangent line to a parabola, under
the three primary conditions (140), can all be easily made on
the principles of projection, as has been shown in connection
with the ellipse.
PROBLEM LXXII.
To construct a tangent at a given point on a parabola.
For the first, which is the most common and important case,
we have the following very simple constructions from Pl. XII.,
Fig. 105.
First Method. By reason of the tangents tS and to, to the
138
DESCRIPTIVE GEOMETRY.
circle MA, whose radius Mt is tangent to the parabola, we see
that the latter tangent bisects the angle between the radius vector
and the diameter, tn, through the given point produced.
Second Method.-Observing that AM = MS, draw the diame-
ter, ¿S, of the parabola through the given point t, and bisect the
distance, SA, included on the tangent, MS, at the vertex, between
the axis and this diameter, and Mt will be the required tangent.
Third Method.-Pl. XII., Fig. 106. Since the tangent at
bisects Fam, while xFxm, the latter lines are two sides of a
rhombus, Famb, of which the tangent is a diagonal. Hence,
Fb = xm = Dy; and therefore Db
But Fa
=
Fy.
Da, hence ab ay. That is, make abay, and
xb will be the tangent at æ.
198. Calling my the ordinate of x, and yb its subtangent, this
construction expresses the theorem that the subtangent, by, of
the parabola is bisected at the vertex.
The Hyperbola.
PROBLEM LXXIII.
To construct the cone of revolution containing a given hyper-
bola, and whose base shall have a common tangent with the
hyperbola
Let H, Pl. XIII., Fig. 111, be the horizontal projection of
the hyperbola, which is in the plane, AA'B'.
At any point, pp', of the hyperbola, conceive a tangent, pm,
meeting the horizontal trace of the plane of the curve at m.
With centre, m, and radius, mA, describe a circle to which two
tangents can be drawn from p. These tangents, pq and py, will
meet the axis at the same points, F and f, wherever p is taken,
as in the ellipse.
PI
This done, project e at e', and I at I', by which I'B' and e'B'
can be drawn, on which project v at v', and V at V', respect-
ively, and finally draw A'o' and A'V', which will complete the
cone vo', containing II, and the circle vA; and the cone VV',
containing II, and the circle VA.
199. From Pl. XIII., Fig. 111, FAƒB. For let K and
DESCRIPTIVE GEOMETRY.
139
k be the asymptotes (136) of the hyperbola, and let p be at
infinity, then K is its tangent, and m appears as at M, on kO.
But the two tangents to circle MB will then be parallel to kO,
and we shall therefore have FO
but AO
fo,
BO;
whence, by subtraction, FA = fB, as stated.
The figure thus serves as the solution of the problem: to find
the foci of the hyperbola whose asymptotes are given.
THEOREM XVIII.
In the hyperbola the difference of the distances from any point
of the curve to the foci is constant, and equal to the trans-
verse axis.
Let pp', Pl. XIII., Fig. 111, be any given point on the hyper-
bola. Then up-v'p' is the element of the cone vv' through pp',
and vP-v'P' is its revolved position, parallel to the vertical
plane. Vp-V'p' is the element of the cone VV', through
pp'; and, as all the elements of both cones are equally inclined
to the horizontal plane, the revolved position and true lengths,
v'P' and V'p''', are parallel; and their difference, v'h, is constant
for every position of pp'. For the same reason, A'h
v'k;
hence v'P'-V'p'" = A'k, and therefore the horizontal projec-
tions of these lines have the same relation.
or,
That is, vP-Vp" = AB,
vp-Vp = AB, which expresses the theorem.
PROBLEM LXXIV.
To construct a hyperbola, having given its transverse axis and
foci.
Let F and ƒ, and AB, Pl. XIII., Fig. 111, be the transverse
axis and foci of a required hyperbola. Then take any two dis-
tances, as Ad and Bd, estimated in the same direction from A
and B, and whose difference shall thus equal AB, and with each
in succession as a radius, and F and fin succession as centres,
describe arcs, not shown, on each side of AB.
140
DESCRIPTIVE GEOMETRY.
The intersections of the arcs from F, with radius Ad, will
then intersect those from f, with radius Bd, in two points of
the hyperbola; or, in general terms, the intersections of the
arcs, whose centres and radii are both different, will be points
of the curve.
Note that d must not be nearer B than ƒ, and that each pair
of radii will thus give four points.
THEOREM XIX.
The hyperbola has directrices, such that the ratio of the dis-
tances of any point of the curve, from the like focus and
directrix is constant, and greater than unity.
Let the hyperbola HH', Pl. XII, Fig. 107, whose horizontal
projection is the hyperbola II be placed with the circles va
and Va, upon the two cones vv' and VV'.
Draw the horizontal planes, v'p' and V'k, through the verti-
ces, vv' and VV', cutting the plane, aa'b', of the hyperbola in
the lines py—p' and MN—N'.
If now we take any point, xx', upon the hyperbola, H, H',
and through it pass the horizontal plane x'e, we shall have the
line vx-v'x' = v'c.
also, ap' is the true size of ap-x'p', a perpendicular from a
to the directrix pq-p'.
The triangles, v'a'p' and a'x'c', are similar, and, for every
position of xx', evidently give
v'c
x'p'
a constant ratio, greater than 1.
All the elements, as vx-v'x', make the same angle, a, with
the horizontal plane; also, all the positions of xp-x'p' make
the same angle, B, with that plane. Hence
v'c. cos a
x'p'. cos p
VX
a constant ratio > 1.
xp
And the like results can be found for the other branch, which
proves the theorem.
This theorem indicates sufficiently the construction of the
directrices of a given hyperbola, by two projections.
200. The problems of tangent lines to the hyperbola corres-
DESCRIPTIVE GEOMETRY.
141
ponding to Probs. LXV., LXVI., and LXVII., are left as an
exercise for the student, they being very similar to the latter.
The tangent to the curve at a given point, x, Pl. XII., Fig.
107, for example, bisects the angle, fxF, included between the
radius vectors of the point, instead of the supplement of that
angle as in the ellipse. Thus there is always a point of con-
trast in a given property as stated relative to the ellipse and
to the hyperbola.
PROBLEM LXXV.
Having given the transverse axis and foci of a hyperbola, to
find its asymptotes.
Let AB, Pl. XIII., Fig. 116, be the transverse axis, and F
and f, the foci of a hyperbola.
By Prob. LXXIII., and Fig. 111, we see that this problem
reduces to finding that position of m, centre of the circle, mA,
for which the tangents to it, from the foci, are parallel.
Then in Fig. 116, which represents the problem as solved,
afk GFO, hence
=
GFO + Ofg = 180°.
But FM bisects GFO, and fM bisects Ofg; hence
= 90°, whence
MFO + MƒO =
FMf 90°.
Hence describe a circle, FMf, on Ff as a diameter, and note
its intersection, M, with the tangent at either vertex, as B, of
the curve. Then MO will be an asymptote, since it is parallel
to FT and ft, which, being the parallel positions of Fp and fp,
Fig. 111, they and OM will all meet at the infinitely distant
position of p.
Given asymptotes are useful to aid in shaping the curve
through the more remote points found by Prob. LXXIV.
The focal Curves of the Conic Sections.
201. Having now exhibited the properties of the conic sec-
tions, considered separately; and by methods appropriate to
Descriptive Geometry, as the Geometry of Projections, their
relation to each other will next be shown; by which it will be
t
142
DESCRIPTIVE GEOMETRY.
more evident that they are three forms of what is essentially
one thing; or, three members of one family of curves.
202. 1°. We have seen that, having an ellipse, E, E', or a
hyperbola, HH', and its foci, ƒ and F, and vertices a and b of
their horizontal projections, we can envelope each of these
curves, and the circles, fa and Fa, by two cones of revolution,
each of which shall have a vertical axis.
2º. We have also seen that the parabola, P,P', having one
focus, f, and vertex, a, can, with the circle, fa, be enveloped by
one such cone.
3º. It has been thus shown that the horizontal projections of
the vertices of these cones, are the foci of the horizontal pro-
jections, E, HI, and P, of these curves.
4°. Finally, we recollect that in demonstrating the property
of the foci, we found, see Pl. XII., Fig. 106, and Pl. XIII.,
Figs. 110 and 111,
vx, v'x' = xm, x'm', for the parabola;
Vp, V'p' + vp, v'p' = constant for the ellipse; and
vp, v'p'—Vp, V'p' = constant for the hyperbola.
203. That is, the conic sections may be said to have other
foci, out, as well as in, their own planes, since the vertices,
VV' and vv', of the cones of revolution which thus envelope
them, have the same general relation to them, as do the points
f and F, commonly called their foci.
May there not, therefore, be some curve, every point of which
is, in the general sense just illustrated, a focus of the conic
section?
Such a curve, if found, will be called the focal of the conic
section.
THEOREM XX.
An infinite number of cones of revolution may be made to
envelope the same ellipse.
Let EE', Pl. XIV., Fig. 117, be an ellipse in the horizontal
plane of projection.
Project this ellipse, EE', upon a plane, Aɑ'P', in the ellipse
DESCRIPTIVE GEOMETRY.
143
EE,', whose shorter axis, cd, will evidently be equal to CD,
that of the ellipse EE'.
We may find by Prob. LXI., the foci, f, and F,, of the ellipse
E,E,', perpendicular projection of EE' upon the plane Aa'P', and
thence the cone, f-v'a'e', which envelopes the circle ƒ,a and
the ellipse EE', and the cone F-V' (not shown), which en-
velopes the circle Fa, and the same ellipse, EE'.
By projecting the ellipse EE', upon any other planes, per-
pendicular to the vertical plane and through the tangent AA',
other points, v', V", similar to fiv' and F,V', may be found,
which will be the vertices of other cones of revolution envelop-
ing the ellipse EE'.
All the points, v'V', v"V", etc., thus have the same relation
to the original ellipse EE'; and, evidently, any number of them
can be found, as otherwise shown in (177).
THEOREM XXI.
The foci of all the different projections of an ellipse, E', upon
planes containing the tangent at the extremity of its trans-
verse axis, lie on a circle, perpendicular to the plane of
the cllipse, and whose diameter is the distance between the
foci of the ellipse.
As both of the ellipses EE' and E,E,', Fig. 117, are in planes
perpendicular to the vertical plane, a vertical plane, Q, con-
ceived to pass through the transverse axis AB of the ellipse
EE', will be parallel to the vertical plane of projection, and
will contain the transverse axis, ab, of the ellipse E,E',, the ver-
tex, fiv', of the cone; its extreme elements, fa—v'a' and ƒb
-v'B'; and the diameter, ac—a'e', of its base.
Now, the projections of EE' upon any other planes, as AA'b',
perpendicular to the vertical plane, will always be ellipses
whose axes, as a'b', will be in the plane Q, and whose other
axes will all be equal to CD. Hence it will be sufficient to ex-
amine the relations of the lines in this plane Q.
See now Pl. XIV., Fig. 118, where E' represents an ellipse.
situated like E' in Fig. 117.
Describe the circle C on the transverse axis ab (or 2a) as a
diameter. From a, draw any chord, ab, which will be the
144
DESCRIPTIVE GEOMETRY.
transverse axis of the ellipse E,, if it be greater than the con-
jugate axis, 26 (not shown) of the ellipse E; and E, will be the
perpendicular projection of E upon the plane PaP'.
The conjugate axis of E₁, being 26, construct its foci, f, and
F₁, as in Fig. 117. Then make fe = fa, erect a perpendicular,
fv, to ab₁, and it will cut be in v, the vertex of the cone of re-
volution containing the ellipse E. Likewise aV, parallel to be,
will cut the perpendicular FV, to ab, in V, the vertex of
another cone of revolution, Vae,, through E.
Now among all the ellipses, projections of E upon planes
through the tangent Pa-a to the ellipse E, one, as E,, whose
transverse axis, ab,, is equal to 26, the conjugate axis of E, will
be a circle. Hence the series of cones of revolution containing
E and the circles in the planes PaP', etc., circles tangent to E
at a, and whose centres are the foci of E, E, etc., will ter-
minate in a cylinder of revolution, whose base is E,, since the
foci of E, unite in its centre, 02.
Other chords, as ab', symmetrical with ab,, ab, etc., relative
to ab as an axis, can be drawn, giving other points as v, and
V₁ of the locus of the exterior foci, so called, which thus occur
symmetrically in sets of four.
1
Next, on o, o, the projecting line of o upon the plane PaP',
and thence perpendicular to ab,, take o,cb, the semi-conju-
gate axis of the ellipses E and E. Join f, and c; alsoƒ, and o.
Then 0,0 b.
ao = a = fc (Prob. LXI.)
аο₁ = α₁ = a. cos α.
αι
0₁f₁ =
2
2
-f² ; 001
2
2
2
√ao²-ao,²
1
2
1
of₁ = √of +00, "= √(a,”—b²)+(a²—a,
2
2
2
√a*—b²=of
That is, of is a constant, whatever chord from ab to ab, in-
clusive, of the circle C be taken; which proves the theorem.
THEOREM XXII.
The focal of the ellipse is a hyperbola, whose vertices are the
foci of the ellipse, whose foci are the vertices of the ellipse;
and whose plane is perpendicular to the ellipse.
The vertices, v and v; V and V₁, exterior foci of the ellipse
E, Pl. XIV., Fig. 118, occur, as we have seen, in sets of four.
Ъ
B
B
44.
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48.
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f
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45.
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49.
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P/
a
P
51.
50.
P
R
i
P
C
46.
47.
ла
G d
A
e L
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54.
Ly
DESCRIPTIVE GEOMETRY.
145
The circle, as ab₂, projections of E, and on each side of ab,
are the limits of ellipses whose transverse axes are in the verti-
cal plane through ab. IIence they are the limits, beyond which
no further constructions like those of the last two problems can
be made. But as the positions of v and V, corresponding to
these circles, are at an infinite distance, being, for E,, the inter-
section of ʊ, and 00,, it is apparent already, that the focal of E,
locus of all its exterior foci, is a curve of two equal, opposite,
and infinite branches, symmetrical with respect to ab.
ol,
2
To proceed: the triangles, afo and acb, are similar, and give,
eb abfo ao
whence,
eb: 2a :: Va² -b²: a
eb = 2 Val' = ƒF.
That is, eb is constant. But ebxb―ve = vb-ra. Thus the
focal II is a hyperbola, whose transverse axis is fF, and whose
foci are a and b; as stated in the theorem.
204. Finally, f,v bisects the angle arb, which establishes this
general theorem.
The axes of all the cones of revolution containing a given
ellipse, are tangent to the focal hyperbola of that ellipse; and
the vertex of each cone is the point of contact of its axis with
this focal hyperbola.
The complete conception of the focal requires one more
theorem, as follows:
THEOREM XXIII.
If we take any two points whatever on the focal, one on each
branch, the sum of the lines from these points to any point
of the ellipse is constant, wherever that point is taken.
This theorem is readily demonstrated by means of Theo. IX.
Imagine an ellipse, E; and its focal hyperbola H; then let p be
any point on the ellipse, and e and V any points, vertices of
cones of revolution containing E, and taken arbitrarily, one on
cach branch of IIII. (Theo. XXII.)
Then the element vp of the cone v consists of two parts
(see Vn, or Vm, Pl. XI., Fig. 94), vt, from v to t, on the circle
of contact, e, of the cone v and sphere s; and tp from t to p.
The former is evidently constant for every position of t on the
circle s, and tp = pf when s is the sphere tangent to E at ƒ.
f.
10
146
DESCRIPTIVE GEOMETRY.
=
Likewise, the element, Vp, of the cone V, consists of two
parts: VT, from V to T, on the circle of contact C, of the
cone V and sphere S, and Tp. But VT is constant, and Tp
pF. And pf + pF is constant, = ab.
Ilence (vt + tp) + (VT + Tp) (vt + pf) + (VT +pF) is
constant wherever p is taken on the ellipse.
=
205. We shall only give the enunciation of the correspond-
ing propositions relative to the hyperbola and parabola, leaving
their demonstrations, which are quite analogous to the preced-
ing, to the student.
The focal of an hyperbola is an ellipse whose plane is per-
pendicular to that of the hyperbola, and whose foci and verti-
ces are respectively the vertices and foci of the hyperbola.
The focal of a parabola is an equal parabola, turned in an
opposite direction, in a plane perpendicular to that of the given
one, and the focus of each is the vertex of the other.
Determination of the Conic Sections by Points.
THEOREM XXIV.
If a hexagon, inscribed in an ellipse, be such that it has two
pairs of opposite parallel sides, the two remaining sides
will be parallel.
Let abcdef, Pl. XIX., Fig. 151, be the hexagon, in which ab
and de are parallel, and be and ef are parallel. Then will ed
and fa be parallel. For the angles, abc and def, are equal, being
included by parallel sides. Therefore they subtend equal arcs,
aec and fbd, and hence are also inscribed in equal arcs, abe and
def.
The latter equal arcs with the remainders, cd and fa, make
the entire circumference. Hence bisect these remainders at g
and h, and add their halves thus found to abc and def, and
gbh geh = 180°. Hence gh is a diameter, bisecting af and
cd, and it is therefore perpendicular to the chords af and cd.
Hence these chords are parallel.
Finally, by (126) the foregoing is true for any ellipse and
its inscribed hexagon, formed by projecting the given circle
and hexagon upon some plane.
DESCRIPTIVE GEOMETRY.
147
THEOREM XXV.
If any hexagon be inscribed in a conic section, the three points
of intersection of its opposite pairs of sides produced
will be in the same straight line.
Let v, Pl. XIX., Fig. 153, be the vertex of any one of the
cones of revolution, of which the given conic section, S, may
be taken as the base, and let abcdef be any hexagon inscribed
in S, and whose opposite sides produced meet at x, y, and z.
Draw the elements va, rb, vc, vd, ve, and vf of the cone; also
the lines va, vy, and vz.
Now to demonstrate that the line xy also contains 2, suppose
the cone to be cut by a plane, P, parallel to the plane v-xy.
This plane will (168) cut an ellipse, E, from the cone; and
from the planes v-af and v-ed, the chords af, and c,d₁,
parallel to væ; since va is the intersection of those planes, and P
is parallel to it. Likewise, the plane P cuts from the planes
v-fe and v-be, another pair of chords, fe, and b,c,, parallel to
vy; and lastly, the same plane, P, cuts the planes v-de and
v--ab in chords, de, and a,b,, which, with the four preceding
chords, form a hexagon, inscribed in the ellipse E.
1
But since four of these chords exist as two pairs of parallel
sides, the two, de, and a,b,, are parallel, also (Theo. XXIV.), to
each other, and hence to ez. That is, vz is in the plane ray,
parallel to the plane P containing E. Hence z is in the trace
ay of that plane, according to the theorem; and as no dis-
tinctive property of the curve S has been employed, it might
have been any conic section.
包
​This theorem is known as that of the hexagram of Pascal.
THEOREM XXVI.
A conic section is determined by five given points, or their
equivalent.
Let a, b, c, d, e, Pl. XIX., Fig. 152, be five given points.
The lines ab, be, cd and de are four sides of the inscribed
hexagon, the remaining vertex of which is required. The
148
DESCRIPTIVE GEOMETRY.
theorem will be proved by showing that any, and as many points
as we please, can be found from the five that are given, and
that they will agree with the last theorem.
z
Produce ab and ed to meet as at ; draw any line, ex, limited
at a by cb produced, and draw the line xz. Note y where de
produced meets xz, draw ya; aud f, where it meets ex, will be
a sixth point, evidently forming, with the given ones, such a
hexagon as was described in the last theorem. By proceeding
in like manner with any five of the six points now found, or
only by new positions of the line ex, any number of other points
may be found, each of which, with the five given, will form a
hexagon agreeing with the condition of the theorem.
It is sufficient for present purposes to mention that if the five
given points are so situated that an ellipse could not contain
them all, the curve would be one of the other conic sections.
The figure of this theorem serves equally for the problem:
To construct a conic section, having given five points of it.
By (179, 184,) the diameter, axes, and foci can be found.
EXAMPLE.-Take the five given points in a variety of different positions.
The Unity of the Conic Section.
206. The constructions, hitherto given, while showing the
conic section in its variety, also indicate its essential unity; by
which is meant that each of its properties is, in general, with
only slight modification in statement, true for all its forms.
207. This unity seems probable in advance, since the conic-
section is really the intersection of two surfaces of the same
kind, a plane being (96) a particular case of the cone, and both
surfaces being of uniform properties.
Thus, the conic section has-
1°. Two axes of symmetry.
2º. Is a curve of two branches.
3°. Each having a focus and directrix.
4°. The distances from any point of the curve to which have
a constant ratio.
5°. The tangent bisects the angle included by lines from its
point of contact to the two foci.
208. In the ellipse, both axes cut the curve; the branches are
DESCRIPTIVE GEOMETRY.
149
real, finite, and united, forming a closed curve; the foci are
between the directrices; the ratio named, taking the distance
to the focus as the numerator is < 1; and the tangent bisects
the external angle between the focal radii.
In the parabola the conjugate axis is imaginary; the branches
are separate; one, real, and finite, forming an open curve; and
the other is an imaginary point at infinity. There is but one real
focus and directrix, the ratio named = 1; the other focus is at in-
finity, making all its radius vectors parallel to the axis, and per-
pendicular to the directrix; and thus the tangent is said to bisect
the angle between the radius vector and the perpendicular to
the directrix, from its point of contact.
In the hyperbola, one axis cuts the curve; the branches are
real, infinite, and separate, forming open curves; the ratio
named is > 1; the directrices are between the foci, and the
tangent at any point bisects the internal angle formed by the
radius vectors of that point.
Like parallel statements can be made of all the properties of
the conic section, often simply called, the conic.
209. The essential unity of the conic section is further and
usefully seen in the common property of the bisection of the
angle between the focal radius vectors of a given point of any
conic section, by the tangent at that point, and in the conse-
quent easy construction for the following general problem,
which will be useful in important future operations.
PROBLEM LXXVI.
To draw a tangent to any one of the conic sections from a
given exterior point.
First, to an ellipse.-Let P, Pl. XIII., Fig. 112, be the given
point. With the major axis, AB, as a radius, and either focus,
as F', as a centre, describe an are, ab. With P as a centre, and
the distance to the other focus, F, as a radius, describe an are
cutting the former at a and b. Draw aF and 6F, then PT and
Pt, perpendicular respectively to aF and F, will bisect them,
and be the required tangents.
For T, for example, like every point of PT, is equidistant
150
DESCRIPTIVE GEOMETRY.
from F and from a.
That is, Ta TF, hence TP bisects the
=
angle FTa. But, adding F'T to each,
F'T+Ta F'T + TF = AB.
=
That is, TP bisects the angle between F'T produced and TF,
hence it is the tangent at T.
Second, to a hyperbola.-The above explanation applies
equally to Pl. XIII., Fig. 114.
Third, to a parabola.—Pl. XIII., Fig. 113. With the given
point P as a centre, and PF as a radius, F being the focus,.
describe an arc cutting the directrix D at a and b. Then, as
the other focus of a parabola is an imaginary point at infinity
on the axis, draw at and bT parallel to the axis VF, and T and
t will be the points of contact of the required tangent, from P,
since, by construction, every point of PT, for instance, is equi-
distant from F and b, whence T, being common to PT and the
curve, is their point of contact.
PROBLEM LXXVII.
To draw a line tangent at once to two conic sections.
This problem is most conveniently solved by the method of
foci, or trial curves.
Case 1st. A tangent to two hyperbolas. Let AB, Pl. XIV.,
Fig. 119, be the transverse axis, II the centre, and F and F' the
foci of one hyperbola; and CD, II,, and f and f the axis,
centre, and foci of another hyperbola; and let it be required to
draw a common tangent to these two curves.
By any of the methods previously given or indicated, sce
(179, 3°; 188; 200, or Prob. LXV.), draw several tangents, as
t, t₁, t, to either branch of either curve, as (in the figure) to
the branch E, of the hyperbola CD.
If, now, the tangent tp were also tangent to the hyperbola
AB, it would bisect the angle formed by joining its point of
contact on that curve with F and F'. Therefore draw the chord
pF', and O perpendicular to it at its middle point, describe the
circle of radius OF', and note its intersection, m, with pt pro-
duced. Then ptm bisects at m the angle between F'm and Fm,
the halves of this angle being measured by a half of the equal
arcs F'p and Fp of the circle OF'F.
DESCRIPTIVE GEOMETRY.
151
2
Making a similar construction with the tangents t₁, etc., and
observing that, as t, p, is parallel to the conjugate axis IIp, the
point p, will be at infinity on IIp produced, and t, p, will be an
asymptote to the trial curve, this curve, mno, is readily sketched
through a few carefully constructed points, some of which, as o,
should, for greater accuracy at t, be found near the branch E,
containing the required point of contact.
The curve mno being thus the locus of the vertices of angles
standing on FF' and bisected by tangents to E, its intersection,
t, with E, is the contact, on E,, of the required tangent to E
and E,. At t, find the tangent, t,T, to E by Prob. LXXVI.
Case 2d. A tangent to two ellipses. The construction will
be essentially the same as that just given.
Case 3d. A tangent to an ellipse or hyperbola, and to a para-
bola. Let II, Pl. XIV., Fig. 120, be one branch of a hyperbola,
and P a parabola, whose vertex is V.
This case is most conveniently solved by a trial curve formed
by means of the principle that the subtangent (198) to a para-
bola is bisected at the vertex of the curve.
Thus, draw several tangents, as tm, tn, to, to the hyperbola
by any convenient method, and make Va= VA, V6 VB,
etc. Also, note m, n, etc., where perpendiculars to the axis OV
meet these tangents. Hence mno is the locus of points on the
tangents to II, such that the subtangents Aa, etc., are bisected
at V. IIence T,, where mno crosses P, gives V8 = VS, when
T,T is drawn from T, tangent to II. IIence TT, is the common
tangent required.
EXAMPLES.-1. Vary the relative positions of the given curves; including
the case in which their similar axes shall be parallel.
Ex. 2. In Fig. 120, find points of mno within P; taking P so that Vs shall
be larger than in Fig. 120.
PROBLEM LXXVIII.
Having given a cylinder of revolution, by its diameter and
axis, to find its intersection by a plane perpendicular to the
axis, and a tangent to the curve.
In Space.-By revolving the given plane either into, or paral-
lel to a plane of projection, the revolved position of its inter-
152
DESCRIPTIVE GEOMETRY.
section with the axis will be the centre of a circle which will
be the revolved position of the required section. Then, by
operations of counter development (77), the projections of the
same section can be found.
In Projection.-Pl. IX., Fig. 84. 1°. The four initial points.
Let Oo--O'o' be the axis, and AB and j'l' the horizontal and
co-parallel diameters; and let PQP' be the given plane. By
Prob. XXI., the axis pierces this plane (construction not shown)
at OO'. This point appears at O", by revolving PQP' about
PQ, into the plane H, whence the circle O"B", where O"B" =
OB, is the revolved position of the required section.
A'B' is the vertical projection of AB, and the horizontal pro-
jection, jl, of j'l' may be found, as in previous cases, giving four
points of each projection of the section.
2°. Other points are found from the revolved section O"B".
Thus RSR is a plane perpendicular to PQ, and intersecting
PQP' in a line from a to R' (Prob. XVII.), which appears at
aR", when RSR' is revolved back about SR as a ground line
into the plane H (27).
