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right by Cornell University
Library 1991.'ELEMENTS
OP
DESCRIPTIVE GEOMETRY,
WITH
THEIR APPLICATION
TO
SPHERICAL TRIGONOMETRY, SPHERICAL PROJECTIONS,
AND WARPED SURFACES.
BY CHARLES DAVIES, LL.D.
AUTHOR Or ARITHMETIC, ELEMENTARY ALGEBRA, ELEMENTARY GEOMETRY, PRACTICAL
GEOMETRY, ELEMENTS OP SURVEYING, ELEMENTS OF DESCRIPTIVE AND ANA-
LYTICAL GEOMETRY, ELEMENTS OF DIFFERENTIAL AND INTEGRAL
CALCULUS, AND A TREATISE ON SHADES, SHA-
DOWS, AND PERSPECTIVE.
NEW YORK:
A. S. BARNES & BURR, 51 & 53 JOHN STREET.
SOLD BY BOOKSELLERS, GENERALLY, THROUGHOUT THE UNITED STATES.
1859.
Í4 .*	.DAVIES’
COURSE OF MATHEMATICS
DAVIES’ FIRST LESSONS IN ARITHMETIC,
DESIGNED FOR BEGINNERS.
DAVIES’ ARITHMETIC,
DESIGNED FOR THE USE OF ACADEMIES AND SCHOOLS.
KEY TO DAVIES’ ARITHMETIC.
DAVIES’ ELEMENTARY ALGEBRA :
Being an introduction to the Science.
KEY TO DAVIES’* ELEMENTARY ALGEBRA.
DAVIES’ ELEMENTARY GEOMETRY.
This work embraces the elementary principles of Geometry. The reasoning is plain
and concise, but at the same time strictly rigorous.
DAVIES’ PRACTICAL GEOMETRY,
Embracing the facts of Geometry, with applications in Artificer’s Work, Men-
suration, and Mechanical Philosophy.
DAVIES’ BOURDON’S ALGERRA,
Being an abridgment of the work of M. Bourdon, with the addition of practica.
examples.
DAVIES’ LEGENDRE’S GEOMETRY and TRIGONOMETRY,
Being an abridgment of the work of M. Legendre, with the addition of a Treatise
on Mensuration of Planes and Solids, and a Table of Logarithms and
Logarithmic Sines.
DAVIES’ SURVEYING,
With a description and plates of, the Theodolite, Compass, Plane-Table, and
Level—suso, Maps of the Topographical Signs adopted by the Engineer
Department—an explanation of the method of surveying
the Public Lands, and an Elementary Treatise ou
Navigation.
DAVIES’ ANALYTICAL GEOMETRY,
Embracing the Equations of the Point and Straight Line—of the Conic Sec*
tions—of the Line aad Plane in Space—also, the discussion of the
General Equation of the second degree, and of Surfaces
of the Second Order.
DAVIES’ DESCRIPTIVE GEOMETRY,
With its application to Spherical Projections.
DAVIES’ SHADOWS and LINEAR PERSPECTIVE.
DAVIES’ DIFFERENTIAL and INTEGRAL CALCULUS.
Entered, according to Act of Congress, in the year 1834,
By CHARLES DAVIES,
in the Clerk’s Office of the District Court of the United States, fur the Southern District of
New York.
A oxPREFACE.
This Treatise on Descriptive Geometry has been prepared
for the use of the Cadets of the Military Academy. In sub
mitting it to the public, the author prefers no claim to inven-
tion or discovery. It has been his object to furnish a useful
text book ; and if this end be attained, he will have no cause to
regret his labours.
The study of the Mathematics, whether considered as in-
troductory to its sister science, Mechanical Philosophy, or as
a salutary and invigorating exercise of the mind, is equally
worthy of attention. The useful and important results to
which it leads, the mutual dependence of its parts, and the
concise and satisfactory reasoning in the development of its
principles, recommend this study, as well to the practical man,
who learns only what he can suçcessfully apply, as to the love*
of science, who explores all its departments in search of new
facts and interesting truths.
The subject of Descriptive Geometry, which is treated of
in these Elements, has not, as yet, been considered in this
country as a necessary part either of a polite or practical
education. It has been taught in the Military Academy since
1817, but has not found its way into other Seminaries with a
rapidity at all proportionate to its usefulness. The progress of
science, like that of truth, is always slow ; yet it compensates forV
PREFACE.
its want of velocity in the steadiness of its advancement and
the certainty of its success. In France, Descriptive Geometry
is an important element of a scientific education ; it is taught
in most of the public schools, and is considered indispensable
to the Architect and Engineer. Its intimate connexion with
Civil Engineering and Architecture, and the facilities which it
affords in all graphic operations, render its acquisition desirable
to those who devote themselves to these pursuits.
The author is by no means indifferent to the reception which
this work may meet with from the public ; yet, he will not com-
plain of a rigid criticism, if it shall appear that he has been
instrumental in diffusing a knowledge of an interesting and
useful branch of science.
Military Academy,
West Point* December, 1098.TABLE OF CONTENTS.
CHAPTER I.
First Principles
9
CHAPTER n.
Of the conventional manner of making the Projections of Lines,
and the Traces of Planes in the different Angles ; how the
given and required parts are distinguished from those which
are used merely to aid in the construction. Solution of some
of the principal Problems on the Right Line and Plane •	•	19
CHAPTER ni.
Of Lines and their Tangents
CHAPTER IV.
Of Curved Surfaces ; how they are represented on the Planes
of Projection,	Of the Projections of Curved Linesÿ and
their Tangents...................................	.	,	40vi
TABLE OF CONTENTS.
CHAPTER V.
Of Tangent Planes to Single-curved Surfaces .	.	.	.	.	45
i
CHAPTER VI.
fif Tangent Planes to Surfaces of Revolution........55
CHAPTER Vn.
Of the Intersections of Curved Surfaces and Planes ; Of Tan-
gent Lines to the Curves of Intersection ; and of the develop-
ment of Surfaces on Planes................... 68
CHAPTER Vm.
Of the Intersections of Curved Surfaces.......... .	84
»
CHAPTER IX.
Practical Problems.................... ,	08
CHAPTER X.
Construction of Spherical Triangles...............104
CHAPTER XL
Spherical Projections. Fundamental Principles.....115TABLE OF CONTENTS.
VK
CHAPTER XII.
Pago
Of the Orthographic Projection....... .	.119
CHAPTER Xm.
Of the Stenographic Projection .	...	.	... 128
COMPLEMENT.
Warped Surfaces............... .... 144ELEMENTS
or
DESCRIPTIVE GEOMETRY®
*
CHAPTER I.
FIRST PRINCIPLES.
§ 1. The object of Descriptive Geometry is twofold: first,
to represent with accuracy all geometrical magnitudes on planes;
and secondly, to construct all graphic problems involving three
dimensions.
§ 2. The representation of a geometrical magnitude on a
plane is called its projection, and the plane on which the repre-
sentation is made is named the plane of projection.
§ 3. In Descriptive Geomètry two planes of projection are
used, and to simplify the constructions, they are taken at right
angles to each other.
§ 4. If one plane be taken horizontal, the other will be ver
tical, and this position of the planes enables us to conceive
most readily how objects are situated in space when their
projections are known.*
§ 5. The planes are called, respectively, the hoi'izontal plane
of projection, and the vertical plane of projection. Their line
of intersection, which is horizontal, is called the ground line9
or common intersection ; and each plane is supposed to extend
indefinitely in the direction of, and from this line of intersection.
* Space is indefinite extension, in which all bodies are situated. The abso-
lute position of bodies cannot be determined, but their relative positions may be,
either by referring them to each other, or to objects whose places are assumed
In Descriptive Geometry all bodies are referred to the planes of projection.10
DESCRIPTIVE GEOMETRY.
§ 6. PL 1. Fig. 1. When, therefore, a line, as AB, is assumed
for the common intersection of the planes of projection, it is the
intention simply to point out the line, and not to limit its ex-
tension.
§ 7. Let AB be the ground line, and the plane of the paper
the horizontal plane of projection. The. vertical plane passes
through AB, and is perpendicular to the plane of the paper.
Suppose the vertical plane of projection to be turned or
revolved around AB as an axis, or hinge, till it coincides with
the plane of the paper. There are two ways in which this
revolution can be made : first, we can so revolve the vertical
plane, that the part which is above the horizontal plane shall
fall in front of the ground line AB ; the part of the vertical
plane which is below the horizontal plane will, in that case, fall
beyond the ground line AB ; or, secondly, it can be so revolved
that the part which is above the horizontal plane shall fall beyond
the ground line, the part which is beneath the horizontal plane
will then fall in front of tfce ground line AB. The latter
method will be used. The part of the paper which is be-
yond the ground line will then represent that part of the ver-
tical plane of projection which is above the horizontal plane,
and also, that part of the horizontal plane which is behind
the vertical plane : and the part of the paper which is in front
of the ground line will represent that part of the vertical plane
which is below the horizontal plane, and that part of the hori-
zontal plane which is in front of the vertical plane.
§ 8. There are four diedral angles formed by these planes.
First, the angle above the horizontal, and in front of the vertical
plane ; second, the angle above the horizontal and behind the
vertical plane ; third, the angle behind the vertical, and beneath
the horizontal plane ; fourth, the angle beneath the horizontal,
and in front of the vertical plane
§ 9. Any line of a plane, about which the plane is made to
turn, or revolve, is called the axis of revolution.
§ 10. In revolving a plane about an axis, like a door, for
example, on its hinges, all the points and lines of the plane
preserve their relative positions.FIRST PRINCIPLES.
11
§11. If from any point of a plane, a line be drawn per-
pendicular to the axis, and the plane be then revolved, the
point will describe the circumference of a circle—the radius
of this circle is equal to the perpendicular let fall on the axis,
and the plane of the circle is perpendicular to the axis, since
the axis is perpendicular to all the radii. If, therefore, through
any point of a revolving plane, a plane be drawn perpendiculai
to the axis, the point will continue during the revolution in the
perpendicular plane. All the points of the axis remain fixed
during the revolution.
§ 12. If from any point in space, a perpendicular be let fall
on the horizontal plane, the foot of the perpendicular is the
horizontal projection of the point. If, in like manner, a perpen-
dicular be drawn to the vertical plane, the foot of the perpen-
dicular is the vertical projection of the point. These perpen-
diculars are called the projecting lines of the point.
§ 13. PI. 1. Fig. 1. Let AB be the ground line, and C the
horizontal projection of a point. * Since the horizontal projec-
tion of a point is tlje foot of a perpendicular passing through the
point (12), the point of which C is the horizontal projection
is any point of the right line drawn perpendicular to the hori-
zontal plane at C. Let C' be the vertical projection of the
same point. Tliat the point may answer these two conditions
at the same time, it must be in a line perpendicular to the hori-
zontal plane at C, and in a line perpendicular to the vertical
plane at C', and these lines intersect, since they pass through
the same point. Conceive a plane to be drawn through the
projecting lines of this point. It will be perpendicular to both
the planes of projection, since it contains lines respectively per-
pendicular to these planes; it will consequently be perpen-
dicular to their intersection, that is, to the ground line. This
plane will then intersect the planes of projection in two lines at
right angles to each other, and perpendicular to the ground line
at the same point. When the vertical plane is revolted about
the ground line, to coincide with the horizontal plane, the ver-
tical projection of the point continues at its distance from the
xis, and in a thiid plane passing through the point perpendicu12	/	DESCRIPTIVE GEOMETRY.
lar to the axis (11) ; and after the revolution, it will be found
in the intersection of this third plane with the horizontal plane ;
that is, in a line through C perpendicular to the ground line AB.
Hence, when the planes of projection are revolved to coincide,
the vertical and horizontal projections of a point, are in a line
perpendicular to the common intersection, or ground line.
§ 14 We may remark, that the distance from the vertical
projection of a point to the ground line, is equal to the dis-
tance of the point in space from the horizontal plane, and that
the distance from the horizontal projection to the ground line, is
equal to the distance of the point in space from the vertical
plane. That i§, C'D is equal to the height of the point above
the horizontal plane, and CD to its distance from the vertical
plane. •
§ 15. All points in the first and second angles are projected
on the vertical plane above the ground line ; and all points in
the third and fourth angles, below it. Points situated in the
first and fourth angles are horizontally projected on that part
of the horizontal plane which is in front of the vertical plane ;
and points situated in the second and third angles are pro-
jected on that part of the horizontal plane which is behind the
vertical plane.
§ 16. Let AB be the ground line, C the horizontal, and C' the
vertical projection of a point ; C'D is its distance above the
horizontal plane, and CD its distance from the vertical plane ;
the point is then in the first angle. If C be the vertical, and
E the horizontal projection of a point, it is situated in the second
angle, CD is its height above the horizontal plane, and DE its
distance behind the vertical plane. If C" be the horizontal,
and C the vertical projection of a point, the point is in the
third angle,—C"D is its distance behind the vertical plane,
and CD its distance beneath the horizontal plane. If C"' be
the horizontal projection, and E the vertical projection of a
point, the point is situated in the fourth angle, in front of the
vertical plane a distance equal to DC'", and beneath the hori-
zontal plane a distance equal to DE.
§ 17 All points situated in one of the planes of projection areFIRST PRINCIPLES.	13
their own projections on that plane, and are projected on the
other plane into the ground line.
§ 18. The two projections of a point determine its position in
space. For, let C and C' be the projections of a point. Erect
ut C a perpendicular to the horizontal plane, it will pass through
the point of which C is the horizontal projection. Draw also
at C' a perpendicular to the vertical plane ; this perpendicular
will intersect the perpendicular to the horizontal plane, before
drawn, and their point of intersection is the position of the
point in space.
§ 19. When it is necessary to refer to a point in space, given
in position by its projections, instead of saying, the point whose
horizontal projection is C and vertical projection C', we say,
simply, the point (C, C').
§ 20. Two lines which intersect, or are parallel, determine
the positipn of a plane passing through them. If, then, the
lines in which a plane intersects the planes of projection are
known, the plane itself is given in position. It is by means
of these lines, which are called traces, "that we are enabled to
show, on the planes of projection, the position which planes
have with each other in space.
§ 21. The line in which a plane intersects the horizontal plane
is called its horizontal trace ; and the line in which it intersects
the vertical plane, is called its vertical trace.
§ 22. If a plane be parallel to either of the planes of pro-
jection, it Will have but one trace, which will be on that plane
to which it is not parallel.
§ 23. If a plane be parallel to the ground line, and not to
either plane of projection, it will have two traces, both of which
will be parallel to the ground line, else they would meet it,
in which case the plane itself would meet the ground line. If
a plane be not parallel to the ground line, it will meet it in a
point ; this point is in the vertical trace of the plane, since it
is in the vertical plane of projection ; it is in the horizontal
trace, since it is in the horizontal plane of projection ; and
hence, when a plane is not parallel to the ground line, its
traces will both intersect it at the same point.DESCRIPTIVE GEOMETRY.
14
§ 24. The horizontal projection of a right line is the hori
zontal trace of a plane passing through the line and perpen-
dicular to the horizontal plane. The vertical projection of a
right line is the vertical trace of a plane passing through the line
and perpendicular to the vertical plane. These planes are called
the projecting planes of the line.
§ 25. The projection of a right line on either plane of pro-
jection, is made up of the projections of all the points of the
line. For, if perpendiculars be drawn from all the points of a
right line, to either plane of projection, they will be contained
in the projecting plane of that line, and will pierce the plane
of projection in the trace of the projecting plane which con-
tains them.
§ 26. The two projections of a line determine its position in
space.	/
Let NM (PI. 1. Fig. 2.) be the ground line, AB the hori-
zontal, and A'B' the vertical projection of a right line. If a
plane be drawn through AB perpendicular to the horizontal
plane, it will be the projecting plane of the line, and will there-
fore contain it. If through A'B' a plane be drawn perpen-
dicular to the vertical plane, this plane, being the other project-
ing plane of the line, also contains it. Hence, the line of which
AB, A'B' are the projections, is the line of intersection of these
two planes, and since the planes are determined in position,
their intersection is also determined. If the horizontal projec-
tion only be given, the line is somewhere in a plane passing
through the horizontal projection and perpendicular to the
horizontal plane, but its position in this projecting plane is not
determined. So, when the vertical projection only is given,
the line may have any position in the plane passing through
the projection and perpendicular to the vertical plane.
§ 27. If a line be parallel to one of the planes of projection,
its projection on the other plane is parallel to the ground line#
for the projecting plane of the line is parallel to that plane of
projection to which the line is parallel.
§ 28. Ij a line be perpendicular to one of the planes of pro-
jection. its projection on that plane is a point ; for, the project-RIGHT LINE AND PLANE.	16
mg lines of all the points coincide with the given line. Art.
24 does not apply to this case.
§ 29. When we have occasion to refer to a line in space,
instead of saying, the line of which AB is the horizontal, and
A'B' the vertical projection, we say, the line (AB, A'B').
§ 30. The projections on the same plane of parallel lines are
parallel : for, the projecting planes containing the given par-
allel lines, and being perpendicular to the same plane (24), are
parallel ; hence their intersections by the plane of projection
are also parallel. But these intersections are the projections
of the lines (24), therefore the projections on the same plane
of parallel lines are parallel.
§ 31. PI. 1. Fig. 3. Let AB be the horizontal, and A'B'
the vertical trace of an oblique plane. If this plane were per-
pendicular to the horizontal plane, and had the same horizontal
trace AB, the vertical trace would pass through A, and he
perpendicular to the ground line. If the plane were perpen-
dicular to the vertical plane, and cut the ground line, its hori-
zontal trace would he perpendicular to thy ground line. If
the plane were perpendicular to both planes of projection,
both its traces would be perpendicular to the ground line. A
line situated in such a plane is not determined in position by its
two projections (26). When we wish to designate a plane
whose horizontal trace is AB, and vertical trace AB', we say,
the plane (AB,AB').16
DESCRIPTIVE GEOMETRY.
CHAPTER II.
Of the conventional methods of making the projections of lines
and the traces of planes in the different angles ; how the
given and required parts are distinguished from jthose which
are used merely to aid in the construction. Solution of some
of the principal problems on the right line and plane.
§ 32. In every projection there is some point at which the
eye is supposed to be situated, and from which the projec-
tion, or drawing, should present the same appearance as is
presented by the objects which it is made to represent.
§ 33. In the projection now used, which is named the Or*
thograpiiic, or Orthogonal projection, the eye is supposed
to be at an infinite distance from the plane on which the pro-
jection is made, and the drawing or representation is supposed
to be viewed from that position of the eye.
§ 34. The position of the eye is generally taken in tne first
angle; hence, all objects situated within this angle can be
seen, but objects in either of the other angles are concealed
by the planes of projection. Lines that are given, or required,
are made full if they can be seen, but are dotted if concealed
by other objects or by the planes of projection. Auxiliary
lines, or lines used to aid in the construction of a problem, are
always dotted.
§ 35. The traces of given or required planes are made full
in the first angle, unless they pass under bodies which prevent
them from being seen, in which case they are broken. But
when, as in Fig. 3, the horizontal trace BA is produced behind
the vertical plane, or the vertical trace B'A is produced below
the horizontal plane, the parts AC, AC', so produced, are made
broken, as in the figure. The traces of auxiliary planes are
always broken.RIGHT LINE AND PLANE.
17
§ 36 Right lines and planes are indefinite ; and the projec-
tions and traces which are made in the figures, are only the
parts intercepted between given points.
PROBLEM I.
A light line being given bíj its projections, it is required to de-
termine the points in which it pierces the planes of projection,
§ 37. PI. 1. Fig. 4. Let CF be the ground line, AB the hori-
zontal, and A'B' the vertical projection of the given line.
Produce the horizontal projection AB till it intersects the
ground line at D. At the point D, erect in the vertical plane
the perpendicular DD' to the ground line—DD' is the verti-
cal trace of the plane which projects the line on the horizontal
plane. Produce the vertical projection of the line till it inter-
sects the perpendicular at D', and this point of intersection is
the point at which the line pierces the vertical plane. To find
the point at which it pierces the horizontal plane, produce the
vertical projection till it intersects the ground line at C. From
this point, draw in the horizontal plane the perpendicular CC
to the ground line—CC' is the horizontal trace of the plane
which projects the given line on the vertical plane ; the point
C', in which it intersects the horizontal projection of the line
produced, is the point at which the line pierces the horizontal
plane.
First, to prove that the line pierces the vertical plane at D"
Every line of a plane pierces the planes of projection in the
traces of the plane. The given line must then pierce the ver-
tical plane somewhere in the line CB', the vertical trace of
the plane which projects it on the vertical plane, and some-
where in the line DD', the vertical trace of the plane which
projects it on the horizontal plane ; hence, the line pierces
the vertical plane at D', their point of intersection. For the
same reasons it follows, that the line must pierce the horizon-
tal plane in CC', the horizontal trace of the plane which pro-
jects the line on the vertical plane, and in BC\ the trace of
2í8
DESCRIPTIVE GEOMETRY.
the plane which projects the line on the horizontal plane—
hence it pierces it at C', their point of intersection. The point
D' is above the horizontal plane the distance DD', and the
point C' is behind the vertical plane the distance CC'.
PROBLEM II.
To find the length of a line joining any two points m spaces
given by their projections.
§ 38. PI. 2. Fig. 1. Let (A, A') and (B, B') be the given
points.
If a line pass through a point in space, the projections of
the line will pass through the projections of the point ; there-
fore, AB is the horizontal, and A'B' the vertical projection of
the line joining the given points.
Revolve the plane which .projects the line on the horizontal
plane around its horizontal trace till it coincides with the
horizontal plane of projection. The point (A, A) will fall in a
line drawn through A, perpendicular to AB (II), and at a dis-
tance from A equal to its height above the horizontal plane.
The point (B, B') also falls in a perpendicular to the trace AB,
and at a distance from the point B equal to its height above
the horizontal plane. Having made AD and BC respectively
equal to the heights'of the points above the horizontal plane,
draw DC, which will be the length of the line sought. The
point F, in which the line joining the points pierces the hori-
zontal plane, being in the axis, remains fixed during the revo-
lution ; the line CD produced should, therefore, pass through this
point. A similar construction might be made on the vertical
plane.
§ 39. We may here remark, that the projection of a line
on either plane is less than the line itself, unless the line be
parallel to the plane on which it is projected. For, if through
D a line DC' be drawn parallel to AB, it will be less than
DC, the length of the given line, and equal to AB, its pro-
jection on the horizontal plane; and the same may be shown
when it is projected on the vertical plane.
1RIGHT LINE AND PLANE.
19
Since the angle formed by a line and plane is measured by
the angle included between the line and its projection on the
plane, the angle CFB, or its equal CDC', is equal to the angle
which the line (AB, A'B) makes with the horizontal plane.
Hence, if a right-angled triangle he constructed, having the
angle at the base equal to the angle which the line makes with
the plane, the hypothenuse will be to the base, as the length of
the line t.Q the length of its projection. We also conclude, that
the length of a line is equal to the hypothenuse of a triangle
whose base is the projection of the line, and whose perpendicular
is equal to the difference between the perpendiculars let fall from
the two extremities of the line, on the plane of projection..
§ 40. If a line be parallel to a plane, its projection on such
plane will evidently be equal to the line itself : for the line, its
projection, and the two projecting perpendiculars through its
extremities form a rectangle, of which the line and its projection
are opposite sides.
§ 41. The length of the line may also be determined from
its projections, thus : Revolve the plane which projects the line
on the horizontal plane, about the perpendicular to the hori-
zontal plane at A, till it becomes parallel to the vertical plane :
the line from this position will be projected on the vertical
plane in its true length (40). In this revolution, all the points
of the projecting plane describe, about the vertical axis, arcs
of horizontal circles. The foot of the perpendicular to the
horizontal plane at B describes the arc BG, on the horizontal
plane, about A as a centre. Project the point G into the ver-
tical plane, and erect at I the perpendicular IH to the ground
line ; 1H will be the vertical projection, from its revolved posi-
tion, of the perpendicular to the horizontal plane at B. But
m the revolution the point (B,B) neither approaches to, nor
recedes from, the horizontal plane ; its vertical projection
must then be always found in a parallel to the ground line
through B'. This parallel intersects the perpendicular III at
H; H is then the vertical projection of the point (B,B') from
the position which it has when the projecting plane of the line
is revolved parallel to the vertical plane. The point (A,A') re20
DESCRIPTIVE GEOMETRY.
mains fixed, being in the axis, and A'H is the vertical projection
of the given line from its revolved position. This line is evi-
dently equal to CD.
PROBLEM III.
To pass a plane through three points in space, given by then
projections ; the points not being in the same right line.
§ 42. Pl. 2. Fig. 2. A plane can always be passed through
three points, and this condition determines its position.
For, conceive two of the points to be joined by a right line,
and through this line let any plane be drawn. Let the plane
be revolved about this line until it embraces the third point ; if
the plane be revolved either way from this position, it will no
longer contain the third point ; hence, there is but one position
in which it will pass through the three given points ; or, in other
words, only one plane can be draum through three points.
Let FH be the ground line, (A,A'), (B,B'), (C,C') the given
points.
Conceive the points to be joined by the right lines (AB, A'B'),
(AC, A'C'), (BC, B'C'), the projections ofthese lines will pass
through the projections of the points respectively. Since these
lines are lines of the required plane, the points in which they
pierce the planes of projection are points of the traces of the
required plane. Therefore, the point E, in which the line
(AB, A'B') pierces the vertical plane, is one point of the ver-
tical trace; and the point D, in which the line (AC, A'C')
pierces the vertical plane, is a second point of the vertical
trace ; hence, DEF is the vertical trace of the required plane
The line (BC, B'C') pierces the horizontal plane at G, which
is a point of the horizontal trace, and F is another point (23),
therefore, GF is the horizontal trace of the required plane, and
(FG, FD) is the plane containing the three given points. The
point I, in w'hich the line (BC, B'C') pierces the vertical plane,
is also a point of the vertical trace, and should be found in
order to verify the construction. The points in which theRIGHT LINE ANI> PLANE.
21
lines (AB, A'B'), (AC, A'C') pierce the horizontal plane are
points of the horizontal trace, and will be found in the trace
FG, if the construction be correctly made.
PROBLEM IV.
Ha ving given one projection of a point of an oblique plane, it is
required to find the other projection, and the position of the
point after the plane shall have been revolved to coincide vrith
either plane of projection.
§43, PL 2. Fig. 3. Let A|^lje*the ground line, (AE, AD')
the given plane, arçd C the ho^zorffa^ projection of the point.
Erect at C a perpendicular tc> the horizontal plane ; the point
. in which this perpendicular pierces the oblique plane is the
only point of that plane which is horizontally projected at C.
Through the point in which the line pierces the oblique plane
conceive a line to be drawn	parallel	to its	horizontal trace.
This parallel is a line of	the	oblique	plane,	is parallel to the
horizontal plane, and its horizontal projection CD passes
through C and is parallel to AE, the horizontal trace of the
oblique plane (30).
Let DD' be drawn in	the	vertical	plane,	perpendicular to
the ground line AB ; the	point D', in	which	it intersects the
vertical trace of the oblique plane, is the point in which the line
drawn parallel to the horizontal trace pierces the vertical plane,
since it must pierce the vertical plane in the trace AD', and
also in DD', the vertical trace of its projecting plane (37).
The line D'C', drawm through D' parallel to the ground line,
is the vertical projection of the line of which CD is the hori-
zontal projection (27). The vertical projection of the required
point is in the line C'D, it is also in the perpendicular from C
to the ground line (13) ; hence it is at C', their point of inter-
section. If the vertical projection were given, the horizontal
projection could be determined by a similar construction.
Let the oblique plane be revolved around AE till it coin-
cides with the horizontal plane. The point (C,C') will fall inDESCRIPTIVE CÌEOMETRY.
22
lhe trace of a plane drawn through it perpendicular to the axis
AE(ll), and at a distance from the point E equal to the
hypothenuse of a triangle whose base is EC, and altitude the
height of the point above the horizontal plane. Making CF,
on the line CD, equal to this altitude, and joining E and F,
gives EF for this hypothenuse. With E as a centre, and
radius EF, describe a semicircle, and the points F', F", in
which it intersects the trace F'F", are the points sought. If
the plane be revolved towards the vertical plane, the point
(C,C') falls at F' ; if from the vertical plane, at F". A similar
construction would determine the position of the point (C,C')
should the plane be revolved about its vertical trace to coin-
cide with the vertical plane.
PROBLEM V
To show how two lines which intersect in space are situated in
projection ; and secondly, to find the angle which they make
with each other.
§ 44. PI. 3. Fig. 1. Let A'B' be the ground line, AC,
BC the horizontal, and A'C', B'C' the vertical projections of
the lines.
As the point of intersection is common to the two lines, its
horizontal and vertical projections will be found in the hori-
zontal and vertical projections of each of the lines. The point
C, in which their horizontal projections intersect, is, conse-
quently, the horizontal projection of the point in which the lines
intersect, and the point C', in w'hich their vertical projections
intersect, is the vertical projection of the same point. The
points C and C', being the projections of the same point, are
contained in the same perpendicular to the ground line (13).
Jf, therefore, two lines intersect in space, the points in which
their projections intersect will he contained in a perpendicvlai
to the ground line.
Secondly, to find the angle which the lines make with each
otherRIGHT LINE AND PLANE.
23
The two lines intersecting, a plane can be drawn containing
them. If this plane be revolved about its horizontal trace till
it coincides w7ith the horizontal plane, or about its vertical trace
till it coincides with the vertical plane, in either of the revolu-
tions the lines will not change their relative positions ; hence,
the angle which they make in space is equal to the angle they
will make after the revolution.
The lines pierce the horizontal plane at A and B ; hence, AB
is the horizontal trace of their plane. Through the point C
draw CD perpendicular to the trace AB. This line is the
horizontal trace of a plane passing through the point (C,C)
and perpendicular to AB. When the plane of the two lines
is revolved about its horizontal trace to coincide with the
horizontal plane, their point of intersection falls at C", a dis-
tance from D equal to C'G, the hypothenuse of a triangle
whose perpendicular FC' is equal to the height of the point
(C,C') above the horizontal plane, and whose base FG is equal
to CD, the distance of its horizontal projection from the
axis (11). But A and B, being in the axis, remain fixed ; there-
fore, AC" and BC" are the lines in their revolved position, and
AC"B is the angle included between them. A similar con-
struction would determine the angle on the vertical plane ; it
would only be necessary to revolve the plane of the lines around
its vertical trace till it coincided with that plane.
PROBLEM YJ.
Two oblique planes being given by their traces, it is required tc
find the two projections of their line of intersection.
§ 45. PI. 3. Fig. 2. Let AB be the ground line, and (AC
AD), (BC, BD) the given planes.
Since the line of intersection is a line of the plane (AC, AD),
it must pierce the horizontal plane in the trace AC, and the
vertical plane in the trace AD. As the line of intersection is
iso a line of the plane (BC, BD), it will pierce the horizontal
•lane in the trace BC, and the vertical plane in the trace BD.24
DESCRIPTIVE GEOMETRY
Hence, the intersection of the two planes pierces the horizonta*
plane at C, the point in which their horizontal traces intersect,
and the vertical plane at D, the point in which their vertical
traces intersect. We have, then, only to find the projections of
this line. The point C is its own projection, on the horizontal
plane (17), and the point D, being in the vertical plane, is hori-
zontally projected in the ground line at D' (17) ; therefore, CD'
is the horizontal projection of the intersection. Projecting C
into the vertical plane at C', determines C'D, the vertical pro-
jection of the intersection.
PROBLEM VII.
To find the angles included between an oblique plane and trie
planes of projection.
§ 46. PI. 3. Fig. 3. Let AB be the ground line, and (AD,
AC) the given plane.
If a plane be drawn perpendicular to the horizontal trace
of the oblique plane at any point, it will be perpendicular to the
horizontal plane, and to the oblique plane ; and will consequently
intersect these planes in lines perpendicular to their common
intersection at the same point. The angle included between
these lines is equal to the angle contained by the planes.
Let DC', drawn perpendicular to AD, be the horizontal trace
of such a plane. As this plane is perpendicular to the hori-
zontal plane, its vertical trace C'C is perpendicular to the
ground line at C'. Let this plane be revolved around DC'
till it coincides with the horizontal plane ; the point C falls at
C", in a perpendicular to DC', and at a distance from C' equal
to C'C, its height above the ground line. Draw DC", and it
will be the intersection of the oblique and perpendicular planes,
in its revolved position, and the angle C'DC" is equal to the
angle which the oblique plane makes with the horizontal plane.
§ 47. If the perpendicular plane be revolved about its ver-
tical trace CC', till it coincides with the vertical plane, the point
D will describe, in the horizontal plane, the arc DD' about C'RIGHT LINE AND PLANE.
25
as a centre, and will fall at D' ; Ciy will be the revolved posi-
tion of the line of intersection of the perpendicular and oblique
[ilanes, and C'D' the revolved position of the intersection of
the perpendicular and horizontal planes; hence, CD'C' is
equal to the angle which the oblique plane makes with the
horizontal plane. This angle is evidently equal to the angle
C'DC". The angle which the plane makes with the vertical
plane, is found by a construction similar to either of those
just given.
PROBLEM VIII.
A plane being given by its traces, and a line not parallel to the
plane by its projections, it is required to find the point in
which the line pierces the plane.
. § 48. PI. 3. Fig. 4. Let AB be the ground line, (AD, AD')
the given plane, and (EC, E'C') the given line.
If any plane be drawn through the line, it will intersect the
given plane in a right line ; this line will contain the point in
which the given line pierces the given plane. The point in
which the given line meets this line of intersection is, there-
fore, the point sought. Let the plane which projects the line
on the horizontal plane be the one drawn through it. This
plane intersects the oblique plane in a line which pierces the
horizontal plane at a, and the vertical plane at D' (45) ; and
a!D' ¡s its vertical projection. But sinçe this line of intersec-
tion, and the given line, intersect in space, the intersection p'
of their vertical projections is the vertical projection of their
intersection (44) ; p' is therefore the vertical projection of the
point in which the line pierces the plane ; and p is its hori-
zontal projection, since the horizontal projection is in the hori-
zontal projection of the line, and in a perpendicular to the
ground line through p\
The point p might be found without demitting the perpen-
dicular to the ground line from p'. For the plane which pro-
jects the given line (EC, E'C') on the vertical plane, intersects2G
DESCRIPTIVE GEOMETRY.
the oblinue plane (AD, AD') in a line of which Dò' is the hori-
zontal projection, and the point p, in which Dò' intersects the
horizontal projection of the given line, is the horizontal projec-
tion of the required point.
PROPOSITION IX. THEOREM.
If a line be perpendicular to an oblique plane, the projections
of the line are respectively perpendicular to the traces of the
plane ; that is, the horizontal projection to the horizontal trace,
and the vertical projection to the vertical trace.
§ 49. For, the plane which projects the line on the horizontal
plane is perpendicular to the oblique plane, since it contains a
line perpendicular to it : it is also perpendicular to the horizontal
plane, and is therefore perpendicular to their intersection, that
is, to the horizontal trace of the oblique plane. Since the
horizontal trace of the oblique plane is perpendicular to the
projecting plane of the given line, it will be perpendicular to
its horizontal trace, that is, to the horizontal projection of the
given line (24). It may be shown, in a similar manner, that the
vertical projection of the line is perpendicular to the vertical
trace of the oblique plane.
§ 50. The converse of this proposition is also true, that is, if
the projections of a line are respectively perpendicular to the
traces of a plane, the line in space is perpendicular to the plane.
For, the projecting planes of the line will be respectively per-
pendicular to the traces of the oblique plane, and therefore
perpendicular to the oblique plane ; hence, their intersection,
which is the line, will be perpendicular to the oblique plane.
§ 51. If two lines are perpendicular to each other, and are
projected on a plane to which one of them is parallel, their
projections will also be at right angles. For, through the line
which is parallel to the plane on which the projection is made,
conceive a plane to be drawn perpendicular to the other line ;
its trace will be parallel to the line through which the plane is
drawn, since the line is parallel to the plane of projection. But• /. i ,V(rni.írRIGHT LINE AND PLANE.
27
the projection of the line through which the plane is passed
will be parallel to the trace of the plane, since they are parallel
in space; and as the projection of the other line is perpen
dicular to the trace (49), it will be perpendicular to any line
parallel to the trace, and consequently to the projection of
the line through which the plane is drawn.
PROBLEM X.
To draw from a given point a line perpendicular to a given
plane ; to find the point in which it pierces the plane, and
the length of the perpendicular.
§ 52. PI. 4. Fig. 1. Let AB be the ground line, (D,Dr) the
given point, and (AC, AC') the given plane.
