N
,.■
-v*
V
s
/
V-
''
y
^
w.
M
1
/.
-''
N
<
X
s
\
T
—
e
^
—
^
\
s^
i
/
>J
V
\
^
Lf
^
/
/
f
\
\^
^
L
_
-^
Fig. 366. Pig. 367. Fig. 368. FiQ. 369.
In reproducing the above diagrams take the small squares as of 0-3 inch side,
7. The ends of a right prism are regular pentagons of 1*25 inches side, and the
altitude of the prism is 1-5 inches. This prism rests with one of its rectangular
faces on the ground and with its ends inclined at 45° to the vertical plane of pro-
jection. Draw the plan and elevation.
8. Draw the plan of a right square prism when a diagonal of the solid is
vertical. Side of base, 1*25 inches. Altitude, 2 inches.
9. A front elevation of a Maltese cross is given in Fig. 370. The thickness of
the cross is 0*7 inch. Draw the plan and from it project an elevation on a ground
line making 50° with XY.
10. Draw an elevation of the solid given in Fig. 371 on a ground line inclined
at 60° to XY.
11. Draw a plan of the solid given in Fig. 371 when the line RS is vertical.
12. Draw an elevation of the solid given in Fig. 372 on a ground line inclined
at 60^ to XY.
H
X
Y
-—5
7^
i —
ffl
Fig. 370.
Fig. 371.
Fig. 372.
Fig. 373.
In reproducing the above diagrams take the small squares as of 0*4 inch side.
13. Draw a plan of the solid given in Fig. 372 when it is tilted about one edge
of its base through an angle of 30°.
14. The solid given in Fig. 373 is formed out of two right square prisms.
Draw an elevation of this solid on a ground line inclined at 60° to XY.
15. Draw the plan of the solid given in Fig. 373 when it is tilted about one
edge of its base through an angle of 30°.
16. Draw the plan of a right hexagonal pyramid when it rests with one
triangular face on the ground. From this plan project an elevation on a ground
line inclined at 45° to the plan of the axis of the pyramid. Side of base, 1*25
inches. Altitude, 2 inches.
17. An isosceles triangle, base 1 inch, sides 1*5 inches, has its base parallel to
XY. This triangle is the elevation of one of the triangular faces of a right
CHANGING THE PLANES OF PROJECTION
197
hexagonal pyramid. Side of base of pyramid, 1 inch. Altitude, 2 inches. Com-
plete the elevation of the pyramid and from it project the plan.
18. Two elevations of an inkstand are given in Fig. 374. From the right-
hand elevation project a plan on XY and from this plan project an elevation on a
ground line inclined at 50° to XY.
19. Draw the plan of the inkstand (Fig. 374) when the edge AB is vertical.
TTl
_ 71] 1 j:-.;
u t
.,!..i
L i :
iL :±__
...Y
- / - --r K.
/^ sX
,. - , ^z s^
/{L £_Z S_
xt:::±:__ _:i
Fig. 374.
Fig. 375.
In reprodiLcing the above diagrams take the small squares as of 0*3 inch side.
20. Two elevations of a trestle are given in Fig. 375. From the right-hand
elevation project a plan on XY and from this plan project an elevation on a
ground line inclined at 60^ to XY.
21. Draw a plan of the trestle (Fig. 375) when the line joining the points R
and S is vertical.
CHAPTER XV
PLANES OTHER THAN THE CO-ORDINATE PLANES
162. Representation of Planes — Traces of a Plane. — Planes
other than the co-ordinate planes are represented by the lines in
which they meet the latter. The lines in which a plane meets the
co-ordinate planes or planes of projection are called the traces of that
plane, the intersection with the vertical plane being called the vertical
trace, and that with the horissontal plane the horizontal trace.
The line in which one plane meets another is also called the trace
of the one plane on the other, but when the traces of a plane are
mentioned its traces on the planes of projection are generally understood.
Planes occupying various positions in relation to the planes of
projection are shown in pictorial projection in the lower parts of
Figs. 376 to 384. The upper parts of the same Figs, show how the
same planes are represented by means of their horizontal and vertical
traces when the planes of projection are made to coincide as explained
in Art. 135, p. 166.
V.T.
NO
HORIZONTAL
TRACE
NO
VERTICAL
TRACE
X Y
H.T
>
X 1
X
32
Fig. 376.
Fig. 377.
Fig. 378.
Fig. 379.
Referring to Figs. 376 to 384 separately. Fig. 376 shows a hori-
zontal plane. . Fig. 377 shows a plane which is parallel to the vertical
plane of projection and is therefore also a vertical plane. Fig. 378
shows a plane which is perpendicular to both planes of projection and
PLANES OTHER THAN THE CO-ORDINATE PLANES 199
is therefore also a vertical plane, and it is also perpendicular to the
ground line. Fig. 379 shows a plane which is perpendicular to the
vertical plane of projection and inclined to the horizontal plane.
Such a plane is generally called an inclined plane. Fig. 380 shows a
Fig. 380.
Fig. 381.
Fig. 382.
Fig. 383.
plane which is inclined to the vertical plane of projection and is
perpendicular to the horizontal plane and is therefore a vertical plane.
Fig. 381 shows a plane which is inclined to both planes of projection
but is parallel to the ground line. Figs. 382 and 383 show planes
which are inclined to both planes of projection and to the ground line.
Fig. 384 shows a plane containing the ground line XY and inclined
to both planes of projection. The horizontal and vertix;al traces
coincide with XY and these traces do not fix
definitely the position of the plane. In this
case a trace on another plane of projection is
necessary, say on a vertical plane of which Xj Yi
is the ground line.
A perpendicular plane is one which is at
right angles to one or other or both of the
planes of projection (Figs. 376, 377, 378, 379,
and 380).
An oblique plane is one which is inclined to
both planes of projection (Figs. 381, 382, and
383).
It is obvious that in the case of a perpen-
dicular plane P, its trace T on the plane of
projection Q, to which the plane is perpen-
dicular, is an edge view of the plane, and the
projections on Q of all points and lines on P
will lie on T.
It is not difficult to see that if the traces of a plane meet
X H.T&V.T. V
Y
V
.Fig. 384.
one another their point of intersection is on the ground line, and if the
200
PRACTICAL GEOMETRY
traces are parallel to one another they must be parallel to the ground
line.
163. Angle between Two Planes. — The angle between two
planes, or the inclination of one plane to another, is the angle between
two straight lines drawn from any point in their line of intersection,
at right angles to it, one in each plane.
Referring to Fig. 385, which is a pictorial projection, AB and CD
are two planes and CE is their line of
intersection. FH is a straight line in the
plane AB and at right angles to CE. FK
is a straight line in the plane CD and at
right angles to CE. If HF be produced
to L, then the angle between the planes
AB and CD is either the angle HFK or
its supplement the angle LFK.
Of the two angles which one plane
makes with another it is generally the
acute angle which is taken as the angle
between the planes. In the case of the angle between two adjacent
plane faces of a solid it is generally the angle within the solid which is
understood, whether it be acute or obtuse.
The angle between two planes is called a dihedral angle.
164. Inclinations of a Plane to the Planes of Projection. —
A plane is given by its traces in Fig. 387. The same plane and the
planes of projection in their natural positions are shown in oblique
projection in Fig. 386. Referring to Fig. 386, iVB is a line in the
Fig. 386.
Fig. 387.
given plane and at right angles to its horizontal trace H.T. AO is a
line in the horizontal plane and at right angles to H.T. By definition
the angle OAB is the angle which the given plane makes with the
horizontal plane. 0, the true value of this angle, is found by the
following construction, shown in Fig. 387. Take a point o in XY and
draw oa at right angles to H.T. With o as centre and oa as radius
describe the arc aAi to cut XY at Aj. Draw oh' perpendicular to XY
to meet V.T. at b'. Join h'Ai. The angle oAjt' is the true value of 0.
PLANES OTHER THAN THE CO-ORDINATE PLANES 201
lb is obvious that the triangle oA^h' in Fig. 387 is the true shape of
the triangle OAB in Fig. 386. A similar construction gives the angle
(j), the inclination ot the given plane to the vertical plane of projection.
The arc aA^ (Fig. 387) or the arc AAj (Fig. 386) may be looked
upon as part of the outline of the base of a right circular cone whose
axis is ob' (Fig. 387) or OB (Fig. 386). The base of this cone is on
the horizontal plane, and the given plane is tangential to the slant
surface of this cone. Hence 6 which is the inclination of the slant
surface of the cone to its base is also the inclination of the given plane
to the horizontal plane. Similarly <^ is the base angle of another right
circular cone whose base is on the vertical plane of projection and
whose axis is in the horizontal plane, and the given plane is tangential
to the slant surface of this cone. Hence cutting XY at
Cj and a line through o at right angles to XY at d. With centre o
Fig. 388.
Fig. 389,
and radius oA^ describe the arc A^a. With centre o and radius oC^
describe the arc Q^c' . A tangent da to the arc A^a is the required
horizontal trace, and a tangent Vc' to the arc C^c' is the required
vertical trace. If these traces meet, their point of intersection must
be on the ground line. The sphere and enveloping cones having been
drawn there will, in general, be four planes which will be tangential
to both cones. The student who has mastered the problem of the
preceding Art. should now have little difficulty in understanding the
solution just given if he carefully studies Figs. 388 and 389. It only
remains to emphasize the point that the two cones are made to envelop
the same sphere in order to ensure that they shall have a common
tangent plane.
The second method of solving this problem depends on the following
theorems. (1) The inclination of a plane to one of the planes of pro-
jection is the complement of the inclination of its normal to that plane
of projection. (2) The trace of a plane on one of the planes of pro-
jection is perpendicular to the projection of its normal on that plane
of projection. Hence the construction. — Draw (Art. 144, p. 176) the
projections of a line inclined at 90° — ^ to the horizontal plane and
90° ~ <^ to the vertical plane of projection. The horizontal and vertical
traces of the required plane will be perpendicular to the plan and
elevation respectively of this line. If the traces meet, their point of
intersection must be on the ground line. If the traces do not meet
within a convenient distance, or if they are parallel to the ground line,
care must be taken to ensure that the traces are the traces of the same
plane.
166. Given one Trace of a Plane and the Inclination of
the Plane to one of the Co-ordinate Planes, to determine the
PLANES OTHER THAN THE CO-ORDINATE PLANES 203
other Trace. — In Figs. 390 and 391, 6 the inclination of the plane to
the horizontal plane is supposed to be given.
In Figs. 392 and 393,
J^^
\f
yp'
/" y
i iX
J^
-^
HJ.\yS^
HT.2.
-Y X
Fig. 404.
Fig. 405.
A plane may be fixed by the projections of three points in it, and
the intersection of two planes fixed in this way may be found without
using the traces of the planes on the planes of projection. Referring
to Figs. 406 and 407, aa\ hV ^,
and cc' are the projections
of three points in a plane.
The triangle a&c, dVd is ob-
viously in this plane. Jc?',
ee', and ff are the projec-
tions of three points which
are in another plane. The
triangle def^ d'e'f is obvi-
ously in this second plane.
•It is required to find the
intersection of these two
planes.
Referring to Fig. 406,
a horizontal plane at the
level of the point ff is
taken. This plane cuts
the plane of the triangle
abc, a'h'd in the line gh, g'h'.
This same plane cuts the p .-g
plane of the triangle def,
d'e'f in the line fk, fh'. pp' the point of intersection of the lines gh^
PLANES OTHER THAN THE CO-ORDINATE PLANES 207
g'h', and fh, f'Tz is obviously a point in the intersection of the two
given planes.
Fig. 407.
A vertical plane of which hi is the horizontal trace is taken. This
plane cuts the plane of the triangle abc^ a'h'c' in the line hl^ 67'. This
same plane cuts the plane of the triangle def^ d'e'f in the line ww, m'n'.
q^ the point of intersection of the lines 6/, h'V and mn, m'n' is obviously
another point in the intersection of the two given planes. That part
of the line of intersection of the two given planes which lies within
both of the given triangles is shown thicker, and this thicker line is the
intersection of the given triangles.
Fig. 407 shows the various planes and their intersections in oblique
projection.
When an auxiliary cutting plane is required in order to find the
intersection of two given planes, this auxiliary plane may be placed in
any number of positions, and the student must exercise his judgment
and ingenuity in determining a convenient position, that is, a position
which will lead to points of intersection of lines which will fall within
convenient distances and which can be found with accuracy. Lines
intersecting at very acute angles should be avoided if possible.
170. Distance between two Parallel Planes.— Parallel planes
have parallel traces, the vertical and horizontal traces being parallel to
one another respectively.
In Figs. 408 and 409, V.T.I, and V.T.2. are the vertical traces and
H.T.I, and H.T.2. are the horizontal traces of two parallel planes. It
is required to find the distance between these planes. Cut the given
planes by a vertical plane whose horizontal trace oc (Fig. 409) or
OC (Fig. 408) is perpendicular to the given horizontal traces. This
plane will cut the given planes in two parallel lines the true distance
between which will be the required distance between the given planes »,
208
PRACTICAL GEOMETRY
Referring to Fig. 409, with o as centre and radii oa and oc describe
arcs to cut XY at A^ and C^ respectively. Draw oh'd' at right angles
Fig. 408.
Fig. 409.
to XY cutting the given vertical traces at h' and d' as shown. Join
Ai6' and C^d'. The perpendicular distance between A^h' and C^d' is the
distance between the two given planes.
The theory of the foregoing construction is that the vertical plane
COD (Fig. 408) is supposed to turn about its vertical trace OD, which
is perpendicular to XY, carrying with it the lines of its intersection
with the given planes, until it is in the vertical plane of projection.
The true distance between these lines can then be measured.
The auxiliary cutting plane may be, quite as conveniently, taken
perpendicular to the given vertical traces, and the student should draw
the figure with the cutting plane so taken.
171. To draw the Traces of a Plane which shall be parallel
to a given Plane and at a given distance from it. — The solution
of this problem is so obvious, if the preceding one has been understood,
that no detailed description of it need be given. It may be pointed
out however that two planes may be placed at a given distance from a
given plane, one on each side of it.
Exercises XV
1. Planes are given by their traces at (a), (6), (c), [d), [c), and (/) in Fig. 410.
Determine in each case the inclinations of the plane to the planes of projection.
Fig. 410.
PLANES OTHER THAN THE CO-ORDINATE PLANES 209
2. What are the true angles between the traces of the planes (a), (6), and (c),
Fig 410 ?
3. Find the true angles between the traces of the planes (d), (e), and (/),
Fig. 410.
4. The vertical trace of a plane is inclined at 45° to XY and the plane is
inclined at 60° to the horizontal plane : draw both traces of the plane.
5. The vertical trace of a plane is inclined at 50° to XY and the plane is
inclined at 50° to the vertical plane of projection : draw both traces of the plane.
6. The horizontal trace of a plane is inclined at 45° to XY and the plane is
inclined at 60° to the horizontal plane : draw both traces of the plane.
7. The vertical trace of a plane is parallel to XY and at a distance of 2*5 inches
from it. The plane is inclined at 55° to the horizontal plane. D^aw the traces of
this plane. .
8. Draw the traces of a plane whose inclinations to the horizontal and vertical
planes of projection are 55° and 60° respectively.
9. Draw the traces of a plane which is inclined at 50° to the horizontal plane
and 40° to the vertical plane of projection.
10. The true angle between the traces of a plane is 60° and the vertical trace
is inclined at 35° to XY. Draw both traces of this plane.
11. Draw the traces of a plane which is inclined to the horizontal plane at 50°,
the true angle between the traces being 70°.
12. The true angle betwee.n the traces of a plane is 60° and the plane is
equally inclined to the planes of projection. Represent this plane by its traces.
13. Pairs of planes are given by their traces at (a), (6), (c), and (d) Fig. 411.
In each case show the plan and elevation of the line of intersection of the planes.
Fig. 411.
14. Find the plan and elevation of
the point which is situated in each of
the three planes given in Fig. 412.
15. Determine the intersection of
the plane triangle abc, a'b'c'- (Fig. 413)
with the plane triangle def, d'e'f. The
dimensions given in Fig. 413 are in
inches.
Fig. 412.
Fig. 413.
210 PRACTICAL GEOMETRY
16. Two parallel planes are parallel to the ground line. The horizontal trace
of the first is 1-5 inches below XY and that of the second 2 inches below XY.
The vertical trace of the first is 2 inches above XY. Find the vertical trace of
the second plane and the distance between the two planes.
17. Referring to Fig. 410, take each of the given planes and show two planes
parallel to the given plane and 0*7 inch distant from it.
18. Two parallel planes have an inclination of 50°, and their horizontal traces
are 1 inch apart. Determine a vertical plane which cuts these planes in lines
which are 1'5 inches apart.
CHAPTER XVI
STRAIGHT LINE AND PLANE
172. Lines and Points in a Plane. — Theorem I. A straight
line contained hy a given plane lias its traces on the traces of the plane.
The traces of the line are points in the planes of projection, but they
are also points in the plane containing the line, therefore the traces of
the line must lie on the intersection of the plane with the planes of
projection, that is, on the traces of the plane.
Referring to the pictorial projection, Fig. 414, LA and LB are the
traces of a plane and AB is a straight line contained by the plane
ALB. The points A and B are the traces of AB.
Theorem II. A straight line which is parallel to one of the planes
of projection and is contained hy a given plane is parallel to the trace of
that plane on the plane of projection to which the line is parallel.
Referring to the pictorial projection, Fig. 414, CD is a horizontal line
lying in the plane LAB. CD is parallel to LA the horizontal trace of
the plane. Hence cd the plan of CD is parallel to LA. c'd\ the
elevation of CD, is parallel to XY. This line has only one trace C
which is a vertical trace. EF is a line parallel to the vertical plane
of projection and lying in the plane ALB. EF is parallel to LB the
vertical trace of the plane. Hence ef the elevation of EF is parallel
to LB. ef, the plan of EF, is parallel to XY. This line has only one
trace E which is a horizontal trace. '
Fig. 415 shows the various lines referred to above by ordinary
plan and elevation.
It is obvious that the distance of CD from the horizontal plane of
projection is equal to the distance of its elevation from XY, also the
distance of EF from the vertical plane of projection is equal to the
distance of its plan from XY.
A knowledge of the foregoing principles will enable the student to
solve a number of problems.
For example. — Given the traces of a plane (Fig. 415) and the plan
p of a point contained by the plane, to find the elevation of the point.
Through p draw ab to cut the horizontal trace of the plane at a and
XY at h. Consider ah as the plan of a line lying in the plane. The
elevation a'h' of this line is found as shown, and a'h' must contain p'
the elevation required. Instead of taking the line ah, a'h' lying in the
212
PRACTICAL GEOMETRY
plane and containing the point P, the horizontal line cd, c'd' lying in
the plane and containing the point P may be taken, or the line ef, e'f
Fig. 414.
Fig. 415.
parallel to the vertical plane of projection may be used to find p'. If
p' is given |> may be found by working the foregoing construction
backwards.
Again, given the traces of a plane and the distances of a point in
the plane from the planes of projection to find its plan and elevation.
Place in the plane (Fig. 415) a horizontal line cd, dd' whose distance
from the horizontal plane of projection is equal to the given distance
of the point from the latter plane. Next place in the plane a line ef,
e'f whose distance from the vertical plane of projection is equal to the
given distance of the point from the latter plane, p the intersection
of cd and ef is the plan, and p the intersection of c'd' and e'f is the
elevation required.
In another set of problems it is required to find the traces of a
plane which contains given points or lines. Three points which are
not in the same straight line will fix a plane, also two parallel straight
lines or two intersecting straight lines, or one straight line and one
point not in that line. These problems are solved by making use of
the facts that if two points are in a plane the straight line joining
them is in that plane and the traces of this line are on the traces of
the plane.
173. To draw the Traces of a Plane which shall contain
a given Line and have a given Inclination to one of the
Planes of Projection. — Let ah, a'V (Fig. 417) be the given line, and
let the given inclination of the plane be to the horizontal plane.
Draw the projections of a cone having its vertex at the vertical
STRAIGHT LINE AND PLANE
213
trace of the given line, its base on the horizontal plane of projection,
and its slant side inclined to its base at the given angle {$).
A line through the horizontal trace of the given line, to touch the
base of the cone, will be the horizontal trace of the plane required,
and a line through the point where this horizontal trace meets the
ground line and through the vertical trace of the given line will be
the vertical trace of the plane required. The foregoing construction
will be readily understood by referring to the pictorial projection
shown in Fig. 416.
Fig. 416.
Fig. 417.
If the horizontal trace of the given line falls outside the base of
the cone two tangents from this point to the base of the cone can be
drawn and there will therefore be two planes which will satisfy the
given conditions. In Figs. 416 and 417 only one of the two possible
planes is shown. If the horizontal trace of the given line falls on the
circumference of the base of the cone only one tangent from this point
to the base of the cone can be drawn and there will then be only one
plane which will satisfy the given conditions, and this plane will have
the same inclination as the given line. If the horizontal trace of the
given line falls inside the base of the cone no tangent from this point
to the base of the cone can be drawn, which shows that the problem is
impossible when the inclination of the plane is less than the inclination
of the line.
In the construction just explained it has been assumed that the
horizontal and vertical traces of the given line have been accessible, it
now remains to be shown how the problem is to be dealt with when
these points are inaccessible.
Referring to Fig. 419, the horizontal and vertical traces of the
given line ah, a'h' are supposed to be inaccessible.
Draw the projections of two cones having their vertices at points
cd and dd' in the given line, their bases on the horizontal plane,
214
PRACTICAL GEOMETRY
and their slant sides inclined to their bases at the given angle (0).
The horizontal trace of the required plane is a common tangent to the
bases of the two cones. The vertical trace of the plane may be found
by taking a point ee' in the horizontal trace of the plane and joining
this point with two points (say cc and dd') in the given line. The line
Fig. 418.
Fig. 419.
joining the vertical traces of these two lines is the vertical trace of the
plane required. A reference to the pictorial projection (Fig. 418) will
make the construction clear.
It should be noticed that although four tangents may be drawn to
the bases of the cones shown in Figs. 418 and 419 only the outside
tangents, and not the cross tangents, can be the horizontal traces of
planes which are tangential to the two cones.
If the inclination of the required plane be given to the vertical
plane of projection instead of to the horizontal plane the bases of the
cones must be placed on the former plane instead of on the latter.
174. In a given Plane to place a Line having a given
Inclination to one of the Planes of Projection. — The inclina-
tion of the line must not exceed the inclination of the plane.
Let the given inclination be to the horizontal plane.
.Take a point C (Eig. 421) on the ground line and draw Ch' making
the given angle 6 with XY and meeting the vertical trace of the
given plane at h'. Draw h'h perpendicular to XY meeting the latter
at h. With centre h and radius 6C describe an arc of a circle to cut
the horizontal trace of the given plane at a. Draw aa' perpendicular
to XY meeting it at a', ah is the plan and ah' the elevation of the
line required.
The correctness of the construction is obvious, for AB (Fig. 420)
is evidently on the given plane and it is also on the surface of a cone
whose base is on the horizontal plane and whose base angle is 6.
The construction is applicable whatever be the positions of the
traces of the given plane.
STRAIGHT LINE AND PLANE
215
If the given inclination be to the vertical plane of projection
instead of to the horizontal plane, the construction will present no
difficulty to the student who has followed that just given.
Fig. 420.
Fig. 421.
175. To draw the Projections of a Line which shall pass
through a given Point, have a given Inclination, and be parallel
to a given Plane. — By the construction of the preceding Art.
place in the given plane a line having the given inclination. Through
the plan and elevation of the given point draw lines parallel to the
plan and elevation respectively of the line which has been placed in
the given plane. These will be the projections required.
That the line whose projections have thus been found fulfils the
given conditions is not difficult to see ; for it is parallel to a line
which has the given inclination, and must therefore have itself the
given inclination. Also it can never meet the given plane, for if
it did it could not be parallel to any line in that plane, therefore
it is parallel to the given plane.
176. To draw the Traces of a Plane which shall contain
a given Point and be parallel to a given Plane. — The horizontal
and vertical traces of the plane required must be parallel to the
horizontal and vertical traces respectively of the given plane.
Fig. 422,
Fig. 423.
pp' (Figs. 422 and 423) is the given point and H.T. and V.T. are
the traces of the given plane.
216 PRACTICAL GEOMETRY
The projections of a line passing through 'pp and parallel to the
given plane are first drawn. The traces of this line are on the traces
of the plane required.
In Fig. 422 a horizontal line pq, p(][ parallel to the given plane
is shown. The vertical trace of this line is a point on the vertical
trace of the plane required, and ;the point where the vertical trace of
the plane meets XY is a point on the horizontal trace of the plane.
In Fig. 423 a line ah, a'h' is placed in the given plane and the
projections rs, r's' of a line parallel to ah, a'h' are drawn through p
and p' respectively. The traces of rs, r's' are on the traces of the
plane required.
The construction shown in Fig. 423 is used when the vertical
trace of a horizontal line through pp' parallel to the given plane is
inaccessible, or when the horizontal trace of a line through pp' parallel
to the vertical plane of projection and parallel to the given plane is
inaccessible.
177. In a given Plane to place a Line which shall be
parallel to and at a given distance from another given
Plane not parallel to the first. — Draw the traces of a plane
parallel to the second given plane and at the given distance from it
(Art. 171, p. 208). The line of intersection of this third plane with
the first given plane is the line required.
178. To draw the Projections of a Line which shall pass
through a given Point and be parallel to two given inter-
secting Planes. — The required projections will pass through the
projections of the given point, and be parallel to the projections of
the line of intersection of the two given planes.
179. To draw the Traces of a Plane which shall contain
a given Line and be parallel to another given Line. — Draw
the projections of a line to intersect the first given line and be parallel
to the second. The plane containing these two intersecting lines is
the plane required.
180. To draw the Traces of a Plane which shall contain
a given Point and be parallel to two given Lines. — Draw the
projections of two lines to pass through the given point and be parallel
respectively to the two given lines. The plane containing the lines
passing through the given point is the plane required.
181. Intersection of a Line and Plane. — Let LMN (Fig. 425)
be a given plane and ah, a'h' a given line. It is required to show the
projections of the point of intersection of the line and plane.
Draw the traces LR and RN of a plane, preferably perpendicular
to one of the planes of projection, to contain the given line. Find the
line of intersection of this plane with the given plane. The points
p and p' where ah and a'h' meet the plan and elevation respectively
of the intersection of the planes are the projections required.
Referring to the pictorial projection (Fig. 424) it will be seen
that AB and LN being in the same plane LRN they must intersect
at some point P unless they happen to be parallel. Again, since LN
is the line of intersection ,of the planes LMN and LRN the line LN
STRAIGHT LINE AND PLANE
217
isin the plane LMN, therefore the point P is in the plane LMN. But
P is in AB, therefore the point P is the point of intersection of AB
and the plane LMN. If AB is parallel to LN then AB is parallel to
the plane LMN.
Fig. 424.
Fig. 425.
182. To draw the Projections of a Line which shall pass
through a given Point and be perpendicular to a given
Plane. — If a line is perpendicular to a plane the projections of the
line are perpendicular to the traces of the plane, the plan to the
horizontal and the elevation to the vertical trace.
The construction is therefore as follows. Through the plan of the
point draw a line at right angles to the horizontal trace of the plane,
and through the elevation of the point draw a line at right angles to
the vertical trace of the plane ; these will be the projections required.
183. To draw the Traces of a Plane which shall contain
a given Point and be perpendicular to a given Line. — Let
pp (Fig. 426) be the given point and a6, a'h' the given line.
Through p draw pq perpendicular to ah, meeting XY at q. Draw
q(J perpendicular to XY, meeting a parallel to XY through p' at q'.
Through (^ draw LM perpendicular to a'h', meeting XY at M. Through
M draw MN perpendicular to ah. LM and MN are the vertical and
horizontal traces respectively of the plane required.
In the foregoing construction it has been assumed that the line
through p perpendicular to ah meets XY within the paper, but as
this is not always the case another construction will now be given
which will apply to any case.
Let pp' (Fig. 427) be the given point and ah, a'h' the given line.
On ah as a ground line make another elevation, ^/, of the given
point P, and a^h^ of the given line AB. Through pi draw c^d per-
pendicular to a-lh-l, meeting ah at d. c^d is the trace on the vertical
plane, of which ah is the ground line, of the plane perpendicular to AB
and containing the point P. It is obvious that the point d is a. point
218
PRACTICAL GEOMETRY
in the horizontal trace of the plane required, and that this trace will
be determined by drawing a line through d perpendicular to ah.
Produce ah to meet XY at c and draw cc/ pei-pendicular to ah
jting c^d at c/. The point c/ is an auxiliary elevation of a point
meetmar
' y
Y X
Fig. 426.
which is in the required plane and also in the vertical plane of which
XY is the ground line. If therefore cd be drawn perpendicular to
XY and made equal to ccl^ c' will be a point in the vertical trace
required which will be perpendicular to a'h'.
An examination of the figures will show that the plane determined
satisfies the given conditions.
184. To draw the Traces of a Plane which shall contain
a given Point and be perpendicular to two given Planes.—'
The plane whose traces are required will be perpendicular to the line
of intersection of the two given planes. Hence one construction is —
find the line of intersection of the given planes, and then by the
construction of the preceding article determine the traces of the plane
which contains the given point and which is perpendicular to this line
of intersection.
Another construction is — determine the projections of two lines
which shall pass through the given point and be perpendicular one to
one plane and the other to the other. The plane containing these
two intersecting lines is the plane whose traces are required.
185. To draw the Projections of a Line which shall pass
through a given Point and meet a given Line at right angles.
— Determine (Art. 183, p. 217) the traces of a plane which shall
contain the given point and l3e perpendicular to the given line. Find
(Art. 181, p. 216) the projections of the intersection of this plane and
the given line. The line joining the plan of the given point with the
plan of the intersection of the line and plane will be the plan, and the line
joining the elevations of the same points will be the elevation required.
STRAIGHT LINE AND PLANE
219
186. To draw the Traces of a Plane which shall contain
a given Line and be perpendicular to a given Plane. — Deter-
miDe the projections of a line to intersect the given line and be
perpendicular to the given plane. Find the traces of a plane to
contain these two intersecting lines. These will be the traces required.
187. Given the Inclinations of two intersecting Lines and
the Angle between them, to draw their Projections and the
Traces of the Plane containing them. — Draw C^A and CiB
(Fig. 428) making the angle AC^B equal to the given angle between
the lines. From a point A in CjA draw AD making the angle CjAD
equal to a, the given inclination of one of the lines. Draw CjD per-
pendicular to AD. With centre C^ and radius C^D describe an arc
DEF and draw BF to touch this arc and make the angle C^BF equal
to (3, the given inclination of the other line. Join AB. Consider the
triangle AC^B to be on the horizontal plane and imagine this triangle
to rotate about the side AB until the point Cj is at a distance equal
to CjD or C^F from the horizontal plane. Denoting the new position
of the point C^ by C (see the pictorial projection, Fig. 429), the lines
Pig. 429.
CA and CB will be inclined to the horizontal plane at angles equal to
a and /3 respectively, and their plans will be equal in length to AD
and BF respectively. Hence, if with centre A and radius AD the arc
Dc be described, and if with centre B and radius BF the arc Be be
described, meeting the former arc at c, Ac and Be will be the plans of
the lines required.
AB will be the horizontal trace of the plane containing the lines
CA and CB.
An elevation of the lines on any vertical plane can easily be
obtained, since A and B are on the horizontal plane and the distance
o£ C from the horizontal plane is known. In Fig. 428 an elevation is
shown on a ground line perpendicular to AB. The distance of c from
220
PRACTICAL GEOMETRY
XY is equal to C^D or CiF. Nc is the vertical trace of the plane
containing the lines CA and CB.
Instead of using the arcs T>c and ¥c to find the point c, this point
maybe determined as follows: Draw XY perpendicular to AB and
produce AB, if necessary, to meet XY at N. Draw C^C/ perpendicular
to XY. With centre N and radius NC/ draw the arc C/c' to meet at
c a parallel to XY at a distance from XY equal to C^D or C^F. A
perpendicular to XY from c' to meet a parallel to XY from C^ deter-
mines the point c.
The student should after drawing the figure ABFC^DA on a piece
of paper cut it out and fold it along the lines ACj and BC^ until the
edges C^D and C^F coincide. Placing this model so that the edges
AB, BF, and DA are on the table, he should then have no difficulty
in fully understanding this important problem.
188. Given the Traces of a Plane, to determine the Trace
of this Plane on a new Vertical Plane.— Let MN (Fig. 431) be
the horizontal trace of a plane, and let LM be the trace of this plane
on a vertical plane of which XY is the ground line ; it is required to
find the trace of this plane on a vertical plane of which X^Y^ is the
ground line. Let MN meet XjY^ at S.
Fig. 430.
Take a point p in the new ground line and consider this as the plan
of a point lying in the given plane LMN. Find the elevation p of
this point on the ground line XY. Draw jpp^' perpendicular to XjYi
and make pp/ equal to the distance of p' from XY. Sp/ is the vertical
trace required. For, P is a point in the given plane and also a point
in the new vertical plane, therefore pi must be a point in the new
vertical trace. Also the point S, where the horizontal trace meets
XjYj, is evidently a point in the new vertical trace, therefore the line
Sp/ is the trace required.
In the case where the new ground line does not meet the horizontal
trace of the plane within a convenient distance, a second point in the
required vertical trace may be found in the same way as the first
point pi'.
STRAIGHT LINE AND PLANE 221
189. To rabat a given oblique Plane, with any Points
or Lines on it, into the horizontal Plane.— Let LMN (Fig. 432)
be the given oblique plane, and
let ahc, a'h'c' be a triangle lying
on this plane. Take a new
ground line XiYi at right angles
to MN, the horizontal trace of
the given oblique plane. Find
(Art. 188, p. 220) ST the verti-
cal trace of the given plane on
the new vertical plane of which
X^Yj is the ground line. This
line ST will be the edge view
of the given plane and a/c/fe/
the elevation of the triangle
ACB on the new vertical plane
will be on ST as shown.
Now imagine the plane TSN Fig. 432.
to turn about SN as an axis.
The point A will describe an arc of a circle whose plane is perpendicular
to SN, and will therefore be vertical, and its plan will be a straight
line aAj perpendicular to SN. The elevation (on XiYj) of this arc
will be an arc of a circle a/ A/ whose centre is S. The point A/ on
XjYi is the elevation of the point A when the plane with A on it has
been brought into the horizontal plane. Hence a perpendicular from
A/ to Xi Yj to meet aA^ determines Aj the plan of A in the new position.
In like manner the points B^ and Cj may be determined, and these
points Aj, Bi, and C^ being joined a triangle is formed which is the
triangle ABC brought into the horizontal plane by turning it
about SN.
It is evident that the triangle AiBjCi is the true form of the
triangle ABC.
The construction given above will be found of great use in solving
numerous problems.
If any line be drawn in the triangle AjB^Cj, or if any figure be
built up on that triangle, it is obvious how this line or figure may be
turned about SN into the plane TSN and their projections in the new
position found.
Exercises XVIa
In reproducing the figures which are shown on a squared ground take the small
squares as of half-inch side.
1. The traces of a plane are given in Fig. 433, also one projection of each of
three points A, B, and C contained by the plane. Find the true form of the
triangle ABC.
2. A plane is given by its traces in Fig. 434, the plan abc of a triangle lying
on the plane is also given. Draw the elevation of the triangle. Show also the
222
PRACTICAL GEOMETRY
plan and elevation of a point which is in the given plane and 1-25 inches from
each of the planes of projection.
Fig. 433.
"
h'
y
5^
7
K
.^•>
y
/^
—
~
/
y
^
>
*^
■*x
„>
y
r
y
y
/^
/
fl
a
JUt
^
<*
/
/
c nu
sl
^a
"
^
c ^
V
s
:j
^
r
's:
^
5*.
—
^
^
•^
V
s
f
M
^<
N
<
_
[\
L
S
ii
^
Fig. 434.
Fig. 485.
Fig. 436.
3. a! and h' (Fig. 435) are the elevations of two of the angular points of a
triangle lying on the plane whose traces are given, and c is the plan of the other
angular point. Draw the plan and elevation of the triangle and find its true
form.
4. Draw the traces of the plane which contains the given triangle ahc^ a'h'c'
(Fig. 436). Show also th3 traces of the plane which contains the line a&, a'b' and
is parallel to the ground line.
5. The horizontal trace of a plane is given in Fig. 437, also the plan and eleva-
tion of a point P in the plane. Draw the vertical trace of the plane. Show also
the plan and elevation of a point Q which is in this plane, and which is 2 inches
from P and 1 inch above the horizontal plane. Assume that the point where the
given horizontal trace meets the ground line is inaccessible.
—
—
R
a
—
d
I'
—
-^
c a'
h' -
^
of
H
R
1
' —
s
h
^
*N,
\
^
s
—
V
V
V
N
^
\
-
c
b *!
■
u
\
1
a
/
a
r
s
w
/
-
u
>.
li
ft
—
s
s
/
v
V,
ih.
^
_?
k
H.T.
^s
Fig. 437.
Fig. 438.
Fig. 439.
Fig. 440.
Fig. 441.
li
6. The horizontal trace of a plane is given in Fig. 438, also the elevation a'h'
of a line AB contained by the plane. The length of AB is 3 inches. Draw the
plan of AB and the vertical trace of the plane. Assume that the point where the
horizontal trace of the plane meets the ground line is inaccessible.
