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Cockbum^ J, Roy
Brief synopsis of the
course of lectures in
descriptive geometry
501
■«
Brief Synopsis of the Course
of Lectures in
Descriptive Geometry
AS ARRANGED FOR THE SECOND YEAR
FACULTY OF APPLIED SCIENCE AND ENGINEERING
UNIVERSITY OF TORONTO
J. ROY COCKBURN, B.A.Sc.
Associate Professor of Descriptive Geometry
1 PUBLISHED BY
'i- THE UNIVERSITY OF TORONTO ENGINEERING SOCIETY
I 1923
Brief Synopsis of the Course
of Lectures in
Descriptive Geometry
AS ARRANGED FOR THE SECOND YEAR
FACULTY OF APPLIED SCIENCE AND ENGINEERING
UNIVERSITY OF TORONTO
J. ROY COCKBURN, B.A.Sc.
Associate Professor of Descriptive Geometry
PUBLISHED BY
THE UNIVERSITY OF TORONTO ENGINEERING SOCIETY
1923
501
- *•<•.
SYNOPSIS OF THE COURSE OF LECTURES IN
DESCRIPTIVE GEOMETRY AS ARRANGED
FOR THE SECOND YEAR
By J. Roy Cockburn, B.A.Sc,
Lecturer in Descriptive Geometry, University of Toronto
Gexeratiox and Classification of Lixes
Any line may be considered to be generated by the continuous
motion of a point and may be considered to be made up of an
infinite number of infinitely small straight or right lines.
The law which directs the motion of the point determines the
nature of the line.
Lines may be classified as follows:
(1) Right or straight lines in which the direction of motion of
the point is always the same.
(2) Curved lines in which the direction of motion of the
generating point is changing continuously.
Curved lines may be divided into two classes:
(a) Curves of single curvature having all points lying in a
single plane.
(b) Curves of double curvature having no four consecutive
points lying in a single plane.
In problems in descriptive geometry, curves like other lines are
usually represented by their projections.
If a curve of single curvature have its plane perpendicular to
a plane of projection, its projection on that plane is a straight line.
If the plane of the curve be parallel to the plane of projection,
the projection will be of exactly the same form as the curve itself.
The projection of a curve of double curvature can never be a
straight line or of the same form as the curve itself.
Tangents and Normals to Curves
A straight line is tangent to another line when it contains two
consecutive points.
Two curves are tangent to each other when they contain two
consecutive points or have at a common point a common tangent.
If a straight line be tangent to a curve of single curvature it
will be contained in the plane of the curve.
A normal to a curve is a line at right angles to the tangent.
The normal in the plane of the curve is generally considered to
be the normal.
The most important lines of single curvature are the conic
sections, which may be defined as follows:
A conic is the locus of a point which moves in a plane so that
its distance from a fixed point is in a constant ratio to its distance
from a fixed line. This ratio is usually denoted by the letter e.
\{ e = l the curve is a parabola.
If e<l the curve is an ellipse.
If e> 1 the curve is a hyperbola.
To Plot the Parabola
In Fig. 1 let F be the given point and CD the given line. Draw
FD at right angles to CD. Bisect FD at the point A. Take any
point M to the right of A. With F as centre and radius FP = DM
describe a circle cutting perpendicular PMPi at P and Pi. P and
Pi are points on the parabola.
To Plot the Ellipse when given the Major and Minor Axes
In Fig. 2, AAi and BBi are the given axes of the ellipse, inter-
secting at the point O. With centre 0 and radius OB describe a
circle and with the same centre and radius OA describe another
circle.
Draw OHE any common radius making an angle 6 with AAi.
Draw EG perpendicular to ^^i and HP perpendicular to EG
and intersecting it at the point P. P is a point on the ellipse.
X =a cos 0 y =hs\nd
a?
= cos2e
yi.
~r =sin2 Q
+
62
= 1
Second Method of Drawing the Ellipse when given the
Major and Minor Axes or a Pair of Conjugate Diameters
In Figs. 3 and 4, AAi and BBi are a pair of conjugate diameters.
Upon these diameters describe a parallelogram as shown.
B PC
Divide the semi-diameter OA into any number of equal parts,
and the side AC into the same number of equal parts.
Straight lines are drawn as shown, and their intersections are
points on the required curve.
To prove that the curve is an ellipse—
OA and A C are divided at the points E and // in the same ratio,
fit
Let this ratio be denoted by
I.e.
CH
11
m
n
OE
HA n EA
In the similar triangles BFP and BCH.
m
CH^_BC m-\-n
FP ~
or
BF^' b-y
and in similar triangles BiOE and BiDP
DP ^ B.D
OE ~ BiO
or
X
h+y
m
m-\-n
a
Equating the values of
m
ni -f n
we have
a(b-y)
bx
bx
a{b-hy)
a^{b^-y'^)=^b^x^
Dividing each side by a*i&" we get
1 -
or
a^
b'-
1. The equation to an ellipse.
Third Method of Describixg the Ellipse whex given the
Major and Minor Axes
In Fig. 5 let AAi be the major axis and BBi the minor axis
intersecting it at the point 0.
On a piece of paper measure off the length Pi? = the semi-major
axis, and P^ = the semi-minor axis.
If the point Q be kept on the major axis and the point R on
the minor axis, then the point P will be on the circumference of the
ellipse.
Let d = the angle between PR and BBi
x = a sin 6 y = b cos 6
y-
— +
a- o~
= sin2e+cos2^ = l.
Method of Describing the Hyperbola
The hyperbola may be generated by moving a point in the
same plane so that the difference of its distances from two fixed
points shall be equal to a given line.
In Fig. 6 let Fand Fi be the two given points (the foci) and FFi.
the given line so placed that FV=FiVi and in the same straight
line as FFi.
With Fi as centre and any radius Pi il/ greater than Pi F describe
an arc MiM.
With P as centre and radius Pilf equal to FiM— VVi describe
a second arc intersecting the first arc at the points Mand Mx.
M and Mi are then points on the curve.