Then take any points, as B'E"D", project them upon SR, as
at B''', etc.; revolve them to b'", e'", d'"'; thence project them
upon SR", at b", e", d". Next, when SR", the vertical trace
of RSR' revolved about SR, is revolved about GL, it will appear
at SR', and Sʊ' SU", etc., as indicated by the arcs "b", with
S as a centre.
=
Finally, B, E, D, etc., are at the intersection of the arcs of
counter revolution, B'B, etc., and the projecting lines b''B, etc.,
whence B', E', D', etc., are projected upon the parallels l'A'B',
e'E', d'D', etc.
210. Otherwise: As there are various minor modifications in
the order and manner of this form of solution-having found
the horizontal projection BEDJC, as described-produce JE,
for example, to the ground line, thence erect a perpendicular
to GL, which will meet P'Q in a point of the parallel e'E', on
which E' may be found as before.
211. 3°. To draw a tangent to the curve, at any point of right
section of a cylinder in an oblique position, as V, V', V''.
Draw the tangent first in development at V". Being in the
plane of the curve, its horizontal trace, Y, is in PQ the like trace
of the plane of the curve. Ienee YV is its horizontal projec
tion; and, projecting Y at Y, its vertical projection is Y'V'.
a
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DESCRIPTIVE GEOMETRY.
153
Thus we may find tangents to plane intersections by counter
development (77), as well as by the method of (161).
EXAMPLES.-12. Revolve the plane PQP' into the plane V, and then proceed
as before.
Ex. 2°. Let the axis of the cylinder cross the first angle.
Ex. 3°. Construct the tangents at EE' and JJ'.
Ex. 4°. Revolve the plane PQP' about AB as an axis, in order to find the
curve.
Ex. 5. Revolve PQP' about C'D', and parallel to the plane V, and then find
the curve.
212. Resuming (161), it is here restated in the form in which
it is actually applied.
Each auxiliary plane, A, parallel to the axis of a cylinder, or
containing the vertex of a cone, cuts from the given intersect-
ing plane a line, L, and from the given cylinder, or cone, two
elements, E and E,, unless it be a tangent plane. The intersec-
tions of these elements with the line L, are thus common to the
given plane and the cylinder, or cone. That is, they are
points of their intersection.
The auxiliary planes may also be made to cut the cylinder, or
cone, in circles.
PROBLEM LXXIX.
Having given a cylinder of revolution by its trace, to find its
intersection by any plane.
In Space. When the trace is given, auxiliary planes may be
used according to (212), since the intersections of their traces.
with the traces of the surface will be the traces of the elements
contained in those planes. These planes may have any position
parallel to the axis of the cylinder, but, as will be seen, positions
perpendicular to a plane of projection are most convenient.
In Projection.-Pl. IX., Fig. S4. It will be evident by in-
spection, on comparison with Prob. XXI., that hh' and kk' are
the intersections of the given plane, TUT', with the elements
contained in the vertical auxiliary plane ET'; and that r
and nn' are the intersections of TUT' with the elements con-
tained in the perpendicular auxiliary plane TIT.
154
DESCRIPTIVE GEOMETRY.
EXAMPLES.-1'. Reconstruct the intersection of the cylinder by the plane
PQP', by the method just explained.
Ex. 2'. Let the vertical trace only of the cylinder be given.
Ex. 3. Construct a tangent line to the intersection at any point.
Ex. 4°. Find two points by an auxiliary plane passed in any manner, not per-
pendicular to a plane of projection, and through the axis, or any element
(Prob. XI., second).
PROBLEM LXXX.
Having given a cylinder of revolution, by its axis and diame-'
ter, to find its intersection with any plane, by means of
auxiliary planes, and to draw a tangent to the curve.
In Space. The required intersection can be found by means
of auxiliary planes, placed as in the last problem, without hav-
ing the trace of the cylinder.
Either projection of the curve can be found first, according
to the position chosen for the auxiliary planes, as explained in
the last problem.
In Projection.-Pl. XV., Fig. 126. 1° Points of the right sec-
tion or base. Let Ox-O'' be the axis, and AB and 20'G' the
diameters which show their real size. Make aa"ya', and
Ox" will be the axis when revolved about Ox, and into the
horizontal plane. Then C"D", perpendicular to Ox", will be
the projection of the corresponding position of the base, giving
C"OC" for the angle made by the plane of the base with the
horizontal plane. Also the circle on AB represents the base as
revolved about AB, and into the horizontal plane.
These preliminaries effected, take Ox, for the time, as the
ground line of the horizontal plane and the vertical plane con-
taining Ox-Ox" and CD-C"D". Then take any point as
H'"', project it at h, revolve it into the primitive plane of the
base at ", in the vertical plane on Ox, whence it is projected
at II in the parallel plane, IIII", whence it came. J and II
are symmetrical relative to CD; and so are E and F, which are
similarly found, beginning with E"". C"" revolves to C", and
is thence projected at C.
The vertical projections, II', J', etc., of II, J, etc., are at
heights above the ground line KT, equal to the heights of their
auxiliary vertical projections, shown at 'r, etc., above Ox.
DESCRIPTIVE GEOMETRY.
155
Thus we may find any points of the base, each of which is
a point of an clement.
2º. Points of the required curve.-Let PQP" and P'QP'''
be the horizontal and vertical traces of the given cutting plane.
Then let IIKP" be a vertical plane containing the elements
through HIII', and S (the latter not shown). This plane cuts
PQP" in the line aK-P""a', which intersects the clements
contained in the same plane, HKP""', at 'b, and at a point on
the element not shown.
Other points may be similarly found; only observe that as
the auxiliary planes HK, Dm, etc., are parallel, their traces on
the given plane are parallel. Thus, projecting m, f, etc., at m',
f', etc., we draw m'n', etc., parallel to Pa', and hence need
find the vertical trace of but one of the auxiliary planes.
Again: P'I'I is an auxiliary perpendicular plane, containing
the element through J'J, and cutting the given plane in the
line P'I'-I. This line and this element intersect at jj', another
point of the required curve, and any other points could be
found in like manner. The plane P'I'I contains another ele-
ment, which can be found by taking a point symmetrical with
J', relative to O'G'.
3°. Visibility.—All points on elements which stand on the
upper half, ACB, of the base, are visible in horizontal projec-
tion.
Likewise, all points on the elements on the front half of the
cylinder, and between the extreme elements, as G'g', of the ver-
tical projection, are visible in vertical projection.
The invisible points, in each projection, are dotted.
See Prob. CXIII. for constructions of the axes of the pro-
jections of the intersections, if desired in this and subsequent
similar cases.
4°. To draw a tangent at any point of the section.—Take the
point b'. The tangent (147) is the intersection of the given
plane, PQP', with the tangent plane along the element Hb—
I''. This element pierces the horizontal plane at T, one point,
therefore, of the horizontal trace of the tangent plane. The
trace t of the tangent line at II, shown in revolved position at
II'''t, is in the trace AB of the plane of the base, and is another
point of the horizontal trace, TP", of the tangent plane. Then
P" is one point of the intersection of the tangent plane and
the given plane. Hence Pb is the horizontal projection of the
156
DESCRIPTIVE GEOMETRY.
tangent line; and, projecting P" at q', q'b' is its vertical pro-
jection.
EXAMPLES.-1°. Complete the vertical projection of the base.
Ex. 2. Find the curve of intersection when PQP" is the vertical, and P'Q the
horizontal trace of the given plane.
Ex. 3. Construct the tangent at nn'.
Ex. 4. Take the cutting plane so as to make some or all of the points of
intersection outside of the first angle.
213. A full set of problems of the plane and cylinder will
embrace two cases intermediate between Probs. LVI. and
LXXX., as in the first two examples following, together with
minor variations on all of them.
EXAMPLES.-1'. Let the cylinder be vertical, and the given plane oblique to
both planes of projection.
Ex. 2. Let the given plane be perpendicular to one of the planes of projec-
tion, and the cylinder oblique to both planes of projection.
Ex. 3. In Prob. LXXX., let the auxiliary planes be perpendicular to the ver-
tical plane.
Ex. 4. In Ex. 1' let them be horizontal.
Ex. 5. In Ex. 1 let the auxiliary planes, all vertical, have any three differ-
ent positions relative to the vertical plane of projection.
Ex. 6. In Prob. LXXX., let the axis of the cylinder cross any other angle
than the first, and let both traces of the cutting plane make acute angles with
the ground line.
PROBLEM LXXXI.
Having given any cone of revolution, by its axis and a diame-
ter, to find its intersection by a plane, and a tangent to
the curve.
In Space. By taking the auxiliary planes used in solving this
problem, by (212) in a uniform manner, the work may be
abridged. Thus, let them all be perpendicular either to the
plane H, or the plane V, and contain the vertex of the cone.
They will then all intersect each other in a line through the
vertex, and perpendicular to H, or V, respectively. Hence
their traces on the given plane will all pass through the trace
of this line on that plane.
Further than this the solution is the same as for Prob.
LXXX.
DESCRIPTIVE GEOMETRY.
157
In Projection.-Pl. XV., Fig. 127. 1°. The projections of
the base. The projections of the two given diameters afford,
as usual, four points of the base. The other points here found
are the extremities of the shorter axes of the projections of the
base. And as in each of the last two problems an auxiliary
plane of projection (58, 3°) was used, one perpendicular, and
the other parallel to the axis of the cylinder, we have here, for
further illustration, revolved the cone till its axis became paral-
lel to a plane of projection (58, 2° ).
Thus, by revolving OV-O'V' about a vertical axis at O0'
to the position OV"-O'V", parallel to the vertical plane,
V"O'L is the true inclination of the axis to the plane H.
Hence, drawing OI" OB, and so that IOI" shall be the com-
plement of V''O'L, we have IOI" as the inclination of the base
to the plane H, and I" the revolved position, around On as an
axis, of the extremity of the shorter axis; whence, by counter
revolution in the are I'I, we find I, and I' at a height equal to
II". Then OJ OI.
=
O'ë
Likewise, revolve any convenient portion, Oz-O'', of the
cone's axis, about a perpendicular (48) axis at OO' and into
the plane H, at O'e""-Or". Then "OV" is the inclination
of the axis to the plane ▼; and O'K", = O'C', and drawn to
make the angle K"O'K' equal the complement of "OV",shows
the inclination of the base to the plane V, and gives K" the
revolved position of the required point K' into that plane, about
O'K' as an axis; whence, by counter revolution, K" appears at
K', and K is at a distance from the vertical plane V'O, equal
to K/K".
O'F' = O'K', and thus having both axes of each projection
of the base, it can readily be drawn.
2º. The common point of the traces on the given plane.—Let
PQ, Q'P' be the given secant plane. Then, first, by the general
method of (161) we should take any element of the cone, pass
any auxiliary plane through it by Prob. XI., find a required
point where this element met the line cut by this plane from the
given plane, and repeat this operation for a sufficient number of
elements. But as in Probs. LXXIX. and LXXX., the aux-
iliary plane would be most conveniently perpendicular to a
plane of projection, so here they should all be perpendicular
to the same plane of projection. Let them be vertical. They
will then all contain a vertical line, V-v'V', through the vertex,
158
DESCRIPTIVE GEOMETRY.
and their traces on PQ,Q'P' will therefore all (Prob. XI.) con-
tain the trace of V-v'V' on PQ,Q'P'. (See also Pl. XVI.,
Fig. 129.)
To find this point (Prob. XX.) draw any line, as VMG—M'G',
containing it, and lying in the plane PQQ'P'; then V, S', the
intersection of GM-G'M' with the vertical line V-v'V', is the
required point.
3°. Points of the required curve.-The vertical auxiliary
plane VGG', cuts from the given plane the line VG-S'G' just
found, and from the cone, the two elements (162) VII—V'II',
and one not shown, which intersect VG-S'G' in two points of
the curve, of which one is 'h.
In like manner, any number of points can be found, only that
having the point V, S' common to the system of lines cut by the
auxiliary planes from PQQP', it is only necessary to draw the
horizontal traces of additional auxiliary planes. Thus the
plane Vn gives the trace Vn-n'S' on the given plane, and the
two elements VI-V'I' and VJ-V'J' of the cone, which meet
Vn-n'S' at i'i and j'j, two more points of the curve.
Planes should, in practice, be passed through the extreme
elements of both projections.
cone.
4°. To construct tangents to the intersection of the plane and
Let be a given point of contact. II, in revolving
about AB into the plane H, appears at II". Then II"N, tan-
gent to the revolved base at II", is the revolved position of a
tangent at IIII' on the element hV-'V'. IIence N, its inter-
section with AB-A'B' produced, is a point of the horizontal
trace of the tangent plane along hV-h'V'. The trace s's of
this element is another point, hence Ns is this trace, and RR'
its intersection with the horizontal trace, PQ, of the given
plane, is a point of the required tangent line, which is therefore
R/-R'.
5°. Particular tangents.-The tangent at II' is horizontal,
and parallel to AB-A'B'. Those at and ce' are in vertical
tangent planes, and hence coincide, in horizontal projection,
with the extreme elements of the cone, which are themselves
apparently, in horizontal projection, but not really in space,
tangent to the given curve (151).
Likewise, the tangents, as at l', are in perpendicular tangent
planes, and therefore their vertical projections coincide with
those of the extreme elements.
DESCRIPTIVE GEOMETRY.
159
EXAMPLES.-1°. Let L, LL2 be the secant plane.
Ex. 2°. Let all the auxiliary planes be perpendicular to the plane V.
Ex. 3°. Let the cone be taken with its vertex downward.
Ex. 4°. Let the cone's axis be a bi-parallel ( 48 ).
Ex. 5°. Let it cross the first angle.
Ex. 6. Let the angle L,LG be obtuse.
Ex. 7. Let all the auxiliary planes contain the cone's axis.
214. Construction of intersections by tangents.—First, the
tangent line to the intersection of any plane with a cone or cyl-
inder, is the intersection of the plane of the curve with a tan-*
gent planc to the cone or cylinder. Second, the intersection of
these two planes can be found by Prob. XVII., without first
having the curve. IIence any number of tangents to the curve
can be found before finding the curve itself. Thus there is
another method of constructing the curve, which may be called
the method by tangents.
215. As a tangent marks the direction of the curve at its
point of contact, each point of contact of a known tangent is
as serviceable as two ordinary points in aiding to sketch the
curve. And the point of contact of each tangent can be found
beforehand, since it is the intersection of the tangent line with
the element of contact of the corresponding tangent plane.
Intersections of pairs of developable Surfaces.
216. Intersections of prisms and pyramids with each other,
and with cylinders and cones, are properly included here, as is
evident from the following table.
The intersection of any line on one surface with any line on
the other will be a point common to both, and hence a point of
their intersection. But, for convenience of construction, these
lines should generally be elements of the surfaces.
217. In combining these developable surfaces in pairs, they
may therefore be arranged either with reference to their cha-
racter, as plane sided or curved, or with reference to the
direction of their elements, as parallel or converging. Combin-
ing both of these bases of classification, we have at once the
following:
160
DESCRIPTIVE GEOMETRY.
GENERAL TABLE OF INTERSECTIONS.
COMBINATIONS OF BODIES.
Plane Sided
with
Plane Sided.
Plane Sided
with
Single Curved.
Single Curved
with
Single Curved.
Prisms
Prisms
1. Both having parallel
elements.
Cylinders
with
Prisms.
with
Cylinders.
with
Cylinders.
2. Both having
con-
verging elements.
Pyramids
with
Pyramids.
Pyramids
with
Cones.
Cones
with
Cones.
3°. One having parallel,
and the other con-
verging elements.
Prisms with
Prisms
Cones.
Cylinders
with
Pyramids.
with
Pyramids
with Cylinders.
Cones.
THEOREM XXVII.
The construction of the intersection of two developable surfaces
having a vertex, depends on the direction of their elements.
218. Both bodies having parallel elements.—First, we know
by Prob. XV., that any number of planes can be parallel to
any two given straight lines. Second, that all the elements of
each body are parallel to its axis, and hence to each other.
Hence any plane, parallel at once to the axes or elements of
both bodies, will contain elements of each. Then the points
where the elements cut from one surface meet those cut from the
other, by this plane, will be common to both of the given sur-
faces. That is, they will be points of their required intersec-
tion.
219. Bodies having converging elements.-First, any num-
ber of planes can be passed through two given points. Second,
any plane cutting a cone or pyramid through its vertex, cuts its
surface in two straight elements.
Hence any plane, passed through the vertices of both bodies,
will contain elements of each, and the intersections of the cle-
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DESCRIPTIVE GEOMETRY.
161
ments of each surface, contained in this plane, will be points
common to both surfaces, and therefore will be points of their
required intersection.
220. One body with parallel, and the other with converging
elements. First, any number of planes can be passed through
a given point, and parallel to a given line. Second, any plane
cutting a cone, through its vertex; or a cylinder, parallel to its
azis, will cut rectilinear elements from each.
Ilence any plane, situated as just described, will cut rectili-
near elements from each body, and the points in which the
elements cut from one body meet those cut from the other, by
this plane, are common to both of the given surfaces, and are
consequently points of their required intersection.
221. The three foregoing solutions for the three general
cases, may be thus condensed into one, which applies to the
entire table of intersections.
Pass auxiliary planes through both bodies, in such a manner
that each plane shall contain rectilinear elements of each body;
then the points in which the elements cut from one body, inter-
sect those cut from the other body by the same plane, will be
points common to the surfaces of both bodies, and hence will
be points of their intersection.
222. From the last theorem it is now evident that in every
problem of the Table (217), there are two parts; first, the
initial step, consisting of the location of the auxiliary planes;
which differs for each of the three general cases, represented
by the horizontal columns of the table; second, the construction
of the points of intersection, which is the same for all cases.
223. When the given bodies are plane sided, the auxiliary
planes may intersect them as described in Theor. XXVII., or
in any manner whatever, since these planes will always inter-
sect the faces of the bodies in straight lines.
224. Varieties of the problems indicated in the table can be
made to any extent, simply by varying the positions of the axes
of the bodies; first, with respect to each other; and second,
with respect to the planes of projection.
11
162
DESCRIPTIVE GEOMETRY.
t
2
2
THEOREM XXVIII.
The curve of intersection of two cylinders, etc., is tangent to
those elements of each which are cut from it by an aux-
iliary plane tangent to the other.
225. It follows from (147,) that the tangent line at any point
of the intersection of two developable surfaces, is the intersec-
tion of two planes, each tangent to one of the surfaces along
the element through that point. Then let S and S, be any two-
developable surfaces of revolution which intersect. Let T be
a plane, tangent to S along the element E, and cutting S, in an
element E,, whose intersection with E will be a point, p, of the
curve of intersection of S and S,. Then the element E, will
be tangent to this curve at p. For the tangent at p must be
the intersection of T with a plane T, tangent to S, along the
element E. But the intersection of these planes is E, itself,
which proves the theorem.
226. When both of the tangent auxiliary planes to one of
two developable surfaces S, intersect the other S,, it is evident
that there will be two separate curves of intersection, one of
entrance, and one of exit. Otherwise there will be but one
curve.
PROBLEM LXXXII.
To construct the intersection of two right cylinders, one of
which is vertical, and the other, parallel to the ground line.
In Space.-Vertical planes, parallel to the ground line, would
evidently cut rectilinear elements from both the cylinders,
whose intersections would be points of the intersection of the
cylinders.
But such planes need not be used since it follows from the
position of the cylinders, that the horizontal projection of their
intersection is known at once, being so much of the horizontal
projection of the vertical cylinder, as lies between the foremost
and liindmost clements of the horizontal cylinder, supposing
the vertical cylinder to be the larger one.
Since the plane
DESCRIPTIVE GEOMETRY.
163
of the base of the horizontal cylinder is a profile plane, an ap-
plication of Prob. XLI. will be necessary.
In Projection.-Pl. X., Fig. 92. 1°. The highest, lowest,
foremost, and hindmost points.—sdB-A'B' is the vertical
cylinder, and EF-H'F' the horizontal one.
Fp-F'p' is the highest element of the horizontal cylinder,
and being in the vertical plane of the axes, meets the extreme
elements s-A'A" and B-B'B' of the vertical cylinder, at the
points ss" and Bf". Likewise the lowest element, Fp-H'F",
determines the points ss' and Bƒ".
The foremost element, cg-c'g', intersects the vertical ele-
ments at d and h in the points dd' and hh'. The hindmost ele-
ment, EC-c'g', gives the hindmost points yd' and kh'.
2°. To find intermediate points.-Take any line, as Tu, which
will be the common horizontal projection of two elements.
These clements pierce the circumference of the base Ec in two
points, whose common horizontal projection is T. Revolve
this base about its vertical diameter till it becomes parallel to
the vertical plane, when it will take the position pe—p'e""H′;
the points T will appear at a, and, in vertical projection, at a'
and ". In the counter revolution, a and b describe horizon-
tal ares whose vertical projections are a'T' and '', giving T'
and '. These being points of the assumed elements, the verti-
cal projections of these clements are T'u' and 'm'. These ele-
ments intersect the clements at t and g of the vertical cylinder,
at tt' and te'; and at qm and qq'. Therefore dsy-s'd's" and
hBk—ƒ"h'ƒ" are the projections of the required curves of in-
tersection.
Q
The back half, as sy of each curve, coincides in vertical pro-
jection with its front half, because the axes of the cylinders
intersect. Otherwise, these portions would appear separate, and
the invisible one would be dotted.
EXAMPLES. —1`. Let the axes make an acute angle with each other, one of
them remaining vertical.
Ex. 2. Let the plane of the axes be vertical, but oblique to the plane V.
Ex. 3. Let the axes, not in the same plane, be parallel to any vertical plane.
Ex. 4. Let the axes be in one profile plane.
Ex. 5. Let the axes be in two profile planes.
Ex. 6. Let one axis be a bi-parallel, and the other a co-parallel.
Ex. 7. Let the horizontal cylinder be the larger one.
164
DESCRIPTIVE GEOMETRY.
THEOREM XXIX.
Two equal cylinders of revolution, whose axes intersect, will
intersect in two ellipses.
When the plane of the axes of the cylinders is parallel to a
plane of projection, this theorem becomes sufficiently obvious
experimentally, by observing that the projection of the inter-
section on that plane will always be two straight lines, as if
drawn at "f" and s'f", Pl. X., Fig. 92.
But the theorem can be directly proved from Pl. XVI., Fig.
132, which represents a projection of two equal cylinders whose
axes we will suppose to be in the plane of the paper, considered
as horizontal. Here elements, IIE and IIB, which really inter-
sect, are at equal heights above the paper; but, as the cylinders
are equal, DE = AB; DF = AC, etc., which obviously makes
GHIL a straight line, and therefore the horizontal projection
of an ellipse.
The like is true of MIN, and the same demonstration would,
on drawing the figure, evidently apply to the case where, other
things remaining the same, the axes should make any angle
with each other.
PROBLEM LXXXIII.
To construct the intersection of two cones of revolution whose
ares are horizontal, and vertical, and in a plane parallel
to the vertical plane.
In Space.-By (219) all the auxiliary planes must contain
the line L, or VV₁, Pl. XVI., Fig. 133, which joins the vertices
of the cones. Hence their traces on the plane of the cone's
bases, must contain the traces of the line L upon these planes,
and the intersections of these traces with the respective bases
will be points of the elements contained in the auxilim y planes.
In this problem the bases are in different planes, hence we
must find the traces of L upon both of these planes.
Pl. XVII., Fig. 135. Let vABE-v'A'D' be the vertical cone,
and Vde-V'c'U' the horizontal one. The line joining their ver
DESCRIPTIVE GEOMETRY.
165
tices is ¿V—'V'; and it pierces the horizontal plane at NN';
and the plane M'QP of the base de-c'U', at MM'. When M'QP
is revolved about its vertical trace and into the vertical plane
of projection, MM' will be found at M", and the base ed—c'U'
at e'c''d"U". Then M'a' is the revolved position of the trace
of a plane tangent to the horizontal cone. In the counter revo-
lution, a' goes to a, and aN is the horizontal trace of this tan-
gent plane. This trace does not intersect the base of the ver-
tical cone, from which we conclude that as the plane of the axis
is a plane of symmetry for both cones, the vertical cone enters
the horizontal cone in one curve and leaves it in another.
In this example the planes NH and NL, which are tangent
to the vertical cone, happen to contain the extreme elements
d''V"—e'V', and Ve-V'd of the horizontal cone. Otherwise
two planes would have been convenient on each side of the
central plane NJ, one of them tangent to the vertical cone, and
the other containing an extreme element of the horizontal cone.
This being understood, we may begin with the auxiliary plane,
whose horizontal trace is NL and whose trace on the revolved
plane PQM', is M'L'. This plane contains the element Fv-
Fe of the vertical cone, and the elements eV"—¿'T'—¿V,
and 7"V"—IV-V of the horizontal cone. The intersection
of the former element with the two latter gives the points s'i'
on the lower curve, and ss' on the upper one. V-V', which
kV
contains ss', is found by a counter revolution of " in the hori-
zontal are l'—""k, around M'Q as an axis.
The intersection of other intermediate elements is found in
the same way, as may be seen by the figure. The extreme ele-
ments, as seen in vertical projection, intersect at i'v', etc.
Either the vertical or the horizontal projection of the re-
quired intersection may be found first. Thus, s'", being the
intersection of e'V and F'e', may be projected down into Fr or
¿V, as a check upon, or without reference to, the intersection
s", of the two latter with each other.
From the situation of the cones, it is evident, that in the ver
tical projection, the curves on the back of the cones are exactly
covered by the equal curves seen in front. In the horizontal
projection, the larger curve is wholly dotted, being hidden, and
the upper curve is seen, and is inked full.
GENERAL EXAMPLE.-Repeat for the cones the examples of Prob. LXXXII.
166
DESCRIPTIVE GEOMETRY.
THEOREM XXX.
When two cones of revolution, with different vertices, have two
common tangent planes, they will intersect in two plane
curves.
Let the ellipse achf, Pl. XVII., Fig. 137, be the common
base of two cones of revolution found as in Prob. LIX., and
whose vertices are VV' and vv'. The point t is the horizontal
trace of the line joining the two vertices, and te and tf are the
like traces of the common tangent planes. Hence ce' and fe are
points of the second branch of the intersection of the two cones,
and p' is another.
Now it is so obvious by experiment, that as the cones approach
the given condition, their intersection, projected on a plane
parallel to that of their axes, will approximate to, and finally
become two straight lines, as bn and mp, Fig. 136, that, without
giving the general proof for all cases, the following is sufficient
for the case of cones of revolution. Take planes as M'N' and
E'D' on opposite sides of a'b'. Each will cut two ellipses from
the two cones; one of which, as E'D, will be larger, the other,
e'd', smaller than a'b'. Then as these planes revolve towards
ab' about the common chord of contact, ef, of the common tan-
gent planes, the four ellipses finally coincide in a''.
Now, in the absence of any reason why they should not in
like manner coincide again in an ellipse in the plane p'e'e, when
revolved towards that plane, we conclude that this plane con-
tains a second ellipse common to both cones. For if it be said
that the plane a'b' cuts the line V'' beyond both vertices, and
cc'p' between them, it is evident that the vertices VV' and vo'
might have been constructed on opposite sides of a'b' ; in which
case, the plane of ab would have cut the line of the vertices as
V'v' is now cut by cc'p'.
PROBLEM LXXXIV.
To find points of the intersection of two cylinders of revolu-
tion, given by their horizontal traces; their axes being
oblique to both planes of projection.