7. Draw the traces of the plane which contains the line ah, a'h', and the point
cc' given in Fig. 439.
8. The plan and elevation of a line AB are given in Fig. 440. This line cuts
a plane whose horizontal trace is given at a point 1*5 inches from B. Draw the
vertical trace of the plane.
9. Draw the traces of the two planes which contain the line ah, a'h' (Fig. 441)
and which are inclined at 60^^ to the horizontal plane. Assume that points to the
left of aa' and points to the right of hh' are inaccessible.
10. The horizontal trace of a vertical plane makes an angle ofi 45° with XY.
Draw the elevation of two lines contained by this plane, one line being inclined
at 30° to the vertical plane of projection and the other inclined at 60° to the
horizontal plane. The first line intersects the ground line and the second inter-
sects the first at a point 1-5 inches from the vertical plane of projection.
11. The vertical and horizontal traces of a plane make angles of 60^ and 45°
STRAIGHT LINE AND PLANE
223
V
H.T.
Fig. 442.
L
/}
/
UL
\
/
\
\
/
f
M
\
%
Cb
\
V
/
\
/
1
\
N
-yw
Hi
\^^^
7
^.'<.
c
/ "
-H'
a
\
\
.d
\
^x«
c
<-"'
-^h
Fig. 443.
Fig. 444.
respectively with XY. Draw the projections of a line lying in this plane, inclined
at 30^ to the horizontal plane and passing through a point 1 inch from each of
the planes of projection.
12. Taking the plane given in Fig. 435 place in this plane a line inclined at
30° to the horizontal plane, the portion of the line between the traces of the
plane to be 2-5 inches long.
13. Draw the projections of a line passing through the given point pp' (Fig.
442) having an inclination of 45° to the horizontal plane and parallel to the given
plane.
14. Draw the traces of a plane which shall contain the given point pp' (Fig.
442) and be parallel to the given plane.
15. In the given plane
LMN (Fig. 443) place a line
which shall be parallel to
the plane PQR and at a
distance of 0-5 inch from it.
16. Draw the projec-
tions of a line which shall
pass through the point aa'
(Fig. 443) and be parallel
to each of the given planes.
17. The vertical and
horizontal traces of a plane
make angles of 60° and
45"^ respectively with XY.
Draw the plan and eleva-
tion of a line perpendicular to this plane, meeting it at a point A, and intersecting
XY at B so that AB is 1-5 inches long.
18. A line inclined at 35° to XY is both horizontal and vertical trace of a
plane. A point P in XY is 2-5 inches from the point where the plane cuts XY.
Find the perpendicular distance of P from the plane.
19. The projections of two non-intersecting lines AB and CD are given in
Fig. 444. Draw the traces of the plane which contains the line CD and is
parallel to AB. Show also the plan and elevation of the projection of AB on this
plane.
20. Draw the traces of the plane which bisects the line,a6, a'b' (Fig. 440) at
right angles.
21. Draw the traces of the plane which shall contain the point aa' (Fig. 443)
and be perpendicular to each of the given planes.
22. Draw the projections of a line which shall pass through the point cc'
Fig. 439) and meet the line ab, a'b' at right angles.
23. Show by its traces a plane perpendicular to the plane of the triangle abc,
a'b'c' (Fig. 436) and bisecting the sides AC and BC.
24. Two intersecting straight lines containing an angle of 60° are placed so
that one of them is inclined at 35° while the other is inclined at 45° to the
horizontal plane. Find the inclination of the plane containing the two lines.
25. A person using a theodolite at a station point A observes the angular
elevations of two objects P and Q above the horizon to be 35° and 65° respectively.
The horizontal angle (or azimuth) between their directions, that is between
vertical planes containing them, measures 75°. Find the angle PAQ subtended
by the two objects at the place A, as would be measured by a sextant. [b.e.]
190. Inclination of a Line to a Plane. — The inclination of
a given line to a given plane is the angle which the line makes with its
projection on the plane. Let PQ be the given line and let it intersect
the plane at Q. Let PR be a normal to the plane, intersecting the
plane at R. If Q and R be joined, then the line RQ is the projec-
tion of PQ on the plane and the angle PQR is the inclination of the
line PQ to the plane. But since the angle PRQ is a right angle the
224
PRACTICAL GEOMETRY
angle QPR is the complement of the angle PQR. Hence, if from a
point P in the given line a normal PR be drawn to the given plane
(Art. 182, p. 217) the true angle between this normal and the given
line may be found as explained in Art. 147, p. 177, and the comple-
ment of this angle is the inclination required.
It is obviously unnecessary to find the points in which the given
line PQ and the normal PR intersect the given plane.
191. To find the Angle between two Planes. — Referring to
the pictorial projection, Fig. 445, AB and CD are two planes intersect-
ing in the line CE. FH is a third
plane which is at right angles to the
line CE. The plane FH intersects the
planes AB and CD in the lines KH
and KL respectively. Since CE is
at right angles to the plane FH it
follows that CE is at right angles to
each of the lines KH and KL. Hence
the angle between the planes AB and
CD is the angle between the lines KH
and KL. If from a point P in the
plane FH perpendiculars PR and PQ
be drawn to KH and KL, then, the angle QKR is the supplement of
the angle QPR, and if RP be produced to S the angle QKR is therefore
equal to the angle QPS. It may be proved that the lines PR and PQ
are perpendicular to the planes AB and CD respectively.
Hence, to find the angle between two given planes, draw, from any
point, two perpendiculars, one to each plane, and find the angle between
these perpendiculars. The acute angle between these perpendiculars
will be equal to the acute angle between the planes.
Another method is as follows. Let LMN and LRN (Fig. 447) be
the two given planes. Draw ZN, the plan of the line of intersection
Fig. 445.
Fig. 446.
Fig. 447.
of the planes. Draw ACB at right angles to IN intersecting MN, /N,
and RN at A, C, and B respectively. With centre / and radii ZC and
STRAIGHT LINE AND PLANE 225
LN describe arcs CCg and NNg cutting XY at Cg and Ng. Join LN^
and draw C2P2 perpendicular to LNg. On ZN make CV^ equal to CgPg.
Join AP, and BP^. The acute angle between the lines AI^ and EPj
is the acute angle between the planes LMN and LRN. In this con-
struction AB is the horizontal trace of a plane which is perpendicular
to the line of intersection of the given planes. This plane intersects
the given planes in two lines .AP and BP (see the pictorial projection,
Fig. 446 ) and the acute angle between these lines is the angle between
the given planes. The lines AP and BP together with the line AB
form a triangle of which APjB is the true form.
The points A and B where the line perpendicular to ZN meets the
horizontal traces of the given planes may be on the same side of ZN
instead of on opposite sides as in Fig. 447. In particular cases AB
may be parallel to the horizontal trace of one of the given planes and
the point A or the point B will then be at an infinite distance from C,
and APi or BP^ will be perpendicular to ZN.
192. To draw the Traces of a Plane which shall make a
given Angle with a given Plane and contain a given Line in
it. — Let LMN (Fig. 447) be the given plane, and ZN the plan of the
line in it which the required plane is to contain.
Draw AC perpendicular to ZN, meeting the latter at C and the
horizontal trace of the given plane at A. With centre Z and radii ZC
and ZN describe the arcs CCg and NNg cutting XY at Cg and Ng. Join
LN2 and draw C0P2 perpendicular to LNg. On ZN make CP^ equal to
C2P2. Join AP^. Draw P^B making the angle AP|B equal to the
given angle. Let P^B meet AC or AC produced at B. Join NB and
produce it to meet XY at R. Join LR. NR and RL are the traces
of the plane required.
If the student has understood the construction of the preceding
Art. referring to Fig. 447, he should have no difficulty in satisfying him-
self as to the correctness of the construction given above for finding
the plane LRN.
193. To draw the Traces of a Plane which shall make a
given angle with a given Plane, and contain a given Line not
contained by the given Plane. — Let LMN (Fig. 448) be the given
plane. This plane is shown as an inclined plane, that is a plane inclined
to the horizontal plane but perpendicular to the vertical plane of pro-
jection. If the plane is given as an oblique plane it should be converted
into an inclined plane by the construction of Art. 188, p. 220.
ah, a'h' is the given line.
Take two points in the given line and consider them to be the
vertices of two cones whose bases are on the given plane, and whose
slant sides are inclined to their bases at the given angle (/?).
It is evident that a plane which touches these two cones will fulfil
the conditions of the required plane.
Turn the given plane about its horizontal trace until the plane, with
the circles which are the bases of the cones, comes into the horizontal
plane. Draw the tangent CjN to these circles when in this position.
Now turn the plane with the tangent on it back to its former position.
226
PRACTICAL GEOMETRY
The plane which contains the lines AB and CN is the plane required ;
its traces are NR and ST. There are two planes which satisfy the
given conditions but only one
of them is shown in Fig. 4-1:8.
194. To draw the Traces
of a Plane which shall con-
tain a given Point, have a
given Inclination, and make
a given Angle with a given
Plane. — Conceive a cone hav-
ing its vertex at the given point,
its base on the horizontal plane,
and having a base angle equal
to the angle which the required
plane is to make with the hori-
zontal plane. Conceive also a
second cone having its vertex
at the given point, its base on
the given plane, and having a
base angle equal to the angle
which the required plane is to
make with the given plane. It
is evident that a plane which
touches these two cones will
fulfil the conditions of the pro-
blem.
Referring to Fig. 449, vv' is
the given point, and LMN is the given plane, which is assumed to be
perpendicular to the vertical plane of projection, v'a'h' is the elevation
of the cone whose base is on the horizontal plane and whose base angle a
is equal to the angle which the required plane is to make with the
horizontal plane, v'c'd! is the elevation of the cone whose base is on
the given plane and whose base angle ^ is equal to the angle which the
required plane is to make with the given plane.
Let the surface of the second cone be produced to cut the horizontal
plane of projection. The curve of section, or horizontal trace of the
cone, will be a " conic section " and it may be found as follows. Draw
t'r perpendicular to v'e', the elevation of the axis of the cone, meeting
v'e' produced at t', XY at s and v'd! produced at r. With t' as centre
and t'r' as radius describe an arc, and through s' draw s'H.^ perpendicu-
lar to t'r' to meet the arc at Ho. Draw s's perpendicular to XY to meet
a parallel to XY through v at 8. On ss' make sli equal to s'Hg. h is a
point on the horizontal trace of the cone. Another point will be at a
distance below 8 equal to sh. By taking different positions for the line
t'r' any number of points on the horizontal trace of the cone may be
found.
The base of the cone whose axis is vertical will be its horizontal
trace and will be a circle whose centre is v and diameter equal to a'h'.
The horizontal traces of all planes which touch the two cones will
Fig. 448.
STRAIGHT LfNE AND PLANE
22^
be tangents to the horizontal traces of these cones. OQ is one of these
tangents meeting XY at Q. Through v draw vp parallel to OQ to
/^
\
A
/:\\
v^
/
/l\\
V. .^
V\^k^
^^
^^'
Y '
h\\ \i
)¥
I ^
\/
1 \y
N /.
■ Y
-^f
k '
pA
x
s
^
» V
Fig. 449.
leet XY at p. From p draw pp perpendicular to XY to meet a line
through v' parallel to XY at p'. Join p'Q. The plane p'QO will satisfy
the conditions of the problem.
If the horizontal trace OQ does
not meet XY within a convenient
distance, the vertical trace may-
be found by the construction shown
in Fig. 419, p, 214.
I'he drawing of the conic section
which is the horizontal trace of
the cone whose base is on the given
plane may be avoided by the fol-
lowing construction. Conceive a
sphere to be inscribed in this cone.
The elevation of this sphere will
be a circle touching the lines v'd
and v'd\ as shown in Fig. 450, its
centre being on v'e'. Next con-
ceive a cone to envelop this sphere
and have its axis vertical and its
base angle equal to a. u' is the
elevation of the vertex of such a
cone and the circle shown whose
centre is w is its horizontal traee, p ,k^
228
PRACTICAL GEOMETRY
A plane which touches the cones whose vertices are at vv' will also
touch the cone whose vertex is at uu', and the horizontal trace of the
plane will be a tangent to the two circles shown in the plan. A little
consideration will show that if the horizontal trace of the required
plane is be a cross tangent to the circles the vertices vv' and uv! must
be on opposite sides of the horizontal plane of projection as shown in
Fig. 450.
In the particular case where the required plane is to be perpen-
dicular to the given plane the cone whose base is on the given plane
will become a straight line perpendicular to the given plane, and
the horizontal trace of the required plane will pass through the hori-
zontal trace of this line and touch the base of the other cone.
195. To draw the Projections of a Line which shall pass
through a given Point and meet a given Line at a given Angle.
— Let jpp' (Fig. 451) be the given
point and aZ>, a'h' the given line.
Determine (Art. 172, p. 211) the
plane LMN which contains the
point P and the line AB. Rabat
this plane with the point P and
line AB into the horizontal plane
(Art. 189, p. 221). Pi and A^fc are
the rabatted positions of P and AB
respectively. Through Pi draw
PiCi making with A^ an angle
equal to the given angle. Through
Ci draw CjC perpendicular to MN
to meet ah at c. Draw cc perpen-
dicular to XY to meet ah' at c'.
The lines pc and p'c' are the pro-
jections of the line required.
If the required line is to be perpendicular or parallel to the given
line there is only one solution, in other cases there will be two
lines which will satisfy the given conditions. In Fig. 451 only one
of the two lines which may be drawn to satisfy the given conditions
is shown.
196. To find the Traces of a Plane which shall contain
one given Line and make a given Angle with another given
Line. — Let AB denote the first given line and CD the second.
Determine the projections of a line EF which shall be parallel to
CD and meet AB at a point E.
Take E as the vertex and EF as the axis of a cone whose semi-
vertical angle is equal to the given angle. Find as in Art. 194 the
horizontal trace of this cone. A tangent from the horizontal trace of
AB to the horizontal trace of the cone will be the horizontal trace of
the plane required. Having found the horizontal trace of the plane,
its vertical trace may be determined from the condition that the plane
is to contain the line AB.
If the horizontal trace of AB falls outside the curve which is thQ
Fig. 451.
STRAIGHT LINE AND PLANE
229
horizontal trace of the cone, there will be two solutions of the problem ;
if it falls on the curve there will be one solution only, and if it falls
inside the curve there can be no solution.
197. To draw the Traces of two Planes which shall be
perpendicular to one another and have given Inclinations.
— Let the given inclinations be o\vi'
to the horizontal plane and let n\«<^L
them be a and ^. Draw MN
(Fig. 452), the horizontal trace
of one of the planes, at any angle
to XY (preferably a right angle),
and determine its vertical trace
LM from the condition that the
inclination of the plane is the
given angle a. Take any point
pj} in this plane (preferably in
its vertical trace) and determine
a line pq^ p'q' perpendicular to
the plane. Any plane which
contains this perpendicular will
be perpendicular to the plane
LMN. Determine also a cone
having its vertex at pp', its base
on the horizontal plane and
having a base angle y8 the given inclination of the second plane. A
line through the horizontal trace of pq, p'q to touch the base of this
cone will be the horizontal trace ST of the second plane. The vertical
trace RS of the second plane is easily found from the condition that
the plane is to contain the line pq, p'q'.
198. To draw the Traces of three Planes which are
each perpendicular to the other two, having given the
Inclinations of two of the Planes. — Determine by the preceding
Article the traces of the two planes whose inclinations are given.
Next determine the line of intersection of these two planes. The
third plane will be perpendicular to this line and its horizontal and
vertical traces will be perpendicular to the plan and elevation of the
line respectively.
199. Projections of a Solid Right Angle. — A solid right
angle is formed by three planes which are mutually perpendicular.
These three planes will intersect in three straight lines which are each
perpendicular to the other two, and the solid angle is represented on
paper by the projections of these three mutually perpendicular straight
lines.
Any three intersecting straight lines on the paper may be taken
as a projection of three lines which are mutually perpendicular. Let
oa, ob, and oc (Fig. 453) be the plans of three lines which are mutually
perpendicular.
Since each line is perpendicular to the other two it will be per-
pendicular to the plane containing the other two, and therefore the
230
PRACTICAL GEOMETRY
plan of each will be perpendicular to the horizontal trace of the plane
of the other two. Also the horizontal trace of the plane containing
any two of the lines will pass through the horizontal traces of these
lines.
It is evident that the form of any elevation of the lines will not be
affected by the position of the horizontal plane of projection ; the
horizontal plane of projection may therefore be taken at any level.
Take any line ab, perpendicular to oc, as the horizontal trace of the
plane containing the lines OA and OB. A perpendicular from h to oa
to meet oc at c will be the horizontal
trace of the plane containing OB
and OC, and ac will be the hori-
zontal trace of the plane of OA
and OC.
Produce co to meet ah at d, and
consider od as the plan of a line
lying in the plane AOB. Since OC
is perpendicular to the plane AOB,
the angle COD will be a right
angle.
Conceive the triangle COD to
revolve about CD as an axis until it
comes int^ the horizontal plane.
The point will describe an arc of
a circle whose plan will be a line
through perpendicular to cd, and
the point O will, when brought into
the horizontal plane, occupy such
a position that the lines drawn from
it to c and d will contain a right angle. Hence on cd describe a
semicircle to cut the line oO^ perpendicular to cd at O^. oO, will be
the distance of the point O from the horizontal plane containing the
points A, B, and C. Having found this distance, an elevation of the
lines on any vertical plane can be drawn. In the figure an elevation
is shown on XY.
This problem is one of great importance in connection with iso-
metric and axometric projections and it is further considered in
Chap. XXI.
200. To find a Point on a given Plane such that the sum
of its Distances from two given Points shall be the least
possible. — Let LMN (Fig: 454) be the given plane and aa', hh' the
given points. These given points are on the same side of the given
plane. Determine the projections of a line to pass through aa' and
be perpendicular to the plane LMN. Pind the point cc' where this
perpendicular meets the plane LMJST. Determine a point dd' in
ac, a'c' produced such that CD is equal to AC. The projections d
and d' of D are such that cd = ac, and cd' = a'c. Join hd and b'd'.
Find the point ee where the line hd, h'd' intersects the plane LMN.
ee' is the point required.
Fig. 453.
STRAIGHT LINE AND PLANE
231
LMN (Fig. 455) be the given plane
of the line AB on the plan ah as a
construction of Art. 16, p. 20, the
The student should have no difficulty in proving the correctness of
the above construction after referring to the corresponding problem in
plane geometry discussed in
Art. 9, p. 8.
It is easy to show that the
lines ae, a'e' and he, h'e' are
equally inclined to the plane
LMN, and hence if a ray of
light from aa' impinges upon
the plane LMN and on re-
flection from the plane passes
through hh', the incident ray
must strike the plane at ee'.
201. To find the Locus
of a Point which moves on
a given Plane so that the
Ratio of its Distances from
two given Points shall be
equal to a given Ratio. — Let
and aa' and hb' the given points.
Draw a new elevation a/?>/
ground line. Determine by the
circle whose centre is o^' and
radius o/c?/ which is the locus of
a point which moves in the plane
of the new elevation and whose
distances from a/ and />/ are to
ratio.
^. ... ..... ..... .. 2:1.
The point o/ is in ai'6/ produced.
Find o the plan and o' the eleva-
tion on XY of the point of which
o/ is the elevation on ah as a
ground line. The surface of a
sphere whose centre is at oo and
whose radius is o/(Z/ will contain
all points in space whose distances
from aa' and hh' are to one an-
other in the given ratio. The
circle which is the section of this
sphere by the plane LMN is the
locus required. In Fig. 455 the
plane LMN being perpendicular
to the vertical plane of pro-
jection, the circle which is the
section of the sphere by the plane
LMN will have the straight line c'd' for its elevation and the ellipse cd
for its plan.
20.2. To determine a Line which shall pass through a given
one another
In Fig. 455
in the given
this ratio is
Fig. 455.
232
PRACTICAL GEOMETRY
Point and intersect two given Lines.^ — Let P be the given point
and AB and CD the given lines.
First method. Determine the traces of the plane containing the
point P and the line AB. Determine also the traces of the plane
containing the point P and
the line CD. The line of
intersection of these two
planes is the line required.
Second method. Fig. 456.
Determine the projections of
two lines, PKL and PMN,
intersecting AB at points K
and M. Let these lines meet
the vertical plane containing
CD at the points L and N.
The line LN will be the in-
tersection of the plane con-
taining P and AB with the
vertical plane containing CD.
Let LN meet CD at Q. PQ
is the line required.
203. To determine the common Perpendicular to two
given Lines which are not in the same Plane. — Let AB and CD
(Figs. 457 and 458) be the two given lines. Determine a line FEH
Fig. 456.
Fig. 457
Fig. 458.
to intersect AB at E and be parallel to CD. The plane containing
AB and FEH will be parallel to CD. The traces of this plane are
KL and MN.
STRAIGHT LINE AND PLANE 233
Take any point P in CD and determine the normal PQ to the
plane KLMN, meeting the plane at Q. The length of PQ is the
shortest distance between the lines AB and CD. Determine a line
through Q parallel to CD to meet AB at R. This line will lie in the
plane KLMN. A line RS parallel to QP and meeting CD at S will
be the common perpendicular to the lines AB and CD.
It is evident that the line QR, produced both ways, is the projection
of CD on the plane KLMN.
204. To determine the Centre and Radius of a Sphere the
Surface of which shall contain four given Points. — It should
first be observed that no three of the given points must be in the same
straight line, and if all four lie in the same plane they must lie on the
circumference of a circle. In the latter case the problem is indeter-
minate, as any number of spheres can be found on the surface of which
the points will lie.
Let A, B, C, and D be the given points. Determine by the con-
struction of Art. 183 a plane to bisect the line AB at right angles.
Since every point in this plane is equidistant from the points A and B
it must contain the centre of the required sphere. In like manner
the centre of the sphere must lie on the planes bisecting BC and CD
at right angles. Therefore the point of intersection of these three
planes must be the centre required, and the distance of this point from
any one of the given points will be the radius of the sphere.
As there are four given points, six lines can be got by joining them,
and the point of intersection of three planes bisecting at right angles
any three of these six lines not lying in the same plane will be the centre
of the sphere.
Another metlwd of finding the centre of the sphere is to determine
the centre of the circle containing three of the given points and then
determine the normal to the plane of this circle through its centre.
This normal will contain the centre of the required sphere. Taking
another three of the points and proceeding in the same way another
normal is found which contains the centre of the sphere, and the point
of intersection of these two normals is the centre required.
205. Trihedral Angles — Spherical Triangles. — The angle
formed by three planes meeting at a point is called a trihedral angle.
The angles between the lines in which the three planes intersect are
the sides or faces, and the dihedral angles between the planes are the
angles of the trihedral angle. Referring to Pig. 459, the three planes
OAB, OBC, and OCA form at O a trihedral angle. The plane angles
AOB, BOC, and COA are the sides or faces of the trihedral angle
referred to, and the dihedral angles between these faces are its
angles.
If a sphere be placed with its centre at 0, the three planes which
form the trihedral angle will intersect the surface of the sphere in
three great circles, which will intersect one another and form what
is called a spherical triangle. In Fig. 459, ABC is a projection of a
spherical triangle.
The sides and angles of a spherical triangle are the sides and
234
PRACTICAL GEOMETRY
Fig, 459.
angles of the trihedral angle which it subtends at the centre of the
sphere on the surface of which it is drawn. Hence the solution of a
spherical triangle is the solution of a trihedral angle.
If from a point S within
the trihedral angle (Fig.
459), perpendiculars SP,
SQj and SR, be let fall on
the faces OBC, OCA, and
OAB respectively of the
trihedral angle, these per-
pendiculars will form the
edges of another trihedral
angle S, which has the
remarkable properties that
its angles are the supple-
ments of the sides of the
trihedral angle O, and its
sides are the supplements
of the angles of the trihe-
dral angle O. The trihedral
angles and S are there-
fore called suj)plementary
trihedral angles, either be-
ing the supplementary trihedral angle of the other.
If from the point O (Fig. 459) perpendiculars OA', OB', and OC
be drawn outwards from and to the faces OBC, OCA, and OAB
respectively to meet the surface of the sphere at A', B', and C, then
A', B', and C are poles of the great circles BC, C A, and AB respectively,
and great circles of the sphere through A' and B', B' and C', and C
and A' form another spherical triangle A'B'C which is called the polar
triangle of the triangle ABC. It is obvious that the trihedral angle
subtended by A'B'C at the centre of the sphere is equal in every
respect to the trihedral angle at S. Hence the polar triangle A'B'C'
and the triangle ABC are supplementary.
The following notation is used to denote the elements of a trihedral
angle or spherical triangle. The three angles are denoted by the
capital letters A, B, and C, and the three sides by the italic letters a,
h, and c, the sides a, b, and c being opposite to the angles A, B, and C
respectively. The elements of the supplementary trihedral angle or
the polar triangle are denoted by the same letters accented ; thus the
angles are A', B', and C, and the sides a', V, and c'.
The relations between the elements of two trihedral angles which
are supplementary or between a spherical triangle and its polar triangle
are expressed by the following equations, all angles being measured in
degrees —
A' = 180 - a a' = 180 - A
B' = 180-6 6' = 180 - B
C = 180 - c c' = 180 - C
From the above equations the following are obtained —
STRAIGHT LI:N^E AND PLANE
235
a = 180 - A' A = 180 - a'
6 = 180 - B' B = 180 - V
= 180 - C C = 180 - c'
Any three of the six elements of a trihedral angle or spherical
triangle being given, the other three can be found. There are in all
six cases to consider. — I. Given the three sides. II. Given two sides
and the angle between them. III. Given two sides and the angle
opposite to one of them. IV. Given the three angles. V. Given one
side and the two adjacent angles. VI. Given one side, an adjacent
angle, and an opposite angle.
By means of the properties of the supplementary trihedral angle,
cases IV, V, and VI may be reduced to cases I, II, and III respectively.
Thus, in solving case IV, find by the solution of case I the angles of
the trihedral angle whose sides are the supplements of the given angles,
then the supplements of the angles found will be the sides required.
Or, using symbols, given A, B, and C, then a' = 180 — A,
6' = 180 — B, and c' = 180 — C; now find, by the construction for
case I, A', B', and C', then a = 180 - A', & = 180 - B', and
c = 180 - C
In case V suppose a, B, and C given, then A' = 180 — a,
6' = 180 — B, and c' = 180 — C. Now, by the construction for case
II, find a', B', and C', then A = 180 - a', 6 = 180 - B', and
c = 180 — C. In like manner case VI is reduced to case III.
The solutions of cases I, II, and III will now be given.
Case I. Given the three sides a, h, and c of a trihedral angle to
determine the three angles A, B, and C. Construct the three sides a, h,
and c, as shown in Fjg. 460, forming the development of the faces of
Fig. 460.
the trihedral angle. Consider this development to be lying on the
horizontal plane. Mark off equal lengths ONj and ON.^ on the outer
lines of the development. Draw IS^n perpendicular to OL, and N.^w
236 PRACTICAL GEOMETRY
perpendicular to OM to meet N^n at n. Join On. The three lines
OL, OM, and 0/i will form the plan of the trihedral angle, the face h
being on the horizontal plane. In finding the point n the faces a and
c have been supposed to turn about the edges OL, and OM respectively
until the lines ONi and ON^ coincide. It is evident that the point N^
describes an arc of a circle whose plan is the straight line N^n ; also,
the point N2 describes an arc of a circle whose plan is the straight line
N2W ; and since N^ and Ng are equidistant from O, n must be the plan
of the point where they meet, and On is the plan of the edge between
the faces a and c.
To determine the angles A and C, draw ?iw/ perpendicular to NjW,
and nn.J perpendicular to N3W. With Q as centre, and QNi as radius,
describe an arc to cut w?^/ at ??/, and with R as centre, and RNo as
radius, describe an arc to cut nnj at nj. Join w/Q and n^R. w/Qn is
the angle C, and w^'Rw is the angle A. It is evident that when the
face a is brought into its natural position, the line N^Q on that face
will remain perpendicular to OL ; also the line wQ on the face h is
perpendicular to OL ; therefore the angle w/Qw, which is the angle
between these lines, will be equal to the angle between the faces a
and b, that is, the angle C. In like manner the angle n-^Hn is equal
to the angle A.
To find the angle B, draw N^L perpendicular to ON^ to meet OL
at L, and draw N2M perpendicular to ON^ to meet OM at M. With
centre L, and radius LNj, describe an arc to cut On at N. Join N to
L and M. LNM is equal to the angle B. It is evident that when
the faces a and c are brought into their natural positions, the lines
LNi and MNg lying on these faces will be perpendicular to their line
of intersection, and therefore the angle between them, which is
obviously equal to the angle LNM, will be equal to the angle B.
The student will more clearly understand the foregoing constructions
if, after drawing the figure shown, he cuts it out along the lines LN^,
NjO, ON2, and N.^M, and then folds the triangles LN^O and MN.p
about OL and OM respectively until the lines ON^ and ON^ meet one
another.
Case II. Given two sides
of a trihedral angle, and the
angle betioeenthem, to determine
the third side and the other
angles. Let a and h be the
given sides, and C the angle
between them. Construct the
two sides a and h on the hori-
zontal plane as shown in Fig.
461. From a point Nj in ONj
draw NjQw perpendicular to
OL and intersecting it at Q.
Draw Qw/ making the angle
w/Qn equal to C. Make Q^i'
equal to QNj, and draw w/w perpendicular to Nin to meet the latter at ».
STRAIGHT LINE AND PLANE
237
Join On. The lines OL, OM, and On will form the plan of the trihedral
angle, the face b being on the horizontal plane.
To determine the side c, draw nNa perpendicular to OM, and with
centre O and radius ONj describe an arc to cut wN.2 at N^. Join ON2.
N2OM is the side c brought into the horizontal plane.
The angles A and B may now be determined exactly as in case I.
If l^e student has understood the constructions in case I, he should
have no difficulty in demonstrating the above constructions for
case II.
Case III. Given two sides of a trihedral angle, and the angle opposite
to one of them, to determine the third side and the other angles. Let a an^
h be the given sides
and A the given angle. ^^
As before, construct
the given sides on the
horizontal plane as
shown in Fig. 462.
Take a ground line
XY at right angles to
OL, cutting OL at Q,
ONi at Ni and OM at
M. From Q draw QT
at right angles to OM
meeting it at T. With
Q as centre, and QT as
radius, describe the arc
TT2 to meet XY at T2.
Draw T2V making the
angle QT2V equal to the angle A and meeting OL at V. Join MV.
MV is the vertical trace of the face c, OM being its horizontal trace.
Now let the face a revolve about OL. Since OL is perpendicular to
XY the elevation of the circle described by Nj will be a circle having
a radius equal to QNj, with Q as centre. Let this circle cut MV at n'.
Join n'Q. The line n'Q, is the vertical trace of the face a when it is
in its natural position, and OQ is its horizontal trace ; therefore, if
a perpendicular be let fall from n' to XY to meet XY at w, the line Ow
will be the plan of the intersection of the faces a and c. Hence the
lines OL, OM, and Ow form the plan of the trihedral angle, the face h
being on the horizontal plane. The third face c may now be deter-
mined as in case II and the angles B and C may be found as in
case I.
The arc Njw' may meet the line MV at two points (it does so in
Fig. 462), and there are then two solutions, that is, there are two
trihedral angles which satisfy the conditions of the problem. This is
then called the amhiguous case.
206. Circumscribed and Inscribed Circles of Spherical
Triangles. — A plane containing the three angular points A, B, and
C of a spherical triangle (Fig. 463) will intersect the surface of the
sphere in a circle which is the circumscribed circle of the spherical
Fig. 462.
238
PRACTICAL GEOMETRY
triangle. A right circular cone, having its vertex at O, the centre of
the sphere, and having its curved surface touching the three faces
of the trihedral angle subtending _~-..
the spherical triangle at O, will
intersect the surface of the
sphere in a circle which is the
inscribed circle of the spherical
triangle. The inscribed cone is
determined by first drawing a
sphere inscribed in the trihedral
angle. The cone envelops this
sphere.
Exercises XVIb
1. Find the angle between the
pairs of planes shown at (a), (&), (c),
and (d), Fig. 464. Show also in each
case the traces of the plane which
bisects the angle between the given
planes.
Fig. 464.
2. The horizontal and vertical traces of a plane (1) make angles of 90^ and
45'^ respectively with XY. A line inclined at 40" to the vertical plane of projec-
tion lies in this plane and the distance between its horizontal and vertical traces
is 3 inches.- This line is the intersection of a plane (2) with the plane (1), the
inclination of the plane (2) to the horizontal plane being 60°. Draw the traces of
the planes (1) and ^2) and find the angle between the planes.
3. OA and OB (Fig. 465) are two straight lines lying on the horizontal plane.
OC is another line of which Oc is
the plan. The inclination of 00
to the horizontal plane is 50-. De-
termine the angle between the
planes AOC and BOG, also the angle
between the planes AOC and AOB,
and the angle between the lines
OA and 00.
4. ab (Fig. 466) is the plan of a
line lying in a plane of which H.T. is the horizontal trace. The inclination of
AB to the horizontal plane is 50°. Draw the traces of second plane to contain
AB and make an angle of 60" with the first plane.
5. The vertical and horizontal traces of a plane (1) make angles of 40° and 45°
respectively with XY. A point in the vertical trace of this plane and 1-5 inches
Fig. 465.
Fig. 466.
* STRAIGHT LINE AND PLANE
239
above XY is the elevation of a line AB. Draw the tra-ces of a plane (2) to contain
the line AB and make an angle of 75*^ with the plane (1).
6. Draw the traces of a plane (1) which is inclined at 45^ to the horizontal
plane and is perpendicular to the vertical plane of projection. Then draw the
traces of a plane (2) which is inclined at 65° to plane (1) and inclined at 75° to
the horizontal plane.
7. The traces of a plane LMN are both inclined at 45° to XY. A point P in
this plane is 2 inches above the horizontal plane and 1 inch in front of the vertical
plane. Draw the projections of a line passing through the point P, inclined at
30° to the horizontal plane and inclined to the plane LMN at 60°.
8. (the is the plan of a triangle, ah = 2-1 inches, be = 2*9 inches, and
ca = 1-4 inches. The heights of the points A, B, and C above the horizontal
plane are, 0, 2*5, and 1-25 inches respectively. Find the plan of a point D in AB
such that the angle ADC is 50°.
9. A line AB of indefinite length passes through a point G which is in the
vertical plane of projection and 2 inches above XY. AB is inclined at 40° to the
horizontal plane and its plan makes 30° with XY. A point P is 1*5 inches in
front of the vertical plane and 1 inch above the horizontal plane, and its plan is
at a perpendicular distance of 1-8 inches from the plan of AB. Draw the plan
and elevation of an equilateral triangle which has one angular point at P and the
opposite side on AB.
10. aMN6 is a quadrilateral, the angles at M and N being right angles.
MN = 1-5 inches, aM = 1 inch, and 6N = 2*5 inches, ab is the plan of a horizontal
line which is 0*8 inch above the horizontal plane. MN is the horizontal trace
of a plane which contains the point A. Draw the plan of a line which lies in the
plane AMN and makes an angle of 75° with AB.
11. Draw the traces of a plane which shall contain the line AB (Exercise 17)
and be inclined to the line CD at an angle of 15°.
12. One plane is inclined at 60° to the horizontal plane and 45° to the vertical
plane. Another plane is perpendicular to the first and inclined at 70° to the
horizontal plane. Draw the traces of the two planes.
13. The horizontal and vertical traces of a plane are in one straight line
inclined at 40° to XY. A second plane is perpendicular to the first and inclined
at 60° to the horizontal plane. Draw the traces of the two planes.
14. Three planes are mutually perpendicular. One is inclined at 50° and
another at 45°. Find the inclination of the third plane.
15. H.T. (Fig. 467) is the horizontal trace of a plane which contains the point
whose plan is c. The height of C above the horizontal plane is 1 inch, a and b
are the plans of two points whose heights above the horizontal plane are 1*5 inches
and 2 inches respectively. Find the projections of a point P on the given plane
such that AP + BP shall be a minimum.
<^
f
-Y
ft
C
b
ftf
Fig. 467.
Fig. 469.
16. Taking the same plane and the same points as in the preceding exercise,
draw the projections of the locus of a point P which moves on the plane in suck
a manner that the ratio of AP to BP shall be equal to the ratio of 8 to 2.
240
PRACTICAL GEOMETRY
17. ah and cd (Fig. 468) are the plans of two straight lines. The heights of
the points A, B, C, and D above the horizontal plane are, 1'5, 0-5, 1-0, and 2-5
inches respectively. Draw the plan and elevation of the common perpendicular
to the lines AB and CD and find its true length.
18. a, b, c, and d (Fig. 469) are the plans of four points. The heights of
A, B, C, and D above the horizontal plane are, 2 5, 0'25, 2-6, and 0-6 inches
respectively. Determine the plan and elevation of the sphere whose surface
contains the points A, B, C, and D.
19. Taking the lines given in exercise 17 and Fig. 468, determine the locus of
a point which moves on the horizontal plane so as always to be equally distant
from these lines.
20. The common perpendicular EF to two straight lines AB and CD is 1 inch
long and its inclination is 40^. The inclination of AB is 35^ and the angle
between the plans of AB and CD is 90°. Draw the plans of AB, CD, and EF, and
find the inclination of CD.
21. A person standing in the corner of a quarry notices that on one vertical
face the line of a particular bed slopes downwards from the corner at an angle of
30^ with the horizontal, whilst on an adjacent vertical face the corresponding
angle is 48°. The angle between the two facps is 110°. What is the true dip or
inclination of the strata ? [b.e.]