The Helix
A useful example of a line of double curvature is the Helix^
which may be defined as follows:
The Helix is a curve generated by a point which has a uniform
motion of rotation about a fixed axis and a uniform motion of
translation parallel to that axis.
Note- -See drawing entitled "Helix". y^
X
Generation and Classification of Surfaces
Surfaces may be classified as follows:
A. Surfaces which can be generated by straight lines :
(a) Planes.
(b) Surfaces of revolution- — -cylinders, cones, hyperboloid of
revolution of one sheet.
(c) Warped surfaces.
B. Surfaces which can be generated only bycurves.
Surfaces of revolution — spheroid, paraboloid and hyper-
boloid^of revolution of two sheets.
To Prove that the Surface Generated by the Revolution of
A Straight Line About Another Straight Line not in the
SAME Plane is a Hyperboloid of Revolution of One Sheet,
i.e., to Prove that its Meridian Section is a Hyperbola.
(Fig. 7.)
Fm. 7.
Let c^p^ and cp he the vertical and horizontal projections of
the straight line CP which is parallel to the vertical plane and
inclined at an angle 6 to the horizontal plane.
Let e^d^ and ed be the vertical and horizontal projections of the
line ED which is perpendicular to the horizontal plane.
If the line CP revolve about D E it will cut from a plane con-
taining DE a pair of curves which we can prove are the two branches
of a hyperbola.
Consider the point P on CP.
If CP be revolved to cut the plane containing D E and parallel
to the vertical plane, the point Pr where P passes through the
plane will be a point on the curve cut from the plane by the generat-
ing line CP.
8
Let the shortest distance between the generating line CP and
the axis DE = a and let — • =tan 6.
a
Let 0 and C be the extremities of the shortest line between CP
and DE.
The actual distance from 0 to P is OP; from 0 to C = a.
Let the distance from Pr to DE = x.
Let the vertical distance that Pr is above 0 = y.
y
The distance from d^e^ to p^ =
tan (9
,9
\,
Then OP' = a'+ ^.0+y- (OP, )-=-^'+y
tan^'-
ora'tan-e-^y-^x'-tan'-d
Substitute — for tan 6.
a
In a- tan- d+y~ = x- tan^ ^
3,2 :^2 ^2 2
b- a- a- 0-
V. Tangent Planes to Curved Surfaces
Problem 1
7 0 /j«<i //ze projections of a right circular cone when given lis
dimensions and its position relative to the planes of projection. (Fig. 8.)
Let the position of the centre of the base be given by its pro-
jections 0^ and 0 and the inclination of the axis of the cone to the
vertical plane be a and the inclination to the horizontal plane be /3.
First, find the projections of the axis o^Vi^ and ovi when it makes
an angle of a with the vertical plane and is parallel to the horizontal
plane. Also find the projections of two diameters of the base, one
parallel to the horizontal plane, ai^^i^ and a^bi being the required
projections, and the other perpendicular to the horizontal plane
Ci^di'- and Cidi being the required projections.
Next, find the projections of the axis o^V2^ and ov^ when it makes
an angle of iS with the horizontal plane and is parallel to the vertical.
With centre 0^ and radius o^Vi^ describe the arc Vi^v^ and
through the point V2^ draw the line v^^v^ parallel to the ground line,
to intersect the arc Vih^ at the point v^. The point v^ is the vertical
projection of a vertex in the required position. Its horizontal
projection v is found in the same way.
Next, find the vertical projections, oS ¥, c^ and d^ of the ex-
tremities of the two diameters AB and CD and their horizontal
projections a, b, c and d.
"jK
a^b^ and c^d^ are conjugate diameters of the ellipse which is
the vertical projection of the base and ab and cd of the ellipse
which is the horizontal projection of the base.
Describe the ellipses by the method given previously.
Draw tangents from the projections of the vertex to the corre-
sponding projections of the base obtaining the outlines of the pro-
jections of the cone.
r
ir/n:
n
Problem 2
To find the projections of a right circular cylinder ivhen given its ^.
dimensions and its position relative to the planes of projection. (Fig. 9) . ^ /
Find the projections of the base and axis of the cylinder by U
the same construction as used in Problem 1 for the cone.
Complete the projections of the cylinder as illustrated in Fig. 9
A
Problem 3
To find the traces of a plane tangent to the surface of a given cone
and passing through a given point on the surface of the cone. (Fig.
10.)
Let the projections of the cone be as found in Problem 1.
10
To assume the projections of a point on the surface, first
assume one projection, in this case the vertical projection, (p^),
and then find the horizontal projection, as follows:
Join the vertical projection p^ to the vertical projection of the
vertex v^, obtaining the vertical projection of the element PV,
through the point P, and find the vertical projection q^ of the
point Q where this element intersects the base.
Find g the horizontal projection of point Q, and join g with
V, obtaining the horizontal projection of the element through P.
The horizontal projection of P is on this line gv where it is inter-
sected by a line through p^ perpendicular to the ground line.
To find the traces of a plane containing the point P and tangent
to the cone, proceed as follows:
Find the projections of a line QR tangent to the base at the
point Q. That is, draw gV^ tangent to the vertical projection of
the base of cone and qr tangent to the horizontal projection of base.
11
Find the traces of the plane containing the lines PV and QR
and they will be the traces of the required plane.
^_ J
Wim. la
Problem 4
To find the traces of a plane tangettt to the surface of a given
cylinder and passing through a given point on the surface of the
cylinder. {Fig. 11.)
Let the projections of the cylinder be as found in Problem 2.
To assume the projections of a point on the surface, first assume
one projection, in this case the vertical projection {p^) and then
find the horizontal projection as follows:
Draw a straight line p^g} through p^ parallel to oh^ obtaining
the vertical projections of the element PQ and find the vertical
projection g^ of the point Q where this element intersects the base.
Find 2 the horizontal projection of Q and draw pg through q
parallel to ov obtaining the horizontal projection of the element PQ.
The horizontal projection of "P" is on this line at p.