The horizontal projection of the line must pass through D,
and be perpendicular to AC, since the line is perpendicular to
the plane (AC, AC') (49). The vertical projection must pass
through D' and be perpendicular to AC'. The lines DF and
D'F' drawn through D and D', respectively perpendicular to
the traces AC, AC', are the projections of the perpendicular
sought. Having determined the projections of the line, the
point (F,F) in which it pierces the plane (AC, AC') is found
as in Prob. 8. The length D"F" of the perpendicular is found
as in Prob. 2. To find the shortest distance between a point
and plane, we have only to draw’ a perpendicular to the plane
and find its length.
PROBLEM XI.
To draw through a given point a plane perpendicular to a
given line.
§ 53. PI. 4. Fig. 2. Let A'R be the ground line, (AD, A'D7)
the given line, and (E,E') the given point.
As the required plane is to be perpendicular to the line, the
aces of the plane must be respectively perpendicular to its28
DESCRIPTIVE GEOMETRY.
projections (49) ; we know then the directions of the traces of
the required plane. But the plane is to pass through the point
(E,E). Therefore, through the point (E,E') conceive a line to
be drawn parallel to the horizontal trace of the required plane.
This line will be horizontal, and also a line of the required
plane. Its horizontal projection passes through E, and is per-
pendicular to AD, for the line in space being parallel to the
horizontal trace of the required plane, its horizontal projection
is parallel to this trace (30) ; that is, perpendicular to AD. The
line, therefore, drawn through E, perpendicular to AD, is the
horizontal projection of the line through (E,E'), the vertical
projection of which passes through E' and is parallel to the
ground line. This line pierces the vertical plane at F, which is
a point of the vertical trace of the required plane. Through
this point draw C'FR perpendicular to A'DJ and it will be the
vertical trace of the required plane. Through the point R, in
which this trace intersects the ground line, draw RC per-
pendicular to AD, and it will be the horizontal trace of the
required plane. If through the point (E,E') a line were drawn
parallel to the vertical trace of the required plane, it would
pierce the horizontal plane at G, which is a point of the hori-
zontal trace of the required plane : this point will fall in the line
RC as before drawn, if the construction be correct.
PROBLEM XII.
To find the shortest distance between a point and line given by
their projections.
§ 54. The length of the perpendicular from the point to the
line is the distance sought. This perpendicular is contained
m a plane passing through the point and perpendicular to the
line. If, then, a plane be drawn through the point and perpen-
dicular to the line (53), and the point in which it cuts the line
be determined (52), the distance between this point and the
given point is the distance required.
§ 55. The problem can be solved otherwise, thus. DrawRIGHT LINE AND PLANE.
29
a plane through the right line and point. Let this plane be
revolved about its horizontal trace till it coincides with the
horizontal plane, or about its vertical trace till it coincides with
the vertical plane. Find the position of the point and line after
either of these revolutions, and draw through the point thus
revolved a perpendicular to the revolved line. This will be
the true length of the perpendicular sought, since the point and
line do not change their relative position in the revolution of
their plane.
PI. 4. Fig. 3. Let A'B be the ground line, (AC, A'C')
the given line, and (D,D') the given point.
First, to draw a plane through the point and line. Through
the point (D,D') draw a line parallel to (AC, A'C'), its two
projections are respectively parallel to AC, A'C', and it pierces
the horizontal plane at F. The given line pierces the hori-
zontal plane at A, therefore AFO is the horizontal trace of a
plane passing through the two parallels, which plane contains
the given point and line. Let this plane be revolved about
its horizontal trace AO till it coincides with the horizontal
plane. The point (D,D') falls at D", in a perpendicular
drawn through D to the trace AO (II), and at a distance
from the point O equal to D'B, the hypothenuse of a triangle
whose perpendicular D'P is equal to the height of the point
above the horizontal plane, and base PB equal to DO, the
distance of the horizontal projection of the point from the
axis. As F remains fixed, being in the axis, FD" is the revolved
position of the parallel line. But lines in the same plane,
which are parallel before revolution, are parallel after (10).
Draw, therefore, through A the line AM" parallel to FD", and
we have AM" for the position of the given line revolved on the
horizontal plane. Through D" draw D"M" perpendicular to
AM", and it will be the perpendicular required. Making a
counter revolution, or bringing the plane back to its first posi-
tion, the point D" will be horizontally projected at D, and the
point M" at M, since the point M" revolves in a plane perpen-
dicular to AO, and must, after the counter revolution is com-
pleted, be horizontally projected in the line AC. The vertical30
DESCRIPTIVE GEOMETRY
projection of the point of which M is the horizontal, is in a
perpendicular to the ground line through M, and also in the line
A'C' ; hence it is at M' their point of intersection. The line DM
is then the horizontal, and D'M' the vertical projection of the per-
pendicular, and M"D" is its true length. The plane might have
been drawn through the point and line by joining the given
point and any point of the line, and drawing a plane through
this and the given line.
PROBLEM XIII.
To measure the angle between two oblique planes.
§ 56. The angle between two planes is measured by the
angle included between two lines, one in each plane, and both
perpendicular to the common intersection at the same point.
If a plane be drawn perpendicular to the common intersection
of the two planes, at any point, it will intersect the planes in
lines perpendicular to the common intersection, and when the
angle between these two lines is determined, the angle between
the planes will be known.
PI. 4. Fig. 4. Let HP be the ground line, and (HC, HC')
(AC, AC') the given planes.
The intersection of these planes pierces the horizontal plane
at C, and the vertical plane at C', and CD is its horizontal
projection. If we suppose a plane drawn perpendicular to this
intersection, its horizontal trace will be perpendicular to CD,
the horizontal projection of the intersection (49). Let FG,
perpendicular to CD, be the horizontal trace of such a plane.
The lines in which this plane intersects the oblique planes,
pierce the horizontal plane at F and G ; and these two lines,
together with FG, form a triangle of which FG is the base.
The vertical angle of this triangle is equal to the angle included
between the planes, and the vertex of this angle lies in their
line of intersection. It is, then, only necessary to find this
angle. The line joining the point N, and the vertex of the
vertical angle of the triangle, is perpendicular to the commonRIGHT LINE AND PLANE.
31
intersection of the oblique planes, since it is contained in a
plane perpendicular to this intersection ; it is also perpen-
dicular to FG, since FG is perpendicular to the projecting
plane of the intersection of the oblique planes, and this pro-
jecting plane contains the line drawn from N. The length of
this line is, therefore, the shortest distance from the angular
point to the line FG, and if this length were known, by revolv-
ing the plane of the triangle about FG as an axis, till it coin-
cides with the horizontal plane, we could determine the posi-
tion of the angular point, and consequently, the magnitude of
the angle.
To find the length of this line, let the plane which projects
the intersection of the oblique planes on the horizontal plane,
be revolved around its horizontal trace CD, till it coincides
with the horizontal plane. The point C' falls at C", and as C
remains fixed, CC" is the revolved position of the intersection.
But the required line from N was perpendicular to the inter-
section before, and consequently will be perpendicular to it
after the revolution. If, therefore, NI" be drawn perpendiculai
to CC", it will be equal to the distance of the vertex of the
vertical angle of the triangle from the base FG. Let now the
plane of the triangle be revolved about its base FG, till it coin-
cides with the horizontal plane. The vertex of the vertical
angle falls in DC (11), and at a distance from N, equal to NT
(11) ; it falls therefore at I'. But since F and G remain fixed,
being in the axis, draw Fi' and GI', and FI'G is equal to the
angle included between the oblique planes.
If from I" we draw I"0 perpendicular to CD, the point O is
the horizontal projection of the angular point ; and by joining
it with F and G, we obtain the horizontal projection of the
angle FI'G.
The line NI" can be found by another construction. Let
the plane which projects the intersection of the oblique planes
on the horizontal plane, be revolved about its vertical trace
DC', till it coincides with the vertical plane. The points C
and N describe arcs of circles in the horizontal plane, around
D as a centre, and fall at P and N' ; hence, C P is the revolved32
DESCRIPTIVE GEOMETRY.
position of the intersection of the oblique planes. From N't
let N'l be drawn perpendicular to C'P ; it is evidently equal
to the line drawn from N, perpendicular to the intersection of
the oblique planes. With N' as a centre, and radius N'l,
let the arc IB be described ; then, with D as a centre, and
radius DB, let the arc Bí' be described ; the point I', in which
this arc intersects DC, is the position of the angular point of
the triangle, when its plane is revolved into the horizontal
plane. The radius N'l is evidently equal to N1', and also to NI".
PROBLEM XIV.
To find the angle which a line makes with a plane.
9 57, The angle which a line makes with a plane, is the angle
which the line makes with its projection on the plane. If from
any point of the line, a perpendicular be drawn to the plane,
the foot of the perpendicular is one point of the projection
of the line on the plane. If the given line be produced till it
meets the plane, the point of meeting will be another point of
the projection of the line. Conceive the projection to be
drawn. The given line, the perpendicular to the plane, and
the projection of the line on the plane, form a right-angled
triangle, and calling the projection of the line the base, the angle
at the base is the angle sought ; this angle is readily found when
the vertical angle is known.
PI. 5. Fig. 1. Let AB' be the ground line, (AC, AC') the
given plane, and (BD, B'D') the given line.
From any point of the given line, as (D,D'), let a line be
drawn perpendicular to the given plane ; its projections will pass
through the points D, and D', and be respectively perpendicular
to the traces AC, AC' (49)—DE is the horizontal, and D'E' the
vertical projection of this perpendicular. We will now find
the angle between this perpendicular and the given line, and
then between the given line and the plane. The perpendicular
pierces the horizontal plane at E, and the given line pierces
it at B : therefore, BEG is the horizontal trace of their plane.RIGHT LINK AND PLANE.
33
Let this plane be revolved about BG till it coincides with the
horizontal plane. The point (I).D') falls at D", and the points
E and B remain fixed ; therefore, BD "E is equal to the angle
included between the lines. From E, draw EF perpendicu-
lar to ED", and EFD" will be equal to the angle which the
hne (DB, D'B) makes with the plane (AC, AC').
PROBLEM XV.
To pass a plane through a given line, and parallel to another
given line.
§ 58. If through any point of the line, through which the
plane is to be passed, a line be drawn parallel to the other
line, the plane drawn through these two lines will be the plane
required.
PI. 5. Fig. 2. Let AB be the ground line, (DE, D'E') the
line through which the plane is to be drawn, and (NC, N'C') the
line to which it is to be parallel.
From any point, as (D,D'), of the line through which the
plane is to be drawn, conceive a parallel to be drawn to (NC,
N'C') ; its projections DI, DT are respectively parallel to CN,
C'N', and the point I', in which this parallel pierces the vertical
plane, is a point of the vertical trace of the required plane.
The point E' is a second point of this trace ; hence ET'A is the
vertical trace, and AD the horizontal trace of the plane con-
taining (DE, D'E'), and parallel to (NC, N'C').
§ 59. If the point A, in which the vertical trace meets the
ground line, were not on the paper, a point of the horizontal
trace might be found thus : through any point of the line (DE,
D'E'), as (F,F'), conceive a line to be drawn parallel to the
vertical trace ET—its vertical projection will be parallel to
this trace, and its horizontal projection FQ parallel to the
ground line (27). This line will pierce the horizontal plane at
Q, which is therefore a point of the horizontal trace : the trace
can then be drawn through D and Q.
334
DESCRIPTIVE GEOMETRY.
PROBLEM XVI.
It lì required to find the shortest distance between two lines, not
in the same plane ; or, to dra c a line that shall be perpen-
dicular to them both.
§ 60. If a plane be drawn through one line, and parallel to
the other, the shortest distance between this plane and the line
to which it is parallel, will be equal to the distance sought.
If the line, to which the plane is drawn parallel, be pro-
jected on the parallel plane, the projecting perpendiculars of
its different points will be equal to each other, and equal, also,
to the shortest distance between the two lines. But since the
line is parallel to the plane, its projection on the plane is
parallel to itself ; and as the given lines are not parallel, this
projection will intersect the line through which the parallel
plane is drawn. Tue projecting perpendicular which passes
through this point of intersection, is perpendicular to the two
given lines, and, is, therefore, the Une sought.
PI. 5. Fig. 3. Let AB' be the ground line, (DG, D'G') one
of the given lines, and (CB, C'B') the other.
Through the line (DG, D'G') let a plane be drawn parallel
to (BC, B'C') ; AD and AG' are its traces.
It is now required to project the line (CB, C'B') on this plane.
From any point of the line, as (C,C'),draw a perpendicular to
the plane (52) ; this perpendicular pierces the plane in the
point (F,F') ; and this is one point of the projection of the line
on the parallel plane (AD, AG'). But since the trace of the
projecting plane on the parallel plane and the line (CB, C'B')
are parallel, their projections are parallel (30) ; therefore FO,
drawn parallel to CB, is the horizontal projection of this trace.
But, as this trace and the line (DG, D'G') intersect, the point
O, in which their horizontal projections intersect, is the hori-
zontal projection of their point of intersection, and O' is the
vertical projection of the same point. If at the point (0,0')RIGHT LINK AND PLANE.
35
a perpendicular be drawn to the parallel plane (AD, AG'), it
will be contained in the plane which projects the line (CB, C'B')
on the plane, and will consequently intersect the line (CB, C B')
This perpendicular to the oblique plane at (0,0') is perpen-
dicular to, and intersects, both of the given lines; and its
projections OP, O'P' are respectively perpendicular to the
traces of the oblique plane (AD, AG ). The length of the line
(OP, O P') can l)e found as in Prob. 2.
§ 61. The solutions of the foregoing problems may be varied,
by changing the positions of the given parts ; and in some cases,
the constructions will be quite different from those which have
been given.
All the principles necessary to solve any problem involving
the right line and plane, have, however, been developed,
and the student who would be skilful in the application of
these principles, must apply them to a great variety of cases.
A few examples are given, to show how the data of the problems
may be varied, and to lead the student to propose cases to
himself.
Io. In Problem III. let the points be so situated that a line
joining two of them shall be parallel to the ground hue.
2\ In Problem IV. let the oblique plane be parallel to the
ground line.
3°. In Problem V. suppose one of the lines to be parallel to
the ground line.
4°. In Problem VI. let the planes be parallel to the ground
line.
5 . In Problem VIII. suppose the line parallel to the ground
line.
6°. In Problem X. let the plane be taken parallel to the grou nd
line.
7°. In Problem XII. suppose the line parallel to the horizontal
plane.
8 ’. In Problem XIII. let the planes be parallel to the ground
line.
9°. In Problem XIV. suppose the plane to be parallel to thé
ground line.
3*36
DESCRIPTIVE GEOMETRY.
10°. In Problem XV. let the line through which the plane is
drawn be parallel to the ground line.
IIo. In Problem XVI. suppose one of the lines to be parallel
to the ground line.
12°. Let it be required to draw a plane through a given point,
and parallel to a given plane.
CHAPTER III.
OF LINES AND THEIR TANGENTS.
§ 62. For the purposes of Descriptive Geometry, lines may
oe divided into three classes.
Io. The right line, which does not change its direction be-
tween any of its points.
2°. Curved lines whose points are in the same plane, which
are called curves of single curvature.
3°. Curved lines whose points are not in the same plane,
which are called curves of double curvature.
§ 63. Lines may be generated by the motion of points : the
conditions which govern this motion fix their different posi-
tions, and determine the class to which the lines generated
belong.
PI. 6. Fig. 1. Suppose, for example, that a point should
move from C, with the conditions of continuing in the plane of
the paper, and at the same distance from the line AB ; it would
evidently generate a right line, passing through C, and parallel
to AB.
PI. 6. Fig. 2. If a point move from B, with the conditions
that it shall not depart from the plane of the paper, and be
constantly at the same distance from a fixed point A, it will
generate the circumference of a circle, a curve of single cur-
vature. If the point were subjected to the first condition only
it would still generate a curve of single curvature, unless the
point were to move in a right line. If the point B were subS/, ni. VLINES AND THEIR TANGENTS.
:ì7
jected io the second condition, to the exclusion of the first, it
would generate a curve of double curvature, which would lie
on the surface of a sphere whose centre is A, and radius AB.
Other curves, both of single and double curvature, may be gene-
rated, by changing the conditions which fix the different posi-
tions of the generating point. The generating point is called
tiie generatrix of the line*
* Note.—If two points A and B (PI. 6. Fig. 5) be taken in the plane of the paper,
and a point H be moved around them, with the conditions that it shall not depart
from the plane of the paper, and that AH-(-BH be equal to a constant quantity,
the point H will describe a curve called an ellipse. The fixed points A and B
are called foci. The line DABE, passing through the foci, is called the trans
verse axis, and its extremities D and E the vertices of the axis. The point C,
the middle of DE, is called the centre of the ellipse, and CO perpendicular to
DE at the point C, the semi-conjugate or semi-lesser axis.
To describe the curve mechanically, fix the two extremities of a thread, whose
length is greater than AB, at the points A and B. Bear apin close against the
thread, and carry it round, its point will describe the ellipse. If the transverse
axis DE, and the foci A and B, be given, points of the curve may be found thus:
take any portion of the transverse axis, as DP ; with this distance as a radius,
and the point A as a centre, describe the arcs t and o ; with the remaining part
EP of the transverse axis, as a radius, and the point B as a centre, describe the
arcs q and s ; the points in which these arcs intersect those before described are
points of the curve. After the arcs t and o are described from the centre A, it
is most convenient to place the dividers at B, and describe with the same radius
the arcs m and n ; and after having described the arcs s and q, from the centre B,
let the dividers be placed at A, and the arcs r and p be described ; their intersec-
tions with the arcs m and n are points of the curve.
When the point H comes into the position of the point O, the lines AH and
BH are equal to each other ; and since their sum is equal to the transverse axis
DE, either of them is equal to the semi-transverse axis CD. If, therefore, the
two axes be given, the foci are easity found. For, take either vertex, as 0, of
the conjugate axis as a centre, and the semi-transverse axis DC as a radius,
and describe the arc of a circle, the points A and B, in which it cuts the trans
verse axis, are the foci of the ellipse.
PI. 6. Fig. 6. If a right line EF, and a point D, be taken in the plane of the
paper, and a point, as G, be so moved in this plane that its distance from D be
constantly equal to its distance from EF, that is, GD equal to GI', and ID
equal to IE, the point G will describe a curve, called a parabola. The line EF
is called the directrix of the parabola, the point D the focus, the line AD, per-
pendicular to the directrix, the axis, and the point B, in which the axis intersects
the curve, the vertex of the axis. Points of lhe curve may be found thus : take
any point of the directrix, as E and draw ED to th« focus. Draw also El, per¿8
DESCRIPTIVE GEOMETRY.
§ 64. PI. 6. Fig. 3. Let EDD' be any curve concave towards
AC. Through any point, as 13, let a right line BD' be drawn,
cutting the curve in the points B and D'. The generatrix of
the curve, in its different positions, occupies all the points be-
tween B and D'. Let the point 1)' be moved towards the
point B. The chord BD' approaches the tangent to the curve
at B, and becomes the tangent when the point D'occupies the
first position which the generatrix assumes on departing from
B, towards D' ; because, in this position of the point D', no
point of the curve lies between it and B, consequently BD'
does not intersect the curve, and is therefore tangent to it. In
this last position of the point D', which is denoted by D, the
points B and D are called consecutive points,
§ 65. If a point be taken in a curve of double curvature, the
right line joining this and its consecutive point will be tangent
to the curve.
For, the right line is determined in position, since it passes
through two given points, and does not cut the curve, since no
part of a curve lies between consecutive points. A right Urie is
therefore tangent to any line, when itjJasses through two conse-
cutive points of the line.
§ 66. If a line be tangent to a right line, it coincides with it
throughout, and is the same line.
§ 67. If a line be tangent to a curve of single curvature, it is
pendicularto the directrix, and at the »'o}nt D make the angle EDI equal to the
angle DEI, the point I, at which the lines El and DI intersect, is a point of the
curve. In the same manner any number of points may be found.
PV 6. Fig. 7. If two points A and D be taken in the plane of the paper,
and a point C be moved, with the conditions that it continue in the plane of the
paper, and that the difference between the distances AC and DC be a constant
quantity, the point C will describe a cune //GC, called an hyperbola. A curve
identical with p'GC can be described around the point A, by drawing lines DC'
and AC', and making their difference equal to the same constant quantity.
These two curves are called opposite hyperbolas. The points A and D are
called foci ; the line FB is named the transverse axis ; and the points F and
B, in which it intersects the curves, are the vertices of the axis, or vertices of
the hyperbolas ; the point E, the middle of FB, is the centre ; and the line EO,
perpendicular to FB, is called the semi-conjugate art?SURFACES.	39
contained in the plane of the curve, since the consecutive points
through which.it passes are in the plane of the curve.
§ 68. If a right line be tangent to a curve of double curva-
ture, it makes the same angle with a line, or plane, drawn through
the point of contact, as the curve makes with the same line, or
plane.
§ 69. Two curves are tangent to each other, when a line tan-
gent to one, at a common point, is tangent also to the other.
§ 70. As no part of a curve lies between two consecutive
points, the distance between them, measured on the curve, is
equal to nothing. Considered then with respect to their distance
apart, measured on the curve, they are regarded as the same
point. The line AB, Fig. 3, is then to be considered as tangent
to the curve BD', at the point B.
CHAPTER IV.
Of surfaces—Their generation—How they are represented on
the planes of projection—Of the projections of curved lines
and their tangents.
§ 71. Surfaces are generated by lines moving according to
some mathematical law. A line which by its motion generates
a surface, is called the generatrix ; and the lines of the surface
which are determined by the different positions of the genera-
trix, are called elements of the surface.
When the generatrix of a surface begins to move from any
position, the first position which it takes determines an element
consecutive with the first position of the generatrix, and the two,
that is, the first and second positions of the generatrix, are called
consecutive elements.
§ 72. Although there is an infinite number of surfaces having
different properties, yet, for the purposes of Descriptive Geome-
try, they may be divided into four classes.
Io. The plane surface, or plane, which is generated by a right40
DESCRIPTIVE GEOMETRY.
line moving along another right line and continuing parallel to
itself.	#
2°. Surfaces which may be generated by a right line, having
its consecutive positions in the same plane ; such are called
single-curved surfaces.
3°. Surfaces which can only be generated by curves ; such
are called double-curved surfaces.
4°. Surfaces which may be generated by a right luie, when
the consecutive positions are not in the same plane ; such are
called warped surfaces.
§ 73. If any curve be taken in space, and an indefinite right
line be drawn through any point of it, and then be moved around
the curve, constantly touching the curve and parallel to its first
position ; the surface generated is called a cylindrical surface,
the moving line the generatrix of the surface, and the curve
around which it moves the directrix.
If the directrix of the cylinder were to move along the gene-
ratrix, parallel to itself, all its points would continue in the sur-
face ; hence, a cylindrical surface can be generated by a curve,
moving parallel to itself. The cylindrical surface can therefore
be generated in two ways, and has two generatrices, a right
line and a curve ; the directrix of the first generation is the
generatrix of the second, and reciprocally. If the curve have
a centre, the right line drawn through it, parallel to the right-
lined elements, is called the axis of the cylinder.
§ 74. If through a point, not in the plane of a curve, a right
line be drawn touching the curve, and be produced indefinitely
in both directions, if the right line be then moved around the
curve, continuing to pass through the point, the surface gene-
rated is called a conic surface, the fixed point the vertex of
the cone, and the curved line the directrix.
That part of the surface which lies below the vertex is called
the lower nappe, and the part of the surface which lies above
the vertex the upper nappe of the cone. If the directrix were
to move towards the vertex, decreasing according to a certain
law, or from the vertex, increasing according to a certain law,
its points would continue in the surface of the cone. The surSURFACES.
41
face of the cone can then be generated by a curve ; it has there-
fore two generatrices, a right line and curve. If the curve
have a centre, the line drawn through the centre and vertex
is called the axis of the surface. The cylinder and cone are
surfaces of the second class, that is, single-curved surfaces.
§ 75. As the rectilinear generatrices of these surfaces are
indefinite, the surfaces are also indefinite. When it is necessary
to consider any finite portions of them, they are intersected by
planes. The curves formed by the intersection of such planes
with the surfaces, are called bases ; the upper plane is named
the plane of the superior base, and the lower plane the plane
of the inferior base.
§ 76. A cylinder, whose rectilinear elements are perpendicular
to the plane of its inferior base, is called a right cylinder ; and
if this base be a circle, a right cylinder with a circular base.
Such a cylinder has all its rectilinear elements at the same
distance from the axis, is the kind of cylinder treated of in
geometry, and may be generated by the revolution of a rec-
tangle about one of its sides. Cylinders are generally named
from their inferior bases. If the inferior base be a circle,
ellipse, hyperbola, or parabola, the cylinder takes the name of
a cylinder with a circular, elliptical, hyperbolic, or parabolic
base, and is either right or oblique according as its rectilinear
elements are perpendicular, or oblique, to the plane of the
base.
§ 77. A right cone is one whose axis is perpendicular to its
base. If the base be a circle, such cone is a right cone with
a circular base ; it can be generated by the revolution of a
right-angled triangle about one of its legs, and its rectilinear
elements make equal angles with the axis. This is the kind
of cone treated of in geometry. The cone, like the cylinder,
takes particular names from its inferior base ; that is, it is a
cone with a circular, elliptical, parabolic, or hyperbolic base,
according as its inferior base is a circle, ellipse, parabola, or
hyperbola.
§ 78. We shall consider, at present, those surfaces of the
third class which can be generated by the revolution of a cun e42
DESCRIPTIVE GEOMETRY.
of single curvature about an axis in its own plane. Such sur*
faces are called surfaces of revolution, *
§ 79. Any plane passing through the axis of a surface of revo-
lution is called a meridian plane, and its intersection with the
surface a meridian curve. Every plane perpendicular to the
axis intersects the surface in a circle, since every point of the
revolving generatrix describes a circle around the axis. Let the
curve EBDD' (PI. 6. Fig. 3) be revolved about AC, it will
generate a surface of revolution.
If a circle of an indefinitely small radius were moved from
the point E, its radius increasing according to a certain law,
its centre continuing in, and its plane perpendicular to AC, this
circle will also generate the surface of revolution. A surface
of revolution, therefore, has two generatrices, a meridian curve,
and a circle whose plane is perpendicular to the axis of the
surface.
§ 80. The surfaces generated by the revolution of the circle
the ellipse, the parabola, and the hyperbola about their axes, are
called respectively the surface of the sphere, of the ellipsoid
of the paraboloid, and hyperboloid. The fourth class of sur
faces is discussed in the Complement.
§ 81. The projection of a curve on either plane of projection
is the base of a cylindrical surface passing through the curve
and perpendicular to the plane on which the projection is made.
This cylinder is called the projecting cylinder of the curve.
§ 82. A curve of single curvature, whose plane is perpen-
dicular to either plane of projection, is projected on that plane in
a right line, since the projecting cylinder becomes the plane
of the curve. Both projections of a curve of double curvature
are always curved lines.
§ 83. Two projections of a curve determine its form and posi-
tion. For, the projections of a curve are made up of the pro-
* A surface of revolution is a surface generated by a line moving around a
right line as an axis ; the points of the moving line describing circles whose
centres are in the axis, and whose planes are perpendicular to it. It is there-
fore proposed to discuss only one variety of this class of surfaces.SURFACES.
43
jections of all its points ; the points are fixed in position when
their projections are known ; and a curve is known in form
and given in position wdien all its points are determined. It is
plain that this reasoning does not apply to the case in which the
curve is of single curvature and its plane perpendicular to the
ground line.
PI. 6. Fig. 4. Let AB, A'B' be the projections of a curve.
If a cylinder perpendicular to the horizontal plane be drawn
through AB, it will pass through the curve in space of which
AB is the horizontal projection. If the cylinder which projects
the curve on the vertical plane were drawn, its intersection with
the cylinder that projects the curve on the horizontal plane is
the curve in space ; but this intersection is given in position,
since the cylinders are given ; hence the two projections of a
curve determine it inform and position*
§ 84. A plane is tangent to a surface when 'there is at least
one point common to the plane and surface, through which, if
any number of planes be draum, the sections made in the plane
will be tangent to the sections made in the surface.
§ 85. Two surfaces are tangent to each other when all the sec-
tions of the one made by planes passing through a common point
are respectively tangent to the sections of the other made by the
same planes ; or, when a plane which is tangent to the one, at a
common point, is also tangent to the other.
§ 86. A plane which passes through the consecutive rectilinear
elements of a cylindrical surface is tangent to the surface. For,
if the cylinder be intersected by any plane, the consecutive
elements will pierce the plane in the curve in which the plane
intersects the surface, and in consecutive points of that curve.
The right line passing through these consecutive points is tan-
gent to the curve (67) ; but this line is also the intersection of
the plane of the consecutive elements and the cutting plane ;
and as the same may be shown for any intersecting plane, it
follows that the plane of consecutive elements is tangent to the
cylinder (84). In the same manner it may be shown, that a
plane passing through the consecutive rectilinear elements of a
conic surface is tangent to the surface.44
DESCRIPTIVE GEOMETRY.
§ 87. As no part of a surface lies between consecutive ele-
ments, the distance between them, measured on the surface, is
equal to nothing. The consecutive elements, therefore, con-
sidered with respect to their distance apart, measured on the
surface, are to be regarded as the same line. The plane of
consecutive elements is then to be considered as tangent to the
surface along one element only.
§ 88. It follows from the definition of a tangent plane (84),
that all right lines passing through a point of contact, and tan-
gent to lines of the surface, are contained in the tangent plane ;
hence this plane is the locus, or place, of the right lines tangent
to all the curves which lie on the surface and pass through the
point of contact. But two right lines, which intersect, deter-
mine the position of a plane. If, therefore, through any point
of a surface two elements be drawn, and, at their point of inter
section, a tangent to each, the plane of these tangents is tangent
to the surface at their point of intersection.
§ 89. A plane tangent to a surface which has rectilinear ele-
ments will contain that element which passes through the point
of contact For, if a right line be drawn tangent to this element
of the surface, it will be a line of the tangent plane (88) ; but
this tangent is the element itself ; hence a tangent plane always
contains the element of the surface passing through the point of
contact
§ 90. If a right line be tangent to a curve in space, the projec-
tions of the line are respectively tangent to the projections of the
curve. For, if through the two consecutive points of tangency two
lines be drawn perpendicular to either plane, they will be common
both to the plane which projects the right line and to the cylin-
der which projects the curve ; therefore, the plane and cylinder
will be tangent to each other (86). The lines in which they are
intersected by either plane of projection are therefore tangent
to each other (84) ; but these lines are respectively the pro-
jections of the curve and tangent ; the projections of tangent lines
are therefore tangent to each other.
§ 91. Surfaces are represented on the planes of projection
by the projections of their elements. The horizontal projec.TANGENT PLANES
45
lion of a single-curved surface is generally made by projecting
on the horizontal plane its inferior base, and the elements of
contact of two planes tangent to the surface and perpendicular
to the horizontal plane. The vertical projection of the surface
is generally determined by projecting on the vertical plane its
inferior base, and the elements of contact of two planes tangent
1o the surface and perpendicular to the vertical plane.
§ 92. The horizontal projection of a surface of revolution is
the intersection by the horizontal plane of a cylinder perpen-
dicular to this plane and tangent to the surface. The vertical
projection is the intersection, by the vertical plane, of a cylin-
der perpendicular to the plane and tangent to the surface. A
surface of revolution, having its axis perpendicular to the hori-
zontal plane, may also be projected by projecting on the hori-
zontal plane some one of its horizontal sections, and on the
vertical plane the meridian curve which is parallel to this
plane.
CHAPTER y.
OF TANGENT PLANES TO SINGLE-CURVED SURFACES.
PROBLEM XVII.
To draw a plane through a given point of a conical surface,
tangent to the surface.
§ 93. PL 6. Fig. 8. Let the circle BFEG in the horizontal
plane be the base of the cone, A the horizontal, and A' the
vertical projection of the vertex, AL" and A'L'" the projections
of the axis. The lines A F and AE are the traces of two planes
tangent to the cone and perpendicular to the horizontal plane ;
they are also the projections of the elements of contact of the
planes and cone ; and AEGF is the horizontal projection of the
cone (91). The line G'a is the vertical projection of the base46
DESCRIPTIVE GEOMETRY.
of the cone ; A'G', AV are the vertical traces of the two planes
tangent to the çone and perpendicular to the vertical plane ;
hence, A'GV is the vertical projection of the cone. The part
FDE of the circumference of the base of the cone is dotted ;
for, being under the surface, it cannot be seen. Let C be the
horizontal projection of the point of the surface through which
the plane is to be drawn. The vertical projection of this point
cannot be assumed ; for, it being a point of the surface of the
cone, its projecting lines must intersect on the surface. If we
suppose a line to be drawn perpendicular to the horizontal plane
at C, it will pierce the surface of the cone in two points ; and
these are the only points of the surface which are horizontally
projected at C.
Through this perpendicular and the vertex of the cone let a
plane be drawn ; ACB is its horizontal trace, and the two ele-
ments in which it intersects the surface of the cone pierce the
horizontal plane at D and B. The projections of these elements
on the vertical plane are A'B' and A'D' ; and the points C' and
0", in which these projections intersect the vertical projection of
the perpendicular from C, are the only points of the surface
which are horizontally projected at C.
Let it then be required to draw a plane through the point
(C,C'), and tangent to the surface : the plane will contain the
element (AC, A'C') passing through this point (89). This
element pierces the horizontal plane at B, which is a point of
the horizontal trace of the required plane. But the trace must
be tangent to the base of the cone (84) ; therefore BK, drawn
tangent to the base of the cone, is the horizontal trace of the
tangent plane. The element of contact (AB, AB') pierces the
vertical plane at L' ; and since this is a line of the tangent plane,
L' is a point of its vertical trace : hence KL' is the vertical trace
of the tangent plane.
If the point K were not used, the vertical trace could be
found thus : through any point of the element of contact, as
(C,C'), let a line be drawn parallel to the horizontal trace of the
tangent plane ; this line pierces the vertical plane at I, which
is therefore a point of the vertical trace of the tangent plane.TANGENT PLANES.
'»7
The point L' being previously determined, the trace L'l can
be drawn. Had it been required to draw the plane through
(C,C"), its traces would have been constructed in a similar
manner.
PROBLEM XVIII.
To draw a plane through, a given point without the surface of a
cone, and tangent to the surface.
§ 94. PI. 7. Fig. 1. Let AnGm be the horizontal, and A'C'D
be the vertical projection of the cone, (A,A') its vertex, and
(E,E ') the given point through which the plane is to be drawn.
As every plane tangent to a cone passes through the vertex,
if the points (A,A'), (E,È') be joined by a right line, this right
line (EA, E'A') will be a line of the required plane ; and the
point F, at which it pierces the horizontal plane, will be a point
of the horizontal trace. Two lines can be drawrn through the
p int F, tangent to the base of the cone, either of which will
be the horizontal trace of a plane passing through the point
(E,E'), and tangent to the cone : hence there are two planes
that can be drawn, either of which will answer the conditions
of the problem.
Draw the tangent FG, and produce it to I : FI is the hori-
zontal trace of one of the tangent planes. The vertical trace
is determined by drawing through the vertex of the cone, or
any other point of the element of contact, a line parallel to the
horizontal trace : the point H, at which this line pierces the
vertical plane, being joined with I, determines IH, the vertical
trace. That part of it which is concealed by the vertical pro-
jection of the cone, is made broker (35). The line (AG, A'G')
is the element of contact.48
DESCRIPTIVE GEOMETRY.
PROBLEM XIX.
To draw a plane parallel to a given line and tangent to the sur-
face of a cone.
§ 95. PI. 7. Fig. 2. Let AGC be the horizontal, A'C'D' the
vertical projection of the cone, and (IH, TH') the given line.
Through the vertex of the cone let the line (AE, A'E') be
drawn parallel to the given line. This parallel is a line of the
required plane, since the plane must pass through the vertex
<A,A'), and be parallel to the line (IH, ITT). The point E, at
which the line (AE, A'E') pierces the horizontal plane, is a
point of the horizontal trace of the tangent plane, and two tan-
gent planes can be drawn answering the conditions of the
problem, since two lines can be drawn through E tangent to the
base of the cone.
Let either, as the tangent EG, be taken for the horizontal
trace of the tangent plane ; (AG, A'G') is the element of con-
tact ; and the point N', at which it pierces the vertical plane, is a
point of the vertical trace. The line (AE, A'E') being a line ot
the tangent plane, the point F', at which it pierces the vertical
plane, is a second point of*the vertical trace : hence F'N' is
the vertical trace of a plane parallel to (IH, I'H'), and tangent
to the surface of the cone. If we draw through E the tangent
EP, it will be the horizontal trace of the second plane which is
parallel to the given line (IH, I'H'), and tangent to the surface
of the cone. The vertical trace of this plane is easily con-
structed.
§ 96. This problem becomes impossible when the line
(AE, A'E'), which is drawn through the vertex of the cone and
parallel to the given line, passes within the surface : in this case,
it would pierce the base of the cone within the circle CDG.
If the parallel should become an element of the cone, the
problem would be possible, but would admit of one solution
only.
§ 97. The last three problems would have been constructedTANGENT PLANES.