22. Two inclined faces of rock A, B in a quarry intersect in L. The lines of
steepest ascent on A make 10° with the vertical and go due North. Those on B
make 20° and have a direction 12° North of East. A seam of coal intersects these
faces in lines which both go upwards from L, making 70° with L on the face A
and 65° with L on the face B. Find the direction, and inclination to the
horizontal, of the lines of st'eepest descent of the seam. [b.e.j
23. The plans of four points O, A, B, and C are given in Fig. 470. O is the
centre of a sphere of 2*5 inches radius. A, B,
and C lie on the surface of the sphere, A and C
being above, and B below the level of O. Draw
the plan of the spherical triangle ABC,* and an
elevation on a ground line parallel to ox. Add
the plan and elevation of the circumscribed
circles of the spherical triangle. Lastly, determine
the plan and elevation of the polar triangle.
24. Solve, by construction, the following
trihedral angles or spherical triangles. The
results, as obtained by calculation, correct to the
nearest tenth of a degree, are added.
(1) Given, a = 40°, 6 = 50°, c = 60°.
Results. A = 47-9°, B = 62-2°, C = 89-1°.
K-r
o-2L^^
J.
.2-3- -
a^
0-3
I I
I I
Fig. 470.
(2)
(3)
Given,
Results.
Given,
Results.
(4) Given,
Results.
(5) Given,
Results.
or,
(6) Given,
Results.
or,
(7) Given,
a = 70°, b = 80°, c =
A = 69-7°, B = 79-4°,
a = 45°, b = 55°, C =
A = 58-4°, B = 80-7°
a = 60°, 6 = 70°, C •=
A = 50-5°, B = 56 9°,
a = 45°, b = 56°. A =
90°.
C = 93-7°.
60°.
c = 46°.
120^
c = 103-6".
:45°.
c = 75-1°, B = 55°, C = 104-9°,
c = 15-5°, B = 125°, C = 15-5°.
a = 60°, A = 70°, C = 90°.
b = 39-1°, c =-- 67-2°, B = 43-2°,
b = 140-9°, c = 112-8°, B = 136-8°.
A = 120°, B = 45°, C = 35°.
Results, a = 78-7°, b = 53-2°, c = 40*5°.
(8) Given, A = 130°, B = 120°, c = 90°.
Results, a = 126°, b = 113-9°, C = 108-7°,
CHAPTER XVII
SECTIONS OF SOLIDS
207. Sections of Solids. — Many objects of which mechanical
drawings have to be made are of such a form that their construction
is not completely shown by outside views only. The construction of
the interior of a house, for instance, cannot be seen from the outside.
In order to exhibit the interior of such an object it is imagined to be
cut in pieces by planes and these pieces are then represented separately.
But in representing an object of comparatively simple form, the
addition of a sectional drawing of it often adds very much to the illus-
tration of it, although such a sectional drawing may not be absolutely
necessary for the complete representation of the object.
That surface which is produced when a plane cuts a solid is called
a section, and if that part of the solid which is below or behind the
cutting plane is also shown on the projection of a section the projection
is called a sectional plan or a sectional elevation. But in the application
of these terms to architectural and engineering drawings the term
section is often used in the same sense as sectional plan or sectional
elevation.
The projection of a section is distinguished in various ways, one
way, used in this work, is by drawing across it parallel diagonal lines
at equal distances apart. These lines are called section lines.
If the true form of a section is required it must be projected on a
plane parallel to that of the section.
In this chapter sections of solids having plane faces are considered.
Sections of the sphere, cylinder and cone are considered in chapter
XVIII.
208. Section of a Prism by a Plane perpendicular to one
of the Planes of Projection. — A plane section of a prism will be a
rectilineal figure whose angular points are at the points where the
plane of section cuts the edges of the prism. Since the plane of section
is perpendicular to one of the planes of projection one of the projections
of the section will be a straight line coinciding with the trace of the
plane of section on the plane of projection to which the plane of section
is perpendicular.
The determination of sections of a right square prism is illustrated
by Fig. 471. The plan (1) and the elevation (2) represent the prism
242
PRACTICAL GEOMETRY
when standing with one end on the horizontal plane, PQ is the
horizontal trace of a vertical plane of section. This plane of section
cuts two of the vertical faces
of the prism in vertical lines of
which the points a and b are
the plans, and a'a' and b'h' per-
pendiculars to XY, the eleva-
tions. The same plane inter-
sects the ends of the prism in
horizontal straight lines of
which ah is the plan and a'h'
and a'h' the elevations. The
rectangle a'b'h'a' is the com-
plete elevation of the section.
RS is the vertical trace of
another plane of section. This
second plane of section is per- Fig. 471.
pendicular to the vertical plane
of projection. On the plan (1) this second section appears as the figure
cdec. Another plan (3) of this second section is shown on a ground
line XiYj parallel to RS. The plan (3) shows the true form of the
section of the prism by the plane RS. The whole solid is also shown
in plan (3), the part above the section by the plane RS being repre-
sented by thin dotted lines
209. Section of a Pyramid by a Plane perpendicular to
one of the Planes of Projection. — The first paragraph of the pre-
ceding Article on the section
of a prism applies also to the
section of a pyramid, and all
that need be done here is to
give an example which the
student should work out.
In Fig. 472, (1) is the plan
and (2) the elevation of a right
hexagonal pyramid when stand-
ing on the horizontal plane
with two sides at right angles
to XY. (3) is a plan of the
pyramid when one triangular
face is horizontal. The plan
(3) is projected from the eleva-
tion (2) as shown. (4) is an
elevation projected from (3) on
a ground line X0Y2 parallel to
XY. The straight" line PQ is
the horizontal trace of a plane
Fig. 472.
of section which is perpendicular to the plane of the plan (3). PQ
is parallel to XY and passes through c, the plan (3) of one angular
point of the base of the pyramid. The section of the pyramid by the
i
SECTIONS OF SOLIDS
243
plane PQ is shown projected on to each of the other views of the
pyramid, and in each view the part of the solid between the vertex
and the section is shown by thin dotted lines. The side of the hexa-
gonal base may be taken 125 inches long.
210. Sections of Mouldings. — A moulding is an ornament of
uniform cross section which may be formed on a piece of a structure
or it may be a separate piece attached to the structure for ornament
only, or it may be the main part of the structure as in a picture frame.
Mouldings are of frequent occurrence in cabinet making, joinery, and
masonry.
The principal problem in the geometry of mouldings is the determina-
tion of the cross section of one moulding which will mitre correctly
Fig. 473.
Fig. 474.
with another when faces of the one are in different planes from corre-
sponding faces of the other.
Given the cross section of a moulding the determination of any other
section is a simple problem and is worked exactly as for a prism. In
Fig. 473, (1) is the true form of the cross section of a straight piece of
moulding. (2) is a plan of the moulding, the long edges being horizontal.
PQ is the horizontal trace of a vertical plane of section, and (3) is the
elevation of the section PQ on a ground line X, Yj parallel to PQ.
The same moulding is shown in Fig. 474. (1) is the cross section,
(2) is a side elevation, and (3) is a plan when the long edges are parallel
to the vertical plane of projection but inclined to the horizontal plane.
RS is the horizontal trace of a vertical plane of section, a'h' is the
projection of the section RS on the plane of the elevation (2), and a/fc/
is a projection of the same section on a plane parallel to it and which
therefore shows the true form of the section.
244
PRACTICAL GEOMETRY
In Fig. 474 the projections (1), (2), (3), and (4) have been drawn
in the order in which they are numbered, but a study of the figure will
show that if the section a/6/ by the plane RS be given instead of the
cross section (1) the latter may be found by working backwards.
In Fig. 475 the section a! is the cross section and a is the plan of a
straight piece of moulding whose long edges are horizontal and which
is fixed to the face of a vertical wall whose plan is rs. h is the plan
and h' the elevation of another straight piece of moulding whose long
edges are inclined to the horizontal plane and which is fixed to the
face of a vertical wall whose plan is st. The faces of the two walls are
at right angles to one another, and the two mouldings intersect in a
vertical plane, whose horizontal trace, or plan, is the straight line ns
which bisects the angle between sr and st. A moulding such as hh'
///////////////A
"T'^y.*-- -/»
<
Fig. 475.
Fig. 476.
whose long edges are inclined to the horizontal is called a raking
moulding.
The problem is now to find the true form of the cross section of the
raking moulding. The condition which the raking moulding must
satisfy is that its section by the plane ns of the joint with the other
moulding must be the same as the section of the other moulding by
that plane. From this condition it follows that the corresponding
longitudinal edges or lines on the two mouldings must intersect in the
plane of the joint. The projections of the longitudinal lines on the
raking moulding may therefore now be drawn and the cross section
c' determined as shown.
The case illustrated by Fig. 470 differs from that illustrated by
Fig. 475 in that the angle rat between the faces of the walls in Fig. 47G
is greater than a right angle. «,', the cross section of the horizontal
SECTIONS OF SOLIDS
245
Fig. 477.
Fig. 477 shows a curved
moulding, is shown rabatted into the vertical plane of projection. It
will be seen that in order that the elevations of the longitudinal lines
on the raking moulding may be drawn the elevation n's' of the joint
must first be determined.
Two straight 'mouldings, of the same cross section, may be mitred
correctly together if they are
attached to a plane surface, or if
they are attached to plane sur-
faces inclined to one another,
provided that the longitudinal
lines of the two mouldings are
perpendicular to the line of in-
tersection of the plane surfaces
to which they are attached.
Two curved mouldings of the
same cross section but of different
curvatures and attached to a
plane surface may be jointed to-
gether but the joint is a curved surface
moulding A jointed to a straight moulding B both having the same
cross section C. The two mouldings are supposed to be attached to
the plane of the paper. The projection of the joint on the plane of
the paper is the curved line DEF determined as shown.
211. Geometry of Rafters. — Practical problems on the sections
of prisms occur in timber construction, and illustrations will now be
given of these problems as applied to a timber roof frame.
Fig. 478 shows plan and elevation line diagrams of part of what is
called a hipped roof. The various members of the frame of this roof
are mainly of rectangular cross section. vUj v'u' is the ridge piece, vr, v'r'
is the hip or hip rafter which passes from one end of the ridge piece to
the top corner of two intersecting walls, cc?, c'd' is one of the jacic
rafters which lie between the hip rafter and the wall plate on the top
of one of the walls, ef ef is one of the common rafters which lie
between the ridge piece and a wall plate.
The dihedral angle between the two roof surfaces intersecting in
the line VR is determined by the construction of Art. 191, p. 224, or
by that construction slightly modified as shown here, vv^' is drawn
at right angles to vr and is made equal to the height of v' above /.
Joining 7;/r determines the true length of VR. mn is drawn at right
angles to vr to meet the horizontal traces of the roof surfaces at the
level of R at m and n. mn is the horizontal trace of a plane at right
angles to VR. This plane intersects the roof surfaces whose line of
intersection is VR in two straight lines the angle between which is
the angle required, os' is drawn at right angles to v^'r and oS is made
equal to os'-. Joining S to m and n determines mSw the dihedral angle
required.
At h and k are shown two forms of the cross-section of the hip
rafter to an enlarged scale. TP and TQ are parallel to Sm and Sr?.
respectively. This determines the angle ^ which the carpenter requires
246
PRACTICAL GEOMETRY
in backing off* the top face of the rafter to bring it into the planes or
plane of the roof surface.
Fig. 479 shows in detail a plan (1), a side elevation (2) and a front
elevation (3) of the upper part of a jack rafter where it joins the hip
rafter. All the rafters have their side faces vertical, and XoY., is the
plan of the vertical face of the hip rafter to which the jack rafter is
Fig. 478.
Fig. 479.
attached. This end of the jack rafter requires to be bevelled in two
directions in order to lit against the face of the hip rafter. The angles
for these bevels are marked a and /?. The angle y8 is shown on the side
elevation (2). To determine a the upper surface of the rafter is brought
into a horizontal position by turning it about a horizontal axis through
B at right angles to XY, as shown in the plan (1) and elevation (2).
The true form of the end of the rafter is shown at (4).
Exercises XVII
1. The solid whose projections are given in Fig. 480 is cut in two by a vertical
plane whose horizontal trace is HT. Draw an elevation of the part to the left
of the plane of section on a ground line parallel to HT.
2. The solid whose projections are given in Fig. 481 is cut in two by a vertical
plane whose horizontal trace is PQ. Draw an elevation of the larger portion on
a plane parallel to the plane of section,
3. The solids whose projections are given in Fig. 482 are cut by a vertical plane
SECTIONS OF SOLIDS^
247
whose horizontal trace is RS. Draw an elevation of the portions whose plans
lie between RS and XY on a ground line parallel to RS.
4. The plan of a right square prism with a rectangular hole through it is given
in Fig. 483. A side of the square base is on the horizontal plane. Draw the
s
^
Jl\
^>.
^
"^
y. "^
A^
^w
s
X
1
s
Y
1
^\
■J
uk
21
^
2
S
7
s
/
V
jZ
\
s
/
^
'
s
/
X
\
/
Y
JT"
^H,^
.
1
s.
LLI
tJ
^
-
-
Fig. 480. Fig. 481. Fig. 482. Fig. 483.
In rep7oducing the above diagrams take tlie small squares as of 0-3 inch side.
a'
V
elevation on XY and show on this elevation the section of the solid by the vertical
plane of which HT is the horizontal trace. Determine also the true form of the
section.
5. A right square prism, side of base 1 inch, altitude 3 inches, is cut into two
equal parts by a plane. The true form of the section is a rhombus of 1-5 inches
side. Draw a plan of one part when it stands with its section face on the
horizontal plane.
6. A pyramid has for its base a square of 2 inches side which is on the
^horizontal plane. The altitude of the pyramid is 2*5 inches and the plan of the
vertex is at the middle point of one side of the plan of the base. This pyramid
is cut into two portions by a horizontal plane which is 1 inch above the base.
Draw the plan of the lower portion.
7. A right hexagonal pyramid, side of base 1 inch, altitude 2 inches is cut by
a plane which contains one edge of the
base and is perpendicular to the opposite
face. Determine the true form of the
section. ,
8. The plan and elevation of one of the
rectangular faces ABCD of a right hexa-
gonal prism are given in Fig. 484, AB and
CD being on the ends of the prism. Com-
plete the plan and elevation of the prism
and show on the elevation the section by
the vertical plane whose horizontal trace
isPQ.
9. The plan and elevation of one of the
triangular faces VAB of a right square
pyramid are given in Fig. 485, V being the
vertex of the pyramid. Complete the plan
and elevation of the pyramid and show on
the elevation the section by the vertical
plane whose horizontal trace is RS.
10. A horizontal moulding has (e), Fig. 486, for its cross-section. Determine
the true form of a vertical section which is inclined at 45° to the longitudinal
lines of the moulding.
11. A moulding whose cross-section is given at (/) in Fig. 486 is placed with
its back on the vertical plane of projection and its longitudinal lines inclined at
80° to the horizontal plane. Draw the plan and elevation and show the true
form of a section by a vertical plane whose horizontal trace is inclined at 60°
to XY.
12. Two mouldings of the same cross-section (c), Fig. 486, are fixed to the face
J
"^
7
f
f
J
./
^d
f
cJ
\
X
\f
?;
y
I
p
J
/
Q
A
/
/
V
N
/
c
Fig. 484.
Fig. 485.
Tahe small squares 0'4 inch side.
u^
I^RACTICAL GEOMETRY
of a vertical wall. One moulding is horizontal and the other rakes at 30° to the
horizontal. The two mouldings join correctly together. Determine the true
form of the joint section.
(a) (b) (c) (d) (c) if) (g)
m
Fig. 486.
In reproducing the above sections take the small squares as of \ inch side.
13. A horizontal moulding mitres correctly with a raking moulding. Inclina-
tion of raking moulding to the horizontal, 25°. The backs of the mouldings are
in vertical planes which are at right angles to one another as
shown in plan in Fig. 487. Select from Fig. 486 a cross- ^
section for the horizontal moulding and then determine the"
cross-section of the raking moulding.
14. Same as exercise 13 except that the angle between
the vertical planes is 120° instead of 90°.
15. Two raking mouldings on vertical walls which are at
right angles to one another as in Fig. 475, p. 244, mitre cor-
rectly. Both mouldings are inclined at 20^ to the horizontal.
Select a cross-section for one of the mouldings from Fig. 486
and then determine the cross-section of the other.
16. Same as exercise 15 except that the angle between the walls is 120°
instead of 90°.
17. A, C, and E (Fig. 488) are three pieces forming part of a timber frame,
each piece being of rectangular cross-section. The projections a' and c are
incomplete. Complete the projections indicated and draw an elevation on a
ground line inclined at 45° to XY.
18. A piece of timber having plane faces is shown in plan and elevation in
Fig. 489. Draw two other elevations, one on a ground line perpendicular to XY,
and the other on a ground line parallel to the longer sides of the plan. Determine
the true form of the cross-section of the piece.
Fig. 487.
X
^
-^
\
-
N
ov
s.
s
f^
^
^
X
y
>
^
X
V
y
Y
T
1
S
a
:
c
\e, 1
^
',m^
^\\^\v\\\\\\\\\\\\\\\\\\\\''
if
\
s
y"
^
y
^
r
^<;
r
>s
\
^
^
^^
^
X
\
(^
\
v^
r
a
^
4^
\
\
^
X
X.
X
>^
Fig. 488. Fig. 489. Fig. 490.
In reproducing the above diagrams take the small squares as of half-inch side,
19. Fig. 490 shows the plan of a piece of timber A, of uniform square cross-
section, which lies between the faces RT and ST of two vertical walls. The
longitudinal edges of A are inclined at 20° to the horizontal, rising from R to S.
B is another piece of timber of uniform square cross-section. The longitudinal
edges of B are horizontal. The piece B fits into a notch on A, the greatest vertical
depth of this notch being equal to half the thickness of A. Draw two elevations
of A and B, one on a ground line parallel to rs, and the other on a ground line
parallel to rt. Determine all the bevels for the ends and the notch of the piece A.
CHAPTER XVIII
THE SPHERE, CYLINDER, AND CONE
212. Surfaces and Solids. — When a surface encloses a space
and that space is filled with solid matter the whole is called a soli<^,
and when the surfaces and solids have certain definite shapes they
Imve certain definite names. But in speaking of certain solids and
their surfaces the same name is frequently used to denote the solid
and its surface. For example, in speaking of the sphere, the cylinder,
and the cone the solids bearing these names may be referred to or it
may be that it is their surfaces that are referred to, although the
words " solid " or " surface " may not be used. In general the omission
of the words " surface of" when a solid is mentioned by name will not
load to any confusion when it is the surface that is referred to. For
example, by " the curve of intersection of two cylinders" is obviously
meant " the curve of intersection of the surfaces of two cylinders."
213. The Sphere. — The sphere may be generated by the revolution
of a semicircle about its diameter which remains fixed. The surface
of a sphere is also the locus of a point in space which is at a constant
distance from a fixed point called the centre of the sphere. The constant
distance of the surface of the sphere from its centre is the radius of
the sphere. A straight line through the centre of the sphere and
terminated by the surface is a diameter of the sphere.
The orthographic projection of a sphere is always a circle of the
same diameter as the sphere.
214. Plane Sections of a Sphere. — All plane sections of a
sphere are circles. When the plane of section contains the centre of
the sphere the section is a great circle of the sphere. All sections of
the sphere by planes which do not contain its centre are called small
circles.
A portion of a sphere lying between two parallel planes is called a
zone^ and a portion lying between two planes containing the same
diameter is Called a lune. See Fig. 491.
The determination of a plane section of a sphere is illustrated by
Figs. 492 and 493. In Fig. 492 the plane of section is perpendicular
to the vertical plane of projection and has V.T. for its vertical trace.
In Fig, 493 the plane of section is inclined to both planes of projection
and has V.T. for its vertical trace and H.T. for its horizontal trace.
In both cases points on the projections of the section may be found
250
PRACTICAL GEOMETRY
as follows. PQ parallel to XY is taken as the vertical trace of a
horizontal plane. This plane cuts the sphere"" in a circle whose plan is
a circle concentric with the plan of the sphere and whose diameter
is equal to the portion of PQ within the circle which is the elevation
of the sphere. This same plane cuts the given plane of section in
a straight line. The points rr in which the plan of this line intersects
the plan of the circle are the plans of points on the required inter-
section, and r'r' the elevations of these points are on PQ. In like manner
by taking other positions for PQ any number of points on the projections
of the required intersection may be found.
Since the required intersection is a circle and the projections of a
circle are ellipses the axes of these ellipses may be found and the
ellipses then be drawn by the trammel method (Art. 45, p. 41).
Referring to Fig. 492 a'a' the intercept of V.T. on the elevation of
; LUNE
Fig. 491.
Fig. 492.
the sphere is the elevation of the circle which is the section of the
sphere by the given plane. Projectors from a' and a! to meet a parallel
to XY through o the plan of the centre of the sphere determines aa
the minor axis of the ellipse which is the plan of the required section.
The major axis bisects the minor axis at right angles and has a length
equal to a!a'.
Referring to Fig. 493, draw aah through o at right angles to H.T.
aah is the horizontal trace of a vertical plane which intersects the
sphere in a great circle and the given plane of section in a straight
line, and A A the portion of this line within the great circle has for
its plan the minor axis of the ellipse which is the plan of the required
section. Let this vertical plane with the line AAB and the great
circle in it be turned round about a vertical axis through O until it is
parallel to the vertical plane of projection. AjAjBi is the elevation of
the line AAB in its new position and the elevation of the sphere is the
THE SPHERE, CYLINDER, AND CONE
251
elevatioQ of the new position of the great circle. This determines
the levels of the points A and A on the given plane of section and the
plans a and a are then readily found, aa is the minor axis of the
ellipse, the major axis of which is equal to AjAj. By a similar
construction, using a plane perpendicular to the vertical plane of
projection with its vertical trace through o' at right angles to V.T., the
axes of the ellipse in the elevation may be found. It should be
remembered that the major axes of these ellipses are parallel to the
horizontal and vertical traces respectively of the given plane of section
and the minor axes are perpendicular to these traces.
By taking a horizontal section of the sphere and given plane of
section through the centre of the sphere the points 8 and s where the
ellipse in the pla-n touches the circle which is the plan of the sphere
are determined (Figs. 492 and 493). These points determine the
limits of the portions of the required section which are on the upper
and lower halves of the sphere respectively. In like manner (Fig. 493)
a section of the sphere and given plane by a plane parallel to the
vertical plane of projection and through the centre of the sphere
determines n' and w' where the ellipse in the elevation touches the
circle which is the elevation of the sphere. And these points determine
the limits of the portions of the required section which are on the front
and back halves of the sphere respectively.
Fig. 494.
Fig. 495.
Useful exercises on the projection of plane sections of the sphere
are illustrated by Figs. 494 and 495. In Fig. 494 the sections, except
the middle one, are small circles of the sphere, while in Fig. 495 they are
all great circles. The different views in each Fig. should be drawn in
the order in which they are numbered. The sphere may be taken,
say, 2-5 inches in diameter.
Another useful exercise is illustrated by Fig. 496. This is a
hexagonal nut with a spherical chamfer. The various curves, other
than those representing the hole in the nut, are plane sections of a
sphere whose centre is O and radius R. In this example the various
252
PRACTICAL GEOMETRY
Fig
views are arranged according to the American system (p. 169). In
the elevations (2 ) and (3) the elliptic arcs a' and h' are usually drawn
as arcs of circles. The
arc c in the elevation
(3) is a true arc of a
circle whose centre is d.
In working this example
the following dimen-
sions may be taken. —
Diameter of screw, 1*75
inches; width of nut
across the flats, 2*75
inches; height of nut,
1'75 inches; radius of
spherical chamfer, 2*1
inches.
215. Projections
of Points on the Surface of a Sphere. — Suppose that one pro-
jection of a point on the surface of a sphere to be given and that it
is required to find the other. Take a plane parallel to one of the
planes of projection and containing the point. The section of the
sphere by this plane will have for one projection a circle and for the
other a straight line, and the given projection of a point will lie on
one of these and the other on the other.
216. Intersection of a Straight Line and a Sphere.— Take a
plane perpendicular to one of the planes of projection and containing the
given line. Draw a projection (A) of the section of the sphere by this
plane on a plane (B) parallel to it. Draw also on the plane (B)
another projection (C) of the given line. (A) is a circle which
intersects (C) at points which are projections of the points of inter-
section of the line and sphere. From these the projections of the
points of intersection on the original planes of projection may be
found.
217. The Cylinder. — The cylinder may be generated by a straight
line which moves in contact with a fixed curve and remains parallel
to a fixed straight line. If the fixed curve is a circle whose plane
is at right angles to the fixed straight line, the cylinder is a right
circular cylinder, and a straight line through the centre of the fixed
circle at right angles to its plane is the axis of the cylinder. The fixed
circle referred to above is a normal section of the right circular cylinder,
and its diameter is the diameter of the cylinder.
When a cylinder is limited in length by two planes which intersect
all positions of the generating line, the sections of the cylinder by these
planes become the ends of the cylinder, and when these ends are at right
angles to the generating line the cylinder is a right cylinder. In
speaking of a right circular cylinder, when its ends are considered, the
ends are generally understood to be at right angles to the axis. But
a right circular cylinder may have its ends inclined to its axis at an
angle which is not a right angle.
THE SPHERE, CYLINDER AND CONE
253
218. Projections of a Right Circular Cylinder. — Starting
with the cylinder in the position in which its axis is vertical, its plan
is a circle and its elevation a rectangle as shown at (1) and (2) in
Fig. 497.
Taking a second ground line XiY, inclined to rV the first elevation
of the axis of the cylinder, and projecting from (2), a new plan (3) is
obtained. In this new plan (3), rs, the plan of the axis of the cylinder,
is parallel to XiYi and at a distance from it equal to the distance of
y,s in (1) from XY. Also the plans of the ends of the cylinder are equal
ellipses whose major axes are at right angles to rs and equal in length
to the diameter of the cylinder. The minor axes bisect the major axes
at right angles and their lengths are found by projectors from (2) as
shown.
Taking a third ground line X2Y2 inclined to rs in the plan (3), r's',
a new elevation of the axis of the cylinder is first obtained, the
distances of r' and s' in (4) from XgYg being equal to the distances of
Fig. 497.
r' and s' respectively in (2) from XjY,. Before going further with the
elevation (4), take another ground line X^Y., parallel to rs in (4) and
draw rs a new plan of the axis of the cylinder as shown in (5). The
distances of r and s in (5) from X.^Y.^ are equal to the distances of r and
8 respectively in (3) from XgYg. The new plan (5) being on a plane
parallel to the axis of the cylinder will be a rectangle as in (2) and
may therefore be now drawn. The axes of the ellipses in (4) are
next determined by projectors from (5) as shown, just as the axes of
the ellipses in (3) were obtained by projectors from (2).
Instead of finding the axes of the various ellipses which are the
projections of the ends of the cylinder and then constructing the
ellipses in the usual way, a number of points on the ends may be pro-
jected as is shown for one point N.
If rs in (3) and r's in (4) be given and also the diameter of the
cylinder, it is obvious that by drawing the projections (2) and (5) the
projections (3) and (4) may be completed.
219. Plane Sections of a Right Circular Cylinder. — All
254
PRACTICAL CxEOMETRY
sections of a right circular cylinder by planes at right angles to its
axis are circles of the same diameter, and all sections of the curved
surface by planes parallel to the axis are parallel straight lines. All
sections of the curved surface by planes inclined to the axis are ellipses
whose centres are on the axis of the cylinder.
Referring to Fig. 498, PQ is the vertical trace of a horizontal plane
cutting the inclined cylinder whose axis SSj is parallel to the vertical
plane of projection. S and Sj are the centres of two spheres inscribed
in the cylinder and touching the plane of section at F and.Fj respec-
tively. The points F and Fi are the foci of the ellipse which is the
section of the cylinder by the given plane of section, a'a' the portion
of PQ lying within the elevation of the cylinder is the elevation of the
major axis of the ellipse and from this the plan aa, is projected. The
Fig. 498.
Fig. 499.
minor axis libx of the plan of the ellipse is equal to the diameter of the
cylinder.
Planes through S and Sj at right angles to the axis of the cylinder
intersect the given plane of section in straight lines XM and XiMj
which are the directrices of the ellipse (Art. 35, p. 32).
Fig. 499 differs from Fig. 498 in that the axis of the cylinder is
inclined to both planes of projection instead of being inclined to the
horizontal plane only. The intercept of J*Q within the elevation of
the cylinder does not now give the major axis of the ellipse, but the foci
and the minor axis are determined as before, but by making ca and ca^
each equal to hf, the major axis is determined.
The determination of the section of a cylinder, whose axis is
THE SPHERE, CYLINDER, AND CONE
255
inclined to both planes of projection, by an oblique plane is illustrated
by Fig. 500. AB is the axis of the cylinder and H.T. and V.T. are
the traces of the plane of section.
The ellipse uv, which is the
horizontal trace of the cylinder, is
the section of the cylinder by the
horizontal plane of projection and
it may be determined by the con-
struction shown in Fig. 499, or by
the construction shown in Fig.
503.
Taking mw, parallel to ah, as
the horizontal trace of a vertical
plane, this plane intersects the
given plane of section in a line of
which m'n' is the elevation. This
same vertical plane intersects the
cylinder in two straight lines
whose elevations c'd' and c/dfj' are
parallel to a'b', the points c' and c/
being projected from c and Cj, the
points where mn cuts the ellipse uv.
The points r' and s' where m'n
intersects c'd' and f/c?/ are the
elevations of points on the required
section. Projectors from r' and s
to meet mn determine points r and s on the plan of the required
section.
By taking other vertical planes parallel to the axis of .the cylinder
other points on the required
section may be found.
It should be noted that
all the lines of intersection of
the assumed vertical planes
with the given plane of section
will be parallel.
The student should work
out this example to the dimen-
sions marked on the figure.
220. Circular Sections
of a Right Elliptical
Cylinder. — A right elliptical
cylinder is one in which a
section by a plane at right
angles to the axis of the
cylinder is an ellipse. In Fig.
501, (?«) is the plan and (y) the elevation of such a cylinder, the axis
of the cylinder being perpendicular to the vertical plane of projection,
and the major axis of a section at right angles to the axis is vertical.
Fig. 500.
Fig. 601.
Fig. 502.
256
PRACTICAL GEOMETRY
The elevation (v) shows the true form of the normal section and is an
ellipse whose semi-major axis is o'a'.
Taking a point o on the plan of the axis as centre and a radius
equal to o'a' aji arc of a circle is described cutting the outlin-e of the
plan which is parallel to the plan of the axis at r and s. The straight
line rs is the horizontal trace of a vertical plane which will cut the
cylinder in a circle whose diameter is equal to the major axis of the
normal section. An elevation (w) of the circular section on a ground
line XiYi parallel to rs is shown. All sections of the curved surface
of the cylinder by planes parallel to one circular section are equal
circles.
The property of a right elliptical cylinder of having circular
sections makes it possible to bore such a cylinder with an ordinary
boring bar as shown in Fig. 502. The axis of the boring bar is set at
an angle to the axis of the cylinder to be bored, being such that
cos = y., where d is equal to the minor axis and D is equal to the
major axis of the normal elliptic section of the cylinder. While
the boring bar is rotated the cylinder is moved in the direction of its
axis, or if the cylinder is stationary the boring bar is moved bodily in
the direction of the axis of the cylinder. It is evident that with the
arrangement sketched in
Fig. 502 the length of
cylinder which may be
bored is limited, the limit
depending on the angle
and the diameter of the
boring bar.
221. Cylinder En-
veloping a Sphere. —
A cylinder which en-
velops a sphere will have
its axis passing through
the centre of the sphere
and the curve of contact
will be a "great circle"
of the sphere whose plane
is perpendicular to the
axis of the cylinder.
Referring to Fig. 503,
oo' is the centre of the
sphere and mn, m'n' is the
axis of the cylinder.
The sphere and the direc-
tion of the axis of the
cylinder are supposed to
be given. In problems
requiring a cylinder to envelop a sphere the cylinder is generally of
indefinite length and the ends are not required to be shown but
i
X i
\'^
m
Fig. 503.
THE SPHERE, CYLINDER, AND CONE
257
usually the trace of the surface on one of the planes of projection is
necessary. Neglecting the ends, the projections of the cylinder will
consist of the tangents to the projections of the sphere parallel to the
projections of the axis of the cylinder. Fig. 503 shows the construc-
tion for finding the horizontal trace of the cylinder and the plan of the
circle of contact. An elevation of the cylinder and sphere is drawn
on mn as a ground line, w/w is the new elevation of the axis of the
cylinder and 0/ that of the centre of the sphere. The points a and 6,
where the tangents to the circle which is the new elevation of the
sphere, parallel to wi/w, meet mn, are the extremities of the major axis
of the ellipse which is the horizontal trace of the cylinder. The
minor axis of this ellipse passes through n, the horizontal trace of the
axis of the cylinder, and is equal to the diameter of the cylinder or
sphere.
The line c/d/, passing through 0/, and perpendicular to m/»i is the
new elevation of the circle of contact. Perpendiculars from c/ and d^'
to mn determine cd, the minor axis of the ellipse which is the plan of
the circle of contact. The major axis of this ellipse is a diameter of
the circle which is the plan of the sphere and is at right angles to mn.
The elevation of the circle of contact is obtained by making an
auxiliary plan of the cylinder and sphere on a plane parallel to the
axis of the cylinder, say on m'n' as a ground line. The construction
is similar to that already described for the plan. The vertical trace
of the cylinder may also be obtained from this same auxiliary plan.
222. Projections of Points on the Surface of a Right
Circular Cylinder. — Suppose that one projection, say the plan r of
a point R, on the surface
of a right circular cylinder
to be given and that it is
required to find the other.
Take an elevation of the
cylinder on a vertical
plane parallel to its axis.
Take any point 00' (Fig.
504) on the axis of the
cylinder as the centre of
a sphere inscribed in the
cylinder. Through r draw
a straight line rs parallel
to the plan of the axis of
the cylinder and assume
this to be the horizontal
trace of a vertical plane
which is parallel to the axis of the cylinder. This vertical plane
cuts the sphere in a circle whose diameter is equal to ac the intercept
of rs on the plan of the sphere. The elevation of this circle is a
circle with its centre at o'. This same vertical plane cuts the curved
surface of the cylinder in, two straight lines whose elevations are
tangents to the circle just mentioned and are parallel to the elevation
S
Fig. 504.
Fig. 505.
258 PRACTICAL GEOMETRY
of the axis of the cylinder. There are two points on the surface of
the cyUnder having the point r for their plan. The elevations of
these points are on the projectors from r and on the lines which are
the elevations of the lines of intersection of the assumed vertical
plane and the cylinder.
Another solution of the problem is shown in Fig. 505. A sphere
is inscribed in the cylinder as before and a second cylinder, with its
axis vertical, is made to envelop this sphere. The two cylinders inter-
sect in two ellipses whose elevations on the vertical plane parallel to
the axes of the cylinders are the straight lines mV and s't'. The points
whose plan is r are evidently on the intersection of the two cylinders
and their elevations are found as shown.
223. Intersection of a Straight Line and a Right Circular
Cylinder. — First obtain from the given projections of the line and
cylinder projections of them, on planes parallel and perpendicular
respectively to the axis of the cylinder. Next take a plane parallel
to the axis of the cylinder and containing the given line. This plane
will intersect the curved surface of the cylinder in two straight lines
which will intersect the given line at the points of intersection required.
Working backwards to the original projections the required projections
of the points of intersection of the line and cylinder are found.
224. The Cone. — The cone may be generated by a straight line
which moves in contact with a fixed curve and passes through a fixed
point. The fixed point is the vertex of the cone. If the fixed curve is
a circle and the fixed point is on the straight line which passes through
the centre of the circle and is at right angles to the plane of the circle,
the cone is a right circular cone. The fixed circle is a circular section
of the right circular cone. The straight line joining the vertex of a
right circular cone to the centre of a circular section is the axis of the
cone. The right circular cone may also be defined as the surface
described by a straight line which intersects a fixed line at a fixed
point and moves so that its inclination to the fixed line is constant.
The fixed line in this case is the axis of the cone.
If a right circular cone terminates at one end at the vertex and
at the other end at a circular section, that circular section is called
the hase of the cone.
225. Projections of a Right Circular Cone. — Starting with
the cone in the position in which its axis is vertical, its plan is a
circle and its elevation is an isosceles triangle as shown at (1) and (2)
in Fig. 506.
Taking a second ground line XjYj inclined to v'r' the first elevation
of the axis of the cone, and projecting from (2), a new plan (3) is
obtained. In this new plan (3), vr the plan of the axis of the cone,
is parallel to X,Yi and at a distance from it equal to the distance of
vr in (1) from XY. Also the plan of the base of the cone is an ellipse
whose major axis is at right angles to vr and equal in length to the
diameter of the base of the cone. The minor axis bisects the major
axis at right angles and its length is found by projectors from (2) as
shown.