To find the traces of a plane containing the point P and tangent
to the surface of the cylinder, proceed as follows :
12
'm. m
Find the projections of a line QR tangent to the base at the
point Q. That is, draw q^r^ tangent to the vertical projection of
the base of cylinder and qr tangent to the horizontal projection of
the base. ^ -
Find the traces of the plane containing the lines PV and QR,
viz., LM and MN, and they will be the traces of the required plane.
C^ey^.
To find the traces of a plane Tangent to the surface of^a given cone
and passing through a given point ivithout the surface of the cone.
{Fig. 12.)
Let the projections of the cone be as found in Problem 1. Let
;v' and X be the projections of the given point through which the
plane must pass. JLoin x^ to v^ and v with r^ obtaining the pro-
jections of the line AT, which joins the point X to the vertex ot '
...^ r
i
13
^^i\A
y^
the cone. Find thf tmrps of thf" plnn^ of the base of the cone,
viz., NW and LS. Find the projections r^ and / of the point R
where the Hne XV intersects the plane NWLS (the plane of the
base). Find the projections g}n^ and gn of the Hne QN, which
passes through the point R and is tangent to the base of the cone
at the point Q. Find the traces of the plane containing the lines
XV and QN, viz., LM and YN. These are the traces of the re-
quired plane.
4
K
>K
Note— A cylinder may be considered to be a particular case
of a cone in which the vertex is at an infinite distance from the
base. Therefore, the solutions given for Problems 1, 3 and 5
apply equally well to the solutions of the corresponding problems
relating to the cylinder.
To find the traces of a plane tangent to the surface of a given
cylinder and passing through a point withont the surface of the cyli'nder.
{Fig. 13.)
Let the projections of the cylinder be as found in Problem 2.
Let x^ and xhe the projections of the given point through which
the plane must pass.
Draw x^u'^ parallel to o^v^ and xu parallel to ov obtaining the
!^ i /! i / ¥ V
\ \ !\ i\
rm^ im^
projections of the line XU which is parallel to the axis OV.
Find the traces of the plane of the base of the cylinder, viz., NZ
and LS.
Find the projections r'^ and r of the point R where the line XU
intersects the plane ZNLS (the plane of the base).
15
Find the projections g'w' and cpt of the line QN which passes
(rough the point R and is tangent to the base of the cyhnder at
le point Q.
Find the traces of the plane containing the lines XU and QN,
z., Zilf and YM.
Problem 7
To find the traces of a plane tangent to the surface of a given cone
id parallel to a given straight line. {Fig. 12.)
Let ef and e^p- be the projections of the given line EF.
Find the projections vx and v^x^ of a straight line VX passing
irough the vertex of the cone and parallel to the given straight
ne EF.
Find the traces of the plane containing VX and tangent to the
irface of the cone as in Problem 5.
Problem 8
To find the traces of a plane tafigent to the surface of a given
dinder and parallel to a given straight line. {No figure given for this
ro hlem^'^ / ,' )■ y^ j*
<^- Find the traces of a plane containing the given line and parallel
) the axis of the given cylinder.
Find the projections of the line of intersection of this plane
ith the plane of the base of the cylinder.
Find the projections of a line parallel to this line and tangent
) the base of the cylinder.
Find the traces of a plane containing this tangent and an element
f the cylinder as in Problem 6, Fig. 13.
16
Problem 9
To find the traces of a plane which is tangent to the surface of a
given sphere and passes through a given point on the surface of the
sphere. {Fig. I4.)
Let the centre of the sphere be given by its projection o^ and 0.
In order to assume the projections of a point on the surface, pro-
ceed as follows:
Assume ni^ the vertical projection of a point M on the surface
of the sphere. To find the horizontal projection of the point M,
join ni^o^, obtaining the vertical projection of the radius MO
through the point M. Revolve the sphere about a vertical line
through 0 until MO becomes parallel to the vertical plane. Wi^
is the vertical projection of M in this position. The horizontal
projection of MO is then the line mio parallel to the ground line,
the point m being on a line through Wi^ and perpendicular to the
ground line. Revolve the sphere back to its original position,
finding the horizontal projection m.
Find the traces WX and XF of the plane passing through the
point M and perpendicular to the line MO and these will be traces
of the required plane.
{ Problem
To find the projections 'of -ihe circle' of contact of a cone which
envelops a given sphere and has its vertex at a given point. {Fig.^Jo.)
17
Let z^^ and z) be the projections of the vertex (F) of the cone
and 0^ and o of the centre (0) of the sphere.
■■»• Pass a vertical plane through the vertex V so as to cut the
sphere in the circle whose horizontal projection is the straight
line ab.
Revolve this plane until parallel to the vertical plane of pro-
jection. The new horizontal projection of the circle cut from the
sphere is the line aibi and the point Ci midway between Oi and bi
is the horizontal projection of the centre of the circle.
— — The vertical projection of this circle is the circle described
about Ci^ as centre and with a radius equal to Ciai.
-The tangent v^Xi^ and v^yi^ to this vertical projection of the
circle are the vertical projections of tv/o elements on the required
tangent cone.
18
The points Xi^ and yi^ are the vertical projections of two points
on the required circle of contact.
The horizontal projections of these points are Xi and yi.
'Revolve the sphere back to its original position obtaining the
horizontal projections x and y and the vertical projections x^ and yK
In the same way, by taking other auxiliary planes, a sufficient
number of points on the circle of contact can be found.
Problem 11
7"o find the traces of a plane containing a given line and tangent
to the surface of a given sphere. {Fig. 16.)
Let a^b^ and ab be the projections of the given line AB, and
0^ and 0 the projections of the centre 0 of the given sphere.
Find the traces LM and MN of a plane LMN containing the
point 0 and perpendicular to the line AB.
Find the projections c^ and c of the point C where AB intersects
the plane LMN. ^-.,..^-^-^ .^.^
Revolve the plane LMN about its horizontal trace MN until
it coincides with the horizontal plane. The point C falls at Ci
and 0 at Oi. With centre Oi and radius equal to the radius of the
sphere describe a circle obtaining the line cut from the sphere by
the plane LMN. — di>
Draw DCi, Ei tangent to this circle.
Revolve the sphere back to the original position. The point
D being on the axis of revolution does not change its position.