In Space. If at any assumed point two lines be drawn
DESCRIPTIVE GEOMETRY.
167
respectively parallel to the axes of the cylinders, any plane
parallel to the plane of these lines, and intersecting the cylin-
ders, will be parallel to both of the axes, and will therefore con-
tain elements of each cylinder, which will intersect in points
common to both surfaces, that is, in points of the required
intersection.
227. Then let or and nu be the axes of two intersecting cylin-
ders, Pl. XVI., Fig. 130, which represents a model of two such
cylinders, and if rs be a line parallel to nu, from any point
assumed on or, and if s be its horizontal trace, os will be the
horizontal trace of a directing plane, through or and parallel to
nu; and any plane dg, parallel to the plane os, will by (218)
cut cylinder o in the elements at d, e, which will intersect
those cut at ƒ and g from the cylinder n, in the points h, k,,
and m.
and
Q.
Thus each plane will generally give four points of the curve
of intersection, unless, like abe, it be tangent to one of the
cylinders, as at a, when it will give but two points, as p
If it be tangent to both bodies, it will give but one point.
228. It only remains to notice that the auxiliary planes should
not be distributed at random, but so as to contain the extreme
elements as seen in both projections, since at these elements
the curve passes from the visible to the invisible side of a sur-
face. Also, the two planes as abe which are tangent to either
body and cut the other should be used, since the curve is wholly
included between them.
In Projection.-Pl. XVI., Fig. 131. 1°. The construction
of the cylinders and of the directing plane.—Designating the
two cylinders and their horizontal traces by the centres of the
latter, let the ellipses, OO' and NN', be the given traces or
bases of the cylinders OO' and NN'. Then tangents at a and
d; and b and e; extremities of the shorter axes of these bases,
will complete the horizontal projections of the cylinders.
Their vertical projections consist of tangents from e' and f',
and from g' and ' to the circles having N' and O' respectively,
as centres, and radii equal respectively to Nb and Oɑ.
Next, drawing the axes through 00' and NN', and parallel
to the elements just drawn, take any point on either axis, as
RR' on the axis OR-O'R', and through it draw RS-R'S'
parallel to the other axis, NU-N'U'. (See also, Fig. 130.)
The horizontal trace of this line is S'S, hence SO is the hori-
168
DESCRIPTIVE GEOMETRY.
A
zontal trace (Prob. XV.) of a plane through the axis of 00'
and parallel to that of NN'.
2°.-The determination of the number of curves, and the
construction of their points.-Draw first the extreme planes,
A and I, parallel to SO; each of them being tangent to one
(but not to the same) body and secant to the other, there will
be one curve (226). The tangent plane, A, contains the element
10 of cylinder O, and the elements 10 and 19 of cylinder N,
which, being followed up, intersect the former at the points
10 and 19 of the curve.
The plane, B, cuts two elements from each cylinder, 11 (18)
and 1 (9) from cylinder O, and 11 (9) and 1 (18) (either num-
ber at each point may be used to designate the point) from
cylinder N. These intersect at the four points, 1, 9, 11, and 18,
of the curve. Likewise, any number of points can be found.
229. But note, that to avoid mistakes, the planes should be
drawn one at a time, and all the points given by each, noted,
in both projections, before another plane is drawn.
The planes B and II contain those extreme elements in hori-
zontal projection on which any required points are found.
The planes C, D, E and G, contain the like extreme elements
as seen in vertical projection.
The initial plane, F, is not used, in this case, since the rest
afford points enough, even without finding all the points
determined by each plane.
230. The vertical projections of the points can be found in
either of two ways; by projecting them upon the vertical pro-
jections of the elements containing them, on either one of the
cylinders; or by noting them, each, as the intersections of the
vertical projections of both the elements containing it.
PROBLEM LXXXV.
To connect the points, found in a required intersection of two
cylinders, etc., and to determine its principal features.
In Space.-See Pl. XVI., Fig. 180 and 181. First, we may
begin at any point in any plane. Second, the next point must
be in the adjacent plane, since the curve is of double curvature.
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DESCRIPTIVE GEOMETRY.
169
Third, in pencilling the curve from point to point, we follow
the corresponding elements on the bases.
In Projection.-1°. The connection of the points.-Beginning
at any point at pleasure on the curve, as 1 in the plane B, the
next point must be in an adjacent plane. Then, observing that
1 is the intersection of the elements 1 and 1, one on each cyl-
inder, and in the same plane B, pass, on the bases, to elements
2 and 2, in plane D, and hence, with pencil, join 1 with 2 on the
curve;
thence to elements 3 and 3, and join 2 with 3 on the curve.
66
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Now, being in a tangent plane, I, we recross the same elements,
just noted on cylinder O, but at their intersections with new
elements, on cylinder N. Hence on the bases, we return, on
O, through 4, 3, 2, etc., but we advance still on N. Thus,
pass to elements 6 and 6, and join 5 with 6, on the curve,
thence
7 66 7,
66
6
7 (C
and so on, to the point 10 on elements 10 and 10, in the tangent
plane A, thence pass to elements 11 and 11, and join 10 with 11,
on the curve; thence pass to elements 12 and 12, and join 11
with 12, on the curve, and so on to point 15, on elements 15
and 15, in the tangent plane, I, again. Now, progressing on
cylinder N and returning on cylinder O; theuce pass to ele-
ments 16 and 16, and join 15 with 16, on the curve; thence
pass to elements 17 and 17, and join 16 with 17, on the curve;
and so on to point 19 on elements 19 and 19 in the tangent
plane, A, once more; thence to elements 1 and 1, and join 19
with 1, the point of beginning.
Thus, on the curve, we have progressed uninterruptedly for-
ward from 1 around to the same point again.
The connection of the points in vertical projection, is made
very simple by numbering each, as found, with the same num-
ber that denotes its horizontal projection. The points can then
be connected, without reference to the traces of the cylinders,
simply by following the order of the numbers.
2º. The visibility of the curve.-The condition for the visi-
bility of any point is, that both of the elements containing it
shall be visible at that point. The curve will then be visible
each way from the point taken, until it runs into the invisible
portion of one or the other of the given surfaces.
170
DESCRIPTIVE GEOMETRY.
Thus take the point 16. It is visible, being the intersection
of the elements 16 and 16, each of which is on the upper, or
visible half of its own cylinder, as seen in horizontal projection.
Hence the curve is visible from 16 to 18, where it runs under
the cylinder O; and from 16 to 14, where it runs under cylin-
der N. All the rest of the curve is invisible in horizontal pro-
jection.
Again, take the points 10 and 18. These are both visible in
vertical projection, because each is seen to be at the intersection
of elements, 10 and 10, and 18 and 18, lying on the front half
of the cylinders as seen in vertical projection. Hence the curve
is visible in vertical projection, from 18 to 2, and to 17, where
it runs behind cylinder N, and from 10 to 8 and to 12, where it
also runs behind the same cylinder.
3°. The visibility of the extreme elements.-Beginning at the
traces and following up the elements, there are three cases.
First. An extreme clement of one surface may intersect a
visible element of the other. In this case it will be visible up
to that intersection. Thus the extreme element at ff", of
cylinder N, as seen in vertical projection, intersects the visible
ones from 8 and 12 on cylinder O. It is therefore visible, in
vertical projection, from f" to 8; is invisible in passing through
cylinder O, from 8 to 12, and reappears at 12.
Second. An extreme element of one body may intersect an
invisible element of the other. In this case, it will disappear
at its apparent intersection with the extreme element of that
other body. Thus the extreme element at d in horizontal pro-
jection, intersects the invisible element at 11 (9) of cylinder
N, and therefore disappears at k and reappears at point 18
where it intersects the visible element 1 (18).
Third. An extreme element of one body may not intersect.
the other body at all. In this case its visibility can be deter-
mined by comparison with that of the like elements of the other
body. Thus the extreme element at a, of cylinder O, does not
intersect cylinder N. But the extreme clement at e, of cylinder
N, intersects the invisible element 4 (6) of cylinder () at the
point 6, and hence disappears at n, from which we see that the
extreme clement, a, is visible throughout.
4°. The predetermination of the character of the intersection.
Suppose the cylinder O alone to be given, and the lines RS-
R'S' which will determine only the direction of the axis of N.
DESCRIPTIVE GEOMETRY.
171
Then draw two traces, A and J, tangent to the base of O, and
parallel to OS. If now we assume the base of N, either be-
tween, but not tangent to, A and J, or, so that both of the two
traces tangent to O shall intersect it, there will be two curves.
If N be taken, as in the figure, so that one of these traces, as J,
does not intersect it, there will be one curve. And on this case
note, that had the base N been nearer the ground line, so that
the points b and a would both have fallen between the planes A
and I, there would have been a fuller curve with a loop and
visible portion about the point 5.
If plane A, for example, had been tangent to both cylinders,
I being, as now, secant to one of them, the curve would have
really crossed itself at the intersection of the elements of con-
tact of the plane A.
Finally, if N had been tangent to both A and J, the inter-
section would have become two ellipses crossing each other
(Theor. XXIX.).
The point where a curve thus really crosses itself, is called a
double point (112).
The only feature peculiar to each of the three cases (Theo.
XXVII.) having been there explained, the detailed explanation
of the last problem and of this one, applies equally to two cones,
or to a cylinder and cone; so that the following examples can
now be solved without further explanation:
EXAMPLES.-1. Find the intersection of two cones of revolution, the axis
of each being oblique to both planes of projection. See (219) for the initial
step, then find and connect the points as in Probs. LXXXIV. and LXXXV.
Ex. 2. Find the intersection of a cylinder and cone, each oblique to both
planes of projection (220) for the initial step.
Ex. 3°. In Prob. LXXXIV. let there be two curves. Prob. LXXXV., 4.
Ex. 4. In Prob. LXXXIV. take N so that the plane A, Pl. XVI., Fig. 131,
shall be tangent to it, and secant to 0.
Ex. 5. In Prob. LXXXIV., let plane A be tangent to both bodies.
Ex. 6'. In Prob. LXXXIV., let the axes of both cylinders cross the first
angle.
Ex. 7. Let SO, Pl. XVI., Fig. 131, be nearly perpendicular to the ground
line. That is, take the ground line perpendicular to its present position.
Ex. 8. Take two cones so that there shall be two curves.
Ex. 9. Take two cones so that the curve shall cross itself.
Ex. 10. Take two cones so that one tangent auxiliary plane to each shall
cut elements quite near each other from the other.
Ex. 11. Repeat Ex. 8, with a cylinder and cone.
Ex. 12'.
9°,
Ex. 13°.
10˚,
172
DESCRIPTIVE GEOMETRY.
Ex. 14°. Find the intersection of a cylinder, parallel to the ground line, with
a vertical cone.
Ex. 15. Find the intersection of a cone, whose axis is parallel to the ground
line, with a vertical cylinder.
Ex. 16. In each of the last two examples, let the horizontal axis be oblique
to the plane V.
PROBLEM LXXXVI.
To construct the intersection of a prism and a pyramid, each
having any position in space.
In Space. This problem from column (3) in the table of
intersections, is introduced partly to illustrate the fact that
when both the bodies given in the problem are plane sided, the
auxiliary planes may be passed through them in any position
whatever, since they will still cut right lines from the surfaces
of both bodies.
Let the auxiliary planes contain the edges of the prism; then
the points in which these edges meet the lines cut from the
faces of the pyramid by the auxiliary planes containing them,
will be points of the intersection of the prism and pyramid.
Neither projection, alone, of the prism determines its shape;
hence, any three parallel lines being assumed as the horizontal
projections of the edges of the prism, any other three may be
assumed as their vertical projections, and any one of the former
may be taken as being, in space, the highest or the lowest edge.
In Projection.-Pl. XVIII., Fig. 141. VABC-V'A'B'C'
are the projections of the pyramid; and as, bt, and cu are the
horizontal, and a'd', b'e' and e'f' the vertical projections of the
edges of the prism; sa being assumed as the highest of the
former three lines.
Let the auxiliary planes through the edges of the prism be
vertical; then their horizontal traces will coincide with the hori
zontal projections of those edges. Project the points m and s,
horizontal projections of the points in which the vertical auxili-
ary plane, as, cuts the edges AB and AV of the pyramid, into
the vertical projections A'B' and A'V' of the same edges, giving
m's', the trace of the auxiliary plane containing the edge as-
a'd', upon the face ABV—A'B'V' of the pyramid. Hence, d'
is the vertical projection of the intersection of the edge as—a'd'
DESCRIPTIVE GEOMETRY.
173
of the prism with the face A'B'V' of the pyramid. The other
points, e' and f', are found similarly, and all are then connected
and projected into the horizontal projections of the edges of
the prism to which they belong, d' at d, e' at e, etc.
The auxiliary planes being parallel to each other, their traces,
m's', n't', etc., on the same face of the pyramid, must be paral-
lel. Also, s', t' and ' must be points of the parallel intersec-
tions of these same auxiliary planes with the face ACV-A'C'V'
of the pyramid. The other points, as v', of these intersections,
are found, as were m', n', etc., by projecting up the points, as v,
in which the horizontal traces of the auxiliary planes cut the
horizontal projection of the edge AC of the pyramid.
The remainder of the construction of the intersection of the
prism with the face ACV-A'C'V' of the pyramid, involves no
operations different from those already explained.
Visibility. In Fig. 141, the line d'f'-df joining the highest
and foremost elements of the prism will be visible, the foremost
point being also the lowest. In Fig. 142, where the foremost
edge, b', is between the highest and lowest, the visible edges, in
vertical projection, will be a's and e'u, the real portions of the
lines joining the highest and lowest points with the foremost one.
EXAMPLES.—1'. The edge AV-A V' might happen to be nearly in a plane
perpendicular to both planes of projection. If so, points, as s and t, could not
be accurately projected to determine s' and t. But A" V-A¨V', represents the
edge AV—A´V', revolved about a vertical axis at V, till parallel to the plane V.
In this revolution, t revolves to t, and can be more accurately projected at t´´,
from which, in the counter revolution, it returns in the horizontal arc, tt-
t't', to the point t'. Construct a case of this kind.
Ex. 2. The case shown in Fig. 142, may arise, in which one or two of the
edges of the prism do not intersect the pyramid at all. The figure is only a
sketch, in which is supposed to be the foremost edge of the prism, so that
it will be in the same plane with th, the foremost trace of an auxiliary plane
on the face ABV of the pyramid. Now, the and bb', being in the same plane,
must intersect, which they do, at b', but the being a line of the plane face.
ABV, the point b is in the indefinite plane of which ABV is but a limited por-
tion, hence is joined with a' and e', the points in which the other two edges
of the prism meet the face ABV. Furthermore, the same edge, ll', being in
the same plane with f', a line of the face ACV, pierces the indefinite plane of
that face at o, which should be joined with n and p, the points in which the
remaining edges of the prism pierce the same face, ACV, of the pyramid. Let
this, or a similar case, be completed in projection.
Ex. 3. Let one or more of the edges of the prism intersect the base of the
pyramid.
Ex. 4°. Let each body be regular, and with four or more sides.
174
DESCRIPTIVE GEOMETRY.
PROBLEM LXXXVII.
To draw a tangent to the intersection of two developable sur-
faces of revolution at a given point of the curve.
In Space. This problem has one general solution for all
cases. Thus, the required tangent is the intersection of the two
tangent planes, one to each surface, along the element contain-
ing the given point. The latter point is thus one point of the
tangent, and the intersection of either pair of traces of the
two planes, is another point; hence only one trace of each tan-
gent plane need be found.
In Projection.-Pl. X., Fig. 92. 1°. To draw a tangent line
at the point, tt', of the intersection of a vertical and a bi-paral-
lel (48) cylinder. t-e't' is the element of the vertical cylin-
der, containing the point of tangency. tn is the horizontal
trace of the tangent plane along this element, and as the plane
is vertical, tn is also the horizontal projection of the tangent
line, and n is the horizontal projection of the point in which it
pierces the plane of the axes. T-t'T' is the element contain-
ing the point of tangency and lying in the horizontal cylinder.
By revolving the base of this cylinder about its vertical diame-
ter p-p'II', the point TT' takes the position T", and T'P',
tangent to the revolved position of the base at T", is the re-
volved position of the trace of the tangent plane to the hori-
zontal cylinder upon the plane of the base of that cylinder.
But the plane of the base cE=II', and the plane of the axes,
are at right angles to each other, and p-H'P' is their intersec-
tion, analogous to the ground line. The tangent plane to the
horizontal cylinder is perpendicular to its base, and T'P' is its
trace on the plane of that base, hence P'n', perpendicular to
P'I', is the trace of this tangent plane on the plane of the axes.
P'n'-pn and tn being lines of this tangent plane, and n the
horizontal projection of their intersection, n' is its vertical pro-
jection, and n't' is the vertical projection of the tangent line
required.
2°. To construct a tangent line at the point tt' of the curve
of intersection of two right cones, Pl. XVII., Fig. 133.
The clement of the vertical cone, at tt', pierces the horizon-
tal plane at G, and GP is the horizontal trace of the tangent
DESCRIPTIVE GEOMETRY.
175
plane along this element. On the horizontal cone Vm-V'm'
is the element through the given point of tangency tt', and its
auxiliary vertical projection is V"m". Then m'p" is the trace
on the plane of the base of this cone, of the tangent plane
along this element. This trace pierces the horizontal plane in
a point whose revolved position is p', and whose true position
is p; therefore p is a point of the horizontal trace of this plane.
To find another point, draw through any point, s'', of the
element m'V"-m'V' a line "z"-'' parallel in space to the
trace m'p"; it will be parallel to the plane of the base of the
horizontal cone, will be a line of the tangent plane to this cone.
It pierces the horizontal plane at the point z", 2′, z.
Hence pz is the horizontal trace of the tangent plane to the
horizontal cone. Joining, where the horizontal traces of the
two tangent planes intersect, with t the horizontal projection of
the point of tangency, we have rt for the horizontal projection
of the required tangent line.
The point being in the horizontal plane, is vertically pro-
jected in the ground line, and r't' is the vertical projection of
the tangent line.
3°. To construct a tangent line at any point of the curve of
intersection of two cylinders, each having any position. Pl.
XVI., Fig. 131. Let tt' be the given point of tangency, it is
therefore one point of the required tangent line. The elements
of the cylinders, containing the point tt', pierce the horizontal
plane at T and T", hence TP and T'P are the horizontal traces
of the tangent planes along these elements, and Pt is the hori-
zontal projection of the required tangent line. P, being pro-
jected into the vertical plane at P', we have P't' for the verti-
cal projection of the tangent line.
4°. To draw the tangents at the multiple point (112) of a .
curve of intersection of two developable single curved surfaces.
In this case the tangent planes, one to each surface, and
whose intersection would be the required tangent in any other
case (225) give no tangent, as they should not, since they coin-
cide in one, which, moreover, contains the two tangents which
can always be drawn to a multiple point. It is therefore neces-
sary to apply to the projections, separately, considered as plane
curves, the method of trial curves as follows:
Let ATBC, Pl. XI., Fig. 101, be any curve having a multiple
point T, at which a tangent is to be drawn. Through T draw
176
DESCRIPTIVE GEOMETRY.
any number of secants of that portion, as ATP, to which a
tangent is to be drawn. Also, describe an arc Ad with centre
T, and radius equal to any one of the chords, as TP. Then lay
off from this arc and on each secant, a distance equal to the
chord of the given curve, made by the same secant, and in the
same direction from the arc, that the chord is from T, and thus
form the trial curve mtnoT, whose intersection, t, with the
are, and the point T, determine the required tangent tT, at T.
For having, as just directed, made co = TO; bn = TN;
am TM, etc., the curve mtn is the locus of distances from
the arc Ad, equal to the respective chords reckoned from T.
Hence at t is a point of the secant whose chord, beginning at
T, reduces to the point T itself. Hence this is the tangent
at T.
A similar construction would evidently give the tangent at
T to the portion DTC, of the curve, or to any point of any
curve.
5°. To draw a tangent to the curve in Pl. XVI., Fig. 131,
and parallel to a given plane. As the most useful case, let the
tangent be parallel to a plane of projection; for example, to
the plane H.
The horizontal traces of the tangent planes whose intersec-
tion is the required tangent line, will be parallel to that tangent
and to each other. But the elements of contact of these tan-
gent planes, being in some one of the auxiliary planes, such as
are used in the construction of the curve, the traces of these
planes will be tangent to the bases of the given surfaces at
their intersection with the trace of that auxiliary plane. Thus
the problem reduces to drawing parallel tangents to the ellipses
O and N, and whose points of contact shall be on a line paral-
lel to OS.
This can be done most simply and with practically entire
accuracy, by the method of trial curves. Thus, to avoid con-
fusion of lines, let the ellipses, O and N, Pl. XVI., Fig. 134,
represent O and N, Pl. XVI., Fig. 131. Then draw tangents
to either, as O, at P, Q, R, etc., and chords AA,, BB,, CC,, etc.,
respectively parallel to them at A, B, C, etc. At some point.
of ellipse N, the chord will become the required tangent. To
find it, make each of the above chords the common base of two
triangles which, since they must be similar, as will be evident
on completing the figure, may most conveniently be equilateral.
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DESCRIPTIVE GEOMETRY.
177
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​Their vertices are a and a,, b and b₁, c and c,, etc., which, when
joined, give the curve abcb,, whose intersection with the ellipse
N gives d, the point through which, if a line parallel to OS be
drawn, it will give the points of contact, d and T, of the re-
quired parallel tangents.
These tangents, being the horizontal traces of the tangent
planes containing the required horizontal tangent line, the
intersection of the elements at T and d will be the point of
contact of this tangent, whose horizontal projection will be
parallel to the traces TM and dN, and whose vertical projection
will be parallel to the ground line.
THEOREM XXXI.
There is the same kind of variety in the intersection of two
cones as in that of a plane and cone.
This theorem might be inferred from the fact that a plane
is only a particular case of a cone of revolution (50). But it
may be directly demonstrated as follows:
Let C and C,, Pl. XVI., Fig. 133, be two cones.
Let every
point of C, move in the direction of the line V,V, joining their
vertices, until V, and V coincide, and call this transported posi-
tion of C,, not shown, C.
Case 1°. If now, as in the figure, if C, were shown, the hori-
zontal traces of cones C and C, have no point in common, it
shows that these cones have no element in common, and hence
that no element of the cone C, is parallel to any of the cone C.
Hence all the elements of C and C,, respectively, which inter-
sect, will do so at a finite distance, and thus the intersection of
these cones will be a closed curve belonging therefore, in a
general sense, to the class of elliptical intersections.
Case 2°. If the traces of the given cone C, and auxiliary cone
C, are tangent to each other, these cones will have one element
of contact. Consequently, the given cones C and C, have two
parallel elements, one on each, which intersect at infinity; and
they therefore give a curve having one point at infinity like a
parabola. This curve belongs therefore, in a general sense, to
parabolic intersections.
Case 3°. Finally, if the traces of the cones C and C,, have a
12
178
DESCRIPTIVE GEOMETRY.
common chord, it shows that these cones have two elements in
common. Hence two elements of C are parallel to two of C
and these elements intersect, giving points at infinity, as on a
hyperbola. These elements are thus asymptotes to the curve,
which belongs to the general class of hyperbolic intersections.
EXAMPLE.-Construct each of the cases of the above theorem. [The points
will be found and connected just as in the ordinary case, Prob. LXXXV; the
special feature of the work being only the use of the test cone C2. The parallel
elements are regarded as intersecting at infinity in both directions, and accord-
ingly the points, as 14 and 16; or 9 and 11, Pl. XVI., Fig. 131, met, in ap-
proaching near to either of the tangent auxiliary planes, as explained in Prob.
LXXXV., will be towards opposite ends of the same asymptote.]
231. In the case of the cone and cylinder, if the auxiliary
line through the vertex of the cone, and parallel to the axis of
the cylinder, happens to be itself an element, K, of the cone,
the tangent plane along this element and cutting the cylinder,
will contain those elements of the cylinder, which will contain
points at infinity on the required intersection, that is, at their
intersection with the parallel element of the cone. Thus there
may be intersections having infinite branches for a cylinder and
cone, as well as for two cones.
232. It is the tangent plane along the element K, whose
intersections with the cylinder will be the elements, E and E,,
which are asymptotes of the curve of intersection of the given
surfaces, because these asymptote elements, as will be imme-
diately evident on making the construction, will be the inter-
sections of the tangent plane to the cone on K, and the tangent
planes to the cylinder along E and E,, as they should be by
( 225 ).
EXAMPLE.-Construct the intersection of a cylinder and cone, so situated as
to give a curve with an infinite branch.
233. Various peculiar appearances in the intersections of
cylinders and cones in certain cases, will now give interest to
the following special principles relating to the projections of
curves of double curvature.
Let DC, Pl. XVII., Fig. 138, be a curve of double curvature,
cut at the points a and T by the plane PQ, which contains the
tangent line 1 L to the curve DC at the point T. If now this
plane turns about TL as an axis, cutting the curve successively
DESCRIPTIVE GEOMETRY.
179
at other points as a,, a₂, etc., it will become the osculatory plane
at T, when a comes to coincide with T.
A curve is not generally crossed by its tangent line, as it is
at a point of inflexion; but the plane PQ, when osculatory,
does cut the curve DC at T, since TD will then be below it,
and the part TC, above it.
234. It may happen that the osculatory plane at a certain
point does not cut the curve at that point. Thus, the curve
DC might have been shaped in the region about T, as at DC,,
so that PQ should cut it in two points a and a, at the same
time. Then by turning PQ about TL, a and a, would together
coincide with T, and the curve DC, would be wholly below
PQ.
This is just what occurs at the points where the intersection
of two cylinders, etc., is really tangent to an element, as at 10,
or 15, Pl. XVI., Fig. 131, in which case the next points, on
each side of such a point, are found on one element.
THEOREM XXXII.
When an element of a cylinder, or cone, is tangent to a curve
of double curvature upon the surface, the osculatory plane
of the curve, at the point of contact, is a tangent to the
surface.
Let the element VT, Pl. XVII., Fig. 139, of the cone VATB,
be tangent at T to the portion ATB of a curve of double curva-
ture, formed by the intersection of this cone with another curved
surface.
By (145) as the planes TVA or TVB revolve about the ele
ment TV as an axis, the variable points A and B will fall at T.
But when this happens the planes will become osculatory to the
curve at T, and tangent to the cone along TV, since it then
contains no other clement of it.
In other words, whenever one of the auxiliary planes used in
finding an intersection, as in Theo. XXVII., is tangent to one
body, and secant to the other, it is an osculatory plane of the
curve at the points determined by the elements contained in it.
180
DESCRIPTIVE GEOMETRY.
THEOREM XXXIII.
The projection of a curve of double curvature may present a
point of inflexion, or a cusp.
Let DOC-D'O'C', Pl. XVII., Fig. 140, be a curve of double
curvature, cut at OO', by the horizontal plane. of projection,
which is its osculatory plane. Its vertical projection therefore
shows a point of inflexion at O'; which is thus seen upon a
plane perpendicular to the osculatory plane.
Let G,L, be the ground line of a new vertical plane, perpen-
dicular to the tangent 00". Then a" and D" are the new
projections of the points aa' and DD' of the curve, and a"O"
D" is the new projection of DOC=D'O'C'.