THE SPHERE, CYLINDER, AND CONE 259
Taking a third ground line XoYg inclined to vr in the plan (3),
rV, a new elevation of the axis of the cone is first obtained, the distances
of v' and r' in (4) from XgYg being equal to the distances of v' and /
respectively in (2) from XiYj. Before going further with the eleva-
tion (4), take another ground line X3Y3 parallel to v'r' in (4) and
draw vr a new plan of the axis of the cone as shown in (5). The
distances of v and r in (5) from X3Y3 are equal to the distances of v
and r respectively in (3) from X^Yg. The new plan (5) being on a
plane parallel to the axis of the cone will be an isosceles triangle as
X
Fig. 506.
in (2) and may therefore now be drawn. The axes of the ellipse in
(4) are next determined by projectors from (5) as shown, just as the
axes of the ellipse in (3) were obtained by projectors from (2).
Instead of finding the axes of the various ellipses which are the
projections of the base of the cone and then constructing the ellipses
in the usual way, a number of points on the base may be projected as
is shown for one point N.
If vr in (3) and v'r' in (4) be given and also the diameter of the base
of the cone it is obvious that by drawing the projections (2) and (5)
the projections (3) and (4) may be completed.
226. Plane Sections of a Right Circular Cone. — In what
follows only the curved surface of the cone is considered, the base
being supposed to be beyond the limits of the section or the part of
the section represented. When the term '* cone " is used " right
circular cone " is to be understood.
In studying the sections of the cone it is necessary to consider that
the straight line which generates the surface is of unlimited length
and consequently that the vertex of the cone is not at one end of the
generating line. The cone then consists of two sheets generated by the
parts of the generating line which are on opposite sides of the vertex.
The two sheets of a cone are exactly alike and the axis of one is the
axis of the other produced. The two sheets of a cone are shown in
Fig. 510.
The following are the various sections of the cone. (1) Two
260
PRACTICAL GEOMETRY
straight lines when the plane of section passes through the vertex of
the cone. (2) The circle when the plane of section is perpendicular to
the axis of the cone. (3) The ellipse when the plane of section cuts
all the generating lines on the same side of the vertex. (4) The
hyperbola when the plane of section cuts both sheets of the cone and
does not pass through the vertex. (5) The parabola when the plane
of section is parallel to a tangent plane to the cone, or, to give a
common but less exact definition, when the plane of section is parallel
to the slant side of the cone.
The circle (2) is a particular case of the ellipse (3), and two straight
lines (l)area particular case of the hyperbola (4). As the plane of
Fig. 507.
Fig. 508.
section turns round from a position which gives an ellipse to a position
which gives an hyperbola it passes through the position which gives
a parabola. Sections (1) and (4) are those which lie on both sheets of
the cone.
The sections which will now be considered are, the ellipse, the
parabola, and the h)^perbola.
Taking the elliptic section first and referring to Fig. 507, a cone,
whose axis VS is inclined to the horizontal plane but is parallel to the
vertical plane of projection, hi cut by a horizontal plane of which PQ
is the vertical trace. S anvd Sj are the centres of two spheres
inscribed in the cone and touching the plane of section at F and F,
THE SPHERE, CYLINDER, AND CONE 261
respectively. The points F and Fj are the foci of the ellipse which is
the section of the cone by the given plane of section, a'a^y the portion
of PQ lying within the elevation of the cone, is the elevation of the
major axis of the ellipse and from this the plan aa^ is projected. The
minor axis h\ of the plan of the elUpse bisects aa^ at right angles at c.
The length of the minor axis may bo determined by taking / as centre
and a radius equal to ac and describing arcs to cut hh^ at h and h^.
Planes containing the circles of contact between the cone and
the inscribed spheres intersect the given plane of section in straight
lines XM and XjMi which are the directrices of the ellipse (Art. 35,
p. 32).
Fig. 508 differs from Fig. 507 in that the axis of the cone is
inclined to both planes of projection instead of being inclined to the
horizontal plane only, w'w/ the intercept of PQ within the elevation of
the cone does not now give the major axis of the ellipse, but the foci
are determined exactly as before. A projector from n' to the plan
will however be a tangent to the plan of the ellipse. From / draw fe
at right angles to this tangent, meeting it at e. With centre c, the
middle point of ffy, and with a radius equal to ce describe arcs to cut ff^
produced at a and a^. Then aa^ is the major axis of the plan of the
ellipse. This follows from the fact that the foot of a perpendicular
from a focus of an ellipse on to a tangent lies on the auxiliary circle
(Art. 42, p. 39). The minor axis hh^ may now be determined as
before.
Consider next a parabolic section, which is illustrated by Fig. 509.
The cone is placed so that it is tangential to a horizontal plane and
the axis is parallel to the vertical plane of projection. PQ is the
vertical trace of a horizontal plane of section. S is the centre of the
sphere which is inscribed in the cone and which touches the plane of
section. Only one such sphere can be drawn in this case. The
point F at which the inscribed sphere touches the plane of section is
the focus of the parabola and A is the vertex.
The plane of the circle of contact between the cone and the
inscribed sphere intersects the plane of section in the straight line XM
which is the directrix of the parabola. It will be found that ax is
equal to af. The parabola may now be constructed as described in
Art. 34, p. 30, or as follows. Take a point o' on the elevation of the
axis of the cone. Draw o'd! at right angles to v'o' to meet the outline
of the elevation of the cone at d' and PQ at /. With o' as centre
and o'c?' as radius describe an arc to meet a straight line r'R^ parallel
to v'o' at Rp Through r' draw a projector to the plan cutting the
plan of the axis of the cone at t. On this projector make >' ' / \ ^^
\/ / ^yi \ ^^
\/^'Ao ^-s^
v ^ .^^ ^ / ^^w.
v--"'^ ' ^ / ^
>^
V ^ ^ /
^\^
^"v.
^*"\^
\ ' ^ '
^v
\ ' \n
>— , ^^
— -^
/^^ "y^-r^-—.^
T^i— -^
\
Jr-ky] \ \ 1
^-V
N. \ / \ \ yXiV^/ 1 ^ /
^ ■' \
\
\
^^k/yP^ \ "- ^ - / X
"■^v^..^^^ ',
\
\
\
/IHvy ) ^ ^
7^
\
\
\
\/-'\K)J
Fig. 511.
elevation of the cone and sphere is drawn on the plan vn of the axis of the
cone as a ground line, v^n is the new elevation of the axis of the cone
and o/ that of the centre of the sphere. The points a and I where the
tangents from v/ to the circle which is the new elevation of the sphere
meet vn are the extremities of the major axis of the ellipse which is the
horizontal trace of the cone. The minor axis of this ellipse must of course
bisect ah at right angles, but it should be noticed that the middle point
of ah is not at n the horizontal trace of the axis of the cone. To deter-
mine the length of the minor axis of the ellipse which is the horizontal
trace of the cone the construction is as follows. Through e the middle
point of ah draw ,^/e/i/ at right angles to v^n meeting v^n at gx and
264-
PRACTICAL GEOMETRY
Vyh at 111. With centre gl and radius glh^ describe the arc A// and
draw ef at right angles to eA/ to meet this arc at /. ef is equal in
length to the semi-minor axis. The theory of this construction is that
a section of the cone perpendicular to its axis has been taken whose
trace on the plane of the auxiliary elevation is glh^. The true form of
this section is a circle a portion of which is shown turned round into
the plane of projection. The chord of this circle drawn through e per-
pendicular to eA/ gives the greatest width of the cone at the level of the
horizontal plane.
The line c/c?/ joining the points of contact of the tangents from v^
to the circle which is the auxiliary elevation of the sphere is the
auxiliary elevation of the circle of contact. Perpendiculars from c/
and d^ to vn determine c and d the extremities of the minor axis of the
ellipse which is the plan of the circle of contact. The major axis of this
ellipse is equal to the diameter of the circle of contact and is therefore
equal to c/ . The section of the cylinder by this plane is a circle equal to
the base circle. The plan of this circle is a circle whose centre is o.
Draw this circle, and from /, the point of intersection of PQ and ST,
draw the projector r^er cutting the circle at r and r which are points
on the plan of the section. In like manner other points may be found.
The plan of the section may however be constructed from its axes aa^
and hhy by the trammel method.
Referring next to the upper plan (3) which shows the true form of
the section, aa-^^ the major axis, is parallel and equal to a'a/. The
semi-minor axis ch is equal to the radius of the base circle and er in (3)
is equal to er in (1).
A section by a plane LM which is perpendicular to the principal
section and makes with the axis an angle Q equal to the angle between
the axis and the base is called a sub-contrary section and is a circle equal
to the base circle.
Planes of section parallel to the base or to a sub-contrary section
are called cyclic planes because the sections by these planes are
circles.
Since a section of an oblique cylinder by a plane at right angles to
its axis is an ellipse the oblique cylinder is evidently also a rigid
elliptical cylinder (Art. 220).
231. The Oblique Cone. — The oblique cone may be generated by
a straight line which moves in contact with a fixed circle and passes
through a fixed point which is not on the straight line through the
centre of the circle at right angles to its plane. The fixed point is the
vertex of the cone and the straight line joining the vertex to the centre
of the fixed circle is the axis of the cone. A section of the oblique
cone by a plane perpendicular to the plane of the fixed circle and
containing the axis of the cone is called the principal section.
An oblique cone is shown in Fig. 515 by a plan (1) and elevation
(2). The fixed circle is on the horizontal plane of projection and
may be called the base of the cone. The axis of the cone is parallel
to the vertical plane of projection. The principal section is therefore
also parallel to the vertical plane of projection.
PQ is the vertical trace of a plane of section which is perpen-
dicular to the vertical plane of projection. The section by this plane
PQ is an ellipse whose major axis is equal to a'a/. Two plans of this
section, which are ellipses, are shown, the lower one (1) on XY as a
ground line, and the upper ono (3) on PQ as a ground line. The latter
plan shows the true form of the section.
To find points on the lower plan (1) of the section, take a horizontal
THE SPHERE, CYLINDER, AND CONE
267
plane whose vertical trace is ST and which cuts the axis of the
cone at od . The section of the cone by this plane is a circle whose
diameter is the intercept of ST on
the elevation of the cone. The
plan of this circle is a circle whose
centre is o. Draw this circle and
from r', the point of intersection
of PQ and ST, draw the projector
r'er cutting the circle at r and r,
which are points on the plan of
the section. In like manner other
points may be found. The above
construction when applied to the
point c' which is the middle point
of dal gives the minor axis hh^ of
the ellipse which is the plan of the
section.
Referring next to the plan (3)
which shows the true form of the
section, aa^, the major axis, is
parallel and equal to dal. The
semi-minor axis cb in (3) is equal
to cb in (1), also er in (3) is equal
to er in (1). Fig. 515.
A section by a plane LM
which is perpendicular to the principal section and makes with the
axis an angle B equal to the angle between the axis and the base is
called a suh-contrary section and is a circle.
Planes of section parallel to the base or to a sub-contrary section
are called cyclic planes because sections by these planes are circles.
As in the case of a right circular cone, a section of an oblique cone
by a plane which is parallel to a tangent plane to the cone is a para-
bola. Also a section of an oblique cone by a plane which cuts both
sheets of the cone is an hyperbola.
Exercises XVIII
1. The circle (Fig. 516) is the plan of a sphere resting on the horizontal
plane. H\Ti and HoTg, parallel to XY, are the horizontal traces of two parallel
planes. The plane of which H,Ti is the horizontal trace contains the centre of
the sphere. Draw the plan and elevation of the zone of the sphere which lies
between these two planes.
2. The parallel lines HiTi and HgTg (Fig. 517) are the horizontal traces of
two planes which pass through the centre of a sphere resting on the horizontal
plane, the given circle being the plan of the sphere. Draw the plan and elevation
of the lower lune of the sphere lying between the given planes.
3. The circle (Fig. 518) is the plan of a sphere resting on the horizontal plane.
HiTi inclined at 45° to XY and HjTj perpendicular to XY are the horizontal
traces of two planes which contain the centre of the sphere. Draw the plan and
elevation of the lower portion of the sphere which lies between these two planes.
4. The given circle (Fig. 519) is the plan of a sphere, centre C. The points
a, a, b, are the plans of points A, A, B on its upper surface. N and S are the
upper and lower ends of a vertical diameter.
268
PRACTICAL GEOMETRY
Draw the plan of a figure on the surface made up of three arcs of great circles
joining NA, NA, AA, and the arc of a small circle ABA.
Draw the stereographic projection of this figure on the horizontal plane
X >- - v^- Y X , ^ ^\ Y X-^
Fig. 516.
Fig. 517.
through G. That is, find the plane section of a cone, vertex S, of which the
figure is a spherical section. [b.e.]
5. The horizontal and vertical traces of a plane make angles of 30^ and 45*^
respectively with XY. A hemisphere of 1*25 inches radius has its base on this
plane and touching the planes of projection. Draw the plan and elevation of the
hemisphere.
6. A square of 1*5 inches side and a circle of 1-25 inches radius, the centre of
the circle being at the centre of the square, form the plan of a sphere with a
square hole through it. From this plan project two elevations, one on a ground
line XY parallel to a side of the square and the other on a ground line XjY,
parallel to a diagonal of the square. Also from the second elevation project a
plan on a ground line XgYg making an angle of 45° with XjYi.
7. A cylinder 2 inches in diameter and 3 inches long has its axis horizontal
and inclined at 45° to the vertical plane of projection. The cylinder is cut in
halves by a vertical plane which is inclined at 60° to the axis of the cylinder and
15° to the vertical plane of projection. Draw the elevation of one of the halves
of the cylinder.
8. rs, r's' (Fig. 520) is the axis of a hollow cj'linder, whose external diameter
is 2*1 inches and whose internal diameter is 1-4 inches. Draw the plan and
elevation.
9. A plan and an elevation of a hollow half cylinder, with one end closed, are
>*
r
—
—
\
s.
>.
s,
^
N
r^
♦;
H
•
T
j(
Y
y.
V
__
!::
i>
s
Fig. 520.
Fig. 521.
Fig. 622.
Fig. 523.
given in Fig. 521. Draw these projections and from the plan project an elevation
on a ground line inclined at 60° to XY.
10. A plan {u) and elevations {v) and [w) of a solid are given in Fig. 522.
Draw these projections and from the plan project an elevation on a ground line
parallel to rs. Also from the elevation {v) project a plan on a groimd line perpen-
dicular to rs.
11. A plan (1) and an elevation (2) of a right circular cylinder with two longi-
tudinal grooves in it are given in Fig. 523. Draw these projections and add others
THE SPHERE, CYLINDER, AND CONE
269
corresponding to (3), (4), and (5) in Fig. 497, p. 253. The angle between X^Y^
and XY to be 60°, and the angle between X^Y^ and X,Yi to be 45°.
12. A right circular cylinder of indefinite length has its axis vertical. A
plane whose horizontal and vertical traces make angles of 30° and 46° with the
ground line cuts the cylinder. Draw the elevation of the section and determine
its true form.
13. A split wrought-iron collar is shown in Fig. 524. SS is a vertical section
plane. Draw a sectional elevation on a ground line parallel to SS, the portion
A in front of the section plane being supposed removed. [b.e.]
Note. Fig. 524 is to be reproduced double size.
Fig. 525.
14. Referring to Fig. 525, (u) is a plan and (v) a side elevation of a quarter of
a ratchet wheel having 24 teeth of the form shown, (w) is an elevation on a
vertical plane inclined at 30° to the side of the wheel. Draw these projections
for the whole wheel to the dimensions given.
15. A sphere 2-2 inches in diameter rests on the horizontal plane and touches
the vertical plane of projection. The plan of the axis of a cylinder which
envelops the sphere makes 45° with XY and the elevation of the axis makes 60°
with XY. Draw the horizontal and vertical traces of the surface of the cylinder,
Fig. 527.
Fig. 528.
Fig. 529.
also the plan and elevation of the circle of contact between the sphere and
cylinder.
It. WW, m'n' (Fig. 526) is the axis of a right circular cylinder 1-5 inches in
270
PRACTICAL GEOMETRY
diameter, r is the plan of a point on the upper surface of the cylinder and s' is
the elevation of a point on the front surface. Find r' and s. Find also the plans
and elevations of the points of intersection of the line ah, a'h' with the surface of
the cylinder.
17. The elevation of a solid made up of two truncated right circular cones is
given in Fig. 527. Draw the plan of the solid. Show also the plan of the section
of the solid by a plane which is perpendicular to the vertical plane of projection
and which has VT for its vertical trace.
18. A right circular cone, base 2*5 inches in diameter and altitude 2 inches,
lies with its slant side on the ground. Draw the plan of the cone and show the
plan of a straight line which lies on the surface of the cone and is inclined at 45"^
to the ground.
19. The plan and an elevation of a right circular cone, whose axis is vertical
and which has a triangular hole through it, are given in Fig. 528. Draw these
projections and add an elevation on a ground line inclined at 45° to XY.
20. A plan and an elevation of a square nut with a conical chamfer are given
in Fig. 529. Draw these projections and add an elevation on a gromid line
parallel to one diagonal of the plan. Also, project from the given elevation a new
plan on a ground line inclined at 60° to r's'.
21. An equilateral triangle v'a'h', of 2 inches side, is the elevation of a right
circular cone, a'h' the elevation of the base being parallel to XY. A circle
inscribed in the triangle is the elevation of a plane section of the cone. Draw
the plan of the section.
22. A right circular cone, base 2 inches diameter, altitude 2*5 inches, lies
with its slant side on the horizontal plane. The cone is divided by a vertical
plane which passes through the centre of the base and whose horizontal trace is
inclined at 30° to the plan of the axis. Draw the elevation of the larger part of
the cone on a plane parallel to the plane of section.
23. vr, v'r' (Fig. 530) is the axis of a right circular cone, vv' being the vertex
and rr' the centre of the base. The cone rests with its slant side on the hori-
FiG. 532.
Fig. 533.
zontal plane. Draw the plan and elevation of the cone. Show the projections of
a point on the surface of the cone which is 0*75 inch above the horizontal plane
and 1-75 inches distant from the vertex of the cone.
24. The given circles a, h, and c (Fig. 531) are each 2 inches in diameter.
These circles touch one another, a and h touch XY and the straight line HT is
tangential to a and c. a is the plan of a cone, altitude 2'5 inches ; h is the plan
of a cylinder, length 4 inches, and c is the plan of a sphere. All these solids
stand on the horizontal plane. HT is the horizontal trace of a plane which cuts
the solids and is inclined at 45° to the horizontal plane. Show in plan and
elevation the sections of the given solids by the given plane.
25. A solid cut from a hollow cylinder is given in Fig. 532 by a plan and an
elevation. Draw an elevation on a ground line which is perpendicular to XY,
also an elevation on a ground line which is inclined at 45^ to XY.
26. Show the points of intersection of the given line ah, a'h' (Fig. 533) with
the surface of the given cone.
THE SPHERE, CYLINDER, AND CONE
271
27. A sphere 1-5 inches in diameter rests on the horizontal plane and its
centre is 1*5 inches in front of the vertical plane of projection. A point on the
horizontal plane, 3'75 inches in front of the vertical plane of projection and 2-25
inches from the plan of the centre of the sphere is the vertex of a cone which
envelops the sphere. Draw the vertical trace of the cone, also the plan and
elevation of the circle of contact between the cone and sphere.
28. A circular section of an oblique cylinder is 2 inches in diameter and the
axis of the cylinder is inclined at 40° to this section. Draw the true form of the
section of the cylinder by a plane perpendicular to its axis,
29. An oblique cylinder is given in Fig. 534. Draw the vertical trace of the
Pig. 534.
Fig. 535.
Fig. 536.
cylinder, that is, determine the section of the cylinder by the vertical plane of
projection. Draw also the plan and elevation of a sub-contrary section.
30. An oblique cone is given in Fig. 535. Draw the plan and also the true
form of the parabolic section of this cone by a plane whose vertical trace is PQ
and which is perpendicular to the vertical plane of projection. Draw also the
elevation of the hyperbolic section by the vertical plane whose horizontal trace
isRS.
31. Draw the plan and elevation of the section of the oblique cone (Fig. 536)
by the plane whose traces are H.T. and V.T.
CHAPTEK XIX
SPECIAL PKOJECTIONS OF PLANE FIGURES AND SOLIDS
232. Projections of a Figure whose Plaiie and a Line
in it have Given Inclinations. — First determine the traces of a
plane containing the figure. The working of the problem is much
simplified by first assuming this plane perpendicular to the vertical
plane of projection. In this position the elevation of the figure will
be a straight line coinciding with the vertical trace of the plane.
Afterwards any other elevation may be obtained by Art. 157, p. 191.
Draw L'M (Fig. 537) making with XY an angle equal to the
given inclination of the plane of the figure. Draw MN the horizontal
trace at right angles to XY. In the
plane L'MN place a line RS having
the inclination a of the line of the
figure given (Art. 174, p. 214).
Now imagine the plane L'MN,
with the line RS upon it, to rotate
about its horizontal trace MN until
it comes into the horizontal plane.
The point S being in MN will remain
stationary while the point R will
describe an arc of a circle in the
vertical plane of projection with M
as centre. Hence, if with M as
centre and Mr' as radius the arc
r'Rj be described, meeting XY at
Rj, RjS will be the position of the
line RS when that line is brought
into the horizontal plane.
Mark off on R^s a length AjEj equal to the length of that line of
the figure whose inclination a is given, and on AjBi construct the
given figure. Note that the line AB of the figure is not necessarily
one of its sides, but may be any line whatever in the plane of the
figure and occupying a definite position in relation to the figure.
Next imagine the figure thus drawn on the horizontal plane to rotate
about MN until its inclination is 6. All points in the figure will
describe arcs of circles whose plans will be straight lines perpendicular
to MN and whose elevations will be arcs of circles having their
centres at M and having radii equal to the distances of the points
SPECIAL PROJECTIONS
273
from MN. Thus the point Cj will describe an arc of a circle whose
plan is CiC perpendicular to MN and whose elevation is the arc c-ld of
a circle whose centre is M. c is of course in a straight line through
d at right angles to XY.
As a verification of the construction it should be noticed that if
any line, say C^B,, of the figure constructed on the horizontal plane
be produced, if necessary, to meet MN it will do so at the point t
where the plan he of the same line in its inclined position meets it.
It must be borne in mind that no line of the figure can have a
greater inclination than that of its plane, that is, a must not be
greater than 6.
233. Projections of a Plane Figure, the Inclinations of
two Intersecting Lines in it being given. — Determine by Art.
187, p. 219, the traces of the plane containing the lines whose inclina-
tions are given, taking the horizontal trace at right angles to the
ground line. Eotate this plane, with the lines upon it, about its hori-
zontal trace, as in the preceding Article, until it comes into the hori-
zontal plane. On the lines thus brought into the horizontal plane
construct the given figure and proceed to determine the plan and ele-
vation of the figure in its inclined position exactly as in the preceding
Article.
234. Projections of a Plane Figure having given the
Heights of three Points in it. — Note that the difference between
the heights of any two points must not exceed the true distance
between the points.
Let A, B, and C be the points whose heights are given. '
Construct on the horizontal plane a triangle A^BiCj (Fig. 538)
equal to the triangle ABC. With centre Aj and radius equal to the
height of the point A describe the circle
EFH. With centre Bj and radius equal to
the height of the point B describe the circle
KLM. With centre Cj and radius equal to
the height of the point C describe the circle
NPQ. Draw HL to touch the circles EFH
and KLM, and produce it to meet A^Bj
produced at R, Also draw EP to touch
the circles EFH and NPQ, and produce it
to meet A^Ci produced at S. RS is the
horizontal trace of the plane which will
contain the points A, B, and C.
Draw XY perpendicular to RS meeting
it at O. Draw A^a/ perpendicular to XY
to meet it at a/. With centre O and radius
Oa/ describe an arc of a circle to meet at
a' a parallel to XY which is at a distance
from XY equal to the height of the point
A. Oa! is the vertical trace of the plane
which will contain the points A, B, and C.
The theory of the above construction for finding the traces of the
T
CCy
538.
274
PRACTICAL GEOMETRY
plane containing the points A, B, and C is similar to that of the con-
struction given in Art. 187, p. 219, which should be again referred to.
It is evident that the angles A^RH and AjSE are the inclinations of
the straight lines A^ and AC respectively.
The plan abc and the elevation a'h'c of the triangle ABC are
next determined as shown, the construction being the same as in
Art. 232.
If ABC is not the complete figure given, a figure equal to it
must be built up on the triangle A^BiCp Then the projections of the
remainder of the figure are obtained in the same way as the projections
of the part ABC.
235. Projections of a Plane Figure when it has been
turned about a Horizontal Line in it till the Plan of an
Opposite Angle is equal to a given Angle. — Let ABCD denote
the given figure, AC the horizontal line about which it is turned, and
B the angle whose plan is to be equal to a given angle.
Construct the figure aB^cDi (Fig. 539) equal to the given figure
ABCD. On ac describe a segment of a circle (Art. 14, p. 18) to
contain an angle equal to that into which
the angle B is to be projected.
As the figure revolves about AC the
point B will describe an arc of a circle
whose plan is a straight line through B^
perpendicular to ac. Let this perpendicular
meet the arc of the segment of a circle
which has been described on ac at the
point b. Join ah and he. Then ahc is the
plan of the angle B when the figure has
been turned as required.
Draw XY at right angles to ac, meeting
ac produced at a'. Draw B^?)/ at right
angles to XY to meet it at 6/. With
centre a' and radius a'6/ describe the arc
6/6', and through 6 draw a perpendicular
to XY to meet this arc at 6'. a'b' is the
vertical trace and aa' is the horizontal trace
of the plane containing the figure ABCD
when it occupies the position required.
From Di draw DjC?/ perpendicular to XY
to meet it at d^'. With centre a' and radius a'd^' describe the arc d^^d'
to meet h'a' produced at d'. Draw d'd perpendicular to XY to meet
a line through D^ parallel to XY at d. Join ad and cd. abed is the
plan and d'h is an elevation of the figure ABCD as required.
236. Projections of a Solid when a Plane Figure on it
and a Line in that Figure have given Inclinations. — The
plane figure may be a face of the solid or it may be a section of it.
Considering the most general case, the first step is to determine the
projections of the plane figure as in Art. 232, p. 272. On the rabat-
jnent of the plane figure on the horizontal plane the feet of the
SPECIAL PROJECTIONS
275
perpendiculars from the angular points of the solid on the plane of
the figure must next be located. The projections of the feet of these
perpendiculars are next found, and the projections of the perpen-
diculars can then be drawn. The projections of the angular points of
the solid are therefore determined.
Two examples are illustrated by Figs. 540 and 541.
In the first example (Fig, 540) a right square prism, side of base
1*5 inches, altitude 1"8 inches, is shown when its base is inclined at 50*^
and one side of that base is inclined at 20°. The plan ahcd and eleva-
tion a'b'c'd' of the square base are first determined as in Art. 232, p. 272.
In this case the feet of the perpendiculars from the angular points of
the solid on the plane of the base are at the angular points of that base.
Hence it is now only necessary to draw from a', h\ c\ and d! perpen-
diculars to the vertical trace of the plane of the base and make them
1-8 inches long in order to obtain the elevations e', /', g', and h' of the
other angular points of the solid. The plans of these perpendiculars
are at right angles to the horizontal trace of the plane of the base and
are therefore parallel to XY.
Fig. 540.
Fig. 541.
In the second example (Fig. 541) a cube of 1-5 inches edge is
shown when the plane of two diagonals of the solid is inclined at 60°
and one of these diagonals is inclined at 40°. The plane of the two
diagonals divides the cube into two right triangular prisms, the
triangular ends of which are each equal to one half of a face of the
cube. The common face of these two triangular prisms is a rectangle
of which two opposite sides are edges of the cube while the other two
276 PRACTICAL GEOMETRY
sides are diagonals of opposite faces of the cube and the diagonals are
diagonals of the solid.
Tlie plan ahde and elevation a'h'd'e' of the rectangle which is the
section of the cube by the plane of two of its diagonals are determined
by Art. 232, p. 272. In this case the feet of the perpendiculars from
the other angular points of the cube on to the plane of two diagonals
are at the middle points of the longer sides of the rectangle ABDE
and the lengths of £hese perpendiculars are each equal to the half of a
diagonal of a face of the cube. Hence, perpendiculars to the vertical
trace of the plane of the above mentioned rectangle from the middle
points of a'h' and d!e' and equal to half of a diagonal of a face of the
cube determines the elevations c', c', /', and /' of the other corners of
the cube. The plans of these perpendiculars are parallel to XY.
237. Projections of a Solid when two Intersecting Lines
connected with it have given Inclinations. — Determine by
Art. 187, the plane containing the intersecting lines whose in-
clinations are given. Rabat this plane with the lines upon it into
the horizontal plane by turning it about its horizontal trace. On
these lines thus brought into the horizontal plane complete the
plan of the solid and then proceed exactly as in Art. 236.
238. Projections of a Solid when the Heights of three
Points in it are given. — Determine by Art. 234, the plane con-
taining the three points whose heights are given. Rabat this plane
with the points upon it into the horizontal plane by turning it about
its horizontal trace. About these points thus brought into the hori-
zontal plane complete the plan of the solid and then proceed exactly as
in Art. 236.
239. Projections of a Solid when two Faces A and B,
which are at Right Angles to one another, have given
Inclinations. — If one of the two faces referred to is a base denote it
by A. First determine L'MN, the plane of the face A, the horizontal
trace MN being perpendicular to XY. Next determine by Art. 194,
p. 226, a plane L'PN perpendicular to the plane L'MN and inclined at
an angle equal to the given inclination of the face B.
Find LN the line of intersection of these two planes. Now rabat
the plane L'MN with the line LN upon it into the horizontal plane by
turning it about MN. On this line thus brought into the horizontal
plane construct the face A, that side of the face A which is adjacent
to the face B being made to coincide with the line, and then proceed
exactly as in Art. 236.
240. Projections of a Solid when two Faces A and B,
which are not at Right Angles to one another, have given
Inclinations. — If one of the two faces is a base denote it by A.
First determine L'MN, the plane of the face A, the horizontal trace
MN being perpendicular to XY. Next determine by Art. 194, p. 226,
a plane L'PN having an inclination equal to that of the face B and
making with the plane L'MN an angle equal to the angle between the
faces A and B. The procedure is now the same as in the second
paragraph of the preceding Article.
SPECIAL PROJECTIONS 277
Exercises XIX
1. Draw the plan of an equilateral triangle of 2*5 inches side when its plane
is inclined at 45°, one side inclined at 30°, and one angular point on the hori-
zontal plane. From this plan project an elevation on a ground line parallel to
the plan of that side qi the triangle which is inclined at 30°.
2. Draw the plan of a square of 2*5 inches side when its plane is inclined at
60° and one diagonal is inclined at 45°. What is the inclination of the other
diagonal ?
3. A regular hexagon ABGDEF of 1-25 inches side has the side AB horizontal
and the plan of the diagonal BD is 2 inches long. Draw the plan and from it
project an elevation on a ground line parallel to ab. What is the inclination of
the plane of the hexagon ?
4. The plane of a square of 2 inches side is inclined at 60° and the plan of one
diagonal is 2 inches long. Draw the plan of the square.
5. The horizontal trace of a plane makes an angle of 45° and the vertical trace
makes an angle of 50° with XY. A regular pentagon of 1-5 inches side lies on
this plane with one side inclined at 30° to the horizontal plane. Draw the plan
and elevation of the pentagon.
6. A square of 2 inches side has one side inclined at 45° and an adjacent side
inclined at 30°. Draw the plan of the square and an elevation on a ground line
parallel to the plan of the diagonal which is inclined at 45°.
7. ABC is a triangle. AB = 2 inches, BC = 2*75 inches, and CA = 3 inches.
Draw a plan of this triangle when the sides AB and BC are inclined at 40° to the
horizontal plane.
8. A regular pentagon of 2 inches side has one side inclined at 30° and a
diagonal through one end of that side inclined at 40°. Draw the plan.
9. ABC is an equilateral triangle of 2*5 inches side. A is on the horizontal
plane, B is 1 inch and C is 1*5 inches above the horizontal plane. Draw the plan
and an elevation on a ground line parallel to ab.
10. Draw the plan of a square of 2 inches side when the heights of its centre-
and two angular points above the horizontal plane are 2 inches, 1-5 inches, and
0'75 inch respectively.
11. Draw the plan of a regular pentagon of 1*5 inches side when one side is on
the horizontal plane and the plan of the opposite angle is 120°.
12. A 60° set-square revolves about its shortest side,, which is horizontal,
until the plan of the opposite angle is 60°. Find what is then the inclination of
the plane of the set-square.
13. ABC is a triangle, AB = 1*5 inches, BC = 3 inches, and CA = 2 inches.
Draw the plan of this triangle when the side AB is horizontal and the plan of the
angle C is 20°.
14. Draw a triangle oab, oa = 1 inch, ob ^ 1*25 inches, and a6 = 1 5 inches.
o is the plan of the centre and ab is the plan of one side of a regular hexagon.
Complete the plan of the hexagon.
15. Draw a triangle abc. ab = 1 inch, be = I'd inches, and ca = 2 inches,
ab and be are the plans of adjacent sides of a regular octagon. Complete the plan
of the octagon.
16. Draw ab 2 inches long. From the middle point c of o6 draw cd 3 inches
long and making the angle acd = 60°. These are the plans of two intersecting
straight lines which are at right angles to one another. The points A and D are
each 0-5 inch above the horizontal plane. Find the height of the point B.
17. a is a point on a straight line HT. ab is a straight line 2 inches long
making an angle of 30° with HT. ab is the plan of one edge of a regular tetrahe-
dron of 2-5 inches edge. HT is the horizontal trace of the plane of a face of the
tetrahedron containing the edge AB. Complete the plan of the tetrahedron and
draw an elevation on a ground line parallel to HT.
18. Draw the plan of the solid given in Fig. 373, p. 196, when the base is in-
clined at 45° and one side of that base is inclined at 30°. From this plan project
an elevation on a ground line parallel to the horizontal trace of the plane of the
base.
278 PRACTICAL GEOMETRY
19. Draw the plan of the solid given in Fig, 372, p. 196, when the base is in-
clined at 50° and one diagonal of the base is inclined at 40°, From this plan
project an elevation on a ground line parallel to the plan of the diagonal which
is inclined at 30°.
20. A right pyramid has for base a regular pentagon of which the diagonals
measure 2-5 inches. The vertex is 2 inches above the base. Draw the plan and
elevation of the pyramid, with its base in a plane inclined at 55° to the vertical
plane and at 60° to the horizontal plane ; one diagonal inclined at 30°, and one
end of that diagonal in the vertical plane. [b.e.]
21. Two diagonals of a cube of 2 inches edge are inclined at 35° to the hori-
zontal plane. Draw the plan of the cube.
22. A right prism 3 inches long has for its ends regular hexagons of 1*25
inches side, AB is an edge of one end and BH is one of the long edges. Draw the
plan of this prism when the heights of the points A, B, and H above the hori-
zontal plane are 0*2 inch, 1 inch, and 2-5 inches respectively.
23. ABC is the base of a pyramid and V its vertex. AB = 2*5 inches, AC =
2 inches, BG = 2-75 inches, AV := CV = 3 inches, and BV = 2-5 inches. Draw
the plan of the pyramid when A is 1-5 inches, B 2*5 inches and C 1 inch above
the horizontal plane. [b.e.]
24. Draw the plan of a cube of 2 inches edge when one face is inclined at 55°
and another face is inclined at 75°.
25. One face of a regular tetrahedron of 2*5 inches edge is inclined at 45° and
another face is inclined at 70°. Draw the plan of the tetrahedron.
CHAPTER XX
HORIZONTAL PROJECTION
241. Figured Plans. — If the plan of a point is given, and
also its distance from the horizontal plane, the position of the
point is fixed without showing an elevation of it. The distance of
the point from the horizontal plane is shown by placing a number
or index adjacent to its plan. For example, if the point is 6
units above the horizontal plane the figure 6 is placed adjacent to its
plan, and if the plan is also lettered the figure is placed to the right of
the letter and at a slightly lower level thus, a^. If the negative or
minus sign be placed in front of the index this denotes that the given
distance is below the horizontal plane. Thus, a point whose plan is
marked — 6 or a.^ is 6 units below the horizontal plane. A plan
such' as has been described is called a figured plan or indexed plan.
If the plan of a straight line is given and also the figured
plans of two points in it, it is obvious that the line is completely
fixed.
In horizontal projection points and lines are shown by their figured
plans. The form and position of a curved line can only be shown in
horizontal projection by giving the figured plans of a sufficient number
of points in it.
242. Scales of Slope. — It has already been shown that planes
may be represented by their traces on the co-ordinate planes. In
horizontal projection, planes are represented by their scales of slope.
The scale of slope of a plane is really the figured plan of a straight
line lying in the plane and perpendicular to the horizontal trace of the
plane. If a straight line be given and it is understood that a piano
whose horizontal trace is perpendicular to the line contains the line
then it is obvious that the plane is completely
fixed. To show that the given line represents a
plane and not simply a line a second and thicker
line is ruled near to and parallel to it. It is
usual to place the thicker of the two lines to the
left of the other when looked at by a person
ascending the plane.
Fig. 542 shows the connection between the
scale of slope ah and the traces L'M, MN of a
plane. The numbers at the difierent points of
the scale denote the distances of these points
from the horizontal plane.
280
PRACTICAL GEOMETRY
Fig. 643.
243. Applications of Horizontal Projection.^ — Most of the
problems on points, lines and planes can be worked as conveniently by
horizontal projection as by ordinary plan and elevation, but it may be
observed that although the solution of a problem may be given by a
figured plan only the working may involve constructions which are
equivalent to the drawing of one or more elevations.
As the principles involved in the solution of problems on points,
lines, and planes are generally the same whether they are solved by
plan and elevation or by horizontal projection, a selection of a few
problems only will be taken in this chapter to illustrate the method of
horizontal projection.