Its vertical projection is the point d^ on the ground line.
The line cW^ is the vertical projection of the line CD and cd
the horizontal projection.
Find the traces of the plane containing the lines AB and CD.
This last step is not shown in Fig. 16.
r
Intersections of Surfaces by Straight Lines'
Problem 12 ^ i//{
To find the projectidns-of thepDints'tnwhich the surface of a given
cone is intersected by a given straight line. {Fig. 17.) /,.
Let the projections of the cone be found as in Problem 1 and
let w^T^and wx be the projections of the given line WX. -
Find the traces of the plane of the base of the cone, viz., M^
and LS.
— Find the traces of a plane containing line WX and the vertex
of the cone, viz. ^ Pi? and PQ.
— Find the projections of the lines of intersection of the two
planes MLS and RJPQ, viz., g}r^ and gr. The line ^i? intersects
the circumference of the base of the cone in the points Y and ^^Jf
of which the projections are y^, z^ amd y, z. 4j
Ui
19
-P
V.
"-^ The lines joining the points Y and ':^' to the vertex of the cone
are the elements cut from the cone by the plane QPR. The points
E and F where the given line WX intersects these elements are
the required points.
The projections of the points £ and Fare the points g^/^ande,/.
4
\J
.r-^3 q'
To find the projections of the poi?tts in which the surface of a given
sphere is intersected by tlie given straight line. {Fig. 18.)
Let 0^ and o be the projections of the centre 0 of the given
sphere, and let d^¥ and ab be the projections of the given line AB.
20
Find the traces LAf and MN of the horizontal projecting plane
of the line AB.
This plane LMN cuts the sphere in a circle whose horizontal
projection is the line g^p (part of MN).
Revolve the sphere and plane LMN about the horizontal
trace MN until the plane coincides with the horizontal plane of
projection. C, the centre of the circle cut from the sphere by
LMN falls at Q. The line AB falls at ABv and the points Wi
and Xi where ABi cuts the circumference of the circle cut from
the sphere by LMN are the required points. Revolve the sphere
and plane back to the original position finding the projections w'S
x^ and w, x of the points \V, X where the line AB intersects the
surface of the given sphere.
il
Intersection of Curved Surfaces by Planes
General Method of Solution
Both the given curved surface and given plane are intersected by a
number of auxiliary planes.
Each plane cuts each surface in a line or lines.
21
The points of intersection of these lines are points on the
required Hne of intersection of the surfaces.
The auxiliary planes are so chosen that the lines cut from the
given curved surface are as simple as possible.
In the case of the cyHnder, the auxihary planes should be
taken either parallel to the elements of the cylinder, in which
case they would cut the surface along straight lines; or parallel to
the base, in which case they would cut the surface in lines similar
and parallel to the circumference of the base.
In the case of the cone, the auxiliary planes should be taken
either passing through the vertex, in which case they would cut
the surface along straight lines; or parallel to the base, in which
case they would cut the surface in lines similar and parallel to
the base.
Problem 14
To find the projections of the line of ifitersection of a right circular
cone by a plajie, and to determine the exact form of the line of inter-
section. {Fig. 19.)
Let the projections of the cone and traces LM and MN of the
plane be as given in Fig. 19.
Note — -The given plane is in this case perpendicular to the
vertical plane of projection. If it were not in such a position, it
22
and the cone could be rev'olved about the axis of the cone until
the plane would become perpendicular to the ^'ertical plane of
projection.
Solution of Problem
Take a plane parallel to the base of the cone and intersecting
the surface in the circle WXYZ of which the projections are the
straight line w^x^ and the circle w, bxc.
The vertical projection of the required line of intersection is
the line a^d^e^ (the same line as the vertical trace of the plane).
The vertical projection of the line cut from the cone by the auxiliary
plane WX is the line w^x^ and the vertical projection of the line
cut from the plane LAIN by the auxiliary plane is the line b^c^
(its vertical projection being a point), and its horizontal projection
is the line be.
The points b and c are points on the horizontal projection of
the required line of intersection.
By taking a sufficient number of auxiliary planes, a sufficient
number of points on the horizontal projection of the line of inter-
section may be found.
In order to find the form of the line of intersection, the cone
is revolved about some convenient axis until the given plane LMN
becomes parallel to the horizontal plane of projection.
To prove that the intersection of a right circular cone by a plane is
the locus of a point whose distance from a fixed point is in a constant
ratio to its distance from a fixed line. {Fig. 20.)
Let the line LABM be the line cut from the given cone by the
given plane and let the line DAH be a line cut from the givert plane
by a plane through the axis of the cone and perpendicular to the
given plane.
Consider a small sphere to be placed within the surface of the
cone touching it along the circumference of the circle through the
points G and E and touching the given plane at the point F.
Let DE be a line in the plane of the circle GE and in the plane
containing the axis of the cone and line DAH. It intersects the
line DAH in point D.
Let DC be a line through point D and perpendicular to both
Z)£and DA.
Let point B be any point on the line LABM.
Let BK be the circumference of a circle through B, whose plane
is perpendicular to the axis of the cone. Its plane intersects the
given plane in the line BH.
Let ^C be a line through B parallel to DAH.
VK is the element through points E and A.
VB is the element through point B.
The triangles AED and AKH are evidently similar and the
side AH is to the side AK as the side AD is to the side AE.
. . AH AD DH , .
'^^'''ak = ae = ek (^)
23
'^5
O^,
DH = CB
and VB ^ VK and VE = VG
. : . EK=BG, and BG = BF, because both are tangent to the same
sphere at the points G and F
. :. EK = BF and AE = AF.
Substitute in equation (a).
CB for DH, BF for EK and AF for AE, obtaining the equation
CB ^ AD
BF ~ A F
But AD and AF are both constant for all positions of B on
the line LABM.
. : . the ratio of BF to CB is constant for all positions of B, that
is, the curve LABM is the locus of a point whose distance from the
fixed point /^ is in a constant ratio to its distance from the fixed
line DC.
If the given plane be parallel to an element of the cone, the line
BF equals BC and the ratio is therefore equal to unity, and the
cur\'e is a parabola.