Again, if D,C,-D,'C,' be another curve of double curva-
ture, not cut at OO' by the horizontal plane which is supposed
to be its osculatory plane, then O' will not be a point of inflex-
ion; and the projection of the curve on a vertical plane at
G,L,, perpendicular to the tangent 00", will appear as at
b"O""¿".
235. Finally, D"O"a" and b″O""" are cusps, of the first and
second species (112) respectively, for O"O" is an element of
the projecting cylinders of both curves upon the parallel verti-
cal planes, at G,L, and G,L, and the horizontal plane, as the
osculatory plane of the curve is tangent to this cylinder (Theo.
XXVIII.), hence its traces, GL, and G,L,, are tangents to the
bases of the cylinder, which are the projections, D'O"a" and
b"O""c", of the curves.
و
236. It follows from the forms of the bases of the projecting
cylinders just described, that the section of either cylinder by
any plane cutting the element O"O"" will be a cusp. That is, if
the curved directrix, as DC-D'C', of a cylinder, or a cone, be
tangent to one of its elements, as O"O", such an element is
called the cuspidal edge of such cylinder, or cone.
237. When a plane of projection is perpendicular to any
line, which cuts a curve of double curvature in two points, there
will be a loop or leaf in the projection of the curve on that
plane, and the projection of the two supposed points will be a
DESCRIPTIVE GEOMETRY.
181
double point (112) in the projection of the curve; and if by
moving the secant parallel to itself, it somewhere becomes a
tangent, the loop will reduce to a cusp point.
238. The two arms, O""b" and O""c", of a cusp of the second
species, may sometimes coincide, as in the case of the projec-
tions of the intersection of the two cylinders, Pl. X., Fig. 92. In
the horizontal projections, for example, d and y are such cusp
points, then called points of arrest, or stop points, beyond which
the portions dh and yk are parasites (113) of the real portion
dsy.
239. Asymptotes being like other tangents, except in the
distance of their points of contact; like principles to the fore-
going are true for them. Thus, the projection of the asymptote
on a plane, P, perpendicular to it, is a point which is also the
projection of the infinitely distant point of contact of the
asymptote. When any plane, Q, containing the asymptote, A,
and cutting the curve in a point a, is revolved about A till a
becomes infinitely distant from P, it becomes the osculatory
plane at infinity, and its trace, as in Fig. 140, becomes tangent
to the projection of the given curve having an infinite branch.
240. If the asymptote be oblique to the plane P, its projec-
tion on that plane will be an asymptote to the like projection
of the curve.
241. If the projection of a curve has an asymptote, the pro-
jecting cylinder of the curve, must necessarily have a plane,
asymptote to it, which is the projecting plane of the asymptote.
But, as seen in the vertical projection of the intersection of the
two cylinders, Pl. X., Fig. 92, the projection of what is a closed
curve in space, may be an open curve, which, if continued to
infinity by a parasite portion might have an asymptote. Thus
even a closed curve in space may, including its parasite, be in-
finite in projection, and may have an asymptote.
242. Thus the resemblance of the vertical projections of the
intersections of the two cylinders, Pl. X., Fig. 92, and of the
two cones, Pl. XVII., Fig. 135, to a hyperbola is obvious. It
is further explained by observing that two intersecting straight
lines really are one case of the hyperbola (170, 5°), and that
182
DESCRIPTIVE GEOMETRY.
two such lines are just what these intersections become, when
the two bodies have, in common, two tangent planes.
We therefore infer, what will be more strictly proved in the
second part, that when these surfaces have a common meridian
plane, the projection of their intersection upon this, or a paral-
lel plane, will be a hyperbola.
B-DEVELOPMENTS.
1-Development of Plane Intersections.
243. In case of developments by transposition, since a curve,
and its tangent, revolve as a rigid system about a fixed axis,
neither their form nor relative position will be disturbed, and
the tangency will therefore remain as before revolution.
244. Having seen, in previous constructions of the ellipse,
the convenience of having the axes given, while they are fre-
quently unknown in the elliptical intersections of planes with
cylinders and cones, it is desirable for the sake of accuracy in
developing such intersections, to find their axes. This can be
done purely by Descriptive Geometry as follows:
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PROBLEM LXXXVIII.
To find the axes of the elliptical intersection of a plane with a
cylinder, or a cone; both being oblique to both planes of
projection.
In Space.-Project the cylinder, Pl. XV., Fig. 126, and the
cone, Pl. XV., Fig. 127, each upon a plane parallel to its axis,
and perpendicular to the given cutting plane. The new pro-
jections of these figures will thus be reduced to the case shown
in the vertical projections in Pl. X., Figs. 90 and 91, so that the
transverse axes will be immediately known. Then the conju-
gate axes of the section of the cylinder will be equal to the
diameter of the cylinder; while that of the conic section will
be perpendicular to the transverse axis of the ellipse at its mid-
dle point, and limited by the surface of the cone.
DESCRIPTIVE GEOMETRY.
183
In Projection. Find the axes of the elliptical section of a
cylinder and of a cone as above indicated.
PROBLEM LXXXIX.
To show the true form and size of the intersection of a develop-
able surface by a plane, together with its tangent line,
having given their projections.
In Space. In the various applications of this gencral prob-
lem, the axis (55), about which the given figure is revolved,
should be taken in the plane of the given curve, and either
coinciding with, or parallel to one of its traces. When the
curve is symmetrical, having an axis parallel to either plane of
projection, it is quite convenient to take that axis as an axis of
revolution.
In either case, the construction consists in finding the true
distances of the points of the curve to be revolved, from the
axis of revolution.
In Projection.-1°. To find the true size of the intersection
of the plane and pyramid.-Pl. X., Fig. 89.
Revolve the cutting plane, PQP', about one of its traces as
an axis, till it coincides with the plane of projection containing
that trace.
Let this plane be revolved about its horizontal trace. As the
construction is very simple, and the same for all points of the
intersection, it is only shown for the point aa'. The perpen-
dicular distance of the point, aa', in space, from the trace PQ,
is the hypothenuse of a right angled triangle, whose base is ar
-the perpendicular distance of the horizontal projection, a,
from the axis PQ-and whose altitude is the vertical height,
equal to a's, or a's', of the point in space above a.
Hence, construct this hypothenuse and lay it off from, on
ar produced, and "", the development of aa', will be deter-
mined. This hypothenuse is laid off on ar produced, because,
the axis PQ being horizontal, the are described about it by ca',
will be in a plane perpendicular to it, through aa', and will
therefore be horizontally projected in the horizontal trace ad'"
of this vertical plane of revolution.
EXAMPLE. Let the cutting plane be revolved about its vertical trace, P'Q,
and into the vertical plane of projection.
*
184
DESCRIPTIVE GEOMETRY.
2º. To find the true size of the curve of intersection of the
plane and right cylinder, Pl. X., Fig. 90, and to develope the
tangent to it.-The transverse axis, AB-a'b', of the curve,
being parallel to the vertical plane, a'b' is its true size. The
plane of the curve being perpendicular to the vertical plane,
mq, oC, etc., parallel to its horizontal trace PQ, are therefore
the true distances of points of the curve from the axis. Hence,
after revolving the curve about a'b', till parallel to the vertical
plane of projection, mq, etc., will appear at T't, Nn', etc., per-
pendicular to a'b'; giving t, N, etc., for points of the required
development.
The tangent at TT' intersects the axis a'b'-AB at ss', a
point which remains fixed during the revolution; hence, sup-
posing the point TT' to revolve to the position t, the develop-
ment of the tangent will be s't.
3°. To find the true size of the intersection of a right cone,
with a plane perpendicular to the vertical plane of projection,
and to show the revolved tangent to it. Pl. X., Fig. 91.-The
transverse axis, pk-p'h', parallel to the vertical plane, is taken
as the axis of revolution. Then, as in the last example, p'h' is
its true size, and tb and gb, etc., are the true distances of points
of the curve from this axis, and are laid off on perpendiculars,
t't'', t'q', to the true size, p'k', of the axis, and from the points,
t', etc., vertical projections of the points, b, etc.
The tangent at tt' intersects the axis pk-p'k' at ss', a point
which, therefore, remains fixed during the revolution; and
supposing the point of tangency, tt', to revolve upward, s't''
will be the revolved position of the tangent st—s't'.
The preceding will enable the student to readily work out
the following:
EXAMPLES.-1°. Show the true size of the intersection of the plane and
cylinder with the tangent, Pl. XV., Fig. 126, upon the plane H.
Ex. 2°. Show the true size of the intersection of the same plane and cylin-
der, on the plane V.
Ex. 3. Show the true size of the intersection of the plane and cone, Pl. XV.,
Fig. 127, on the plane H.
Ex. 4. Show the true size of the intersection of the same plane and cone on
the plane V.
Ex. 5. Show the true size of the intersection of the plane and pyramid, PL
XVIII., Fig. 141.
Ex. 6'. Show the true size of the intersection of the plane and cylinder,
described in Prob. LXXIX.
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DESCRIPTIVE GEOMETRY.
185
2-Development of Cylinders and Cones, and of their
Intersections.
245. If a developable surface be made tangent to a plane, and
then rolled upon that plane till the original element of contact
again coincides with the plane, a portion of the plane will have
been rolled over, equal to the area of the convex surface of the
rolling body. This plane area is the development of the sur-
face.
246. When a surface is developed, any curved line upon it
will be developed into some curve, whose length will be equal
to that of the original curve.
A curve of double curvature is generally developed in this
way, that is, by developing a surface which contains it. Plane
curves may be developed in the same way, or by being unrolled
upon a tangent line. Both modes would be development by
transformation.
The latter method is direct, and gives a development of the
same form, for any given curve, viz., a straight line. The
former is indirect, and will give a different form of develop-
ment of the same curve, for every different surface containing
that curve.
THEOREM XXXIV.
If a line be tangent to any curve traced on a developable sur-
face, it will continue tangent to that curve as found on the
development of the surface.
Every curve traced on a developable surface, crosses each
clement of the surface at a certain angle, and this angle will
evidently not be changed by the process of development, either
of a cylinder, Pl. VII., Fig. 74, or of a cone. Now as the
surface may be developed upon any plane which is tangent to
it, let the cylinder, Fig. 74, be developed upon the tangent
plane, LKt, along the element EE, containing the point of con-
tact, t, of the tangent line Kt with the curve, abc, traced upon
the cylinder. Then, as the angles made by the curve and its
tangent with the element EE are unchanged by development,
186
DESCRIPTIVE GEOMETRY.
their relation to each other at t is unchanged also. IIence Kt
will continue tangent to abe in development.
Indeed this would be true in the same way for a cone or a
cylinder, and upon whatever tangent plane either should be
developed.
If the curve abc should cross all the elements at the same
angle, its development would be a straight line, coinciding with
its tangent, since both have the same direction at t, and that
direction is constant for both in development.
PROBLEM XC.
To develope the surface of a right cylinder of revolution,
together with one of its oblique sections and a tangent to
that section.
In Space. By reference to an actual cylinder rolling upon
a plane, it will appear, that the length of the development—
that is, the perpendicular distance between two successive con-
tacts of the same element, will be equal to the circumference
of the base. The lengths of the elements, from the base to the
oblique section, will be equal to their vertical projections, the
cylinder being supposed to stand upon the horizontal plane.
In Projection.—1°. The general development.-Pl. XVIII.,
Figs. 143, 144. ABC and A'C'A"C" are the projections of the
obliquely truncated cylinder. Place the longest element, A'A',
in the plane of the paper at AA'. On CC, perpendicular to
AA', lay off AB AB on the plan, AE AE on the plan,
etc. (246). Make BB' and EE' each B'B' on the elevation,
etc. Join the points C', B', A', E', etc., thus found, and we
shall have the development of the oblique section, and C'A'C'
CC for the development of the convex surface of the cylinder.
=
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2°. The tangent.-The development of the tangent at TT',
will (Theo. XXXIV.) be tangent to the development at T".
The horizontal distance Tn, from the element containing the
point of tangency, to any horizontal plane, 'n", is called a sub-
tangent. From n"" lay off n""n" Tn, and join n" and T",
and n"T" will be the development of the tangent.
The tangents to the development at A', and the two points,
C', will be parallel to CC.
DESCRIPTIVE GEOMETRY.
187
3°. The points of inflexion.-The tangents at FF' and GF'
mark the declivity, meaning the greatest declivity, of the cut-
ting plane, A'C', which is perpendicular to the vertical plane of
projection. The tangents at AA' and CC' are horizontal or
have no declivity. That is, in passing in either direction from
FF', or GF', the tangents to the oblique section, A'C', of the
cylinder become more and more nearly horizontal; and the
like being obviously true for the development, F' and G', on
the development, are its points of inflexion.
THEOREM XXXV.
The development of the oblique section on the development, of a
cylinder of revolution, is a sinusoid.
The sinusoid is a curve such that if its abscissas, or successive
horizontal distances from F', FB, FA, etc., Fig. 144, be
equal to successive arcs, FB, FA, etc., Fig. 143, reckoned from
a common initial point, F, its ordinates, bB', aA', etc., or corre-
sponding perpendicular distances from the line of abscissas, will
be equal to the sines, F"B", F"A", etc., of these arcs; or pro-
portional to them-the former being the usual definition.
But, the successive horizontal distances, FT, FC, etc., FB,
etc., Fig. 144, from the line FF', are equal to the successive
ares, FT, FC, etc., of the base of the cylinder, reckoned from
F; and the horizontal distances, f't', f'c', f'b', etc., reckoned
from F', Fig. 143, are respectively the sines of these arcs. But,
it is clear that the corresponding ordinates, or distances of T',
C', B', etc., from the horizontal line, a'd', are proportional to
these sines, and that they are equal to the ordinates of the
corresponding points on the development, reckoned from the
same line F-F'G'. Hence the curve C'AC' is as described.
PROBLEM XCI.
To develope the surface of a right cone of revolution, also an
oblique section of it, and its tangent.
In Space.-All the elements of a cone of revolution are
equal. Hence, when the cone is rolled out upon a tangent
188
DESCRIPTIVE GEOMETRY.
plane, its base will develope into a circular arc, having the ver-
tex for its centre, and the length of an element for its radius.
By laying off on this arc the circumference of the base, and
connecting the extreme points so found with the vertex, the
resulting sector of a circle will be the development of the sur-
face of the cone.
To obtain the development of the oblique section, find the
true distances from the vertex to the points of the oblique sec-
tion, and lay them off from the vertex of the development
along the developments of the same elements.
The development of the tangent is found in much the same
manner as in the last problem.
In Projection.—1°. The general development.-Pl. XVIII.,
Fig. 145. VNPO-V'M'N' are the projections of the cone,
and AB is the vertical projection of the oblique section, made
by the secant plane, ABT, perpendicular to the vertical plane.
With any point, V", as a centre, and a radius V'M', describe
the arc N'N', on which lay off N"O" NO; O"M" = OM,
etc., and join these points with the vertex; they will be de-
velopments of the elements VN, VO, etc. Then for the
development of the oblique section, make V'B' and V"B"
each equal to V'B, since the elements V"N" and V"N' are the
developments of the element VN-V'N', along which the cone
is cut in order to unfold it upon a plane. Likewise V"A"
V'A. Any other distance, as V'C, is found in its true size by
revolving it about the axis of the cone till it becomes parallel
to the vertical plane, when, in a right cone with a circular base
it will be found, as at V'e, in the vertical projection, V'M', of
an extreme element. V'c is now laid off at V"C" on the de-
velopment of the element V'C. E", c', etc., being determined
similarly, and connected, we shall have the development of the
oblique section.
For the tangent, make t"T", the sub-tangent, tangent to
N'N" at t" and equal to T, and join T" and ", which will give
the development of the tangent at E.
The tangents at N" and B"; M" and A"; N' and B', are
parallel.
2°. The points of inflexion.-Returning to Prob. XC., the
cylinder and secant plane form a system, in which the tangents
to the curve A'C' make larger angles with the elements of the
cylinder, the further they are taken from FF' and GF'. But
DESCRIPTIVE GEOMETRY.
189
น
at these points, only, the secant plane is a normal plane to the
surface of the cylinder, since it contains a perpendicular to that
surface at these points. This then being the distinctive relation
of the secant plane at the points whose developments are points
of inflexion, we proceed to find the same thing for the cone,
viz., the points where the secant plane, ABT, is normal to the
surface. Pl. XVIII., Fig. 145.
At ƒ', where the secant plane cuts the axis, draw f'g', per-
pendicular to V'N', and make it the generatrix of an auxiliary
conic surface, which will be normal to that of the given cone,
at all points of the circle, g'h', of the intersection of the two
cones. The secant plane intersects this cone in the two ele-
ments projected in fl'. Hence it is normal to the given cone
at the two points, hh', and mh', on the elements VI-V'I' and
VJ—V'I'. Hence h" and m", developments of these points,
will be the required points of inflexion of the developed plane
section, AB, of the given cone.
EXAMPLES.-1°. Note what relations of the angles AVƒ', Aƒ 'V', and 90°;
and what values of AV'f' will give separate, or united, tiro, or no points of in-
flexion for the different conic sections. To do this, it will be enough to sketch
the vertical projection, only, of the cone.
Ex. 2°. Develope a cone having a given parabolic section.
Ex. 3°. Develope a cone with a given hyperbolic section; and find the points
of inflexion of the developed section.
Ex. 4. Develope a cone, when the angle at the vertex is greater than 90°.
Ex. 5. Develope the cone, and its plane section, shown in Pl. XV., Fig. 127.
Ex. 6. The axis of the cone being vertical, show the true lengths of its ele-
ments on H.
190
DESCRIPTIVE GEOMETRY.
CHAPTER III.
WARPED SURFACES OF REVOLUTION.
247. A WARPED SURFACE is generated by a straight line,
moving so that its consecutive positions are not in the same
plane.
248. The only motions of revolution which a straight line
can have are:
1º. Round an axis which is parallel to it.
2°.
3°.
66
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intersects it.
is not in the same plane with it.
The surfaces generated in the two former ways, are the cylin-
der, and the cone of revolution. These are developable. If
then there be a warped surface of revolution, it will be gene-
rated in the latter way.
THEOREM XXXVI.
The surface, generated by the revolution of a straight line,
around an axis not in the same plane with it, is warped.
To prove this theorem, it is only necessary to show that the
consecutive elements of the surface neither intersect nor are
parallel.
See, then, Pl. XIV., Fig. 121, where OZ is a vertical axis, and
AB the generatrix, not in the same plane with OZ.
Now since the generatrix does not intersect the axis, evory
point of it moves in revolving; therefore its consecutive posi-
tions do not intersect.
Again the horizontal, OC, is the common perpendicular to,
or shortest distance between, AB and OZ. Let it be considered
as a revolving arm carrying AB with it, at its extremity C, as
1
DESCRIPTIVE GEOMETRY.
191
both revolve about OZ. Then the different positions of AB,
being obviously perpendicular to the different positions of OC,
evidently are not parallel.
Thus, as the consecutive elements, that is, the consecutive
positions of the generatrix AB, neither intersect, nor are parallel,
the surface, which they form, is warped.
THEOREM XXXVII.
The warped surface of revolution has two sets of elements, each
element of either of which intersects all those of the others.
First. Let the vertical OZ, Pl. XIV., Fig. 121, be the axis of
the surface, ACB the generatrix, and OC the least distance from
ACB to OZ.
Next let A'B' be a line so taken that the plane of AB and
A'B' shall be parallel to OZ, and so that AB and A'B' shall
make equal angles with the horizontal plane through OC.
Under these conditions, any two points, as A and A', one
on AB, and one on A'B', and both in the same horizontal plane,
will be equidistant from OZ, and will therefore describe the same
horizontal circle, as AA'II, when OC, AB and A'B' revolve
together as an invariable system, about OZ as an axis. Hence the
surface generated by the revolution of AB, and composed of all
the positions of AB, is identical with that generated by the
revolution of A'B', and composed of all the positions of A'B'.
Second. Hence as any position of AB lies wholly in the sur-
face composed of all the positions of A'B', and is not parallel to
them, it intersects them all. By revolving AB, 180°, one of its
positions will be seen to be parallel to A'B', and it then inter-
sects A'B' only at infinity.
The like is evidently true of any position of A'B'; and thus
the theorem supposes the surface to be extended indefinitely on
both sides of the circle OC.
249. The two sets of elements thus existing are distinguished
as those of the first and the second generation; and we say that
any element of either generation intersects all those of the other
generation.
OC, the common perpendicular between the generatrix AB,
192
DESCRIPTIVE GEOMETRY.
and axis OZ, is evidently the least of the radii eh, df, etc., of the
horizontal circles described by points of AB and A'B' in their
revolution. The circle of radius OC is called the circle of the
gorge; and the like term applies, in all cases, to the smallest
interior opening of a surface.
A-PROJECTIONS.
a-Projections of Forms.
PROBLEM XCII.
To construct the projections of the warped surface of revolution.
In Space. The fundamental position of any surface of revo-
lution, from which the projections of all other positions may be
derived, is that in which the axis is a vertical, or, a perpendicular.
Let the axis in this problem be vertical. The radii of the
gorge will then be horizontal, and all the elements being perpen-
dicular in space to those radii, will lie in planes parallel to the
axis and tangent to the gorge, and will be determined by their
intersections with the circle of the gorge, and with any other
parallel (14) of the surface.
In Projection.-Pl. XIV., Fig. 122. Let A-A'X' be the
axis of the surface, and AB-A'B' the radius of the gorge. All
the radii of the gorge being horizontal, the elements, being per-
pendicular to them in space, will be so in horizontal projection
(Theo. II.), and hence will there be tangent to the gorge. Thus
DE-D'E' and DE-D"E" are those elements, one of each
generation, which are parallel to the plane V.
Their inclinations, A"D'D" and A"D'D', to the horizontal
plane being equal, their horizontal traces, D'D and D'E, will
describe the same horizontal circle; which will also contain the
like traces of all the other elements; and will be also the hori-
zontal projection of an equal circle at a distance above the gorge,
equal to A'A".
250. Every tangent, as UV, to the gorge is the horizontal pro-
jection of two elements, one of each generation, UV-U'V' and
UV-U"V". Likewise any line in the vertical projection, as
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DESCRIPTIVE GEOMETRY.
193
U"K", is the vertical projection of two elements, one of each
generation, U"K"-UK, and U”K”—FH.
PROBLEM XCIII.
To represent the projections of the warped surface of revolution
by means of its meridian curve of apparent contour.
In Space. From the description of the surface already given,
it is obvious that its meridian curve is convex towards the axis
of the surface. And as all the elements except the pairs at DE
and OS, Pl. XIV., Fig. 122, are partly on opposite sides of the
meridian plane MN, of the apparent contour as it is seen in ver-
tical projection, it is evident that this contour will simply be a
tangent to the projections of those elements. That is, it consists
of the sum of their consecutive intersections, when they become
consecutive.
In Projection.-Draw numerous elements, and very carefully,
in the neighborhood of UK and FII, and, like these, symmetri-
cal relative to the meridian plane MN. Their intersections, as
II' and II", will be points of the required meridian curve.
THEOREM XXXVIII.
The meridian curve of the warped surface of revolution is a
hyperbola.
1º. Pl. XIV., Fig. 122. As all those pairs of elements as
UK (250) which are in vertical planes, intersect each other on
the gorge, as at au', their vertical projections proceeding from
every point, as a', of B'C', are symmetrical relative to B'C'.
Hence the meridian curve, to which these projections are tan-
gent, is symmetrical relative to B'C'.
2º. As the same result evidently occurs in like form on the
left of A'X', the two opposite branches of the curve are also
symmetrical relative to A'X'.
3°. The elements DE and OS, meeting the plane MN only at
infinity, their vertical projections are tangent to the curve only
at infinity. That is, these projections are asymptotes to the curve.
13
194
DESCRIPTIVE GEOMETRY.
The meridian curve is thus seen to consist of two equal, oppo-
site and infinite branches; having two axes of symmetry at right
angles to each other, and a pair of asymptotes through its centre.
All these properties are characteristic of the hyperbola, but—
4°. If we consider any element as JP, its segment be, between
DE and OS, is bisected at d. But the vertical projections of be
are tangent to the meridian curve at the vertical projections of d,
and the vertical projections of DE and OS are the asymptotes
D'E' and D'E". That is, the meridian curve is such, that the
tangents to it, limited by the asymptotes, are bisected at their
points of contact. This distinctive property, added to the others
(1°—3°), proves the meridian curve to be a hyperbola.
251. From the preceding theorems we have three definitions of
the warped hyperboloid of revolution. It may be generated:
1°. By the revolution of a straight line, L, around an axis not
in the same plane with L.
2°. By a straight line moving upon three fixed straight lines;
elements of the other generation (Theo. XXXVII.).
3°. By the revolution of a hyperbola about its conjugate axis.
252. Also, we see that the surface is divided symmetrically
into two nappes by the gorge, as a cone is by its vertex. Ilence
the surface is called the warped hyperboloid of revolution; also
called the hyperboloid of revolution of two united nappes, in
distinction from the hyperboloid of revolution of two separate
nappes, which is generated by the revolution of a hyperbola, as
AEF-HDI, Pl. XIV., Fig. 121, about its transverse axis DE,
and which would obviously consist of two separate parts, such
that DE would be the least distance between them.
The warped hyperboloid is generally called that of one nappe,
but improperly, in view of its close analogy with the cone.
PROBLEM XCIV.
Having given one projection of a point on a warped surface of
revolution, to find the other projection of the same point.
In Space.-Every point of the surface being the intersection
of two elements, one of each generation, its projections will be
the intersections of their projections. Otherwise: the point
J
DESCRIPTIVE GEOMETRY.
195
being imagined as on any other line of the surface, its projections
will be on those of that line.
In Projection.-First. The horizontal projection given.
1º. The method by elements. Let c, Pl. XIV., Fig. 122, be
the given horizontal projection of a point on the surface. Two
tangents to the gorge, cD and cP, will (250) be the horizontal
projections of the two elements through e. By (Prob. XCII.),
find their vertical projections, and their intersections, either with
each other, or with the projecting line cc'e' from c, will be the
vertical projections, c' and c', of the two points, of which c is the
horizontal projection.
2°. The method by parallels. Let e be the given projection
of the required point, and suppose DE-D'E', and DE—D"E",
as may often be the case, to be the only elements given. The
are with radius Ae cuts the pair DE at ce' and ce", points of the
two parallels, ec-e'c' and ec-e"c", on which e is projected at e'
and e" the required points.
3°. The method by meridians. Revolve the points e to ƒ,ƒ”
and ff" on the meridian curve, whence it returns in the arcs fe
and f'e' to e' and e".
This construction is thus evidently that of the method by
parallels, only modified by determining the parallels by their in-
tersections f'f" with the meridian of apparent contour, instead
of by their intersections, c' and c', with an element.
Second. The vertical projection given.—In the method by
elements, simply draw tangents to the meridian of apparent con-
tour (Theo. XXXVIII.).
In the methods by parallels merely make the same construc-
tions as above, only in inverse order.
EXAMPLES. —1°. Find the horizontal projection of e' by each of the three
methods.
We add the following general examples for further practice
in constructing the projections of the warped hyperboloid in
different positions, on account of its uses in the mechanical arts:
Ex. 2°. Let the axis of the warped hyperboloid be a perpendicular.
Ex. 3°. Let it be a co-parallel.
Ex. 4°. Let it be a bi-parallel.
Ex. 5°. Let it be oblique to both H and V.