244. Simple Problems on the Straight Line.— Let aAs
(Fig. 543) be the figured plan of a straight line AB. From a and b
draw perpendiculars to ah, and make them
respectively equal to the indices of a and b.
The line AjBj joining the tops of these per-
pendiculars will have a length equal to the
true length of AB. The angle between A^Bi
and ah will measure the inclination of AB,
and the point where AjB^ meets ah will be
the horizontal trace of AB. It is evident that
the index of the horizontal trace of a line is
O. If one of the indices is positive and the other negative the per-
pendiculars aAi and feB^ must be drawn on opposite sides of ah.
In the above construction A^Bj maybe looked upon as an elevation
of AB on ab as a ground line, or it may be taken as the line AB
rebatted on to the horizontal plane about ab as an axis.
The true length and inclination of AB may also be found by
drawing 5P perpendicular to ab and equal to the difference between
the indices of a and b. The line aP will then be the true length of
AB, and the angle Pah will be the inclination of AB.
To determine a point c in ab which shall have a given index, say
6*5, make 6Q equal to 6*5, draw QC^ parallel to ab to meet AjB^ at
Cp A perpendicular C^c to ab determines the point required.
To determine the index of a given point c in ab, draw cC^ per-
pendicular to ah to meet A^Bj at C^. The length of cCi is the index
required.
To draw a line through a given point c^ (Fig. 544) parallel to a
given line «2^9 > draw through c a line cd parallel to aS, and make
cd = ab. The index of d will be 7 greater than
5 the index of c because the index of & is 7
greater than the index of a. If the line cd be
produced in the opposite direction, and ce be
made equal to ba, then the index of e will be 7
less than 5 the index of c because the index
of a is 7 less than the index of b. The index
of e will therefore be — 2, or the point e is 2
units below the horizontal plane.
245. Inclination of a Given Plane.— Let ab (Fig. 545) be
HORIZONTAL PROJECTION
281
I
the scale of slope of the plane. Since the long lines of the scale of
slope are at right angles to the horizontal trace of the plane, the
inclination of these lines must be the same as
the inclination of the plane. Hence if 6Bi be
drawn perpendicular to ah, and made equal to
the difference between the indices of h and a,
the angle B^afe will measure the inclination of the
plane.
246. Two Problems on Parallel Planes.
— (1) To determine a plane to contain a given point p^ (Fig. 646),
and be parallel to a given plane ah. Since the horizontal traces of
parallel planes are parallel it is clear n ,
that their scales of slope which are at ^ | i | |~"^
right angles to these traces, must also 5 | 10 15
be parallel. Draw, therefore, the long Pj
lines cd of the required scale of slope ^
parallel to ah^ and in any convenient ^ j'q jj
position. Through p draw a line pq^ -^^^ -^g
at right angles to cd. pq will be
the plan of a horizontal line lying in the required plane, and q will
therefore have the same index as p. The scale of slope cd must be
graduated in the same way as ab ; that is to say the difference
between the indices of a given length on cd
must be equal to the difference between
the indices on an equal length of ab.
(2) To determine the distance between
two parallel planes ah and cd (Fig. 547).
If the traces of these planes on a vertical
plane perpendicular to their horizontal traces,
and therefore parallel to their scales of slope,
be drawn, the distance between these traces
will be the distance between the planes.
ah is taken as the ground line, and ac'
and hh' are drawn perpendicular to ah. hV p^^ ^^rj
is made equal to the index of h, and ad equal
to the index of c. As in the example shown in the figure, the index
of a is 0, ah' is the vertical trace of the plane whose scale of slope is
ah, and as the other plane is parallel to this one, a line c'd! parallel to
ah' will be its vertical trace on the assumed
vertical plane. The distance between ah' and
c'd' is the distance between the given planes.
247. The Plane containing three
given Points. — Let a^, , hy^ , and
(Fig.
648) be the given points. Find by Art. 244
a point d in ah having the same index as c.
Join cd. cd is the plan of a horizontal line
lying in the plane containing the three given
points; therefore the scale of slope of the
plane containing these three points must be perpendicular to cd.
282 PRACTICAL GEOMETRY
Where cd cuts the scale of slope determines a point on it having
the same index as c or d. If through a a line be drawn parallel to cd
to meet the scale of slope, a point is determined on the latter having
the same index as a. These two points on the scale of slope having
been found, the scale may be graduated if required.
248. In a Given Plane to place a Line having a Given
Inclination so that it shall contain a Given Point in the
Plane. — Let ah (Fig. 549) be the given plane
and pao ttie given point. Through r, any point
in the scale of slope not having the same index
as jp, draw rq perpendicular to ah. Draw pQ^,
in any convenient direction, say parallel to rq,
and a linepPi at right angles to jpQ^. Make
2?Qi equal to the difference between the indices
of p and r, and draw PiQi making the angle
PiQiP equal to the given inclination. With Fig. 549.
centre p and radius pQ^ describe an arc cutting
rq at q. po^qi^ is the line required. Except when the given inclination
is the same as the inclination of the given plane there are obviously
two straight lines which will satisfy the given conditions.
249. Intersection of Two Given Planes. — Let ah and cd
(Fig. 550) be the given planes. Through a and 6, any two points in
the scale of slope ah, draw the lines ap and
hq at right angles to ah. Through points
c and d in cd, having the same indices re-
spectively as a and h, draw lines perpen-
dicular to cd. The line op is the plan of a
horizontal line lying in the plane ah. cp is
the plan of a horizontal line lying in the
plane cd. Since these lines have the same
indices, they are in the same horizontal
plane, therefore they will intersect at a
point of which p is the plan. Therefore ^20 is a point in the inter-
section of the two planes. In like manner, q^^Q is a point in the
intersection, therefore the line p2o(Zio is the intersection of the given
planes.
If the given scales of slope are parallel, the above construction will
evidently fail, because the horizontals of both planes will be parallel,
and therefore never meet. In this case a third plane may be taken,
not parallel to either of the given planes, and its intersection with
each of them found by the method just given. This determines two
lines whose intersection with one another will be a point in the
intersection required. It is evident that the intersection of the given
planes in this case will be a horizontal line, therefore a line perpen-
dicular to the given scales of slope through the point found determines
the intersection of the given planes.
If the third plane mentioned above be taken perpendicular to each
of the given planes, it will be a vertical plane, and the plans of its
intersection with the given planes will coincide, and an elevation of
HORIZONTAL PROJECTION' 283
them must be drawn to determine the point where they meet. This
elevation is best taken on the third plane itself.
If the given scales of slope are nearly parallel, the horizontals of
the planes will meet at a very acute angle, and it is more accurate to
find two points in the required intersection by the method explained
for planes whose scales of slope are parallel, that is, by cutting each of
the given planes by two other planes.
250. Intersection of a Line and Plane. — Let aj)^^ (Fig. 551)
be a given line and cd a given plane. Through any two points a and
6 in the given line, draw the parallel lines ar and
hq in any convenient direction. Through c and
d, points on the scale of slope having the same
indices as a and h respectively, draw the hori-
zontals cr and dq^ meeting the parallels ar and hq
respectively at r and q. Join rq. The point P
where the line rq or rq produced meets ah or ah
produced is the plan of the point where the given
line meets the given plane.
The theory of the above construction is as
follows — «28*'28 ^^d ?>ia5i3 are horizontals of a plane
containing the given line. The line r^^ia ^^
evidently the intersection of this plane with the
given plane. The point V is therefore a point in both planes and also
in AB, therefore it is the intersection required.
The intersection may also be found by taking an elevation of the
line and plane on a ground line parallel to the scale of slope.
251. The Normal to a given Plane through a given Point.
— The plan of a line which is perpendicular to a plane is at right angles
to the horizontal trace of the plane, and will therefore be parallel to its
scale of slope.
To figure the plan of the line (which of course passes through the
figured plan of the point), determine the trace of the plane and the
elevation of the point on a vertical plane parallel to the scale of
slope. Through the elevation of the point draw a perpendicular to the
trace of the plane, this perpendicular will be the elevation of the
normal and from it the plan may be figured.
252. The Plane through a given Point perpendicular to a
given Line. — Taking the figured plan of the line as a ground line,
determine the elevations of the point and line. Through the elevation
of the point draw a line perpendicular to the elevation of the given
line. This perpendicular will be the vertical trace of the required
plane. The scale of slope of the plane will be parallel to the given
figured plan of the line, and it may be graduated from the vertical trace
found as above.
253. Contour Lines. — The plan of a portion of the earth's
surface is made to show the form of that surface very clearly, by
drawing on it the sections of the surface by a series of horizontal
planes, at equal distances from one another. These sections are called
contours or contour linen. It is evident that the relative closeness of the
284 PRACTICAL GEOMETRY
contour lines shows the relative steepness of the different parts of the
surface, the surface being steepest where the contour lines are closest
together. It is usual to affix to the contour lines their heights above
some fixed horizontal plane.
254. Intersection of a Plane and a Contoured Surface.—
Fig. 552 shows the scale of slope of a plane and the contoured plan of
Fig. 652.
a surface. To determine the intersection of the plane and the surface,
draw the plans of a number of horizontal lines lying in the plane, and
having the same indices as the contour lines. The points where these
lines meet the contour lines having the same indices, are points in the
intersections required. The complete intersection is obtained by drawing
a fair curve through the points thus found.
255. Contouring a Surface from its Equation. — A formula
or equation which contains three variables is represented graphically
by a surface and in representing this surface on paper a system
of contour lines may be necessary. As an example take the
, ^^ (62800 - Si^)d:'v , . , . , . . .
formula, H = 9^0000 which gives the relation between the
horse-power H transmitted by a cotton rope and the diameter d of
the rope in inches and the velocity v of the rope in feet per second,
allowing for the stress in the rope due to centrifugal force.
For each diameter of rope the relation between H and v is shown
by a plane curve, and if the curves for ropes of different diameters be
constructed these curves will be contours of the surface which
represents the original formula. The contours of the surface correspond-
ing to diameters of J inch, 1 inch, 1^ inches, 1^ inches, 1| inches, and
2 inches are shown at (a) Fig. 553 between the limits v = 0, and v = 1 20.
To construct these curves it is first necessary to construct, by calculation,
the table given on the opposite page.
HORIZONTAL PROJECTION
285
Horse-power H, for different values of v and d.
V
Diameter d.
1
^
1* 1 U
1
If
2
10
1-5
2-7
4-2
61
8-3
10-9
20
30
5-4
8-4
121
16-4
21-4
30
4-4
7-8
12-2
17-6
24-0
31-4
40
5-7
10-1
15-8
22-7
30-9
40-3
50
6-8
12-0
18-8
27-0
36-8
48-1
60
7-6
13-6
21-2
30-5
41-5
54-3
70
8-2
14-6
22-9
32-9
44-8
58-6
80
8-5
15-2
23-7
34-1
46-4
60-7
83-5
8-6
15-2
23-8
34-2
.46-6
60-8
90
8-5
151
23-5
33-9
46-1
60-3
100
8-0
14-3
22-3
32-1 .
43-7
57-0
110
7-1
12-7
'19-8
28-5
38-8
50-7
120
5-8
10-2
16-0
230
31-3
40-9
In addition to the velocities at intervals of 10 feet per second
given in the first column of the table it will be noticed that the
velocity 83-5 feet per second is given. This velocity 83 5 is the
velocity at which the horse-power for any given rope is a maximum.
The velocity 83-5 is obtained by calculation as follows. The horse-
power for any rope will evidently be a maximum when 62800 v — 3v'^
is a maximum. Let y = 62800?; — 3v\ Differentiating this, -=^
= 62800 — %^ and yisa, maximum when 62800 — 9v^ = o, that is when
3v = V62800; or V = 83-5.
Assuming that the curves at (a) Fig. 553 are in vertical planes,
horizontal sections of the surface at levels whose intervals are 10
horse-power are shown at (6) Fig. 553 between the limits d = ^
and d = 2,
The student should work out this example to the following scales.
H, 1 inch to 10 horse-power, d, 1 inch to J inch diameter, v, 1 inch
to 20 feet per second. In addition to the contours shown at (a)
Fig. 553 the student should add the curves for ropes of | inch,
IJ inches, If inches. If inches, and 1~ inches diameter. Also in
addition to the contours shown at (p) he should add the curves for
5, 15, 25, 35, 45, and 55 horse-power. Lastly, on a base line parallel
to MN he should construct the contours of the surface showing the
relation between H and d for velocities at intervals of 10 feet per
second. These contours are vertical sections of the surface parallel
to MN just as the curves at (a) are vertical sections by planes parallel
to LM.
280
PRACTICAL GEOMETRY
10 20 30 40 50 60 70 80
Velocity V 83-5
90 100 ilO 120
N
10 20 30 40 SO 60 70
WjfcUy V «3-5
Fig. 553.
90 too 110 120
10 20 30 40 50 60 70
Velocity V.
Fig. 564.
90 100 110 120
HORIZONTAL PROJECTION
287
An oblique parallel projection of the surface which has just been
considered is shown in Fig. 554. The student should have no difficulty
in making such a projection after he has studied Chapter XXI. This
projection may be very readily drawn on squared paper, taking the axis
for (Z at 45° to the axis for v.
When the student draws the oblique parallel projection shown in
Fig. 554 he should put in the additional contours suggested with
reference to Fig. 553.
Exercises XX
s^
Note. Unless otherwise stated the unit for the indices is Vg or 0*1 of an inch, so
that an index 25 denotes a height of 2-5 inches above the horizontal plane.
1. Two points a and b in the plan of a straight line are -2 inches apart. The
index of a is 3 and the index of b is 11. Determine, (1) the index of a point c in
ab which is J inch from a ; (2) a point d..^ in the plan of the line ; (3) a point e^
in the plan of the line ; (4) a point /-j in the plan of the line ; (5) the true length
of EF.
2. Draw a triangle a^ 6-4 Cig {ab = 2-7 inches, 6c = 2*2 inches, ca = 1*7 inches).
Find a point d in be whose index is 8, and show the figured plans of two
straight lines passing through B and C, and parallel to AD.
3. A straight line making an angle of 30"^ with the ground line is both hori-
zontal and vertical trace of a plane. Show the scale of slope of this plane.
4. Show the scales of slope of two parallel planes inclined at 50°, the distance
between the planes being 0*7 inch.
5. Determine the scale of slope of the plane of the triangle given in exercise 2,
also the true shape of the triangle.
6. An equilateral triangle of 2*5 inches side has
its angular points indexed 3, 12, and 15. Show the
figured plans of the bisectors of the angles of the
triangle of which the given triangle is the plan.
Draw also the ellipse which is the plan of the cir-
cumscribing circle of the triangle.
7. Draw the scale of slope of a plane whose
inclination is 60° and show the figured plan of a
triangle ABC which lies in this plane. The inclina-
tions of AB and BC are 30"^ and 45° respectively,
the indices of a, b and c are 3, 24, and 8 respec-
tively.
8. A plane is inclined at 35° to the horizontal,
the lines of steepest upward slope going due East.
A second plane has an inclination of 52°, the direc-
tion of the lines of upward slope being due North.
Represent the planes by scales of slope, unit for
heights 0-1 inch. Show the figured plan of the inter-
section of the planes. Find and measure the angle
between the planes. [b.e.]
9. Fig. 555 is a plan, drawn to a scale of 1 inch
to 200 feet, showing at a, b two places A, B on a
hill-side, the surface of the ground being an inclined
plane represented by a scale of slope, heights being indexed in feet. Draw Fig.
555 to a scale of 1 inch to 100 feet and add the plan of a zigzag path connecting
A and B, made up of three straight lengths of a constant inclination equal to
half that of the hill, and deviating equally on each side of a straight line through
A and B. Ascertain and measure in feet the total length of this path. [b.e.]
^^ ._
140
120
iOO
-80
60
-40
20
Fig. 555.
288
PRACTICAL GEOMETRY
10. Two lines AB and CD are given by their figured plans in Fig. 556. Draw
the scale of slope of a plane which will
contain the line CD and make an angle of
15° with the line AB.
11. The plans of six points a^ h^, Cj^, d^^,
C20, and /as, taken in order, are situated at
the angular points of a regular hexagon of
IJ inches side. Find the intersection of the
plane containing A, C, and F, with the plane
containing B, D, and E, and state its in-
clination.
12. Draw a square a^^ b^^ c^^ d^ of 2 inches
side. Determine the plan of the sphere on
whose surface the points A, B, C, and D are
situated.
13. Draw the figured plan of the common perpendicular to the straight lines
AC and BD of the preceding exercise.
14. ^50 6eo Cgo ^50 (Fig. 557) is the figured plan of a straight roadway which
is to be made, partly by cutting, and partly by embankment, on the ground
—
—
—
—
y
■yy
K
—
—
_•
y
a
f^
^
K
To
H
*
H
*■
V
J
«r
'C3
Fig. 556.
Fig. 557.
whose surface is given by its contoured plan. The sloping faces of the cutting
and embankment are to be inclined at 40°. Draw the plan of the intersections of
the faces of the cutting and embankment with the surface of the ground.
Show also, on EF as ground line, vertical sections at LL, MM, and NN.
N.B. — Fig. 557 is to be enlarged four times. The unit in this exercise is 1 foot.
Fig. 658.
HORIZONTAL PROJECTION
289
15. The surface of a piece of ground is given by contours at vertical intervals
of 5 feet (Fig. 558), the linear scale of the plan being 1 inch to 100 feet. A road
is to be cut at the given heights, the face of the cutting on each side having a
slope of 38° to the horizontal. Draw Fig. 558 to a scale of 1 inch to 50 feet and
complete the plan of the finished earthwork. [b.e.]
1 1 h ([
1 1 1 /Vti
t J- ^ -r tttU
t r ^11 ^ , TZutif
/„7 l.*^^ Vi t-tttW
Ml t:l u t tl ttu-
"t^^'U.' L o / , -'f
±1 z -/:--:#/-,- i--i- / ;/ s
113 ^jLJ Lr. J*.._j4ia_
4LI 4 1 ^( 1 ^ s
jzti 1 '^^4^1 ' t-?-4-
"Vr-g-ii: iiirsirs « "
M
(a)
Fig. 559.
16. You are given, in Fig. 559, the plan of the surface of a piece of , ground,
contoured in feet, the linear scale being 1 inch to 100
feet. The curve pp is the centre line of a road, 20 feet
wide, which is to be made partly by cutting and partly
by embankment, the former having a slope of 45^ and
the latter one of 38° to the horizontal. Draw Fig. 559
to a scale of 1 inch to 50 feet and complete the plan of the
finished earthwork within the limits of the data. [b.e.]
17. Represent by contour lines, as in the example
worked out in Art. 255, the surface whose equation is
M =- — ^. The axes to be arranged as shown at' (a), (h),
t "~ oU
and (c), Fig. 560. v denotes velocity in feet per minute,
and t denotes temperature in degrees Fahrenheit. Scales.
— For V, 1 inch to 50 feet per minute. For <, 1 inch to
20°. For M, 2 inches to 1. The limits of v and t are
shown at (a) and (6). j; —
Fig. 560.
CHAPTER XXI
PICTORIAL PROJECTIONS
256. Pictorial Projections. — Since the orthographic projection
of a line only shows its true length when the line is parallel to the
plane of projection, it is usual, in making working drawings of an object,
to arrange it so that as many of its lines as possible are parallel to at
least one of the co-ordinate planes. The object is then in a simple
position. Fig. 561 shows a plan and two elevations of a rectangular
solid when the solid is in the simplest possible position in relation to
the planes of projection. These projections are very easily drawn, but,
although they represent the solid completely, they are not at all
" pictorial," and a special training is necessary before the observer can
form from them a correct mental picture of the object.
FRONT ELEVATION. END ELEVATION
Fig. 561.
Fig. 562.
Now let the solid be tilted over as shown by the plan a and eleva-
tion a'. Fig. 562, and let a new elevation a/ be drawn in the manner
explained in Chap. XIV. It will now be observed that the elevation
a/ by itself gives a much better idea of the form of the object than
any of the other projections shown in Figs. 561 and 562, but unless a
much simpler method of drawing such a pictorial projection than that
shown in Fig. 562 can be devised, and unless some simple method of
PICTORIAL PROJECTIONS
291
measuring the various dimensions of the object direct from the drawing
can be given, such a pictorial projection would be of little value as a
substitute for ordinary working drawings.
In the subsequent articles of this chapter, methods are described
which enable the draughtsman to make pictorial projections resembling
that shown at a/ in Fig. 562 with almost as little trouble as is required
for the simple projections of the kind shown in Fig. 561, and it will be
found that the dimensions of the object represented may be determined
from these pictorial projections as easily as from the ordinary
projections.
Fig. 563 shows an isometric projection of the solid represented in
Figs. 561 and 562, and Fig. 564 shows an oblique parallel projection of
Fig. 663.
Pig. 564.
the same solid, and a glance at either of these pictorial projections
conveys to the mind at once a clear picture of the object. Since no
lines other than those shown require to be drawn, and since the
dimensions of length, breadth and thickness may be measured directly
on these projections, it will be seen that such drawings may often be
most useful.
The theorems upon which the methods of this chapter depend
are : —
(1) Parallel lines have parallel projections.
(2) Parallel lines have equal inclinations to the same plane.
(3) Parallel lines and lines which are equally inclined to the plane of
projection have the lengths of their projections to the same scale.
In stating the above theorems it is assumed that the projectors
are parallel to one another.
257. Isometric Projection. — When a solid is rectangular each
edge is parallel to one or other of three lines or axes which are
mutually perpendicular, and when the solid is placed so that these
axes are equally inclined to the plane of projection their projections
make angles of 120"^ with one another, and the projections of the
edges of the solid may be measured with the same scale since the edges
are parallel to lines which are equally inclined to the plane of projec-
tion. Hence follows the simple construction shown in Fig. 565 for
making an isometric projection of a rectangular solid. The edges of
the solid meeting at one angular point are taken as axes and their
292
PRACTICAL GEOMETRY
plans oa, oh, and oc are drawn by the aid of the T-square and 30° set
square as shown. The lengths of the plans of the edges are marked
off with the isometric scale, the
construction of which is explained
in the next article.
The axes referred to in this
article are called isometric axes
and a plane containing any two
of them or a plane parallel to
this plane is called an isometric
plane. The word isometric means
equal measure. It should be
noted that only those lines which
are parallel to one or other of
the isometric axes are isometric
lines.
258. The Isometric Scale.
— Let oa, oh, and oc (Fig. 566) be the projections of three lines OA,
OB, and OC which meet at O and are mutually perpendicular and
which are equally inclined to the plane of projection ; it is required to
find the ratio of say 06 to OB.
Draw he at right angles to oa. On he as diameter describe a
semicircle cutting oa at O^. 0^ is the true length of the line OB of
which oh is the projection. For the proof of this construction see Art.
199, p. 229, on the projection of a solid right angle.
Fig. 565.
a
Fig. 566,
ISOMETRIC SCALE.
ORDINARY SCALE.
Fig. 567.
Fig. 568.
If ao be produced to cut he at d, then it is easy to show that,
angle dhO^ = angle dO^b = 45° ; also that angle dho = 30°, and
angle dob = 60° ; hence angle ohO^^ = 15°. Then it follows that
oh : O16 : : a/2 : ^3, that is, the length of the isometric projection of a
line is to the true length of the line as ^2 is to a/^. Hence to draw
an isometric scale corresponding to a given ordinary scale it is necessary
to get two intersecting lines whose lengths are to one another as;
V2 to V3.
Fig. 567 shows one construction for the isometric scale. EF is the
ordinary scale. Angle FEH = 15°. Angle EFH s= 45°. EH is the
isometric scale, the sub-divisions of which are obtained by drawing lines
through the sub-divisions on EF parallel to FH a§ ibown. \ti w^l be.
PICTORIAL PROJECTIONS
293
seen that the triangle EFH in Fig. 567 is similar to the triangle hOiO
in Fig. 566.
Another construction for the isometric scale is shown in Fig. 568.
Angle FHK = 90°. FH = HK. HE = FK. The ordinary scale is
made on EF and the isometric scale on EH as shown. If FH = 1,
HK = 1, and FK = V^ = EH. Hence EF = ^3 and EH : EF
::V2:V3.
In actual practice in making drawings in isometric projection, the
isometric scale is seldom used. The lengths of the isometric lines are
marked off directly from the ordinary scale. The result of this is that
the isometric drawing is larger than it would be if an isometric scale
were used in the ratio of a/ 3 to a/ 2.
259. Projection of any Given Figure lying in an Iso-
metric Plane.— Let ABCDE (Fig. 569) be the given figure, the
plane of the paper being the plane of the figure, and let OX and OZ be
two lines at right angles to one another in that plane. Let the plane
XOZ be tilted up until the axes OX and OZ and the axis OY perpen-
dicular to OX and OZ are equally inclined to the plane of projection.
The projections ox, oz, and oy (Fig. 570) will be isometric axes. It is
required to add to the projections of the axes the projection of the
figure ABCDE.
From the extremities of the straight sides of ABCDE draw parallels
to OX to meet OZ at Ai, B,, C^ and Dj. Also, from a sufficient
number of points on the curved
side AED, draw parallels to
OX to meet OZ as shown in
Fig. 569.
The isometric projections
aa^, bbij cci, etc. of the lines
AAi, BBi, CCj, etc. can now be
determined and these determine
the projections a, b, c, etc. of
the points A, B, C, etc. Hence
the required projection of the
given figure can be completed.
260. Projection of a
Circle lying in an Isometric
Plane. — The method of the preceding article may be applied to find
the projections of a sufficient number of points on a circle lying in an
isometric plane, and a fair curve drawn through the points thus deter-
mined is the projection required. This is shown in Fig. 571, where two
diameters of the circle at right angles to one another are taken as two
of the three axes. Since, however, the orthographic projection of a
circle is an ellipse, a better construction is to find the axes of the
ellipse and then determine a sufficient number of points on it by means
of a trammel as explained in Art. 45, p, 41.
Referring to Fig. 572, take two diameters of the circle at right
angles to one another as the axes OX and OZ. Draw the inscribed
square ABCD having two sides parallel to OX and two sides parallel
Fig. 569.
294
PRACTICAL GEOMETRY
^A^^^' ^^ ^^® isometric projection, aocy the projection of the diameter
AOC will evidently be perpendicular to oy the third isometric axis.
a
Fig. 571.
Fig. 572.
aoc is therefore the projection of the horizontal diameter of the circle
and will therefore have a length equal to the true diameter. Con-
sequently aoc must be the major axis of the ellipse which is the
projection of the circle. Also hod must be the minor axis of the ellipse.
It is easy to show that the angle hao is 30°.
Hence, having found o the isometric projection of the centre of the
circle, the major and minor axes of the ellipse which is the isometric
projection of the circle are found as follows. — Draw aoc perpendicular
to the third isometric axis, and make oa equal to oc equal to the true
radius of the circle. Draw ah inclined at 30° to oa, or draw ah parallel
to the isometric axis ox. Draw hod perpendicular to aoc to meet ah at h.
Make od equal to oh. aoc and hod are the required axes of the ellipse.
If the ellipse is the projection of a circle connected with an object
which is to be projected isometrically without using an isometric scale,
then oa must be made greater than the true radius of the circle in the
ratio of the ordinary scale to the isometric scale, that is, in the ratio of
V3 : V2. .
ELEVATION.
Fig. 573.
261. Isometric Projection of an Object which is not
Rectangular. — If three axes, mutually perpendicular, be taken in
PICTORIAL PROJECTIONS 295
relation to any object, the object is said to be projected isometrically
when the three selected axes are projected isometrically ; but only
those Jines on the object which are parallel to the selected axes will
be projected truly isometrically.
The general construction is to locate a sufficient number of points
on the outline of the object by perpendiculars from them on to the
planes containing the axes referred to above, and then find the isometric
projections of these perpendiculars. For example, take the case of the
pyramid shown in Fig. 574 by ordinary plan and elevation. Take a
vertical axis OZ through the vertex V and two axes OX and OY at
right angles to one another and in the horizontal plane of projection.
Fig. 573 shows these three axes projected isometrically and the projec-
tion of the pyramid built up on these axes.
262. Axortietric Projection.^ — In axomeinc projeciion the three
principal axes of a rectangular object are not all equally inclined to
the plane of projection, and although all those lines which are parallel
to one axis have their projections drawn to the same scale, those which
are parallel to another axis will, in general, have their projections
drawn to a difierent scale. Thus in the axometric projection of a
rectangular box, if no two axes have the same inclination to the plane
of projection, one scale is required for measuring its length, another
for its breadth, and a third for its depth. These three scales are
determined in a manner similar to that for the isometric scale shown
in Figs. 566 and 567. The angles at E and F, Fig. 567, must of course
be found from the figure corresponding Ao Fig. 566. The angle at E
(Fig. 567) must be made equal to the angle O^to, Fig. 566, and the
angle at F equal to the angle hO^o, The scale thus found will be the
scale for measuring lines parallel to oh.
It will be observed that isometric projection is a particular case of
axometric projection.
263. Oblique Parallel Projection. — In parallel projection the
projectors from the different points of an object to the plane of
projection are parallel to one another. In orthographic projection the
projectors are perpendicular to the plane of projection and are therefore
parallel to one another ; hence, orthographic projection is a particular
case of parallel projection. It is usual, however, to restrict the term
parallel projection to the case where the parallel projectors are inclined
or oblique to the plane of projection. By parallel projection is therefore
meant oblique parallel projection.
A special and useful case of parallel projection is that in which
the plane of projection is parallel to one of the principal faces of the
object to be projected, and the projectors are inclined at 45° to the
plane of projection. It follows that all faces of the solid which are
parallel to the plane of projection will have for their projections
figures which are of the same size and shape as the faces themselves,
also all lines on the object which are perpendicular to the faces just
mentioned will have their projections of the same lengths as the lines
themselves.
In Fig. 575, ab is the ordinary or orthographic plan, and a'b' the
296
PRACTICAL GEOMETRY
corresponding elevation of a rectangular solid placed with two parallel
faces parallel to the vertical plane of projection and with two other
parallel faces horizontal. The solid being in this simple position in
relation to the planes of projection, another projection of it, A'B', is
made on the vertical plane of projection, the projectors being parallel
to one another but inclined at 45° to that plane of projection. The
elevations of these projectors are shown inclined to XY at 30°, but
they may be inclined at any other angle. Having fixed the inclinations
of the elevations of the projectors to XY the direction of their plans is
found from the further condition that the true inclination of the
projectors to the plane of projection is 45°.
X-\ 1^
Fig. 575.
Fig. 576.
It will now be seen that the projection A'B' may be drawn directly
without any preliminary plan and elevation. The rectangle A'D' is
first drawn, being of the same size and shape as the front face of the
solid. A'C is then drawn at 30° (or any other convenient angle) to
A'E', and its length is made equal to the true length of the edge of
which it is the projection. The projection may now be completed by
drawing parallel lines as shown.
Fig. 576 shows parallel projections of solids drawn as described
above, and the student should draw these, full size, to the dimensions,
in inches, marked on them.
If desired, the projections of those lines on the object which are
perpendicular to the front face, which is parallel to the plane of
projection, may be drawn to a scale which is either
smaller or larger than that used for the projection
of the front face, and the resulting projection will
be a correct parallel projection of the object. This
simply means that the inclination of the parallel
projectors to the plane of projection is taken either Fig. 677.
greater or less than 45°. For example, referring
to Fig. 577, if the face A'D' is drawn full size the length A'C may be
drawn half full size and the complete parallel projection of the solid
Ci
A'
IPn
b^B
PICTORIAL PROJECTIONS
297
will then be as shown. When different scales are used the scales
;jhould be stated on the drawing.
When a face of the object perpendicular to the front face contains
a figure made up of curved lines, or lines which are not parallel to the
edges of that face, the projection of such a figure is determined by
using co-ordinates in the manner described in Art. 259 for isometric
projection. Also a parallel projection of an object of any form not
rectangular may be made by using co-ordinates as described in Art. 261
for isometric projection.
Exercises XXI
Note, The pictorial projections asked for in the following exercises are to be
drawn, as far as possible, directly without first drawing the object in ordinary plan
and elevatimi.
1-12. Ordinary orthographic projections of various objects are given in Figs.
578 to 589. Represent these objects in isometric projection.
For the dimensions of the objects shown in Figs. 578 to 589 take the sides of the
small squares as half an inch.
Fig. 578.
^' 1 1 -^ K k' "^-1 fr \a^ AT].
J i^E ij ^>o .i- i 4lvt
z ^ ^ r z\ \ //\\
1 1 r'^ ' > 1
j" 1 '11]
r\\/y
i--'-lt -~ X-^k^iiS^
oF.. X loK x] y oi.i 1 X VA^
Fig. 579. Fig. 580. Fig. 581. Fig. 582.
oyi\ — X -I — p — f-x
\o\ X _J__^_L_ac
^T^^
Wn
rEi-r::
S /
^^-^ _
J>^
-3
T^?^
-r
1
^
1 ■ ^
\l
^^
s.
i
^'-
/ \
\
-O-
v-j
X
i
V,
T^
/
±:
u^±
Fig. 583. Fig. 584.
Fig. 585.
Fig. 586.
298
PRACTICAL GEOMETRY
a'
X'
^
^-"i "W
S '^
a
c-i ,.:S..
z'
\ f
Mt\
SB^/\i
mu
^
^-i
\y /-^ / /
_ 1
g^gg
^ /\
c-l
W
X'
V
s.
V
s
^
^
,,*^
^
-
^rf**
^
^
^^
X'
X
f
^
Fig. 687.
Fig. 588.
Fig. 589.
13-24. Kepresent the objects shown in Figs. 578 to 589 in axometric pro-
jection, the axes to be arranged as shown in Fig. 590.
25-36. Represent the objects shown in Figs. 578 to 589 in oblique parallel
projection, the axes to be arranged as shown
in Fig. 591. The scale is to be the same
for all the axes of projection.
37. The axes of two equal cylinders (2
inches in diameter) and their common per-
pendicular are in isometric projection.
Show the intersection of the cylinders with-
out using an auxiliary plan or elevation,
(a) when their axes intersect, (&) when
their axes are \ inch apart where they are nearest together.
38. The same as exercise 37 except that the projection of the axis of one
cylinder is parallel to ox and the projection of the axis of the other is parallel to
07/, Fig. 590.
39. Represent the cylinders referred to in exercise 37 in oblique parallel pro-
jection. The projection of the axis of one cylinder to be parallel to ox and the
projection of the axis of the other to be parallel to oy, Fig. 591. Show the inter-
section of the cylinders. The scale is to be the same for all the axes of projection.
40. A solid made up of a hemisphere and part of a circular cylinder is
shown in Fig. 592. The centre of the base of the hemisphere is on the axis of the
cylinder. Draw an isometric projection of this solid.
41. Fig. 593 shows a locomotive crank axle with the
fillets at the junctions of the various cylindrical parts with
one another and with the crank arms left out. Draw an
isometric projection of this axle to a scale of 1^ inches to
a foot.
42. Draw an oblique parallel projection of the crank
axle shown in Fig. 593 to a scale of IJ inches to a foot.
The axes of projection to be arranged as shown in Fig.
591, ox being the projection of the axis of the axle.
43. The relation between the pressure, P (in lb. per
square inch), the volume, V (in cubic feet), and the abso-
lute temperature, T (in degrees Fahrenheit) of 1 lb. of air
is given by the equation PV = 0*37 T. Fig. 594 shows,
in oblique parallel projection, contours of this surface,
between pressures of 10 and 100 lb. per square inch, at intervals of 500° from
500° to 2500°, the planes of these contours being parallel to the front pressure-
FiG. 592.
PICTORIAL PROJECTIONS
299
volume plane. Draw this Fig. to the following scales. — Pressure, 1 inch to 20
lb. per square inch. Volume, 1 inch to 4 cubic feet. Temperature, 1 inch
to 600^.
"~~"
_.-
_
K-7-^
K- 9 —
*A*
Jl
.4.
4^
^-l!--*^
K4^.
k4^
k4*
*- 9 -*h-7-H
1
-00-
* ■ ■
.
\
t
' — '
Fig. 593.
Add the contours whose planes are parallel to the bottom volume-temperature
plane at intervals of 20 lb. per square inch. Also, add the contours whose planes
r"A -y'2S00
VOLUME. 18-5
Fig. 594.
are parallel to the left hand pressure-temperature plane at intervals of 4 cubic
feet.
Note that the curves in Fig. 594 are rectangular hyperbolas points on which
may be found by the construction given in Art. 50, p. 47.
CHAPTER XXII
PERSPECTIVE PROJECTION
264. Difference between Perspective and Orthographic
Projection. — In perspective, conical, or radial projection all the projec-
tors converge to a point, while in orthographic projection they are per-
pendicular to the plane of projection. The perspective projection of
an object represents it as it would actually appear to the eye of an
observer placed at the point to which all the projectors converge.
In perspective projection, the plane of projection, called the picture
plane, may be situated anywhere but it should meet all the projectors,
otherwise the projection or picture as it is called would be incomplete.
Generally the picture plane is assumed to be placed in a vertical
position between the eye of the observer and the object, and perpen-
dicular to the direction in which the observer is looking.
The point to which all the projectors converge is called the station
point or point of sight.
265. Direct Method of drawing a Perspective Projection.
— The following method of drawing a perspective projection of an
object follows directly from the definition of perspective projection,
and it will be seen that it is simply the determination of the plane
section of a pyramid or a combination of pyramids which have a
common vertex.