If the given plane cut all the elements of the cone, the line BF
is evidently less than BC and the ratio is then less than unity and
the curve is an ellipse.
If the given plane is not parallel to an element and does not cut
all the elements, BF is evidently greater than BC and the ratio is
greater than unity and the curve is a hyperbola.
24
Problem 15
To find the projectio?is of the line of intersection of a given sphere
by a given plane. (Fig. 21.)
Let 0^ and o be the projections of the centre 0 of the given
sphere and LM and MN the traces of the given plane LMN.
^, Find the projections p^ and p of the point P, where a vertical
line through centre O intersects the plane LMN.
To find the projections of P. take a plane containing the point
0 and parallel to the vertical plane ot projection. It intersects
TTie plane JAIN in the line PQ of which pq (parallel to the ground
line) is the horizontal projection, and p^g^ (parallel to the vertical
trace LM) is the vertical projection.
*A
N, M
^ali.
The projections p^ and p of the point where PQ intersects the
vertical line through 0 are then readily found.
— Revolve the sphere and the plane LMN about the vertical line
OP until it becomes perpendicular to the vertical plane of pro-
jection. In this position, their line of intersection is vertically
projected in the straight line Ui^hi- and horizontally projected in
the ellipse aihiC\d\.
— Revolve the sphere and plane back to the original position
obtaining the projections of the points A, B, C and D, viz., a}, b^,
<"' and rf' and a, b, c and d.
The ellipses passing through these points are the required pro-
jections of the line of intersection.
v^
V i.
25
Y-
^"^v\j^:^
■r
Intersectiox of Curved Surfaces ^, . ,. . ,,
General Solution a ^ /^
Both surfaces are intersected by a system of auxiliary planes.
Each auxiltary'"^ prane will cut from the given surfaces, lines the
intersection of which wijl be points on the required line of inter-
section.
The auxiliary planes should be so chosen as to cut from the
given surfaces the simplest lines; straight lines if possible or the
circumferences of circles, or other regular curves.
Problem 16
To find the projections of the line of intersection of two cylinders.
{Fig. 22.)
Let l^m^ and Im be the projections of the axis LM of one cylinder,
and n^o^ and no the projections of the axis NO of the other cylinder.
Find the traces on the planes of the bases of the two cylinders
of a plane containing any point P and parallel to both axes.
Note — The planes of the bases of the cylinders in this case both
coincide with the horizontal plane of projection; therefore, the
line QR is the required trace upon the plane of the bases.
26
QR is the horizontal trace of a plane containing any point P
and parallel to the axes of both cylinders; that is, it contains the
lines PQ and PR which are respectively parallel to the axes of the
cj'linders.
Any other plane parallel to the axes of the cylinders will have
its horizontal trace parallel to QR. The horizontal trace of such
a plane is ABCD cutting the circumferences of the bases in the
points A, B, C, D, and the plane must cut the cylinders in the
elements which pass through these points, viz., in the elements
A WX, B YZ, C YW and DZX.
The points of intersection of these elements WXY and Z are
points on the required line of intersection. Their projections are
the points w^, x^, y^ and z^ and w, x, y and z.
The projections of a sufficient number of points on the required
line being found, the curve may be plotted as shown in the figure.
Problem 17
To find the projections of the lines of intersection of two cones.
{Fig. 23.)
The solution of this problem is similar to that of Problem 12,
with this one exception: thcjauxiliary planes must all pass through
the vertices of the two cones insTead of being parallel to the axes.
27
Problem 18
To find the projections of the lines of intersection of a cylinder and
a cone. {Fig. 2Jf.)
The planes are taken parallel to the axis of the cylinder and
passing through the vertex of the cone; that is. thev must contain
a straight line through the vertex parpillel >n the axis of the cylinder.
Development of Curved Surfaces
If a single curved surface be rolled over on any tangent plane
until each of its elements has come into this plane, the portion of
the plane thus touched by the surface and limited by the extreme
elements will be a plane surface equal to the given surface and is
the development of the surface.
Awarped- surface, or surface of double curvature, cannot be
devejopedT^ ~ "^
In order to determine the position of the different rectilinear
elements of a single curved surface as they come into the tangent
plane or plane of development, it will be necessary to find some
curve upon the surface which will develop into a straight line,
or a circle or some simple known curve upon which the distances
between these elements can be laid off.
For illustrations of the above principles, refer to the drawings,
"Intersection and Development of Cylinder and Cone" and
"Intersection and Development of Pipes".
Shades and Shadows
Light is transmitted along straight lines radiating in every
direction from each point of a lummous body.
28
If, therefore, an opaque body be placed between a source of
light and a surface, it will cut off a portion of the light and a shadow
will be cast on the surface and the part of the body hidden from
the source of light will be in shadow.
The line of shade on a body is the line separating the illuminated
part from tne part in shade.
The line of shadoiu is the outline of the shadow.
Rays of sunlight are sensibly parallel^ecause the sun is a
great distance away.
Cc
I
7| /^jUj. - l/«
Cti
In descriptive geometry, the rays of light are generally assumed
to take such a direction that thfik. prQ_kctioiis on either plane
make angles of 45 with the ground hne.
The line of shade on a body m.ay be assumed to be traced out
by passing a tangent ray around the body, keeping its direction
constant and always in contact with the body. The locus of the
point of contact is the line of shade.
The line of shade having been traced out, the shadow of the
body may then be found by passing rays of light through its
various points and finding their intersections with the surface
upon which the shadow falls.
The Jijie of shadow ij thus an oblique projection of the line
of shade.
'""nTHnPig. 25, the lines joining the points A2B2C2Y2Z2W2 are the
lines of shadow cast upon the vertical plane of projection by the
29
rectangular prism ABCDWXYZ and the lines joining the points
WAiBiCiYZ are the lines of shadow cast upon the horizontal plane
of projection.
The lines of shade would be the lines joining A, B, C, Y, Z, W.
In Fig. 26, the circle ABCD has its plane parallel to the hori-
SP«
zontal plane of projection. Its shadow on the horizontal plane is
therefore the equal circle AiBiCiDi and its shadow on the vertical
plane, the ellipse A2B2C2D2.