49
ín nearly the same manner, had the surfaces been cylindrical
instead of conical. Indeed, we may consider the cylinder as
a cone whose vertex is at an infinite distance from the base ;
for, if the vertex of a cone be supposed to move from the base,
the rectilinear elements will make a less and less angle with
each other ; and when the vertex is removed to an infinite dis-
tance, these elements become parallel, and the cone becomes a
cylinder.
§ 98. In order to vary the constructions, and to apply the
principles used in drawing tangent planes to single-curved sur-
faces, so as to embrace the greatest variety of cases, the posi-
tions of the cylinders to which tangent planes are to be drawn,
will be so chosen as to require new applications of the prin-
ciples which have already been developed.
PROBLEM XX.
To draw a plane through a given point of a cylindrical surface,
tangent to the surface.
§ 99. PI. 7. Fig. 3. Suppose the cylinder to be a right
cylinder with a circular base, having its axis parallel to the ground
line. Let AEFB be the horizontal, and CLGD the vertical
projection of the cylinder, and I the horizontal projection of th^
point of the surface through which the tangent plane is to be
drawn.
If a perpendicular be erected to the horizontal plane at I, it
will pierce the surface of the cylinder in two points, both of
which are horizontally projected at I. Through this perpen-
dicular let a plane MHP be drawn perpendicular to the axis of
the cylinder. This plane will intersect the surface of the cylinder
in a circle, and H is the horizontal projection of its centre.
The two points in which the perpendicular at I intersects this
circle, are the two points in which it intersects the surface of the
cylinder.
Let this plane be revolved about its horizontal trace MP, till
it coincides with the horizontal plane. The centre of the circle,
4
#50
DESCRIPTIVE GEOMETRY.
in which it intersects the surface of the cylinder, falls at H':
with this point as a centre, and a radius equal to the radius of
the base of the cylinder, describe the circle ENFN'. But IN
is the revolved position of the perpendicular to the horizontal
plane at I ; therefore N' and N are the revolved positions of
the two points in which this perpendicular pierces the surface
of the cylinder.
Let it be required to draw the tangent plane through the
upper point, which in its revolved position is at N. Through
N draw the tangent ONM to the circle NEF ; this tangent is
the intersection, revolved, of the required tangent plane and
the plane MP. Let now the plane of the circle ENF be re-
volved back into its primitive position, the point M remains
fixed, being in the axis, and the point O describes the arc OO'
in the vertical plane ; the tangent line will, therefore, pierce the
planes of projection at M and O', which are consequently points
of the traces of the required plane. But, since the rectilinear
elements of the cylinder are parallel to the ground line, and the
plane must be tangent along an element, it follows that the plane
will be parallel to the ground line. Therefore, the parallels to
the ground line through M and O' are the traces of the re-
quired plane. If the plane had been drawn through the point
(1,1") its traces would have been constructed in a similar
manner.
PROBLEM XXI.
To draw aplane through a given point, without the surface of a
cylinder, tangent to the surface.
§ 100. PI. 7. Fig. 4. Let the axis of the cylinder be parallel
to the ground line, and the projections of the cylinder as repre-
sented in the figure ; and let (1,1') be the given point through
which the plane is to be drawn.
If through the point (1,1') any plane be drawn intersecting the
surface of the cylinder in a curve, and if through the same
point two lines be drawn tangent to this curve, each line will7
J. I .StolitSr.TANGENT PLANES.	51
be a line of a plane which can be drawn through the point (1,1'L
and tangent to the cylinder.
Let AP be the horizontal trace of a plane passing through the
point (1,1') and perpendicular to the axis of the cylinder; this plane
will intersect the surface of the cylinder in a circle whose centre
is horizontally projected at p, and vertically projected at p'.
Let this plane be revolved around AP till it coincides with the
horizontal plane ; (p,p'), the centre of the circle, falls at p", and
the point (1,1') at I". Describe the circle, and let the tangent
I"0" be drawn. Revolving this plane back into its primitive
position, the point A remains fixed, being in the axis ; and the
point B describes the arc BB' in the vertical plane about the
centre P : the tangent to the circle, therefore, pierces the
planes of projection at A and B', which are points of the traces
of the required tangent plane. But the traces are parallel to
the ground line ; therefore, the parallels to the ground line,
drawn through the points A and B', are the traces of the re-
quired plane.
After the counter revolution of the plane, the point O" is
horizontally projected at O, and vertically at O' : the lines drawn
through these points, parallel to the ground line, are the projec-
tions of the element of contact.
Had a second tangent line been drawn through the point I"
to the circle whose centre is p'\ it wTould have been a line of a
second plane through the given point (1,1'), and tangent to the
surface of the cylinder.
PROBLEM XXII.
To draw a plane parallel to a given line, and tangent to the
surface of a cylinder,
§ 101. PI. 8 Fig 1. We shall again use a right cylinder
with a circular base, and take its axis parallel to the ground
line. Let ABGD and EFG'R be the projections of the cylin-
der, and (IH, I'll') the given line to which the plane is to lie
parallel.
4*52
DESCRIPTIVE GEOMETRY.
Since a tangent plane to a cylinder is tangent along a recti-
linear element, it is parallel to the axis. The required plane
must then be parallel to the axis as well as to the line (IH, I'H').
A plane drawn through (1H, I'H'), and parallel to the axis
of the cylinder, is, therefore, parallel to the required tangent
plane.
But, since the axis of the cylinder is parallel to the ground
line, the plane parallel to it is also parallel to the ground line :
therefore, IT and H'Q', drawn through the points I and H',
parallel to the ground line, are the traces of this plane. If, for
a moment, we suppose the tangent plane to be drawn, and make
an intersection by a plane perpendicular to the axis of the
cylinder, this plane will intersect the cylinder in a circle, the
tangent plane in a line tangent to the circle, and the parallel
plane through (IH, I'H'), in a line parallel to that tangent.
Let MS be the horizontal trace of such a plane, and suppose
it to be revolved around this trace to coincide with the hori-
zontal plane. The centre of the circle, in which the cutting
plane intersects the surface of the cylinder, falls at P, and the
line of intersection with the parallel plane takes the position TQ :
now, if the line MP'F be drawn parallel to TQ and tangent to
the circle, it will represent, in its revolved position, the line
which would have been cut from the tangent plane had that
plane been drawn. The tangent MF is, therefore, a line of the
required tangent plane ; M is a point of its horizontal, and F'
a point of its vertical trace ; and as th 3 traces are parallel to
the ground line, the parallels drawn through these points are the
traces of the required plane.
As two lines can be drawn parallel to TQ and tangent to the
circle whose centre is P, it follows that two planes can be
drawn parallel to the line (IH, I'H'), and tangent to the cylinder.
§ 102. We cannot, in general, draw a plane through a given
line, and tangent to a single-curved surface ; for, in the conic
surface the tangent plane always passes through the vertex,
and if a plane be drawn through the given line and the vertex
of the cone, it will, in general, intersect the surface, though it
may be tangent to it. If the given line contain the vertex andTANGENT PLANES.	53
does not pass within the surface, a plane can always be drawn
through it tangent to the surface of the cone.
As a tangent plane to a cylindrical surface contains an element
of the surface, and as a plane can be drawn through two lines
only when they intersect, or are parallel, it follows that aplane
cannot be drawn through a right line, and tangent to a cylin-
drical surface, unless the line touch the surface or le parallel to
its rectilinear elements.	•
§ 103. The shortest distance between two lines in space, which
are not in the same plane, is readily found by means of the
cylinder and its tangent plane.
PI. 8. Fig. 2. Let (AB, A'B') be one of the lines, and (CD,
C'D') the other.
Draw a plane through the line (AB, A'B') parallel to the line
(CD, C'D'). This is done by taking a point (B,B') of the line
(AB, A'B'), and drawing through it a line parallel to (CD, C'D') ;
B'C' is its vertical and BF its horizontal projection, and AF is
the horizontal trace of the parallel plane. If now we suppose
(CD, C'D') to be the axis of a right cylinder with a circular
base, to which the plane just drawn shall be tangent, the radius
of the base will be equal to the distance between the axis of the
cylinder and tangent plane, which distance is equal to the dis-
tance between the given lines.
To find the radius of the base of this cylinder. Through the
axis (CD, C'D') let a plane be drawn perpendicular to the hori-
zontal plane ; its trace DCG intersects the trace of the parallel
plane at G, and these planes intersect in a line parallel to (CD,
C'D'). Let this plane be revolved around DG till it coincides
with the horizontal plane. Any point of tne axis of the cylinder
as (1,1'), falls in a perpendicular to, and at a distance from, the
axis of revolution DG, equal to its height above the horizontal
plane: making II" equal to this distance, and drawing Cl",
determines the revolved position of the axis. But the axis ot
the cylinder, and the line in which the vertical plane intersects
the parallel plane are parallel ; they are therefore parallel after
revolution ; hence GP, drawn parallel to Cl", is the revolved
position of this line of intersection.54
DESCRIPTIVE GEOMETRA.
At the point C let a plane be drawn perpendicular to the
axis of the cylinder ; its trace CL is perpendicular to CD, the
projection of the axis (49). This plane intersects the vertical
plane through the axis of the cylinder in a line perpendicular to
the axis at C. If then CH be drawn perpendicular to Cl",
the revolved axis, it will represent this intersection revolved
into the horizontal plane ; and II is one point of the inter-
section of the tangent plane and the plane perpendicular to
the axis of the cylinder. Let now the plane perpendicular
to the axis of the cylinder be revolved about its horizontal trace
LC till it coincides with the horizontal plane. The point L is
one point of the intersection of this plane and the tangent plane,
and remains fixed ; the point H, which is another point, falls at
H' a distance from C equal to CH, its distance in space : LH'
is then the intersection of these planes revolved into the hori-
zontal plane. Let CN be drawn perpendicular to LH' ; it will
be the revolved position of the perpendicular from C to the
tangent plane, and is therefore the radius of the base of the
cylinder. Making a counter revolution of the plane about LC,
the point N returns in a perpendicular to the axis ; and NE,
drawn parallel to CD, is the horizontal projection of the element
of contact.
Since (AB, A'B') is a line of the tangent plane, and is not
parallel to (CD, C'D'), it is consequently not parallel to the
element of contact; it will, therefore, intersect this element;
and E is the horizontal and E' the vertical projection of their
intersection. If from this point a line be drawn perpendicular
to the tangent plane, it will intersect and be perpendicular to
both the given lines : ED, E'D' are its projections, and CN is
it? length.&TANGENT PLANES.
55
CHAPTER VI.
OP TANGENT PLANES T.) SURFACES OF REVOLUTION.
PROBLEM XXIII.
To dì aw a plane tangent to a surface of revolutiont at a given
point of the surface.
§ 104. Let the surface of the paraboloid be the surface of
revolution to which the tangent plane is to be drawn. As the
planes of projection can be assumed at pleasure, let the hori-
zontal plane be taken perpendicular to the axis of the surface ;
the vertical plane will then be parallel to it.
PI. 9. Fig. 1. Let A be the horizontal and gg' the vertical
projection of the axis ; EHF the circle in which the surface is
intersected by the horizontal plane ; E'F' the vertical projection
of this circle ; and E'g*F' the vertical projection of the meridian
curve, whose plane is parallel to the vertical plane of pro-
jection.
Having made the projections of the surface, let the point C
be assumed for the horizontal projection of the point at which
the plane is to be drawn tangent. The vertical projection of
this point cannot be assumed, for its projecting lines must inter-
sect on the surface. To find it, erect at C a perpendicular to the
horizontal plane ; the vertical projection of the point in which
this perpendicular meets the surface is the point required.
Through this perpendicular draw the meridian plane I A, and
let it be revolved about the axis of the surface till it becomes
parallel to the vertical plane. In this revolution, C, the foot
of the perpendicular, describes the arc Cl) on the horizontal
plane, and D'D" is the vertical projection of the perpendicular
from its revolved position. The meridian curve having become
parallel to the vertical plane, its vertical projection is the curveTESCRIPTIVE GEOMETRY.
0(3
F/gF'; and the point D', in which it intersects the vertical pro-
jection of the perpendicular, is the revolved position of the point
at which the perpendicular to the horizontal plane at C pierces
the surface. Making the counter revolution, D' returns in the
arc of a horizontal circle, of which DC Is the horizontal and
D'C' tne vertical projection ; C' is, therefore, the vertical pro-
jection of the point of the surface of which C is the horizonta*
projection: hence, (C,C') is the point at which the plane is tj
be tangent to the surface. If a line be drawn tangent to the
meridian curve at the point (C,C'), and a second line be drawn
tangent at the same point to the horizontal circle passing
through (C,C'), the plane of these two tangents is tangent to
the surface at the point (C,C') (88).
If, when the meridian plane ICA is parallel to the vertical
plane, the line A'D'L/ be drawn tangent to the meridian curve
at the point D', this line, being in the plane of the curve, re-
volves back with it, and continues tangent to the curve at the
point D'. In this counter revolution the point A', in which the
tangent intersects the axis of the surface, remains fixed ; and
when the point (D,D') comes into the position (C,C'), A'CT is
the vertical and AC I the horizontal projection of the tangent
line. The line (AC, A'C') being a line of the required plane,
the point I, in which it pierces the horizontal plane, is a point
- of its horizontal trace. If at (C,C') a right line be drawn tan-
gent to the horizontal circle passing through this point, it wall be
a horizontal line, and perpendicular to the radius of the hori-
zontal circle drawn through (C,C') ; therefore its horizontal
projection CM is perpendicular to AC, the horizontal projection
of the radius ; and its vertical projection C'M' is parallel to the
• ground line (27). Since this line is horizontal, and a line of the
tangent plane, the horizontal trace of the tangent plane is par-
allel to it, and, consequently, to its horizontal projection (30).
The line IN, therefore, drawn through the point I, parallel to
CM, is the horizontal trace of the required plane. The tan-
gent to the horizonta] circle at (C,C') pierces the vertical
plane at M' : hence, NM' is the vertical trace of the tangent
planeTANGENT PLANES.
57
§ 105. We may remark, that the tangent to the horizontal
circle at (C,C') is perpendicular both to the radius passing
through (C,C') and to the perpendicular demitted from this point
to the horizontal plane : hence, it is perpendicular to the plane
of these two lines ; that is, to the meridian plane passing through
the point (C,C'). But the tangent plane contains this tangent
to the horizontal circle ; therefore, it is perpendicular to the
meridian plane : and as the same may be shown for any posi-
tion of the point of contact, we conclude that a tangent plane
to a surf ace of revolution is perpendicular to the meridian plane
passing through the point of contact.
§ 106. The construction just made answers for all surfaces
of revolution, with this slight difference, that the perpendicular
to the horizontal plane at C may pierce the surface in several
points. We find the vertical projections of these points by the
methods already shown : assume either of them for the one at
which the plane is to be tangent, and make the construction as
in the last problem.
§ 107. Second Solution. Having determined the vertical
projection of the point, as before, let the meridian plane passing
through it be revolved about the axis of the surface till it becomes
parallel to the vertical plane, and draw the tangent A'D'L',
whichfc in this position pierces the horizontal plane at L. Let
the meridian plane be revolved about the axis of the surface ;
the meridian curve generates the surface, and the tangent line
the surface of a cone tangent to the surface of revolution. The
point (A,A') is the vertex of the cone, and LIQ is its base.
The curve of contact of the cone and surface is the horizontal
circle described by the point (D,D') ; for, the point (D,D'),
throughout the revolution, is common to both the surfaces.
Every plane tangent to this cone will be tangent to the sur-
face of the paraboloid, and the plane which is tangent along the
element passing through (C,C') will be tangent to the surface
at the point (C,C'). This element pierces the base of the cone
at I ; IN, tangent to the base of the cone, is the horizontal .race
of the tangent plane ; and NM' is its vertical trace. This isr
the same plane as before determined.58
DESCRIPTIVE GEOMETRY.
§ 108. If two surfaces of revolution. having a common axis%
are tangent to each other, their curve of contact is the circum-
ference of a circle whose plane is perpendicular to the axis.
For, if a plane be drawn through the common axis, it will inter-
sect each surface in a meridian curve (79), and these curves will
be tangent to each other (85). If, then, the plane of these curves
be revolved around their common axis, each will generate the
surface to which it belongs ; the point of contact will generate
the circumference of the circle of contact of the surfaces ; and
the plane of the circle is perpendicular to the axis, since the
axis is perpendicular to all the radii. If one of the surfaces of
revolution be that of a cylinder or cone, similar reasoning would
show, that the curve of contact is the circumference of a circle
whose plane is perpendicular to the common axis.
§ 109. If through any point in the plane of a circle a line be
drawn tangent to the circle, and through the same point a line
be drawn to the centre ; if the plane be then revolved about
this latter line as an axis, the circumference of the circle will,
generate the surface of a sphere, and the tangent line the sur-
face of a cone which will be tangent to the sphere. The
elements of the cone, intercepted between the vertex and the
points in which they touch the sphere, are equal to each other;
and the line drawn to the centre of the sphere is the axis of
the cone. Hence, if any point be taken without the surface of
a sphere, and through this point a system of lines be drawn tan-
gent to the sphere, they form the surface of a right cone having a
circular base ; and the line drawn from the vertex to the centre
of the sphere is the axis of the cone.
§ 110. It has been shown, that a plane which is passed
through a given point, without a single-curved surface, and tan-
gent to the surface, is determined in position. Similar condi-
tions do not determine the plane when the surface is of double
curvature. For, through the point, conceive any number of
planes to be passed intersecting the surface in curves. From
the assumed point let lines be drawn tangent to the curves ;
these lines will form the surface of a cone tangent to the double-
curved surface. There may be an infinite number of planesTANGENT PLANES.
59
drawn tangent to this cone, each of which w.ll be tangent to the
irface ; therefore,from any point without a double-curved surface,
M infinite number of planes may be drawn tangent to the surface.
$ 111. It has also been shown, in Chap. V., that a plane is
letermined in position when it is drawn parallel to a given line,
and tangent to a single-curved surface. Similar conditions do
not determine the plane when the surface is of double curva-
ture, For, let the surface be intersected by any number of
planes parallel to the given line, and let tangents be drawn to
these curves also parallel to the given line ; this system of par-
allels forms the surface of a cylinder whose right-lined elements
are parallel to the given line, and this cylinder is tangent to the
double-curved surface. Every plane tangent to this cylinder
is also tangent to the double-curved surface and parallel to the
given line ; and as an infinite number of planes can be drawn
tangent to the cylinder, it follows that an infinite number of
planes may be drawn parallel to a given line and tangent to a
double-curved surface.
§ 112. It has been shown, (102), that a plane cannot, in
general, be drawn through a given line and tangent to a single-
curved surface ; but it is always possible to draw a plane
through a given line which shall be tangent to a double-curved
surface, provided the line does not meet the surface. For, sup-
pose the surface to be circumscribed by a tangent cylinder,
whose right-lined elements are parallel to the given line. A
plane can be drawn through the line, and tangent to this
cylinder (102). But this plane will also be tangent to the
double-curved surface : hence, a plane can always be drawn
through a given line and tangent to a double-curved surface
PROBLEM XXIV.
To draw aplane through a given line, and tangent to the surjace
of a sphere.
§ 113. PI. 9. Fig. 2. Let the centre of the sphere be in the
ground line at C, and (AB, Al the line through which the
plane is to be drawn.GO
DESCRIPTIVE GEOMETRY.
If any poi it of this line be made the vertex of a cone tangent
to the sphere, and a plane be drawn through the line and tan-
gent to the cone (102), the plane so drawn will be tangent to
the sphere, and will consequently be the plane required.
• Let (A,A') be the point chosen for the vertex of the cone.
Let the lines Aí and AL be drawn tangent to the circle of in-
tersection of the sphere and horizontal plane ; join the points I
an i L . nd draw AC to the centre of the sphere. The line
A / bisects the angle I AL, and is perpendicular to IL at the
point P. If the horizontal plane be supposed to revolve about
AC as an axis, the semicircle N'IN will generate the sphere,
and the right-angled triangle I PA a cone tangent to it ; and
since the surfaces have a common axis, the plane of theii
circle of contact is perpendicular to that axis ; that is, to the
horizontal plane, since the axis AC is a line of that plane. The
line HID' is the horizontal trace of the piane of the circle oi
contact of the cone and sphere ; and if this plane be revolved
about its trace HL till it coincides with the horizontal plane,
the circle of contact will be represented by the circle IG"L.
If we find the point in which the line (AB, A'B') pierces this
plane, and from that point draw a line tangent to the circle of
contact of the cone and sphere, this tangent wi 1 be the trace,
on that ; lane, of a plane which will contain the given line, and
be tangent to the cone. To find this point, produce AB, the
horizontal projection of the given line, till it meets the trace at
D', and erect the indefinite perpendicular D'F in the plane
HI D' : the line (AB, A'B') will pierce the plane somewhere in this
perpendicular (37), and therefore the point will be horizonlally
projected at D'. Through D' draw the indefinite perpendicular
DD' to the ground line, and produce A'B', the vertical projec-
tion of the me, till it meets the perpendicular at E. The point
E is the vertical projection of the point of which D' is the hori-
zontal projection (13), and DE is the distance of the point above
the horizontal plane (14). If, then, D'F be made equal to ED,
F will be the point in which the line pierces the plane HID'.
Through F draw the tangent FG"H to the circle IG 'L ; it will
be the uvice, on the plane of the cones base, of a plane containTANGENT PLANES.
61
mg the given line and tangent to the cone. The tangent line
pierces the horizontal plane at H ; and since the tangent plane
passes through the vertex A, AH is its horizontal trace. The
vertical trace, on the vertical plane of projection, is found by
drawing through any point, as (0,0'), of the line (AB, A'B'), a
parallel to the horizontal trace ; the point Q', in which it pierces
the vertical plane, is a point of the vertical trace ; and since
B' is also a point, B'Q' is the vertical trace of the tangent plane.
To find the projections of the point of contact. When the
plane of the cone’s base is revolved into the horizontal plane,
the point of contact is ,at G". In the counter revolution, G"
describes the circumference of a circle whose plane is perpen-
dicular to the axis HL ; and when the plane of the cone’s base
becomes perpendicular to the horizontal plane, G is the hori-
zontal projection of the point of contact. The height of the
point of contact above the horizontal plane is equal to GG" ;
drawing through G an indefinite perpendicular to the ground
line, and making g*G' equal to GG", gives G' for the vertical pro-
jection of the point of contact.
As two lines can be drawn from the point F tangent to the
circle IG"L, it follows that two planes can be drawn through
the given line and tangent to the sphere. As it wrould confuse
the figure to draw them both, only one is determined ; the other
is left to be constructed by the student.
§ 114. All lines drawn tangent to a sphere at any point of
the surface are perpendicular to the radius passing through that
point : hence, the tangent plane which contains these lines is
perpendicular to the radius passing through the point of con-
tact. The projections of the radius drawn through the point
of contact pass through the projections of this point, and are
respectively perpendicular to the traces of the tangent plane
(49). We can verify the construction just made, by drawing
CG, and examining whether or not it be perpendicular to AH ;
CG' ought also to be perpendicular to the vertici trace Q'B.62
DESCRH’TIVE geometry.
PROBLEM XV.
2b draw a plane through a given right line and tangent to tht
surface of a sphere, by means of two cones.
§ 115. PI. 10. Fig. 1. Let (AB, A'B') be the given line,
C and C' the projections of the centre of the sphere, and the
circles described around these points as centres, the projec-
tions of the sphere.
If any two points of the given line be taken as the vertices
of two cones which are drawn tangent to the sphere, a plane
tangent to these cones will contain the given line and be tan-
gent to the sphere.
First, through the centre of the sphere let the plane B'C' be
drawn parallel to the horizontal plane, and take the point (B,B'),
in which the given line meets this plane, for the vertex of the
first cone. Through the centre of the sphere let the plane DC
be drawn parallel to the vertical plane, and take the point (0,0%
in which the given line pierces this plane, for the vertex of the
second cone.
If from the point (B,B') we conceive two lines to be drawn
in the horizontal plane B'C', and tangent to the great circle in
which this plane intersects the sphere, they will be those right-
lined elements of the tangent cone which lie in this plane, and
will be horizontally projected in the tangent lines BE and BF
(90). The axis of the cone is horizontal, and passes through
the centre of the sphere (109) ; the plane of the circle of con-
tact being perpendicular to the axis is perpendicular to the
Horizontal plane, and FE is its horizontal projection. The axis
of the cone, whose vertex is (D,I)'), is parallel to the vertical
plane of projection ; and the plane of the circle of contact, being
perpendicular to the axis, is consequently perpendicular to the
vertical plane.
We wish now to draw a tangent plane to these two cones.
The planes of their bases intersect in a right line, of which EF
is the horizontal projection and GH the vertical projection (24).TANGENT PLANES.
63
The points in whi^h this right iine pierces the surface of the
sphere, are the points in which the circumferences of the circles
of contact of the cones and sphere intersect, and are, therefore,
common to the surface of the sphere and to the surfaces of
both cones. If at either of these points a plane be drawn tan-
gent to the sphere, it will be tangent to both cones, will contain
the given line, and will therefore be the plane required.
To find the points in which this line of intersection pierces
the surface of the sphere, it will first be necessary to find tne
point in which it pierces the horizontal plane B'C', and secondly,
the point in which it pierces the vertical plane DC. It pierces the
horizontal plane B'C' in the line in which this plane intersects
thè plane of the base of the cone whose vertex is (B,B') ; that
is, in the line of which EF is the horizontal projection. It also
pierces the horizontal plane B'C' in its intersection with the
base of the cone whose vertex is (D,D') ; that is, in a line per-
pendicular to the vertical plane at a'. The horizontal projec-
tion of this line is n a ; therefore, the line in which the bases of
the cones intersect, pierces the horizontal plane B'C' at the point
(a,a'). The base of the cone whose vertex is (D,D') intersects
the vertical plane DC in a line of which GH is the vertical pro-
jection ; and the base of the cone whose vertex is (B,B') inter-
sects the same plane in a line of which ƒ is the horizontal and
f'g' the vertical projection ; therefore, the line of intersection
of the cones’ bases pierces the plane DC in the point (ƒ,#').
Let the plane of the base of the cone whose vertex is (B,B')
be revolved about its intersection with the plane B'C' till it be-
comes parallel to the horizontal plane. The point (a,a% being
in the axis, remains fixed ; the point (ƒ,£*') falls in a perpendicular
to EF, and at a distance from ƒ equal tof’g\ its height above the
plane B'C'. Making^* equal to f'g\ and drawing a g, a g is the
revolved position of the intersection of the bases of the cones.
If a. circle be described on EF as a diameter, it will be the base
of the cone whose vertex is (B,B') in its revolved position. The
intersection of the cones’ bases intersects this circle in the points
d" and c" ; these are the revolved positions of the points in
which it pierces the surface of the sphere. Making the counter64
DESCRIPTIVE GEOMETRY.
revolution about EF, the points d" and c" describe arcs of ver-
tical circles about the axis, and are horizontally projected at d
and c. Since these points are in the base of the cone whose
vertex is (D,I)'), they are vertically projected at the points d!
and c'. A plane drawn tangent to the sphere, through either
of the points, will contain the given line ; and conversely, a
plane drawn through either of the points and the given line, will
be tangent to the sphere. The lines Cc, C'c' are the projections
of the radius of the sphere passing through the point (c,c').
But the horizontal trace of the tangent plane must pass through
A, the point in which the given line pierces the horizontal
plane, and be perpendicular to Cc ; therefore AN, drawn per-
pendicular to Cc, is the horizontal trace of the plane tangent to
the sphere at the point (c,c'). The point I, at which the given
line pierces the vertical plane, is a point of the vertical trace ;
hence IP, drawn perpendicular to C'c', is the vertical trace of
the plane which contains the given line, and touches the sphere
at the point (c,c').
The second plane, which is tangent to the sphere at the point
(d,d'), is determined in another way, thus : through the point
(d,df) let a line be drawn parallel to the given line ; its projec-
tions are parallel to the projections of the given line (30), and
it pierces the horizontal plane of projection at the point Q ; AQ
is therefore the horizontal trace, and IMV the vertical trace of
the second tangent plane to the sphere which contains the
given line. The projections of the radius passing through
the point (d,d') should be respectively perpendicular to these
traces.
§ 116. If the centre of the sphere were placed in the ground
line, and the points at which the given line pierces the planes of
projection taken for the vertices of the tangent cones, the con-
struction would not differ materially from the one already
made. Let the construction be made when the sphere has this
position.
§ 117. Second method by which the points of contact may be de-
termined. If through the given line we conceive two planes to
be passed tangent to the sphere, a plane drawn through theTANGENT PLANES.
65
centre of the sphere and perpendicular to these tangent planes
will be perpendicular to their intersection ; that is, to the given
line. The perpendicular plane will also contain the radii drawn
through the points of contact, will intersect the sphere in a groat
circle, and the tangent planes in two lines tangent to this circle.
The tangent lines will intersect at the point in which the per-
pendicular plane is pierced by the intersection of the tangent
planes ; that is, where it is pierced by the given line.
If, therefore, through the centre of the sphere a plane be
passed perpendicular to the given line, and the point in which it
cuts the given line determined, and through this point two
lines be drawn tangent to the great circle in which the plane
intersects the sphere, they will be lines of the required tan-
gent planes ; the points in which they touch the circle are the
points at which the required planes will be tangent to the
sphere.
PI. 10. Fig. 2. Let C and C' be the projections of the centre
of a sphere ; (AF, A'F') the given line. First, to draw the plane
through the centre of the sphere, and find the point at which it
is pierced by the given line. Since the required plane through
the centre of the sphere is to be perpendicular to the given line,
it will be perpendicular to both the projecting planes of the line.
Through the centre (C,C') let two lines be drawn, one perpen-
dicular to the plane which projects the given line on the hori-
zontal plane, the other perpendicular to the plane which projects
the given line on the vertical plane ; these are lines of the re-
quired plane, and determine its position (20). The two points
in which they pierce the plane that projects the given line on
the horizontal plane, determine the intersection of this project-
ing plane with the plane through the centre of the sphere ; and
the point in which this intersection meets the given line, is the
point in which the given line pierces the plane through the
centre of the sphere.
The line CD, perpendicular to AD, is the horizontal projec-
tion of the line drawn through the centre (C,C'), and perpen-
dicular to the plane which projects the given line (AF, A'F') on
the horizontal plane. The line C'D' is the vertical projection of
566
DESCRIPTIVE GEOMETRY.
this line, and (D,D') is the point at which it pierces the project-
ing plane.
The line drawn through (C,C') perpendicular to the plane
which projects the line (AF, AT') on the vertical plane, being
parallel to the vertical plane, CE is its horizontal projection, and
E is the horizontal projection of the point in which it pierces
the plane which projects the given line (AF, A'F') on the hori-
zontal plane. The vertical projection of this line is found by
drawing through C' the line C'E' perpendicular to AT' ; the
point E', in which this projection intersects the perpendicular
to the ground line through E, is the vertical projection of the
point in which the second line through (C,C') pierces the plane
which projects the line (AF, AT') on the horizontal plane.
Therefore D'E' is the vertical projection of the line in which
the plane drawn through the centre of the sphere intersects the
projecting plane of the line (AF, A'F'). This intersection
meets (AF, AT') in the point (F,F') ; hence (F,F') is the point
at which the plane through the. centre of the sphere cuts the
given line. Let the perpendicular plane be revolved about
(CD, C'D'), its intersection with the horizontal plane passing
through the centre of the sphere, till it coincides with this
plane. The circle in which the plane intersects the sphere
becoming parallel to the horizontal plane, is projected into the
circle whose centre is C : the point (F,F') falls at G ; DG being
made equal to F'g', the hypothenuse of a triangle whose base
Ig' is equal to DF. From the point G draw the two tangents
GNO" and GL" : O" and L" are the revolved positions of the
points of contact. In the counter revolution of the plane, the
point N remains fixed, the point 07 describes the arc of a circle
perpendicular to the axis, the point G returns to the point
(F,F'), FNO is the horizontal projection of the tangent, and O
is the horizontal projection of the point of contact. The point
F is vertically projected at F', and N at N'; hence, F'N'is the
vertical projection of the tangent line, and O' the vertical pro-
jection of the point of contact.
To find the projections of the other point of contact. As
he line GL" does not intersect, on the paper, the axis DC aboutTANGENT PLANES.
67
which the plane is ^evolved, a different construction from the
one just made becomes necessary. In the line GL" take any
point, as A", and draw the line DA" ; and note the pointk” in which
DA" intersects the line GO". In the counter revolution the
point k" describes the arc of a circle perpendicular to CD, and
when the revolution is completed, is horizontally projected at k ;
and since D remains fixed, Dkh is the horizontal projection of
the line which passed through A" and A". In the counter revo-*
lution the point h" describes the arc of a circle of which A"A
is the horizontal projection. Hence A is the horizontal pro-
jection of A" when it is revolved into its true place ; FAL is
then the horizontal projection of the tangent GL", and L is the
horizontal projection of the point of contact. The vertical
projection of the point of which k is the horizontal, must be
found in the line OT', and also in a perpendicular from k to the
ground line ; hence it is at A'. Since the point of which D is
the horizontal projection is vertically projected at D', D'/c'A' is
the vertical projection of the line of which DAA is the horizontal
projection. Drawing from A a perpendicular to the ground
line determines A', the vertical projection of the point of which
h is the horizontal projection. The line F'A'L' is therefore the
vertical projection of the line of which FL is the horizontal
projection, and L' the vertical projection of the point of contact
determined by the tangent (FL, F'L').
The methods of constructing the tangent planes after the
points of contact are found, have already been shown. If the
centre of the sphere were taken in the ground line, the con-
struction would be similar in evéry respect to the one already
given.
§ 118. If we suppose a cylinder, having its axis parallel to the
given line, to be drawn tangent to the sphere, the curve of con-
tact will be a great circle whose plane is perpendicular to the
axis of the cylinder or given line. If two planes be drawn
through the given line and tangent to the cylinder (102), they
will also be tangent to the sphere. The point at which the
given line pierces the plane of the circle of contact of the
cylinder and sphere, is the point from which the tangent lines
5*68
DESCRIPTIVE GEOMETRI'.
to the base of the cylinder are drawn ; for this point is common
to the traces, on the base of the cylinder, of both the tangent
planes. If the centre of the sphere be in the ground line, the
construction is somewhat simplified, and the reader will find it
an interesting problem to draw the tangent planes when the
sphere has this position. The general problem, to draw a plane
through a given line, and tangent to a surface of revolution, is
solved in the Complement.
CHAPTER VII.
OP THE INTERSECTIONS OF CURVED SURFACES AND PLANES ; OF
TANGENT LINES TO THE CURVES OF INTERSECTION ; AND OF
THE DEVELOPMENT OF SURFACES ON PLANES.
§ 119. The intersection of a curved surface by a plane is, in
general,determined by intersecting the curved surface and plane
by auxiliary planes, so chosen that the lines in which they
intersect the surface and plane shall intersect each other;
the points in which the right lines cut out of the plane intersect
the curves cut out of the surface, will be points of the intersec-
tion of the surface and plane. The auxiliary planes should be
so taken as to intersect the surface in its most simple ele-
ments, recollecting that the right line is a more simple ele-
ment than the circle, and the circle than either of the conic
sections.
§ 120. A tangent line to the curve of intersection of a plane
and surface is contained in the plane of the curve (67) ; it is
also contained in the tangent plane to the surface at the point
(88) : hence it is the intersection of these planes. If therefoie,
it be required to draw a line tangent to the curve of intersection
of a plane and surface, we have only to dram a plane tangent
to the surface at the point, and determine its intersection with
the cutting plane.INTERSECTIONS OF SURFACES.
69
§ 121. A plane being tangent to the surface of a cylinder along
one of its rectilinear elements (87), if the cylinder be rolled on
this plane, it will continue to be tangent to it, and the consecutive
right-lined elements will successively come into contact with the
plane. When it has rolled once over, every element, that is,
the whole surface of the cylinder, will have been in contact with
the plane. The portion of the plane touched during the revo-
lution is equal to the surface of the cylinder, and is called the
development of that surface.
§ 122. A plane is tangent to a cone along a right-lined ele-
ment. If the vertex of the cone remain fixed, and the cone
be rolled around on its tangent plane, its elements will succes-
sively coincide with the plane ; and after they shall all have
coincided, the part of the plane included between the extreme
lines is called the development of the surface of the cone.
§ 123. If we suppose the cylinder and cone to be limited
by planes, the curves in which these planes intersect the surfaces
become lines on the developments of the surfaces.
§ 124. Double-curved surfaces cannot be developed. For,
as a plane touches a double-curved surface in a point, if the
surface be rolled around on its tangent plane, its successive
contacts will form a line ; hence, the surface will not develop
itself on a plane. There exists, therefore, this striking dif-
ference between single-curved surfaces and double-curved sur-
faces : the former can he developed on aplane ; the latter cannot
This difference has been made, by some authors, the basis of
classification of curved surfaces ; arranging into one class the
developable surfaces, and those which are not dev^boable
into another.70
DESCRIPTIVE GEOMETRY.