196
DESCRIPTIVE GEOMETRY.
b-Projections of Tangencies.
THEOREM XXXIX.
Every tangent plane to a warped hyperboloid of revolution cuts
the surface in two elements, one of each generation, and is
tangent only at their point of intersection.
Resuming (148) let us take the tangents to the meridian,
and to the parallel at any given point, II', of the hyperboloid,
Pl. XIV., Fig. 122, in order to determine the tangent plane
at II'.
Now, the parallels being convex towards surrounding space,
the tangent to the parallel at II' (see also CG, Pl. XIV., Fig.
121), is tangent to the exterior side of the surface. But the
meridians being convex towards the axis, the tangent at II' to
the meridian is tangent to the interior side of the surface.
Hence, while these lines, being two tangents at the same point,
determine the tangent plane at that point, yet, being on oppo-
site sides of the surface, the plane containing them is also a
secant plane.
Again: out of all possible sections through II', two and only
two, viz., the elements are straight. By (149) their tangents
coincide with these lines themselves. Ilence the tangent plane
at any point is most simply determined by the two elements
through that point. And as a plane through UK-U"K" cuts
all the elements of the other generation at their intersection with
UK-U"K" (Theo. XXXVII.), and as the like is true for
FH-U"K", the plane containing both of these elements inter-
sects the hyperboloid along these lines, so as to be within the
surface while in the opposite angular spaces IIIK and UIF, and
exterior to it, within the angles HIU and KIF.
The tangent plane is thus tangent only at a point, as II.
253. Resuming the principle of (Prob. XIV.), a tangent plane
to the warped hyperboloid at a given point, is determined by
that point, with one upon each of any two of the tangents at
that point, all of which are in the tangent plane.
The tangent plane having only a point of contact, may be
DESCRIPTIVE GEOMETRY.
197
determined by that as a required point and any two given exte-
rior points, and hence may be drawn through a given line, and
tangent to the hyperboloid.
A plane is also determined by three points in being tangent
to the surface at a required point, and parallel to a given plane.
Hence a plane may be drawn, tangent to the hyperboloid and
parallel to a given plane.
Any number of tangent planes may evidently be drawn
through a given point, or parallel to a given line.
254. The plane through any element will (Theo. XXXIX.)
contain an element, of the other generation, whose horizontal
projection will join the points in which any two parallels cut
the traces of the given plane on the planes of those parallels.
By this principle we can find the point of contact of a plane
containing any element.
The tangent plane at any point on the circle of the gorge is
parallel to the axis of the hyperboloid.
The tangent plane at infinity in any meridian plane, contains
elements which are parallel to that plane, and it makes the same
angle with the plane of the gorge that those elements do. Thus
the angles made by different tangent planes with that of the
gorge, lie between the limits of 90° and the angle just described.
255. Since the tangent plane to the warped hyperboloid con-
tains the two elements through the point of contact, the tangent
plane to the surface and containing a given line, will be deter-
mined by this line and one or the other of these elements.
Hence-
First. Two tangent planes can be drawn through the given line.
Second. We find the peculiar property, that no tangent plane
can contain a given line, unless that line intersects the hyper-
boloid.
PROBLEM XCV.
To construct a tangent plane to a warped hyperboloid of revo-
lution, at a given point.
In Space.—The required plane will by (252) be determined
by the two elements, or by any other two tangents at the given
198
DESCRIPTIVE GEOMETRY.
point, or by any lines which intersect any pair of such tangents.
The only readily found tangents besides the elements, are those
belonging to the parallel and the meridian through the given
point.
In Projection. The traces of the required plane will be de-
termined by those of any two of the lines just described as
determining the plane itself. The construction is therefore left
to the student.
The axis of the surface being vertical, the horizontal trace of
the tangent plane will be perpendicular to that of the meridian
plane through the point of contact (Theo. VII.).
EXAMPLES.-1°. Construct the traces of a tangent plane, at any point on the
lower nappe of a vertical warped hyperboloid.
Ex. 2. Construct the traces of a tangent plane, at any point on the upper
nappe.
Ex. 3°. Construct the traces of a tangent plane at infinity on a given meri-
dian.
Ex. 4°. Construct the traces of a tangent plane through a given point ( 253 ).
Ex. 5°. Construct the traces of a tangent plane, parallel to a given line
(253).
PROBLEM XCVI.
To construct a plane, tangent to a warped hyperboloid of revo-
lution, and through a given line.
In Space.-Let the given line, L, revolve around the axis, A,
of the given hyperboloid II. It will generate a new hyperbo-
loid of revolution II,, having A for its axis. The required tan-
gent plane, T, containing L, will be tangent to II, at some point
of the element L. The plane T is also by requirement tangent
to H. But the tangent plane to a surface of revolution is per-
pendicular to the meridian plane through its point of contact,
and as II and II, have a common axis, they will have a common
meridian plane, M, containing the point of contact of T with
each. IIence the trace of T on M will be a common tangent, t,
to the meridian curves contained in M. Hence t and the given
line L will determine the required tangent plane.
In Projection.-Pl. XIV., Fig. 123. Let the given hyper-
boloid be given by its gorge, of radius Oa-O'a', and the ele-
ment E-O'E', and let AB-A'B' be the given line. Draw
DESCRIPTIVE GEOMETRY.
199
the perpendicular (Theo. II.) Oq-O'q' from the axis 0—0′0″
to this line, and it will be the radius of the gorge Ob-O'b' of
the auxiliary hyperboloid.
By Prob. XCII., construct several elements of each hyper-
boloid in the neighborhood of the common meridian plane, OC,
parallel to the plane V, and find the two meridian curves in OC
by finding the intersections of these elements with this meridian
plane. These curves are D'a'H', for the given hyperboloid, and
C''C" for the auxiliary one.
By Prob. LXXVII., construct the common tangent to these
meridian curves, and, by the solution "in space," its points of
contact, c'e' and tt, will be the revolved positions of the
points of contact of the required tangent plane with the two
hyperboloids.
But the real position of c'e'" is on AB-A'B'; hence revolve
it to ce on that line; and ec' will be the point of contact on
the auxiliary hyperboloid; Oe will be the common meridian
plane (M) containing the contacts on both hyperboloids, and
hence t't" is found in its true position at tt' in this plane Oc.
Finally, the tangent plane (T) being perpendicular to the cour-
mon meridian plane Oe, its horizontal trace, PQ, through B, is
perpendicular to Oc. Its vertical trace joins Q with the vertical
trace of any convenient line of the tangent plane, as P', that of
the horizontal tp-t'P', or P'', that of AB-A'B'.
256. If this problem is impossible in any case, the fact will
be indicated by the meridian curves being so situated, relatively,
that no common tangent to them can be drawn.
Note that the arc d''e intersects AB twice. We determine
which of these intersections is real, in space, by noting whether
c'e' is above or below the gorge through q'. IIere it is below,
and ec' is therefore the real position of e'e"".
257. Complete graphical study.-On account of the many
principles and problems applied in this construction, it is one of
those which may be well selected for full construction on a large
scale as in the following—
EXAMPLE.-Make the figure fill a whole plate, as large as those of this
volume, by taking the ground line in the direction 0-00"; then make the
detailed construction of both meridian curves, of their common tangent, and
of its exact point of contact with each curve, and finally of the second tangent
plane to the given hyperboloid and through AB-A B'; which may be found
by drawing the second common tangent, to D'a II and to the opposite branch
(not here shown) of the hyperbola C'b'C".
200
DESCRIPTIVE GEOMETRY.
PROBLEM XCVII.
To construct a plane tangent to a warped hyperboloid of revo-
lution, and parallel to a given plane.
In Space.-First method. Find the meridian H, contained in
a meridian plane, M, perpendicular to the given plane P. The
tangent to II and parallel to the intersection of M and P will
serve to determine the traces of the required plane T, since the
traces of the latter will be parallels to the traces of P, drawn
through those of the tangent to II.
Second method. Make any point of P the vertex of a cone,
C, of revolution, whose elements make the same angle with the
horizontal plane H, that those of the hyperboloid do, the axis of
the latter being vertical.
The elements cut from this cone by P will be parallel to those
of the hyperboloid, whose intersection with each other will be
the point of contact of the plane containing them; which plane
will, by construction, be parallel to the given plane P.
In Projection. The construction of both methods is left to
the student.
258. Of the conic asymptote to the warped hyperboloid.-
Having found (Theo. XXXVIII.) that the projections, as D'E',
and D'E", Pl. XIV., Fig. 122, of the pairs of elements hori-
zontally projected in DE and OS, are tangent to the like projec-
tions, as K'C'K', of the meridian curve contained in the plane
MN, parallel to the planes DE and OS, it follows that the pro-
jections, MAN-D'E', and NAM-D"E", of the same pairs of
elements upon the plane MN, will be the asymptotes to the
meridian contained in that plane.
259. This being so, and the axis A-A'X' being vertical, the
infinitely distant points of contact (above and below the gorge)
of these asymptotes with the meridian curve will generate the
equally distant horizontal circles of contact of the cone generated
by the revolution of the asymptotes about the axis A—A'X',
with the hyperboloid, generated by the revolution of the meri-
dian curve K'C'K' around the same axis.
The vertex of this cone is evidently AA", the centre of the
gorge, and its circles of contact with the hyperboloid being at
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DESCRIPTIVE GEOMETRY.
201
infinity, above and below the gorge, it is called the conic asymp-
tote, or asymptote cone of the hyperboloid.
Therefore, conversely to (258) any meridian plane containing
the common axis of the hyperboloid and its asymptote cone,
will cut from the former a meridian hyperbola; and from the
latter, a pair of elements which will be the asymptotes to this
hyperbola.
THEOREM XL.
Every plane, tangent to the conic asymptote of a warped hyper-
boloid of revolution, contains two elements of the hyper-
boloid; one of each generation, and both parallel to the
element of contact with the cone.
Let ag, Pl. XIV., Fig. 122, be any diameter of the gorge.
The elements, aK and gK,, at its extremities, are parallel, and
of different generations.
That element, Al, of the asymptote cone, which is perpen-
dicular to ag is parallel to aK and gK, as all are equally inclined
to the plane of the gorge. Hence aK, Ah, and gK,, being paral-
lel to each other, and perpendicular to ag, are in the same plane,
whose trace on the plane of the gorge is ag. But ag is also the
trace on the latter plane, of the tangent plane to the conic
asymptote, along the element Ah, which proves the theorem.
260. The director (S) of a warped surface is a surface to
which all the elements of the warped surface have some simple
and uniform relation. Thus, since every element of the hyper-
boloid is thus parallel to some element of the asymptote cone,
the latter may also be called the cone director of the surface.
Also, any plane perpendicular to its axis may be called its
plane director, since all the elements of the hyperboloid make
a constant angle with this plane.
261. Analogy of the hyperboloid to the cone, cylinder, and
plane. As the generatrix AB, Pl. XIV., Fig. 121, may be
moved parallel to itself along the radius CO of the gorge, CO
may be of any length, for a given value of the angle ACG.
When C, approaching O, comes to coincide with it, the hyper-
*
202
DESCRIPTIVE GEOMETRY.
boloid becomes the asymptote cone of revolution, generated by
ON.
Again; CO being constant, the angle ACG may vary from
0° to 90°. In the former case, the hyperboloid becomes the
plane, generated by the revolution of CG about OZ; in the
latter, it becomes the cylinder, generated by the revolution of
CE around OZ.
Thus the plane, the cylinder, the cone, and the warped hyper-
boloid of revolution, are the four forms of related surfaces of
revolution formed by the revolution of a straight line around
an axis; and such that each is a particular case of all which
succeed it in the order named.
262. Problems of tangency of other than plane surfaces, tan-
gent to a given surface, are seldom required, and can always be
solved by means of a common tangent plane to the two surfaces.
Thus to place a given cone, tangent at a given point, to a given
hyperboloid, it would be made to touch, along one of its ele-
ments, the tangent plane to the hyperboloid at the given point.
c-Projections of Intersections.
263. The problems of intersection belonging here will con-
sist of the intersections of the hyperboloid of revolution of
two united nappes (252) with planes, and with other ruled sur-
faces (15).
As every ruled surface is generated by a straight line, the sum
of whose successive positions constitutes the surface, the construc-
tion of the whole group of intersections just indicated reduces
to that of finding the intersection of the hyperboloid with a
given straight line.
As a plane is straight in every direction, the operations re-
quired in every case of the latter problem, can be illustrated in
finding the intersection of the hyperboloid with a plane-a
problem more apparently practical than that of its intersection
with a single line.
Every straight line which meets the warped hyperboloid at
any point, intersects both of the elements at that point (Theo.
XXXVII.) in the same point. Hence it is in the same plane
with each of these clements.
DESCRIPTIVE GEOMETRY.
203
THEOREM XLI.
When two concentric surfaces are cut by a plane in two curves,
one surrounding the other, the inner one being a conic sec-
tion, and so that any chord of the outer one, which is tan-
gent to the inner one, is bisected at its point of contact, the
two curves are similar and similarly placed conic sections.
Let c, Pl. XIV., Fig. 124, be any point upon the conic sec-
tion E, and let the tangent to E at c,, cut the surrounding curve
E, at a, and b,; suppose that ca, cb, and that a like relation
holds for every point of E.
1
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Through O, the centre of E, draw Oa,, cutting E at a; draw
ab, parallel to a,b,, cutting E at b; and draw Oc,; then Oc,
bisects ab (178). Join O and b, and Ob will cut the tangent
ab, in a point B,, so that as ac = be, we must have ca, c,B,.
But there is given the relation ca, c,b,. Hence B, and b, are
identical, and Oa : Oa, :: Ob : Ob. But this relation being
true for every position of e, since it follows from the relation
1
=
cb,, which was to be true for every position of c,, the
curves E and E, are similar and similarly placed. That is, they
are similar and concentric conic sections.
THEOREM XLII.
Every plane section of the warped hyperboloid of revolution,
is a conic section, of the same kind as that cut from its
asymptote cone by the same plane.
Imagine the hyperboloid II (or see any of the figures of this
surface on Pls. XIV. and XIX.), and its asymptote cone A.
Any plane P will cut A in a conic section C, and will cut II in
a curve C,, which will enclose C within it.
Then suppose any point p upon the conic section C, and con-
ceive a plane T, tangent to A along the element E which contains
p. This plane T will (Theo. XL.) contain the centre, o, of the
gorge, G, of the hyperboloid H, and will contain two elements,
of different generations, e, and e, parallel to each other, and to
the element of contact, E, of the cone A.
204
DESCRIPTIVE GEOMETRY.
Now by (148) the plane T will cut the secant plane P in a
tangent, t, to the conic section C at the point p; and this tan-
gent being in the same plane, T, with e, and e, will cut them at
points m and n, which will evidently be points of the curve of
intersection, C₁, of P with H, since t is in the plane P. Now as
the line E is equidistant from e, and e, (Theo. XL.) pm = pn.
Hence by (Theo. XLI.) the curves C and C, are concentric and
similar conic sections.
264. The numerous following methods of finding the inter-
section of a warped hyperboloid with a plane, which arise from
the many elementary properties of the surface, will serve to give
special insight into the subject of intersections generally.
Some of them will illustrate the obvious general principle,
that the intersection of two surfaces of revolution, having a com-
mon axis, will be a circle perpendicular to that axis; since such
a circle would be generated by the point of intersection of the
meridian curves.
PROBLEM XCVIII.
To find the intersection of the warped hyperboloid of revolu-
tion, by a meridian plane, and by any other planes parallel
to it.
In Space.-Besides the construction of the meridian curve
made in Prob. XCIII., on the principles of tangency relating to
the hyperboloid, all the curves here required can be found by
noting the intersections of the given plane with the parallels of
the surface; and one point of each parallel will be the intersec-
tion of its plane with any given element of the surface.
In Projection.—In Pl. XX., Fig. 160, let the hyperboloid be
given by its gorge, of radius OC-O'C', and one element
Nd-O'd'.
Assume any convenient number of points, aa', gg', dd', etc.,
on the given element, and through them draw the parallels of
radii, Oa-p'a', Od-m'd', etc. Now let AO, DM, dh and cq,
be the horizontal traces of a series of planes, all parallel to the
plane V. We then immediately have AA', hh', ce', etc., as
DESCRIPTIVE GEOMETRY.
205
points of the upper half, CA-C'A', of one branch of the meri-
dian hyperbola.
The exactly similar construction of the remaining intersections
is obvious on inspection.
265. By (Theo. XXXVIII.) we know also that the sections
of the hyperboloid made by the planes DM and cq are hyper-
bolas, and we see that of these sections, that contained in the
plane kd, which is tangent to the circle of the gorge, is the par-
ticular case consisting of two intersecting lines.
266. It is now further evident that planes, which, like DM,
contain a chord of the gorge, intersect the hyperboloid in hyper-
bolas whose transverse axes, as O'C' or O'H', are perpendicular,
in direction, tọ (—O'Z′ that of the surface; while a plane like
cq, exterior to the gorge, cuts the surface in a hyperbola, whose
transverse axis = 20'' is parallel to that of the surface.
267. The difference, d'A', between the true length Od, and
the vertical projection m'd', of the radius of a parallel, contin-
ually decreases as we recede from the gorge. But this differ-
ence would not vanish till dd' should be at infinity from N on
the element Nd, when Od and OA being infinite, A would be
straight, and perpendicular to OA. It thus appears again (Theo.
XXXVIII.) that OA-O'd' is the asymptote to the curve CA
—C'A', which, being symmetrical relative to a pair of reetan-
gular axes at 00', is thus again shown to be a hyperbola.
In the same manner it can evidently be shown that c'd' or
¿'D' will vanish at infinity, and thus that O'd' (projection of
Nd) is the common asymptote to all the hyperbolas cut from the
hyperboloid by planes parallel to the tangent plane hd.
The following theorem generalizes this property for all cases.
THEOREM XLIII.
Every hyperbolic plane section of the warped hyperboloid of
revolution, has asymptotes parallel to the elements contained
in the parallel tangent plane.
Let O be the vertex of the asymptote cone, A, of a warped
hyperboloid of revolution, II, and centre of its gorge. Pass a
plane, P, through O, so as to contain two elements, E and E,, of
the cone A. By (Theo. XL.) a tangent plane, T, to the cone A
206
DESCRIPTIVE GEOMETRY.
along the element E will contain a pair of parallel elements, G
and y, one of each generation, of the hyperboloid II.
Likewise a tangent plane, T,, to cone A along the element E,
will contain a pair of parallel elements, G, and g,, of the hyper-
boloid II. (The figure can be sketched, if not readily imagined.)
These two planes will intersect each other in a line L through
the vertex O. Hence, those pairs of these elements which are
(Theo. XXXVIII.) symmetrical, relative to the meridian plane,
M, through L, will meet at some points, p and p, of the line L,
and (Prob. XCIII.) p and p, will be those points of the meridian
curve of H, in which L pierces the surface II. IIence the tan-
gent planes, Q and Q,, to the hyperboloid at p and p₁, being de-
termined by pairs of elements, each pair parallel respectively,
to. E and E,, will be parallel to each other and to the plane P.
Hence, finally, every plane P,, parallel to P, will cut the cone
A in a hyperbola h, and the hyperboloid H in a hyperbola h,
whose common centre is the intersection of P, and L, and
whose asymptotes, a and a, will be respectively parallel, a to the
lines G, g and E; and a, to G₁, J, and E¸.
1
268. If the plane P, in the last problem is without the tangent
planes Q and Q,, the hyperbolas h and h, will be similar and
similarly placed, that is, in the opposite angles formed by the
asymptotes. But if P, be between the planes Q and Q,, the curves
hand h,, still concentric, will be inversely similar, or situated in
the adjacent angles formed by their asymptotes.
PROBLEM XCIX.
To find the intersection of a warped hyperboloid of revolution
by a plane, and a tangent to the curve at a given point.
In Space.-Any one point of the required intersection, will
be the intersection of some line of the plane with some line of
the hyperboloid; both being in the same auxiliary surface.
Any straight line of the plane may be taken, since it is straight
in every direction.
The simplest lines of the hyperboloid are its elements, its
parallels, and its meridians.
DESCRIPTIVE GEOMETRY.
207
Those auxiliary surfaces should be chosen, which are simplest
in relation to the given surfaces. These will be planes, cutting
the hyperboloid in any of its simplest sections; concentric cones,
cutting it in parallels (263 ), and auxiliary hyperboloids, cutting
it in parallels, or in hyperbolas.
In Projection:
I. REGARDING THE HYPERBOLOID AS A RULED SURFACE.
1º. The auxiliary surfaces, the projecting planes of ele-
ments.-Let AEZ-A'B', Pl. XX., Fig. 161, be the gorge, and
be-b'c', one element of a hyperboloid, and PQP' a cutting
plane. Then, by Prob. XXI., this element pierces that plane
at ce', one point of the required intersection. By repeating this
operation on a sufficient number of elements, given as in Pl.
XIV., Fig. 122, and joining the points found, we shall have the
complete intersection.
2°. The auxiliary surfaces, tangent planes to the hyperbo-
loid.-Every tangent plane, T, to the hyperboloid contains two
elements of it, and their intersection with the cutting plane, P,
will be in the line of intersection of P with T.
3°. The auxiliary surfaces, tangent planes to the asymptote
cone. Let the circle O-O', Pl. XIX., Fig. 148, be the
gorge, and do—a'B", the generatrix, which determine the given
hyperboloid, and let PQP' be the given secant plane. Circle
OA, and u̸'l'—O'—A"B" are the projection of the asymptote
cone. P is the horizontal trace of a plane, tangent to this
cone along the elements Ob-0'7', and ce is its parallel trace on
the parallel plane, c'h', of the gorge; ed-e'd' and es-e's' are
the elements parallel to Ob-O'b' and cut from the hyperbola
by the tangent plane Pb. The point P" is therefore one point
of the intersection of the planes PQP' and Pb. Another point
is hh', the intersection of the traces, hn-h'n' and ch-c'h, of the
same planes on the plane of the gorge. Hence P-7 is the
intersection of the two planes, and ' and ss', its intersections
with the elements cd and es, are when those elements pierce the
given plane PQP', and therefore points of the required intersec-
tion of PQP' with the hyperboloid.
By taking other like tangent planes to the conic asymptote,
other points of this curve can be found.
208
DESCRIPTIVE GEOMETRY.
II. REGARDING THE HYPERBOLOID AS A SURFACE OF REVOLUTION.
4°. The auxiliary surfaces, horizontal planes.-Let the hy-
perboloid, Pl. XX., Fig. 162, be given by the gorge whose
radius is A¿-A', and the generatrix YN-Y'N', and let PQP'
be the given cutting plane.
Then let ' be the vertical trace of any horizontal plane,
which will therefore cut a parallel from the hyperboloid, and
a horizontal from the plane PQP', whose intersections will be
two points of the required curve.
One point of the parallel is u'u, where the plane x'q' cuts
the generatrix YN-Y'N'. This parallel is therefore the circle
Au. The horizontal is 'q-xy, and it meets the parallel in
the two points, 27' and ss', of the required curve. Other points,
as mm', oo', etc., may be similarly found.
This method is evidently inapplicable in seeking the highest
and lowest points. These, however, can readily be found by
other methods.
5°. The auxiliary surfaces, meridian planes.-In Pl. XIX.,
Fig. 149, let the hyperboloid be given by its gorge, with radius
ОA—O'A', and generatrices mq-m'O' and qm—q'O'; and let
PQP' be the given cutting plane.
First. By (Prob. XCIII) or (265) find the meridian curve
C'A', D'B' in the meridian plane AB.
Second. By (Prob. XX.) find the point O, c', where the given
plane PQP' cuts the axis, 0-0″O', of the hyperboloid.
Then any line, as OP-c'u', will be the intersection of the
plane PQP' by an auxiliary meridian plane, as OP. The latter
plane will cut from the hyperboloid a meridian hyperbola,
whose vertical projection, when revolved about the axis O-
O'O', till parallel to the plane V, will coincide with C'A' and
D'B'. The like revolved position of the line OP-c'a' is Oa"-
c'a'"', and this intersects the revolved meridian curve at d'"d",
and e'"' (e" not shown), which are revolved positions of two
points of the required curve of intersection. Their primitive
positions, found by counter-revolution about O—0″O', to the
primitive position, OP-c'a', of the line containing them, are
the points dd', and ce'.
6°. The auxiliary surfaces, concentric cones.-This method
is exhibited in Pl. XX., Fig. 162, in finding the highest and
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DESCRIPTIVE GEOMETRY.
209
lowest points, or the extremities of the transverse axis of the
intersection of the plane PQP' with the hyperboloid, given by
its gorge, iy-iy', and one element, YN-Y'N'.
The meridian plane AD, perpendicular to PQP', is a common
plane of symmetry to PQP' and to the hyperboloid, and thence
to the required curve also, and thus evidently contains the trans-
verse axis of that curve.
By (Prob. XX.) PQP' cuts the axis, A-A"A', of the hyper-
boloid at Al'; and the horizontal trace of the plane AD, at DD'.
Hence DA-D'b', the intersection of these planes, is the indefi-
nite transverse axis. It only remains to find its limits. For
this purpose revolve DA-D'' about A-A"A' as an axis, and
it will generate the cone whose vertex is Al' and base is the
circle of radius AD. This cone is concentric with the hyperbo-
loid, and intersects it in two parallels, one point of each of which
is on the given element YN-Y'N'. Hence we have to find
where YN-Y'N' intersects the cone. To do this, pass a plane
through this element and the vertex. It will be determined by
YN-Y'N' and the parallel line, l'e'-Ac, through the vertex.
IIence Ye is its horizontal trace, and it cuts from the cone the
elements RA and SA which, being thus by construction in the
same plane with YN-Y'N', cut it at e and l, points of the
parallels of radius Ae and Ak which intersect DA-D'V' at ƒƒ''
and dd', the required extremities of the transverse axis of the
intersection of PQP' with the hyperboloid.
By proceeding in the same manner to generate concentric
cones by the lines cut from PQP' by other meridian planes,
other points of the required curve could be found, and such
construction would be more rigorously geometrical than that of
Pl. XIX., Figs. 149, 150, where the hyperbola is used as a line
of construction (41, 3°).
III. REGARDING THE HYPERBOLOID AS HOMOLOGOUS WITII ITS
ASYMPTOTE CONE, OR CONE DIRECTOR.
7°. The auxiliary surface, the cone director, to whose in-
tersection with the given plane, the required curve is made
similar.—In Pl. XIX., Fig. 150, let the hyperboloid be given
by its gorge, axis, generatrix ab-a'O', and meridian curve
parallel to the plane V, and let PQP' be the cutting plane.
By Prob. LVII. find the axes, only, fk-f'k' and gh-g' of
14
DESCRIPTIVE GEOMETRY.
210
the section of the cone director (260), a'b'—O', whose base is the
circle Oc. Then, as ep-e'p' is the transverse axis of the similar
section (Theo. XLI.) of the hyperboloid, contained in PQP',
make its shorter axis a fourth proportional to fk, gh and ep;
and thence by Prob. XLIII. construct the required curve.
8°. Construction of the tangent.-Pl. XX., Fig. 162. Let qq'
be the given point of contact. The required tangent at q' is
the intersection of the given plane PQP' with the tangent plane
at qq'. But the latter plane is determined by the two elements
drawn from q, tangent to the gorge. Thence the horizontal
trace, MT, of the tangent plane can be found, and thence tq—
t'q', the required tangent line at 17'.