Perspective
g^ is its projection upon a plane
Tl^e£ersp
(usually^aT vertical plane) when the point of sight is at a finite
distance from the plane of projection jwhich__is com^monly called
the picture plane.
The picture plane is usually taken between^ the object and
the point of sight. '
The point of sight is the point from which the object is viewed
and is the vertex of a cone from which all projecting lines or rays
to different points on the object radiate. Hence, the perspective
of an object is said to be a conical projection.
The orthographic projection of the points of sight 5 on the
picture plane is the centre of the picture.
A horizontal line through the centre of the picture is the horizon.
It is the trace on the picture plane of a horizontal plane through
the point of sight.
TJie^vanuhhig point of a line is the perspective of a^ point at
infinity on the line, and is found by drawing a line t^rougTi the
point^of .sight. 5 parallel to the line and finding where it intersects"
uTe picture plane. This last point is the vanishing point.
1
30
The measuring point for a given line is the \ anishing point of
anotHer line equally inclined to the given line and to the picture
plane.
A line perpendicular to the picture plane is called a perpen-
dicular. Another definition for the centre of the picture is thFs:
It is the \ anishing pomt for all perpendiculars. The term diagonal
rs~applied to a horizontal line inclined at an angle of 45° to the
picture plane.
The_distance point or point of distance is a point on the horizon
distant from the centre of the picture a length equal to the distance
of the point of sigh tJ5) from the picture plane. It is the vanishing
point for all diagonals.
In Fig. 27, a method of drawing a perspective is shown.
The rectangular prism is given by its orthographic projection
on the horizontal and vertical planes, and it is required to find
its perspective on the plane PP, which is perpendicular to the
ground line, the point of sight 5" being given by its horizontal and
vertical projections.
Join the various corners of the prism to the point of sight S
by straight lines, and find where these straight lines intersect the
plane PPi.
The plane PP is then revolved about its vertical trace until it
coincides with the vertical plane of projection. The figure OibiCidi
W\XiyiZi is the perspective of the given prism.
0
^
<
r' b'
d
• c
'""-+.- ' "■
'■^--.^
I ■
1
1
1
""•■■-.."*
''^^5;;5gH£::,,,,.
p
i
1
^1
F
"%
tj; ^
ife-;:v-.
1
1
1
1
I
1
1
s'
1
1
.,,.-.V-;^/^;^'*
1
1
1
I
- 2,
.„K
^
,Xj.^.__._______.,
:l'-^'^
;r^-''
T?r
■■t.w»- ■'■",. ,--'*..--"'
1
i--''i--""| i----"'' .--'■'
1
1
'^■'■■;i-'i-';--i't""
1
w
-xh::--'-
z
1
""''...-'Y i 1 ;
-
jbx
1 i 1 ;
! \ \ \
■ \ \ ^ \
/
:::::;^
-i- A \ W
J
^^^ A^^:^^-.-
/ \ ^v;-^ •-- —
! ~--.
—
.S..~---'-^"-~^^^
aw
4- -
■"/ ^~"r----
/
.-■
P
X
V
^
m
W.
31
The general method of drawing perspectives of objects is to
find the perspectives of Hnes rather than of isolated points.
The general method of drawing the perspective of a given line
is illustrated in Fig. 28.
Let a^b^ and ab be the orthographic projections on the vertical
and horizontal planes of projection, and s^ and 5 the orthographic
projections of the point of sight on the same planes.
We might find the perspective of the line AB on the vertical
plane, by finding the perspectives of its extermities A and B,
simply by joining A and B to the point of sight and finding where
the lines this found intersect the vertical plane.
The following method is, however, the one generally used.
If a line SD be drawn through the point of sight 5 and a point at
infinity on the given line AB, it will be a line parallel to the given
•line and the point where this pierces the picture plane, viz., D
will be the perspective of the point at infinity on the line and
is known as the vanishing point for AB and, consequently, the
vanishing point for all lines parallel to AB.
If the point C where the given line AB intersects the vertical
plane be found, the point C will be its own perspective, and the
line joining C and D will be the perspective of the portion of the
line AB reaching from the point C to infinity.
CD is known as the indefinite perspective of the given line AB.
To find the perspective of the point A on the given line AB,
assume the horizontal and vertical projections of any line AE
passing through A; then find the indefinite perspective of AE,
viz., EG. The point ai where the lines CD and EG intersect, is
the perspective of the point A. In the same \vay,»-the perspective
&i of the point B is found.
',V-1
.<;■
5
The measuring point is the vanishing point for lines equally
inclined to the picture plane and the given line; that is, if the
auxiliary line AE made the same angle with the picture plane (in
this case the vertical plane of projection) as with the given line
AB, its vanishing point Q would be the measuring point for AB.
32
In Fig. 29, let 5' and S'- be the horizontal and vertical projection
of the point of sight.
Let ABCD (wxyz) be the horizontal projection of a rectangular
prism, the base ABCD resting on the horizontal plane of projection.
The horizontal trace of a plane containing the vertical edge BX
and the point of sight is the line Bs. The vertical trace of the plane
is b, X, . ' . h, X, is the perspective of the edge BX. The perspective
of the edge AB may be found as follows.
Produce AB to intersect the vertical plane at point E. If the
point of sight S be joined to a point at infinity on AB by a straight
line, the straight line will be parallel to AB.
Therefore, find the projection of a straight line passing through
5 and parallel to AB. It intersects the vertical plane at point R.
The point R is the perspective of a point on AB at infinity, and
is known as the vanishing point for the line.
The line joining E and R is the indefinite perspective of the
line AB.
The perspective of point B is the point ^i, the intersection of
hi, Xi, and ER.
BK is a line equally inclined to the given line AB and the
vertical plane . ' . EK = EB. The indefinite perspective of BK,
\ 0/ i
• •'/ /
s
Sh i^Wsi
viz., BMP, is found by obtaining its vanishing point (called the
measuring point for AB) and joining K and MP.
hi is the intersection of ab and KMP. The indefinite perspective
of BC, viz., bu r,, may be found by obtaining its vanishing point
VP-z and joining the point H to VP2. The perspective of the
point B may be found by the intersection of the perspectives of
any two of the lines BX, AB, CB, or BK.