PROBLEM XXYI.
To find the intersection of a plane with the surface of a right
cylinder having a circular base ; to draw a tangent line to
this curve at any point ; to find the curve and its tangent in
their own plane ; and to develop the cylindrical surface.
§ J25. PI. 11. Fig. 1. Let the horizontal plane be taken
perpendicular to the axis of the cylinder, and the vertical plane
of projection at right angles to the cutting plane. The circle
AgBfis the horizontal projection of the cylinder; EA'B'Zf is
its vertical projection, and (DE, DE') is the intersecting plane.
Let the surface of the cylinder and the cutting plane be inter,
sected by a system of auxiliary planes parallel to the axis of the
cylinder and perpendicular to the vertical plane : FC, ab,fig*
and k h are the horizontal traces of such planes. These planes in-
tersect the plane (DE, DE') in lines which are perpendicular to
the vertical plane at the points D', d\ &c. ; the points in which
these lines intersect the elements cut out of the surface of the
cylinder are points of the curve of intersection. Since the
cylinder is a right one, and the curve sought lies on its surface,
the curve is horizontally projected into the circle AgBf; but,
as FC is the horizontal projection of the line which is perpen-
dicular to the vertical plane at D', the points n and C are the
horizontal projections of two points of the required curve, and
D'is their vertical projection. The points a, b, ƒ, g, k, and h
are the horizontal projections of other points of the curve deter-
mined in the same manner ; their vertical projections are seen
byr inspecting the figure. Let the plane AB be drawn through
the axis of the cylinder and perpendicular to the cutting plane :
it intersects the cutting plane in a line which divides the curve
symmetrically ; this line, therefore, passes through the centre,
and is, furthermore, the longest diameter of the curve : it is
called the transverse axis, and the points (A,E') and (B,G) in
which it meets the curve, the vertices of the transverse axis.in
/ /INTERSECTIONS OF SURFACES.
71
To draw a tangent line to this curve at any point, as (C,D').
Pass a plane tangent to the cylinder along the element con-
taining the point (C,D') ; the intersection of this tangent plane
and the cutting plane is the tangent line required (120). The
line NC is the horizontal trace of the tangent plane; it is also
the horizontal projection of the tangent line, and N'D' is its ver-
tical projection.
To find the curve in its own plane. Let the plane of the
curve be revolved around the transverse axis till it becomes
parallel to the vertical plane ; it will, from this position, be
projected on the vertical plane in its true dimensions. The ver-
tices (B,G) and (A,E') remain fixed, being in the axis ; the points
(C,D'), («.D'). (b,df), (0/^)» continue at their respective dis-
tances, DC, Dn,db, and d a, from the axis ; therefore, making
in the vertical plane the lines D'C', DV, d'b\ and d’d respect-
ively equal to these lines determines points of the curve. Hav-
ing found a sufficient number of points, let the curve be de-
scribed through them ; this curve is the curve of intersection
of the cylinder and plane. The tangent line to the curve at the
point (C,D') intersects the transverse axis produced at (N,N' );
and as this point remains fixed during the revolution, the tan-
gent line assumes the position N'C'.
To develop the surface of the cylinder, suppose it to be so
placed on the plane of the paper that the element (A, A'E')
shall have the position AE' (Fig. 1. n) ; the plane of the paper
will then be tangent to the cylinder along this element. Let
the cylinder be divided into two equal parts, by a plane passing
through the element of contact and perpendicular to the plane of
the paper; and let one half of the cylinder be rolled out to-
wards B', the other towards B. The base of the cylinder being
perpendicular to the plane of the paper, its circumference will
be developed into the right line BAB'. From the point A lay
off Ak equal to the arc Ak on the base of the cylinder, and at
the point k erect a perpendicular to BB' equal to the height
above the horizontal plane of that point of the curve which is
horizontally projected at k. Making kf equal to the arc k f>
fo fa, an=an, nB'=wB, and laying off on the other side of72
DESCRIPTIVE GEOMETRY.
the point A the distances Ah, hg, &c., respectively equal to the
corresponding arcs on the base of the cylinder, erecting at all
these points perpendiculars to the line BB', making these per-
pendiculars equal to their corresponding elements, and drawing
a curve through their extremities, gives the curve of intersec-
tion on the development.
To find the position of the tangent line. When the point
C comes into the line BB', the element through C comes into
the plane of the paper. But the plane of the paper being con-
stantly tangent to the cylinder, the tangent plane through any
element will coincide with the plane of the paper at the moment
the element comes in contact with it. But the subtangent N"C,
being perpendicular to the element through C, falls in the line
BB' when the element through C comes into the plane of the
paper. But the tangent line pierces the plane of the base of
the cylinder at N" ; therefore, laying off CN equal to CN'
and drawing NC, determines the tangent line on the develop-
ment.
§ 126. If a right line be tangent to a curve in space, and the
curve be developed on a plane, the line will be tangent to the de-
veloped curve. For, suppose the surface of a cylinder to be
passed through the given curve : the right line being tangent to the
curve, passes through twro consecutive points (65), and the ele-
ments of the cylinder passing through these points are also con-
secutive ; when the surface of the cylinder is developed, these
elements are consecutive lines, their extremities are consecutive
points of the developed curve, and a right line passing through
these points is tangent to the curve. But this line occupies the
position which the tangent line in space assumes, since it passes
through the two points which fix the position of the tangent in
space.
§ 127. The surface of any right cylinder may be developed
in the same w'ay as we have developed the right cylinder with
a circular base. For, the plane of the base being perpendicular
to the tangent plane on which the development is made, the
base of the surface will be developed into a right line ; and by
laying off on this right line parts of the base equal to the dis-INTERSECTIONS OP SURFACES.	73
tances between the elements, we shall obtain the development
of the surface.
§ 128. The finding a right line equal to the length of a given
curve is called the rectification of the curve. In rectifying a
curve we cannot, of course, take the exact lengths of the small
arcs, but must use their chords instead of them. The smaller
the arcs are taken, the nearer will the chords coincide with
them, and, consequently the nearer will the right line, which
is the sum of these small chords, be equal to the length of the
curve which it is taken to represent.
PROBLEM XXVII.
To find the intersection of aplane with the surface of a right cone
having a circular hase ; to draw a tangent line to the curve ;
to find the curve in its own plane ; and to develop the surface
of the cone.
§ 129. PI. 11. Fig. 2. Let (C,C') be the vertex of the cone ;
FPEI the horizontal projection of the cone ; C'P'Q' its vertical
projection ; and (AB, AB') the cutting plane to which the ver-
tical plane of projection is taken at right angles.
Let the cone and plane be intersected by a system of planes
through the vertex and perpendicular to the vertical plane of
projection : they will intersect the cone in right-lined elements,
and the cutting plane in right lines ; the intersections of these
latter lines with the elements are points of the curve. Let
C'D be assumed for the vertical trace of one of these planes :
DF, perpendicular to the ground line, is its horizontal trace
(31) ; this trace intersects the circumference of the cone’s base
in the points E and F ; CF and CE are the horizontal projec-
tions of the elements in which the plane intersects the surface
of the cone, and C'D is their vertical projection. The line in
which the auxiliary plane intersects the cutting plane (AB, AB')
being perpendicular to the vertical plane, is vertically projected
at ƒ', and fg is* its horizontal projection. The points g and ƒ74
DESCRIPTIVE GEOMETRY.
in which the horizontal projection of this line intersects the hori-
zontal projections of the elements before found, are the hori-
zontal projections of two points of the required curve, and f' is
their vertical projection. The points (njri), (mjri), (G,G'), &c.
are found in a similar manner. The plane PaCb, perpendicular
to the cutting plane and containing the axis of the cone, deter-
mines the transverse axis of the curve («6, a'b') ; the points
(«,«') and (<b,b') are the vertices of the axis. Having deter-
mined as many points as are necessary, let the curve a g bn be
described through them : this is the horizontal projection of the
required curve, and a'b' is its vertical projection.
To draw a tangent line to the curve at any point, as (G,G').
Let a plane be drawn tangent to the cone along the right-lined
element, passing through the point (G,G'). The line NI, drawn
tangent to the base of the cone, is the horizontal trace of the
tangent plane ; N is one point in which the latter plane inter-
sects the cutting plane,and (G,G') is another point: hence NG
is the horizontal projection of the tangent line, and its vertical
projection is AB', the vertical trace of the cutting plane.
To find the curve and tangent in their own plane. Let the
plane of the curve be revolved about its vertical trace till it
coincides with the vertical plane : the points of the curve will
fall at their respective distances from the axis (10) ; that is, the
distances of their horizontal projections from the ground line
(13). Drawing through the points b' and a' perpendiculars to
AB', layingoff b'b" and da" respectively equal to the distances of
the points ¿/and a from the ground line, determines the positions
of the vertices and of the transverse axis after the plane of the
curve is revolved to coincide with the vertical plane. In the
same manner the points n', ƒ", g' and m" are determined ; and
through these points the curve is described. The position oí
the tangent is easily found ; for the point N falls at N', and
joining this point with G" determines the tangent line N'G".
To develop the surface of the cone, and trace on the develop-
ment the curve in which it is intersected by the plane (AB,
AB'). Let the development be made on the plane which is
tangent to the cone along the element (CP C'P'). SupposeINTERSECTIONS OP SURFACES.
75
the cone to be placed on the plane of the paper, the vertex at
C (Fig. 2. ?i), the point (P,P') at P ; the plane of the paper is
then tangent to the cone along the element (CP, C'P'). Let us
suppose the cone to be divided by a plane passing through this
element of contact perpendicular to the plane of the paper, and
let one half of it be developed from P towards Q', the other from
P towards Q. As all points of the circumference of the base
are equidistant from the vertex, they will in the development
be equidistant from the point C, and will consequently be
found in the circumference of a circle described with the centre
C and radius CP, equal to C'P' the slant height of the cone.
Let a part of this circle be described. From P lay off PE
equal to the arc PE of the base of the cone ; also EA'=EÆ',
h'k—h'k', kQ/=k'Q ; and on the other side of the point P lay
off PF equal to the arc PF of the base of the cone, FA=FA,
hl=hl, IQ'=IQ, Through C and the extremities of these
arcs let lines be drawn ; these lines are the several positions
which the elements of the cone take on the development of its
surface.
To trace the curve. If the distance of each point of the
curve from the vertex of the cone be ascertained and laid off
from the point C, the curve traced through these points will be
the curve required. On the element CP lay off Ca=CV ; on
CF, C/=CV; on Ch, Cn=C'p; on Cl, CG=C <7 ; on CQ',
C6=C'fc' : lay off equal distances on the corresponding elements
on the other side of CP; the curve bGnb, drawn through these
points, is the curve sought.
To find the position of the tangent line. The line IN (Fig. 2)
is perpendicular to the element of the cone passing through
(G,G') ; and when this element takes the position CGI (Fig.
2. n) on the development of the surface, the tangent plane
becomes the plane of the paper, and the line IN (Fig. 2) pre-
serving its position with Cl, is tangent to the curve Q'PQ ;
hence, if we make NI=NI, and join N and G, GN will be the
tangent line.
§ 130. The curves described in the note to Art. 63, and
named conic sections, can be obtained by intersecting a right76
DESCRIPTIVE GEOMETRY.
cone with a circular base by planes having different positions
with its elements. If the cutting plane be oblique to the plane
of the base, but intersect all the elements on the same nappe,
as in the last example, the curve will be an ellipse. If the
plane be parallel to one of the elements, the curve of intersec-
tion is a parabola. If it make a greater angle with the plane
of the base than the elements of the cone make with the base,
it will intersect both nappes of the surface, and the curves are
opposite hyperbolas. To prove that these curves have the
same properties as those described in the note to Art. 63
belongs rather to Conic Sections than to Descriptive Ge-
ometry.
PROBLEM XXVIII.
To find tlie intersection of a plane with the surface of a cylinder,
the plane and cylinder having any position with each other
and with the planes of projection ; to draw a tangent line to
the curve ; and to find the curve in its oum plane.
§ 131. PI. 12. Fig. 1. Let the ellipse BGDC be the base
of the cylinder. Let ABODE be the horizontal projection of
the cylinder, and A'C'B"E' its vertical projection, and (DF, DF')
the cutting plane.
If the cylinder be intersected by a system of auxiliary planes
parallel to the axis, they will intersect the surface of the cylinder
in right-lined elements, and the cutting plane in right lines ;
the points in which these lines intersect are points of the re-
quired curve. To render the construction as simple as possible,
let the auxiliary planes be taken perpendicular to the horizontal
plane. Let F"w" be the horizontal trace and FT" the vertical
trace of one of these planes ; Ff' is the vertical projection of
its intersection with the cutting plane (DF, DF'). But the
right-lined elements in which this auxiliary plane intersects the
surface of the cylinder pierce the horizontal plane at h and g ;
projecting these points on the vertical plane at h' and g', and
drawing through h' and g' parallels to the vertical projection
of the axis of the cylinder determines the vertical projectionsINTERSECTIONS OP SURFACES.
77
of these elements ; the points ri, k', in which they intersect the
line Ff', are the vertical projections, and n and k the horizontal
projections of the two points of the required curve which are
determined by the vertical auxiliary plane F "ri’. If another
auxiliary plane, as o"C, be passed, it will intersect the cutting
plane (DF, DF') in a line parallel to (F"f, Ff') ; the vertical
projection of this intérsection is, therefore, parallel to Ff' ; and
the same for all intersections determined by vertical auxiliary
planes. The points (o,o'), (m,m') are determined by the plane
o"C.
The part of the curve which is made full in horizontal pro-
jection, is the part which lies above the elements of contact of
the tangent planes which are perpendicular to the horizontal
plane ; and the part of the curve which is made full in vertical
projection, is the part which lies on the surface of the cylinder
in front of the elements of contact of the tangent planes which
are perpendicular to the vertical plane. The points at which
the full part of the curve terminates and the dotted part begins,
in either projection, are readily found by drawing auxiliary
planes through the elements of the surface which contain
these points. The points (0,0') and (p,//) are determined by
drawing auxiliary planes through the elements (Co", C'A')
and (BA, B'l).
To draw a tangent line to the curve at any point, as
Draw a tangent plane to the cylinder along the element passing
through the point (m,m') ; liG is its horizontal trace, and Hw
the horizontal projection of its intersection with the cutting
plane (DF, DF'), and consequently, the horizontal projection
of the tangent line to the curve at the point	(120).
The line H'm' is the vertical projection of this tangent, and
both its projections are respectively tangent to the projections
of the curve.
To find the curve and its tangent in their own plane. Let the
plane of the curve be revolved about its horizontal trace DF till it
coincides with the horizontal plane ; find where the different points
of the curve fall, and the curve described through these points
will be the curve in its own plane. In the example we are78
DESCRIPTIVE GEOMETRY.
considering, the trace FD is perpendicular to the horizontal pro-
jection of the axis of the cylinder ; so that, after the revolution
of the cutting plane, the points of the curve will be found in
the traces of the auxiliary planes. The point (n,rí), for ex-
ample, is found at n", a distance from the point ƒ equal to the
hypothenuse of a triangle whose base is fn, and wrhose perpen-
dicular is equal to the altitude of the point above the horizontal
plane. The point (k,kf) is found at /c", the point (m,m') at m'\
and so for other points. The point H of the tangent line re-
mains fixed, being in the axis ; Hm” is the revolved position
of the tangent line.
If the trace DF were not perpendicular to the horizontal
projections of the elements of the cylinder, we could determine
the curve in its own plane thus ; find the position of the point
(F",F) after the cutting plane shall have been revolved to coin-
cide with the horizontal plane, and join this point with the point
f ; this line would be the revolved position of the intersection
of one of the auxiliary planes with the cutting plane ; and since
all such intersections are parallel, they will be parallel after
they are revolved. Take any one of the points, as q, in which
the trace of an auxiliary plane intersects the trace of the cut-
ting plane ; through this point draw a parallel to the revolved
position of the intersection of the auxiliary and cutting plane
which passes through the point ƒ ; from o and m let perpen
diculars be drawn to the trace FD : the points in which they
intersect the line through q are the points of the curve in its
own plane.
If it were required to develop the surface of this cylinder
it would be necessary, first, to intersect it by a plane perpen-
dicular to the axis ; a right cylinder would thus be formed
the surface of which could be developed as in Art. 125 ; and
any curve resulting from the intersection of a plane with the
surface of the cylinder could be traced on the development, as
in Prob. 26.INTERSECTIONS OP SURFACES.
79
PROBLEM XXIX.
To find the intersection of a plane with the surface of a cone, the
plane and cone having any position with each other and with
the planes of projection ; to draw a tangent line to the curve :
and to find the curve in its own plane.
§ 132. PI. 12. Fig. 2. Let VòCA be the horizontal and
V'C'B' the vertical projection of the cone, and (DI,DE) the cut-
ting plane.
If the cone be intersected by a system of planes through its
vertex, these planes will intersect the surface in right-lined
elements, and the cutting plane in right lines ; the points in
which these lines intersect the elements are points of the required
curve. We will take the system of planes through the vertex
perpendicular to the horizontal plane ; all the planes will then in-
tersect each other in a line perpendicular to the horizontal plane ;
and the point in which this line pierces the cutting plane
(DI, DE) is a point common to the traces of all the auxiliary
planes on this intersecting plane. To find this point. The
projecting line of the vertex (V,V') pierces the cutting plane
(DI, DE) in a point of which V is the horizontal projection ;
the vertical projection of the point is V" (43) : hence (V,V")
is the point through which the traces of the auxiliary planes
pass. Take VF for the horizontal trace of one of the auxiliary
planes ; the elements in which it intersects the surface of the
cone, as well as the line in which it intersects the plane (DI,
DE), are horizontally projected into the trace VF ; V'C', \'q'
are the vertical projections of the elements, and V"F' is the ver-
tical projection of the intersection of the planes ; hence/' and d!
are the vertical projections, and ƒ and d the horizontal projections
of the points in which the intersection of the planes meets the ele-
ments of the cone : hence ( ff) and {d,dj are two points of the
curve. By taking any other plane, the points of the curve which
it would determine are found in the same manner. The points80
DESCRIPTIVE GEOMETRY.
a and c, where the part of the curve which is made full in hori-
zontal projection joins the part which is dotted, are constructed
as in the cylinder, by taking auxiliary planes tangent to the
surface. The part of the curve which is to be made full in
the vertical projection is limited by the elements (VC, V'C')
and (VB,V'B') ; the points ƒ' and g' are determined by the aux-
iliary planes through these elements. The portion of the curve
which lies on the front part of the cone is made full, the other
dotted.
To draw a tangent line to any point of the curve, as (hyhr).
Draw a tangent plane to the cone along the element passing
through this point : \k is its horizontal trace, and Ih is the hori-
zontal projection of its intersection with the cutting plane ; for,
I is the point in which the traces of the planes intersect, and
both the planes pass through the point (h,h'). The vertical pro-
jection of the tangent is found by projecting the point I into the
vertical plane at I' and drawing FA'.
To find the curve in its own plane. Let the plane of the
curve be revolved around its horizontal trace DI till it coincides
with the horizontal plane. The point (V,V") falls at v ; and,
as the intersections of the auxiliary planes and the plane (DI,
DE) pass through the point (V,V"), they will, when revolved,
pass through the point v. But the points in which the intersec-
tions meet the trace DI remain fixed ; therefore, the revolved
positions of these intersections are easily drawn. The line Fv
is the revolved position of the intersection of the plane (DI, DE)
and the auxiliary plane VF. But as the points of the curve
revolve in planes perpendicular to the axis of revolution (11), the
intersections/" andd", of the perpendiculars ff” and dd" with
the line Fv, are points of the curve when its plane is revolved
to coincide with the horizontal plane. The points A", c", g'
and a" are determined in the same manner. The line I A" is the
position of the tangent line when the plane of the curve is re-
volved to coincide with the horizontal plane.
§ 133. These points might also be determined by the general
method of finding the hypothenuse of a triangle whose base ia
the distance from the horizontal projection of the point to the7?
/ / . C/..INTERSECTIONS OF SURFACES.
81
axis, and perpendicular, the altitude of the point above the hori-
zontal plane, and then laying off these several distances from
the axis on the perpendiculars drawn through the horizontal
projections of the several points.
§ 134. The surface of this cone cannot be developed by the
method used in Prob. 27 to develop the surface of a right cone
with a circular base. For the lengths of the elements, meas-
ured from the vertex to the base, are unequal : hence, the
curve of the base traced on the development will not be a circle ;
therefore, the positions of the different elements cannot be de-
termined as in Prob. 27. The manner of making this develop-
ment will be shown hereafter.
PROBLEM XXX.
To find the intersection of a plane and surface of revolution ; to
draw a. tangent line to the cuiwe ; and to find the curve in its
oum plane.
§ 135. PI. 13. Let the surface to be intersected be the sur-
face of an ellipsoid. Let the horizontal plane of projection be
taken perpendicular to the axis ; let A be its horizontal and A'B
its vertical projection. The circle whose centre is A and radius
Ag is the horizontal projection of the surface, the ellipse
A'F'BG is its vertical projection, and (CD, CD') is the cutting
plane.
If a system of horizontal planes be drawn, they will intersect
the surface in horizontal circles ; and the cutting plane (CD,
CD') in lines parallel to its horizontal trace ; the points in wrhich
these lines and circles intersect are points of the required
curve.
The transverse axis of any curve resulting from the intersec-
tion of a plane and surface of revolution is the line of intersec-
tion of the cutting plane, and the meridian plane perpendicular
to it ; the points in which this line pierces the surface are the
vertices of the transverse axis, or vertices of the curve. First,
to find the axis and vertices. The line AE, drawn at right
682
DESCRIPTIVE GEOMETRY.
angles to CD, is the horizontal trace of the meridian plane per-
pendicular to the cutting plane. The intersection of these two
planes pierces the horizontal plane at E, and intersects the
axis of the surface at the point in which the axis pierces the
cutting plane (CD, CD') ; that is, at the point of which A * is
the vertical projection (43) : therefore, E'A" is the vertica pro-
jection of the intersection of the planes ; the horizontal projec-
tion is the trace AE. To find the points in which this line inter-
sects the surface, let the meridian plane AE be revolved about
the axis of the surface till it becomes parallel to the vertical
plane. The meridian section of the surface is, from this posi-
tion, projected on the vertical plane into the ellipse A'F'BG,
and the line of intersection of the planes into the line E"A" ;
for, the point (A,A"), being in the axis, remains fixed, and the
point E describes in the horizontal plane the arc EE', and E" is
the vertical projection of E'. In this position the line of inter-
section pierces the meridian curve at ƒ'" and d!" ; in the counter
revolution these points describe arcs of horizontal circles about
the axis of the surface ; and when the meridian plane takes its
primitive position, are vertically projected at ƒ' and d\ and
horizontally projected at ƒ and d. Therefore (ƒ,ƒ') and (d,d')
are the vertices of the curve, and (fd,f'd') its transverse axis.
To find other points of the curve, intersect by horizontal
planes, and let F'G be the vertical trace of one of them. Since
the circle in which this plane intersects the surface is horizontal,
its horizontal projection is an equal circle ; and as its centre is
in the axis of the surface, A is its horizontal projection. The
point F' is horizontally projected at F : hence AF is the hori-
zontal projection of the radius, with which let the circle be
described about the centre A. The line in which the plane
F'G intersects the plane (CD, CD') pierces the vertical plane
at H', and is parallel to the horizontal trace CD ; therefore its
horizontal projection Hm is parallel to the horizontal trace CD.
But this horizontal projection intersects the horizontal pro-
jection of the circle in the points n and m ; n and m are, there-
fore, two points of the horizontal projection of the curve ; and
n', w! are their vertical projections. In the same manner anyINTERSECTIONS OF SURFACES.	83
number of points may be found. If it were required to tind
the points in which the curve intersects, on the surface of the
ellipsoid, the circle whose plane passes through the centre,
take the plane of this circle as an auxiliary plane ; it determines
the points (g,g') and (p,p'); the horizontal projection of the curve
is tangent to the horizontal projection of the surface at the
points g and p The part of the curve which lies above the
horizontal plane through the centre of the ellipsoid is made full
in horizontal projection ; the part which lies below it is dotted.
The points in which the curve determined by the plane (CD,
CD') intersects the meridian curve parallel to the vertical plane,
are horizontally projected at i and h, and vertically at i' and /*' ;
the vertical projection of the surface is tangent to the vertical
projection of the curve at the points i' and h'. The part of
the curve which lies in front of the meridian plane is made
full in vertical projection ; the part which lies behind it is
dotted.
To draw a tangent line to the curve at any point, as (m,m').
Let the surface be circumscribed by a tangent cone whose ver-
tex shall be in the axis of the surface, and whose contact with
the surface shall be the horizontal circle containing the point
(m,m'). The line F'G is the vertical projection of this circle.
A plane tangent to the cone along the element passing through
the point (m,m') will be tangent to the ellipsoid at this point,
and its intersection with the cutting plane is the tangent re-
quired. The tangent line to the ellipse A'F'BG at the point P
is the element of the cone which is parallel to the vertical plane ;
it pierces the horizontal plane at N ; and a circle described with
A as a centre, and radius AN, is the intersection of the cone and
horizontal plane. The element of the cone passing through (m,m!)
pierces the horizontal plane at M, and PM is the horizontal
trace of the tangent plane. This plane intersects the plane
(CD, CD') in the line (Pm, P'm'), which is therefore the tan-
gent line sought.
To find the curve in its own plane. Let the plane be re-
volved about its vertical trace CD' till it coincides with the ver-
tical plane ; the points will fall at their respective distances
6*84
DESCRIPTIVE GEOMETRY.
from the axis. To find the position of the horizontal trace oí
the plane of the curve, as also that of the tangent line, after the
plane is revolved to coincide with the vertical plane, take any
point of the horizontal trace, as P, and project it into the ground
line at P'. Through the point P' draw a perpendicular to D'C
produced ; with C as a centre, and radius CP, describe the arc
PP" ; CP" is the revolved position of the trace CD. For, the
point (P,P') revolves in a plane perpendicular to the axis D'C,
and continues at the same distance from the point C ; therefore
it must be found, after the revolution, in the perpendicular P'P",
and also in the arc PP", and consequently at P". The line
P"m" is the position of the tangent line in the plane of the
curve.
CHAPTER VIII.
OF THE INTERSECTIONS OF CURVED SURFACES.
§ 136. The general method of determining the lines in which
surfaces intersect, is to intersect them by auxiliary surfaces :
these auxiliary surfaces intersect the given surfaces in lines ;
the points in which these lines intersect are points of the inter-
sections of the given surfaces. The auxiliary surfaces should
be of that class, and so chosen in position, that the simplest lines
of section may be determined on the given surfaces ; else a
plain and simple problem might be rendered complex and dif
ficult.
§ 137. A line is, in general, drawn tangent to the intersec-
tion of two surfaces at any point, by drawing two planes through
the point, respectively tangent to the surfaces ; the intersection
of these planes is the tangent line.INTERSECTIONS OP SURFACES*
85
PROBLEM XXXI.
To find the intersection of the surfaces of tico cylinders, and to
draw a tangent line to the curve.
§ 138* PI. 14. Let (CA', C'D) be the axis of one cylinder,
(AB, A'B') the axis of the other. The manner of making their
projections having already been explained, we will suppose
them constructed as in the figure.
If through the axis of one cylinder a plane be passed parallel
to the axis of the other, such plane will, if it cut the other
cylinder, intersect the surfaces of both cylinders in right lines,
since it is parallel to their rectilinear elements ; and all planes
parallel to it, if they intersect the surf ices of the cylinders, will
also intersect them in right-lined elements.
Draw a plane through the axis (CA\ C'D) parallel to the
axis (AB, A'B') (58) : CE is the horizontal trace of this plane ;
its vertical trace could be easily found, but is not used in the
construction. As the system of auxiliary planes is to be parallel
to this plane, their traces will be parallel to CE. The plane
whose trace is ECX intersects the cylinder whose axis is (AB,
A'B') in two elements which pierce the horizontal plane at V
and T ; it intersects the cylinder whose axis is (CA', C'D) in
two elements which pierce the horizontal plane at M and X.
Since these four elements are in the same plane, and are not
parallel, they intersect each other; that is, each element of one
cylinder intersects both elements of the other cylinder. The
horizontal projections of these elements are respectively paral-
lel to the horizontal projections of the axes of the cylinders, and
the points 6, d, p, and q. in which they intersect, are the hori-
zontal projections of four points of* the required curve. The
vertical projections b\ d\ //, and qf are found either by project-
ing the elements of both cylinders on the vertical plane and
determining their points of intersection, or by projecting the
elements of one cylinder only and determining the points in86
DESÇKIPTIVE GEOMETRY.
which these projections intersect the perpendiculars to the
ground line drawn through the points b, d,p, and q. By drawl-
ing lines parallel to CE, and considering them as the traces of
auxiliary planes, any number of points of the curve are found
in the same manner. Let the auxiliary plane RN be drawn
tangent to the cylinder whose axis is (AB, A'B') ; it determines
two points of the curve (ƒ,ƒ') and (g,g'). Through the element
( fc,f'c') conceive a plane to be drawn tangent to the cylinder
whose axis is (CA', C'D) ; this plane intersects the plane RN,
tangent to the other cylinder, in the element (fcf'c') : hence,
this element is tangent to the curve of intersection of the two
cylinders at the point (ƒ,ƒ') (137) ; and the projections of the
element (cfcf') are respectively tangent to the projections of
the curve (90). In the same manner it may be shown, that
the element (Rg*, R'g*') is tangent to the curve of intersection
of the cylinders, and that its projections Rg, R’g' are tangent
to the projections of the curve. If the auxiliary plane FGE
be drawn tangent to the cylinder whose axis is (CA', C'D), it
is easily proved that the elements (Gz, GV) and (Ha, HV) are
tangent to the curve of intersection of the cylinders ; and hence
their projections Gz, H a, G V, and H are tangent to the pro-
jections of the curve.
The surfaces of these cylinders intersect in one curve only,
for, when they intersect in two curves the smaller cylinder will
enter the larger in a curve returning into itself, and leave it in
a curve also returning into itself. When this is the case, if tw o
tangent planes be drawn to the lesser cylinder, parallel to the
auxiliary secant planes, they will cut the larger cylinder ; but
if they be drawn to the larger cylinder, they will not intersect
the smaller. Hence, if two tangent planes be drawn to either
cylinder, parallel to the auxiliary secant planes, and one of them
cuts the other cylinder and one does not cut it, there will be but
one curve of intersection. But when both the tangent planes in-
tersect the other cylinder, or when neither of them intersects it,
there will be two curves.
As there is some difficulty in tracing this curve, and deter-
mining what part of its horizontal and what part of its verticalINTERSECTIONS OF SURFACES.
87
projection ought to be made full, and what part of each ought
to be dotted, we shall examine it a little in detail. We will
begin at the point (¿/,ö'), and proceed around in the direction
{?>?')> (hji'), (ƒ,ƒ'), (b,b’), (z,z’)f (t,x’), (d,d'), (g,g'), (pp') and
(w,w'). First, with respect to the horizontal projection. The
point (a,a') is on the upper portion of the surfaces of both
cylinders. At v thé curve passes to the under part of the sur-
face of the cylinder whose axis is (CA', C'D), thence along
through the points q, h, f, b, to u; at which point it returns to
the upper part of the surface. From u it continues on the
upper surfaces of the cylinders to t ; at this point it passes to
the under surface of the cylinder whose axis is (AB, A'B'), and
continues on the under surface through d and g, to fc" : at this
point it returns to the upper surface, and passes through p, w,
and a. The points v and u are determined by the auxiliary
plane through S ; the point i, by the plane through It ; and the
point k", by the auxiliary plane passing through the element
whose horizontal projection is tangent to the curve at this point
The part of the curve between u and t, as well as the part be-
tween /c"and v, is made full ; the remaining portion is dotted (34).
To trace the vertical projection of the curve. From b' to a
point near d\ the curve is made full, lying on the front por-
tion of the surfaces of the two cylinders. At d! (determined
by the auxiliary plane XE), it passes to the back surface of the
cylinder whose axis is (CA', C'D) ; at i! it comes to the front
part of the surfaces of both cylinders, is made full to x, and dotted
from x to b\
To draw a tangent line to the curve at any point, as (hji').
The elements of the cylinders which, by their intersection,
determine this point, pierce the horizontal plane at P and s ;
PQ and sQ are the horizontal traces of the planes tangent to
the cylinders along these elements : the intersection of these
planes is the tangent line required (137) ; Qh is its horizontal
and Q7i' its vertical projection.*8
DESCRIPTIVE GEOMETRY.
PROBLEM XXXII.
To find the. intersection of the surfaces of two cones, and to draw
a tangent line to the curve.
§ 139. PI. L5. Let (A,A') and (V,V') be the vertices of the
cones ; EGaD the base of one cone, and LcXn the base of the
other.
If a system of cutting planes be drawn through the vertices
of these cones, each plane will, in general, intersect each cone
in two right-lined elements ; the points in which these elements
intersect are points common to the two surfaces. All these
planes will pass through the line joining the vertices of the
cones : hence, the point in which this line pierces the plane of
the bases of the cones is a point common to the traces of all
the auxiliary planes. Let it be remarked, that the cone whose
vertex is (V,V') intersects the upper and lower nappes of the
cone whose vertex is (A,A'). The line joining the vertices of
the cones pierces the horizontal plane at B ; this point is com-
mon to the traces of all the auxiliary secant planes.
From B draw the two tangents Bb and BC to the base of the
cone whose vertex is (V,V') ; they are the traces of two planes
tangent to that cone, and V6, YC are the horizontal projections
of the elements of contact. The plane whose trace is EB6 in-
tersects the cone whose vertex is (A,A') in two elements of
which A ad and EAg* are the horizontal projections; and the
points d and g, in which they intersect Yb, are the horizontal
projections of two points of the required curves ; d belongs to
the lower and g to the upper nappe of the cone whose vertex
is (A,A'). The plane whose horizontal trace is ¿BC intersects
the cone whose vertex is (A,A') in two elements of which tAh
and A/D are the horizontal projections; and the points/and
A, in which they intersect VC, are the horizontal projections of
two points of the required curves ; the point ƒ belongs to the
lower and the point h to the upper nappe of the cone whose/4,
14-
Jtf : l .	0y\INTERSECTIONS OF SURFACES.
S9
vertex is (A,A'). The vertical projections of the elements in
which the plane whose trace is EB6 intersects the cone whose
vertex is (A,A'), are the lines E'A,gr and A’a!d!\ and d! and g' are
the vertical projections of the points determined by this plane.
Drawing perpendiculars to the ground line through the points
ƒ and h, and projecting on the vertical plane the elements
which pierce the horizontal plane at D and t, determines the
vertical projections of the two points of the curve contained in
the plane whose trace is ¿BC. Thus, four points of the curves
(ƒ,ƒ'), (g,g'), and (d,df) are found.
The elements of the cone whose vertex is (A,A') that pass
through these points, are tangent to the intersections of the
cones. For, through either of them, as the element (kA, lì!A'),
conceive a plane to be passed tangent to the cone whose vertex
is (A,A') ; this tangent plane intersects the plane tangent to the
other cone along the element whose horizontal projection is
VC, in the element (/¿A, A'A') : hence, this element is tangent
to the curve of intersection of the cones (137) ; consequently,
its projections are respectively tangent to the projections of the
curve (90), and the same may be shown for each of the other
elements.
The auxiliary plane xBL determines the points (n,uf) and
(5,s') in the lower curve, apd the points (Q,Q') and (M,M') in
the upper. Thus every auxiliary plane which intersects both
cones determines two points of each curve. The lower curve
intersects the horizontal plane at the points c and n. As we
do not use the upper nappe of the cone whose vertex is (VjV'),
we make the part of the curve which lies on the upper part of
the surface of the lower nappe full. The horizontal projection
of the lower curve is dotted, being entirely concealed by the
upper nappe of the cone whose vertex is (A,A'). The portions
of the curves which can be seen in vertical projection are made
full on the vertical plane ; the other parts are dotted. The
bases of both cones are dotted, since they are concealed by the
upper nappe of the cone whose vertex is (A,A'). Of the ele-
ments which show the vertical projections of the cones, the
parts seen are made full, the parts concealed dotted.90
DESCRIPTIVE GEOMETRY.
To draw a tangent line to the upper curve at the point
(0,0'), and to the lower curve at the point (&,&'). Let aplane
be drawn tangent to the cone whose vertex is (V,V') along
the element passing through these points, NXH is its horizontal
trace ; this plane contains both the required tangents. Let a
plane be drawn tangent to the other cone along the element
(Ah, A'A') ; NI is its horizontal trace, and it intersects the tan-
gent plane before drawn in the line (Nk, Nwhich is tangent
to the lower curve at the point (k,k') (137). The line Nk is
the horizontal and N'h' the vertical projection of the tangent.