EXAMPLE.-Let the given point of contact be above the gorge.
PROBLEM C.
To find the intersection of any ruled surface, not plane, with
a warped hyperboloid of revolution.
In Space.-As every ruled surface is composed of straight
elements, the problem will be sufficiently solved by finding
where a straight line pierces the given surface.
There are three cases in the relative position of a straight line
and the warped hyperboloid.
1º. It may be in a meridian plane, M, and hence will intersect
the axis.
2º. It may be in a plane, M,, parallel to M, and containing a
chord of the circle of the gorge.
3º. It may be in a like plane, M, but which is exterior to the
gorge.
In each case the line pierces the hyperboloid in points of the
hyperbola contained in the plane, M, M₁, or M,; points which,
however, can be found by using only straight and circular con-
struction lines (41, 3°) as will now be shown.
In Projection.-Case 1°. This was solved in Prob. XCIX.
(5º, 6º).
Case 2°. Let MN-M'N', Pl. XX., Fig. 158, be the given
line, after revolution till parallel to the plane V, about the axis,
DESCRIPTIVE GEOMETRY.`
211
A—A'A" of the given hyperboloid; whose gorge is of the radius
AF A"X', and whose generatrix is GH-G'H'.
=
The vertical plane through MN-M'N', will cut the given
hyperboloid in a hyperbola whose transverse or real axis is
xy-x'y', and one of whose asymptotes is MN-G'H'.
Revolve this hyperbola about a vertical axis at a, and it will
generate an auxiliary hyperboloid whose rectilinear generatrix
will be gh-G'H', and whose gorge will be Exy-x'y'. The
required intersection, being now the intersection of MN-M'N'
with the new hyperboloid, HI,, is reduced to that of Prob.
XCIX., 7°.
Thus the circle of radius aN, is the base of a cone, concentric
with II, and whose vertex is aE'. Drawing aC-E'C' parallel
to the generatrix gh—G'H', we find Cg the horizontal trace
of a plane containing this generatrix, and the elements ab and
ac of the cone. These elements meet gh-G'H' in 'H' and
kh' revolved positions of the required points. Their real posi-
tions are the intersections of the primitive position of MN-
M'N' with the parallels through II' and
'.
Case 3°.-Pl. XX., Fig. 159. Let db-d'b' be the gorge,
and Ga-G'A" the generatrix of the given hyperboloid, and
Rq-a'Q', the given line, where Rq is exterior to the given
gorge.
The vertical plane through Rq now cuts the given hyperbo-
loid in a hyperbola whose axis is a vertical line at R, and whose
revolution about that axis would be a hyperboloid of a different
kind from the given one.
Instead, therefore, of directly finding the intersection of Rq-
a'Q' with this new hyperboloid, we may proceed inversely, and
find the intersection of Ga-G'A" with the hyperboloid H,,
generated by the revolution of Rq-a'Q' around the given axis
A-A'A'; since II, and the given hyperboloid, II, intersect
in horizontal circles, evidently containing the intersections of
Rq-a'Q' with II, and of Ga-GʻA" with H₁. That is, these
intersections are of equal heights.
The vertical plane through Ga-G'A' cuts the hyperboloid
II, in a hyperbola whose real axis is ay-x'y', and thus this case
is reduced to the preceding one. This hyperbola generates a
third hyperboloid, II,, whose generatrix is DQ-a'Q', and
which, by Case 2°, is intersected by Ga-G'A" at mm' and ce'.
But since the given hyperboloid, and the one whose genera-
212
DESCRIPTIVE GEOMETRY.
trix is Rq-a'Q', intersect in horizontal circles; and since Rq-
a'Q' pierces the given hyperboloid in the same horizontal circle
in which Gɑ-G'A" pierces the second auxiliary hyperboloid;
and since mm' and ce' are points of these circles, Rq-a'Q" (the
given line) pierces the given hyperboloid at rr' and nn'. These
points could have been found directly by transference from r''op'
and n'n', for either pair of them, as r''r' and rr', is in the same
horizontal plane, and the two generatrices, Rq—a'Q' and DQ—
a'Q', are in the same plane, perpendicular to the vertical plane,
hence the intersections of these two lines with the horizontal
plane m'r', are in a perpendicular, rr''—r'.
B-DEVELOPMENT.
269. Since warped surfaces are undevelopable, the only opera-
tion of development here, is one of transposition (43) in find-
ing the true size of the intersection of a plane, or of a develop-
able surface, with the hyperboloid.
The true size of the intersection of PQP', Pl. XX., Fig. 162,
with the hyperboloid there given, is shown at f'"'q"d"o", by re-
volving the curve and its tangent about the trace P'Q of its
plane and into the plane V (Prob. LXXXIX). It could have
been revolved about either trace, or about any line parallel to
either trace of PQP'.
270. The development of the intersection of the hyperboloid
with any developable surface would be made by developing the
latter surface.
271. Complete graphical study of Prob. XCIX. Making the
figure fill a plate not smaller than those of this volume, all the
methods of Prob. XCIX. may be applied on one figure; dif-
ferent methods being applied in different quarters of the curve;
and including the construction of one or more tangent lines,
and the true size of the intersection, in this case, and in the
following-
EXAMPLES.-Rule the area of the intersection in all these cases, with parallel
shade lines, and let the body be truncated by the plane so as to more clearly ex-
hibit the curve, and show the true size of the curve.
Ex. 1. Find an elliptic section lying wholly on the upper nappe.
DESCRIPTIVE GEOMETRY.
213
Ex. 2. Let both traces of PQP' make acute angles with the ground line, on
the same side.
Ex. 3. Find a parabolic section of the hyperboloid.
Ex. 4°. Find a hyperbolic section of the same, by a plane cutting the gorge.
Ex. 5. Find a hyperbolic section made by a plane not cutting the gorge.
Ex. 6°. Apply method 6° to finding the points on the meridian curve parallel
to plane V.
Ex. 7°. Take PQP' in any of the positions shown in Prob. VII.
Raccordment.
272. Ruled surfaces may be tangent to each other in two
ways (152) at a point, or along an element, or on a curve of
contact. In the second case they will have a common tangent
plane at every point of their line of contact. Such contact is,
especially in case of warped surfaces, distinguished as raccord-
ment.
Thus two hyperboloids may raccord along a common ele-
ment.
THEOREM XLIV.
When two warped hyperboloids have a common tangent plane
at each of three points of the same element, they will
raccord along the element.
Let G, Pl. XXI., Fig. 163, be an element, common to two
hyperboloids, H and H,, and let p, PP, be any three points of
it. Let P, P, and P, be the three parallels of the hyperboloid
H, taken through p, p, and p,; and let C, C, and C,, respective-
ly, be the three curves cut from H, by the planes of these paral-
lels. If now H and H, be so situated that these three pairs of
curves shall have common tangents, as t, t, and t,, the element
C in moving on the three parallels P, P, and P,, will be entirely
separate from its positions in moving on C, C, and C..
Hence any plane, intersecting G, will cut H and H, in the
same manner as at P, P., P. Hence H and H, are tangent at
all points of G.
But note that, as the axes of H and H, cannot here be in the
same plane, the tangents to parallels, h and h,, cut from H and
H₁, respectively, and through the same point on G, cannot coin-
cide with each other.
214
DESCRIPTIVE GEOMETRY.
PROBLEM CI.
To construct two warped hyperboloids in raccordment.
In Space.-A warped hyperboloid being determined by its
axis, gorge and one element, the two hyperboloids will thus be
determined, the element being common to both.
In Projection. From the sketch, Pl. XIX., Fig. 147, a full
plate diagram can be constructed, showing the meridian curves,
and any desired number of elements.
Let AB-A'B' be one axis, ab-a'b' the other, and og-o'g'
the common element; and ON-O'N' equal to O-O'T', and
On-Q'n' equal to O-Q'T', the radii of the gorges of the two
hyperboloids.
The radius, BF, of the base of the larger body is then the
real size of Bg-B'g'; and bf, the radius of the base of the
smaller body, is the true size of bg—b'g'.
Having these bases, their vertical planes may be taken as new
planes of projection on which to show their circular form, as at
EB'F and da'c. Thence the vertical projections of the bases
can be found. Their axes are B-H'I', and EF-2B′E', for
the larger body, and b-h' and ef-2b'f' for the smaller one;
together with the opposite bases whose centres are at A' and ɑ'.
Any number of elements can now be found; first in horizon-
tal projection, and from the circular projections, as in Prob.
XCII., and then in vertical projection.
The vertical projections of the apparent contours will still be
tangents to the like projections of the elements, and will, on
construction, evidently consist of two equal, opposite, and infi-
nite branches, having two axes of symmetry, as P'T' and A'B'
for the larger body, as in Prob. XCII. Hence they are hyper-
bolas.
The horizontal trace, RS, of the vertical plane of the contour
just mentioned of the lower hyperboloid is through O (hence
called a diametral plane) and parallel to the tangent at t, the
extremity of the diameter Ot, conjugate to RS. The asymptotes
to this contour, are the elements cut from the conic asymptote
by the plane RS, where By is the radius of the base of this cone.
EXAMPLES.-1. Construct the case in which the contact, og, is in the sym-
metrical position g7, relative to AB.
DESCRIPTIVE GEOMETRY.
215
Ex. 2°. Construct the case when it has a position like either og or g₁ q on the
lower half of the larger body.
Ex. 3. Construct the case after revolving a'b' 180° about og-o'g', making
the smaller body inscribed within the larger.
Ex. 4'. Construct the case when the angle AOɑ between the axes, and in-
cluding og, is obtuse; in which case the bodies will intersect in two elements,
besides the element of contact.
Ex. 5. Ink, in heavy lines to show their forms clearly, thin frusta of both
bodies, included between planes of right section through the same points of
the element og.
Other Warped Surfaces.
273. Leaving, now, the only warped surface of revolution, the
existence and general character, only, of other warped surfaces,
may be indicated here, the fuller discussion of them being
placed by the plan of this work in Part II.
Taking the second definition (251) as a point of departure,
let A, B, and C, Pl. XXIV., Fig. 176, be any three lines, straight
or curved, and of single or double curvature, assumed at plea-
sure in space. Make any point of either of them, as a on A,
the vertex of a conic surface, a-bed, whose base or curved
directrix (90) shall be either of the others, as B. This cone
will intersect any auxiliary surface containing the remaining
line C, in some curve as D. But D and C, being on the same
auxiliary surface, will intersect, as at e; and D being also in the
conic surface, e is where the conic surface cuts C. Hence ane is
the only element, E, of the conic surface, which intersects A, B,
and C. By taking any number of other positions of a, other
lines like E could likewise be found.
274. Now if a plane be passed through E so as to cut A, B,
and C in points consecutive with a, n, and e, it follows from the
arbitrary form and position of A, B, and C, that these points
could not generally be in the same straight line. And converse-
ly, for the same reason, if points consecutive with a, n, and e, be
so taken as to be in the same straight line (133), that line could
not generally be in the same plane with E.
Whence we have this general definition: A warped surface
may always be generated by a straight line moving upon three
fixed lines taken as directrices.
275. Since the consecutive elements are thus not in the same
216
DESCRIPTIVE GEOMETRY.
plane, it follows that a plane passed through any one of them
must generally cut all the others on each side of it, thus forming
a curve which will cross the given element.
Then let E, Pl. XXIV., Fig. 179, be an element of a warped
surface, and let C be the curve cut from the surface by some
plane containing E. By (147, 149) the tangent plane at t is
determined by E as tangent to itself (135) and T the tangent
to C at t.
276. From this we learn three things-
1º. That the tangent plane to a warped surface has only a
point, and not an element of contact.
2º. That every plane containing an element of a warped sur-
face, is generally tangent to the surface at some point of that
element, viz., the one at which it intersects the curve cut from
the surface by that plane.
3º. That the tangent plane to a warped surface is also a secant
plane.
277. Of particular warped surfaces, the plan of this volume
permits the further mention of such only, as those in which the
actual motion of the generatrix may be resolved into component
motions, one of which is a motion of revolution around a fixed
axis. The following are the most common and useful:
278. First. The hyperbolic paraboloid.—Pl. XXIV., Fig. 180.
Let PQ be any plane, AB a perpendicular to it, and CD a line
oblique to PQ. Then if the line AC while ascending along AB,
and remaining parallel to PQ, also revolves around AB so as
always to touch CD, it will generate a simple form of this sur-
face, as seen in the cow-catcher of a locomotive.
279. Second. The helicoid.—Pl. XXIV., Fig. 181. Let ADS
be a cylinder of revolution. If a point, A, moves on the convex
surface of this cylinder so as to cross its elements BB', etc., at
a constant angle, it will generate the curve called a helix. If
now from all points of this helix lines be drawn intersecting the
axis, ax, at any constant angle, they will form the warped sur-
face called a helicoid. If the constant angle be acute, the heli-
coid is called oblique; if the angle be a right angle, the surface
is called a right helicoid. Triangular, and square-threaded
screws are examples respectively of the use of oblique, and of
right helicoids.
280. From the definition of the helix, it is evident that if the
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DESCRIPTIVE GEOMETRY.
217
cylinder containing it be developed, the development of the
helix, Fig. 181, will be a straight line, ac', which will coincide
with the development of its tangent, A'D', since both make the
same constant angle with any plane perpendicular to the axis of
the cylinder.
281. Third. The conoid.-Pl. XXIV., Fig. 182. Let PQ
be a square in which the circle C is inscribed, and let ab be ver-
tically over, and equal to the diameter AB. Let PR be a plane
perpendicular to PQ, and let cfg, hkm, CoD, etc., be planes
parallel to PR and cutting ab at g, m, o, etc. Then eg and ƒg;
hm and km, etc., are elements of the simplest form of the warped
surface called a conoid, and which may be described as a cone,
whose vertex, o, is expanded into a line ab equal to a diameter
of its base.
Calling Aa the generatrix, it thus has an angular motion
within the range CoD, around ab as an axis, while it advances
along that line, and remains parallel to PR.
Other forms of the preceding, together with other warped
surfaces, will be appropriately treated in Part II.
BOOK II.-DOUBLE CURVED SURFACES OF
REVOLUTION.
282. A DOUBLE CURVED SURFACE is one on which no straight
line can be drawn.
The most important division of such surfaces for the purposes
of descriptive geometry, is into doubly-convex surfaces, and con-
cavo-convex surfaces.
Thus let the curve abc, Pl. XXI., Fig. 164, generate a double
curved surface of revolution by revolving about AB as an axis.
Both the meridians (14), as bc, and the parallels of the portion
generated by be, will be convex towards exterior space, and that
portion will therefore be doubly convex. But in the portion
generated by ab, the parallels will be convex, and the meridians
concave towards exterior space, and that portion will therefore
be concavo-convex.
283. As sufficient examples of double curved surfaces of each
kind, we shall here consider first, those generated by the revolu-
tion of any conic section about whichever of its axes meets it;
and second, those generated by any closed plane curve as a circle
about an axis exterior to it and in its own plane.
The former surfaces are all doubly convex. The latter, called
an annular torus, has a concavo-convex portion.
284. The surface generated by the revolution of an ellipse
about one of its axes, is called an ellipsoid of revolution. Since
an ellipse has two axes, there are two forms of ellipsoid of
revolution, first, that generated by revolving the ellipse about
its longer axis, and called a prolate ellipsoid, or spheroid; and,
second, that formed by its revolution about its shorter axis, and
called an oblate ellipsoid, or spheroid.
When the two axes are equal, the ellipsoid becomes a sphere.
DESCRIPTIVE GEOMETRY.
219
4
285. The surface formed by revolving a parabola about its
one axis is a paraboloid of revolution. That formed by revolv-
ing a hyperbola about its transverse axis is a hyperboloid of re-
volution of two separate nappes, in distinction from the hyper-
boloid of two united nappes (252) which is warped, and formed
by revolving the hyperbola about its conjugate axis, or the one
which does not intersect it.
286. An ellipsoid, etc., of revolution, may also be generated
by a circle. Thus, let S be any conic section, one of whose axes
is a.
Then let a variable circle C move, so that, first, its centre
shall remain on a; second, its plane shall be perpendicular to
a; and third, one of its diameters shall always be a chord of S.
The circle will thus generate a surface of revolution, of which
its successive positions will be parallels.
287. The regular pyramid of any number of sides, is evi-
dently analogous to the cone of revolution. Likewise, when a
polyedron may be inscribed in a double curved surface, as the
regular polyedrons of elementary geometry may be inscribed in
a sphere, that is, with its vertices in sets, each set in a parallel
of the surface, it is analogous to the double curved surface.
A-PROJECTIONS.
a-Projections of Forms.
288. In treating one double curved surface alone, its axis of
revolution, or principal axis, may always be made perpendicular
to a plane of projection. In this case, the projections of the
surface will consist of a parallel, on the plane to which the axis
of the surface is perpendicular; and of a meridian on the other
plane.
289. In a given combination of surfaces, some one, or more,
of them may be placed with its axis oblique to both planes of
projection.
In such a case, first, the body may be revolved till brought
into the same relation to the planes of projection, that was
described in the last article; or, second, new planes of projection
may be assumed having the same simple positions relative to the
220
DESCRIPTIVE GEOMETRY.
surface; or, third, the surface may be directly represented by
the projections of all its axes; or, by the projections of a suffi-
cient number of its parallels, or of its meridians, tangent to
all of which a curve may then be sketched, which will represent
the visible boundary, or apparent contour of the surface.
PROBLEM CII.
To construct the projections of double curved surfaces of
revolution.
In Space.-10. The sphere.-The two projections of a sphere
will always be two equal circles, whose centres will be on the
same perpendicular to the ground line; except when the centre
of the sphere is in the ground line, when the two projections
will coincide.
2°. The ellipsoid.-The axis being vertical for example, the
projection on H will be that of the greatest parallel. That
on V, that of the co-parallel meridian.
3°. The paraboloid, and the hyperboloid of separate nappes.
The greatest parallel of each of these being at infinity, they are
represented on one plane of projection by the meridian which
is parallel to it; and on the other, by the parallels in which they
intersect it.
4°. The annular torus.-This is represented by its greatest
and least parallels, and by that meridian which is parallel to a
plane of projection.
In Projection.-See Pl. XXII., Figs. 169, 172, 173, for the
projections of a sphere; Pl. XXI., Fig. 167, for that of an
ellipsoid; Pl. XXI., Fig. 166, for that of a paraboloid; and
Pl. XXIII., Fig. 175, for that of an annular torus, the axis of
revolution being vertical in each case. The torus is generated
by the revolution of the meridian circle, BC-C'D'B'II', about
the axis A-A'A". The portion of the surface without the
concentric cylinder of revolution, whose radius is AE-A"E',
is doubly convex, that within is concavo-convex (282).
EXAMPLES. -1°. Construct by (289), the projections of an ellipsoid of revo-
lution, whose axis is oblique to both planes of projection.
Ex. 2°. Construct by (289) the projections of an annular torus, under the
same condition.
DESCRIPTIVE GEOMETRY.
221
290. The construction of either projection of any point on a
double curved surface of revolution, when its other projection
is given, is made precisely as in Prob. XCIV., 2° and 3º.
EXAMPLE.—In Pl. XXIII., Fig. 175, find the vertical projections of the points
m, and the horizontal projections of the points qʻ.
PROBLEM CIII.
To construct the projections of a regular dodecaedron.
In Space. The regular dodecaedron is a polyedron which is
bounded by twelve equal and regular pentagons. Hence
First. Assuming any one of the pentagons as a base, each of
its sides is an edge of five adjacent pentagons, whose planes
make equal angles with the assumed pentagon. These form
half of the body.
Second. The vertices of the dodecaedron are in four parallel
planes, those of the opposite equal and parallel bases, and two
intermediate planes, which are equidistant from the adjacent
bases, and which contain the remaining vertices of the body.
If, then, we represent three adjacent pentagons as revolved
into one plane of projection, and then fold up two of them,
about their edges of intersection with the third, as axes, till they
reach their true relative positions, as faces of the dodecaedron,
we shall determine their vertices as seen in projection, so as to
be able to construct the projections of the remaining vertices.
In Projection.-Pl. XXI., Fig. 165. Let the four parallel
planes above mentioned be taken; one, to coincide with the
horizontal plane, and the other three, parallel to it.
1º. The horizontal projection.-Let the base abcde be in the
plane H, and with one of its sides parallel to the ground line;
and let the adjacent and equal pentagons, whose edges are de
and ca, be represented as already revolved into that plane, at edf
and aer.
When these pentagons are counter-revolved about de
and ca as axes, the points g and g' will revolve in arcs whose
projections are gH and g'II, perpendicular, respectively, to de
and ea, and which intersect at that vertex of the dodecaedron
whose horizontal projection is II. IIe is, therefore, the horizon-
tal projection of an edge of the body. As He, produced, passes
222
DESCRIPTIVE GEOMETRY.
through O, the projections of similar edges, as dp and aQ, may
be drawn from the remaining angles of the base.
To find the horizontal projection of f. The lines fu and gk,
being parallel, are respectively proportioned to their projections
Fu and Hk. Hence, construct Fu, a fourth proportional to gk,
H and fu, and the point F will be determined as the hori-
zontal projection of f. Having gone thus far, it is obvious that
as He, FE, dp, etc., are equal in space, and equally inclined to
the plane H, and as the bases are evidently inscribed in the
same circle, of centre O, the points H, F, p, etc., are vertices of
a regular decagon.
2º. The vertical projection. The points a, b, etc., of the lower
base, will be vertically projected in the ground line. To find
the height of the horizontal plane containing the points of which
II is one, consider that Hh is the horizontal projection of gh,
then revolve the vertical plane whose horizontal trace is gk, about
that trace as an axis. H will fall in a perpendicular to gk, and
at a distance g''H, equal to its height above the horizontal plane,
and found by drawing the arc gg", with centre 7, till it intersects
the indefinite perpendicular Hg". Moreover, II and Fu are
in a plane which is perpendicular to the vertical plane, gk,
through II, hence ko uf is the projection of uF on this ver
tical plane, and no is the true height of F, and of other similar
points, above the horizontal plane.
=
Hence, make a'Q' Hg", and a'q' no, and draw the traces
II'Q' and F'q'. Also q'D' a'Q', which locates the trace of
=
the plane of the upper base. Project the alternate points, II,
Q, etc., into the vertical trace II'Q' of their plane, the interme-
diate points F, P, etc., into the vertical trace F'P' of their plane,
and the upper base, AD, into its plane E'C'. Lastly, con-
nect the points thus found, and dot the concealed edges.
291. Peculiar properties.-Some interesting properties follow
from the value, 108°, of the angles of the pentagon, and from
the proportional division of the parallels fu and gk, by FH.
The angles as deu being 108°, geg′ = 36°, and hence the angles
at g and g = 72° each, as marked. IIence fgg'r is one straight
line; but, as fg is the revolved position of FH, about de as an
axis, fy, FII and de must meet at a common point, which is no
other than; for cuf 90°, and ufr = 54°, hence fru = 36°.
But 36° is just the angle which g'r as the side of a pentagon
makes with the chord, as re, of two of its sides; and, as dr de
DESCRIPTIVE GEOMETRY.
223
produced, and fr = fg, produced, the angles fru and gre are
identical. These properties serve to check errors in construction.
EXAMPLE.-From the position just given, construct the projections when the
axis, O-a'D' is oblique to either, or both, H and V.
b-Projections of Tangencies.
292. Since no straight line can be drawn upon a double
curved surface, there can be only a point of contact between it
and a tangent plane.
Therefore, regarding every plane as determined by three
points, either given, or implied in other equivalent conditions
(253), the fundamental given conditions for drawing tangent
plancs to double curved surfaces, are the same as in the case of
warped surfaces, viz., at a given point; through a given line,
and parallel to a given plane.
293. Through a point in space, an infinite number of planes
can be drawn, tangent to a double curved surface. For, an in-
finite number of secant planes may be drawn through the given
point; and to each of the curves cut from the surface by these
planes, a tangent line may be drawn through the given point.
These lines will, collectively, form a cone to which an infinite
number of tangent planes can be drawn, which will all be tan-
gent to the double curved surface, since the cone and double
curved surface will be tangent to each other.
On the other hand it will be remembered that, at most, but
two planes can be drawn through a point in space, and tangent
to a single curved surface.
294. An infinite number of planes may also be drawn, tan-
gent to a double curved surface, and parallel to a given line.
For, through some point of the surface a tangent line can be
drawn parallel to the given line. An indefinite number of
such tangents will form a cylinder which will be tangent to
the double curved surface. Now an infinite number of planes
can be drawn tangent to this cylinder and parallel to the given
line, and all of them will be tangent to the sphere.
295. It is generally impossible, as has been seen, to draw a
plane, tangent to a single curved surface, and through a given
line. Two tangent planes can however always be drawn
through a line and tangent to a double curved surface, unless
224
DESCRIPTIVE GEOMETRY.
that line pierces the surface, or is tangent to it. For a tangent
cylinder being drawn, as above described (294), two planes can
be drawn tangent to it and through the given line.
296. If two surfaces of revolution having a common axis are
tangent to each other, they will have a circle of contact whose
plane will be perpendicular to the common axis. For, this
circle will be generated during the revolution of the meridian
curves, by the point of contact, as t, t,, t,, Pl. XXII., Fig. 171, of
those meridian curves which are in the common meridian plane.
All these parallels of contact are evidently perpendicular
to the axis AX.
297. Comparing the different cases, it is clear that a cylinder
can only be tangent on a maximum or a minimum parallel, as
at t,, or D; while a cone, or a double curved surface, can be
tangent on any parallel of the given double curved surface.
The tangent cone at the vertex, as A, of the given surface,
reduces to a tangent plane at that point.
298. Since any two tangent surfaces will have a common
tangent plane at their point, or element of contact, problems
of other than plane surfaces, tangent to given double curved
surfaces, may conveniently be solved by means of auxiliary
tangent planes.
299. Conversely, as will soon appear more fully, tangent
planes to double curved surfaces may often be easily drawn by
making them tangent to some simpler auxiliary surface which
is itself tangent to the given surface (293, 294).
300. Curved surfaces may be tangent to each other in two
ways, first, at a point of contact; second, on a curve of contact
(296). Thus, Pl. XXII., Fig. 170, AtB and CṭD, tangent at t,
may represent two double curved surfaces, having only a point
of contact. Their axes, AB and CD, may, or may not, be in
the same plane, and the bodies may be on the same, or, as in
the figure, on opposite sides of the common tangent plane at t.
301. The problems of tangent planes to double curved sur-
faces, being numerous, varied, and useful, it will be inter-
esting to observe their natural order, the planes being con-
sidered as determined in thought in every case, by three given
or supposed points; and, in fact, by any auxiliary points, con-
structed so as to be in the same plane with the primarily con-
ceived points. (Prob. XIV.)
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DESCRIPTIVE GEOMETRY.
225
PROBLEM CIV.
To construct a tangent plane to a double curved surface of revo-
lution, at a given point of contact.
In Space. The required plane may be determined; first,
by any two tangent lines at the given point of contact (147) or
by any lines which intersect both of them; and, second, by the
normal to the surface at the given point, to which the tangent
plane will be perpendicular.
In Projection.-Pl. XXI., Fig. 166. Adopting the first
solution, and taking the tangents to the parallel, and to the
meridian of the surface chosen for illustration, let the given
surface be a paraboloid of revolution having its axis vertical.