Shadows ix Perspective
A convenient method of obtaining shadows in perspective is
illustrated in Fig. 30.
VPi
z^^
yPof HPofRoys
VP.
'WPof Rays
Fffl
Mi
The method of obtaining the perspective of the shadow of
the prism on the horizontal or ground plane is as follows.
Join the perspective of a point to the vanishing point of rays
and join the perspective of the horizontal projection of the same
point to the vanishing point of the horizontal projections of rays.
The intersection of the two lines is the perspective of the saadow
of the point on the horizontal plane.
V
<
APPENDIX
Methods of Computing Areas, Centres of Gravity, and
Moments of Inertia
1. The Trapezoidal Rule.
Area = ^ ( -|- -}-b + c-^d. . . . + ~J where a,b, c,. . . .z are the
ordinates and h the common interval between them. See Fig. 1.
a
c
*- - /} - -t*- - ff - »<^ -/r -■*
Fig. 1
2. Simpson's First Rule.
^h
Area= -^ (a+4&+2c+4J + 26 + 4/+g)
An approximate proof for this rule is as follows. See Fig. 2.
The area below the curve is practically equal to the area of the
trapezoid.
'm + c\2h
-4- I — I -H- I
Area
= {'_o^\2h /l^m\ 2h /m + c\2h
= (a+2/+2m + c) \^
o
(a+46+f)
Ih
34
35 '
A more exact proof is as follows :
Assuming the equation to the curve to be y = ao-\-aiX-{-a2X'.
See Fig. 3.
Area of narrow strip =ydx.
r2h
whole area =
ydx
or
0
'2h
Jo
(ao-j-aiX+a2X-)dx.
.2h
(1 1 \2'
Qox + -^a,x- + -Y^'i^^ )
which has to be evaluated between the limit x = 2h and .r = 0.
The expression then becomes
(A)
ao2h-\-^^Ah^ -\-~aSh^
Now from the equation to the curve when
x = 0 : y = ao
x = h : y = ao-\-aih-\-a2h'^
x = 2h : y = ao-\-2aih4-'ia2h'^
But calling the ordinates in the ordinary way Vi, y-z, ys.
when x = 0, y = yi,
x = h,y=y2,
x = 2h,y = ys,
therefore ao = yi
ao+aih-i-aoh^ = y2
ao-\-2aih-{-4:a2h^ = ys
. • . ao = yi
fli= 2^(43'2-33'i-j3)
02= 2^2(>'3-2>'2 + 3'i)-
36
Substituting these values of ao, fli and Ui in expression (A)
Area=>'i2// +
4^2 j^
2 '2h
(43'2-33'i-V3) +
8/j3 1 o I ^
1
Indirect Proof for Simpson's First Rule.
Assuming the equation to the curve to be y = afi-\-aiX-\-a2X-.
h
By Simpson's First Rule, Area= -5 (>'i + 43'2+3'3).
o
y\ = a,
y2 = ao-\-ctih-\-a2h-,
y3 = ao-i-2aih-\-4a2h-.
Area= -x (ao+4ao+4al/^^-4a2/^-+ao+2al/^+4a2^^)
o
= o (6ao+6aiA4-8a2/?^)
8
= 2aoh-\-2aih'-\- -^a2¥.
Area =
'2/j
ydx
■2h
(
1
(ao + aiX + a2.%"-) (/x
2h
= I aoX-\- ^aix- +
1 \2'
8
= 2ao/^+2ai/i2+^a2/?^
o
which is the same as expression (B).
3. Simpson's Second Rule.
Area= ^h (a + 36 + 3c + 2(i+3g + 3/+g).
o
Assuming the equation to the curve to be
y = ao-{-aiX-^a2X--{-a3X^. See Fig. 4.
(B>
Y
/*
ys
y*
II
X
Fig. 4
---»)
Z7
Area of narrow strip ydx
ydx
whole area =
I
o
3/!
1 o , 1 .\2''
/I 1 1 \2'
which equals i Oo + ^ ^i-'^" + o «2-v^ + 7 fls-x"* j
which has to be evaluated between the limits .v = 3/? and x = 0.
The expression then becomes
aSh + \ aim- + \ a227¥ + i 81/^^
Z o 4
From the equation to the curve.
When
x- = 0, y = ao = yi
x = h, y = ao-{-aih'\-a2h--\-ash^ — y2
X = 2h, y = ao-{-2aih-^4La2h- -^-Saah^ = yz
X = Sh, y = ao-{-3aih-{-9a2h--\-27ash^ = yi
Solve the above equations for values of Oo, fli, ^2 and as and
substitute these values in the equation for the area and the
result is
3
- area= -^(j'i + 3>'2 + 3>'3+3'4).
Indirect Proof for Simpson's Second Rule.
Assuming the equation to the curve to be
y = ao-\-aiX-\-a2X~-{-asX^.
rsh
Area = {ao-\raiX-{'a2X--\-a3X^) dx
(111 \^''
aox + - aix"^ + o <^i^^ + 4 (^^^^ )
Q 81
= 3aoh-^^aih'--{-9a2¥ + ^ash* (C)
By Simpson's Second Rule,
3
Area = ^h (vi + 33-3 + 33-3 +3-4)
Now3'i = ao,
3'2 = ao + al^+«2/^-^-«3/^^
y3 = ao-i-2aih-\-4:a2h--{-8a3¥,
yi = ao-\-3aih-i-9a2h--^27a3¥.
Substitute these values in expression
3
Area = ^ A (3'i + 33'2-r33'3+3'4)
38
o
+8a3h)-\-ao-^Saih-\-9a2h~-\-27a3h^)
= hi (8ao+12aih-^24:a2h~-j-o4:as¥)
o
9 81
= 3aoh-\- 2 ai^+9a-2A2+ — aghi
which is the same as expression (C).