Drawing a tangent plane along the element (OAG, O'A'G'),
GH is its horizontal trace, and the line in which it intersects
the tangent plane (of which NH is the trace) is tangent to the
upper curve at the point (0,0'). The lines HO and H'O' are
the projections of this intersection, which are respectively tan-
gent to the projections of the curve.
PROBLEM XXXIII.
To find the intersection of two surfaces of revolution whose
axes are in the same filane^ and to draw a tangent line to the
curve.
§ 140. PI. 16. If the axes are in the same plane, they will
either intersect each other or be parallel. We shall first con
sider the case in which they intersect.
Let the horizontal plane be taken perpendicular to the axis
of one surface, and the vertical plane parallel to the plane of
the axes. Let one of the surfaces be an ellipsoid whose axis
is (A,A'C), and the other a paraboloid whose axis is (AB, CB') ;
dB is the horizontal trace of the plane of the axes. The ver-
tical projections of the curves in which this plane intersects the
surfaces are the vertical projections of the surfaces. The large
circle described around the centre A is the horizontal projec-
tion of the surface whose axis is (A,A'C) ; the horizontal pro-
25INTERSECTIONS OF SURFACES.	91
jection of the other surface, not being necessary in the solution
of the problem, is not made.
Before constructing the intersection of the surfaces, we will
remark, that all surfaces of revolution having a common axis
and intersecting each other, intersect in circles whose planes are
perpendicular to the common axis. For, let the surfaces be
intersected by a plane through their common axis, this plane
intersects each surface in a meridian curve ; the points in which
these curves intersect are points of the intersection of the sur-
faces. Let this plane be now revolved around the common
axis ; each meridian curve will generate the surface to which
it belongs, and their points of intersection will describe the
circumferences of circles whose planes are perpendicular to
the axis ; that is, to the common axis of the surfaces. But
these circles are the intersections of the surfaces : hence, two
surfaces of revolution, having a common axis, intersect in circles
whose planes are perpendicular to that axis.
Let the point C, in which the axes of the surfaces intersect,
be the common centre of a system of auxiliary spheres ; every
sphere will intersect each surface in a circle ; the points in
which the circumferences of these circles intersect are points
of the required curve.
With C as a centre, and any radius, as CD, conceive a sphere
to be described. This sphere intersects each of the surfaces
in a circle ; the planes of these circles are perpendicular to the
vertical plane of projection, since the axes of the surfaces are
parallel to this plane. The line DD' is the vertical projection
of one circle, and EE' the vertical projection of the other.
The point F, at which these lines intersect, is the vertical pro-
jection of the line in which the planes of the circles intersect ;
and the points in which this line of intersection pierces the sur-
faces are two points of the required curve. Let the circle of
which DD' is the vertical projection be projected on the hori-
zontal plane ; D is projected at d, and the circle described with
A as a centre, and radius Ad, is the horizontal projection of
the intersection of the sphere and ellipsoid. It is evident that
ff is the horizontal projection of the line in which the planesDESCRIPTIVE GEOMETRY
92
of the two circles intersect ; therefore ƒ and ƒ ' are the hori-
zontal projections, and F is the vertical projection, of two points
of the curve. In the same manner any number of points may
be found. To find the points in which the curve intersects the
circle of the ellipsoid whose plane passes through the centre,
describe a sphere which shall intersect the ellipsoid in its largest
horizontal circle ; this sphere will determine those points ; they
are (m,N) and (ra,N). The points a! and b\ in which the me-
ridian curves parallel to the vertical plane of projection inter-
sect, are horizontally projected at a and b.
To draw a tangent line to the curve at any point, as (ƒ,F).
This line could be determined by the general method of draw-
ing two planes respectively tangent to the surfaces at the point,
and constructing their line of intersection, which would be tan-
gent to the curve. We shall, however, employ another method,
which introduces new principles, and is, perhaps, more elegant.
It is called the method by normals.
A line is perpendicular to a curve when it is perpendicular to
the tangent drawn through the point in which it meets the curve.
This perpendicular is called a normal. A line is perpendicular
to a surface when it is perpendicular to the tangent plane at the
point in which it meets the surface. This perpendicular is
called a normal, and any plane passing through it a normal
plane. If through the point (ƒ, F), at which we wish to draw
a tangent line, two lines be drawn, the one perpendicular to
the surface of the ellipsoid, the other perpendicular to the sur-
face of the paraboloid, the plane of these two lines is a normal
plane to both the surfaces at the point (/,F). But as this plane
contains the normal lines, it will be perpendicular to two planes
drawn through the point (ƒ, F), the one tangent to the ellipsoid,
the other to the paraboloid ; and, consequently, it will be per-
pendicular to their intersection. But, as this intersection is the
tangent line to the curve at the point (ƒ, F), we conclude that
the normal plane to both surfaces, at any point of their inter sec-
twn, is perpendicular to a line tangent to their intersection at the
same point.
Let the point (jf,F) be carried around on the surface of theINTERSECTIONS OP SURFACES.
93
ellipsoid, in a horizontal circle, till it comes into the meridian
curve parallel to the vertical plane at (d,D) ; at this point draw
the normal DH. It is evident that all normal lines which meet
the surface in the circumference of the circle of which DD' is
the projection will intersect the axis at H. In like manner, by
revolving the point (/,F) about the axis of the paraboloid till it
comes into the meridian curve, parallel to the vertical plane, at
E', the normal E'H' can be drawn : hence, FH, FH' are the
vertical projections, and ƒ A, ƒH" the horizontal projections of
the two normal lines to the surfaces at the point (ƒ, F) ; and
HH' is the trace of their plane on the plane of the axes. But
since the plane of the axes is parallel to the vertical plane of
projection, HH' is parallel to the vertical trace of the normal
plane. Since the tangent line passes through the point (ƒ, F),
and is perpendicular to the normal plane, its vertical projection
passes through F and is perpendicular to HH' (49). The nor-
mal plane intersects the plane of the horizontal circle of which
DD' is the vertical projection, in a line parallel to the horizontal
trace of the normal plane. But as the diameter DD' and the
trace HH' are both in the plane of the axes, the point (G,G'),
in which they intersect, is one point of the intersection of the
horizontal and normal planes ; the point ( ƒ, F) is another point :
hence, Gf is the horizontal projection of their line of intersec-
tion ; which projection is parallel to the horizontal trace of the
normal plane. Therefore the line drawn through F, perpendicu-
lar to HG', is the vertical projection of the tangent line, and the
line through/, perpendicular to ƒ G, is its horizontal projection.
Instead of determining the direction of the traces of the normal
plane, by constructing its intersections with the plane of the
axes and the horizontal plane DD', we might construct its
traces on the planes of projection, since we have two lines oí
the plane, viz. (/A, FH) and (ƒ H", FH') ; the points in which
these lines pierce the planes of projection are points of the traces
of the normal plane.
§ 141. If the axes of the surfaces were parallel, their inter-
section could be obtained very easily by intersecting them by
planes, since planes perpendicular to their axes would intersect94
DESCRIPTIVE GEOMETRY.
both surfaces in circles, and the intersections of these circles
would be points of the curve. The tangent line is drawn in
the same manner as when the axes intersect. The construc-
tion is left for the student.
PROBLEM XXXIV.
To find the intersection of a cone and sphere, the vertex of the
cone being at the centre of the sphere ; and to draw a tangent
line to the curve at any point.
§ 142. PI. 17. Fig. 1. Let (C,C') be the vertex of the cone
and centre of the sphere, EGII the base of the cone, and the
circle described with the centre C', and radius C'A the ver-
tical projection of the sphere. The horizontal projection of
the sphere, not being required in the construction, is not made.
Through the vertex of the cone let any number of planes be
drawn perpendicular to the horizontal plane ; they will inter-
sect the sphere in great circles, and the cone in right-lined
elements ; the points in which the elements intersect the circles
are points of the curve.
Let GCH be the horizontal trace of one of the auxiliary
planes. Let this plane be revolved around the projecting line
of the vertex of the cone till it becomes parallel to the vertical
plane of projection. The points G and H describe the arcs
GG' and HH', in the horizontal plane, around C as a centre ;
and the great circle in which the plane intersects the sphere
becomes parallel to the vertical plane. From this position the
great circle is vertically projected into the circle whose centre
is C', and the elements of the cone into the lines C'G" and C'H".
These lines intersect the circumference of the circle at 6" and
A". In the counter revolution these points describe the arcs
of horizontal circles; and when the plane has resumed its
primitive position, they are vertically projected at bf and A', and
horizontally projected at b and h. The points (6,6') and (A,A')
are therefore two points of the required curve. In the same'iÓ.INTERSECTIONS OP SURFACES.
95
manner the points («,«'), (n,ri), (m,m'), (If), (g.g'), (ƒ,ƒ')> (e,e’)
and (c,d) are determined. The part b'c'ef', being seen, is made
full in vertical projection ; the remaining part is dotted. If we
suppose the upper hemisphere, which is not used in the problem,
to be removed, the horizontal projection of the curve should be
made full.
To draw a tangent line to the curve at any point, as (<c,cr).
Let a cone be drawn tangent to the sphere whose circle of
contact shall be horizontal, and also contain the point (c,c').
The line AI'V, drawn tangent to the vertical projection of the
sphere, is one of the elements in which the plane G'CH' inter-
sects the cone. This element pierces the horizontal plane at I ;
and the circle described with the centre C and radius Cl is the
intersection of the cone and horizontal plane. The point (C,V)
is the vertex of the cone. The element of this cone which
passes through (c,c') pierces the horizontal plane at B ; and BD
is the horizontal trace of a plane tangent to the cone along this
element. The element of the cone whose vertex is (C,C')>
that passes through the point (c,c'), pierces the horizontal plane
at E, and ED is the horizontal trace of a plane tangent along
this element. The intersection of this plane with the plane
whose trace is BD is the tangent line to the curve at the point
(c,d). The line Dc is the horizontal and D'c' the vertical pro-
jection of this tangent.
PROBLEM XXXY.
To develop the surface of a cone having any curve for its base,
to trace on this development the intersection of the cone by a
given plane ; and to find, on the development, the position of
a tangent line to the curve of intersection.
§ 143. If a cone be rolled around on any of its tangent planes,
the vertex remaining fixed, all the elements will arrange them-
selves around the fixed point; and when the cone shall have
been rolled once round, all the elements will have been in con-96
DESCRIPTIVE GEOMETRY.
tact with the plane. If the elements of the cone, after coming
in contact with the plane, remained in the plane, it is evident
that the surface of the cone would gradually extend itself on
the plane ; that any line on the surface of the cone would de-
velop into a line, and that the points of this line would be at the
same distance from the common point around which the elements
arrange themselves, as they were in space, from the vertex of
the cone.
The curve in which a sphere, having its centre at the vertex,
intersects the surface of a cone, will, when the surface of the
v>one is developed, become a circle ; the centre of the circle
being the point at which the vertex of the cone is placed, and
the radius equal to the radius of the sphere. With the vertex
of the cone as a centre, and any assumed radius, let a sphere
be described, and the intersection of the sphere and cone deter-
mined. If, then, any twro elements be chosen on the surface
of the cone, and the length of that part of the intersection of
the sphere and cone intercepted between them determined,
this distance laid off on the circle into which the intersection
develops, determines the position of the two elements on the
development of the cone. It is then necessary to find the
length of the curve of intersection of the cone and sphere. To
do this, let us develop the right cylinder which projects this
curve on the horizontal plane, and trace the curve on this
development. Since the cylinder is a right cylinder, its base,
which is the horizontal projection of the curve, will become a
right line on the plane of development (127) ; the elements of
the cylinder will become perpendicular to this right line, and
the curve is traced through the upper extremities of these
elements.
Let the curve in which the cone and sphere intersect each
other be found, as in the last problem ; and let the cylinder
which projects the curve on the horizontal plane be developed
on the plane which is tangent to it along the element that pierces
the horizontal plane at e (Fig. 1). Draw the indefinite right
line LED (Fig. 2) ; erect a perpendicular to it at E, and make
Ee equal to th? height of the point (e,e') (Fig. 1), above theINTERSECTIONS OF SURFACES.	„	07
horizontal plane ; e is one point of the intersection of the cone
and sphere on the plane of development. Laying off EC equal
to the arc ec, CB equal to cb, BA equal to ba, AN equal to an,
NM equal to nm, ML equal to mi ; and on the other side of the
point E make EF equal to ef, FG to fg, GH to gh, IIL to hi ;
LEL will be the base of the cylinder rectified. Erect at L,
H, G, F, &c., perpendiculars to the line LD, and make them
respectively equal to the distances of the corresponding points
in (Fig. 1) above the horizontal plane ; the curve Ihg, &c., tc
l, traced through these points, is the curve of intersection of the
sphere and cylinder on the plane of development. The tangent
line to the curve at the point (c,c') is determined on the plane
of development, by laying off from C (Fig, 2) the distance CD
equal to the subtangent cD (Fig. 1), and joining the points c
and D (Fig. 2).
It is now required to develop the cone, and trace on the
development the curve of intersection with the horizontal
plane of projection. Let the cone be developed on the tan-
gent plane passing through the point (c,d), and let the vertex
of the cone be placed at C (Fig. 3). With C as a centre, and a
radius equal to the radius of the sphere, describe the arc of
the circle Imna, &c., and draw a radius CE to represent the
element (Cc, C'c'). From the point c lay off the arc cb equal
to cb (Fig. 2); and draw CòG ; this is the position, on the plane
of development, of the element of the cone that passes through
the point (5,6') (Fig. 1). By making, in like manner, ba equal
to ba (Fig. 2), an equal to an, &c., and laying off on the other
side of c the arcs ce, ef,fg, &c. respectively equal to their cor-
responding arcs in (Fig. 2), and drawing the lines C a, C n, Ce,
&c., we determine the positions of a series of elements of the
cone on the plane of development. The tangent line being
perpendicular to the radius of the sphere, or to the element of
the cone passing through (c,cf), is perpendicular to this element
on the plane of development, and therefore has the position
cD (Fig. 3).
We will now trace on the plane of development the intersec-
tion of the cone with the horizontal plane. Laying off from C98
DESCRIPTIVE GEOMETRY.
the distance CE equal to the length of the element which passes
through (c,c') (Fig. 1), determines one point. The length of the
element is formed by revolving it about the axis of the cone till
it becomes parallel to the vertical plane, its projection from this
position will be equal to its true length : hence, C'E" is the
length of the element. Thus, the length of any element may
be determined, which, being laid off from C, determines a point
of the required curve. Making cD (Fig. 3) equal to cD (Fig. 2),
and joining E and D, ED is the position which the tangent line
ED (Fig. 1) assumes on the plane of development.
CHAPTER IX.
PRACTICA!, PROBLEMS.
PROBLEM I.
Knowing the distance of a point from three given points, it u
required to find its distance from the plane of these points, and
its projection on the plane.
§ 144. As the line joining any two of these points is known,
and as there are six such lines, the problem may be thus enun-
ciated : having given the six edges of a triangular pyramid, tc
construct the pyramid.
PI. 18. Fig. 1. Let the plane of the three points be assumed
for the horizontal plane of projection, and let ABC be the tri-
angle formed by joining the given points A, B, and C ; and
suppose the lines A, B, and C to be equal respectively to the
distances of the fourth point from the angles A, B, and C.
With the angular points A, B, and C as centres, and radii eqùal
to the distances A, B, and C, let three spheres be described.
As each condition, taken independently of the other two, fixes
the required point on the surface of one of these spheres, the three
conditions, taken together, fix it at their points of intersection.
Since either two of the spheres intersect each other in a circle17
. -Ir ï / li. l/r///PRACTICAL PROBLEMS.
99
perpendicular to the line joining their centres (140), the spheres
whose centres are B and C intersect in a circle perpendicular
to CB, and of which ED is the horizontal projection. The
spheres whose centres are A and C, intersect *in a circle per-
pendicular to AC, and of which GZ is the horizontal projection.
The line in which the planes of these circles intersect is per-
pendicular to the horizontal plane at I ; the two points in which
this line pierces the surface of either of the spheres are com-
mon to the surfaces of the three spheres: either of them,
therefore, will answer all the conditions of the problem. Let
the vertical plane of projection be taken perpendicular to
CB, and project the circle of which ED is the diameter upon
it. The line D'E' is the vertical projection of the diameter ; by
describing the circle on this diameter, and projecting the per-
peüdicular to the horizontal plane at I, the points I' and I", in
which it intersects this circle, are determined. The fourth point
is, therefore, horizontally projected at I, and is above or below
the plane of the other three points a distance equal to NI" or NI'.
Either of these points will answer all the conditions of the prob-
lem, and may be regarded as the vertex of a triangular pyramid,
three of whose edges lie in the horizontal plane, and whose three
other edges are drawn from this point to the angles A, B, and C.
PROBLEM II.
To find the centre and radius of a sphere whose surface shall
pass through four given points ; or, by regarding the four
points as the four angular' points of a triangular pyramid,
the problem may be enunciated, to circumscribe a sphere about
a triangular pyramid.
§ 145. If any chord of a sphere be bisected by a plane per-
pendicular to it, the plane will pass through the centre of the
sphere. Hence, if we bisect the line joining any two of the
given points by a perpendicular plane, this plane will contain
the centre of the required sphere. Bisecting a second line by
a perpendicular plane, this plane will also contain the centre of
7*DESCRIPTIVE GEOMETRY.
100
the sphere. Bisecting in like manner a third line, not m the
plane of the other two, the point in which this last plane cuts
the common intersection of the other two planes is the centre
of the required sphere, being a point equidistant from the four
given points.
PI. 18. Fig. 2. Take the plane of any three of the points
for the horizontal plane of projection. Let A, B, and C be the
three points situated in this plane, and (D,D') the fourth point.
Through L, the middle of BC, let the plane LI be drawn per-
pendicular to BC ; through N, the middle of AB, let the per-
pendicular plane NI be drawn ; these planes intersect in a line
perpendicular to the horizontal plane at I ; this line contains
the centre of the sphere, and its vertical projection I'T contains
the vertical projection of the centre. Through (E,E'), the
middle of the edge of which DC is the horizontal projection, let
a plane be drawn perpendicular to this edge (53) ; GH and
GH' are its traces ; the point in which the plane cuts the line
(1,11") is the centre of the sphere.
To find this point, let a plane be drawn through the line
(I,DI') parallel to the vertical plane ; it intersects the plane
(GH, GH') in a line parallel to its vertical trace, and H'T is
its vertical projection. The point (1,1') is, therefore, the point
in which the line (I,I'T) pierces the plane (GH, GH'), and is,
consequently, the centre of the sphere. A line drawn from
(1,1') to either of the four angular points, A, B, C, or (D,D'),
is the radius of the sphere. Taking the radius passing through
C, IC is its horizontal projection, and I'c its vertical projec-
tion. Revolving the plane which projects it on the horizontal
plane about (I,I"I') till it becomes parallel to the vertical plane,
C describes the arc CC', and the point (1,1') remains fixed ;
I'C" is, therefore, the length of the radius. With this radius,
and about I and 1' as centres, let two circles be described ;
they are the projections of the required sphere.
If the vertical plane of projection had been taken parallel to
(DC, D'c), the plane perpendicular to this line would have been
perpendicular to the vertical plane, and the point (1,1') would
then have been vertically projected in the trace GH'.PRACTICAL PROBLEMS.
iOi
PROBLEM III.
To find the radius of a sphere which shall he tangent to four
given planes ; or, to inscribe a sphere in a given triangular
pyramid.
§ 146. If we bisect any three of the diedral angles of this
pyramid by planes passing through three edges which do not
meet in the same point, the point in which these three bisect-
ing planes intersect, is the centre of the required sphere.*
It is, therefore, required to draw these planes, and determine
their point of intersection.
PI. 18. Fig. 3. Let OAB, in the horizontal plane, be the
base of the pyramid, (D,D') its vertex, DA, DB, and DO, the
horizontal, and D'A', D'0',and D'B', the vertical projections of
its edges. The bisecting planes will be drawn through the
lines AO, AB, and BO.
* For, let AC"B (Fig. 3, n) be the base, and (D,D') the vertex of a triangular
pyramid ; suppose (C,C') to be the point of intersection of the three bisecting
planes passed through the edges AC", AB and BC" ; it is to be proved that this
point is equidistant from the four faces of the pyramid.
Through the point (0,0') let a plane be drawn perpendicular to either edge,
as AC". Now, as the bisecting plane through the edge AC" divides equally
the diedral angle included between the face DAC" and the plane of the base,
its intersection with the perpendicular plane bisects the angle contained by the
line in which the perpendicular plane intersects the face DAC" and the hori-
zontal plane. Let this perpendicular plane be revolved around its horizontal
trace GC till it coincides with the horizontal plane. The point (C,C') falls at
E ; GF, GE, and GC, are the lines revolved, in which the perpendicular plane
intersects the plane of the face ADC", the bisecting plane, and the plane of the
base : EF and EC are the perpendiculars let fall from the point (C,C') on the
face ADC", and on the plane of the base. But, since the angles EGF and
EGC are equal, the angles at F and C right angles, and the side EG common,
the two triangles EFG and ECG are equal, and have the side EF equal to EC.
In the same manner it may be proved, that the length of the perpendicular on
either of the other faces is equal to the length of the perpendicular on the plane
of the base of the pyramid • hence, the perpendiculars drawn from the point
(C,C') to the four planes are equal ; (C,C') is, therefore, the centre, and EC or
EF the radius of a sphere to which the four lanes will be tangent.102
DESCRIPTIVE GEOMETRY.
Through the vertex (D,D') of the pyramid let three planes
be drawn respectively perpendicular to the three edges AO,
AB, and BO ; and let these planes be revolved around their
horizontal tracçs DE, DH, and DG, till they coincide with the
horizontal plane. In revolving the plane ED to coincide with
the horizontal plane, the point at which the vertex of the pyra-
mid falls is determined by drawing DV perpendicular to ED,
and making it equal to the height of the vertex above the hori-
zontal plane. Let V anò E be joined ; EV is the revolved
position of the intersection of the perpendicular plane ED with
the face DOA ; and EE', bisecting the angle VED, is the
revolved position of a line of the bisecting plane. By revolving
the plane DH to coincide with the horizontal plane, the vertex
of the pyramid falls at V", and the line bisecting the angle DHV"
is the revolved position of a line of the bisecting plane passing
through AB. In like manner, by revolving the plane GD to
coincide with the horizontal plane, the vertex of the pyramid
falls at V' ; and Ga', bisecting the angle V'GD, is a line of the
bisecting plane which passes through OB. We have, therefore,
two lines of each bisecting plane, and may therefore conceive the
planes to be drawn. These bisecting planes form a new pyra-
mid, which the horizontal plane intersects in the triangle ABO ;
the edges of this pyramid pass through the points A, B, and O,
and its vertex is the centre of the sphere.
Let this second pyramid be intersected by a plane parallel
to the horizontal plane, and at the distance P'Q above it. This
plane will intersect the faces of the new pyramid in lines
respectively parallel to AO, AB, and OB, which lines will form
a triangle similar to the triangle AOB ; the angular points of
this triangle are in the edges of the second pyramid. Recol-
lecting that Ga', in its position in space, is directly above GD,
we see,that if Ga" be made equal to P'Q, and a"a' drawn parallel
to GD, a' will be the revolved position of one point of the inter-
section of the plane parallel to the, horizontal plane, with the
bisecting plane through OB. The point a', from its position in
space, is horizontally projected at a ; and as the plane parallel
to the base of the pyramid intersects the bisecting plane in aPRACTICAL PROBLEMS.
103
ime parallel to OB, it follows that b'P is the horizontal projec-
tion of this line. . Making sb, in the angle VED, equal to P'Q,
and drawing it parallel to OA, bb' will be the projection of the
intersection of the plane parallel to the base of the pyramid
with the bisecting plane through AO. In like manner, the
parallel to AB is determined ; and thus we have the horizontal
projection of the triangle, in which the plane parallel to the base
of the pyramid intersects the second pyramid. The horizontal
projections of the edges of the second pyramid pass through the
angular points of this triangle, and also through the points A,
B, and O ; drawing them, determines C, the horizontal projec-
tion of the vertex of the second pyramid, or centre of the
sphere. The vertical projection of the centre of the sphere is
in the perpendicular to the ground line through C, and also in
the vertical projection of either of the edges of the second pyra-
mid. The point B is vertically projected at B' ; the point P,
at P': hence, B'P'C' is the vertical projection of the edge
which pierces the horizontal plane at B ; and C' is the vertical
projection of the centre of the sphere. The radius of the
sphere is equal to C'S ; with this radius let circles be described
around C' and C as centres : these circles are the projections
of the sphere. It is evident that this problem is the same as
the problem to find a point equidistant from four planes, neither
two of which are parallel to each other.104
DESCRIPTIVE GEOMETRV
SPHERICAL TRIGONOMETRY.
CHAPTER X.
CONSTRUCTION OF SPHERICAL TRIANGLES.
§ 147. Descriptive Geometry can readily be applied to
the graphic solutions of the several cases of Spherical Trigo-
nometry.
Every spherical triangle being formed by the arcs of three
great circles intersecting each other on the surfaces of a sphere,
it follows that the planes of these circles must intersect each
other at the centre. These three planes form what is called a
spherical pyramid : the centre of the sphere is its vertex ; the
planes of the sides of the spherical triangle are its faces ; and
the lines in which these planes intersect are called its edges.
The edges pass through the angular points of the spherical tri-
angle, and through the centre of the sphere.
§ 148. If at either angle two lines be drawn respectively
tangent to the sides of the spherical triangle, each line will lie
in the plane of the side to which it is tangent (67) ; both the
tangents will be perpendicular to the radius of the sphere pass-
ing through the angular point ; and the angle contained by them
will be the measure of the diedral angle of the two faces of the
pyramid which intersect in the radius passing through the an-
gular point. But since the lines are tangent to the arcs, the
angle which they make with each other is the measure of the
inclination of the arcs ; therefore, the diedral angles of the pyra-
mid are equal to the corresponding angles of the spherical tri-SPHERICAL TRIGONOMETRY.
105
angle, and may be taken to represent them. The angle inclu-
ded between any two edges of the pyramid, being an angle at
the centre of the sphere, is measured by the side of the spherical
triangle contained in the plane of these two lines : hence, the
sides of a spherical triangle measure the angles included be-
tween the edges of the pyramid ; these angles may then be
taken to represent the sides.
Seeing, therefore, that a pyramid can always be formed,
having its diedral angles equal to the angles of a spherical tri-
angle, and the angles formed by its edges equal to the sides of
the triangle, it follows that all the cases which can arise in
spherical trigonometry, will be comprised in the general prob-
lem, to find the remaining parts of a spherical pyramid when any
three parts of it are given.
CASE I.
Having given the three sides of a spherical triangle, to find
the angles : that is, having given the three angles included
between the edges of a spherical pyramid, to find its diedral
angles.
§ 149. PI. 1. Fig. 1. Let ACB be the spherical triangle,
having the sides a, b, and c given.
Make on the plane of the paper the angle DGH equal to the
side c. If the pyramid were entirely constructed, and the two
other faces revolved, the one around GD, the other around
GH, to coincide with the plane of the paper, the third edge
would, as it were, divide and fall into the two lines GC' and
GC", making the angle DGC' equal to the side a of the
spherical triangle, and the angle HGC" equal to the side b If,
therefore, the angle DGC' be made equal to the side a, and
the angle HGC" equal to the side b, they will represent these
sides revolved into the plane of the paper.
It is now required to fold up this pyramid and determine its
diedral angles. When the pyramid is folded up, every point
of the line GC' will unite with a point of the line GC" at the106
DESCRIPTIVE GEOMETRY.
same distance from the point G ; the points F and F, at the
same distance from G, will then become the same point. But,
in turning the p»ane DGC' around GD as an axis, the point F'
revolves in a vertical circle whose centre is B and radius BF' ;
and in revolving the plane HGC" around GH, the point F de-
scribes a vertical circle whose centre is A and radius AF : the
planes of these circles intersect ín a line perpendicular to the
face HGD at E, and GE is the projection of the intersection
of the faces DGC' and HGC". Revolving the plane of the
circle described by the point F' about the trace F'E till it coin-
cides with the plane of the paper, the point of which E is the
projection will fall in a perpendicular to F'E at E', a point of
the arc of the circle described by the point F' : hence, BE' is
the intersection of the vertical plane F'BE with the plane of
the face F'GB ; and, consequently, E'BE is equal to the angle
which this face makes with the face DGH. For the same
reasons E"AE is çqual to the angle which the face HGC" makes
with the face DGH.
It is now required to find the angle w7hich the faces HGC'
and C'GD make with each other. Let a plane be drawn per
pendicular to the edge of which GE is the horizontal projection,
its trace DH is perpendicular to GE, and the lines in which it
intersects the faces pass through the points II and D, and are.
both perpendicular to the edge at the same point. When the
faces are revolved around GD and GH to coincide with the
plane of the paper, this point will, as it were, divide and fall at
C' and C", equidistant from G ; the lines DC' and HC', per-
pendicular respectively to GC' and GC", are the lines in which
the perpendicular plane intersects the planes of the faces. Let
there be taken any two points in the lines GC' and GC" equi-
distant from G, as C' and C", and from these points let lines be
drawn respectively perpendicular to GC' and GC" : then con-
struct a triangle with the lines HD, DC', and HC" ; the angle
HCD will be equal to the angle C of the spherical triangle. It
is plain that the angle EBE' is equal to the angle B of the
spherical triangle, and the angle EAE" to the angle A.SPHERICAL TRIGONOMETRY.
107
CASE II.
Having given two sides and the included angle of a spherical
triangle, to find the other parts.
§ 150. PI. I. Fig. 2. Let ABC be the triangle, c and b the
given sides, and A the given angle.
Make, on the plane of the paper, the angle BHA equal to the
side c, and the angle AHG' equal to the side b : AHG' is the
face, revolved into the plane of the paper, which makes with
the face AHB the angle A. Through any point, as A, of the
edge HA, conceive a vertical plane G AF to be drawn perpen-
dicular to HA ; the angle contained by the lines in which this
plane intersects the faces is equal to the angle contained by the
faces. These lines, however, can only be represented by
revolving the vertical plane which contains them to coincide
with the horizontal plane ; when it is so revolved, let the line
AF" be draw’n, making the angle FAF" equal to the angle A
of the spherical triangle ; the line AF" is a line of the face AHG',
With the centre A and radius AG' let the arc of a circle be
described ; this arc meets AF" at F". Let the plane FAF" be
revolved back into its primitive position, the point F" is hori-
zontally projected at F. If now we suppose the face AHG'
to be revolved around AH till it shall make with the plane of
the paper an angle equal to the angle A of the spherical tri-
angle, the point G' will describe a circle whose plane is perpen-
dicular to HA ; and when the face makes the required angle,
the point G' is horizontally projected at F : hence, HF is the
projection of the intersection of the face AHG' with the un-
known face of the pyramid. Revolving this third face around
HB till it coincides with the horizontal plane, the point of which
F is the projection falls at G, in a perpendicular to HB, and at
a distance from B equal to the hypothenuse of a triangle whose
base is BF and altitude FF" or FF', the height of the point
above the plane of the paper. The angle BHG is, therefore,DESCRIPTIVE GEOMETRY.
108
equal to the side of the triangle opposite the angle A, and F'BI .
to the angle opposite the side b. The angle C, opposite the
side c, can be found as in Case I.
CASE III.
Having given two angles and the included side of a spherical
triangle, required the other parts.
§ 151. PI. 2. Fig. 1. Let ABC be the triangle, A and B the
given angles, and c the given side.
Draw, in the plane of the paper, the lines HA and HB,
making the angle AHB equal to the given side c. At any point
of HB, as B, let a perpendicular Bí be drawn, and let the angle
IBP be made equal to the angle B of the triangle. Conceive
the line Bí' to be situated in the vertical plane whose trace is
FBI, it will then be a line of the face which makes an angle
equal to B with the plane of the paper; the lines Bí' and HB
determine the position of this face. Drawing, in like manner,
a perpendicular AE to the edge HA, and making the angle
EAD equal to the angle A of the triangle, we determine AD, a
line of the face which passes through HA, and makes an angle
equal to A with the plane of the paper. The line AD, whose
position in space is directly over AE, and the line HA, deter-
mine the position of this face. The pyramid is therefore
determined, and the three parts which are unknown are the
required parts of the triangle.
Let the unknown faces be intersected by a plane parallel to
the plane of the paper ; it will intersect the faces in lines respect-
ively parallel to HA and HB. Suppose I"N to be the dis-
tance of this plane from the plane of the paper ; PN is the pro-
jection of its intersection with the face passing through HA.
From B lay offBd equal to I"N, and draw dd' parallel to BS ;
d!S is equal to Bí?, and, consequently, to NI" ; therefore, S is
one point of the projection of the line in which the parallel
plane intersects the second unknown face. But the line ofSPHERICAL TRIGONOMETRY.
109
intersection is parallel to HB ; therefore, its projection which
passes through S must be parallel to HB. The point P is, there-
fore, the projection of a point common to both the unknown
faces : hence, it is the projection of a point of their intersection,
and HPE is the projection of the intersection. Produce AN to
E, and draw the perpendicular ED. With A as a centre and
radius AD describe the arc DG, and draw HG : AHG is equal
to the side of the spherical triangle opposite the angle B. By
producing BS to I, erecting the perpendicular IP, describing the
arc PF with the centre B, and drawing the line HF, we deter-
mine the angle BHF, equal to the side of the spherical triangle
opposite the angle A. The unknown faces might have been
found thus: from P draw a perpendicular POM to the edge
HO, and make OM equal to AI" ; M is a point in the edge of
the unknown face revolved into the plane of the paper. The
point M' is found by making O'M' equal to Bdf. The angle C
opposite the side c, can be found as in Case 1.
CASE IV.
Having given two angles and a side opposite one of themn
required the other parts.
§ 152. PI. 2. Fig. 2. Let ABC be the triangle, A and B the
given angles, and b the given side.
Let HA be the intersection of the known face with the plane
of the paper. Make the angle AHN equal to the side b ; AHN
is the known face revolved on the plane of the paper. At any
point of HA, as A, draw the perpendicular NAI, and make the
angle IAP equal to the angle A of the triangle : AP is the re-
volved position of the intersection of the known face with the
plane NAI. With A as a centre and radius AN let the arc
NI be described. From the point I', in which it intersects the
line AÍ', demit the perpendicular PI on the plane of the paper.
In the vertical plane NAD let. the line I'D be drawn, making
the angle II'D equal to the complement of the angle B of the
spherical triangle ; the angle IDI' is equal to the angle B. I et110
DESCRIPTIVA GEOMETRY.
the right-angled triangle I'D I be revolved around its perpen-
dicular I I ; it will generate a right cone whose elements make
with the plane of the paper angles equal to the angle B. Let
a tangent plane be drawn to this cone through the point H :
this plane will make with the plane of the paper an angle equal
to the angle B. Its trace is HB, drawn tangent to the circle
DB, and BHA is equal to the side c.
The side a is found by revolving II' till it becomes perpen-
dicular to IB, joining I" and B, and describing the arc I"N'
around the centre B. The angle C might be found as in
Case I. If the angle B were obtuse, the tangent plane would
be passed on the other side of the cone ; if it were a right angle,
I'D would coincide with I'l, and the cone would be reduced
to a right line ; in this case the trace of the tangent plane would
pass through the point I.
CASE Y.
Having given two sides and an opposite angle of a spherical
triangle, to find the other parts.
§ 153. PL 3. Fig. 1. Let ABC be the triangle, a and h the
given sides, and A the given angle.
Let the side b of the spherical triangle be placed on the hori-
zontal plane, and the angle AHC be made equal to it. Make
the angle CHF equal to the side a ; CH is the intersection of
the known faces of the spherical pyramid. At any point, as
A, of the line HI, draw the perpendicular GAS, and make the
angle DAD' equal to the angle A of the triangle ; AD' is a line
of the unknown face revolved into the horizontal plane around
GS as an axis. At any point, as C, of the line HC, conceive
a plane to be drawn perpendicular to HC. This plane and the
plane of the lines AD and AD' are perpendicular to the hori-
zontal plane ; their line of intersection is, therefore, perpen-
dicular to it at D. This perpendicular, when revolved into the
horizontal plane around GD, takes the position DD' ; but when
revolved around FI as an axis, it takes the position DD". If aSPHERICAL TRIGONOMETRY.
ill
line be drawn from D" to I, it will represent the revolved posi-
tion of the intersection of the plane FCI with the unknown face
of the pyramid. Let the face CHF be now revolved about
HC ; it will describe a cone which the plane of the. unknown
face will intersect, in general, in two elements ; and when the
face CHF has such a position that the line HF is either of
these elements, all the conditions of the problem are evidently
answered, and the remaining parts of the pyramid are easily
determined. Were the cone described, the plane FCI would
intersect it in a circle of which C is the centre and CF the
radius ; the points E and E", in which this circle meets the line
ID", are the two points at which the elements in question pierce
the plane FCI. As the remaining part of the construction is
the same, whether we take the element that pierces at E or the
one which pierces at E", we can use either, and will take the
one which pierces at E. The line drawn from H to the true
position'of the point E is, therefore, the intersection of the face
CHF with the unknown face of the pyramid, and HE' is the
horizontal projection of this intersection. From the point S,
in which the projection of the intersection meets GAD, draw
SS' perpendicular to AD ; SS' is the height, above the plane
of the paper, of that point of the intersection which is projected
at S. With the centre A and radius AS'describe the arc S'G,
and draw HG ; AHG is equal to the side c of the spherical
triangle.