First draw the orthographic projections of the object, station point,
and picture plane so that the latter is perpendicular to the ground
line XY, as shown in Fig. 595, which illustrates the method applied
to a square prism standing on the ground, oc, o'c' is the picture plane
and 8s' is the station point.
Consider the edge ab, dV of the prism. Draw the plans sa, si and
the elevations s'a', s'b' of the conical or radial projectors of the extre-
mities of the edge under consideration. These radial projectors meet
the picture plane at points of which a and f3 are the plans and a and /3'
the elevations. Through a point o in oc draw on parallel to XY. With
centre o and radii oa and of3 draw arcs of circles to cut on at a^ and p^
respectively. From a^ and /?i draw perpendiculars to XY to meet
lines through a and ^' parallel to XY at A' and B' respectively. A
straight line A'B' will be the perspective projection of the edge ab, a'b'
of the prism. In like manner the perspectives of the other edges may
be obtained as shown.
It is evident that by the first part of the foregoing construction
PERSPECTIVE PROJECTION
301
edge views of the required picture are obtained, one being a plan and
the other an elevation, and by the second part of the construction the
true shape of this picture is determined.
The points c and c' are the feet of the perpendiculars from 8 and s'
respectively to the edge view of the picture plane. If with centre o
and radius oc an arc of a circle be drawn to cut on at fi, and if c^G be
drawn perpendicular to XY to meet the horizontal through a' at 0,
then C will be the position, in relation to the picture A'B', of the foot
of the perpendicular from the station point to the picture plane.
The foot of the perpendicular from the station point to the picture
plane is called the centre of vision. It will be observed that the centre
of vision is the orthographic projection of the station point on the
picture plane.
The horizontal line drawn on the picture plane through the centre
of vision is called the horizontal line.
Fig. 595.
The intersection of the picture plane with the ground plane is
called the ground line or base line.
The plane perpendicular to the ground, and containing the station
point and centre of vision is caUed the vertical plane.
The line in which the vertical plane intersects the picture plane is
called the vertical line.
266. The " School of Art " Method of drawing a Perspec-
tive Projection. — The direct method of drawing a perspective
302
PRACTICAL GEOMETRY
projection, described in the preceding Article, is exceedingly simple in
principle, and may be quickly learned, and is easily understood by a
student possessing a little knowledge of elementary solid geometry.
The direct method may be applied to draw the perspective of any solid
in any position, and in some cases it is the best method to use, but if
the solid has many parallel lines on it the construction can be much
simplified and greater accuracy obtained by making use of a number
of principles which are explained in succeeding articles. These prin-
ciples are applied in what may be called the " School of Art " method
of making a perspective drawing. Afterwards a combination of the
two methods very commonly used in practice will be described.
267. To determine the Perspective of a Given Point. — The
point is given by stating its distances from the ground and from the
vertical and picture planes. The first and second of these distances
determine the position of the orthographic projection of the point on
the picture plane.
In Fig. 596 the plane of the paper is the picture plane, C is the
centre of vision, CH is the horizontal line, CL is the vertical line, and
f' is the orthographic projection of the given point on the picture
plane. On CH make CD equal to the distance of the station point
from the picture plane. Draw p'K parallel to CH, and make p'K
equal to the distance of the given point P from the picture plane.
Join DK and jp'C. The point F where these lines intersect is the
perspectiTe of the point P. It is evident that CP' : Yp' ; : CD : p'K.
Fig. 696.
Fig. 597.
Referring now to Fig. 597 which is a distorted perspective projec-
tion of the lines of Pig. 596 together with the point P and the station
point S. Since SC and Pp' are perpendicular to the picture plane they
are therefore parallel to one another, and the triangles SCP' and Pp'P'
are similar. Hence CP' : Fp' : : SC : Pp' ; but CD is equal to SC and
p'K is equal to P^, therefore CP' : Pp' : : CD : p'K. But F is the point
where SP cuts the ^picture plane and is therefore the perspective of
the point P.
The point D, Fig. 596, is called the 'point of distance.
268. The Perspectives of Parallel Lines converge to a
Point.— Let A'B' and CD' (Fig. 598) be the perspectives of the
parallel lines AB and CD respectively. Let SV be a line parallel to
PERSPECTIVE PROJECTION
303
A', B', and V are in
same straight line.
Fig. 598.
AB and CD, and if it meet the picture plane at all, let it meet it at Y.
SV will be the line of intersection of the planes ABS and CDS.
Since each of the points A', B', and V are in the picture plane,
and also in the plane ABSV they must all lie on the intersection of
these two planes ; there-
fore
the
Similarly C, D', and V
are in a straight line ;
therefore A'B^ and CD',
the perspectives of the
parallel lines AB and CD,
converge to the point V.
The point towards
which the perspectives of
parallel lines converge is
called the vanishing point
of these lines.
The following particu-
lar cases are worthy of notice : —
(a) When the lines are perpendicular to the picture plane, SV will
also be perpendicular to that plane, and V will coincide with the centre
of vision. Hence, the perspectives of lines which are perpendicular to the
picture plane, converge towards the centre of vision.
(h) When the parallel lines are parallel to the picture plane, SV
will also be parallel to that plane ; therefore V will be at an infinite
distance from S, and A'B' and CD' will be parallel to one another.
Hence, the perspectives of parallel lines ichich are also parallel to the
picture plane, are parallel to one another.
(c) When the parallel lines are horizontal, SV will also be hori-
zontal ; therefore V must be on the horizontal line. Hence, the per-
spectives of parallel horizontal lines converge towards a point on the
horizontal line.
{d) When the parallel lines are parallel to the vertical plane, S V
will be in the vertical plane ; therefore V will be on the vertical line.
Hence, the perspectives of parallel lines ivhich are also parallel to the
vertical plane, converge towards a point on the vertical line.
e) The perspectives of vertical lines are parallel to the vertical line.
/) The perspectives of horizontal lines which are parallel to the
picture plane are parallel to the horizontal line.
(cj) The perspectives of parallel horizontal lines which are inclined at
45° to the picture plane converge towards the point of distance.
269. To Determine the Vanishing Point of a Line. —
Take the picture plane and the horizontal planfe containing the
station point for co-ordinate planes, and on these draw a plan cd) and
elevation a'V of the line whose vanishing point is required (Fig. 599).
HC, the horizontal line, will be the ground line for this plan and
elevation.
Draw Cs perpendicular to HC and equal to the distance of the
304
PRACTICAL GEOMETRY
Fig. 599.
station point from the picture plane, s will be the plan of the station
point and C its elevation. Through s draw sv parallel to ah, meeting
HC at V. Through C draw CV parallel
to a'h' to meet the perpendicular from v
to HC at V. The point V is the required
vanishing point, because V is the point in
which a line through the station point
parallel to the original line AB meets the
picture plane.
If the original line is horizontal CV
will coincide with HC, the horizontal line ;
therefore Y is on the horizontal line, and
the plan sv (or sV) makes an angle with
the horizontal line equal to the inclination of the original line to the
picture plane.
270. To mark off on the Perspective of a Line a Part
whose True Length is given. — Let TV (Fig. 600) be the per-
spective of a line whose trace on the
picture plane is T, and whose vanishing
point is V. As in Fig. 599, s is the
plan of the station point, and sv the
plan of the line which passes through
the station point, arid is parallel to the
line of which TV is the perspective.
With centre v and radius vs de-
scribe the arc sSg meeting the hori-
zontal line HC at Sg. VS2 is the real
distance of the vanishing point V from -p^Q^ 500.
the station point. Through V draw
VM parallel to HC, and with centre V and radius VS2 describe the
arc SgM.
Draw TQ parallel to HC. Let A' be a given point in TV. Draw
MA' and produce it to meet TQ at P. Make PQ equal to the given
length. Draw QM meeting TV at B'. The true length of the part
whose perspective is A'B' will be equal to the given length.
The construction is much simplified when the original line is hori-
zontal, for then V is on the horizontal line, and sv is equal to the true
distance of V from the station point
The construction for this case is shown
in Fig. 601.
The theory of the construction
shown in Figs. 600 and 601 will be
understood by reference to Fig. 602,
which is a perspective view of the
ground and picture planes, and the
various lines in their natural positions.
In Figs. 600, 601, and 602, the same
letters denote the same points. S is the station point, TAB the line
whose perspective is TV and V is its vanishing point. T is the trace
M
i
V
y_ / \V-^^
^\
A>
Y
c
A
-•^
T \
\
J
/
T'
V
>
V
H
M J
c
y
/
"/'
'^> \ \
"n
V
Fig. 601.
!►
PERSPECTIVE PROJECTION
305
Fig. 602.
of AB on the picture plane. SV is parallel to TAB, and the per-
spectives of all lines which are parallel to TAB or SV will pass through
V as already explained
in Art. 268, o.nd il- ^"^^
lustrated by Fig. 598. ""
Similarly, the per-
spectives of all lines
which are parallel to
SM will pass through
M, and conversely,
lines whose perspec-
tives pass through M,
and which are in the
same plane, will be
parallel to SM. Now
AP and BQ are lines
whose perspectives
A'P and B'Q pass
through M, and they are in the same plane, therefore AP and BQ
are parallel to SM.
Comparing the triangles SVM and BTQ, TB is parallel to VS, TQ
is parallel to VM, and BQ is parallel to SM, therefore the triangles
are similar. But VM is equal to VS, therefore TB is equal to TQ,
and it follows that TA is equal to TP since AP is parallel to BQ,
therefore AB is equal to PQ. But PQ is a line in the picture plane,
therefore if PQ is made equal to the given length, the construction
given will determine the perspective of the part required.
The point M in Figs. 600, 601, and 602 is called a measuring point
for the vanishing point V. It is evident that each vanishing point
can have two measuring points, one to the right and another an equal
distance to the left of the vanishing point.
When the original line AB is parallel to the picture plane the
foregoing construction fails because the points T and V are "at
infinity." A simple and con-
venient construction in this case
is shown in Fig. 603 where A'E'
is the direction of the perspective,
and A' is the point in it from
which the part is to be marked
off. Take any point M on the
horizontal line. Join MA' and
produce it to meet at P a horizontal line PR on the picture, plane
which is at a height above the ground line equal to the height of
A above the ground. Draw PQ parallel to A'E', and make PQ equal
to the given length. Join MQ meeting A'E' at B'. A'B' is the part
required. The theory of this construction is as follows. PM and QM
are the perspectives of parallel horizontal lines passing through the
points A and B and meeting the picture plane at P and Q. The
figure ABQP in space is a parallelogram and AB is equal to PQ.
X
Fig. 603.
306
PRACTICAL GEOMETRY
But PQ is in the picture plane and may therefore be measured
directly.
271. Examples. — (1)^ square prism, 2 inclies X 2 inches x 4
incites, rests loith one rectangular face on the ground. Nearest corner
J inch to left of vertical plane and 1 inch hehind picture plane. Long
edges inclined, at 50° to picture plane and vanish to left. Station point
3 J inches above ground and 6| inches from picture plane.
Fig. 604 shows this example worked out by the " School of Art "
method. V^ and Y^ are the two vanishing points for the horizontal
edges, determined by the con-
struction explained in Art.
269, and M^ and Mo are the
corresponding measuring
points (see Art. 270). D is
the point of distance (Art.
267).
The point O', the perspec-
tive of the nearest corner on
the ground, is first determined
(Art. 267). Joining O' to V^
and V2 determines the direc-
tions of the perspectives of
the horizontal edges meeting
at O. The lengths OT' and
O'Q' of the perspectives of
these edges are then de-
termined (Art. 270). A
vertical line through O' will
be the direction of the per-
spective of the vertical edge
which has its lower extremity
at O. The length O'R'
marked off by the construction
explained in the latter part of Art.
270.
Fig. 605.
The remainder of tho
PERSPECTIVE PROJECTION
307
construction is obvious. All the construction lines are shown in the
figure.
Fig. 605 shows the same example worked out by a combination of
the " School of Art " method and the direct method explained in Art.
265. HC, the horizontal line, is taken as a plan of the picture plane,
and a plan pq of the solid is drawn, the solid being placed in its proper
position in relation to the picture plane and vertical plane. 8 is the
plan of the station point. Lines from the angular points of the plan
to s intersect HC at points which are the plans of the angular points
of the picture on the picture plane. Vertical lines through these
points on HC will contain the angular points of the picture which is
to be drawn below HC.
Fig. 606.
The vanishing points V, and Vg (Vg is outside the figure) are
determined as before. Produce po to meet HC at t and through t
draw the vertical {I to meet the ground line of the picture plane at
T. T is obviously the perspective of the point where PO produced
308 PRACTICAL GEOMETRY
meets the picture plane, and is also the point itself. The line TV^
will therefore contain the perspective of OP. The points 0' and P
are determined by vertical lines from d and p' to meet T V^ as shown.
Joining O' to Vg determines the line containing the perspective of OQ,
and the point Q' is found by dropping a vertical line from g' to meet
O'K,.
The perspectives of the vertical edges are of course vertical lines.
The length O'R' is determined as in Fig. 604 and the remainder of
the construction is obvious.
(2) A second example is shown worked out in Fig. 606 by a com-
bination of the " School of Art " and direct methods. The per-
spectives of the circular arcs are of course determined by first finding
the perspectives of a number of points in them and then joining
by fair curves. The construction lines for a point P' on one of the
curves are fully shown, others are omitted for the sake of giving
clearness to the figure.
272. Shadows in Perspective. — When a solid which is drawn
in perspective casts a shadow, the perspective of the shadow may be
drawn in the ordinary way from the shadow as determined from the
orthographic projections of the solid as shown in Chap. XXVII., and in
cases where the shadow involves the intersection of curved surfaces
this is generally the simplest method to adopt. In many cases how-
ever the perspective of the shadow can easily be determined from the
perspective of the solid, by applying the constructions explained in
the remainder of this article.
It will only be necessary to consider the perspective of the shadow
of a point. At first all the light will be assumed to come from a
point, the modifications of the constructions for parallel rays of light
being afterwards described.
Let P' be the perspective of the point from which all the rays of
light diverge, and let A' be the perspective of a point whose shadow
is to be determined. The positions of the points P and A in space are
supposed to be known.
(a) To determine the perspective of the shadow cast hy the point A on
the ground. Let M' and N' (Fig. 607) be the perspectives of the feet
of the perpendiculars from A and P on
the ground. Join M'N' and produce it
to meet PA' produced at Q'. Q' is the
perspective of the trace of PA on the
ground, and is therefore the perspective
of the shadow of A on the ground.
(6) To determine the perspective of the
shadow cast by A on a horizontal plane
ivhich is at a given height above the ground.
Produce M'N' (Fig.' 607) to meet the
ground line at T and the horizontal line
at V. Draw the vertical line TTj and
make TTj equal to the given height of the horizontal plane above the
ground. Draw TjV to cut P'A' produced at Q/. Q/ is the perspective
I
PERSPECTIVE PROJECTION
309
Fig. 608.
of the shadow required. TV and T^V are the perspectives of parallel
horizontal lines which are in a vertical plane containing PA, therefore
Q/ is the perspective of the trace of PA on a horizontal plane at a
height above the ground equal to TT^.
(c) To determine the perspective of the shadow cast hy A on a given
vertical plane. Let EF (Fig. 608) be the perspective of the trace of
the given vertical plane on the ground,
and let EF cut WW produced at R'.
Draw the vertical line R'Q' to meet P'A'
produced at Q'. Q' is the perspective of
the shadow required. R'Q' is evidently
the perspective of the intersection of the
given vertical plane with the vertical
plane containing PA, therefore Q' is the
perspective of the trace of PA on the
given vertical plane.
(d) To determine the perspective of the
shadow cast hy A on a given inclined plane. Let EF (Fig. 609) be the
perspective of the trace of the given plane on the grouTid and let
FFi be its trace on the picture plane.
Produce M'N' to meet the ground line at
T and the horizontal line at V. Draw
the vertical line TT^ At any convenient
height draw the horizontal line T^F, to
cut TTi at Ti and FF, at F^. Join T^V
and F^E and let these lines meet at U'.
Let TV meet EF at R'. Join R'U'.
Produce P'A' to meet R'U' at Q'. Q' is
the perspective of the shadow required.
T^V and F^E are the perspectives of
the lines in which a horizontal plane,
whose trace on the picture plane is TjFi, intersects the vertical plane
PM, and the given inclined plane respectively. U' is therefore the
perspective of a point in each of these planes; but R' is the per-
spective of another point in each of these planes, therefore R'U' is the
perspective of the line of intersection of these planes. Hence Q' is the
perspective of the trace of PA on the given inclined plane.
When the rays of light are parallel their perspectives will converge
to their vanishing point which is determined by the construction
explained in Art. 269. This vanishing point may then be taken as the
perspective of a luminous point from which the perspectives of all the
rays of light diverge, and the perspectives of the shadows are determined
as already explained. It should be observed that in this case the
perspective of the foot of the perpendicular on the ground from the
luminous point (now at infinity) is on the horizontal line.
Fig. 609.
310
PRACTICAL GEOMETRY
Exercises XXII.
1. Represent in perspective a point A on the ground plane, 1 foot to the
right of the spectator and 2 feet from the ground line ; also a point B 4 feet
to the left of the spectator, 2 feet from the picture plane, and 3 feet above the
ground plane. Join AB. AB is the nearest side of a rectangle of which two
other sides are horizontal and 4 feet long. Complete the perspective represen-
tation of the rectangle. Position of eye, 11 feet from picture plane and 5 feet
from ground. Scale | inch to 1 foot. [b.e.]
2. A rectangular stone slab, 10 feet long, 8 feet wide, and 6 inches thick,
lies with one face on the ground. The longest edges recede from the picture at
an angle of 60° towards the right, and the nearest corner is 1 foot to the left of
the spectator, and 2 feet from the ground line. Through the centre of the slab
is cut a circular hole, 6 feet in diameter. K.epresent the slab in perspective.
Position of eye and scale as in exercise 1. [b.e.]
3. Two rectangular blocks are shown at Ex. 3, Fig, 612. The lower block
lies on the ground. Draw these blocks in perspective projection. The corner
(m! is to be \ inch to the left of the spectator, and
\ inch from the ground line. The edge a6, a'h' is to
recede towards the right at an angle of 60° to the
picture plane. Position of eye, b\ inches from the
picture plane and 2i inches above the ground.
4. A solid letter R is shown at Ex. 4, Pig. 612.
Draw this in perspective. The corner aa! is to be
on the ground 1 inch to the right of the spectator
and 3 inches from the ground line. The face of the
letter is to be in a vertical plane which is to recede
from the picture plane at an angle of 50° towards
the left. Position of eye, 5^ inches from the picture
plane and 2,} inches above the ground.
5. Put into perspective a point on the ground
plane 5 feet to the right of the spectator, and 3 feet
within the picture, and, receding from the point,
draw a line 10 feet long at right angles with the
picture plane. This line is to be the lowest edge
of a square slab 1 foot thick, having its square faces
inclined to the ground at 60° towards the right. On
the upper face of the slab inscribe a circle, and,
centrally with it, draw another circle 4 feet in dia-
meter, which is to be the base of a right cone having
an altitude of 2 feet. The eye is to be 14 feet from
the picture plane, and 8 feet from the ground. Scale,
5 inch to a foot. [b.e.]
6. A right prism 3| inches high has its base,
which is a regular octagon of 1 inch side, on the
ground. A cylindrical slab 4 inches in diameter and
1 inch thick rests on the prism, the axes of the two solids being in line. Repre-
sent the two solids in perspective. One corner of the base of the prism is to be
\ inch to the right of the spectator and 2 inches from the ground line, and one
of the edges of the base adjacent to that corner is to vanish towards the right at
an angle of 50° with the ground line. Height of eye above the ground, 2^ inches.
Distance of eye from picture plane, 6 inches.
7. Fig. 610 gives the elevation and half plan of an object standing on the
ground plane. Put the whole object into perspective, with the sides of its base
inclined at angles of 45° to the picture plane, and its nearest vertical edge 2 feet
to the right of the spectator, and 2 feet within the picture. The eye is to
be 5 feet from the ground, and 12 feet from the picture. Scale \ inch to a
foot. " [B.E.]
"7(
+
1
1
1
i.P.l"'
/ATION
Fig. 610.
PERSPECTIVE PROJECTION
311
Fig. 611.
8. A vertical section of a square ceiling is shown in Fig. 611, the upper part
being of the form of a shallow square pyramid. Put the ceiling into perspective
with its horizontal edges inclined
to the picture plane at angles of
30° to the right and 60° to the
left, its nearest corner being 12
feet vertically over a point on
the ground plane 3 feet to the
left of the spectator, and 1 foot
within the picture. The eye is
to be 12 feet from the picture,
and 5 feet from the ground. Scale ^ inch to a foot. [b.e.]
9. Draw, by the method of Art. 265, the perspective projection of the solid
shown at Ex. 9, Fig. 612. The edge AB is to be on the ground and parallel to
the picture plane. The axis of the solid is to be inclined at 50° to the ground,
and the nearest point C is to be in the picture plane, and IJ inches to the right
of the spectator. Station point, 4 inches from the picture plane, and 2 inches
above the ground.
10. Draw the perspective projection of the solid shown at Ex. 10, Fig. 612.
The solid is to stand on the ground with the edge AB in the picture plane, and
1^ inches to the right of the spectator. The face ABC is to be inclined at 4.0°
e y\\ II
I \ .. Ex.13.
-r -r ^ / \ \ y ^ H
-- ,nr -::^ = } \ -rVl^T.'IIliS'
EX.3. I -af..... J'V f— t b'-^ ti
zL %..— m.^- —r- 5^-
:t: i: zri-^-zzz"-.- ". r -
H NmJmU U iffe'lm
:f E alr^ h-_-l%7jzz 'r J: - =F -
y /\ \ ^ ^ J Ex.14.
-i^i A--Z- S.J..^-=:.ii
Jl f \ ^
r'\ C! fl^ 1 ^—^
\^EX.4 - - /(-|\ ^ V
vV /NEx.iiy //|\\,«'/k \\
a W ^ '^ ' \ aW. ]i'4llx^44
"^ 7 \^ ilirk ^^^z ^ W-i^
^-K / sUp V ^
a \ / [
1 \^ 1 1 :x.i5.
Fig. 612.
In reproducing the above diagrams the sides of the small squares are to be
taken equal to half an inch.
to the picture plane. Height of station point 2 inches. The position of the
centre of vision, and the distance of the station point from the picture plane is
to be determined from the further condition that the distance between the
vanishing points for the horizontal edges is to be 7j inches.
312
PRACTICAL GEOMETRY
11. Draw the perspective of the splid shown at Ex. 11, Fig. 612. The base
of the solid is to be on the ground, and the nearest edge AB | inch behind the
picture plane and IJ inches to the right. The face ABC is to be inclined at 30^
to the picture plane. Station point 3 inches from picture plane and 2i inches
from ground.
12. A skeleton cube is shown at Ex. 12, Fig. 612. Draw this object in per-
spective when the nearest vertical edge is in the picture plane, and 1 inch to the
left, and a face containing that edge is inclined at 35'^ to the picture plane.
The station point to be 3^ inches from the picture plane and 1 inch above the top
face of the object.
13. Draw the perspective of the object shown at Ex. 13, Fig. 612. The edge
AB to be vertical, and in the picture plane, and h inch to the left. The face ABC
to be inclined at 30^ to the
picture plane. Station
point to be 4 inches from
the picture plane and 1|
inches above the point B.
14. Draw the perspec-
tive of the object shown
at Ex. 14, Fig. 612. The
edge AB to be vertical and
ix). the picture plane, and
directly opposite to the
station point. The point
C also to be in the picture
plane. Station point, 14-
inches above the level of
point C, and 3^ inches
from the picture plane.
15. A gable cross is
shown at Ex. 15, Fig. 612 ;
draw this in perspective.
The edge AB to be vertical
and in the picture plane
and ^ inch to the left.
The face ABC to be in-
clined at 45° to the picture
plane. Station point, IJ inches below the point A, and 4 inches from the picture
plane.
16. The plan and elevation of a wheelbarrow standing on the ground are
shown in Fig. 613 to the scale, J inch to a foot. Represent this object in
perspective using the scale, 1 inch to a foot. The line AB is the horizontal trace
of the central vertical plane of the wheelbarrow. The point A is to be 1 foot
to the left of the spectator and 1 foot from the ground line, and the line AB is
to vanish towards the right at 40° to the picture plane. The eye is to be
6 feet, by scale, from the picture plane and 2i feet above the ground plane, [b.e.]
17. Referring to exercise 14, let the station point be moved 1^ inches
nearer to the object, and let the picture plane be moved parallel to itself a
distance of 5 inches, so that the station point comes between the solid and the
picture plane.
18. Work the example shown in Fig. 606, p. 307, to the dimensions given.
Then take a point 3 inches to the left, 5 inches above the ground, and 3 inches
behind the picture plane. Consider this as a luminous point and determine the
perspective of the shadow cast by the solid on itself and on the ground ; all tho
rays of light to come from the luminous point.
Fig. 613.
CHAPTER XXIII
CURVED SURFACES AND TANGENT PLANES
273. Generation of Surfaces. — Surfaces may be considered as
generated by a line, straight or curved, moving in a definite manner.
Thus, a plane may be generated by a straight line moving parallel to
one fixed straight line and in contact with another fixed straight line.
Again, a sphere may be generated by the revolution of a semicircle
about its diameter which remains stationary.
The moving line which generates a surface is called the generating
line or generatrix of the surface, and a line which serves to con-
strain or direct the motion of the generatrix is called a directrix.
The same surface may be generated in numerous ways, but generally
there are only a few simple ways in which a surface may be generated.
Take the case of the surface of a right circular cylinder. There are
two simple ways in which this surface may be generated: (1) by a
straight line moving in contact with a fixed circle to the plane of
which it remains perpendicular, as shown by the oblique projection in
Fig. 614, where the moving line is shown in twelve diflferent positions :
(2) by a circle moving so that its centre remains on a fixed straight
line to which the plane of the circle is always perpendicular as shown
by the oblique projection in Fig. 615, where XX is the fixed straight
line and 1, 2, 3, 4, and 5 are positions of the moving circle. This
surface may however be generated by an ellipse which moves so that all
points on it travel along parallel lines, but the ellipse must be such
Fig. 615.
Fig. 616.
Fig. 61^
that its projection on a plana at right angles to the direction of its
motion is a circle.
314
PRACTICAL GEOMETRY
The two simple Wcays in which the surface of a right circular cone
may be generated are : (1) by a straight line which passes through a
fixed point (the vertex of the cone) and moves in contact with the
circle which is the base of the cone as shown in Fig. 616 : (2) by a
circle of changing radius which moves with its centre on the axis and
its plane perpendicular to the axis, the radius of the circle being
proportional to its distance from the vertex of the cone as shown in
Fig. 617.
A surface which may be generated by a line, straight or curved,
revolving about a fixed straight line is called a surface of revolution.
The fixed straight line about which the generating line revolves is
called the axis of the surface. Sections of a surface of revolution by
planes at right angles to its axis are circles. Sections by planes
containing the axis are called meridian sections. All meridian sections
are exactly alike.
A surface which may be generated by the motion of a straight
line is called a ruled surface. Ruled surfaces may be divided into
two classes — developable surfaces and twisted surfaces. A developable
surface may be folded back on one plane without tearing or creasing
at any point. The generating line of a developable surface moves
in such a manner that any two of its consecutive positions are in the
same plane. All ruled surfaces which are not developable are twisted
surfaces.
274. Plane Sections of Curved Surfaces. — The way in
which a curved surface is generated being known the projections of
the generating line in any number of positions can be drawn. The
intersections of a given plane with the generating line in each
of these positions can then be determined. This will give a number
of points on the inter-
section of the plane with
the surface, and a fair
curve through them will
be the complete inter-
section required.
When the curved sur-
face can be generated in
a number of simple ways,
that mode of generation
should be made use of
which has the projections
of its generating line the
simplest possible.
Example 1. A sur-
face is generated by a
horizontal line which
moves in contact with
the hne AB (Fig. 618) . ^^^^ g^g.
and the surface of the
cone VCD. To find the, section of this surface by the plane LMN.
CURVED SURFACES AND TANGENT PLANES 315
Take a section of the given cone and plane by a horizontal
plane cutting ab, a'h' at tt'. The plan of the section of the cone is
a circle and tangents to this circle from t are the plans of two
positions of the generating line. The points r and 8 in which
these tangents cut os the plan of the intersection of the assumed
plane of section with the plane LMN are the plans of points on
the section required. Projectors from r and s to meet the
horizontal through i determine / and s' . In a similar manner
any number of points on the required section may be found.
Example 2. Referring to Fig. 619, ah, ah' is a horizontal circle
3 inches in diameter, cd, c'd! is another horizontal circle IJ inches
Fig. 619.
in diameter. The line joining the centres of these circles is vertical
and 21 inches long. Two points move, one on each circle, wdth equal
velocities in opposite directions. A surface is generated by
angular
a straight line w^hich contains the above mentioned
pomts.
316
PRACTICAL GEOMETRY
Twelve positions of the generating line are shown in plan and in two
elevations.
It wiJl be found that horizontal sections of the surface at
levels IJ inches and 4^ inches above the plane of the larger circle
are straight lines e/, e'/' and gh, g'li respectively. These straight
lines are of definite lengths and are at right angles to one another.
It will also be found that horizontal sections at levels other than
that of the circles or the straight lines are ellipses. Two of these
elliptic sections are shown, one at the level m'n' and the other at
the level p'(^. The student should work out this example, full
size.
The surface described in this example is one that occurs in
the science of optics. It is obvious that the surface may also
be generated by a straight line moving in contact with the lines
ab^ a'b', cd, c'd' and one of the circles or one of the ellipses.
Example 3. A surface is generated by a circle whose plane is
horizontal and whose centre moves along the straight line AD
(Eig. 620). The radius of the circle varies so that the circle
V.T. OF PLANE
Fig. 620.
intersects the semicircle ABC. It is required to find the section
of this surface by the given plane which is parallel to XY.
Take a horizontal plane intersecting the arc ABC at M, the straight
line AD at N and the given plane in the straight line PQ. With
n as centre and radius ww describe a circle to cut pq at r and 8.
Perpendiculars to XY from r and 8 to meet pq' determine the points
CURVED SURFACES AND TANGENT PLANES 317
/ and s. rr' and ss' are points on the section required, and in like
manner any number of points may be found.
It is evident that the circle with n as centre and nm as
radius is the plan of the generating circle when it is at the
level p'q.
275. Intersection of Straight Line and Curved Surface. —
Assume a plane to contain the straight line and determine the
intersection of this plane with the curved surface by the method
described in the preceding article. The intersection of the straight
line with the intersection of the assumed plane and the curved surface
will be the intersection required. In selecting a plane to contain the
straight line choose one whose intersection with the curved surface will
have the simplest possible projections or projections which are easiest
to determine.
276. Tangent Planes. — If through a given point on a curved
surface any two lines be drawn on that surface, the plane containing
the tangents to these lines through the given point is the tangent plane
to the surface at that point. If a straight line can be drawn on the
curved surface through the given point, as can be done on all ruled
surfaces, the tangent plane at the given point will contain this line.
The normal to a surface at a point on it is the perpendicular to the
tangent plane at that point.
277. Tangent Plane to a Cone at a Given Point on its
Surface. — V (Fig. 621) is the vertex of the cone and P the given
Fig. 621. Fig. 622.
point on its surface. Join YP and produce it, if necessary, to meet
the horizontal trace of the cone at Q. Through Q draw a tangent to
the horizontal trace of the cone. This tangent will be the horizontal
trace of the required plane. The vertical trace of the plane can be
determined from the condition that the plane contains the vertex of
the cone. The straight line VQ is obviously the line of contact
between the cone and plane.
318 PRACTICAL GEOMETRY
278. Tangent Plane to a Cone through a Given External
Point. — T (Fig. 622) is the vertex of the cone and R the given point.
Join YR. The required plane will contain the straight line YR, and
the horizontal trace of the plane will pass through T the horizontal
trace of YR. The tangent TQ to the horizontal trace of the cone is
the horizontal trace of the required tangent plane and its vertical trace
may be found from the condition that the plane contains the straight
line YR.
If Q is the point of contact of the tangent from T to the horizontal
trace of the cone, then the straight line YQ is the line of contact
between the cone and plane.
When T the horizontal trace of YR falls outside the horizontal
trace of the cone there will be two tangent planes to the cone passing
through the given point R. When T falls on the horizontal trace of
the cone there will be one tangent plane only containing the given
point. If T falls inside the horizontal trace of the cone the problem
is impossible but in that case the point R would be inside the cone.
[Note. In the preceding articles on tangent planes to the cone, use
has been made of the horizontal trace of the cone ; but in practice
it may often be more convenient to take a vertical trace of the cone,
and first determine the vertical trace of the tangent plane, which will
be a tangent to the vertical trace of the cone. For example, if the
base of the cone is circular and in a vertical plane, take this vertical
plane, or a plane parallel to it, as one of the planes of projection and
use the trace of the cone on this plane in determining the tangent
plane. These remarks also apply to problems which follow on tangent
planes to cones and cylinders.]
279. Tangent Plane to a Cone Parallel to a Given Straight
Line. — Through the vertex of the cone draw a line parallel to the
given line. By the construction described in the preceding article,
determine the planes to contain the former line and touch the cone.
These planes are the planes required. In the particular case where
the given line is parallel to a generating line of the cone, there will be
only one tangent plane to the cone parallel to the given line. If the
line through the vertex of the cone parallel to the given line falls
within the cone, the problem is impossible.
280. Tangent Planes to Right Circular Cones. — The methods
described in the preceding articles on tangent planes to the cone are
applicable to any form of cone, it being assumed that the trace of the
surface of the cone on one of the planes of projection is given. If a
trace of the surface of the cone is not given there would necessarily be
sufficient data given to enable a trace on one of the planes of projection
to be found.
When the cone is a right circular cone and its axis is inclined to
one of the planes of projection its trace on that plane will be one of
the conic sections which may be drawn and the methods of the
preceding articles may then be applied. But the drawing of the conic
section which is the trace of the cone may in general be avoided by the
adoption of special constructions.
CURVED SURFACES AND TANGENT PLANES 319
Fig. 623.
Eor example, let 'pp' (Fig. 623) be a given point on the surface of a
right circular cone whose axis vo, v'd is inclined to the planes of
projection. (In general only
one projection of P will be given
and the other will have to be
found as explained in Art. 228,
p. 264). Draw the projections
of a sphere inscribed in the
cone. Join the vertex vv to
fp' and find the point rr
where the line vp, v'p touches
the sphere. A tangent plane
to the sphere at rr' will be a
tangent plane to the given cone
and the line vp, v'p' will be the
line of contact between the
cone and plane. The tangent
plane to the sphere at rr' will
be perpendicular to the radius
or, o'r' and its traces may be
determined as explained in (Art.
183, p. 217).
Further examples of special constructions to avoid drawing the non-
circular trace of a right circular cone, to which a tangent plane is
required, will be found in subsequent articles of this chapter. It
should however be noted that in many cases the drawing of the non-
circular trace of the cone is the most straightforward construction to
adopt, and, particularly when the trace is an ellipse, it will often be
found quicker and more accurate to draw the trace than to adopt
more or less elaborate constructions involving straight lines and
circles only.
281. Planes Tangential to a Given Cone and having a
Given Inclination. — Let 6 be the given inclination of the planes
which are to be tangential to the cone.
Referring to Fig. 624, the cone is given by its vertex vv' and its
horizontal trace. Determine a right circular cone having its vertex
at vv', its base on the horizontal plane, and it base angle equal to 0.
Planes tangential to these two cones are the planes required. In
Fig. 624 there are four such planes, their horizontal traces being
common tangents to the bases or horizontal traces of the cones. The
vertical traces of the planes are found from the condition that the
planes contain the common vertex of the two cones. The vertical trace
of the fourth plane is outside the limits of the £gure.
In Fig. 625 the given cone is a right circular cone whose axis
vo, v'o' is inclined to both planes of projection. The horizontal trace of
this cone may be found as in Art. 227, p. 263, and the construction just
given for Fig. 624 applied. A much simpler construction however is
as follows. Draw a sphere inscribed in the given cone, its centre
being oo'. Determine a cone enveloping this sphere, having its base
320
PRACTICAL GEOMETRY
on the horizontal plane, and its base angle equal to 6. uv! is the
vertex of this cone. Determine also a right circular cone having its
Fig. 624.
Fig. 625.
vertex at vv', its base on the horizontal plane, and its base angle equal
to 6. The two planes (1) and (2) tangential to these two auxiliary
cones are two of the planes required, their horizontal traces being
common tangents to the circles which are the bases or horizontal traces
of the auxiliary cones.
If the auxiliary cone enveloping the sphere be taken with its vertex
u{a^ below the centre of the sphere instead of above it, the other
auxiliary cone being as before, two other tangent planes (3) and (4) to
the given cone and having the inclination 6 are found. In this case
however it is the pair of tangents to the traces of the auxiliary cones
which cross one another between the circles which must be taken.
The vertical trace of plane (4) is not shown.
282. Tangent Plane to a Cylinder at a Given Point on its
Surface. — Let the cylinder be given by its horizontal trace and the
direction of its generating line (Fig. 626), and let the given point on its
surface be given by its plan p. Through^ draw pqr parallel to the plan
of the direction of the generating line, cutting the horizontal trace of the
cylinder at q and r. If P is on the under surface of the cylinder pq
will be the plan of the generating line through P and q will be its
horizontal trace. If P is on the upper surface of the cylinder pr will
be the plan of the generating line through P, and r will be its horizontal
trace. The elevations of these generating lines are found as shown.