4. The Five-Eight Rule or Simpson's Third Rule.
Area = Tq ^^ (^J'l + 83'2 - J's) •
The proof of this rule follows directly from the proof of Simpson's
first rule. See Fig. 5.
/; >« /$. >
Fig. o
By Simpson's First Rule
Area of figure AX.BCD= ^h{AD-{-4XE-hBC).
By the five-eight rule
Area AXED= ^h ■ {5AD+8XE-BC)
and area XBCE= ^ h {5BC-{-SXE-AD).
Adding these two areas
= ^h{4:AD + lQXE-AD)
= ^h(AD-^AXE+BC).
It should be noted that the result found by the formula is the
area between two ordinates only, although three ordinates are
required to find it.
39
The extension of the five-eight rule is as follow:
/5 8-1
12 5 8-1
\ 5 8-1
p//(5vi + 13v2+12v3 + 12v4 + 7v5-V6)
, / 5 , 13 , , , 7 1 >y
(l2-^*'~^ p^ V2+V3 + V4+ p5 Vo- j^3'6 )
The trapezoidal rule is applicable when there are any number
of equidistant ordinates.
Simpson's first rule is applicable only when there are an even
number of equal spaces between ordinates.
Simpson's second rule is applicable only when the number of
spaces between the ordinates is divisible by three.
The five-eight rule is applicable when there are any number of
equidistant ordinates, provided an ordinate beyond the area
required be known.
The position of the centre of graxity of a plane figure or solid
may be readily found by taking moments about any convenient
axis.
The following examples illustrate the method of obtaining the
area and position of the centre of gra\"ity of a plane figure bounded
on one side by a curve.
1. Solution by Simpson's First Rule.
Distances between ordinates = 2' — 0".
No. Length of Simp. Functions Levers Products
Ordinate Mult. of Ordinate
1 1.45 1 1.45 0 0.0
2 2.65 4 10.60 1 10.60
3 4.-35 2 8.70 2 17.40
4 0.45 4 25.80 3 77.40
5 8.50 2 17.00 4 68.00
6 10.40 4 41.60 5 208.00
7 11.85 1 11.85 6 71.10
117.00 452.50
Area = 4(117X2)=78sq. ft.
o
455* 50
Distance of C of G from Ordinate Xo. 1 = ^T^ X2 = 7.72 ft.
11 /
40
2. Solution by Simpson's Second Rule.
No. Length of Simp. Functions Levers Products
Ordinate Mult. of Ordinate
1 L45 1 L45 0 0.0
2 2.65 3 7.95 1 7.95
3 4.35 3 13.05 2 26.10
4 6.45 2 12.90 3 38.70
5 8.50 3 25.50 4 102.00
6 10.40 3 31.20 5 156.00
7 11.85 1 11.85 6 71.10
103.90 401.80
Area = | (103.90X2) = 77.925 sq. ft.
o
Distance of C of G from Ordinate No. 1 = f^^^^ X2 = 7.71 ft.
103.90
Moment of Inertia.
-I
y^dx
The moment of inertia about the axis OX may be found by
Simpson's Rules as follows. Substitute the cubes of the ordinates
for the ordinates.
Multiply these by Simpson's multipliers in the ordinary way
thus obtaining the required functions of cubes.
Multiply the sum of the functions of cubes by \ the common
interval If using Simpson's First Rule, and by f the common
interval if using Simpson's Second Rule.
This gives the value of
yHx. Divide this result by 3 and the
result is the required Moment of Inertia.
L
AMSER'S MECHANICAL INTEGRATOR
Areas
Area of elementary strip = Y</.v
= / sin . ddx.
For a movement dx of Integrator, movement of wheel A (length
of arc on circumference) =dx sin.^,
.'./.(movement of wheel ^) = /sin. Mv = elementary area and
/. (total movement of wheel ^) = / sin . ddx = total area, i.e., if each
1
unit length on the circumference of the wheel be divided into
/ equal spaces, the number of spaces through which the wheeF
turns when the tracing point moves around a plane figure equals
the number of square units contained within the figure.
Moments
Moment of elementary strip about axis OX = -y^dx= -x
{I sin. eydx.
Movement of wheel AI (length of arc on circumference) =</.v
cos 2e = dx(cos~e-sm^d)=dx{l-2 sin^^).
When this is integrated dx = 0 because the tracing point returns
to its original position.
— (2 sin-d)dx or
Total movement of wheel =
= -2
sin^ddx.
If length of divisions on wheel = -y- , reading on wheel = —21
. ,^ , , reading on wheel 1 ,.,
sm^Odx or / ^ — : = ^P
— 4 2
sin-ddx =
Moment of area about axis OX.
By reversing the sense of the graduations on the wheel Af the
moment is given by the expression
/ reading on wheel .
-
41
42
Moments of Inertia
Moment of Inertia = 5
y^dx =
P .sm^d.dx.
because y = l sin d
Reading on A Scale ■■
Reading on / Scale
dx sin d (constanti)
dx sin 3^ (constants)
Sin 30 = sin 6 cos-0-sin¥ + 2 sin d cos^^
= 3 sin 0 cos^^ — sin^0
43
= 3 sin0(l-sin2^)-sin'^
= 3 sin^-4sin3^
dividing both sides by 12
sin 3^ _ sin d _ sin ^d
~W ~ 4 3
sin-^^ sin d sin Sd
• 3 4 12
sin ^6 ,3 _ sin ^ ,3 _ sin Z9 .3
3 4 12
/= sin ddx^-!^\ - sin ^^(^-"^(to)
_ Reading on A Scale/ P\ Reading on / Scale //^\
~ " (Constanti) \T J (Constant^) \T2/
Const. 1 = number of divisions per unit length on circumference
of A wheel.
Const. 2 = number of divisions per unit length on circumference
of / wheel.
t^}'
f
An
Ir)
¥
J
I 0)
J
>
'--^/;
//
>b
fl
r
QA
501
C633
P&ASci
Cockbum, J. Roy
Brief synopsis of the
course of lectures in
descriptive geometry
PLEASE DO NOT REMOVE
CARDS OR SLIPS FROM THIS POCKET
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