To find the angle C : join the points C and E ; ECE' is equal
to the angle C. The remaining angle can be found as in Case I.
If ID" were tangent to the circle FEE", the unknown face
IHG would be tangent to the cone ; in this case there would
be but one result, or one third side of the triangle, which would
answer the conditions of the problem. If the line ID" do not
touch or. cut the circle FEE", the conditions of the problem are
impossible, and then no triangle can be constructed with such
data.112
DESCRIPTIVE GEOMETRY*
CASE VI.
The three angles of a spherical triangle being given, to find
the sides.
$ 154. PI. 3. Fig. 2. Let ABC be the triangle, and A, B,
and C the given angles.
Let G'l be the intersection of two planes at right angles to
each other. Draw a plane perpendicular to the vertical plane,
and making one of the given angles, as A, for example, with the
horizontal plane : bo, be are its traces. If now a plane can be
drawn, making with the plane (bo, be), and with the horizontal
plane, angles respectively equal to the angles C and B, these
three planes will form a triangular pyramid whose died ral angles
will be respectively equal to the angles of the triangle. To
draw this plane, take any point within the angle which the plane
(bo, be) makes with the horizontal plane ; and let this point be
the common vertex of two right cones whose axes are respect-
ively perpendicular to the planes, and whose elements make
with them angles respectively equal to the angles C and B of
the triangle. If a plane be then dravm tangent to these cones,
it will make the same angles with the planes as the elements of
the cones make ; that is, angles equal to the angles C and B of
the triangle : hence, this tangent plane is the plane sought.
Let V be taken for the common vertex of the two cones ;
it is assumed in the vertical plane in order tp render the construc-
tion more easy. Draw Y o' perpendicular to the horizontal
plane, and Yn perpendicular to the plane (bo, be). Make the
angle o'Vd equal to the complement of the angle B, and the
angle nYf equal to the complement of the angle C. Let the
right-angled triangles Yo'd and Yvf be revolved about their per-
pendiculars Yo' and Yn ; they will generate the two right cones
whose elements make, respectively, with the horizontal and
oblique planes, angles equal to the given angles B and C. It is
now required to draw a tangent plane to these cones.
Inscribe a sphere in each cone, and draw a plane tangent toSPHERICAL TRIGONOMETRY.
113
these spheres through the point Y ; this plane will be tangent
to both cones, and, consequently, the plane required. Let the
spheres to be inscribed be equal. At d draw dg perpendicular
to the element Yd ; the point g, in which it meets the axis of
the cone, is the centre of a sphere inscribed in the cone whose
axis is Vo' ; gd is the radius of the sphere, and qd is the ver-
tical projection of its circle of contact with the cone. To
inscribe an equal sphere in the other cone : at the vertex Y,
draw Yk perpendicular to the element Ym, and make it equal
to dg ; draw km parallel to Vn, and from the point m, in which
it cuts the element produced, draw mu parallel to Yk, or per-
pendicular to Ym : u is the centre and um the radius of the
second sphere ; ms is ttye vertical projection of its circle of con-
tact with the cone. It is required to draw a tangent plane to
these spheres through the point Y.
Let the plane (bo, be) be moved parallel to itself till it em
braces the circle of contact of which ms is the projection ; it
then becomes the plane (FD, FG). Let the points u and g he
joined ; ug is a line of the vertical plane. Suppose a cylinder
of which this line is the axis to be tangent to both spheres. A
plane drawn through the point V, tangent to this cylinder, will
be tangent to both the spheres and to both the cones, and will,
consequently, be the plane required. But since this plane is
tangent to all the surfaces, the element in which it touches the
cylinder touches the cones and spheres, and therefore passes
through the points in which the circles of contact of the cylin-
der and spheres intersect the circles of contact of the cones and
spheres ; for these are the only points common to all the sur-
faces. Through g let gt be drawn perpendicular to gu ; it
will represent the vertical projection of the circle of contact of
the sphere whose centre is g with the Cylinder. The plane of
this circle intersects the plane of the circle of contact of the
sphere and cone in a horizontal line, which is vertically pro-
jected at t ; the points in which this line intersects the circle
of contact, are common to the cone, sphere, and cylinder.
The element of the cylinder passing through either of these
points, is the element through which the tangent plane is to be
8114
DESCRIPTIVE GEOMETRY.
drawn. The horizontal trace of this plane must be tangent to
the base of the cone (84) ; the line DiT, drawn tangent to the
base of the cone at the point t, is therefore the horizontal trace.
As the tangent plane is to contain the point Y, its vertical trace
is IVG. The point D is the vertex of the pyramid, and the
angles included between the edges are easily found. Revolve
the plane (DF, FG) into the horizontal plane, around DF as an
axis. The point G falls at G', and DG' is the revolved position
of the edge of the pyramid of which DG" is the horizontal pro
jection : the angle FDG' is equal to the side b of the triangle.
We can verify this result by considering the circle of contact
sm. In the revolution, its centre p describes the arc pp' : with
p as a centre, and pV, equal to ps, as radius, describe a circle ;
the line DG' should be tangent to this circle. The point which
is common to the sphere, cone, and cylinder is vertically pro-
jected at h : this point, in the revolution, describes the arc hh! :
the ordinate h'h" should pass through the point of contact. The
side a is easily found, and may be constructed by the student.
r t . w/, u/ hSPHERICAL PROJECTIONS
115
SPHERICAL PROJECTIONS.
CHAPTER XI.
FUNDAMENTAL PRINCIPLES.
■$ 155. To conceive of the whole surface of the earth, and
the positions of objects situated on it, it is necessary to have
recourse either to artificial globes or to drawings which repre-
sent the earth and the different points of its surface.
§ 156. As it is quite difficult to construct artificial globes, and
indicate on them the different places upon the surface of the
earth, as well as their relative positions, the method by draw-
ings, or the representation on planes, has been generally
adopted.
§ 157. Spherical Projections show the manner in which
these drawings are to be made, so that they shall present to the
eye, situated at a particular point, the same appearance as the
sphere would present if the drawing were removed and the
sphere placed in its stead.
§ 158. Three kinds of projections are generally used to make
these representations, viz. the Orthographic, the Stereographic,
and the Globular.
§ 159. The plane of that circle of the sphere on which the
representation or projection is made, is called the Primitive
Plane ; and the intersection of this plane with the surface of the
sphere, the Primitive Circle.
8*! I Ü
DESCRIPTIVE GEOMETRY.
§ 160. In the Orthographic projection tiie eye is supposed to
be situated in the axis* of the primitive circle, and at an infinite
distance from its plane.
§ 161. In the Stereographic projection the eye is placed at
the pole of the primitive circle, whose plane in this, as well as
in the other projections, is supposed to offer no obstruction to
seeing that part of the sphere which lies beyond it.
§ 162. In the Globular projection the eye is supposed to be
situated in the axis of the primitive circle, without the sur-
face of the sphere, and at a distance from it equal to the sine
of 45°.
§ 163. It has sometimes been found convenient to draw a
cylinder tangent to the earth in the circumference of some one
of its great circles, or a cone tangent to it in the circumference
of one of its small circles ; to suppose the eye to be at the centre
of the sphere, and to project from this point all the circles on
the tangent cylinder, or cone : then, by developing the surface
of such tangent cylinder, or cone, the surface of the sphere will
be reduced to a plane, and we can easily conceive of the dif-
ferent positions of its points. The Orthographic and Stereo-
graphic methods are, however, the most common and the most
useful. We shall treat of these projections only.
§ 164. Before examining in what manner the sphere is to be
projected by these methods, we shall define those points and
circles of the sphere to which particular names have been
given, which are used by geographers in locating places on
the surface of the earth, and which are generally delineated on
maps.
§ 165. The line about which the earth revolves is called its
* The axis of a circle is a line passing through its centre, perpendicular to
its plane ; the points in which this line meets the surface of the sphere are called
the poles of the circle.’ Either pole is at the same distance from every point of
the circumference of the circle ; since, if a line be perpendicular to a plane all
points of the plane equidistant from the foot of the perpendicular are equidistant
from any point of the line. The distance from the pole to any point of the cir-
cumference of a circle, measured on the sphere, is called the polar distance of
the circle.SPIIEItlCAL PROJECTIONS.	117
axis ; the points in which the axis pierces the surface are called
the poles ; one the north, the other the south pole.
§ 166. The circumference of the great circle whose plane is
perpendicular to the axis, is called the equator ; and this cir-
cumference is 90° from the poles.
§ 167. Circles whose planes pass through the axis of the
earth, and which are consequently perpendicular to the equator,
are called meridians. Twelve entire circles, or twenty-four
semicircles, fifteen degrees distant from each other, are gene-
rally represented on the maps of the earth : the semicircles are
called hour circles. Although every point on the surface of
the earth has its meridian, yet, for the sake of convenience, we
shall project the hour circles only.
§ 168. The distance of any point from the equator, measured
on the meridian passing through it, is called the latitude of the
point, and is north or south according as the place is north or
south of the equator.
§ 169. Small circles parallel to the equator are called par-
allels of latitude.
§ 170. The ecliptic is a great circle making an angle of 23°
30' nearly, with the equator ; the points in which it intersects
the equator are called the equinoctial points.
§ 171. The meridian passing through the equinoctial points
is called the equinoctial colure, and the meridian which is per-
pendicular to this last is called the solstitial colure. The points
in which the solstitial colure intersects the ecliptic are called
solstitial points ; the parallels of latitude passing through these
points are called tropics; the one north of the equator the
tropic of Cancer, the one on the south of it the tropic of Cap-
ricorn.
§ 172. The parallels of latitude which are 23° 30' distant
from the poles, measured on a meridian, are called polar circles ;
the one around the north pole the arctic circle, and the one
around the south pole the antarctic circle.
§ 173 The horizon of any place is the circumference of a
great circle whose plane is perpendicular to the radius passing
through the place.DESCRIPTIVE GEOMETRY.
118
§ 174. The elevation of the pole above the horizon of anyplace
is equal to the latitude of that place. For, let (PI. 1. Fig. 1)
HESP be the meridian passing through the place P on the
surface of the sphere ; HO perpendicular to PP', the horizon,
NS the axis, and EQ the equator. The arc NH measures the
elevation of the pole above the horizon, and QP is the latitude
of the place (168). But the arc PNH is equal to 90° (173),
and QPN is also equal to 90° (166) ; taking away the common
part PN, there remains NH equal to PQ ; that is, the elevation
of the pole above the horizon of any place is equal to the latitude
of that place.
We may also remark, that the distance QO from the
equator to the horizon is equal to NP, the complement of the
latitude.
§ 175. The angle included between the planes of twTo circles
is equal to the angle contained by their axes, since the axes
are respectively perpendicular to the planes. This is shown
in Fig. 1, where NS is the axis of the equator, and PP' the
axis of the horizon: it is evident that the angle NP'P, con-
tained by the axes, is equal to the angle HP'E, contained by the
planes.
§ 176. The line of measures of any circle is the line of inter-
section of the primitive plane, and a plane passed through the
axis of the primitive circle and the axis of the given circle.
This latter plane is perpendicular to the planes of both the
circles, since it contains lines respectively perpendicular to
them : hence, its trace on the primitive plane is perpendicular
to the line in which the plane of the given circle intersects the
primitive plane.SPHERICAL PROJECTIONS.
119
CHAPTER XII.
OP THE ORTHOGRAPHIC PROJECTION.
§ 177. In this projection, which is the same as we have used
in Descriptive Geometry to represent geometrical magnitudes,
the eye, or projecting point, being at an infinite distance from
the plane of projection, the projecting lines through the different
points to be projected are perpendicular to the plane of pro-
jection. The manner of making the projections of points and
right lines has already been shown, and no further remarks on
this part of the subject seem necessary.
§ 178. The projections of all circles whose planes are per-
pendicular to the primitive plane are right lines (82).
§ 179. Every circle which is parallel to the primitive plane
is projected into a circle equal to itself ; for the projecting lines
through the different points of its circumference form the sur-
face of a right cylinder ; and the intersections of a right cylin-
der, by parallel planes, are equal.
THEOREM I.
The projections of all circles oblique to the primitive plane are
ellipses.
§ 180. PI. 1. Fig. 2. Let ADB be a circle in the plane of
the paper, AB one of its diameters, and CD a radius perpen-
dicular to AB.
Revolve the plane of this circle around AB till the point
D shall be elevated above the plane of the paper any con-
venient distance, as D'D" ; join C and D" ; the plane of the
triangle D"CD', in its true position, is perpendicular to the
plane of the paper, though it is now revolved into this plane
around the axis CD. The angle D"CD' is equal to the angleDESCRIPTIVE GEOMETRY.
120
which the plane of the circle to be projected makes with the
plane of the paper, and D' is the projection of one point of its
circumference. Now we have this proportion, CD" : CD' : :
radius : co-sine of the inclination of the planes:	At any point
of AB, as N, conceive a plane to be drawn perpendicular to it,
and let this plane be revolved around NI till it coincides with
the plane of the paper ; the point of which I' is the projection
falls at I", making the distance NI" equal to NI : the angle
I"NI is equal to the inclination of the planes ; that is, equal to
the angle D"CD'. But in the triangle NI'T we have NI' :
NI' : : radius : co-sine of the inclination of the planes. Com-
paring this with the previous proportion, we see that the third
and fourth terms of each are the same ; therefore, the first
couplets are proportional ; that is, CD" : CD' : : NI" : NI', or
CD"2 : CD'2 : : NI"2 : NI'2. But CD"2, or CD2, is equal to
AC.CB ; and NI"2, or NI2, is equal to AN.NB : therefore,
AC.CB : AN.NB : : CD'2 : NI'2; that is, the rectangles of the
abscissas are to each other as the squares of their corresponding
ordinates ; and as this is a known property of an ellipse, we con-
clude that the projections of all circles oblique to the plane on
which they are projected are ellipses. The semicircle which Ì3
above the plane of the paper is projected into the semi-ellipse
AD'B ; the semicircle below the plane of the paper, into the
semi-ellipse AFB.
§ 181. We see that the transverse axis AB is equal to the
diameter of the circle, and that the semi-conjugate axis CD' is
equal to the cosine of the inclination of the plane to the radius
of the circle which is projected. If the plane on which the
projection is made should not pass through the centre of the
circle, its projection is still an ellipse : for, conceive a plane to
be passed through its centre pai allei to the primitive plane, the
projections of the circle on these parallel planes are equal
curves. The projection of that diameter which is parallel to
the primitive plane, is the transverse axis of the ellipse ; for
this diameter is projected into its true length, and all the other
diameters, being oblique to the primitive piane, are projected
into lines less than themselves. That diameter of the circleSPHERICAL PROJECTIONS.
121
which is perpendicular to the one that is parallel to the primi-
tive plane, will, in projection, be the conjugate axis of the ellipse.
For, the two lines being at right angles in space, and one of
them parallel to the plane on which they are projected, their
projections are also at right angles (51). The conjugate axis
will evidently lie in the line of measures, since its projecting
plane is perpendicular both to the primitive plane and to the
plane of the projected circle. The length of the conjugate axis
is the same, whether the primitive plane does or does not pass
through the centre of the circle which is projected. The pro-
jections of all circles made by the orthographic method are
either right lines, circles, or ellipses.
§ 182. If the whole surface of a sphere were projected on a
plane passing through the centre* it is evident that each point
within the circumference of the primitive circle would be the
projection of two points of the surface of the sphere, since each
projecting line meets the sphere in two points. In order to
delineate the whole surface, so that each point of projection
shall represent but one point of the surface, we generally pro-
ject that hemisphere which is nearest the eye, and then revolve
the other hemisphere 180°, around a line tangent to the primi-
tive circle, thus bringing it between the eye and the primitive
plane ; and then project it from this position.
PROBLEM I.
To project the circles of the sphere on the plane of the equinoctial
colure.
§ 183. PI. I. Fig. 3. Let SE'NE be the equinoctial colure,
and suppose the eye to be situated above the plane of the paper.
Assume any point, as N, for the place of the north pole, and
draw the line NIS ; NIS is the projection of the solstitial colure,
and S is the projection of the south pole. The equator, being
a great circle perpendicular to the primitive plane, is projected
into EE' perpendicular to NS, and E and E' are the equinoct ial
points. The ecliptic passes through E and E' ; and as it makes122
DESCRIPTIVE GEOMETRY.
with the equator an angle of 23|°, it makes with the equinoc-
tial colure an angle equal to 66£° ; its projection, therefore, is an
ellipse whose transverse axis is EE', and whose semi-conjugate
axis is ID, the cosine of 66 £°, to the radius IE. Laying off
from S, with a scale of chords, or protracter, SC equal to 66^°,
and drawing CD parallel to EE', determines ID, the cosine of
66 £°, and the semi-conjugate axis of the ellipse EDE. The
projections of all the meridians have the common transverse
axis NS ; and laying off from I, on their common line of meas-
ures EE', the cosines of 15°, 30°, 45°, &c. determines the
extremities of their conjugate axes, and having the axes, the
ellipses are easily described. The half of the meridian which
lies above the plane of the paper, and makes with it an angle
of 60°, is projected into the semi-ellipse SFN ; IF, the cosine
of the inclination, is equal to half the radius IE'. The semi-
meridians which lie between the solstitial colure and the semi-
circle NES are projected by laying off the cosines of their
inclinations from I towards E.
The parallels of latitude, being perpendicular to the primitive
plane, are projected into right lines (82). To find the projec-
tion of the arctic circle, lay off from N to A' 23i°, and draw AA'
parallel to EE' ; A'A is the projection of the arctic circle. The
projection of the tropic of Cancer is found by laying off NB
equal to 66^°, and drawing through B a parallel to EE'. The
antarctic circle and the tropic of Capricorn are found in a
similar manner. The projection of the tropic of Capricorn
passes through D, and is tangent to the ellipse EDE' ; for D is
the projection of the southern solstitial point, at which point the
ecliptic touches the tropic in space.
Let the hemisphere which is behind the plane of projection
be revolved 180° around a line through E' parallel to NS ;
this brings the hemisphere in front of the primitive plane. It
now has the same position with the primitive plane as the hemi-
sphere which has been projected, and its projection is made in
a similar manner. The line E'Q is the projection of the remain-
ing half of the equator, and E'D'Q of the remaining half of the
ecliptic. The line N'S' is the projection of the half of the sol-SPHERICAL PROJECTIONS.
123
stitial colure corresponding to NS, and N'GS' the projection
of the half of the meridian corresponding to SFN ; the projec-
tions of the parallels of latitude are also drawn in the figure.
§ 184. If the projection of the sphere were made on the sol-
stitial colure, it would be the same in every respect as the one
just constructed on the equinoctial colure, excepting that the
ecliptic would be projected into the right line CIB passing
through the centre of the primitive circle : for, the solstitial
colure is perpendicular both to the ecliptic and equator. Con-
ceive the line EDE'D'Q to be removed, and the figure will
represent the projection of the sphere on the plane of the sol-
stitial colure.
PROBLEM II.
To project the sphere on the plane of the equator.
§ 185. PI. 2. Fig 1. Let AEBD represent the equator, and
suppose the eye to be placed on the north of it : under this sup-
position the northern hemisphere will be first projected.
The meridians are projected in right lines passing through
the centre N : for the planes of the meridians are perpendi-
cular to the primitive plane and pass through the axis of the
sphere, and the axis is projected at N. The centre N is also the
projection of the north pole. Let A and B be assumed for the
equinoctial points; ANB is the projection of the equinoctial
colure, and DNE of the solstitial colure. To project a meri-
dian making any angle with the equinoctial colure : lay off from
A an arc AF equal to this angle, and through the extremity F
draw the diameter FNN' : this line is the projection required.
Every diameter passing through N is the projection of a meri-
dian. The ecliptic makes an angle of 23with the equator,
and passes through the equinoctial points A and B : hence AB
is the transverse axis, and NI the cosine of 23¿°, the semi-
conjugate axis of the ellipse into which it is projected.
The parallels of latitude being parallel to the primitive plane,
their projections are circles (179) ; N, the projection of theDESCRIPTIVE GEOMETRY.
124
axis of the sphere, is their common centre, and the radii with
which they are described are the sines of their polar distances ;
for the radius of any small circle of the sphere is equal to the
sine of its polar distance. The projections of the tropic of
Cancer and the arctic circle are described about N as a centre,
and with radii respectively equal to the sine of 66¿°, and the
sine of 23|°. The projection of the tropic of Cancer passes
through the point I. By revolving the southern hemisphere
in front of the primitive plane, around a line tangent to the
primitive circle at B, its projection can be made in the same
manner. The remaining half of the ecliptic, the antarctic circle,
whose radius is SQ', equal to the sine of 23|°, the tropic of
Capricorn, whose radius is Sí', and the meridians made in the
figure, are easily recognised.
PROBLEM III.
To project the sphere on the horizon of any place ; that place9
for example, the latitude of which is 45° north.
§ 186. PL 3. Fig. 2. Let ADBC be the horizon, and A and
B the equinoctial points. The elevation of the pole above the
horizon is equal to 45°, the latitude of the place (174).
The equinoctial colure passes through the points A and B,
and makes an angle with the horizon equal to the elevation of
the pole above it ; that is, equal to the latitude of the place.
The line AB is the transverse axis of the ellipse into which it
is projected ; and ON, the cosine of 45°, is its semi-conjugate.
The semi-ellipse ANB is the projection of that part of the colure
which is above the horizon. The point N is the projection of
the north pole, and CN is the versed-sine of the latitude. The
equinoctial colure intersects the planes of the parallels of latitude
in lines parallel to AB, its intersection with the equator ; that
is, in lines parallel to the primitive plane, since AB is a line of
that plane. These lines are diameters of the parallels of lati
tude, their projections are equal to the lines themselves, andSpherical Projections
./ Í'.'.ÍV/Y//. ,SPHERICAL PROJECTIONS.
125
are the transverse axes of the ellipses into which the parallels
are projected ; and the vertices of these axes are all found in
the projection of the equinoctial colure. Let the plane of the
equinoctial colure be revolved around AB till it coincides with
the primitive plane. The pole falls at C. From C lay off 23^°
to E ; the chord EG is the line of intersection, in its revolved
position, of the plane of the equinoctial colure and the plane of
the arctic circle. Let the plane be revolved back again ; E'G'
is the projection of this diameter, and is the transverse axis of
the ellipse into which the arctic circle is projected. In like
manner, making CL equal to (56i°, drawing LP parallel to AB,
erecting the perpendiculars LL' and PP' to AB, determines
L'P', the transverse axis of the ellipse which is the projection
of the tropic of Cancer. The transverse axes of the projec-
tions of any number of parallels of latitude may be found in
the same manner.
To find the conjugate axes. The points in wThich the trans-
verse axes intersect the line CD are the centres of the ellipses.
The planes of the parallels of latitude, being parallel to the
plane of the equator, make the same angle with the primitive
plane ; that is, an angle equal to the complement of the latitude,
or 45°. Find then the cosine of 45° to the radius of the par-
allel ; this cosine is the semi-conjugate axis of the ellipse into
which the parallel of latitude is projected. The conjugate axis
is determined by laying off this distance on both sides of the
centre of the ellipse in the line CD (180). Those parallels of
latitude whose northern polar distances are less than 45° will
be entirely above the horizon, and will therefore be seen, while
those whose polar distances are greater than 45° pass below the
horizon, and therefore a part of them will not be seen. The
tropic of Cancer passes below the horizon at the points b and a.
The equator and ecliptic will be next projected. The line
AB is the transverse axis of the ellipse which is the projection
of the equator. Laying off from O the distance OU equal to
the cosine of 45°, OU is the semi-conjugate axis. The semi-
ellipse AUB is the projection of that part of the equator which
is above the primitive plane. If that half of the ecliptic which9
126
DESCRIPTIVE GEOMETRY.
is on the upper hemisphere lie between the equator and north
pole, it would make with the primitive plane an angle greater
than the angle which the equator makes by 23|° ; but if it lie
between the equator and south pole, it wrill make with the
primitive plane an angle less than the angle which the equator
makes by 23£°. It is taken in the latter position. AB is the
transverse axis of the ellipse into which it is projected, and OU',
equal to the cosine of 211°, is the semi-conjugate axis, and
AU'B is the projection of that part of the ecliptic which is above
the primitive plane.
To project the meridians. Let a tangent plane be drawn
to the sphere at its north pole : this plane will be perpendicular
to the axis, and will intersect the planes of the meridians in lines
making angles of 15° with each other ; the points in which
these lines pierce the primitive plane are points of the traces
of the meridian planes. To pass this plane. Suppose the
plane of the solstitial colure to be revolved around CD till it
coincides with the primitive plane ; the solstitial colure th
coincides with the primitive circle, and the pole falls at h.
Through N' draw the tangent line N'S ; the point S, in which
it meets DC produced, is a point of the trace of the tangent
plane. But, as the plane is to be perpendicular to the axis of
the sphere, its trace must be perpendicular to the projection of
the axis (49) ; therefore SH, drawn perpendicular to DS, is
the trace of this plane. Let this tangent plane be revolved
around its trace SH, from the sphere, till it coincides with the
primitive plane ; the pole falls at N", SN" being made equal to
SN'. Drawing N"R, making the angle SN"R equal to 15°,
determines R, a point of the trace of the meridian plane which
makes an angle of 15° with the solstitial colure ; for N"R is the
revolved position of the intersection of this meridian plane with
the tangent plane, and the point R, being in the trace, remains
fixed. But the point O is another point of the trace of this
meridian plane ; the trace can therefore be drawn. Laying
off the angle SN'R' equal to 30° determines R', a point in the
trace of the meridian plane that makes an angle of 30° with the
solstitial colure. Thus, laying off at N", and from the line8a.	TO»®...
SPHERICAL PROJECTIONS.
N"D, an angle equal to the angle which any meridian plane
makes with the solstitial colure, determines a point of its trace
on the primitive plane. The line N"R" makes an angle of 45°
with N"D : hence, R'C'V is the trace of that meridian plane
which makes an angle of 45° with the solstitial colure, and
C'V is the transverse axis of the ellipse into which the meridian
is projected. In a similar manner, the transverse axis of the
ellipse into which any meridian is projected can be found. It
only remains to find the conjugate axes, and then the ellipses
can be described. To find the conjugate axis of the ellipse
whose transverse axis is C'Y. Through the pole let a plane
be drawn perpendicular to C'Y ; NA' is its horizontal trace.
Let this plane be revolved to coincide with the primitive plane ;
the pole falls at P" ; A'P" is the intersection, revolved, of the
perpendicular and meridian planes, and P"A'N is equal to the
angle included between the meridian and primitive planes.
Find the cosine of this angle to the radius of the primitive
circle, and lay it off from O to S', in the line of measures OS' ;
OS' will be the semi-conjugate axis. Or, the semi-conjugate
axis may be found by a better construction, thus : produce A'N
and OS' till they meet the circumference of the circle in the
points N" and S". Draw the line S"N", and produce it till it
intersects VC'R" at M. Draw a line through M and N ; the
point S', in which it meets OS", is the extremity of the semi-
conjugate axis. The semi-ellipse V S'C' is the projection of that
part of the meridian which is above the primitive plane. In the
same manner any number of meridians can be projected. The
other hemisphere is revolved in front of the primitive plane,
and its projection made in the same way. The figure shows
the circles that are projected.
§ 187. By considering the principles of the orthographic
projection, we see that if the primitive circle be a great circle
of the sphere, all the points of the surface are projected within
it, and that the projection of any point is at a distance from the
centre of the primitive circle equal to the sine of the arc inter-
cepted between the point on the surface of the sphere and either
Dole of the primitive circle.128
DESCRIPTIVE GEOMETRY.
The poles of any circle are projected in its line of measures
at distances from the centre of the primitive circle equal to the
sine of its inclination. They are projected in the line of mea-
sures, since they are projected in the trace of the plane passing
through the poles and perpendicular to the primitive plane (176).
They are projected at distances from the centre of the primi-
tive circle equal to the sine of the inclination ; since the arcs
intercepted between the poles of the primitive circle and
the poles to be projected, are equal to the inclination of the
circles (175)
CHAPTER XIII.
OF THE STEREOGRAPHIC PROJECTION.
§ 188. In this projection, the eye, or projecting point, is sup-
posed to be at the pole of the primitive circle (161).
§ 189. The projection of any point of the surface of a sphere,
is the point in which the line drawn through it and the eye
pierces the primitive plane.
§ 190. The tangent of half an arc is called the semi tangent
of the arc : thus, if the arc be sixty degrees, its semi-tangent is
the tangent of thirty degrees.
§ 191. The polar distance of a point is its distance from that
pole of the primitive circle which is opposite the eye.
THÉOREM I.
The projection of any point of the surface of a sphere is at a
distance from the centre of the primitive circle equal to the
semi-tangent of its polar distance.
§ 192. PI. 2. Fig 2. Suppose a plane to be passed through
the point to be projected and the axis of the primitive circle,
and let the plane of the paper be this plane : AC PB is the cir-SPHERICAL PROJECTIONS.
129
cumference of the circle in which it intersects the sphere.
Let A be the place of the eye, or projecting point, and BC the
trace of the primitive plane ; this trace is perpendicular to AP,
since AP is the axis of the primitive circle. Let D be the point
to be projected. Draw the line AD ; the point D', in which
it pierces the primitive plane, is the projection of the point D.
With A as a centre, and radius AP', let the arc P'E be described.
The angle DAP, being at the circumference, is measured by
half the arc PD ; but P'D' is the tangent of the angle P'AD' to
the radius of the primitive circle ; it is therefore the semi-tan-
gent of the polar distance PD (190).
§ 193. It follows, from the preceding demonstration, that the
projections of all points which have equal polar distances are
equidistant from the centre of the primitive circle : hence, all
circles which are parallel to the primitive plane are projected
into circles ; the radii of the projections of such circles are the
semi-tangents of their polar distances.
§ 194. The tangent of 45° being equal to radius, it follows
that when the primitive plane passes through the centre of the
sphere, all points whose polar distances are less than 90° will be
projected within the circumference of the primitive circle, and
all points whose polar distances are greater than 90° without
it. The polar distances of the points of the primitive circle
being 90°, it follows that they are neither projected without nor
within it, but in it : hence, the primitive circle is its own pro-
jection. The polar distance of the pole opposite the eye being
nothing, this pole is projected at the centre of the primitive
circle ; and the eye, whose polar distance is 180°, is projected
at an infinite distance from the centre of the primitive circle.
It is easily shown from the figure, that all points of the semi-
circle BPC are projected within the primitive circle, and all
points of the semicircle BAC without it. If through the eye,
at A, and any point of the surface of the sphere, a line be drawn,
and the point be then moved along the surface of the sphere
towards A, the line will make a less and less angle with the
primitive plane, and w7hen the point unites with A the line
becomes parallel to the primitive plane and tangent to the
9130
DESCRIPTIVE GEOMETRY.
sphere. Hence the eye is projected on the primitive plane at
an infinite distance from the centre of the primitive circle.
§ 195. If the plane of a circle pass through the eye, the pro-
jection of the circle is a right line. For, the lines which are
drawn from the eye to the different points of the circumference
are contained in the plane of the circle, and therefore pierce
the primitive plane in the line in which it is intersected by the
plane of the circle ; that is, in a right line : hence, the projections
of all circles whose planes pass through the eye are right lines.
The projections of the great circles of the sphere pass through
the centre of the primitive circle.
§ 196. The projections of right lines which pass through the
eye are points. Right lines which do not pass through the eye
are projected into right lines. For, if through the eye lines be
drawn to the different points of the right line to be projected,
they form a plane ; the intersection of this plane with the primi
tive plane is the projection of the line.
THEOREM II.
The projections of all circles oblique to the primitive plane, and
whose planes do not pass through the eye, are circles.
§ 197. PI. 2. Fig. 3. Let the circle to be projected be a
small circle.
Through the axis of the circle and the axis of the primitive
circle suppose a plane to be passed : this plane may be taken
for the plane of the paper. The circle AC OB is its intersec-
tion with the sphere. The primitive circle and the circle to
be projected are perpendicular to this plane. Let A be the
position of the eye, CB the trace of the primitive plane, and ED
the orthographic projection of the circle to be projected. Con-
ceive the circle to be circumscribed by a cone of which A is the
vertex : the intersection of this cone by the primitive plane is
the projection of the circle. It is then to be shown that this
intersection is a circle.
The point D is projected at D\ and the point E at E'. TheSPHERICAL PROJECTIONS.
131
angle AED being at the circumference, is measured by half
the arc ABD ; the angle DD'B, formed by the intersection of
two chords, is measured by half the sum of the arcs AC and BD,
or half the arc ABD : hence, the angles AED and AD'E'
are equal. The triangles AED and AE'D' have the angle
EAD common ; they are consequently equiangular, and have
the angle ADE equal to the angle AE'D' : hence, the intersec-
tion of the cone by the primitive plane is a sub-contrary section,
and therefore a circle.* But this intersection is the projection
of the oblique circle : hence, the projection of every small circle
oblique to the primitive plane is a circle. If the primitive plane
be revolved around CB till it coincides with the plane of the
paper, the projected circle will be represented by the circle
described on the diameter E'D'.
§ 198. Had the line ED passed through the centre O', the
* A cone whose axis is oblique to the plane of its base is called a scalene cone ;
and if it have a circular base, a scalene cone with a circular base. If the surface
of such a cone be intersected by a plane parallel to its base, the section is a
circle ; a cutting plane may be oblique to the plane of the base, in a certain
angle, and still intersect the surface in a circle. Let ABC (PI. *Z. Fig. 4) be
the triangle in which a scalene cone with a circular base is intersected by a
plane passing through its axis and perpendicular to the plane of its base. The
line BC is the orthographic projection of the base on the plane of the paper. If
any plane, as EF, be drawn perpendicular to the plane of the triangle ABC,
and making the angle AEF equal to the angle ABC, or the angle AFE equal
to the angle ACB, this plane EF will intersect the surface of the cone in a
circle, and the section is called a sub-contrary section. Through any point of
FE, as I, let the plane LIM be drawn parallel to the plane of the base BHC.
The plane LIM intersects the plane FIE in a line perpendicular to the plane of
the paper at the point I ; this line is a common ord nate of the two curves in
which the planes intersect the surface of the cone. Since the angle AEF is
equal to the angle ABC, that is, to the equal angle ALI, and the opposite angles
EIM and LIF are equal, it follows that the two triangles LIF and EIM are
equiangular :
Therefore, LI : FI : : IE : IM ;
Hence,	LI.IM = FI.IE.
But, since the plane LIM intersects the surface of the cone in a circle, LI.IM is
equal to the square of the ordinate at the point I : hence FI.IE is also equal to
the square of the ordinate of the curve in which the plane FIE intersects the
surface of the cone : consequently, that curve is a circle.
9*132
DESCRIPTIVE GEOMETRY.
circle projected would have been a great circle. It can be
shown, by similar reasoning, that the cone having such a circle
for its base and A for its vertex, would be cut by the primitive
plane in a sub-contrary section.
.	THEOREM III.
If at any point of the surface of a sphere a line he draum tan-
gent to the sphere, and produced till it meets the primitive
plane, the part intercepted between the point of contact and
the primitive plane is equal to its projection.
§ 199. PI. 3. Fig. 1. At the point of contact conceive a
plane to be drawn tangent to the sphere ; this plane will con-
tain the tangent line (88).
Let a plane be passed through the point of contact perpen-
dicular to the tangent and primitive planes, and take this plane
for the plane of the paper : ABDC is the circle in which it
intersects the sphere, BCE is the trace of the primitive plane,
A the place of the eye, and D the point of tangency. The
plane of the paper intersects the tangent plane in a line tan-
gent to the circle in which it intersects the sphere : hence, DE,
drawn tangent to the circle at D, is the trace of the tangent plane.
Since the tangent and primitive planes are perpendicular to the
plane of the paper, their intersection is perpendicular to it ;
therefore, EF, drawn perpendicular to the plane of the paper
at E, is the intersection of these planes. Every tangent line to
the sphere at D pierces the primitive plane at some point of the
line EF. Let the tangent which pierces it at E' be first pro-
jected. The point D is projected at D', and E' is its own pro-
jection ; therefore, D'E' is the projection of the tangent, and it
only remain^ to be proved that this projection is equal to the
tangent DE' in space. The angle ADE, being formed by a
tangent and chord, is measured by half the arc ACD; the
angle DD'C, being formed by two chords, is measured by half
the sum of AB and CD, or half the sum of AC and CD, or
half of ACD : hence, the triangle EDD' is isosceles, and ED is¿¿s.SPHERICAL PROJECTIONS.