CURVED SURFACES AND TANGENT PLANES 321
A tangent to the horizontal trace of the cylinder at q will be the
horizontal trace of the plane which is tangential to the surface of the
cylinder at P when P is on the under surface, and a tangent to
the horizontal trace of the cylinder at r will be the horizontal trace of
the plane which is tangential to the surface of the cylinder at P when
P is on the upper surface. The generating lines PQ and PR will be
the lines of contact of these planes with the surface of the cylinder,
and by making use of this the vertical traces of the planes may be
found.
Fig. 626.
Fig. 627.
If the given cylinder is a right circular cylinder its trace on one
of the planes of projection is readily found and the method just described
may then be applied. The drawing of the trace of the cylinder, when
that trace is not a circle, may be avoided as follows. Draw an elevation
of the cylinder (Fig. 627) on a vertical plane parallel to its axis. Take
a plane U V W at right angles to the axis of the cylinder cutting that
axis at 0. The section of the cylinder by this plane is a circle whose
centre is O. Draw the rabatment of this circle on the horizontal
plane as shown. The plan of the generating lines through the two
possible positions of P, P being given by its plan p as before, cuts the
rabatment of the circle at m^ and n^. Draw tangents to the rabatment
of the circle at m^ and Wi to meet V W at s and t respectively. • q and r
being the horizontal traces of the generating lines through M and N,
lines qs and rt will be the horizontal traces of the planes required.
The traces of the planes on any vertical plane of projection may be
found from the condition that each plane contains a generating line
through one of the two possible positions of P.
The theory of the foregoing construction is that MS and NT, being
Y
322 PRACTICAL GEOMETRY
taogents to a section of the cylinder at points on the generating lines
through the two possible positions of P, must lie on the planes, one on
each, which are tangential to the cylinder along these generating lines.
283. Tangent Planes to a Cylinder through a given external
Point. — Through the given point draw a line parallel to the generating
line of the cylinder. Let Q be the trace of this line on one of the
planes of projection. Through Q draw the tangents to the trace of
the cylinder on the plane of projection containing Q. These tangents
will be the traces on one of the planes of projection of the planes
required and the other traces may be found from the condition that
the planes contain the given point. i
If the cylinder is a right circular cylinder the drawing of its trace,
when that trace is not a circle, may be avoided by a slight modification
of the construction shown in Fig. 627, p. 321. Let the line through
the given point parallel to the generating line of the cylinder intersect
the plane UVW at L and the horizontal plane of projection at K.
Find Zi the rabatment of L on the horizontal plane. From l^ draw
tangents to the rabatment of the circle which is the section of the
cylinder by the plane UVW. Let these tangents intersect VW at
s and t. K.S and K< are the horizontal traces of the planes required
and the vertical traces may be found from the condition that the planes
contain the given point.
284. Planes Tangential to a Cylinder and Parallel to a
given Straight Line. — Let AB be the given straight line.
Let the cylinder be given by its trace on one of the planes of
projection and the direction of its generating line. Through any
point C in AB draw CD parallel to the generating line of the cylinder.
Determine the traces of the plane containing AB and CD. The
planes required will be parallel to this plane. Tangents to the trace
of the cylinder parallel to the corresponding trace of the plane con-
taining AB and CD will be the traces on one of the planes of
projection of the planes required. The other traces may be found
from the condition that the required planes are parallel to the plane
containing AB and CD.
If the cylinder is a right circular cylinder, the drawing of its
trace, when that trace is not a circle, may be avoided by a modification
of the construction shown in Fig. 627, p. 321. As in the case just
considered, draw CD to intersect AB and be parallel to the generating
line of the cylinder and find the traces of the plane containing AB
and CD. Find the intersection of this plane with the plane UVW.
Let EF be this intersection. Draw the rabatment Cj/i of EF on the
horizontal plane. Draw parallel to Cj/i tangents to the rabatment of
the circle which is the section of the cylinder by the plane UVW. Let
these tangents meet VW at 8 and t. Lines through s and t parallel
to the horizontal trace of the plane containing AB and CD are the
horizontal traces of the planes required. The vertical traces may be
found from the condition that the required planes are parallel to the
plane containing AB and CD.
285. Planes Tangential to a Cylinder and having a given
CURVED SURFACES AND TANGENT PLANES 323
Inclination. — Let 6 be the given inclination, say to the horizontal
plane.
In Fig. 628 the cylinder is given by its horizontal trace and the
direction of its generating lines. Take ah, a'b' a generating line, a
being its horizontal trace. Take a point vv' in ah, a'h' as the vertex
of a right circular cone whose base is on the horizontal plane and
whose base angle is equal to 6. A tangent aL to the base of this
cone will be the horizontal trace of a plane tangential to the cone and
containing the generating line ah, a'h'. The vertical trace LM of this
plane is found from the condition that the plane contains the point vv'.
This plane will also have the given inclination 6. Tangents to the
horizontal trace of the cylinder parallel to ah will be the horizontal
traces of two of the planes required and their vertical traces will be
parallel to LM.
Two other planes (not shown) fulfilling the given conditions may
be found by using the tangent aN to the base of the cone in the same
way that aL was used.
The planes found as above will evidently have the required
inclination and they will also contain each a generating line of the
cylinder whose horizontal trace is at the point where the horizontal
trace of the plane touches the horizontal trace of the cylinder. In
Fig. 628, cd, c'd! and ef, e'f are the lines of contact with the cylinder
of planes (1) and (2) respectively.
Fig. 628.
When the given cylinder is a right circular cylinder the construc-
tion shown in Fig. 629 may be used. In this case a trace of the
cylinder is not required. Take rr' and ss' two points on the axis of
the cylinder as the centres of two spheres inscribed in the cylinder.
Envelop these spheres by cones „whose bases are on the horizontal
324
PRACTICAL GEOMETRY
plane and whose base angles are equal to 9. Tangent planes to these
cones are two of the planes required. The vertical trace of the second
plane falls outside the limits of the figure, cd, c'd' and e/, e'f the
lines of contact of the tangent planes with the cylinder pass through
the points of contact of these planes and the spheres.
Two other planes satisfying the given conditions may be found by
drawing two inverted cones enveloping the spheres and constructing
the tangent planes to them. The base angles of these inverted cones
must of course be equal to 0.
The given angle 6 must not be less than the inclination of the
generating lines of the cylinder.
286. Planes Tangential to a Sphere and containing a given
Line. — First solution. O is the centre of the sphere and AB the given
line. Draw an elevation of the
sphere and given line on a vertical
plane parallel to the line (Fig.
630). Take a plane perpendicular
to AB and containing the centre
of the sphere. LM the vertical
trace of this plane will be at right
angles to a'b' and will be an edge
view of the plane. This plane
intersects the given line at C and
the sphere in a great circle. The
plan of this great circle, which
is an ellipse, is shown but it need
not be drawn. In the plane LM
take the horizontal line OS
through the centre of the sphere.
Obtain the rabatment of the
section of the sphere by the plane
LM and also the point C by
rotating the plane LM about OS until it becomes horizontal. The
rabatment of the section of the sphere will be the circle which is the
plan of the sphere and the rabatment of C will be Cj. Through q
draw the tangent c^ei to the circle which is the plan of the sphere, e^
being the point of contact. Restore the plane LM to its original
position taking with it the point Ci determining e' and e as shown.
CE is a tangent to the section of the sphere by the plane LM, and the
plane containing CE and AB will be a tangent plane to the sphere
at E. The horizontal and vertical traces of the tangent plane will be
perpendicular to oe and o'e' respectively.
Since a second tangent can be drawn from Cj to the circle, the
above construction applied to this second tangent will lead to a
second plane tangential to the sphere and containing AB. This second
plane is not shown.
Second solution. Envelop the sphere by a cone having its vertex
in the given line AB. The planes tangential to this cone and con-
taining AB will also be tangential to the sphere. The horizontal
Fig. 630.
CURVED SURFACES AND TANGENT PLANES 325
traces of the tangent planes required will pass through the horizontal
trace of AB and will be tangential to the horizontal trace of the cone.
By selecting the point in AB for the vertex of the cone so that
the axis of the cone is parallel to one of the planes of projection, the
plane (M) of the circle of contact between the cone and the sphere is
readily found. The base angle of the cone is the inclination of the
planes required to the plane M, and the planes required may be found
by the construction of Art. 193, p. 225.
287. Cones Enveloping Two Spheres. — If one sphere lies
entirely outside the other and the spheres do not touch one another
there will be two cones which will envelop both.
^"-^■^^"^^^^^-Clrcles of contact ^ \J^ cfcordojct
Fig. 631. Fig. 632.
The projections of the cones are obtained by drawing the common
tangents to the circles which are the projections of the spheres.
These tangents intersect the projection of the line joining the centres
of the spheres at the projections of the vertices of the cones. One
cone has its vertex on the line joining the cefttres of the spheres
produced beyond the smaller sphere (Fig. 631), while the other has its
vertex between the spheres (Fig. 632). If the spheres touch one
another externally, one of the cones becomes a plane tangential to the
spheres at their point of contact. If the spheres cut one another there
is only one enveloping cone.
288. Planes Tangential to Two Spheres and having a
given Inclination. — Let the given inclination be to the horizontal
plane. Envelop the two spheres by a cone. Find by Art. 281, p. 319,
the planes tangential to this cone and having the given inclination.
The following construction follows readily from that shown in Fig.
629, p. 323. Envelop each sphere by a cone, axis vertical and base
a«igle equal to 0. Planes tangential to these two cones are the planes
required. There may be as many as eight planes satisfying the given
conditions, depending on and the relative positions of the spheres.
Denoting the cones by A and B, there are two planes for the case
where A and B are upright, two when A and B are inverted, two
when A is upright and B is inverted and two when A is inverted and
B is upright.
289. Planes Tangential to Two Spheres and containing
a given Point. — Envelop the two spheres by a cone. Find by Art.
278, p. 318, the planes tangential to this cone and containing the given
point. Or, proceed as follows. Let V be the vertex of the cone
enveloping the two spheres, and let P be the given point. Joi^ PV,
Find by Art. 286, p. 324, the planes tangential to one of the spheres
326
PRACTICAL GEOMETRY
and containing the line PV. These planes will also be tangential to
the other sphere.
290. Planes Tangential to Three Spheres. — Determine the
vertex of a cone enveloping any two of the spheres, also the vertex of
a cone enveloping another two. A plane containing these two vertices
and touching any one of the spheres will also touch the other two.
If the spheres are entirely external to one another and no two
touch one another there will be eight planes which will touch all three
spheres. Two of these will have all the spheres on the same side,
while the others will have one sphere on one side and two on
the other.
291. Tangent Plane to a Surface of Revolution at a given
Point on the Surface. — Let the axis of the surface (Figs. 633 and
634) be vertical, and let pp be the given point on the surface.
^21
Fig. 633.
Fig. 634.
All tangent planes to the surface at points on it at the same level
as pp' will have the same inclination, and the normals to the surface at
these points will meet on the axis at the same point. Hence if p'r' be
drawn parallel to XY to meet the outline of the elevation (or plane
generatrix) of the surface at /, and a line qr' be drawn perpendicular
to the tangent at r' meeting the elevation of the axis at q, q[p' will be
the elevation of the normal to the surface Q.t pp' . The plan of this
normal will be the line joining p with the centre of the circle which is
the plan of the surface.
A plane L M N through the point p^' and perpendicular to the line
fq, p'(^ will be the tangent plane to the surface a,t pp'. This plane is
also tangential to the right circular cone which envelops the surface
CURVED SURFACES AND TANGENT PLANES 327
of revolution and has for its line of contact- the horizontal circle
through pp'. MN, the horizontal trace of the tangent plane LMN, is
a tangent to the circle which is the horizontal trace of the above-
mentioned cone.
In Fig. 633 the tangent plane meets the surface of revolution at
one point only, while in Fig. 634 the tangent plane also cuts the
surface.
292. Cylinder Enveloping a Surface of Revolution. — The
vertical line a6, a'h' (Fig. 635) is the axis of a surface of revolution
Fig. 635.
whose outline is given, cd^ c'd', is a line which is parallel to the
vertical plane of projection but is inclined to the horizontal plane. It
is required to determine the cylinder which envelops the surface of
revolution and has its generatrices parallel to cd, c'd'. The cylinder is
determined when its line of contact with the surfa-ce of revolution is
found. The construction is shown for finding two points on this line
of contact, one on the upper or concave part of the surface of
revolution, and the other on the lower or convex part.
Take a point rr' on the meridian section of the surface of revolution
which is parallel to the vertical plane of projection. Draw r'o' the
normal to the elevation of the outline of this meridian section at r',
328
PRACTICAL GEOMETRY
and leb r'd meet a!V at d . With d as centre and oV as radius describe
a circle. This circle is the elevation of a sphere inscribed in the
surface of revolution. The line of contact of this sphere and the
surface of revolution is a horizontal circle whose elevation is the straight
line 5V'. A cylinder which envelops this sphere and has its axis
parallel to cd^ c'd' will touch the sphere in a circle of which s'o't',
perpendicular to c'd', is the elevation. Let s'o't' intersect 5'/ at p'y
then p' is the elevation of a point on the line of contact of the surface
of revolution and the required enveloping cylinder. The plan p of
this point is found by an obvious construction which is shown.
A line mn, m'n' through j^p' parallel to cd, c'd' is one of the generatrices
of the cylinder which envelops the sphere and is tangential to the
sphere at the point pp' ; but the surface of the sphere is tangential to
the surface of revolution at this point, therefore the line mn, m'n' is
a generatrix of the enveloping cylinder required.
The outline of a projection of the surface of revolution by projectors
parallel to cd, c'd' will be the projection of the line of contact of the
Fig. 636.
enveloping cylinder determined as above. Such a projection is shown
at (I), on a plane perpendicular to the projectors.
The projection of a surface of revolution on a plane inclined to its
axis may also be obtained directly by first drawing the projections of
a sufficient number of spheres inscribed in the surface and then
drawing the boundary line of all these projections. The " anchor
ring " shown in Fig. 636 is a solid which is easily projected by this
method. The anchor ring may be considered as generated by a sphere
rotating about the axis of the ring, and the projections of this sphere
in a sufficient number of positions being drawn, the curves bounding
them form the projection of the ring.
293. Cone Enveloping a Surface of Revolution. — The vertical
line ah, ah' (Fig. 637) is the axis of a surface of revolution whose
outline is given, cc' is a given point at the same distance from the
vertical plane of projection as ah, a'b'. It is required to determine a
cone which envelops the surface of revolution and has its vertex at cc'.
The cone is determined when its line of contact with the surface of
revolution is found. The construction is shown for finding two points
CURVED SURFACES AND TAKGENT PLANES 329
on this line of contact, one on the upper or concave part of the surface
of revolution, and the other on the lower or convex part.
As in the preceding Art. draw the elevation of a sphere inscribed
in the surface of revolution, having for its line of contact the circle
whose elevation is ^V. A cone which envelops this sphere and has its
A^ertex at the point cc' will touch the sphere in a circle of which the
straight line s't' is the elevation. The point j?' where s't' intersects gV
is the elevation of a point on the line of contact of the surface of
revolution and the required enveloping cone. The plan p of this
point is obtained by an obvious construction which is shown. A line
mw, m'n' through cc' and pp is a generatrix of the cone which envelops
Fig. 637.
the sphere and is tangential to the sphere at pp' . But the surface of the
sphere is tangential to the surface of revolution at this point, therefore
the line mw, w'w' is a generatrix of the enveloping cone required.
294. The Spheroid. — A spheroid may be generated by the
revolution of an ellipse about one of its axes, and is called a prolate or
oblate spheroid according as the axis of revolution is the major or minor
axis of the ellipse.
A spheroid may also be generated by a circle of varying diameter
which moves so that its plane is perpendicular to, and its centre in, a
fixed straight line, the diameter of the circle being regulated by an
ellipse which has one axis coinciding with the fixed straight line. The
fixed straight line is the axis of the spheroid.
330
PRACTICAL GEOMETRY
Fig. 638 shows the plan and elevation of a prolate spheroid whose
axis is vertical. Eight meridian sections and seven circular sections at
equal intervals are shown. A section by a plane LM perpendicular to
the vertical plane of projection and inclined to the horizontal plane is
also shown. All plane sections of a spheroid other than sections at
right angles to its axis are ellipses.
Fig. 639 shows the same spheroid tilted over so that its axis is
inclined at 45° to the horizontal plane but remains parallel to the
vertical plane of projection. The same meridian and circular sections
Fig. 638.
Fig. 639.
which are shown in Fig. 638 are shown in Fig. 639. The student
should draw these projections full size taking the major and minor axes
of the generating ellipse 4 inches and 3 inches long respectively. The
plan in Fig. 639 may be projected directly from the elevation in
Fig. 638 by turning the ground line through 45°. This will save the
labour of redrawing the elevation in the inclined position.
It should be noted that all the curves in the plan, Fig. 639, includ-
ing the boundary line are ellipses. The boundary ellipse in the plan,
Fig. 639, is the horizontal trace of a vertical cylinder enveloping the
CURVED SURFACES AND TANGENT PLANES 331
Hypcrboloid of revolution
of one sheet.
Fig. 640.
Hyperboloid of revolu-
tion of two sheets.
Fig. 641.
spheroid, the straight line i'i' being the elevation of the line of contact
between the spheroid and the enveloping cylinder.
The axes of the various ellipses having been found the curves may
be drawn by the paper trammel method.
When the two axes of the revolving ellipse are equal the spheroid
becomes a sphere.
295. The Hyperboloid of Revolution. — The Tiyperholoid of
revolution may be generated by the revolution of an hyperbola about one
of its axes. If the axis
of revolution is the con-
jugate axis BBi of the
hyperbola (Fig. 640) each
branch of the hyperbola
describes the same sur-
face which is the hyper-
boloid of revolution of one
sheet. If the axis of re-
volution is the transverse
axis AAi of the hyper-
bola (Fig. 641) the two
branches of the hyper-
bola describe separate surfaces and the two surfaces form the hyper-
boloid of revolution of two sheets. Of these two hyperboloids of revo-
lution the one of one sheet is the more important, and when an
hyperboloid of revolution is referred to the one of one sheet will be
understood.
In Fig 642, a' a/ is the transverse axis of an hyperbola whose plane is
parallel to the vertical plane of projection and whose conjugate axis
b'bi is vertical, m'o'm' and n'o'n' are the asymptotes of the hyperbola, and
/' and // are the foci. As the hyperbola and its asymptotes revolve
about the vertical conjugate axis, the hyperbola describes an hyperboloid
of revolution and the asymptotes describe a cone which is asymptotic to
the hyperboloid. Sixteen positions of the moving hyperbola are shown
in the plan (u) and elevation (v).
Sections of the hyperboloid by planes perpendicular to the axis of
revolution are circles, the smallest of which has a diameter equal to the
transverse axis of the hyperbola and is called the collar or throat or
gorge of the hyperboloid.
At (w) in Fig. 642 is shown a plan of the hyperboloid when the
axis of revolution is inclined at 30° to the horizontal plane. In this
new plan only the visible meridian and circular sections shown in the
plan (m) and elevation (v) are shown.
A section of the hyperboloid by a plane is of the same kind, as the
section of the asymptotic cone by that plane or by a plane parallel to
it. For example if a plane cuts the asymptotic cone in an ellipse all
planes parallel to that plane will cut the hyperboloid in ellipses. The
same is true for parabolic and hyperbolic sections, and it is obviously
true for circular sections.
A plane containing the axis of revolution cuts the hyperboloid in an
332
PRACTICAL GEOMETRY
hyperbola and the asymptotic cone in two straight lines which are the
asymptotes of the hyperbola.
A plane parallel to the axis of revolution and tangential to the
collar of the hyperboloid cuts the asymptotic cone in an hyperbola and
the hyperboloid in two straight lines P and Q which intersect on the
collar and which are the asymptotes of the hyperbola. These two
straight lines P and Q are also parallel to the straight lines Pj and Q^
respectively in which the asymptotic cone is cut by a plane containing
the axis of revolution and parallel to the plane of P and Q. Also all
these straight lines P, Pj, Q and Qi are inclined at the same angle to
HYPERBOLOID
OF REVOLUTION
OF ONE SHEET
Fig. 642.
the axis of revolution but if P and Pi slope forwards Q and Qi slope
backwards.
It follows that from any point S on the surface of the hyperboloid
two straight lines can be drawn on the surface, because two planes can
be taken parallel to the axis of revolution and containing S and touch-
ing the collar and each of these planes will intersect the surface in a
pair of straight lines and one of each pair will pass through S.
There are therefore two systems of straight lines which can be
drawn on the surface of the hyperboloid. No two lines of the same
system are in the same plane, but any line of the one system and any
line of the other are in the same plane. The plane which contains a
line of one system and a line of the other is the tangent plane to the
surface at the point of intersection of the lines.
CURVED SURFACES AND TANGENT PLANES 333
Since any line of one system and any line of the other are in the
same plane it follows that any line of one system will intersect every
line in the other system if the lines are produced far enough. Hence
the hyperboloid of one sheet may be generated by the motion of a
straight line which moves in contact with any three lines of either
system.
The hyperboloid may also evidently be generated by the revolution
of a line belonging to one or other of the systems above referred to,
the axis of revolution being the axis of revolution already used. From
this mode of generation the hyperboloid of revolution is also called the
twisted surface of revolution. This mode of generation is illustrated by
HYPERBOLOID
OF REVOLUTION
OF ONE SHEET
Fig. 643.
Fig. 643 which shows the same hyperboloid represented in Fig. 642,
but instead of different positions of the revolving hyperbola different
positions of the revolving straight line are shown. It may be left
as an exercise for the student after drawing the views as shown
in Fig. 643 to add the second system of straight lines on the
hyperboloid.
The hyperboloid of two sheets cannot be generated by the motion of
a straight line, it is therefore not a ruled surface.
296. Hyperboloids of Revolution in Rolling Contact. —
Referring to the plan (u) and elevation (v) in Fig. 644, o'a' is the eleva-
tion and the point o^a is the plan of a vertical axis, o'b' is the elevation
and o^h is the plan of another axis which is parallel to the vertical
334
PRACTICAL GEOMETRY
plane of projection but is inclined to the horizontal plane. The point o'
is the elevation and o^o.^ is the plan of the common perpendicular to these
two axes. o'c' is the
elevation and oc is the D_
plan of a straight line
which is parallel to the
vertical plane of projec-
tion, o'c' makes an angle
a with o'a' and an angle
P with oh'.
If an hyperboloid of
revolution be described
by the revolution of OC
about the vertical axis
and another hyperboloid
be described by the re-
volution of OC about
the inclined axis, these
two hyperboloids will be
in contact along the line
OC, and if one hyper-
boloid be made to rotate
about its axis the other
will also rotate if the
Fig. 644.
frictional resistance between the hyperboloids is sufficient.
The construction of the hyperboloids when the axes and the line of
contact are given is clearly shown in Fig. 644. The view (w) is a pro-
jection of the inclined hyperboloid on a plane perpendicular to its axis.
The common normal to the surfaces of the hyperboloids at C will
be at right angles to the tangent plane at C and this tangent plane
will contain the line OC. Since the line OC is parallel to the vertical
plane of projection the vertical trace of the tangent plane containing
OC will be parallel to o'c', therefore the elevation of the common
normal will be perpendicular to o'c'. But this common normal must
intersect the axis of each hyperboloid. Hence a line m'c'n' at right
angles to o'c' and intersecting o'a' at m' and o'b' at n' will be the eleva-
tion of the common normal to the surfaces at C. The plan men of this
normal is easily found as shown.
The relative angular velocities of the two hyperboloids, when one
rotates the other, will now be determined. To facilitate reference
denote the vertical hyperboloid by A and the inclined hyperboloid by
B. Let a>i and wg be the angular velocities of A and B respectively.
Let Vi be the linear velocity of the point O considered as a point on
the throat of A, and let Vg be the linear velocity of the point O con-
sidered as a point on the throat of B. Let r^ and r^ be the radii of the
throats of A and B respectively.
Draw o'D, o'E, and o'F at right angles to o'a'^ o'V, and o'd respec-
tively. Make o'D equal to Vj and draw DFE at right angles to o'F
meeting o'F at F and o'E at E. Then o'E will be equal to Va , because
CURVED SURFACES AND TANGENT PLANES 335
o'F and ED are evidently the components of Vi along and perpen-
dicular to OC respectively, and since there is rolling motion at O the
component of Vg at right angles to OC must also be equal to o'F. In
addition to the rolling of the one hyperboloid on the other there is
also a sliding motion along the line of contact and the velocity of this
sliding is represented by DE.
fixes n and mn being drawn a projector from c' to meet mn
determines c. Then co parallel to o.J) is the plan required and the
hyperboloids may be drawn as shown.
These hyperboloids are the pitch surfaces of what are called skew
bevel wheels, which are used to transmit motion directly from one shaft
to another when the axes of the shafts do
not intersect and are not parallel to one
another. In practice generally only com-
paratively short frusta of the hyperboloids
are used as the pitch surfaces of the
wheels as shown in Fig. 645. If the
frusta are taken at the bases of the hyper-
boloids, as A and B, they are approxi-
mately conical, and if at the throats, as
Ai and B^, they are approximately cylin-
drical. In these wheels the teeth have
line contact.
297. The Geometry of Cross-Rolls. — Referring again to Fig.
644, if the axes of the hyperboloids are given and it is also given that
the line of contact OC is parallel to the axis of the inclined hyper-
boloid, then o'c' would coincide with o'h' and the inclined hyperboloid
would become a cylinder. But the point nn' would coincide with the
point cc' and the cylinder would have no diameter, that is, it would
become a straight line only.
Fig. 645.
336
PRACTICAL GEOMETRY
If it is desired that the hyperboloid which has become a straight
line should be a cylinder of definite diameter, then another surface of
revolution will have to take the place of the other hyperboloid. It
will now be shown how this other surface of revolution may be deter-
mined.
Referring to Fig. 646, hob is the plan and h'o.Jb' is the elevation of
the axis of the cylinder which is assumed to be horizontal, aoa is the
plan and o/ is the elevation of
the axis of the other surface or
solid which is assumed to be
horizontal and perpendicular to
the vertical plane of projection.
This other solid will be called
the roll, o is the plan and o^o'o^
is the elevation of the common
perpendicular to the two given
axes. Making o^o' equal to the
radius of the cylinder determines
o^o' the radius of the roll at the
throat. The plane of the throat
circle of the roll intersects the
cylinder in an ellipse which
touches the throat circle at oo'.
Any other plane parallel to the
plane of the throat circle will
intersect the cylinder in an
ellipse and the roll in a circle,
and the ellipse and circle will
touch one another. Moreover
all such sections of the cylinder
will be exactly alike.
HT is the horizontal trace
of a plane parallel to the plane
of the throat circle of the roll.
This plane intersects the cylin-
der in an ellipse of which m'v!
is the elevation. A circle with
o/ as centre drawn to touch the ellipse w'w' is the elevation of the
circular section of the roll by the plane HT, and this determines the
points e and / on the horizontal meridian section of the roll. A pro-
jector from s', the point of contact of the ellipse and circle, to HT
determines s a point on the plan uov of the curve of contact between
the cylinder and roll.
The radius of the circular section of the roll at HT and the point
of contact ss' may be found without drawing the ellipse m!n'. It is
evident that if the ellipse m'n! be moved to the right a distance equal
to w on the plan it will then coincide with the ellipse c'd' which is the
elevation of the section of the cylinder by the plane of the throat
circle of the roll. And if the circle e'f be moved an equal distance to
Fig. 646.
CURVED SURFACES AND TANGENT PLANES 337
the right it will then be in the position e/// and will touch the ellipse
c!d' at s/. Hence the circle e//i' may be drawn first and then the circle
e'f which is equal to it. Projecting from s/ to Si and making s-^s equal
to w determines th^ point s. In like manner, by taking other positions
for the plane HT any number of points on the horizontal meridian
section of the roll and any number of points on the plan uov of the
curve of contact may be determined. The ellipse cd' is therefore the
only ellipse which need be drawn but it should be constructed as
accurately as possible, to ensure a good result.
298. The Paraboloid of Revolution. — The paraboloid of revo-
lution may be generated by the revolution of a parabola about its axis.
Sections of the surface by planes parallel to or containing the axis are
equal parabolas. Sections by planes perpendicular to the axis are
circles. All other plane sections are ellipses.
299. The Ellipsoid. — The ellipsoid may be generated by a vari-
able ellipse which moves so that its plane is parallel to a fixed plane
and the extremities of its axes are on two fixed ellipses which have one
common axis and which have their planes at right angles to one another
and to the fixed plane.
Referring to Fig. 647, let the horizontal plane be the fixed plane.
Let oa, o'a' and oh, o'h' be the semi-axes of one fixed ellipse whose plane
is horizontal, and let oh, o'h
and oc, o'c' be the semi-
axes of the other fixed
ellipse whose plane is pa-
rallel to the vertical plane
of projection. A moving
variable ellipse whose axes
are horizontal chords of
these two fixed ellipses
generate an ellipsoid.
The plan of the hori-
zontal fixed ellipse is the
plan of the ellipsoid, and
the elevation of the other
fixed ellipse which is pa-
rallel to the vertical plane
of projection is an elevation
of the ellipsoid. def is
the half plan and d'e'f is
the elevation of the moving
variable ellipse in one
position. OG on the half
plan is equal to o/e' on the
part elevation (y^ which is a projection on a vertical plane at right
angles to the plane of the elevation (m).
All plane sections of the ellipsoid are either ellipses or circles.
If g'o'd', a diameter of the ellipse which is the elevation of the
ellipsoid, be taken as the vertical trace of a plane which is perpendicular
z
ELLIPSOID
Fig. 647.
338 PRACTICAL GEOMETRY
to the vertical plane of projection, the elliptic section of the ellipsoid
by this plane will have one semi-axis equal to o'^' and the other
semi-axis equal to oa. If o'^ is equal to oa then the section is
evidently a circle. This suggests the construction for finding the
plane containing the centre of the ellipsoid and cutting the ellipsoid in
a circle. There are evidently two such planes.
All plane sections parallel to one circular section are circles.
In Fig. 647, g'o'd! is the elevation of one circular section of the
ellipsoid and the chords of the ellipse parallel to g'o'd' are the
elevations of other circular sections. The elevations of the centres
of these circles lie on the diameter wVw' which is conjugate to the
diameter g'o'd!.
As the plane of a circular section moves away from the centre of
the ellipsoid the circle gets smaller and smaller until the plane becomes
tangential to the ellipsoid at mm! or nn' when the circle becomes a
point which is called an umhilic. There are evidently four umbilics on
an ellipsoid.
At (w) in Fig. 647 is shown a projection of the ellipsoid on a plane
parallel to planes of circular sections. On this projection nine circular
sections are shown.
The curve of contact between an ellipsoid and an enveloping
cylinder or an enveloping cone is a plane section of the ellipsoid. The
outline of the projection (w) in Fig. 647 is the trace of a cylinder
which envelops the ellipsoid and is perpendicular to the plane of the
projection ; the diameter i'o'i' of the ellipse (u) is the elevation of the
curve of contact.
The tangent plane to the ellipsoid at any point on it contains the
tangents at that point to any two plane sections of the ellipsoid
through the point. When two of the three axes of an ellipsoid are
equal it becomes a spheroid.
300. The Hyperboloid of One Sheet. — The Jiyperloloid of
one sheet may be generated by a variable ellipse which moves so
that its plane is always parallel to a fixed plane and has the
extremities of its axes on two fixed hyperbolas whose planes are
perpendicular to one another and to the fixed plane and which have a
common conjugate axis'.
Referring to Fig. 648, OA and OB are the semi-transverse axes
of two fixed hyperbolas and OC is a common semi-conjugate axis.
OA and OB are horizontal and OC is consequently vertical. OA
is parallel to the plane of the elevation (v). The moving variable
ellipse is in this case always horizontal, and it will be smallest
when its plane coincides with AOB, and it is then the throat
ellipse or principal elliptic section of the hyperboloid. In this
article when " hyperboloid " is mentioned " hyperboloid of one sheet "
will be understood.
The moving ellipse remains similar to the throat ellipse, hence if
DD and EE are the axes of the moving ellipse in one position de is
parallel to ah.
At (w) is shown an elevation of the hyperboloid on a plane
CURVED SURFACES AND TANGENT PLANES 339
parallel to OB. The true form of one fixed hyperbola is shown
in the elevation (v) and the true form of the other is shown in the
elevation (zc).
The asymptotic cone of the hyperboloid is generated by a variable
ellipse which moves so that its plane is parallel to the throat ellipse
and has the extremities of its axes on the asymptotes of the fixed
hyperbolas. The vertex of this cone is at O. GG and HH are the
axes of this moving variable ellipse in one position. This generating
ellipse is similar to the throat ellipse, hence gh is parallel to ab. The
ellipse which is the section of the asymptotic cone by a plane parallel
/ /
/ /
/ /
/ /
HYPERBOLOID /'/
OF ONE SHEET/'/
Fig. 648.
to the plane of the throat ellipse and at a distance from it equal to
OC is evidently equal to the throat ellipse.
Sections of the hyperboloid and of the asymptotic cone by the same
or by parallel planes are curves of the same kind.
If the transverse axes of the fixed hyperbolas are equal the hyper-
boloid becomes an hyperboloid of revolution.
The- plane of a circular section of the hyperboloid may be found as
follows. Let OA be greater than OB. Then on the elevation (w)
take o' as centre and with a radius equal to oa describe an arc to cut
the hyperbola b'k'e' at k' ; then k'o'k' is the vertical trace of a plane,
perpendicular to the plane of the elevation (w), which will cut the
340
PRACTICAL GEOMETRY
hyperboloid in a circle whose radius is equal to o'h'. All sections
by planes parallel to this plane will be circles. There are evidently
two systems of parallel planes which will cut the hyperboloid in
circles.
The hyperboloid of one sheet may also be generated by a moving
variable hyperbola having a fixed conjugate axis, and having the
extremities of its transverse axis on a fixed ellipse whose plane bisects
at right angles the fixed conjugate axis of the moving hyperbola. The
fixed ellipse is the throat ellipse of the hyperboloid.
Referring to the plan (w) and elevation {v) Fig. 648, the straight
lines RSR and TST lie on the hyperboloid. These lines are in a
vertical plane and their plans coincide and are tangential to the plan
of the throat ellipse. The hyperboloid may therefore be generated by
the motion of one or other of these straight lines. These two lines
belong to two different systems. A line of one system will never meet
any other line of that system but it will intersect every line of the
other system if the lines are produced far enough. Hence if a straight
line moves in contact with any three lines of one of the systems it will
generate the hyperbol'oid.
301. The Hyperboloid of Two Sheets. — The hjperbohid of
two sheets may be generated by a variable ellipse which moves so
that its plane is always parallel to a fixed plane and has the extremi-
ties of its axes on two fixed hyperbolas whose planes are perpendicular
to one another and to the fixed plane and which have a common
transverse axis.
Fig. 649.
Referring to Fig. 649, which is a pictorial projection of one quarter
of an hyperboloid of two sheets, OB and OC are the semi- conjugate
axes of two fixed hyperbolas and OA is a common transverse axis.
OA and OB are horizontal and OC is vertical. The plane containing
OB and OC is the fixed plane referred to in the above definition.
ADE is part of one branch of one fixed hyperbola having OA for its
semi-transverse axis and OC for its semi- conjugate axis ; the plane of
this hyperbola is vertical. AFH is part of one branch of the other fixed
hyperbola having OA for its semi-transverse axis and OB for its semi-
CURVED SURFACES AND TANGENT PLANES 341
conjugate axis ; the plane of this hyperbola is horizontal. The corre-
sponding parts of the other branches of these fixed hyperbolas are
shown to the right in Fig. 649. 1, 2, and. 3 are three positions of the
moving variable ellipse which is always parallel to the fixed plane
BOC. The generating ellipse in any position is a section of the
hyperboloid by a plane parallel to the plane BOC.
The asymptotic cone of the hyperboloid of two sheets is generated
by a variable ellipse which moves so that its plane is parallel to the
fixed plane BOC and has the extremities of its axes on the asymptotes
of the fixed hyperbolas. The vertex of the cone is at O. This
generating ellipse in any position is a section of the asymptotic cone
by a plane parallel to the plane BOC.
All sections of the hyperboloid of two sheets, and of the asymptotic
cone, by planes parallel to the plane BOC are ellipses similar to the
ellipse having OB and OC as semi-axes. Straight lines such as KL
and MN are therefore parallel to the straight line BC.
Any plane containing the common transverse axis of the fixed
hyperbolas cuts the hyperboloid in an hyperbola, and the asymptotic
cone in straight lines which are the asymptotes of the hyperbola.
APQ is such an hyperbolic section, OR being an asymptote. The
semi-conjugate axis of this hyperbolic section is OT the semi-diameter
of the ellipse BTC which is in the plane of section.
The hyperboloid of two sheets may therefore be generated by a
moving variable hyperbola having a fixed transverse axis, and having
the extremities of its conjugate axis on a fixed ellipse whose plane
bisects at right angles the fixed transverse axis of the moving
hyperbola. The fixed ellipse is the ellipse whose axes are the conjugate
axes of the fixed hyperbolas.
The hyperboloid of two sheets cannot be generated by a straight
line and it is therefore not a ruled surface.