The point t being assumed as the horizontal projection of the
point of tangency, t', its vertical projection is found by Prob.
XCIV., and tt' then becomes the given point of tangency.
The line tn-t'n' is the tangent to the parallel at tt'. Its verti-
cal trace is n', which is therefore a point of the vertical trace
of the tangent plane. After revolving the meridian plane CR,
containing the point tt', about C-a'C' as an axis, till it is
parallel to the vertical plane, R'''P-R"C drawn tangent at t'''
to C't'"'A', the revolved position of the meridian curve in the
plane CR, is evidently the revolved position of the tangent to
the meridian curve at tt'. In the counter revolution, P, being
in the axis, remains fixed, and PR'—CR is the primitive posi-
tion of the same tangent. Its horizontal trace is R, which is
therefore a point of the horizontal trace of the tangent plane.
But as the meridian plane, CR, is perpendicular both to the
tangent plane and to the horizontal plane, the horizontal traces
of the meridian and tangent planes will be at right angles to
each other. IIence Ph (Q on the ground line), perpendicular
to CR, and parallel to tn-t'n', is the horizontal trace of the
required tangent plane. p'n' is its vertical trace, but as Q is not
within the limits of the diagram, any line, as UP-—ʊ'P', being
drawn parallel to any known line of the tangent plane, as
RC-R'P, and through a point, as bb' of the known trace, will
be a line of the tangent plane, and the point p', in which it
pierces the vertical plane, will be a second point of the vertical
trace, through which, and n', that trace may be drawn.
15
DESCRIPTIVE GEOMETRY.
226
EXAMPLES. -1°. Let the given point be behind the meridian AC.
Ex. 2°. Let it be on the upper right hand quarter of an ellipsoid having a
vertical axis.
Ex. 3°. On the lower half of the same ellipsoid.
Ex. 4°. On a sphere whose centre is in the ground line.
Ex. 5°. On a torus whose axis is parallel to the ground line.
Ex. 6. On the outer half or nappe of a torus whose axis is vertical.
302. The plane PQP', Pl. XXIII., Fig. 175, illustrates that
when a plane is tangent at a given point, TT', of a concavo-
convex (282) double curved surface, such as the inner portion
of the annular torus, which is generated by the semicircle
D'C'H', it is also a secant plane.
The tangent plane at any point of the torus, exterior to the
cylinder of revolution containing the highest and lowest hori-
zontal circles, E'D' and G'I', would not be a secant plane.
The tangent plane at any point on the greatest or least paral-
lels would be vertical, in Pl. XXIII., Fig. 175, that is, in gen-
eral, parallel to the axis.
Tangent planes perpendicular to the axis would have circles
of contact. This is a particular application of (296).
PROBLEM CV.
To construct a plane, through a given point in space, and tan-
gent to an ellipsoid at a given point on one of its plane
sections.
In Space.-As both a parallel and a meridian can be drawn
through every point of the given surface, only these sections
will be considered.
First. If the given section be a parallel, conceive of it as the
circle of contact of a cone, or sphere of revolution (296). Then
the planes, made tangent to this cone, or sphere, will by (293)
also be tangent to the ellipse, at points of the given parallel.
Second. If the given section be a meridian, conceive of it as '
the curve of contact of a cylinder, whose elements are perpen-
dicular to its plane. Then the tangent planes to this cylinder
(294) will be tangent to the ellipsoid, where their traces on the
plane of the meridian are tangent to that curve.
DESCRIPTIVE GEOMETRY.
227
In Projection.-First. The given section a parallel.
1º. Auxiliary cones.-Pl. XXI., Fig. 167. Let the axis of the
ellipsoid be vertical. The circle whose radius is AB, and the
ellipse, A'M'L', are the projections of the ellipsoid. VV' is the
given point through which the required tangent planes are to be
drawn.
Let L'M'-EgM be the given parallel. L'N' is an element of
the auxiliary cone, tangent to the ellipsoid along this parallel.
Its base, in the horizontal plane, V'N', is the circle YNk. VY
is therefore the trace, upon this plane, of a tangent plane to the
cone. Hence YA is the element of contact of the tangent plane
and the auxiliary cone, and gg', the intersection of this element
with EgM-L'M', is the point of contact, on the given parallel,
of a plane, through VV', and tangent to the ellipsoid.
Draw a circle on VA, as a diameter. From Elementary
Geometry it is evident that a line through V, tangent to any
circle having A for its centre, will have its point of tangency on
the circle described on VA. Hence A is the horizontal pro-
jection of the element of contact of a second tangent plane, to
the auxiliary cone, and hh' is another point of contact, on the
parallel, EgM-L'M', of a plane, tangent to the given ellipsoid,
and through VV'.
When the greatest horizontal section, B'C'-BeC, is the given
parallel, the auxiliary cone becomes a cylinder, and the horizon-
tal traces of the two vertical tangent planes to this cylinder, con-
taining VV', are Ve and Vd-the latter not drawn. Hence ee'
and dd' are the projections of the two points of contact of these
planes on the parallel, BeC-B'C'.
2º. Method by auxiliary spheres.—Let EgM-E'T′ be the
given parallel. Make it the circle of contact of the auxiliary
sphere. E'F', normal to the ellipsoid at E', is the radius of the
sphere, and the circle with radius E'F' is its vertical projection.
V'''q' is the revolved position of one element of the auxiliary
cone, having its vertex at VV' and tangent to the sphere. q'm',
perpendicular to V'F', is the vertical projection of the circle of
contact of the cone and the sphere. II'II is then the revolved
position of one point of the chord of intersection of the given
parallel with the circle of contact q'II'm'. II" is the horizontal
projection of the true position of this point, and GI is the hori-
zontal projection of the chord just mentioned. Hence, G and I,
the intersections of this chord with EgM, the horizontal projec
DESCRIPTIVE GEOMETRY.
228
tion of the parallel, are the horizontal projections of the points-
sought. G' and I' are their vertical projections, on the vertical
projection, E'I', of the parallel.
The traces of the planes through VV', and tangent at GG'
and II', can now readily be found by Prob. CIV.
Second. The given section a meridian. Let PfA be the hori-
zontal trace of any assumed meridian plane. This plane being
vertical, VP, drawn perpendicular to it, is the horizontal pro-
jection, and true length of the intersection of two tangent planes
to the ellipsoid; drawn through VV', and perpendicular to the
meridian plane, PfA. Revolve the plane, PfA, about a verti-
cal axis at A, till it becomes parallel to the vertical plane of pro-
jection. Then the meridian curve in it will be vertically pro-
jected in A'C'D'B', and the point P at P", since the line, VP,
is in the horizontal plane, V'P". Hence draw the tangents,
P'"'e'" and P""""; and ""e" and f"f", will be the revolved
positions of the points of contact of the two tangent planes.
Making the counter revolution, the points, c'e'" and f"f",
will return in the horizontal ares, d'e-e""e' and f"fff", to
the positions ff' and ce', which are the points of contact of the
tangent planes, whose traces can then be easily found.
The highest and lowest points of contact of tangent planes
through VV', are in the meridian plane, VoA. Hence revolve
this plane about the vertical axis at A, when VV' will appear at
V''V''', and V"A-V""b" and V"A-V"""" will be revolved
positions of the traces of these tangent planes, upon the plane
VoA, giving bb"" and o"o"" for the revolved positions; and,
by counter revolution of the meridian plane to the position,
VoA, the points bb' and oo', which are the projections of the
highest and lowest points sought.
The lines, V'a' and V'n', are the vertical traces of planes
through VV', and perpendicular to the vertical plane of projec
tion, hence these planes are tangent to the meridian curve,
BAC-A'B'C'D', at the points, a'a and n'n. Their horizontal
traces will be perpendicular to the ground line.
303. If, Pl. XXI., Fig. 167, straight lines be drawn from
VV' to all the points of contact aa', bb', ce', dd' . . . . hh', etc.,
there found, they will evidently be tangent to the ellipsoid,
since they all lie in tangent planes, at those points, and contain-
ing VV'. Hence they are elements of the circumscribing tan-
DESCRIPTIVE GEOMETRY.
229
gent cone, whose vertex is VV', and whose curve of contact
with the ellipsoid is the curve, abceh—a'b'c'e'h', which joins the
points of contact just mentioned.
This may serve to illustrate how, as mentioned in (298),
other than plane surfaces are made tangent to given curved sur-
faces by means of auxiliary tangent planes, common to both.
The next article affords another illustration.
301. To find the curve of contact of a cylinder with a double
curved surface of revolution.-Let a line be given, or assumed,
to which the elements of the cylinder shall be parallel. If we
place any auxiliary cone with its vertex in the axis of the surface,
that is, so that its curve of contact will be a parallel of the sur-
face, a plane can, by Prob. LIII., be drawn tangent to this
cone and parallel to the given line. The element of contact on
this cone can readily be found, also the point where it meets the
parallei. The latter point being the point of contact of the
plane and double curved surface, will be a point of the curve of
contact of the cylinder and given surface; found, as now seen,
by applying the problem: To draw plane parallel to a given
line, and tangent on a given parallel.
305. The highest and lowest points of this curve, the axis of
the surface being vertical, will evidently be in that meridian
plane which is parallel to the given line, and at the points of
contact of lines parallel to the given line, with the meridian
curve in that plane.
306. Utility of this class of problems.—If VV' (Fig. 167),
be the position of the eye, the contact of the cone, having VV'
for its vertex, with the ellipsoid, will be the curve of apparent
contour of the ellipsoid, as seen from VV'. It will also be the
line of separation between the light and the shade of the ellip-
soid, if VV' be a luminous point.
Again: if the line, L (304), be a ray of light, the curve of
contact of the cylinder parallel to it, will be the boundary of the
shade of the ellipsoid. Also, if L represents the direction of
vision, a right section of the parallel tangent cylinder would be
the projection of the ellipsoid upon the plane of that section.
EXAMPLES.-1°. Construct the curve of contact of a cone having a given
vertex, with an urn of revolution.
Ex. 2. Construct the curve of contact of a cylinder parallel to a given line,
with an annular torus.
230
DESCRIPTIVE GEOMETRY.
PROBLEM CVI.
To draw a plane, tangent to a sphere, and containing a given
line.
In Space.-Two such planes can evidently be found. A
plane, N, through the centre of the sphere, and perpendicular to
the given line, will cut from the sphere a great circle, C, from
the given line, a point,p; from which two tangents can be drawn
to the circle C.
Each of these tangents, with the given line, will determine a
tangent plane as required. Hence there are two such planes.
Pl. XXII., Fig. 169. Let SS' be the centre of the given
sphere, and AB-A'B' the given line.
The auxiliary plane, N, perpendicular to AB-A'B', is deter-
mined by the two lines, SA-S'a' and Sb-S'', drawn through
the centre of the sphere, and parallel to the traces of N. By
(Prob. XXII.) it cuts the plane which projects AB-A'B' upon
the horizontal plane in the line Ab—a'b', thence giving (see Fig.
168) CC', the intersection of Ab-a'b' with AB-A'B', as the
point in which the perpendicular plane, N, through the centre
of the sphere, cuts the given line.
SA-S'a, being horizontal, is a convenient axis, about which
to revolve the perpendicular plane parallel to the horizontal plane
of projection. By such revolution, the great circle in the plane,
N, will coincide with the horizontal great circle of the sphere.
AB, being perpendicular to the axis SA, the point CC' will, after
this revolution of the perpendicular plane, be found at C", at a
distance, CA, from the axis, equal to the hypothenuse, C'e, of
a right angled triangle, whose base is CA, and whose altitude is
C'd, the height of the point CC' above the horizontal plane con-
taining the axis SA-S'a'.
The tangents C'h" and C''t" are evidently the revolved posi-
tions of the tangents, cut from the required tangent planes by
the perpendicular plane; hence, " and t" are the revolved posi-
tions of the points of contact of these tangent planes. The point
n, being in the axis SA, and, therefore, fixed during the revolu-
tion, Ct is the horizontal projection of C''t" after counter revo-
lution, and t't, perpendicular to SA, is the horizontal projection
of the vertical arc in which the point t" revolves, and gives t for
DESCRIPTIVE GEOMETRY.
231
the horizontal projection of one of the points of contact. Pro-
ject n into the vertical projection, S'a', of the axis, at n'; draw
C'n't', and t', projected from t, is the vertical projection of the
same point.
The tangent, C"h", does not intersect the axis, SA-S'a',
within the limits of the diagram, hence through any point, o",
of the tangent, C''t", and any point, f, of the axis, SA, draw an
indefinite line, "fe", and note its intersects with C"". In the
counter revolution, o" revolves in a plane perpendicular to SA,
and is found at o. Then f, being in the axis, remains fixed, and
ofe is the horizontal projection of o"fe", when in its primitive
position. In this counter revolution, e' returns in the arc e'e,
perpendicular to SA, to e on of. Ce is therefore the horizontal
projection of the second tangent line, to which h" returns in the
are h'h, to h, its point of contact with the sphere. Project f
into the vertical projection of the axis at f', o into the tangent,
C't', at o', draw of', and project e upon it at e', and draw C'e',
which is the vertical projection of Ce, the second tangent, cut
by the perpendicular plane from the tangent plane, on which
project hat h'.
The traces of the tangent plane at the point, hh', are PQ and
P'Q. PQ is drawn through the point B, in which the given
line pierces the horizontal plane, and perpendicular to Sh, the
horizontal projection of the radius at the point of contact. Like-
wise P'Q passes through ', where the given line intersects the
vertical plane, and is perpendicular to S'', the vertical projec-
tion of the same radius at the same point of contact, hh'.
The traces of the tangent plane, at tt', may be similarly found.
For greater accuracy in the construction of ce', o"e" should
have made a less angle with the ground line.
In all problems where as here, some of the points of construc-
tion are points of contact of tangents, as t", they should be
exactly determined by means of their radii.
EXAMPLES. -1°. Let AB-A'B' be to the left of the sphere.
Ex. 2°. Let it cross the first angle.
Ex. 3'. Take the projecting plane of AB—A'B' perpendicular to the plane V.
Ex. 4. Revolve the plane N about Sb-S'b' till parallel to the vertical plane.
Ex. 5. Let the centre of the sphere be in one of the planes of projection.
Ex. C. Let the centre of the sphere be in the ground line.
Ex. 7. Let AB-A B be a bi-parallel.
Ex. 8'. Let it be in a profile plane.
Ex. 9. Let the sphere be in any other angle than the first.
232
DESCRIPTIVE GEOMETRY.
PROBLEM CVII.
To construct a plane, containing a given line, and tangent to a
sphere, by means of one auxiliary tangent cone.
In Space.-Make any point of the given line the vertex of a
circumscribed tangent cone of revolution, whose axis will
therefore contain the centre of the sphere, and whose base will
be the circle of contact C, of the cone and sphere.
Two planes can generally be drawn containing the given line
and tangent to the cone. By (293) they will both be tangent
to the sphere, and will be determined by the given line and their
traces upon the plane of the circle of contact C.
In Projection.-Pl. XXII., Fig. 172. Let AB-A'B' be the
given line. Let the plane containing the centre of the sphere
and the axis of the cone be taken for convenience as the horizon-
tal plane of projection, and let the circles at S and S' be, accord-
ingly, the projections of the sphere.
As the horizontal plane cuts the given line at AA', this point
is taken as the vertex of the auxiliary cone, whose axis is AS and
whose extreme elements, in the horizontal plane, are Ae and
Aƒ; ef is, therefore, the diameter and horizontal projection of
the circle of contact of the cone and sphere.
The given line AB-A'B' pierces the plane of the cone's base
at the point CC', for, as that base is vertical, feC is the horizon-
tal trace of its plane, and the vertical plane ABB', which pro-
jects AB-A'B' upon the plane, H, intersects the former plane
in the vertical line, C-K'C' which meets AB-A'B' at CO'.
Revolve the plane of the cone's base about f&C as an axis,
and into the plane H. CC' will describe the vertical arc
whose horizontal projection, CC", = C'K', and is perpendicular
to feC; and C" will be the revolved position of CC', and the
circle fl'a" the revolved position of the cone's base. The tan-
gents C"a" and "b" are then the revolved positions of the traces
of the two tangent planes upon the plane feC; and a" and "
are the revolved positions of their points of contact with the
sphere. Making the counter revolution of the plane feC, the
points a" and b' revolve in vertical arcs, whose horizontal pro-
jections are a'a and b'b, to the positions a and b, which are the
horizontal projections of the points of contact.
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DESCRIPTIVE GEOMETRY.
233
By observing the direction of the revolution, it appears that
aa' is below the horizontal plane at a distance equal to aa",
while bb' is above the same plane at a distance equal to bb'.
Now, since both of the tangent planes contain the given line,
their horizontal traces will pass through A, and their vertical
traces through B'. The horizontal traces will pass through the
points, as N (the other falls out of the plate), in which the tan-
gents C"a" and C"b" intersect Cef. Also, AN is perpendicular
to Sa, and QB' to S'a', the radius of the point of contact aa'.
Similar conditions pertain to the traces AR and RB' of the tan-
gent plane at lʊ'.
EXAMPLE.-Repeat the examples of Prob. CVI. here, except 3° and 4°.
PROBLEM CVIII.
To draw a plane through a given line and tangent to a double
curved surface of revolution, by means of two auxiliary
tangent cones.
In Space.-Make any two points, p and p,, of the given line
L, the vertices of circumscribing tangent cones of the given sur-
face. Two planes, P and P₁, can generally be drawn through
the given line, and tangent to both of the cones. They will
both be tangent also to the given surface; and the point of con-
tact of either of them, as P, may be conceived, either as the point
in which the curves of contact, C and C,, of the two cones, with
the given surface, intersect; or as that in which their elements.
of contact with this plane intersect.
To reduce the labor of finding the curves C and C,, the plane
of each should be perpendicular to one of the planes of projec-
tion. Hence p and p, should be the points in which the given
line L intersects the plane of the greatest parallel, and a meri-
dian plane. The latter are therefore the most convenient planes
of projection.
In Projection.-While the above solution is general, it is most
simply illustrated when the given surface is a sphere.
Hence, Pl. XXII., Fig. 173, let A'S be the ground line, S
the centre of the sphere, and AB-A'B' the given line. A is
the point at which the horizontal plane, and B' is the point at
234
DESCRIPTIVE GEOMETRY.
which the vertical plane cuts the given line. AA' is the vertex
of the tangent cone whose axis, AS, is in the horizontal plane.
The centre of the sphere being in the ground line, one and the
same circle represents both of its projections, hence Ag and Ae
are the extreme clements of the cone whose vertex is AA', and
eg is at once the diameter and the horizontal projection of its
base, or circle of contact with the sphere. Likewise, BB' is the
vertex of the cone whose axis, B'S, is in the vertical plane; B'n
and B'p' are its extreme elements, and p'n' is the diameter and
vertical projection of its circle of contact with the sphere.
Now ge is the horizontal, and d' the vertical trace of the
plane of the vertical circle ge; also, p'n' is the vertical, and c'e
the horizontal trace of the plane of the circle pn; hence, by
Prob. XVII., cd-c'd' is the intersection of these planes, that is,
of the circles of contact.
Revolve the plane edd' about its horizontal trace, eg, and into
the horizontal plane of projection. The circle eg will then ap-
pear at ea"b", the point e, being in the axis, remains fixed, and
the point dd' will revolve in a plane d'dd", perpendicular to the
axis ge, and to the position d", at a distance, d'd, from the axis
equal to the original distance dd" above the axis.
Produce cd", and a"b" will be the revolved position of the
common chord of intersection of the circles of contact. The
extremities a'' and '' of this chord, are the points of intersec-
tion of the circles of contact, and are, therefore, the revolved
positions of the points of contact of the tangent planes.
a'
In the counter-revolution, these points will return in arcs,
projected in a'a and b'b to the points a and b. Their vertical
projections are a and b'. By considering the direction of the
revolution, it will be evident that aa' is above the horizontal
plane at a distance equal to aa", and that b' is below it at a
distance equal to bb". QB' and QA (Q not shown) are the traces
of the plane which is tangent to the sphere at aa', RB' and
RA are the traces of the plane which is tangent at bb'.
Each of these planes is perpendicular to the radius, and its
traces perpendicular to the projections of the radius of its point
of contact. Also, these traces pass through A and B', those of
the given line.
The method of Problem CVI. is direct. That of Problems
CVII. and CVIII., where an intermediate auxiliary tangent
surface is used, is called indirect.
DESCRIPTIVE GEOMETRY.
235
EXAMPLES.-12. Vary the problem, by changing the position of the given
line.
Ex. 2. Vary the last two problems, by taking a different given surface.
Ex. 3. Apply the method of Prob. CV. to any given double curved surface,
a paraboloid of revolution, for example.
307. A plane can be made tangent to a double curved surface
of revolution, and parallel to a given plane, by Prob. XCVII.,
First method.
GENERAL EXAMPLES.-1°. Construct a plane, tangent at once to a sphere
and a cone of revolution.
Ex. 2. Construct a plane, through a given point, and tangent to two spheres.
Ex. 3. Tangent to three spheres.
Ex. 4°. A sphere, tangent to four indefinite planes, so placed as to inclose a
triangular pyramid.
c—Projections of Intersections.
308. The intersections of double curved surfaces of revolu-
tion with each other, or with any other form of surface of revo-
lution, are all found on one general principle, as follows:
Intersect the given double curved surface, D, and the other
given surface, S, by auxiliary surfaces, A, of such form and
position as will cut the given surfaces in their simplest sections.
Then these sections, s and s,, both being in the same surface,
A, will intersect, but one of them being a section of D, and the
other of S, their intersection must be a point common to the
given surfaces, D and S. That is, it will be a point of their
intersection.
What the auxiliary surface, A, shall be in each case, will
depend on the form and position of the two given surfaces,
D and S.
PROBLEM CIX.
To find the intersection of a double curved surface of revolu-
tion with a plane; and the tangent line at any given
point of the curve.
In Space. Since a meridian plane, M, perpendicular to the
236
DESCRIPTIVE GEOMETRY.
given plane, P, is a plane of symmetry relative to both of the
given surfaces, the intersection of these two planes will evi-
dently be a line of symmetry of the required curve, and those
points in which it intersects the meridian curve lying in M, are
its extremities in the required curve.
Other points may be found at the intersections of either paral-
lels or meridians of the given double curved surface, with the
lines cut from P, by the planes of those parallels or meridians.
In this problem therefore, D (308) is a double curved sur-
face; S, a plane; and the auxiliary surfaces, A, are planes.
In Projection.-1°. The highest and lowest points.-Pl.
XXIII., Fig. 174. Let the vertical, C-C'D', be the axis of
the ellipsoid. Let ef-ef' be the diameter of its greatest paral-
lel, then will the circle on ef as a diameter, be the horizontal
projection of the ellipsoid. The ellipse, described on C'D' and
ef' as axes, will be its vertical projection.
PQP' is the cutting plane. IIC is the horizontal trace of
the meridian plane perpendicular to it.
We might now construct the intersection of the perpendicu-
lar and cutting planes, by Prob. XVII., since it is the indefi-
nite axis of symmetry of the required curve; but as this axis
intersects C-C'D, it will do so at the point in which the cut-
ting plane cuts C-C'D'. Hence, C being one projection of
the required point, find by Prob. XX. its other projection,
o'. Then HC and II'o' are the projections of the intersection
of the two planes IIC and PQP', and determine the indefinite
axis of the required curve.
To find the extremities of this axis, that is, the required high-
est and lowest points, revolve the plane IIC about C—C'D' as
an axis, till it is parallel to the vertical plane of projection; the
point of being in the axis, remains fixed, and IIII' revolves to
H"II", giving H"C and II""o' for the revolved position of
the axis. The meridian curve cut from the ellipsoid by the plane
HC will, after revolution, be vertically projected in the ellipse
C'D'e'f' and A″ and B", the points in which II""'o' meets this
curve, are the revolved positions of the extremities of the re-
quired axis, as seen in vertical projection. A and B'"' are the
horizontal projections of these points. In counter revolution,
A"A" proceeds in the horizontal are A""A-A'A' to its true
position AA'; and B""B" in the are B""B-B"B' to BB', and
AB-A'B' is the transverse axis of the curve.
DESCRIPTIVE GEOMETRY.
237
2°. The points on the apparent contours of the ellipsoid.—In
vertical projection. CG is the horizontal trace of the meridian
plane, parallel to the vertical plane. It cuts from the ellipsoid,
the meridian of apparent contour ef-C'f'D'e', and from the
given plane, PQP', the line KC-K'o', which meets the meri-
dian at m'm and ħ'h ( 65 ).
In horizontal projection. The horizontal auxiliary plane
having ef' for its vertical trace cuts from the ellipsoid its
greatest parallel, or apparent contour e'f'—celf, and from the
given plane, PQP', the horizontal F'e-Fe; which intersects
that parallel at bb' and ec'; making with m'm, h'h; AA' and
BB', six points of the required curve.
3°. Other points. Visibility.-Other points may be found
like the last. Thus the horizontal auxiliary plane U'P', gives
the points aa' and tt', on the parallel U'V'-Uat.
Also, as AB-A'B' is an axis of symmetry of the curve, per-
pendiculars, not shown, from mm' and ' to AB-A'B', will
be parallel to at—a't', and will meet the curve in two additional
points equidistant from AB-A'B' with mm' and hh'.
All that part of the curve which is in front of the plane CG,
will be visible in vertical projection. That which is above the
horizontal plane ef', will be visible in horizontal projection.
4. The tangent.-Let this be drawn at the point tt'. It is
the intersection of the given plane PQP', with the tangent
plane to the ellipsoid at tt'. The latter plane may be found by
Prob. CIV., or by (298). By the latter method, G'U', tangent
to the vertical projection of the ellipsoid, at U', is the genera-
trix of the auxiliary cone of revolution, tangent to the ellipsoid
along the parallel Uat-U'V'; and CN, CG, is the horizon-
tal projection of the element containing the point of contact tt'.
Then NT, tangent to the base, CMN, of the cone, is the hori-
zontal trace of the tangent plane. Hence TT, its intersection
with the horizontal trace, PQT, of the given plane, is one point
of the intersection of the two planes. The given point tt' is
another, hence Tt-T't' is the required tangent line; which
also passes through SS' its vertical trace, which is on P'Q' (62).
EXAMPLES.-1. Let the axis of the ellipsoid be a perpendicular.
Ex. 2°. Find both projections of the shorter axis of the curve.
Ex. 3. Let the given plane make acute angles with both Н and V.
Ex. 4°. Substitute an annular torus for the ellipsoid, see Fig. 175; and, 1st.
238
DESCRIPTIVE GEOMETRY.
The line cut from PQP' by the meridian plane AP to intersect the meridian
curve in AP; or, 2d, not to intersect it.
Ex. 5°. In 4°, let PQP' cut, 1st, the gorge; 2d, the outer half of the torus.
Ex. 6. Substitute an oblate, for a prolate ellipsoid.
PROBLEM CX.
To circumscribe a sphere about a given triangular pyramid.
In Space. The plane of each face of the pyramid cuts from.
the sphere a circle which passes through the vertices of the three
angles of that face. The centre of the sphere is in a line per-
pendicular to the plane of such a circle, and erected at the
centre of that circle. If then we construct two such circles, the
perpendiculars to them at their centres will intersect at the
centre of the sphere.