133
equal to ED'. But DE', in space, is the hypothenuse of a
triangle of which DE is the base and EE' the perpendicular :
D'E' is the hypothenuse of a triangle of which D'E is the base
and EE' the perpendicular : as the bases of these right-angled
triangles are equal and their perpendiculars the same line, it
follows that their hypothenuses are equal. But DE' is the tan-
gent line in space, and D'E' is its projection ; therefore, the
tangent line intercepted between the point of contact and the
primitive plane is equal to its projection.
§ 200. Drawing another tangent at the point D, as DF, it
will pierce the primitive plane at F, and its projection D'F is
equal to itself. The angle E'DF, which the tangents make with
each other in space, is equal to the angle E'D'F contained by
their projections. For, in the triangles DE'F and E'D'F the
side DE' is equal to D'E', DF to D'F, and the side E'F com-
mon ; the two triangles are therefore equal, and the angle E'DF
is equal to the angle E'D'F ; that is, the angle contained by the
tangents in space is equal to the angle contained by their pro-
jections.
§ 201. If a right line be tangent to a circle of the sphere, the
projection of the right line is tangent to the projection of the
circle. For, the projection of the circle is the intersection by
the primitive plane of the cone of which the circle is the base
and the eye the vertex ; the projection of the right line is the
intersection of the plane passing through it and the eye, with
the primitive plane : but the plane which projects the line is
tangent to the cone which projects the circle ; their intersec-
tions with the primitive plane are, therefore, tangent to each
other.
§ 202. The angle formed by the arcs of two circles intersecting
on the surface of a sphere is equal to the angle contained by
their projections. For, the angle contained by two arcs inter-
secting on the surface of a sphere, is measured by the angle con-
tained by two right lines drawn tangent respectively to the arcs
at their point of intersection : the angle contained by the pro-
jections of these arcs is also measured by the angle of their
tangents : but the projections of the tangents are tangent to[34
DESCRIPTIVE GEOMETRY.
the projections of the arcs (201) ; and the angle contained by
the projections of the tangents is equal to the angle of the tan-
gents in space : hence, the angle contained by the projections
of the arcs is equal to the angle formed by the arcs on the sur-
face of the sphere.
§ 203. If from the centres of the projections of two circles
‘‘adii be drawn to the points in which their circumferences
ntersect, they will make the same angle with each other as
the two tangents drawn to the circles at the same point, since
they are respectively perpendicular to the tangents.
Hence, the radii drawn through either point in which the
projections of circles intersect, make an angle with each other
equal to that which the circles themselves formed on the sur-
face of the sphere.
THEOREM IV.
y he centre of the projection of a great circle is in the line of
measures y at a distance from the centre of the primitive circle
equal to the tangent of the inclination of the circles ; and the
adius with which the projection is described is equal to the
secant of the inclination.
§ 204. PI. 4. Fig. 1. Let ACA'B be a circle passed through
the axis of the primitive circle and the axis of the circle to be
projected, A the place of the eye, P'"BD' the trace of the
primitive plane, and ED the orthographic projection of the circle
to be projected : P'"BD' is its line of measures (176).
The point D is projected at D', and the point E at E' ; E'D'
is a diameter, and S, the middle of E'D', the centre of the circle
into which ED is projected. With S as a centre, and SE' or
SD' as radius, let the circle be described in the primitive plane.
Since all lines in the primitive plane are projected orthographi-
cally in P'"D',they can only be presented to the eye by revolving
the primitive plane to coincide with the plane of the paper.
Let it be revolved around P'"D' : the points E', S, and D' remain
fixed, being in the axis ; the point directly over O falls at A\
24-SPHERICAL PROJECTIONS.
135
tlie primitive circle coincides with the circle ACA'B, and AD'A'
is the projection of the circle DE. As the primitive circle is
its own projection, it follows that the angle SAO is equal to the
inclination of the circles (203) ; that is, equal to the angle EOB.
But OS is equal to the tangent of the angle SAO, to the radius
of the sphere, and AS is its secant ; therefore, the centre of the
circle AD'A' is in the line of measures at a distance from the
centre of the primitive circle equal to the tangent of the inclina-
tion, and its radius is equal to the secant of the inclination.
THEOREM V.
The poles of a circle are projected in its line of measures ; the
one farthest from the eye, at a distance from the centre of the
primitive circle equal to the semi-tangent of the inclination of
the circles ; the one nearest the eye, at a distance from the centre
of the primitive circle equal to the semi-cotangent of the
inclination.
§ 205. Since the axis of every circle of the sphere passes
through the centre of the sphere, a plane can be drawn through
the poles to be projected and the axis of the primitive circle.
As this plane passes through the eye, the poles will be pro-
jected in its trace ; that is, in the line of measures of the circle
to which they belong (176).
Let DE (PI. 4. Fig. 1) be the orthographic projection of a
circle, PP' its axis, P and P' its poles, and A A' the axis of the
primitive circle. The pole P is projected at P", and the pole
P' at P'". The angle A'OP is equal to the inclination of the
circles (175) ; and the angle A'AP is half this angle, being an
angle at the circumference, and standing on the same arc A'P.
But OP" is the tangent of the angle P"AO ; it is therefore the
semi-tangent of A'P, or the semi-tangent of the inclination.
The angle PAP' is a right angle, being an angle in a semicircle ;
therefore, OAP'" is the complement of OAP, or the comple-
ment of half the inclination of the circles :• consequently, OF",
the tangent of the angle OAP'", or the cotangent of OAP", is136
DESCRIPTIVE GEOMETRY.
the cotangent of half the inclination of the circles, or semi-
cotangent of their inclination.
Since the poles of a great circle and of a small circle parallel
to it are the same, it is evident that the poles of a small circle
are also projected in its line of measures, and at distances from
the centre of the primitive circle equal to the semi-tangent and
semi-cotangent of its inclination.
THEOREM VI.
The centre of the projection of a small circle perpendicular to
the primitive plane is in the line of measures, at a distance
from the centre of the primitive circle equal to the secant of the
circle's polar distance, and the radius of the projection is equal
to the tangent of the polar distance.
§ 206. PI. 4. Fig. 2. Let ADEB, in the plane of the paper
be the circle in which the plane through the axis of the primi-
tive circle and the axis of the lesser circle intersects the sphere,
A the place of the eye, D'B the trace of the primitive plane,
and ED the diameter of the lesser circle to be projected
The extremity D of the diameter is projected at D, the
extremity E at E', and E'D' is a diameter of the projected circle.
Bisect it at G, and suppose the circle to be described in the
primitive plane. Let the primitive plane be revolved around D'B
to coincide with the plane of the paper. The primitive circle will
then coincide with the circle ADEB, and DE'ED' is the projected
circle thus revolved. The lines DO and DG, passing through
D, the point in which the circumferences of the circles inter-
sect, are perpendicular to each other, since the circles are at
right angles in space (203) ; GD, therefore, is tangent to the
circle ADEB. But CD is the polar distance of the small circle,
GD is the tangent, and OG is the secant of this arc ; therefore,
the distance from the centre of the primitive circle to the centre
of the projected circle is equal to the secant of its polar dis-
tance, and the radius with which it is described to the tangent
of the polar distance.SPHERICAL PROJECTIONS#
137
THEOREM VII.
The extremities of a diameter of a small circle oblique to the
primitive plane are projected in its line of measures, at dis-
tances from the centre, of the primitive circle equal to the semi-
tangent of the inclination plus the polar distance, and the
semi-tangent of the difference between the inclination and
polar distance; the projections of these extremities are on the
same side of the centre of the primitive circle when the polar
distance is less than the inclination, and on different sides
when it is greater.
§ 207. PI. 4. Fig. 3. Suppose the plane of the paper to
pass through the axis of the primitive circle, and the axis of
the circle*to be projected; and let ABA'C be the circle in
which it intersects the sphere. Let A be the position of the
eye, AA' the axis of the primitive circle, OP the axis of the
circle to be projected, and HG its orthographic projection ;
PG is its polar distance, and PA' its inclination.
The point H is projected in the line of measures at H', a dis-
tance from the centre of the primitive circle equal to the semi-
tangent of ATI ; that is, the semi-tangent of A'P the inclination,
plus PH the polar distance. The point G is projected at G', a
distance from the centre of the primitive circle equal to the
semi-tangent of A'G ; that is, the semi-tangent of the inclina-
tion AT, minus the polar distance GP: G'H' is a diameter of
the circle into which the circle HG is projected.
For the second case, take a circle parallel to GH, and whose
orthographic projection is DE. The polar distance PD of this
circle is greater than PA', its inclination. It is plain that the
point D is projected at D', and the point E at E'. The line
OD' is the semi-tangent of A'D ; that is, of PD minus PA' :
hence OD' is equal to the semi-tangent of the polar dis-
tance minus the inclination. It is plain, that OE' is equal to
the semi-tangent of ATE ; that is, equal to the semi-tangent
of the inclination A'P, plus the polar distance PE. In theJ38
DESCRIPTIVE GEOMETRY.
second case, therefore, the extremities of that diameter which
is in the line of measures correspond in their positions to the
enunciation of the text.
§ 208. If the inclination of either of the circles, as DE, were
equal to its polar distance, the point D would be at A', and
would be projected at O, the centre of the primitive circle.
Hence, the projections of all small circtes whose polar distances
are equal to their angles of inclination, pass through the centre
of the primitive circle,
§ 209. If the surface of an entire sphere were projected on
the same plane, without changing the position of the eye, that
part of it lying between the eye and primitive plane would be
projected without the primitive circle ; small circles near the
eye would be projected into very large circles, and circles near
the opposite pole would be projected into circles much less
than themselves. Thus, the magnitudes of circles would bear
little proportion to that of their projections ; equal circles of
the sphere would be unequal in projection, and the projection
of the sphere made after this method would rather confuse than
aid the mind in conceiving of its different parts. To remedy,
in some degree, this defect of the stereographic projection, we
generally project the hemisphere between the primitive plane
and the pole opposite the eye ; then revolve the other hemi-
sphere 180° around a line tangent to the primitive circle, and
suppose the eye to be removed parallel to the primitive plane,
till it comes into the axis of this hemisphere after it is revolved :
the hemisphere is then behind the primitive plane, and the eye
m the pole of the primitive circle. If from this position the
hemisphere be projected, we shall have the projection of the
entire sphere on the same plane.SPHERICAL PROJECTIONS.
139
PROBLEM I.
To project the sphere on the plane of the equatot.
§ 210. Let the eye be supposed at the south pole ; the north-
ern hemisphere will then be first projected.
LetFAGB (PI. 4. Fig. 4) be the equator ; the eye is in a perpen-
dicular to the plane of the paper at N, and at a distance from it
equal to the radius NA. The northern hemisphere lies behind
the primitive plane. The north pole is projected at the centre N.
Let A and B be the equinoctial points : AB is the line in which
the plane of the ecliptic intersects the plane of the equator.
The ecliptic passes through the points A and B, and makes an
angle of 23¿° with the equator ; its line of measures is FG (176),
which contains the centre of the circle into which it is pro-
jected (204). At the point A make the angle NAO equal to
23^° ; the line AO will pass through the centre of the circle
into which the ecliptic is projected (203) ; O is, therefore, the
centre of this circle. With this centre, and radius OA, let the
arc AEB be described ; this arc is the projection of that half
of the ecliptic which lies north of the equator. The centre O
might have been found by making NG equal to the tangent of
23i° (204).
The planes of the meridians, passing through the axis of the
earth, must pass through the eye, and will consequently be
projected into right lines (195) ; but, as they are great circles,
their projections pass through the centre of the primitive circle
(L95). The projections of the meridians are, therefore, deter-
mined by laying off from A arcs equal to 15°, 30°, 45°, &c.,
and drawing diameters through their extremities. The line
ANB is the projection of the equinoctial colure, and GNF of
the solstitial colure ; the projections of four other meridians are
also drawn in the figure.
The parallels of latitude, being parallel to the primitive plane,
the radii of their projections are equal to the semi-tangents of
their polar distances (192). To project the arctic circle. Make140
DESCRIPTIVE GEOMETRY.
the angle NFI equal to half of 23£° ; NI is the semi-tangent
of 23¿° to the radius of the sphere. With NI as a radius and
N as a centre, describe a circle ; it will be the projection of the
arctic circle. In like manner, making the angle NFD equal
to half of 66|° determines ND, the radius of the circle into
which the tropic of Cancer is projected. The projection of the
tropic of Cancer is tangent to the projection of the ecliptic at
E. The principles used in projecting the arctic circle and the
tropic of Cancer are equally applicable in projecting any of the
parallels of latitude.
Let the semi-sphere, which is in front of the primitive plane,
be now revolved 180° around a line tangent to the primitive
circle at B. The eye is supposed to be moved parallel to the
primitive plane till it is projected at S. As the hemisphere,
primitive plane, and eye have the same relative positions as
they had in the projection which has just been made, it follows
that the circles of the hemisphere will be projected in the same
manner. The arc BE'A' is the projection of the half of the
ecliptic corresponding to AEB ; G'E F is the projection of the
half of the solstitial colure corresponding to GF ; BA' is the
projection of the half of the equinoctial colure corresponding
to AB ; the small circles described about the centre S are the
projections of the antarctic circle and tropic of Capricorn.
The diameters passing through S are the projections of the
meridians.
PROBLEM II.
To project the sphere on the plane of the equinoctial colure.
§ 211. PI. 5. Fig. 1. Let SBNQ be the plane of the equi-
noctial colure, N the place of the north, and S the plane of the
south pole : Q and B the equinoctial points.
Since the meridians pass through the poles, their projections
pass through the projections of the poles N and S, and their
planes intersect the primitive plane in the line SN. The line
QBR, drawn through 0, the centre of the primitive chnle, perSPHERICAL PROJECTIONS.
141
pendicular to NS, is the line of measures of all the meridians
(176). To project any meridian, as the one, for example, that
makes an angle of 30° with the primitive circle. At either
pole, as S, lay off an angle OSD equal to 30° ; the point D, in
which the line SD meets the line of measures, is the centre of
the projection of the meridian (203). With D as a centre and
radius DS, let the meridian SEN be described. The centre
D could also be found by laying off from O, in the line of
measures, OD equal to the tangent of 30° the angle of inclina-
tion of the meridian with the primitive plane. After the
same manner the other meridians are projected. The line
SON is the projection of the solstitial colure. The equator
passes through the eye, and its projection is the right line QOB.
The ecliptic passes through the points Q and B, and makes an
angle of 66~° with the primitive plane; NS is its line of measures.
If, on the hemisphere which is behind the primitive plane, the
ecliptic lies between the equator and south pole, lay off from O.
in the direction ON, the tangent of 66	; with the extremity
of this line as a centre, and the distance to Q as radius, describe
the arc QFB, which will be the projection of that half of the
ecliptic that lies behind the primitive plane. If, on the hemi
sphere behind the primitive plane, the ecliptic had been situated
between the equator and north pole, the radius of its projection
would have been laid off from O in the direction OS.
The parallels of latitude are perpendicular to the primitive
plane, and SN is their line of measures. In projecting them,
we shall begin with the arctic circle. From N lay off the arc
Nƒ equal to 23^°;. draw NC perpendicular to SN, join O/,
and produce it to C. The line OC is the secant and NC the
tangent of 23^° to the radius of the primitive circle. With 0
as a centre and radius OC describe the arc CC' ; with C' as a
centre and radius equal to CN or C ƒ describe the arc ƒ I ; this
arc is the projection of the arctic circle. We determine, by
similar constructions, ba the projection of the tropic of Cancer,
and cFd, the projection of the tropic of Capricorn (206).
Let the semi-sphere which is in front of the primitive plane
be now revolved behind it, around a line tangent to the primi142
DESCRIPTIVE GEOMETRY,
tive circle at B ; the projections of its different circles can then
be made by constructions entirely similar to those already
given.
§ 212. The projection of the sphere on the plane of the sol-
stitial colure, is made in the same manner as its projection on
the plane of the equinoctial colure, excepting that the ecliptic,
being perpendicular to the primitive plane instead of being
oblique to it in an angle of 66^°, is projected into a right line
passing through the centre of the primitive circle, and making
an angle of 23 with the projection of the equator. Draw-
ing the lines bOd and d'b', making angles of 23|° with QR, the
projection of the equator, and supposing the curve QFBR to
be removed, the figure will represent, the projection of the
sphere on the plane of the solstitial colure.
PROBLEM III.
To project the sphere on the horizon of any place, that place ¡for
example, the latitude of which is 45° north.
§ 213. PI. 5. Fig. 2. Let AEBL be the plane of the hori-
zon. Let the eye be supposed at the lower pole of the primi-
tive circle ; the upper hemisphere will then be first projected.
Assume A and B for the equinoctial points ; AB is the line
in which the plane of the equator intersects the horizon ; the
plane of the ecliptic and the plane of the equinoctial colure also
intersect the horizon in the same line AB. The line EL, drawn
through the centre D, perpendicular to AB, is the line of meas-
ures of the equator, ecliptic, and equinoctial colure. The
equator makes an angle with the primitive plane equal to the
complement of the angle which the axis of the earth makes
with it ; that is, an angle of 45°. The projection of the equator
can therefore be described, and is the arc AFB. Suppose the
ecliptic, on the upper hemisphere, to lie between the equator
and north pole ; it will, in that case, make an angle with the
primitive plane greater by 23° than the angle made by the
equator. The projection of the ecliptic can then be described •SPHERICAL PROJECTIONS.
143
t is the arc AGB. The equinoctial colure makes an angle of
45° with the primitive plane ; and as DE is equal to the tangent
of 45°, E is the centre, and the distance EA or EB the radius
with which its projection ANB is described. The solstitial
colure, being perpendicular to AB the intersection of the
equator and equinoctial colure is,perpendicular to the primitive
plane ; its projection is therefore the diameter LDE. The
point N, in which this line intersects the projection of the equi-
noctial colure, is the projection of the north pole. If the remain-
ing part of the projection of the equinoctial colure be described,
it will meet the line LE in another point, which point would
be the projection of the south pole. If the distance intercepted
between N and the projection of the south pole be bisected at
E> and EH be drawn perpendicular to NE, EH will contain
the centres of the circles which are the projections of the me-
ridians. For, as the meridians pass through the poles, their
projections will pass through the projections of the poles ; there-
fore, the part of the axis intercepted between the projections
of the poles is a common chord of the projections of the meri-
dians ; and the line EH, bisecting it perpendicularly, contains
all their centres. To project any meridian, as the one, for
example, making an angle of 30° with the equinoctial colure.
At N lay off the angle ENH equal to 30° ; the point H, in which
NH intersects EH, is the centre of the projected meridian
(203). With H as a centre and radius HN let the meridian
H'NP be described. The projections of the other meridians
drawn m the figure are made by similar constructions.
To project the parallels of latitude. These parallels are
small circles, and being parallel to the equator have the same
line of measures, and make the same angle with the primitive
plane. To project the arctic circle. Its polar distance is 23i°,
and its inclination 45°. Lay off from L tog*' 68|°, that is, the
polar distance 23j°+45° the inclination, and join g*' and E.
The line Dg* is the semi-tangent of 68|°. Let hf be made
equal to 45°—23¿°=21±°, and draw Ef; Df is the semi-tan-
gent of the inclination minus the polar distance. Let the dis-
tances Dg and Df' be laid off from D on the line of measures141
DESCRIPTIVE GEOMETRY.
EL ; f‘"g" is the diameter of the projection of the arctic circle
(207). Let this diameter be bisected and the circle described.
Ali other parallels of latitude are projected by similar construc-
tions. The projection of the tropic of Cancer touches the pro-
jection of the ecliptic at G, and intersects the primitive plane at
a and b. The tropic of Capricorn intersects the primitive plane
at cl and c. No part of the antarctic circle lies above the primi-
tive plane.
Let, now, the lower hemisphere be revolved 180° around a
line tangent to the primitive circle at B, and then project the
diflerent circles by the methods already explained. Their pro
ections are easily recognised in the figure*■ te
warped surfaces.
i45
COMPLEMENT
OF
DESCRIPTIVE GEOMETRY.
WARPED SURFACES.*
$ 214. We have already defined Warped Surfaces to be
those surfaces which may be generated by a right line moving
in such a manner that its consecutive positions shall not be in the
same plane (72). This class of surfaces is entirely distinct from
the single-curved surfaces, though both are generated by a right
line. In the single-curved surfaces, the consecutive positions
of the generatrix are in the same plane ; in the warped surfaces
they are not; and although this difference in the manner of
their generation may seem unimportant, yet it gives to the
surfaces very different forms, and essentially different pro-
perties.
§ 215. This family of surfaces presents, perhaps, more varie-
ties than any other; we shall examine only the most useful
kinds, and begin with those whose properties are most simple.
§ 2L6. Of warped surfaces which have a plane-direc-
ter. Let there be supposed any two curves in space, and a
line to move along these curves constantly touching them and
continuing parallel to a given plane. Unless the curves have
a particular position with each other, or with the given piane,
the consecutive positions of the generatrix will not be in the
* A portion of this Complement is translated from Valléis excellent Traite
ie la Géométrie "Descriptive.
10146
DESCRIPTIVE GEOMETRY.
same plane : hence, the surface generated will be a warped
surface The plane to which the generatrix continues parallel
is called the plane-directer, and the lines which it touches
directrices.
§ 217. PI. 1. Let (AB, A'B') and (CD, C'D') be the linear
directrices of a warped surface, and (OM, ON') its plane-
directer. It is required to construct any element of the surface ;
the one, for example, passing through the point (A, A') of the
first directi ix.
If through this point, a plane be passed parallel to the plane
directer, and the point in which it cuts the second directrix
(CD, C'D') determined, the line joining this point and the point
(A,A') will be the element required. We could thus deter-
mine the element passing through any point of either directrix ;
but, as drawing the planes and finding the points in which they
cut the directrices is rather tedious, we give the following
method for constructing the elements as more concise and
elegant.
Draw in the plane-directer (MO, ON') any right line, as
(MN, M'N'). Intersect the plane-directer by a system of ver-
tical planes, PM, P a, P b, &c. drawn through any point, P, of
the trace OM : these planes cut the line (MN, M'N') in the
points (M,M'), (a,a'), (b,bf), &c. ; the lines joining these
points and the point (P,P') are the intersections of the vertical
planes with the plane-directer. Through the point (A,A') draw
a system of lines (Ar, A'r'), (Aq, Ay), &c. respectively parallel
to the lines (Pb, P 'b'), (P a, FV), &c. of the plane-directer ;
this system of lines forms a plane passing through (A.A') par-
allel to the plane-directer ; it is required to find the point in
which this plane cuts the second directrix (CD, C'D'). The
system of parallels through (A,A') intersects the surface of the
cylinder which projects the second directrix on the horizontal
plane in a curve of which DCp is the horizontal projection ; the
vertical projection of this curve is found by drawing perpen-
diculars to the ground line through the points r, q, p, &c., and
determining their intersections r\ q', p’, &c. with the vertical
projections of the parallels through (A,A') ; r'q'p' is the verticalwarped surfaces.
147
projection of the curve. The point in which this curve intersec ts
the second directrix is the point in which the second directrix
is cut by the plane passing through (A,A') parallel to the plane-
directer. The vertical projection of the point is in the curve
D'C', and also in the curve r'q'p'z hence, it is at C', their point
of intersection. Drawing through C' a perpendicular to the
ground line, determines C, the horizontal projection of the point.
Therefore, (AC, A'C') is the element of the surface passing
through (A,A'). If we take any point, other than (A,A'), of
either directrix, the element of the surface passing through it
would be determined in a similar manner.
§ 218. To find an element of the surface parallel to any lines
as (Pa, P'a'), of the plane-directer.
Through the several points (d,d'), (e,e'), (ƒƒ')* {gig*), &c. of
either directrix, drawr lines parallel to the given line (Pa, P'a') ;
they form the surface of a cylinder parallel to the line (Pa, P'a') ;
the element of this cylinder passing through the point in which
the second directrix pierces its surface, touches both direc-
trices, and is an element of the surface. This cylinder inter-
sects the vertical cylinder which projects the second directrix
on the horizontal plane in a curve of which kl)p is the hori-
zontal projection ; its vertical projection is determined by find-
ing the points k', ï,j\ h', &c. in which the perpendiculars to the
ground line through the points h,i,j, h, &c. intersect the ver-
tical projections of the parallels : k'i'j'h' is the vertical projec-
tion of this curve. The point (D,D'), in which this curve inter-
sects the second directrix, is the point in which the second
directrix pierces the surface of the cylinder : hence, BD drawn
parallel to Pa, and B'D' drawn parallel to P'a', are the projec-
tions of the required element.
§ 219. If one of the directrices (AB, A'B'), (CD, C'D')
were a right line, the surface that would be generated belongs
to a particular class of warped surfaces called conoids, because
of the analogy existing between them and the surfaces of cones.*
* If all the points in which the elements touch the rectilinear directrix were
brought together into one point, the elements still passing through the points in
10*148
DESCRIPTIVE GEOMETRY.
If the directrix were perpendicular to the plane-directer, the
conoid takes the name of right conoid, and the directrix the
name of the line of striction .* *
If both the directrices are right lines, the surfaces generated,
though a species of conoid, are called hyperbolic paraboloids,
because the curves in which they are intersected by planes are
either hyperbolas or parabolas. The rectilinear directrices are
not in the same plane ; for if they were, the generatrix would
generate a plane, and not a warped surface.
§ 220. The elements of a hyperbolic paraboloid divide the
directrices proportionally.
Let AC and A'C' (PI. 2. Fig. 1) be the directrices of a hyper-
bolic paraboloid ; AA', BB', and CC' three of its elements.
Through A draw AD parallel to CC' ; AD will be parallel to
the plane-directer; and since AA'is also parallel to the plane-
directer, it follows that the plane A'AD is parallel to the plane-
directer and may be taken for it. Demit from the points B',
C', C, and B the perpendiculars B'&', C'c', Cc, and Bb on the
plane A'AD. Since BB' and CC' are elements of the surface,
they are parallel to the plane-directer, and consequently to the
plane A'AD ; therefore, C'c'=Cc, and B'6'=B& : by drawing the
lines A'b'c' and A be two similar triangles are formed, which give
Cc : Bb : : CA : BA, and C'c' : B'ò' : : C'A' : B A' ;
but, on account of the equality of the terms of the first couplets,
we have
CA : BA : : C'A' : B'A' ;
and by division,
CB : BA : : C'B' : B'A' ;
that is, the directrices CA and C'A' are divided proportionally
by the element BB'.
which they touch the second directrix, the surface becomes a conic surface.
And if the vertex of a cone be moved along a right line, and lines be drawn
from its different positions to the points of its base, such lines being parallel to
a given plane, the surface thus formed is a conoid.
* It takes this name because it contains the shortest distance between the
elements, so that the surface is, as it were, cramped or compressed along thisWARPED SURFACES.
149
§ 221. Reciprocally, if the lines AA', BB', and CC* of a
warped surface divide the rectilinear directrices AC and A'C'
into proportional parts, they will he parallel to the same plane,
and consequently elements of a hyperbolic paraboloid of which
that plane is the plane-directer.
Let AD be drawn parallel to CC', and demit the perpen-
diculars Cc, Bò, C'c', and BV on the plane A'AD. Drawing
Abe and A'6'c', we have
AB : AC : : B6 : Cc ;
and	AB' : A'C' :: B'¿>' : C'c';
but, by hypothesis,
AB : BC : : A'B' : B'C'.
By composition,
AB: AC:: A'B: A'C';
therefore,	B¿> : Cc : : B'6' : C'c'.
But the line CC' being parallel to AD is parallel to the plane
A'AD ; therefore, Cc is equal to C'c', and consequently Bb is
equal to B'6'. All lines, therefore, which divide the directrices
proportionally are parallel to the same plane, and consequently
the surface generated by a right line moving with this law is a
hyperbolic paraboloid.
§ 222. If any two elements, as A A' and CC' of a hyperbolic
paraboloid be taken as directrices, and a plane-directer be as-
sumed parallel to AC and A'C', the directrices in the first case, the
surface generated by a right line moving on the new directrices
and parallel to the new plane-directer, is the same surface as is
generated by a right line touching AC, A'C', and continuing
parallel to the plane A'AD. The surfaces are named respect-
ively the hyperbolic paraboloid of the first and second generation.
We shall show that these surfaces are the same, by proving
that all points of any element of the second generation are
points of an element of the first generation, and reciprocally ;
that is, that the paraboloid of the second generation has all its
points common with the paraboloid of the first generation : if
this be proved, they are evidently the same surface.
Let mn (PI. 2. Fig. 2) be any element of the second
generation. In this generation the plane-directer is parallel150
COMPLEMENT.
to AC, A'C', and AA', CC' are the directrices. Draw
from the points C and C' the lines Cc and C'c' parallel to
mn ; and suppose c and d the points in which they pierce
the plane AA'D, drawn parallel to CC'. Let Ac and hid also
be drawn, and join the points c and d : this line is the intersection
of the plane AA'D with the plane passed through the element
CC' and the parallels Cc, win,.and C'c'; it therefore contains
the point n, in which mn intersects AA'. Take, now, upon mn
any point, as O, and conceive a plane to be drawn through this
point parallel to AA'D ; that is, parallel to the plane-directer
of the first generation (220). This plane will cut the direct-
trices AC and A'C', of the first generation, in two points B and
B' ; the line BB' is therefore an element of the first generation.
It is now to be proved that O is a point of the element BB',
and therefore a point of the surface of the first generation. Let
the lines Bfe and B'6' be drawn parallel to mn, and as Cc and C'd
are also parallel to mn, it follows that Bfe and B'6' will pierce the
plane AA'D in the lines Ac and hid. The triangles ABfe and
ACc are therefore similar, as also the triangles A'B'fe' and A'C'c' •
therefore,	AB : AC : : Afe : Ac ;
Since the lines Bfe and Cc are parallel to B'fe' and C'd, and
since the four lines are ail parallel to the plane-directer of the
second generation, the plane of the triangle ACc is parallel to
the plane of the triangle A'C'c', and therefore their intersections
A fee and A'fe'c' with the plane AA'D are also parallel. But as
these intersections are divided proportionally at fe and fe', it fol-
lows that the points fe, n, and fe' are in the same right line : hence,
the plane of the parallels Bfe and B'fe' contains the lines BB' and
mn ; therefore, the point O of the element mn of the second
generation is a point of the element BB' of the first generation.
It may be proved in a similar manner, that any point of an ele-
ment of the second generation is also a point of an element of
the first generation : hence, the hyperbolic paraboloid is suscep-
tible of two generations, as enunciated in the text.
But (220),
therefore,
and
A'B': A'C':: A'fe': A'c'.
AB: AC :: A'B: A'C';
Afe : Ac : : A'fe' : A'c'.WARPED SURFACES.
151
§ 223. From what has preceded we conclude, that if we take
any two elements of the first generation and a plane parallel to
its directrices, these lines and this plane are the directrices and
plane-directer of the second generation ; and, conversely, if we
take two elements of the second generation and a plane parallel
to its directrices, these lines and this plane are the directrices and
plane-directer of the first generation.
§ 224. Of warped surfaces which have three direct-
rices. If we subject the generatrix to the condition of touch-
ing a third directrix, instead of continuing parallel to a plane-
directer, the surface generated is still a warped surface, pro-
vided the directrices have such positions with each other that
the consecutive elements of the surface are not in the same plane.
Let (AB, A'B'), (CD, C'D'), and (EF, E'F') (PI. 3) be the
three directrices of a warped surface ; and let it be required to
find the element passing through any point of either directrix
say the point (M,M') of the directrix (AB, A'B').
Suppose (M,M') to be the vertex of a cone of which the
second directrix (CD, C'D') is the base. If the point (N,N), in
which the third directrix (EF, E'F) pierces the surface of thu
cone, be determined, the line joining (M,M) and (N,N) wil
touch the three directrices, and consequently be an element of
the surface. To find this point, take in the second directrix a
series of points (D,D), {a,a!), (p,b'), (c,d), &c. ; through these
points and the vertex (M,M) draw the elements (MD, M'D),
(Mg, M'g'), &c., and construct the points in which these ele-
ments pierce the vertical cylinder which projects the third
directrix on the horizontal plane : these points are (F,H), (if),
(j,f), (k,k'), &c., and Hij'kTm'n'G is the vertical, and FijklmnE
is the horizontal projection of the curve in which the cone inter-
sects the cylinder: the point (N,N), in which this curve inter-
sects the third directrix, is the point in which the third directrix
pierces the surface of the cone : (MN, M'N) is therefore a line
which touches the three directrices, and is, consequently, an
element of the surface. We can construct in a similar manner
any number of elements, by means^of which we can determine
the contour and projections of the surface152
COMPLEMENT.
§ 225. If the three directrices are right lines, the surface
generated belongs to a particular class of warped surfaces called
hyperboloids of one nappe. Of this family of surfaces we shall
discuss the most useful and interesting variety, viz. the hyperbo-
loids of revolution of one nappe .*
Before generating the hyperboloid of revolution of one nappe
by a generatrix constantly touching three right lines having a
particular position with each other, we shall generate it by a
right line moving around another right line as an axis, and then
show that this surface can also be generated by a right line
touching three linear directrices. Let the horizontal plane be
taken perpendicular to the line which is used as ah axis, and the
vertical plane parallel to the generatrix in any one of its posi-
tions.
Let (A,A'B') (PL 2. Fig. 3) be the axis, and (CD, CD') the
generatrix. The generatrix is to move around the axis in such
a manner that each point of it shall describe a horizontal circle
whose centre is in the axis (A,A'B'). Conceive a line to be
drawn perpendicular to the axis and generatrix (60). Since
the axis is perpendicular to the horizontal plane, this line will
be parallel toit: hence, its horizontal projection is equal to
* If the hyperbola CFp' (PI. 6. Fig. 7. Des. Geom.) be revolved around its
conjugate axis, which is perpendicular to EB at E, it will generate the surface
of a hyperboloid of revolution of one nappe. The convexity of this surface is
turned towards the axis. If the hyperbolas CFp* and C'BC' be revolved around
the transverse axis FB, they will generate two distinct surfaces, but having the
same axis FB ; the two surfaces are called a hyperboloid of revolution of two
nappes. If at either vertex of the transverse axis, as F, a line IFI' be drawn
tangent to the curve, and the parts FI and FF be each made equal to EO, the
semi-conjugate axis, the lines CFC' and C'EH drawn through their extremities
and the centre C, are called asymptotes. The asymptotes continually approach
the curves CFj/ and C'BC\ but never intersect them. If at any point of eithei
curve, as G, a line HGN be drawn tangent to the curve, the part HG inter-
cepted between the point of contact and one asymptote is equal to the part GN
intercepted between the point of contact and the other asymptote. These prop-
erties of the asymptotes and tangent are demonstrated in conic sections, and
are mentioned here only that they may be borne in mind in discussing the prop-
erties of the surface generated by a right line touching three rectilinear direct-
rices.WARPED SURFACES.
153
itself and perpendicular to CD, the horizontal projection of the
generatrix (51). The line AL is the horizontal and L'the vertical
projection of this perpendicular. When the generatrix (CD,C'D')
is revolved around the axis (A, A'B'), the line (AL,L') continues
perpendicular to it and to the axis : hence, the projections of the
generatrix, from its different positions, are perpendicular to the
projections of (AL,L') from its different positions ; that is. per-
pendicular to the extremities of the radii of a circle described
with the centre A and radius AL. Hence, the horizontal projec-
tions of the elements of this surface are tangent to the circle iqoh,
which is the smallest of the circles described by the points of
the generatrix, and is called the circle of the gorge. The con-
secutive elements of this surface are not parallel, for if they
were their horizontal projections would be parallel (30) ; but
the horizontal projections are not parallel, since they are tan-
gent to the circle iqo\j. Neither do the consecutive elements
intersect each other ; for the ir points which are in the same
horizontal plane are separated by arcs of horizontal circles.
The surface, therefore, is a w rrped surface, and it is also a sur-
face of revolution, since the sections by planes perpendicular to
the axis are circles. Let C&G'EE' be the circle described by
the point C, as the element (CD, C'D') moves around the axis ;
this is the circle in which the horizontal plane intersects the
surface.
Through D draw the line (DC, D"C") parallel to the ver-
tical plane, and making the same angle tvith the horizontal plane
as s made by the line (CD, C D') : these lines intersect at the
point (L,L'), and the perpendicular (AL,L') to the one, is also
perpendicular to the other. If the plane of the two lines be
carried around the right cylinder whose axis is the axis of the
surface, and whose base is the horizontal circle oqih, each of
the lines will generate the same surface ; for, if a plane be drawn
perpendicular to the axis (A,A'B') it will cut the lines in points
equidistant from the axis, and in the revolution of the lines these
points describe the same horizontal circle : the point in which
the lines intersect, describes the circle of the gorge. But if two
surfaces have the same axis, and if all sections made in them154
COMPLEMENT.
by planes peipendiculur to this axis are respectively equa.,
the two surfaces coincide throughout and are the same surface :
\ience, the surface we are discussing can be generated by either
jf two right lines at the same distance from the akis and making
(he same angle with a plane perpendicular to it.