In Projection.-Pl. XXIV., Fig. 177. By taking one face.
of the pyramid to coincide with one of the planes of projection,
the solution will be simplified. Let ABC-D-A'B'C'-D' be
the two projections of the pyramid, the face ABC—A'B'C'
being in the horizontal plane. The plane of this face cuts from
the sphere a circle passing through A, B and C, and whose
centre, found by drawing perpendiculars, as O and c0 at the
middle points of any two of the edges of the face, ABC, is O.
O is also the horizontal projection of the line, perpendicular to
this circle at its centre, and containing the centre of the sphere.
By revolving either edge, as AD—A'D', of an adjacent face,
ACD—A'C'D', into the horizontal plane, about AC as an axis,
we shall have the true relative positions of the three angles of
that face, and hence can find the revolved position of the centre
of the circle containing their vertices. DD' will, after revolu-
tion, be found in the trace DD" of its plane of revolution, per-
pendicular to AC, and at a distance from the point AA' equal
to the true length of AD-A'D'. To find the latter distance,
revolve AD—A'D' about a vertical axis at DD' till it takes the
position DA"-D'A'' parallel to the vertical plane of projec-
tion. D'A"" will then be the true length of DA—D'A', and
the radius, from A as a centre, of an are which will intersect
DD" at D", the revolved position of DD'. Draw AD", bisect
it at d" and draw d'e" perpendicular to AD"; it will meet ce",
the perpendicular at the middle point of AC, at e' the revolved
DESCRIPTIVE GEOMETRY.
239
position of the centre of the circle cut from the sphere by the
plane of the face ACD-A'C'D'.
To find the primitive position of e", consider that it must be
at the intersection of dk and cO, which are evidently the posi-
tions of d'k and e'c after counter-revolution about AC as an
axis. Then e is the horizontal projection of e' in its true po-
sition; e' is found on d'', the vertical projection of d'k. Oe
is the horizontal projection of the line at ee perpendicular to
the face DAC-D'A'C', since it is to intersect the vertical line
at O. Its vertical projection e'O' is perpendicular to the trace
of the plane of the face ACD-A'C'D' on the plane V, or on
any parallel vertical plane. Such a trace is found by connecting
the points in which any two lines of the face ACD-A'C'D'
meet the proposed vertical plane. Thus the edges AC—A'C'
and AD-A'D', pierce the vertical plane whose trace is G'L'
at the points PP' and KK', hence P'K' is the vertical projec-
tion of the trace of the face ACD-A'C'D' on the plane G'L',
and is parallel to its trace on the primitive vertical plane.
Hence e'O', perpendicular to P'K', is the vertical projection of
a second radial line of the sphere, and OO' is the centre of the
sphere. The radius of the sphere is evidently the line from
OO' to any vertex, as AA'. The true length of this line is
shown parallel to the vertical plane at Oa'-O'a"". Then with
O'a'"' as a radius and O and O' as centres, describe circles, and
they will be the projections of the sphere in which the given
pyramid may be inscribed.
Let us suppose that the front hemisphere, as seen in vertical
projection, and the upper hemisphere, as seen in horizontal pro-
jection, have been removed. The visible edges of the pyramid
will then be shown as full lines. The vertical projection of the
part of the sphere below the horizontal plane will be shown in
a dotted line.
THEOREM XLV.
Every plane section of an ellipsoid is an ellipse.
Let S, Pl. XIX., Fig. 154, be a sphere, whose centre o is in
the ground line GL. Let C and C, be the great circles cut
from the sphere S by the planes of projection H and V, respec-
240
DESCRIPTIVE GEOMETRY.
tively; and let ou be the vertical diameter of S, through o, and
therefore, in the plane V.
Every meridian plane, P, containing the line ou, will cut the
sphere S in a vertical great circle, or meridian, M.
So much being done, imagine lines parallel to each other, and
oblique to the plane V, in planes perpendicular to GL; from
every point of the circle C. They will form a projecting cylin-
der of C, oblique to the plane V; and its intersection with that
plane will by (Theo. VI.) be an ellipse E, whose centre will be e,
and one of whose axes will be on the line ov; and which will be
the oblique cylindrical projection (126) of C upon the plane V.
Next, by (129) we have from the ordinates, np and nq, paral-
lel to ov, the proportion
np
np: ng:: oa: ov, or
C.
ng
The revolutions of C, and E about on would generate, respec-
tively, a sphere, all of whose meridians would be equal to C₁,
and an ellipsoid, all of whose meridians would be equal to E;
and lines, parallel to ov, and intersecting the sphere, the ellipsoid,
and the plane H, respectively, at P₁, 7, and n₁; P2, I½ and n₂9
etc., would give
np
nq
n,p₁ =
ngi
2
N₂ P₂ etc.,
N₂ J 2
=e, a constant ratio.
2
Hence, take any plane, PQ, cutting from the sphere a small
circle as QR, and transform this plane cylindrically (126) into
the plane PQ, (or, in other words, project the circle QR upon
the plane PQ, by projecting lines parallel to ov, and we shall
have for every point Z on the plane PQ, and corresponding
point f, of the plane PQ,,
118
hf
NQ
NQ
= C.
Hence, as h and Q are on the sphere, and Q, on the ellipsoid,
fis on the ellipsoid. But R,Q, is a section, oblique to the
plane PQ, of the projecting cylinder of the circle QR; hence
R,Q, is an ellipse.
309. In Pl. XIX. Fig. 155, let PQ and RS be any two of a
series of parallel planes, cutting the sphere in circles perpen-
dicular to the diameter de. Then PQ, and RS, are the trans-
formations of the same planes relative to the ellipsoid, by cyl-
indrical projecting lines dd,, etc., parallel to ov.
DESCRIPTIVE GEOMETRY.
241
Hence by the last theorem, parallel plane sections of an ellip-
soid are similar ellipses, whose centres are on a diameter, d, è̟
of the ellipsoid.
1)
THEOREM XLVI.
If a cone be circumscribed, tangentially, about an ellipsoid, its
curve of contact will be an ellipse, whose centre will be on
the line joining the vertex of the cone with the centre of
the ellipsoid.
In like manner (Theo. XLV.), a tangent plane, T, to the
sphere, S, at a point t will be transformed into a tangent plane
T, to the ellipsoid at a point t,, the cylindrical projection of t
upon T,.
Also, a cone, C, tangent to the sphere, S, on a circle of con-
tact, c, will be transformed into a cone, C,, tangent to the ellip-
soid along an ellipse, e,, section of the projecting cylinder of c.
And as the vertex v of cone C, the centre a of its circle of con-
tact, c, and the centre, o, of the sphere are in one straight line,
the same will be true of vertex v, of C₁, the centre, a, of its
ellipse of contact with the ellipsoid, and centre, o, of the el-
lipsoid. (See 303).
THEOREM XLVII.
1
Every plane which is parallel to the axis of a paraboloid of re-
volution, cuts the surface in a parabola, identical in form
with the meridian parabola of the surface.
Imagine an ellipsoid of revolution E, Pl. XIX., Fig. 156,
whose axis of revolution is ab, a vertical line in the vertical
plane of projection, and whose centre is o. Let PQ be any hori-
zontal plane, containing a parallel, pq, which intersects the meri-
dian M, of E, in the points p and q.
Now suppose a to remain fixed on the fixed axis ab, but that
b moves to successive positions b, ....... by on ab produced.
As five points determine a conic section (Theo. XXVI.), the
meridian ellipse, M, may be transformed into a new ellipse E,,
determined by p, q, b,, and the two points to which the known
#
16
242
DESCRIPTIVE GEOMETRY.
tangent at a is equivalent. The conjugate axis, c, d, of M,
will evidently be greater than cd, and will be found by (129).
The ellipsoids, generated by the meridians M and M,, will
intersect in the parallel pq and will be tangent to each other
at a.
Every plane parallel to the meridian plane M,, will cut the
ellipsoid E, in an ellipse, e,, which will be in vertical projection
concentric with, and similar to the ellipse M,.
When, at bn, the axis ab, becomes infinite, M,, still remaining
a conic section determined by bn, P, q and the tangent at a,
becomes a parabola M,; and the ellipsoid, E, becomes a para-
boloid P. Hence the ellipses, e, will become parabolas, P,
concentric with, and similar to the meridian parabola, M.
THEOREM XLVIII.
Two curves are similar when all the tangents to one are parallel
to the homologous tangents to the other.
Every curve may be considered as that form of a certain
species of polygon, in which the number of sides has been
made infinite; thus a circle may be considered as a regular
polygon of an infinite number of sides.
Polygons are similar, when all the sides of one are parallel
to the corresponding sides of the other; any side having a cer-
tain fixed direction for a finite distance. But when the number
of sides is infinite, each reduces to a point whose direction is
represented by that of the motion of the generatrix of the curve
when at that point, that is, by the tangent at that point.
Hence if all the tangents of a curve are parallel to all the
homologous tangents to another curve, the two curves are
similar.
THEOREM XLIX.
Any two similar and concentric parabolas, are also identical,
and may be superposed upon each other.
First. Any two parabolas, like any two circles, are similar.
For, imagine two parabolas, P and P,, to be placed so that their
DESCRIPTIVE GEOMETRY.
243
foci and axes shall coincide, and draw from the common focus,
F, any radius vector cutting the two curves as at m and m,.
Then as the diameters at m and m, are parallel, the tangents,
which bisect the equal angles made by these diameters with
the common radius vector Fm,, must be parallel. And as the
same holds true in like manner for all points of the two curves,
they are similar by the last theorem.
Second. All parabolas being similar, then as two circles in
which the centres and one point of each circumference coincide,
are identical; so two parabolas, having a common centre, axis,
and vertex, will coincide.
THEOREM L.
Every plane section of a paraboloid of revolution, oblique to
the axis of the surface, is an ellipse.
Let every point, a, of the paraboloid P (Theorem XLVII.),
move parallel to the axis of revolution ab₂, and so that aa', Pl.
XIX., Fig. 157, remaining on this axis takes a new position.
aa. The two paraboloids P and P,, original and new positions
of P, will then be concentric and similar. Their parallels, as
C and C,, taken in the same plane will be two concentric cir
cles; and the sections of each made by a plane parallel to the
axis, ab, will be equal and concentric parabolas.
It shall now be shown that if a tangent plane T be drawn to
the paraboloid P, at tt', the point tt' will be the centre of the
evidently closed curve m'n', cut by T from the paraboloid P,.
Imagine any tangent line, 7, at the point tt', and hence in the
plane T; it will cut the curve m'n', and hence the surface P₁,
in two points, which call, h and k. Next pass a plane, Q,
parallel to the axis a-a'a', and through ; it will cut from
the two paraboloids, P and P,, the similar and concentric para-
bolas p and p,, of which p, will pass through h and h, while p
will be tangent to l at tt'.
But (Theo. XLI.) parabolas, like other conic sections, when
concentric and similar, possess the property that the portion of
the tangent to the inner, which is a chord of the outer, is bisected
at its point of contact. That is, th= th; and as I was any
line in T, tangent at tt, the like is true for all positions of l,
244
DESCRIPTIVE GEOMETRY.
and hence tť is the middle point of mn-m'n', or the centre of
the curve m'n'.
Now, in place of a rigorous demonstration of the rest of the
theorem, it will be enough to observe that the mode of deriva-
tion of the paraboloid P from the ellipsoid E, Fig. 156, makes
it especially obvious, that the paraboloid is but a particular case
of the ellipsoid, and that, as it is only the sections parallel to
the axis that become infinite, or open curves, the other sections,
all remaining closed curves as before, will not lose their proper-
ties, and are therefore ellipses.
Besides, let MN be a plane parallel to m'n', cutting the para-
boloids P, and P in the curves MN and RS. Then, as tt' might
have been any point of RS, the line I would then have been a
chord of curve MN, bisected at its point of contact with RS;
so that by (Theo. XLI.) if we knew that RS were a conic sec-
tion, MN would be one also.
Finally, as a point is a particular case of the ellipse, tt' and
m'n' may be regarded, as in case of concentric and similar ellip-
soids, as the extreme members of a series of similar and concen-
tric ellipses contained in a series of planes parallel to m'n'.
We therefore conclude that every oblique section of a para-
boloid is an ellipse, and that parallel planes, as m'n' and MN.
cut a paraboloid in similar and concentric ellipses whose centres
will be on a diameter, as tO, of the paraboloid.
THEOREM LI.
If two surfaces, S and S,, be cut by a plane in two curves, E
and E,, such that E, is a conic section, and that the point
of contact of every tangent to E bisects that portion of this
tangent which is a chord of E,, then the curve, E, is a conic
section, similar and concentric to E,.
Making Pl. XIV., Fig. 124, serve for this theorem, let now
E, be the given conic section, and let E represent the other
given curve. Let a,b, be one of numerous tangents to E, in
which the part, as a,b,, is bisected at the point of contact c,.
Then make the centre of E,, the centre of a conic section, s₁,
concentric and similar to E, and tangent to a,b, at e. Likewise
make the centre of other similar sections, 82, 8, etc., each
DESCRIPTIVE GEOMETRY.
245
tangent at c₂, C, etc., the middle points of chords, a,b,, a,b,, of
E,, which are tangent to E at their middle points.
Now E, being tangent to all of these chords, must enclose
Hence
tangentially, that is, be the envelope of s,, s., S., etc.
these must coincide with each other, and with E; else they
would be concentric but unequal, in which case E would not
be their envelope as it must be. That is, E is, itself, the conic
section, concentric and similar to E,.
THEOREM LII.
The plane sections of the hyperboloid of revolution of two sepa-
rate nappes, and of its conic asymptote, made by the same
plane, are concentric, and similar conic sections.
See Pl. XIX., Fig. 150. If the meridian hyperbola II, of
the warped hyperboloid revolve about A'B' as an axis, it will
generate the double curved hyperboloid, the same surfaces being
those, respectively, of united, and of separate nappes; and the
asymptote O' will, in revolving about the same axis, 'B',
generate a cone, asymptote to the double curved hyperboloid.
If now, this cone and the double curved hyperboloid be cut
by any plane, P, it is to be shown that the sections, XY, of
the cone, and ay of the hyperboloid, are concentric and similar
conic sections.
Imagine any point, t, on the hyperboloid and pass both a
meridian plane, M, and a parallel plane, P,, through it. The
former will contain a meridian II, and the latter a circular paral-
lel, C₁, of the hyperboloid. The tangents, h, and e,, to these
curves respectively, at the same point, t, will determine the
tangent plane, T, at that point.
But the plane M, cuts the conic asymptote in two elements,
asymptotes of the meridian H,, which the tangent h, will meet
at two points, e and e,, such that et = et (187, 2°).
Also the plane P, cuts from the hyperboloid and its conic
asymptote, two concentric circles, C, and C, so that that segment
of the tangent to C, at t, which is a chord intersecting C, at f
and f, is bisected at t, giving tf₁ = tf.
1
Hence, letting t in Fig. 150 represent the point we have
imagined, the tangent plane T, at t cuts from the conic asymp-
246
DESCRIPTIVE GEOMETRY.
2
tote an ellipse E whose centre is t, since ee, and fif, are its axes
since they bisect each other perpendicularly at t.
Hence any line, L, in T, and through t, will be a diameter of
the ellipse E.
Now as the same can be proved for all positions of t let t, t₁,
t, etc., be points of any plane section, S, of the hyperboloid.
The portion of the tangent at each of these points, which is a
chord of the section, S, of the conic asymptote, made by the
same plane which contains S, is bisected at its point of contact
with S,.
But S is a conic section, hence by the last theorem S, is a
similar and concentric conic section.
PROBLEM CXI.
To find the intersection of any two surfaces of revolution
whose axes are in the same plane.
In Space. The solution of this general problem is essentially
the same whether the given axes intersect, either perpendicu
larly, or obliquely, or are parallel; and whether one, both,
or neither of the surfaces be double curved, developable, or
warped.
Thus intersect both surfaces by auxiliary surfaces, such and
so placed, as to cut the simplest sections from the given sur-
faces. Then the two sections, one on each given surface, and
both in the same auxiliary surface, will intersect in points com-
mon to both surfaces.
In Projection.-I. Axes parallel. 1°. A cylinder; and any
double curved surface, cone, or the warped hyperboloid. Let
both axes be vertical.
Then first, horizontal auxiliary planes will cut parallels from
both surfaces whose points of intersection will be required
points.
Or second. Meridian planes, M, of the non-developable sur-
face, being parallel to the axis of the cylinder, will contain cle-
ments of the latter, and meridians of the former. Tho inter-
sections of the elements and meridian contained in the plane,
M, will also be required points. The plane of the axes will be
a common meridian plane.
DESCRIPTIVE GEOMETRY.
247
2º. A cone and a non-developable surface, double curved or
warped, as before.
Employ the common meridian plane, and planes perpendicu-
lar to the plane of the axes, as auxiliary planes.
3°. Two non-developable surfaces, either or both, warped, or
double curved.
Proceed as in Case 2.
II. Axes intersecting.—1°. Particular cases; axes intersect-
ing perpendicularly.
-A cylinder whose aris is horizontal, with a non-develop-
able surface whose axis is vertical. Horizontal auxiliary planes
will cut elements from the cylinder, and parallels from the other
surface, whose use is obvious. Also the common meridian
plane should be used as an auxiliary plane.
b-A cone whose vertex is in the axis of the given non-develop-
able surface. In this case, every meridian plane M, of the latter
surface, which cuts the cone, will cut it in elements, which will
intersect the meridian contained in M, in points of the required
curve. Also, if the horizontal plane through the vertex should
be secant to both bodies, it may be used as an auxiliary plane.
-If either of the given surfaces be a sphere, the auxiliary
surfaces may evidently be planes.
2° The general case. Axes intersecting obliquely.
In Space. Here the point of intersection of the axes is made
the centre of a system of auxiliary spheres, each of which will
(263) cut each of the given surfaces in a parallel of that sur-
face. The pair of parallels thus found by means of any one
sphere, intersect each other in two of the required points.
In Projection.—In Pl. XXIV., Fig. 178, let the given sur-
faces of revolution be an ellipsoid and a paraboloid. Let the
vertical plane of projection be parallel to the plane of their
axes; and let the axis of the ellipsoid be vertical.
The circle Cef and the ellipse A'B'C'D' are the projections
of the ellipsoid. The meridian parabola is the vertical projec-
tion of the paraboloid. The horizontal projection of this sur-
face, not being needed in the construction, is not represented.
E is the intersection of the axes of the given surfaces; n''s'N
is the vertical projection of an auxiliary sphere whose centre is
E. This sphere cuts from the ellipsoid the circle whose projec-
218
DESCRIPTIVE GEOMETRY.
tions are n's-and bst, and from the paraboloid the circle
whose vertical projection is Nr. These two circles intersect in
a common chord which is perpendicular to the vertical plane at
t', and whose horizontal projection, tb, gives, by its intersection
with the circle bst, the two points t and b of the horizontal pro-
jection of the curve. The points a' and p', in which the meri-
dian curves of the two surfaces intersect, are points of the re-
quired intersection. Their horizontal projections are found at
a and p in the horizontal trace, CU, of the meridian plane con
taining them, because this meridian plane is vertical.
The points which are on the greatest horizontal section,
C'D', should be found, they being the limits of visibility in
horizontal projection. To find them, take an auxiliary sphere,
C'D'K, with radius EC', so that it shall intersect the ellipsoid
in the assumed section C'D'. The same sphere will intersect
the paraboloid in the circle qK. The two circles qK and C’D'
intersect in the common chord at f", which gives f', for the ver-
tical, and ƒ and e for the horizontal projections of the two
points on the circle C'D' of the ellipsoid.
The projections can be drawn through any convenient num-
ber of points thus found.
310. Parasite of the vertical projection. The foregoing con-
struction, when not limited to points within the parallels
which determine them, will serve to find some of the points of
the vertical projection produced; points, however, which have
no real existence on the intersection of the bodies, yet which
are useful in sketching its vertical projection.
Thus the extreme sphere of radius EB', which touches the
ellipsoid, contains the parallel GH' of the paraboloid and touches
the ellipsoid at B'. Then G'H', produced, meets the perpen-
dicular B'g' to A' B', at g' a parasite point of a't'p'.
311. It has been shown by analysis that when two "surfaces
of the second order," that is, those like the foregoing, whose
plane sections are conic sections, have a common meridian plane,
the projection of their intersection upon that plane will be a
conic section. Accordingly, a't'p', Pl. XXIV., Fig. 150, is a
conic section, whose species may often be determined, if desired,
by the construction for finding its centre and axes.
Analyse appliquée à la Géométrie des trois dimensions, C. F. A. Leroy.
DESCRIPTIVE GEOMETRY.
249
312. If the axes of two double curved surraces of revolution
are not in the same plane, the auxiliary planes should be made
to cut parallels from one of the surfaces, in which case they
will cut a series of similar sections from the other, if it be a sur-
face whose plane sections are conic sections.
The intersections of cylinders and cones, when their axes in-
tersect, could be found by auxiliary spheres.
GENERAL EXAMPLE.-Construct the figure of each of the cases not con-
structed in Prob. CXI.
PROBLEM CXII.
To construct the tangent line at a given point of any of the
intersections described in Prob. CXI.
In Space.-The required tangent is the intersection of the
two tangent planes, one to each surface, at the given point;
and each of these planes may be constructed by Prob. CIV., or
by an auxiliary cone tangent on the parallel containing the
point of contact, as in Prob. CIX. (the tangent line).
But the obliquity of the axis of one of the surfaces to one of
the planes of projection, in many cases, hinders the application
of this method, and makes the following, called the method by
normals, more convenient.
There can be but one normal to a given surface at a given
point, and the tangent plane, and hence every tangent line,
through its foot, is perpendicular to it. Hence, if at the
given point of intersection, the normal to each surface at that
point be drawn, the plane of these normals, which is the
normal plane at that point, will be perpendicular to both of
the tangent planes at the same point, and hence to their inter-
section, which is the required tangent line.
In Projection.-In Pl. XXIV., Fig. 178, let tt' be the given
point of contact.
Suppose now the meridian curve of the ellipsoid, containing
the point tt', to be revolved about A-B'A' to, the position
A'C'B'; then n'o', the normal to the ellipse A'C'B', at n', the
revolved position of tt' will be the revolved position of the nor-
mal to the ellipsoid at tt'. In the counter-revolution, o' being in
250
DESCRIPTIVE GEOMETRY.
the axis, remains fixed, n' returns to t't, and t'o and tA are the
projections of the normal to the ellipsoid at tt'. Likewise, let the
meridian curve of the paraboloid, containing tt', be revolved
about the axis Ek' till it takes the position FNK. NE' is then the
revolved position, and t'k'-th the projections, of the normal to
the paraboloid. The line o'k' is, therefore, the trace of the nor-
mal plane on the meridian plane, CU, and as this plane is paral-
lel to the vertical plane of projection, o'k' is parallel to the ver-
tical trace of the normal plane; hence, T't', perpendicular to
o'k', is the vertical projection of the required tangent line.
Again, tt' being a point of the normal plane, is one point in
its trace upon the horizontal plane, n'U'. U' is the intersec-
tion of the trace o'U', of the normal plane with the line of inter-
section, n'U'-CU, of the meridian plane, CU, and the hori-
zontal plane, n''U'. Hence, Ut is the horizontal projection of
the trace of the normal plane upon the horizontal plane, n'U'.
Ut is therefore parallel to the trace of the normal plane upon
the plane H, and hence Tt, perpendicular to U', is the horizon-
tal projection of the required tangent line at tt'.
B-DEVELOPMENT.
313. Having shown in many previous problems how to con-
struct elliptical plane intersections by points, and to show their
true size, the next problem is added, to illustrate the convenience
in some cases, of returning from constructions in space to plane
solutions; and as a means of securing greater accuracy in prac-
tical applications.
PROBLEM CXIII.
Having given a pair of conjugate diameters of an ellipse, to
find its axes.
Let OM and ON, Pl. VII., Fig. 76, be given semi-conjugate
diameters. From the extremity, M, of the longer, let fall the
perpendicular, MK, upon the shorter. Set off, on MK, the
semi-diameter ON, at MQ or MP; draw OQ, and OP, and
through the middle points, E and E,, respectively, of these lines,
DESCRIPTIVE GEOMETRY.
251
draw MEP and ME,p,, meeting the arcs, with radii EO and
EO, respectively, at p and p₁. Also describe the arcs Qq, and
Pq,, with E and E,, respectively, as centres, limited by EM, and
E, M. Then OA, on Oq (or Og,), and equal to Mp (or Mp,), and
OB, on Op (or OB₁, on Op,), are the semi-axes required.
Demonstration.-Describe an arc with OA as a radius, and
through M, draw da perpendicular to OA, and meeting the arc
at a point d.
-
Now since E₁₁ = E,P
E,P = OE,, the angle E,7,O= E,O,; and
the right-angled triangles, Maq, and daO, are similar; whence
the angle 4,Ma Oda. Then since the opposite angles, Ma
and E,Md, are equal, Oda = EM; hence the triangle E,Md is
isosceles, and E,ME.
= =
Also, the similar triangles, Maq, and Oda, give
But Mq,
whence
or
Ma: Mq, :: da : Od.
OB, OB, and Od = OA.
=
Ma: OB da: OA,
Ma: da :: OB: OA; which (129) being
a property of the ellipse, proves the construction.
PROBLEM CXIV.
To develop the intersection of any plane with any double curved
surface; also, the tangent to the given intersection.
In Space.-Revolve the given curve till its plane shall be
parallel to, or shall coincide with, one of the planes of projection,
according as the axis is in, or parallel to, the same plane. Each
point of the revolved curve will be found at a perpendicular dis-
tance from the axis equal to the hypothenuse of a right-angled
triangle whose base and altitude are the perpendicular distances
of the projections of the points of the curve, from the respective
projections of the axis.
In Projection.-Pl. XXIII., Fig. 174. Let the intersection
of the plane, PQP', with the ellipsoid be developed. Take the
vertical trace QP' as the axis. Any point, as aa', will be found.
on the perpendicular, p'a'a", to the axis, and at a distance, a'p":
from the axis, equal to the hypothenuse of a right-angled triangle
whose base is a'p" and whose altitude is aa"". Other points of
the development are similarly found.
252
DESCRIPTIVE GEOMETRY.
The point t'', is the revolved position of one point, viz., of the
point of tangency, of the tangent tT-'T'. The point TT', in
which this tangent pierces the horizontal trace PT of the cutting
plane, being behind the vertical plane, revolves in an arc whose
vertical projection is T'T", and is found, after revolution, at a
distance, T", from the axis, equal to the hypothenuse of a right-
angled triangle whose sides are kT' and TT.
PROBLEM CXV.
To develop any double curved intersection, contained in a
double curved surface.
In Space. Through all the points of the curve, draw lines
perpendicular to one of the planes of projection. These lines
will form a right cylinder, whose intersection with the assumed
plane of projection is a right section. This cylinder can, there-
fore, be immediately developed; the right section will develop
into a straight line, and the points of the curve will be found on
the developments of the elements containing them, and at dis-
tances from the development of the base equal to their perpen-
dicular distances from the plane of the base, distances which will
be found in projection in their true size.
By (Theo. XXXIV.) the tangent to the curve will be also
tangent in the development.
EXAMPLE.-Apply this solution to any case of Prob. CXI.
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Fin benthos
CS. Seuism
Sept. 89
Mathemati
QA
501
W 294
1883

UNIVERSITY OF MICHIGAN
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