§ 226. If one of the generatrices remain fixed and the sur-
face be generated by the other, the fixed generatrix will intersect
the moving one in all its positions. Let the generatrix (DC,
D"C") remain fixed, and suppose the surface to be generated
by (CD, C'D'). When the point C is at any point of the
circle CG'ED, as E, the horizontal projection of the generatrix
is determined by drawing EoG tangent to the circle oqiL (225).
Its vertical projection is determined by projecting E into the
ground line at e, and o into the vertical projection of the circle
of the gorge at o', and drawing eo'g. The horizontal projections
of the generatrices intersect at n, and no is equal to nL, since
the lines are tangent to the same circle. But the points (L,L')
and (o,of) are in the plane of the circle of the gorge, and the
generatrices make equal angles with this plane: hence, the
parts of the generatrices of which no and wL are the projections
are equal. The points of the two generatrices of which n is the
horizontal projection are, therefore, at the same distance above
the plane of the circle of the gorge, and consequently above the
horizontal plane ; but their vertical projections are contained
in a perpendicular to the ground line through n (13) : hence, they
are the same point ra', and therefore the generatrices intersect
in space (44), and (w,ra') is their point of meeting. This point
is above the circle of the gorge, and at an infinite distance from
it, when the generatrix (EG, eg) becomes parallel to the ver-
tical plane. When the generatrix (CD, C'D') takes the posi-
tion (E'G', E"G"), it intersects the generatrix (DC, D"C") at
(m,m% a point of the surface below the circle of the gorge.
In the same manner it may be shown, that if the generatrix
(CD, C'D') remain fixed, and the generatrix (DC, D"C") be
revolved, (CD, C'D') would, in all its positions, intersect
(CD, C"D") : hence, we Conclude that the generatrix of the first
generation intersects all the elements of the second generation.WARPED SURFACES.
155
and that the generatrix of the second generation intersects all
the elements of the first generation. If, therefore, any three
elements of the first generation be assumed, and a right line
drawn touching them, this line is the generatrix of the second
generation ; and if three elements of the second generation
be chosen, a right line touching them is the generatrix of the
first generation.
We have now shown that the surface generated by the revo-
lution of a right line about an axis which it does not intersect,
may also be generated by a right line touching constantly three
rectilinear directrices. We should remark, however, that these
directrices must have the same relative position as three elements
of the surface generated by the other method ; that is, they are
at the same perpendicular distance from a fourth line, and the
perpendiculars measuring this distance are contained in the
same plane.
§ 227. To show that this surface is the surface of a hyper-
boloid of revolution of one nappe.
PI. 2. Fig. 4. Let the axis of the surface be perpendicular
to the horizontal plane at A, and let dAc be the trace of a meri-
dian plane to which the vertical plane of projection is taken
parallel. It will be proved that this meridian plane intersects
the surface in hyperbolas, and that the projections of the gene-
ratrices (CD, C'D") and (DC, D'C") on this plane are asymp-
totes of the curves.
The vertical projections of the generatrices are the lines
C'D" and D'C" ; (ab, a'b') is the line in which the meridian
plane intersects the circle of the gorge : this line is the trans-
verse axis, and (a,a') and (b,b') the vertices of the curve in
which the meridian plane intersects the surface. To find other
points of the curve, let the surface be intersected by horizontal
planes; these planes will intersect it in horizontal circles, and
the meridian plane in right lines ; the points in which these
lines intersect the circles are points of the required curve.
Let h'li'Gm' be the vertical trace of one of these planes ; (H,H')
is the point in which it cuts the generatrix (CD, C'D"), and
is therefore one point of the circumference of the horizontal156	COMPÌ.EMKNT.
circle in which it intersects the surface. With A as a centre
and AH as a radius, let the semicircle hYLm be described ;
this is the horizontal projection of a part of the circle in which
the horizontal plane intersects the surface, and thfe points h and
m, in which it meets the projection of the line of intersection
of the horizontal and meridian planes, are the projections of two
points of the required curve. The vertical projections of
these points are h! and m'. The horizontal plane f'F'n' deter-
mines the points (ƒ,ƒ') and (n,n'). Thus, by using horizontal
planes above and below the circle of the gorge, we obtain as
many points as are necessary to describe the curves d'hf'a'd"
and c'm'n'ò'c".
The lines C'D" and D'C" continually approach these curves.
For, the distances Fƒ', H'A', D"d', &c. are equal to the dif-
ferences between the radii AF, AH, AD, &c. and their ver-
tical projections EF', GH', and UD". But these differences
continually diminish ; for the radii AF, AH, and AD make a
less and less angle with the vertical plane as the cutting plane
is removed from the plane of the gorge, and therefore the dif-
ferences between them and their projections constantly
diminish. If the horizontal cutting plane were taken at an infi-
nite distance from the circle of the gorge, the radius AD would
become parallel to the vertical plane ; the points d' and D"
would then coincide, and L'D" would become tangent to the
curve. We see, therefore, that the line L'D" continually ap-
proaches the curve d"af'h'd', and becomes tangent to it at an
infinite distance from a' ; this is the property of a hyperbola and
its asymptote. The same can be shown for the curve c"6'n'm'c'
and the line D'C", and also for the curves and lines below the
circle of the gorge.
Let (QIO, QTO') be any element of the surface intersecting
the meridian plane dAc in the point (p,p') ; (p,p') is a point
of the curve c'b'c”. The line drawn through (p,pf) tangent to
the horizontal circle of the surface passing through this point,
is perpendicular to the vertical plane, and is therefore a line of
the plane which projects the element on the vertical plane.
This projecting plane is consequently tangent to the surface atWARPED SURFACES,
157
the point (p9p') (38) : hence its intersection with the meridian
plane is tangent to the meridian curve. But the vertical pro-
jection of this intersection is the same line as the vertical pro-
jection of the element ; consequently, Q'O', the vertical projec-
tion of the element, is tangent to the curve d'h’d at the point
p\ But Ip and pO are equal ; therefore their vertical projec-
tions I y and p[O' are also equal ; that is, the part TO' of the
tangent intercepted between the lines L/C" and L'C' is bisected
at p', the point of tangency : and as this may be shown for any
other point, it follows that the curve is a hyperbola, and the
lines C'D" and D'C" its asymptotes. The same can be shown
for the other curve. If either of these hyperbolas be revolved
around A'B as an axis, it will evidently generate the surface
from which it has been obtained. We therefore conclude, that
the surface generated by a right line revolving around another
right line which it does not intersect, or by a right line con-
stantly touching three right lines having a particular position
with each other, is the same surface as is generated by the
revolution of a hyperbola around its conjugate axis, and is
therefore properly called a hyperboloid of revolution of one
nappe.
§ 228. Of warped surfaces in general. It is easily per-
ceived from what has preceded, that if a right line be moved
along two curves so that the part of the right line intercepted
between them shall be of a given length, or so that it shall make
with a given plane a constat angle, or make a given angle
with one of the directrices ; or should we move it upon three
surfaces, or upon a curve and two surfaces, or upon two curves
and a surface, or upon two surfaces and making a constant
angle with a given plane, either of these conditions imposed
upon the generatrix would, in general, give a warped surface
of a different kind.
§ 229. We shall now demonstrate a general property of
warped surfaces. 11 is this : every plane passing through any ele-
ment K of a warped surface is, ingenerai, tangent to this surface
at some point of the element K. Suppose the plane to have anyC03IPLEMENT.
158
position, and let K', K", K'", &e. be elements of the sunace on
one side of the element K ; and H, H', H", &c. elements of
the surface on the other side of the element K ; and let these
elements be consecutive with each other, and situated on the
surface in the order in which the letters are written. The plane
through the element K will not, in general, be parallel to these
elements ; it therefore intersects them in a series of consecutive
points k"\ k'\ k\ h, h\ h'\ A"', &c. forming a curve kf,,k"k,hiih,,h,,>;
but since the points kf and h are on different sides of the ele-
ment K, the indefinitely small part k!h of the curve intersects
the element K in a point. Let this point be designated by k ;
we say that the plane through the element K is tangent to the
surface at the point k. For, if at k a line be drawn tangent to
the curve k'"k"k'khh'h"hf", it will be contained in the plane of
the curve (67) ; that is, in the plane passed through the element :
the element also is tangent to the surface at the same point ;
the plane, therefore, containing these tangent lines is tangent
to the surface at the point k of the element K (88).
§ 230. As the curve k,,,k,,k,khh,h,,h,,, varies with the position
of the plane through the element K, it is evident that if this
plane be turned around K as an axis, the point of contact k will
move along this element. From these properties we con-
clude, that every tangent plane to a warped surface is also a
cutting plane : secondly, that if we wish a plane tangent to a
warped surface, we have only to draw it through an element
of the surface ; and thirdly, that the point of contact is the point
in which the element intersects the curve of intersection of this
plane and the surface.
§ 231. There are, however, a few cases in which a plane
through an element of a warped surface is not tangent to it.
Suppose, for example, that the given surface has a plane-
directer, and that the plane through the element were parallel
to the plane-directer ; all the elements being also parallel to
this plane, the curve kf,,k!,kfkhh,h,fhm would not exist, and the
plane through the element wTould not be tangent to the surface.
We have not heretofore spoken of the manner of representingWARPED SURFACES.
J59
warped surfaces on the planes of projection. They are, like
other surfaces, generally represented by the projections of
their elements, and their intersections with one or the other of
the planes of projection.
PROBLEM.
To draw a plane tangent to a hyperboloid of revolution at a
given point of its surface.
§ 232. PI. 2. Fig. 4. Let (A,A'B) be the axis of the surface,
XDC its intersection with the horizontal plane, (xaLb, a'V) the
circle of the gorge, and v the horizontal projection of the point
at which the plane is to be tangent. Its vertical projection
cannot be taken at pleasure (93), but must be constructed.
Through v draw DvC tangent to the horizontal projection
of the circle of the gorge ; this tangent is the projection of two
elements, either of which may pass through the point of which
v is the horizontal projection, according as the point is «above or
below the circle of the gorge. The two elements make the
same angle with the horizontal plane, and pierce it at the points
D and C. Projecting the point of tangency L into the vertical
projection of the circle of the gorge, and the points C and D
into the ground line at C' and D', two points in the vertical pro-
jection of each element are determined, and their vertical pro-
jections C'L'D" and D'L'C" can be drawn. Drawing from v
a perpendicular to the ground line, and noting its intersections
v' and v" with the projections of the elements, we determine
the vertical projections of the two points of the surface of which
v is the horizontal projection ; one point is above the circle of
the gorge and vertically projected at v", the other below it and
vertically projected at v' : these points are evidently those in
which a line perpendicular to the horizontal plane at v pierces
the surface. Let the tangent plane be first drawn at the point
Through v let the line TvxX be drawn tangent to the hori-
zontal projection of the circle of the gorge ; this tangent is theCOMPLEMENT.
160
projection of an element of the surface passing through (v,v') ;
and as (v,v') is below the circle of the gorge, it pierces the hori-
zontal plane at T. The plane containing this element and the
element (DC, D'C") is tangent to the surface at the point (v,v')
(88) : DT is its horizontal trace, and its vertical trace is easily
found.
§ 233. If it were required to draw a tangent plane to the sur-
face at the point (v,v") above the circle of the gorge, it would
only be necessary to determine the plane of the elements of the
two generations which pass through this point. The element
(CD, C'D") pierces the horizontal plane at C ; and since (v,v")
is above the circle of the gorge, the element of which TvX
is the horizontal projection pierces it at X : hence, XC is the
horizontal trace of a plane tangent to the surface at the point
(v,v").
§ 234. The traces DT and XC of the tangent planes are
parallel. For, draw Av and produce it in both directions ;
since the chords DvC and TvX make equal angles with the
diameter passing through their point of intersection, the chords
joining their extremities are perpendicular to this diameter, and
consequently are parallel. This is as it should be ; for the
meridian plane of which vA is the horizontal trace is perpen-
dicular to both the tangent planes (105) ; and being also per-
pendicular to the horizontal plane, its trace is perpendicular to
the traces of the tangent planes.
§ 235. We see, therefore, that to draw a plane tangent to the sur-
face of a hyperboloid of revolution, it is only necessary to deter-
mine the elements of the two generations passing through this
point ; the plane of these elements is the tangent plane required.
Or, find the element of either generation passing through the
given point, and draw through this element a plane perpen-
dicular to the meridian plane of the given point : this plane will
be tangent to the surface.WARPED SURFACES.
161
PROBLEM.
To pass a plane through a given right line, and tangent to a
surface of revolution.
$ 236. PI. 4. Let the surface be that of the ellipsoid. Let
the transverse axis be perpendicular to the horizontal plane at
A and A'B, its vertical projection ; let the circle described with
A as a centre and AE for a radius be the horizontal projection
of the surface, and the ellipse A'G'E'BR' its vertical projection ;
and let (CD, C'D') be the given line.
Suppose the line (CD, C'D') to revolve around (A,A'B) as
an axis ; it will generate the surface of a hyperboloid of revolu-
tion of one nappe. The hyperboloid thus generated, and the
ellipsoid, having a common axis, a meridian plane of the one
will be a meridian plane of the other. Let us suppose, for a
moment, that the plane were drawn through (CD, C'D') tangent
to the surface of the ellipsoid, and that the point of contact were
known.
Through the point of contact conceive a meridian plane to
be passed ; it will be perpendicular to the tangent plane (105),
and will cut the line (CD, C'D'), which is an element of the
hyperboloid, in a point. Since the tangent plane to the ellip-
soid contains an element of the hyperboloid, it will be tangent
to the Hyperboloid at some point of the element (229). But
the meridian plane passing through the point of contact on the
hyperboloid is perpendicular to the tangent plane : hence, the
meridian plane which passes through the point of contact on
the ellipsoid also contains the point of contact on the hypei bo-
loid ; therefore the point of contact on the hyperboloid is where
this meridian plane cuts the given line (CD, C'D'). This me-
ridian plane intersects the tangent plane to both surfaces in a
line tangent to the two meridian curves. Suppose this meridian
plane to be revolved about the common axis of the surfaces till
it becomes parallel to the vertical plane of projection : the me-
ridian curves of the hyperboloid would be projected into the
11COMPLEMENT.
162
hyperbolas Hf''"di' and f'c'o’ ; these curves are the same as the
sections made by a meridian plane parallel to the vertical plane,
and may be determined as in Art. 227 ; the meridian section of
the ellipsoid would be projected into the ellipse A'E'BR', and
the intersection of the tangent and meridian planes would be
projected in a line tangent to these curves. But we can con-
struct these curves without knowing the point of contact. If,
then, we draw G'H' tangent to the two curves, (H,H') and
(G,G') are the revolved positions of the points at which the
tangent plane touches the two surfaces. But the point (H,H'),
in its true position in space, is a point of the line (CD, C'D1) ;
in the counter revolution of the meridian plane this point
describes the arc (H7¿, H'A') of a horizontal circle, and the
point (h,h'), in which this arc intersects the line (CD, C'D'), is
the point at which the plane is tangent to the hyperboloid. The
meridian plane Agh contains the point at which the plane is
tangent to the surface of the hyperboloid, and* also the point at
which it is tangent to the surface of the ellipsoid. The point
(GjG'), in the counter revolution, describes the arc (Gg, G'g')
of a horizontal circle, and the point (g,g'), in which it inter-
sects the meridian plane Agh, is the point at which the plane
through the line (CD, C'D') is tangent to the ellipsoid. Through
the point of contact (g,g') let a line be drawn parallel to the
line (CD, C'D') : the point m, at which it pierces the horizontal
plane, is a point of the horizontal trace ; but C is another point ;
therefore PCmN is the horizontal trace of the tangent plane.
The line drawn through (g,g^ pierces the vertical plane at n
hence, nPQ is the vertical trace of the tangent plane.
If we consider the tangent LÍ', we perceive that it also gives
a point of contact (1,1') on the surface of the ellipsoid. The
traces of the plane tangent at this point are found in the same
manner as were the traces of the tangent plane in the other
case. We see, therefore, that two planes can be drawn through
a given line and tangent to a surface of revolution. The figure
shows the manner in which the hyperbolas in the vertical plane
are constructed.WARPED SURFACES
163
PROBLEM.
To find the intersection of a hyperboloid of revolution of one
nappe with a given plane ; to draw a tangent to the curve,
and, to find the curve in its own plane.
§ 237. PI. 5. Let the horizontal plane be taken perpen-
dicular to the axis of the surface ; let A be the horizontal pro-
jection of the axis, and A'B its vertical projection. Let (CD,
C'D') be the generatrix, (cots, c't') the circle of the gorge, and
(FE, FG") the cutting plane.
Through the axis of the surface and perpendicular to the
cutting plane let a plane be drawn ; the intersection of these
planes determines the transverse axis of the curve, and the
points in which the transverse axis intersects the curve are the
vertices. The line AE, drawn perpendicular to FE, is the
horizontal trace of the meridian plane. The line of intersec-
tion of this plane and the cutting plane (FE, FG") meets the
axis of the surface at the point in which the axis pierces the
cutting plane ; that is, at the point (A,A") (43) ; therefore
AE is the horizontal and E'A" the vertical projection of the
line of which the transverse axis of the curve forms a part.
The points in which this axis pierces the surface are next to
be found. If the line (EAc?, E'A"d') be revolved around the
axis of the surface, it will generate the surface of a right cone
with a circular base ; (A,A") is the vertex of this cone, and the
circle described with A as a centre and radius AE is its inter-
section with the horizontal plane. But the cone and hyperbo-
loid, having a common axis, intersect in circles, the planes of
which are perpendicular to this axis ; and the points in which the
line (eZAE, d'A"E') pierces the surface are the points which
describe these horizontal circles of intersection ; the circles
therefore contain the vertices of the axis. To find the radii
of these circles, it will be sufficient to find the two points in
which any element of the hyperboloid pierces the surface of
the cone, since all the elements pierce the surface of the cone
11*164	COMPLEMENT.
in tb* horizontal circles in which the two surfaces intersect.
Let LjS find the points in which the element (CD, C'D') pierces
the surface of the cone. Draw a plane through the element
(CD, C'D') and the vertex of the cone ; this plane will inter-
sect the surface of the cone in two right-lined elements ; the
points in which the element of the hyperboloid intersects these
elements, are points of the horizontal circles. To draw this
plane. Through the vertex (A,A") of the cone let a line be
drawn parallel to the element (CD, C'D') of the hyperboloid :
its projections are parallel to CD and C'D', and it pierces the
horizontal plane at a : hence, N<zCL is the horizontal trace of
a plane containing the element of the hyperboloid and vertex
of the cone. This plane intersects the cone in two elements, of
which the horizontal projections are AL and NAp ; the points
r and p, in which these projections intersect CD, the projection
of the element of the hyperboloid, are the horizontal projec-
tions of the two points, one in each of the circles, in which the
hyperboloid and cone intersect ; and as the circles are horizontal,
Ar and Ap are the radii of their projections. But as the hori-
zontal projections of the vertices are in the line Ed as well as
in the horizontal projections of the circles, they are at q and d,
the points in which the line Ed intersects the arcs described
with the centre A and radii Ar and Ap. The vertical projec-
tions of these vertices are at q' and d', in the vertical projection
of the line (Ed, E'd'). The vertex (q,q') is below the circle of
the gorge in the lower nappe of the cone, and the vertex
(d,d') is above the circle of the gorge in the upper nappe of
the cone.
To find other points of the curve, intersect by horizontal
planes between the points (d,d') and (</,<?'). Such planes will
intersect the surface of the hyperboloid in horizontal circles,
and the cutting plane in right lines parallel to its horizontal
trace ; the intersections of these right lines with the circles
determine points of the curve. Let h'V be the vertical trace
of a horizontal plane ; this plane cuts the element (CD, C'D')
in the point (v,v'), and intersects the plane (FE, FG") in a line
of which bh, parallel to FE, is the horizontal projection. The30
A .Stout,WARPED SURFACES.	165
«árele described with the centre A and radius Av is the hori-
zontal projection of the circle in which the auxiliary plane inter-
sects the surface of the hyperboloid. The points k and h9 in
which bh intersects this circle, are the horizontal projections of
the two points of the curve determined by the auxiliary plane
h'b', and h' and k' are the vertical projections of these points.
The points at which the horizontal projection of the curve is
tangent to the horizontal projection of the circle of the gorge,
are found by using the plane of the circle of the gorge as an
auxiliary plane ; they are the points o and s, and their vertical
projections are o' and s'. Thus, having found any number of
points, the projections of the curve can be described.
To draw a tangent Hne to the curve at any point, as (h,h').
Draw a plane tangent to the surface of the hyperboloid at this
point (232) ; its intersection with the cutting plane is the tan-
gent required (120). The line mnG is the horizontal trace of
the tangent plane, and (Gh,gh') the tangent line.
Let the plane of the curve be now revolved around its ver-
tical trace FG" ; the points of the curve wiil fall in perpen-
diculars to this trace drawn through their vertical projections,
and at distances from the trace equal to the hypothenuses of
triangles whose bases are the distances from the vertical pro-
jections of the points to the trace FG", and whose perpen-
diculars are equafl to the distances of the horizontal projections
of the points from the ground line. Having found the positions
of the points, let the curve q"h"Od"s"k" be described. The
tangent line takes the position G'h". In making the projections
of the curve, we have supposed the part of the surface above
the cutting plane to be removed ; the horizontal projection of
the curve is therefore made full. In the vertical projection, that
part of the curve is made full which lies in front of the meri-
dian plane nAt.
§ 238. We are next to consider the hyperbolic paraboloid,
and shall begin by examining the manner of representing it on
the planes of projection. If through any element of the sur-
face a plane be drawn perpendicular to the horizontal plane
(the surface being considered indefinite), the plane will be tan-166
COMPLEMENT.
gent to the surface at some point of this element (229) ; the
line drawn through this point of contact perpendicular to
the horizontal plane is tangent to the surface, and is therefore
an element of the tangent cylinder which projects the surface
on the horizontal plane. But the plane which is tangent to the
surface is also tangent to the cylinder: hence, its horizontal
trace is tangent to the base of the cylinder. But the horizontal
trace of the tangent plane is the horizontal projection of the
element through which the plane is drawn : hence, the hxn'i-
zantal projection of every element of the surface is tangent to
the base of the cylinder which projects the surface an the hori-
zontal plane ; that is, tangent to the curve which represents the
projection of the surface. By similar reasoning it may be
shown, that the vertical projection of every element of the sur-
face is tangent to the curve which represents the vertical pro-
jection of the surface. Therefore, if the projections of any
number of elements be determined, and two curves be drawn
respectively tangent to all the elements in each projection, these
curves will represent the two projections of the surface. We
will now show the easiest method of finding the projections of
the elements.
§ 239. PI. 6. Let (AB, A'B') and (CD, C'D') be the direct-
rices of a hyperbolic paraboloid, and (FH, EG) the plane-
directer. This plane cuts the directrix (AB, A'B') in the point
(1,1), and the directrix (CD, C'D') in the point (1,1) ; therefore,
the line joining 1 and 1 in the horizontal plane is the horizontal
projection of an element of the surface, and the line joining 1
and 1 in the vertical plane is the vertical projection of the same
element. Let now a plane be drawn parallel to the plane-
directer, and at any distance from it : LN and L'N' are the
traces of such a plane ; it cuts the directrix (AB, A'B') at the
point (2,2), and the directrix (Cl), C'D') in the point (2,2). If
through these points a line \)p drawn, it will be an element of
the surface ; both its projections are made in the figure. If, now,
a system of planes be drawn parallel to the plane (LN, L'N'),
and at the same distance from each other as this plane is from
the plane-directer ; first, this system of planes being parallel
31
WARPED SURFACES.
167
to the plane-directer, each plane will cut the directrices in two
points, and the line joining them will be an element of the sur-
face ; secondly, since the planes are equidistant, the parts of
the same directrix intercepted between any two of them which
are adjacent will be equal, and the projections of these equal
parts are also equal. If, therefore, on AB, the horizontal pro-
jection of one directrix, the parts from 2 to 3, from 3 to 4, from
4 to 5, &c. be each made equal to the part from 1 to 2 ; and
on CD, the horizontal projection of the other directrix, the parts
from 2 to 3, from 3 to 4, from 4 to 5, &c. be each made equal
to the part between 1 and 2 ; the lines drawn through the cor-
responding points are the horizontal projections of elements of
the surface. The vertical projections of the elements are deter-
mined either by finding the vertical projections of the points 1,
2,3,4, &c. in their corresponding directrices, and joining them ;
or by laying off on A'B' the parts from 2 to 3, from 3 to 4,
from 4 to 5, &c. each equal to the part from 1 to 2 ; and on
C'D' the parts from 2 to 3, from 3 to 4, from 4 to 5, &c. each
equal to the part from 1 to 2 : the lines joining corresponding
points are the vertical projections of elements of the surface.
The cur ve fdeba, drawn tangent to the horizontal projections
of the elements, is the horizontal projection of the surface ;
and the curve g'b'c'df, drawn tangent to the vertical pro-
jections of the elements, is the vertical projection of the
surface.
In making the projections of the elements on either plane,
the parts which are seen are made full, and the concealed parts
dotted. With respect to the horizontal projection, it is evident
that the part of each element which lies below the point at
which the element touches the projecting cylinder, or at which
the projecting plane of the element is tangent to the surface, is
concealed, and the part which lies above this point is seen.
Therefore, the horizontal projection of the element passing
through 3 and 3 is made full from a, the element through 4 and
4 from the point at which it touches the curve, the element 7
and 7 from c, &c. With respect to the vertical projection, the
part of each element which is in front of the point at which itCOMPLEMENT.
168
touches the cylinder that projects the surface on the vertical
plane is made full, and the part which is behind this point is
dotted. Thus, the element passing through 2 and 2 is dotted to
g', the element passing through 4 and 4 is dotted to 0', &c.
The directrices in either projection are seen when the elements
which touch them are seen ; but as they are important lines in
the construction, they have been made full. The horizontal
trace of the plane-directer, excepting the part Fb9 is concealed
by the surface, and is therefore made broken ; the vertical
trace is also concealed by the surface, excepting the part 5G.
It is easy to find the intersection of this surface with a given
plane, since the points in which the elements pierce the plane
are points of the curve. The horizontal plane of projection
intersects the surface in the curve ponml : the element passing
through the points (5,5) and (5,5) pierces it at l, the element
through (6,6) and (6,6) at m, and similarly for the other
points.
PROBLEM.
To draw aplane tangent to a hyperbolic paraboloid at a given
point of the surface.
§ 240. PI. 7. Let (GH, G'H') be the plane-directer, (AB,A'B')
and (CD, C'D') the directrices, and a the horizontal projection
of the point at which the tangent plane is to be drawn.
The vertical projection of the point cannot be taken at plea-
sure, but must be found by construction (93). The plane-
directer cuts the directrix (AB, A'B') in the point (1,1), and the
directrix (CD, C'D') in the point (1,1). Let any plane, as
(EF, E'F'), be drawn parallel to the plane-directer ; it cuts the
directrix (AB, A'B') in the point (2,2), and the directrix
(CD, C'D') in the point (2,2). By laying off the projections of
the parts of the directrices intercepted between these two
planes, and drawing lines through the corresponding points,
we determine the projections of any number of elements.
At the point a conceive a line to be drawn perpendicular toWARPED SURFACES.
169
the horizontal plane ; the point in which this perpendicular
pierces the surface is the only point of the surface which is
horizontally projected at a. Draw through this perpendicular
the vertical plane {cab, bbf) ; this plane intersects the surface in
a curve ; the point in which the perpendicular meets this curve
is the point in which it pierces the surface. This plane cuts
the element drawn through (2,2) and (2,2) at the point {d,d'),
it cuts the element drawn through (3,3) and (3,3) at the point
{f",f”)9 the element drawn through (7,7) and (7,7) at the point
(e,e'), and the element drawn through (9,9) and (9,9) at the
point (c,c') ; thus df'e'a'c' is the vertical projection of the
curve in which the vertical plane (cab, 66') intersects the sur-
face, and the point a’, in which this curve intersects the per-
pendicular to the ground line through a, is the vertical projec-
tion of that point of the surface which is horizontally projected
at a. Having found the point {a,a'), if we draw through it an
element of the first generation, it will ue a line of the tangent
plane (89) ; and if we draw through the point {a,a') an element
of the second generation, it will also be a line of the tangent
plane (89) : hence, the plane of these elements is the tangent
plane required. To find the element of the first generation
passing through the point {a,a'). Let a plane be drawn through
this point parallel to the plane-directer, the point in which it cuts
either of the directrices being joined with the point (a,a') deter-
mines the element sought. Draw in the plane-directer, and
through the same point, any two lines, as {fg,fgf) and {f,b,f'h%
and as two lines determine a plane, a plane drawn through
{a,a’) parallel to ( fg,fg') and (fh,fhf) will be parallel to the
plane-directer. Drawing {ai, a'i') parallel to {fg,f g')> its pro-
jections are parallel, and it pierces the plane which projects the
directrix (CD, C'D') on the horizontal plane in the point {i,i').
Drawing {at,a’t) parallel to {fh,fk'), this parallel pierces the
plane which projects the directrix (CD, C D ) on the horizontal
plane at the point {t,t'). But the lines {ai, a'i') and {at, a't')
determine a plane passing through {a,a') parallel to the plane-
directer, and t'i' is the vertical projection of its intersection with
the plane which projects (CD, C'D') on the horizontal plane :COMPLEMENT.
170
hence, n' is the vertical projection of the point in which it cuts
the directrix (CD, C'D'), and n is the horizontal projection of
the same point. Therefore, {an, a'n') is the element of the first
generation passing through the point {a,a'), and the point L, in
which it pierces the vertical plane, is one point of the vertical
trace of the required tangent plane.
To find the element of the second generation passing through
{a,a'). The plane-directer of the second generation is parallel
to the directrices of the first generation (223). Therefore, if
we draw through {a,a') a plane parallel to the directrices (AB
A'B') and (CD, C'D'), and determine the point in which it cuts
any element of the first generation, regarded as a directrix of
the second generation, this point being joined with {a,a') deter
mines the required element of the second generation. Take
the element passing through (8,8) and (8,8) for one of the di-
rectrices of the second generation, and draw through {a,a') the
lines (aq, a'q') and (<ap, a'p') respectively parallel to the direc
trices of the first generation ; they determine a plane parallel
to the plane-directer of the second generation (223). This plane
intersects the plane which projects the directrix (88, 88) on the
horizontal plane in a line of which prq' is the vertical projection;
the point s', in which p'q' intersects 88, is the vertical projection,
and s is the horizontal projection of the point in which the plane
passed through (a,a'), parallel to the plane-directer of the second
generation, cuts the directrix (88, 88) of the second generation :
hence, {as, a's') is the element of the second generation passing
through the point («,«'). Having thus determined a second line
of the tangent plane, a second point of its vertical trace is easily
found, and the vertical trace LN can be drawn. The horizontal
trace does not fall on the paper, but may be considered as found,
since two lines of the plane are known.WARPED SURFACES.
171
THEOREM.
If tivo warped surfaces, M and N, have the same plane-directer,
an element E common, and two tangent planes also common,
points of contact m and m! being on the element E, the
sui faces will be tangent to each other throughout this element.
§ 241. For, conceive any two secant planes P and P' to be
drawn through the points of contact m and m'. These planes
intersect the surface M in the curves d and d\ and the surface N in
the curves ƒ and ƒ'. Since the surfaces are tangent to each other
at the point m (85), the plane P intersects the tangent plane at
this point in a line tangent to the curves d and ƒ ; consequently,
these curves are tangent to each other at the point m. For
similar reasons the curves ã and ƒ ' are tangent to each other
at the point m'. If, now, we take the two curves d and d' for
directrices, and the common plane-directer for a plane-directer,
the elements of the surface M which pass through the conse-
cutive points of tangency of the curves d and f and d' and ƒ' are
consecutive and belong also to the surface N. If any secant
plane be drawn through a point of the element E, it will inter-
sect the consecutive elements in consecutive points, and the
curve of the surface M and the curve of the surface N will be
tangent to each other, since they have two consecutive points
common. Hence the surfaces themselves are tangent through-
out the element E.
THEOREM.
Any two warped surfaces M and N having an element E com-
mon and three common tangent planes, their points of contact
m9 m\ m", being on the element E, are tangent to each other
throughout this element.
§ 242. The demonstration of this theorem is very analogous
to the preceding. Through the three points of contact m, m\172
COMPLEMENT
and mh conceive three secant planes to be pas< / «et d, d',
and d" be the curves in which they intersect the urface M,
and ƒ, ƒ', and ƒ" the curves in which they intersect the sur-
face N.
Since the surfaces are tangent to each other at the three
points m, and m", it is evident that the curves d and f d and
ƒ', and d" and f" are tangent to each other at the same points.
Let the generatrix of the surface M move on the three curves
d, d', and d" as directrices ; when indefinitely near the element
E, it will pass through the consecutive points of tangency of the
curves d and/, d! and ƒ', and d" and ƒ", and in this position it
is consecutive with the element E : hence, the two surfaces
have two consecutive elements common about the element E ;
therefore, the surfaces are tangent to each other along this ele-
ment : for, if the surfaces be intersected by a plane through
any point of the element E, the sections made in the surfaces
will be tangent to each other.
•	PROBLEM.
A warped surface whose generatrix is parallel to a plane-directer
being given, it is required to draw a tangent plane at a given
point of this surface.
§ 243. Pi. 8. Take the vertical plane of projection for the
plane-directer; let the curves (abcf a'b'c'f) and (ghjm, g'h'mj')
be the directrices, and M the horizontal projection of the point
at which the plane is to be tangent. The vertical projection
of this point must be determined by construction. Draw a
series of planes ag, bh, cj, &c. parallel to the vertical plane of
projection ; they cut the directrices in the points (a,a), (6,ft'),
(c,c'), &c., and (gg*), (VO* (/ƒ)> &c- ; the right lines joining
these points are elements of the surface. Having thus deter-
mined as many elements as are necessary, draw through the
point M the vertical plane wMS ; this plane cuts the elements
before found in the points (n,rc'), (o,o'), (7,/), (if), (s,s'), and the
curve drawn through these points is the curve in which theWARPED SURFACES.
173
plane intersects the surface. The point of which M is the
horizontal projection being a point of the surface and of the
secant plane /iMS, is a point of the curve of intersection : hence,
it is vertically projected in the curve S’i’q'o'rí ; it is also verti-
cally projected in a perpendicular to the ground line through
M; therefore, M' is the vertical projection of the point.
Knowing the point of contact	let a plane be drawn
through it parallel to the vertical plane of projection ; this plane
cuts the directrices in the points (A,A') and	and the line
(AB, AB), drawn through these points, is an element of the
surface, and consequently a iine of the required tangent plane.
Now, of all the planes which can be drawn through (AB, A'B'),
it is required to find the one which shall be tangent to the sur-
face at the point M. To do thi9, let us use an auxiliary sur-
face, the hyperbolic paraboloid. Through the points (A,A')
and (B,B') draw the right lines (AC, A'C') and (BD, B'D )
respectively tangent to the directrices (abf, a'bf') and
(ghm, g'h'm'), and let us suppose a right line to move upon these
tangents, continuing parallel to the vertical plane of projection ;
it is evident that it will generate the surface of a hyperbolic
paraboloid containing the element (AB, A'B') ; the hyperbola
paraboloid is tangent to the warped surface along the element
(AB, A'B'). For, if through the tangent (AC, A'C') and the
element (AB, A'B') a plane be drawn, it will be tangent to both
surfaces at the point (A,A') ; and if through the tangent
(BD, B'D') and the element (AB, A'B') a plane be drawn, this
plane will also be tangent to both surfaces at the point (B,B').
Hence, the surfaces are tangent to each other along the element
(AB, A'B') (241). If, now, a plane be drawn tangent to the
hyperbolic paraboloid at the point (M,M'), this plane will also
be tangent to the given surface, and consequently be the plane
required.
§ 244. The vertical plane having been taken for the plane-
directer, it was not necessary to construct the curve (woS, n'o'S')
in order to find the vertical projection of the point (M,M') ;
for, the point M being given, the plane AB, which contains the
element of the surface in which the point (M,M') is found, is174	COMPLEMENT.
determined, and projecting the points A and B into the vertb
cal projections of the directrices determines A'B', the vertical
projection of the element ; the point M' is therefore known.
It would not be thus if the plane-directer were not parallel to
the vertical plane of projection ; we should then have to use
the first method to determine the vertical projection of the
element.
THE END