302. The Elliptic Tai.YSiholoid.— The elliptic paraboloid may he
generated by a parabola which moves with its vertex in a fixed
parabola, the two parabolas
having their axes parallel,
their planes at right angles
to one another, and their
concavities turned in the
same direction.
Fig. 650 is a pictorial
projection of one quarter of
an elliptic paraboloid. ABE
is the fixed parabola of
which AN is the axis and
whose plane H.P. is hori-
zontal. 1, 2, 3, and 4 are
different positions of the
moving parabola whose plane
is vertical and whose axis is
parallel to AN, and whose vertex is on the parabola ABE.
Fig. 650.
342
PRACTICAL GEOMETRY
Sections of the elliptic paraboloid by planes perpendicular to AN
are ellipses. Four such sections are shown at 1', 2', 3', and 4'. These
ellipses are similar to one another so that lines such as BC and ED
are parallel. Hence the elliptic paraboloid may also be generated by
a variable ellipse which moves with its centre on, and its plane
perpendicular to, the common axis AN of two fixed parabolas ABE
and ACD whose planes are perpendicular to one another, the
extremities of the axes of the ellipse being on the parabolas.
303. The Hyberbolic Paraboloid. — The hyperhoUc paraboloid
may be generated by a parabola which moves with its vertex on a
fixed parabola, the two parabolas having their axes parallel, their
j)lanes at right angles to one another, and their concavities turned in
opposite directions.
Fig. 651 is a pictorial projection of one quarter of a hyperbolic
paraboloid. APC is the fixed parabola of which the vertical line AN
is the axis and whose
plane is the vertical plane ' .r V.P.
V.P. 1, 2, 3, etc. are
different positions of the
moving parabola whose
plane is at right angles
to the plane V.P. and
whose axis is parallel to
AN, and whose vertex
is on the parabola APC.
A section of the para-
boloid by a horizontal
plane through A, the
vertex of the fixed para-
bola, is two straight lines
of which AL is one.
Vertical planes contain-
ing these two straight
lines are asymptotic
planes of the paraboloid.
All other horizontal sec-
tions of the paraboloid
are hyperbolas the
asymptotes of which are the lines of intersection of the planes of
section with the asymptotic planes. Horizontal sections are shown
at the levels C, P, A, and B. It will be found that the hyperbolic
sections above A are on opposite sides of the asymptotes to those
below A.
The projections of the horizontal hyperbolic sections on a horizontal
plane are hyperbolas of which the horizontal traces of the asymptotic
planes are the asymptotes.
It follows that the hyperbolic paraboloid may be generated by a
variable hyperbola which moves with its centre on, and its plane
perpendicular to, the line aAN containing the axes of the two fixed
Fig. 651.
CURVED SURFACES AND TANGENT PLANES 343
parabolas APC and ATB whose planes are at right angles to one
another, whose asymptotes are in fixed planes, and having the
extremities of their transverse axes on the corresponding fixed
parabolas.
Sections of the hyperbolic paraboloid by planes parallel to the
asymptotic planes are straight lines. RS and RT are two such
sections, rs the projection of RS on the plane H.P. is parallel to al
the horizontal trace of one of the asymptotic planes and it will be
found that Rs' the projection of RS on the plane V.P. is tangential to
the parabola APC at R. Also, rt the projection of RT on the plane
H.P. is parallel to the horizontal trace of the other asymptotic plane
and E< the projection of PvT on the plane V.P. is also tangential to
the parabola APC at R. It will also be found that the projections of
these straight lines RS and RT on the plane of the parabola ATB are
tangential to that parabola.
Hence the hyperbolic paraboloid may also be generated by either
of two systems of straight lines, one system being parallel to one fixed
plane and the other parallel to another fixed plane. No two lines of
the same system intersect, but each line of one system will intersect
every line of the other system if they are produced far enough.
These results lead to the following definition. The hyperbolic para-
boloid may be generated by a straight line which moves parallel to a
fixed plane and intersects two straight lines which are not parallel
and do not intersect. From this mode of generation the hyperbolic
paraboloid has been called the twisted plane.
304. Tortuous Curves. — A tortuous curve is one in which no
definite part is in a plane.
If Q, P, and R be points on a tortuous curve, P being between Q
and R, a plane may be found containing these three points. If Q and
R be moved nearer to P then as Q and R move up to P the ultimate
position of the plane containing Q, P, and R is the osculating plane of
the tortuous curve at P, and the ultimate position of the straight line
joining Q and R is the tangent to the tortuous curve at P. The
tangent at P is evidently in the osculating plane at P.
A plane through a point P on a tortuous curve at right angles to
the tangent at P is called the normal plane to the curve at P. Any
straight line passing through P and lying in the normal plane to the
curve at P is a normal to the curve at P. The line of intersection of
the normal and osculating planes of the curve at P is called the
vrincipal normal of the curve at P. The normal through P which is
perpendicular to the osculating plane is called the hinormal to the
curve at P.
The projection of the tangent at a point P of a tortuous or a plane
curve is the tangent to the projection of the curve at p the projection
of P.
344
PRACTICAL GEOMETRY
Exercises XXIII
1. A circle 4 inches i-n diameter is in the V.P. A vertical straight line is 4 inches
in front of the V.P. and its elevation touches the elevation of the circle. A
surface is generated by a horizontal straight line vsrhich moves in contact with the
given line and circle. Draw the elevation of the section of the surface by a plane
parallel to the V.P. and 2 inches in front of it.
2. A right circular cone, base 4 inches in diameter, and altitude 4 inches,
stands with its base on the HLP. A right circular cylinder 2 inches in diameter
and 4 inches long stands with its base on the H.P. and its axis 3 inches away
from the axis of the cone. A surface is generated by a horizontal straight line
which moves in contact with the surface of the cone and the axis of the cylinder.
Draw the elevation of the line of intersection of this surface with the surface of
the cylinder, the plane of the elevation to be inclined at 45° to the plane of the
axes of the cylinder and cone.
3. Three straight lines AB, CD, and EF are given by their projections in Fig.
652. A surface is generated by a straight line which moves in contact with each
g»
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m'
e.
z,
n^
c
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V,
/
"V
V
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s
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y
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y
^
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^
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,
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r
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y
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y
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/
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ft
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c
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7
d
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Fia. 652.
Fig. 653.
of the given lines. Draw the elevation of the intersection of this surface with the
vertical plane whose horizontal trace is H.T. Take the squares of the squared
background in Fig. 652 as of \ inch side.
4. The semicircles ahc and a'h'c' (Fig. 653) are the projections of the half of an
ellipse, de and d'e' are the projections of a vertical line. A horizontal straight
line moves in contact with the curve ABC and the line DE. Draw the plan and
elevation of the curve of intersection of this surface and the plane whose traces
are given. Find also the point of intersection of the line tun, vi'n' with the sur-
face generated as above. Take the squares of the squared background in Fig. 653
as of \ inch side.
5. The circles ahcd and a'h'c' d' (Fig. 654) are the plan and elevation respectively
of an ellipse. A conical surface is generated by a straight line which moves in
contact with the ellipse and passes through the fixed point vv'. Determine the
horizontal trace of the conical surface. Show also the traces of the planes which
are tangential to the conical surface and contain the point rr'»
CURVED SURFACES AND TANGENT PLANES 345
6. xi'a'h' (Fig. 655) is the elevation of a right circular cone and vs is the plan
of its axis, v'f is the elevation of a straight line on the front surface of the cone.
Fig. 654
Fig. 656.
Without drawing an ellipse determine the traces of a plane which touches the
cone along the line VR.
7. The plan and elevation of a cone enveloping a sphere are given in Fig. 656.
Determine the traces of a plane inclined at 60° to the horizontal plane and
tangential to the cone.
8. Determine the traces of a plane tangential to the cone given in Fig. 656 and
inclined at 45° to the vertical plane of projection.
9. Referring to Fig. 656, ^ is the plan of a point lying on the upper surface of
the given cone. Draw the traces of the plane which is tangential to the cone at
the point P.
10. A right circular cone whose vertical angle is 50° lies with its slant side on
the horizontal plane and the plan of its axis perpendicular to XY. Draw the
traces of a plane which touches the cone in a line whose inclination to the
horizontal plane is 40°.
11. A right circular cone, base 2*5 inches diameter and axis 2*5 inches long,
has its base on the vertical plane of projection and its axis 1"5 inches above the
horizontal plane. Draw the traces of the four planes which are tangential to this
cone and inclined to the horizontal plane at 60°.
12. The plan of the vertex of a cone is on the circumference of its horizontal
trace which is a circle 2 inches in diameter. The height of the vertex above the
horizontal plane is 2-5 inches. Draw the scale of slope of a plane tangential to
this cone and inclined at 60° to the horizontal plane.
13. The circle ah (Fig. 657) is the horizontal trace of a cylinder and c^d^^ is
the indexed plan of its axis, p is the plan of a point on the upper surface of the
cylinder. Draw the traces of the plane which is tangential to the cylinder at P.
14. Draw the traces of a plane which is tangential to the cylinder of the
preceding exercise and inclined at 70° to the horizontal plane.
15. Draw the traces of a plane which is tangential to the cylinder of exercise
13 and which contains the line tn^n^ (Fig. 657).
16. Os&2o (Fig. 658) is the indexed plan of the axis of a right circular cylinder
1-5 inches in diameter, r is the plan of a point on the lower surface of the
cylinder. Draw the scale of slope of a plane which is tangential to the cylinder
atR.
17. Draw the scale of slope of a plane which is tangential to the cylinder of
the preceding exercise and which contains the point S33 (Fig. 668).
346
PRACTICAL GEOMETRY
r.J^a E% ^^-^ '" ^^? ^^'^^ P^^"",^^ ^ '*^^^S^* 1^^« ^^ and a sphere whose
fP^alf !?• f^'^'l*^^ *'^'^' °^ *^^ P^^^^^ ^^i^^ ^^ntain the line and are
tangential to the sphere.
llllllll'^^^
^ » AZ Y
-^ '' z
^^ ^^/ V
y -.2v 3
^ ^7 4 ^
-? J 4
^ ^ Jn
r 2 ^
\ .4
I
/•
/
/
/
J*
/
^
/
_
yrv
.,-, Y
JL
V
\ t
\
7fl
" i
^ it
V
;^ A
\
J~2^
^^
^^ J
T
: t
Fig. 657.
Fig. 658.
a,!
Fig. 659.
In reproducing the above diagrams the sides of the small squares are to be taken
equal to a quarter of an inch. The unit for the indices is 0*1 inch.
19. Draw two circles, one 2 inches and the other 1 inch in diameter. The
centres of the circles to be 2 inches apart. These circles are the plans of two
spheres. The centre of the smaller sphere is 1 inch and the centre of the larger
sphere is 2-5 inches above the horizontal plane. Draw the scales of slope of all
the planes which are tangential to the two spheres and are inclined at 70"^ to the
horizontal plane.
20. AB is a straight line whose plan is 3 inches long. A is one inch and B is
2 inches above the horizontal plane. Represent a plane which passes between A
and B, is inclined at 60° to the horizontal plane, and is at perpendicular distances
of 1 inch and 1*5 inches from A and B respectively.
21. vh and he are two straight lines at right angles to one another and each 3
inches long. V is the vertex and VH the axis of a right circular cone whose
vertical angle is 40°. H is on, and V is three inches above the horizontal plane.
C is the centre of a sphere 2 inches in diameter which rests on the horizontal
plane. Represent a plane which is tangential to the cone and sphere.
22. abc is a triangle, ab = 1*5 inches, be = 1"3 inches, and ac = 2*5 inches,
A is 2 '5 inches, B is 1 inch, and G is 2*4 inches above the horizontal plane. AB
is the axis of a right circular cylinder 1"6 inches in diameter. C is the centre of
a sphere 1-4 inches in diameter. Draw the scale of slope of a plane which passes
over the cylinder and under the sphere and is tangential to both.
23. A, B, and C are the centres of three spheres whose radii are, 0"8 inch, 0*4
inch, and 0*6 inch respectively, ab = 1*7 inches, be = 1'3 inches, and ac = 1'2
inches. The heights of A, B, and C above the horizontal plane are, O'S inch,
0*6 inch, and 1-6 inches respectively. Represent a plane which passes over the
spheres A and C and under the sphere B and is tangential to all three.
24. The elevation of a solid of revolution, whose axis MN is parallel to the
vertical plane of projection, is given in Fig. 660. Draw the horizontal trace of a
vertical cylinder which envelops this solid, and show the elevation of the curve of
contact, s' is the elevation of a point on the front surface of the solid. Represent
a plane tangential to the surface at S.
25. The circle ab (Fig. 661) is the plan of a prolate spheroid whose axis is
vertical and 3 inches long, its lower end being on the horizontal plane, cd is the
plan of a straight line, G being 4 inches above and D on the horizontal plane. A
cylinder whose axis is parallel to GD envelops the spheroid. Show in plan and
elevation the curve of contact between the cylinder and the spheroid, r is the
plan of a point on the lower surface of the spheroid. Draw the traces of the plane
which is tangential to the spheroid at R.
CURVED SURFACES AND TANGENT PLANES 347
26. Taking the spheroid and the point C as in the preceding exercise, draw in
' plan and elevation the curve of contact between the spheroid and a cone
7
-/-i^. "
a)
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"
"7
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Q
I,
I
7
-
^^
v, I
--s^^^S
ft
\ 1
\ 1
Jp
Fig. 704.
Fig. 705.
axial cylinders. Although the outer edge of this thread is quite sharp
it will be observed that in the projection (a) it has at the right and
left of the figure the appearance of being rounded. The boundary lines
connecting the projections of the outer and inner helices at (a) are
tangential to these projections and they are practically straight lines.
A section of the nut for this screw by a plane containing its axis is
shown at (h).
Fig. 705 shows at (c) a projection of a right-handed square-threaded
screw on a plane parallel to its axis. The section of the thread by a plane
containing the axis is a square. The top and bottom edges of the thread
form four helices of the same pitch, two on one cylinder and two QO.
HELICES AND SCREWS
367
riiiother, the cylinders being co-axial. A section of the nut for this
screw by a plane containing its axis is shown at (d).
A screw thread of reetangular section is shown in Fig. 706. In
this case the thread is thin compared with its depth and pitch.
A quadruple- threaded screw or worm is shown in Fig. 707. At the
top left-hand corner of the figure is shown the form of the section of
the threads by a plane containing the axis of the screw.
Fig. 706.
Fig. 707.
gr 315. Pitch and Lead of a Multiple-threaded Screw.— It is
usual to take the pitch of one of the threads of a multiple-threaded
screw as the " pitch " of the screw, and the distance between the centres
of two adjacent threads (measured parallel to the axis of the screw) is
then called the " divided pitch " of the screw, as already stated for
helices (Art. 311, p. 362). In America however it is a common
practice to call the pitch of one of the threads of a multiple-threaded
screw the "lead" of the screw and reserve the term "pitch" to what
has been called the " divided pitch." The lead of a screw is of course
the axial distance through which the nut moves for one revolution of
the screw.
316. Helical Springs. — "While a screw has a screw thread in one
piece with a cylinder from which it projects, a helical spring is a screw
thread without the attached cylinder. It therefore follows that in the
case of the spring the absence of the cylinder will expose to view a
greater part of the thread than is seen on a screw.
Fig. 708 shows a helical spring of which the section by a plane
containing the axis of the spring is a square.
A helical spring made of round wire is represented in Fig. 709.
The centre line of the wire forms a helix and the boundary lines of the
projections of the spring are obtained by considering the spring as the
368
PRACTICAL GEOMETRY
envelope of a sphere whose diameter is equal to that of the wire and
which moves with its centre on the helix. On the half plan of the
spring is shown a section by a horizontal plane whose vertical trace is
LN. The outline of the section is tangential to the horizontal sections
of the moving sphere by the given plane of section. A section of the
spring by a plane containing the axis is nearly but not quite circular.
Fig. 708.
Fig. 709.
317. Axial and Normal Sections of Screw Threads.— By
an axial section of a screw thread is meant a section by a plane contain-
ing the axis and by a normal section is meant a section by a plane at
right angles to the central helix of the thread. When the pitch of a
screw thread is small compared with its diameter there is little differ-
ence between an axial section and a normal section, but when the pitch
is large the difference is considerable.
Fig. 710 shows a projection of a part of a screw thread or helical
spring whose central helix has a large pitch angle, the projection
being on a plane parallel to the axis. It is evident that where the
projections of the various helices cut the projection of the axis these
projections are straight and are inclined to the projection of the axis
at angles which are the complements of the pitch angles of the helices.
Also at the points considered the helices are parallel to the plane of
projection.
Still referring to Fig. 710, the axial section of the thread is the
square S, oah is the projection of the central helix of the thread, and ot
is its tangent at o. LoN, at right angles to ot is the edge view of a
plane which is perpendicular to the helix at O. The figure EFHK is
the true form of the normal section at LN. The edges EF and KH
are nearly straight but are arcs of ellipses, being parts of an oblique
section of two cylindrical surfaces. The edges EK and FH are sections
of screw surfaces and are very approximately straight lines. It will be
i
i
HELICES AND SCREWS
369
seen that the thickness of the thread tapers distinctly at a normal
section.
Fio. 710.
Fig. 711.
Fig. 711 shows the construction for finding S the true form of an
axial section when the noi:nial section EFHK is of uniform width, that
is, when the thickness of the thread at a normal section is uniform,
318. Handrails. — The making of handrails for stairs is considered
to be one of the most difficult parts of the craft of the joiner and a
knowledge of the geometry of handrails is essential to their correct and
economical construction. The complete treatment of the subject of
handrailing in all its technical details is beyond the scope of this work
but the fundamental principles and their applications to a few examples
may be studied here with advantage.
The construction of a straight handrail presents no difficulty.
Bamps and level casings are also easily drawn and made. A ramp is the
part of a handrail whose centre line is curved in a vertical plane only.
A level easing is the part of a handrail whose centre line is curved in a
horizontal plane only. A level easing whose centre line is a quarter of
a circle is called a level quarter. The part of a handrail whose centre
line is curved vertically and horizontally is called a wreath. A straight
piece of handrail formed on the end of a wreath is called a shank. It is
the construction of wreaths which is the difficult part of handrailing
and their geometry will now be considered.
The geometry of the centre line of the wreath will first be studied.
Referring to Fig. 712, ah, a'h' and cd, c'd' are the straight centre lines
of two straight pieces of handrail which are to be connected by a
wreath whose centre line is, in plan, a quarter of a circle he. The lines
ah, a'h' and cd, c'd' are called the central tangents of the wreath and in
order that the wreath may be cut economically from a plank it is usual
to arrange that these tangents shall intersect and therefore lie in a
plane. In Fig. 712 the tangents ah, a'h' and cd, c'd' intersect at the
2b
370
PRACTICAL GEOMETRY
point it! and in the example considered the tangents are equally inclined
to the horizontal plane.
The centre line of the wreath lies on the surface of a vertical
circular cylinder, and to the left of the elevation (U) is shown the
development of the vertical surface containing the centre line of the
wreath and the tangents, A^Bj and C^Dj being the developments of the
tangents. The development of the centre line of the wreath is shown
as a straight line BjC^, but it will be seen that AjB^ and C^Di are not
in the same straight line with BjC^. To produce a graceful centre line
" easing curves " should be introduced between the centre line of the
wreath and the tangents. These easing curves may be entirely on the
surface of the cylinder or entirely outside of it or they may be partly
on the cylinder and partly ontside of it.
Fig. 712.
Taking the straight line B^Cj as the development of the centre line
of the wreath the elevation of this centre line will be a helix obtained
by the usual construction or by projectors from the plan and the
development as shown. The plane containing the tangents ah, aV and
cdj c'd' is an important one. If a'b' be produced to meet XY at h' and
a perpendicular h'h be drawn to XY to meet ab produced at h, the
horizontal trace of the tangent ah, a!h' on a plane at the level of cc' is
determined, and he is the horizontal trace of the plane of the tangents
ah, a'h' and cd, c'd'. Any line in the plane of the tangents which is
parallel to he will be a horizontal or level line of the plane. When the
tangents are equally inclined to the horizontal it is easy to prove that
he is parallel to to and that to bisects the angle hoc. This is true
whatever be the magnitude of the angle hoc provided the tangents
intersect and have equal inclinations to the horizontal.
HELICES AND SCREWS
371
An elevation of the tangents on a pl^ne perpendicular to Tic or to
will show the two tangents in one straight line since this elevation is
an edge view of the plane containing them. Such an elevation is
shown to the right in Fig. 712, the plan having been redrawn and
turned round for convenience so that ot is perpendicular to XY. The
plane of the tangents will intersect the cylinder, on the surface of
which the centre line of the wreath is situated, in an ellipse whose
major axis is m'w' and whose semi-minor axis is the radius of the
cylinder. The true form of this ellipse is shown with the tangents of
the wreath in their proper positions in relation to the ellipse. The
student who has reached this stage will not require to be fold how to
construct the ellipse and the lines connected with it shown to the right
in Fig. 712. The line ABECD is the centre line of what is called the
face mould of the wreath, the use of which will be explained later.
Lines o'^ and o'C drawn from the centre of the ellipse to the points B
and C where the tangents meet the ellipse are called the springing lines
of the wreath. The plane containing the tangents will not contain the
true centre line of the wreath but in most cases the true centre line of
the wreath will lie very near to this plane. In the case shown in Fig.
712 the true centre line of the wreath lies partly on one side of the
plane of the tangents and partly on the other crossing it at the point
ee'. In the right hand part of Fig. 712 the elevation of the true centre
line of the wreath is not shown.
The case where the tangents are in vertical planes at right angles
to one another and one of the
tangents
cd.
c'd' is horizontal is
illustrated in Fig. 713. The
student should have no diffi-
culty in dealing with this case
after having mastered the more
general case which has just
been considered. It will be
noticed that the plane of the
tangents has dct for its hori-
zontal trace and the elevation
(U) is an elevation on a plane
perpendicular to the plane of
the tangents ABCD the
centre line of the face mould is
made up of the two tangents
AB and CD and BC the quarter
of an ellipse. B^C^ the develop-
ment of the centre line of the
wreath has been drawn with
an easing curve joining it to
CjDi the development of the
horizontal tangent, this easing curve being entirely on the develop-
ment of the surface of the cylinder on which the central line of the
wreath is situated.
372
PEACTICAL GEOMETRY
If the centre line of the face mould, determined as explained for
Figs. 712 and 713, be taken as the true centre line of the wreath, the
student should have no difficulty in drawing the development and any
required projections of it. The construction of this new centre line is
shown in Fig. 714 for the case illustrated in Fig. 712. The plan and
the positions of the tangents ah, a'b' and cd, c'd' are supposed to be given
as before. The plane of the tangents and the edge view of this plane
shown to the right in Fig. 714 are determined exactly as in Fig. 712.
From the edge view of the plane of the tangents and the plan beneath
it the development AiBiCjDi is projected as shown. The elevation (U)
is projected from the plan and the development, or from the plan and
the edge view as shown.
In making a wreath it is first formed of rectangular or approxi-
mately rectangular cross-section, and in this form it is called a squared
loreaih. The squared wreath is moulded to the required cross-section
by hand with suitable tools. The geometry of the handrail is mainly
the geometry of the squared wreath, and the correctness of the form of
the finished wreath very largely depends on the skill and care with
which the squared wreath is constructed.
The squared wreath resembles to some extent the thread of a
square-threaded screw of large pitch. In one system of constructing
squared wreaths the surfaces of the wreath are described by the sides
of a rectangle which moves with its centre on the centre line of the
wreath, the plane of the rectangle intersecting the centre line at right
angles, and two sides of the rectangle are horizontal. The horizontal
sides of the moving rectangle sweep out the upper and lower surfaces
HELICES AND SCREWS
373
of the wreath but even when the centre line of the wreath is a helix
these surfaces are not true screw surfaces because the generating lines
do not intersect the axis of the cylinder. Also, the other sides of the
moving rectangle do not lie exactly on the cylindrical surfaces. It may
however be assumed with sufficient accuracy for all practical purposes
that the sloping sides of the moving rectangle sweep out portions of
cylindrical surfaces which become the inner and outer vertical surfaces
of the wreath. In that case the developments of these surfaces will be
figures of uniform width. These developments of the inner and outer
vertical surfaces of the wreath are called the inner and outer falling
moulds of the wreath. The development of the vertical section of
the wreath which contains its centre line is called the central falling
mould.
The surface which contains the centre line and the centre lines of
the inner and outer vertical surfaces of the wreath will be described by
the horizontal centre line of the moving rectangle which, when produced',
intersects the axis of the cylinder. This gives the construction for find-
ing the centre lines of the inner and outer falling moulds from the centre
line of the central falling mould.
Fig. 715 shows the constructions for determining the falling moulds
and projections of the squared wreath whose centre line is the same as
that shown in Fig. 712. The tangents ah, a'V and cd, c'd' are first
drawn intersecting at it'. The shanks are then drawn in plan and
elevation. The plan may then be completed. The centre line of the
central falling mould is next drawn as in Fig. 712 and from this the
centre lines of the inner and outer falling -mould are determined as
374
PRACTICAL GEOMETRY
shown. Upon these centre lines the falling moulds are constructed of
uniform width. The elevation (U) may now be projected from the
plan and the falling moulds as shown.
Eig. 716 shows an elevation of the wreath on a plane pei^endicular
to the plane of its central tangents and from this the face mould is
shown projected. The /ace mould is the section of the wreath by the
plane containing the central tangents. The determination of the centre
line of the face mould has already been considered in connection with
Figs. 712 and 713. The inner and outer curves of the face mould are
the elliptic sections of the inner and outer cylindrical surfaces of the
wreath by the plane containing the central tangents.
Thickness of Plank
Fig. 716.
It is not necessary to draw the elevation of the wreath shown in
Fig. 716 in order to determine the face mould. Having found the
centre line ABECD, as previously explained, OE being the semi-minor
and OF the semi-major axis of the ellipse of which the curve EEC is a
part, make EH equal to half the horizontal width of the wreath.
Join EF and draw HK parallel to EF. Then OH is the semi-minor
and OK is the semi-major axis of the ellipse of which the inner curved
portion of the face mould is a part. In lij^e manner the semi-minor
and semi-major axes of the ellipse of which the outer curved portion of
the face mould is a part may be determined, and the elliptic arcs may
then be drawn by the trammel method. The ends of the face mould
are perpendicular to the tangents AB and CD, and the other straight
i
HELICES AND SCREWS
375
edges are tangents to the ellipses and are also parallel to the central
tangents respectively, as shown.
On the elevation of the wreath in Fig. 716 is shown the thickness
of the plank from which the wreath may be cut.
The wreath is cut from the plan either by the " bevel cut " shown at
(m) Fig. 716, or by the " square cut " shown at (n) Fig. 716. The latter
is considered to be the best as it involves less labour and requires less
material.
The angle marked ^ at (w) and (n) in Fig. 716 is called the
" bevel " for that end of the wreath and this angle must be determined
for each end of the wreath before the shanks can be formed and the
'' twist " given to the wreath. The angle ^ is simply the angle
between the vertical faces of a shank and the plane of the central
tangents, and it may be con-
veniently found as follows. Re-
ferring to Fig. 717, at, at' and td,
t'd', intersecting at tt\ are the
central tangents of the wreath, ah
being the ground line for these
projections, hd is the horizontal
trace' and ha' is the vertical trace
of the plane of these tangents.
By the usual construction (Art.
164, p. 200) ^A is the angle be-
tween the plane of the tangents
and the vertical plane containing
the tangent at, a't'. Taking td
as a ground line t'^d is a second
elevation of the tangent td, t'd'
and it is also the new vertical
trace of the plane of the tangents.
The angle <^i„ which is the angle
between the plane of the tangents
and the vertical plane containing the tangent td, t'd', can now be found
as shown. If the given tangents are equally inclined to the horizontal
it is evident that ^^ is equal to ^d-
In Figs. 715 and 716 the central, inner, and outer falling moulds
are all of the same width, measured at right angles to their centre
lines, and the resulting wreath has everywhere an approximately
rectangular cross-section, the cross-section being taken at right angles
to the centre line of the wreath. In another system of construction,
sections of the wreath by planes containing the axis of the cylinder
are rectangles, and the sides of these rectangles which lie on the upper
and lower surfaces of the wreath are horizontal ; the falling moulds
are then of different widths as will be seen from Figs. 718 and 719. Fig.
718 shows the falling moulds, on this second system, for the wreath
illustrated by Fig. 715, and Fig. 719 shows the falling moulds, on this
second system, for the case where the lower central tangent is hori-
zontal. The central falling mould is constructed in each case as
376
PRACTICAL GEOMETRY
before and from this the outer and inner falling moulds are projected
as shown. It will be seen that at corresponding vertical lines the
widths of the moulds, measured veriicalhj, are the same. Having con-
FiG. 718.
2
^
\ /\
(\
"'\
y^
\
A
^
"N
^.
X^
.2. -^
__^
.^rfC ^
y
/ /
1^^
/^
,,>
-^-^
__
>'
-?
L<1_.
—
—
—
■
—
— 1
—
l>
IdT-
- —
---
-
—
—
bJ
g^
t^i
\ — ;
)--€
»
—
;:
1-
^
K4 5 6
^<57rfna^ laUxng Mould
Outside FalUng Mould
Fig. 719.
Inside Falling Mould
structed the falling moulds, any required projections of the wreath
may be drawn as before. But it will easily be seen that only the
central falling mould need be drawn for the pui'pose of obtaining the
projections of the wreath.
319. Screw Propellers. — The geometry of a screw propeller
blade and the construction of the working drawings of it will be more
easily understood after a brief reference to the usual method of pre-
paring the mould for a propeller in the foundry.
Referring to Fig. 720, ABC is a templet, which when laid out fiat
is a rightrangled triangle having the right angle at B. The templet
ABC, having AB vertical, forms part of a vertical cylindrical surface
of which DE is the axis. If a straight line, intersecting the axis DE
and making a constant angle with it, moves in contact with the sloping
edge AC of the templet ABC it will generate a true screw surface.
AE and CD are the extreme positions of the moving line and three
intermediate positions are also shown.
In the foundry a rough bed of brickwork and loam is built up, the
upper face of this bed roughly conforming to the screw surface required.
A vertical spindle is fixed so that its axis coincides with DE. A
sleeve attached to one end of a horizontal arm fits on the spindle and
can slide or rotate freely on it. To the arm is attached a board, called
HELICES AND SCREWS
377
a loam hoard, the lower edge of which, guided by the sloping edge of
the templet ABC, sweeps out the screw surface on the loam as the arm
rotates about the axis of the spindle. The screw surface is then
smoothed by hand. Sharp iron pins attached to the lower edge of the
loam board at intervals cut helices FHJ, KLM, etc. on the screw
surface of the loam. The work described so far is for one blade and
this has to be repeated for each of the other blades if all the blades are
cast together.
The boss of the propeller may be moulded from a pattern inserted
in the usual way or it may be swept out of the loam with a suitable
loam board. The mould thus far prepared is dried and blackwashed.
The next step is to form with loam a piece of the same shape and
size as the required blade. Strips of wood EHJN, PLRQ, etc. are
cut out to the shapes of the required cylindrical sections of the blade
at different radii. One of these strips laid out flat is shown at (a).
These strips or thickness pieces are bent and fixed on edge to the
mould, their lower edges coinciding with the corresponding helices
FHJ, KLM, etc. which have been previously traced on the mould.
The surface of the blade having been outlined on the mould this
surface is covered with loam to the depth of the thickness pieces
FHJN, PLRQ, etc. After drying and blackwashing, the upper part
of the mould can now be built up on the lower part. The upper part
of the mould is taken off and then smoothed and dried. The loam
thickness piece is now removed and the finished halves of the mould
are put together.
A screw surface cannot be developed, but the face of the blade of a
screw propeller is generally a small portion of a complete convolution
of a screw surface and it may be developed approximately.
378
PRACTICAL GEOMETRY
It may here be pointed out that the helices on the face of the
blade are approximately elliptic arcs each being a portion of a plane
section of the surface of a cylinder whose axis is the axis of the screw,
the inclination of the plane being the same as that of the helix. This
is made use of in obtaining the projections of the blade from its
approximate development in the manner to be shown presently.
The shape and area of the developed face of the blade, also the
pitch of the screw surface and the general dimensions of the propeller
are settled by the naval architect in consultation with the marine
engineer. It will now be shown how the drawings of a propeller may
be made, the necessary particulars being given.
The case where the face of the blade is a true screw surface gene-
zLB
TRUE SCREW, LEFT-HAND
PITCH = 1-3 X DIAMETER
Fig. 721.
rated by a straight line moving at right angles to the axis will first be
considered. Referring to Fig. 721, the dotted curve ABC is the given
outline of the developed face of the blade drawn on the end elevation
about the vertical line dz which bisects the developed surface in the
neighbourhood of the boss.
Take a point e on o'^' and with centre d and radius dd describe
the arc dtid cutting the horizontal through d at d. Make do equal to
•p
^ where P is the pitch of the screw surface of the blade. Join od
and produce it. The portion of this straight line od which falls within
the plan of the blade will be, for all practical purposes, the plan of
that part of the helix lying on the face of the blade at the radius dd.
HELICES AND SCREWS 379
Further, od is the horizontal trace of a vertical plane which cuts
the surface of the cylinder of which e'n'd is a part elevation, and the
section is an ellipse an arc of which practically coincides with the por-
tion of the helix just referred to. Obviously od is the length of the
semi-major axis and o'e' is the length of the semi-minor axis of this
ellipse. Make o'D equal to od and taking o'D and o'e' as the semi-
major and semi-minor axes respectively construct DNe' a part of the
ellipse. This ellipse cuts the outline of the development of the face of
the blade at N. A horizontal through N to eut the circle through e'
at n' determines a point on the end elevation of the edge of the blade,
and a vertical from n' to meet od determines the plan of this point.
These points n' and n may be more accurately found as follows. From
N drop a perpendicular to meet the horizontal through o at Wq. Make
on equal to on^. Draw the vertical nn' to meet the horizontal through
N at n'.
Instead of drawing the ellipse it will be sufficient to draw e'N as a
circular arc struck from the centre of cu^ature of the ellipse at e.
The construction for finding this centre of curvature has been given
in Art 51, p. 48, and is shown to the right in Fig. 721. DF is
parallel to o'e' and e'F is parallel to o'D. Fs perpendicular to e'D
meets e'o' produced at s. A circular arc struck from the centre 8 with
a radius se' will practically coincide with as much of the ellipse as lies
within the development of the face of the blade.
Other points on the plan and end elevation of the edge of the
blade are found in the same way. For the side elevation shown to
the left in Fig. 721, make »h/«i' equal to mn and similarly for other
points.
On the elevation to the left in Fig. 721 is shown the maximum
thickness section of the blade. This is not a true plane section of the
blade, but its width at any distance from the axis of the screw is the
maximum thickness of the blade at a radius equal to that distance.
The developed sections of the blade at different radii are shown
between the two elevations. Referring to the second section from
the axis, the straight base NjHi is made equal in length to the elliptic
arc Ne'H. These developed sections give the sizes and shapes of the
thickness pieces such as PLRQ (Fig. 720) referred to in describing
the moulding of a propeller. These developed sections are generally
superimposed on one of the elevations of the blade, but in Fig. 721
they have been moved to one side for the sake of clearness.
A second example is illustrated by Fig. 722 which shows a detach-
able blade with a flange for bolting to a boss keyed to the shaft. In
this example the blade is set back by inclining the generating line of
the screw surface to the axis at an angle less than a right angle.
This is shown on the elevation to the left in Fig. 722 by the tilting
over of the maximum thickness section. Points on the generating
line will describe helices just as before but the various helices will be
displaced in an axial direction each by a definite amount depending on
its distance from the axis. For example, the helix at the radius o'e'
will be displaced by the amount d^^Oi. Hence the plan of this helix
380
PRACTICAL GEOMETRY
will cub the plan of the axis at e such that oe is equal to dj'e/. The
slope 6 of the helix may be obtained as before but in Fig. 722 the
triangle for finding 6 is shown on the elevation instead of on the plan.
neSy the plan of the helix on the face of the blade at radius o'e' is drawn
at right angles to e'T. The plan and the two elevations of the edge
of the blade are otherwise drawn as before, the construction lines being
clearly shown.
It will be seen that for the same diameter of screw the true length
RIGHT- HANDED SCREW
PITCH s|-38x DIAMETER
Fig. 722.
of the blade is slightly increased by giving it a set back, but this is
generally neglected in applying the developed face of the blade to the
elevation as shown in Fig. 722.
In the examples so far considered all the helices on the face of the
blade have had the same pitch. Frequently however the helix at the
root has a less pitch than the helix at the tip and the pitch increases
uniformly from the root to the tip. The screw is then said to have a
radially increasing pitch. Fig 723 shows clearly how the slopes of the
HELICES AND SCREWS
381
helices at different radii are obtained when the blade has no set back.
After the slopes of the various helices have been found, the plan, shown
in Fig. 723, and the elevations (not shown) are determined as before
from the developed face of the blade.
For a screw which has a radially increasing pitch the moulder
Ri= Radius at Tip
Ri= Radius at Root
Pi = Pitch at Tip
Pz' Pitch at Root
Fig, 723.
requires a templet for the root as well as one for the tip for the purpose
of guiding the loam board in sweeping out the face of the blade. The
loam board must also be connected to the arm in such a way that it
can alter its inclination as the arm revolves, or the arm may be simply
forked to embrace the central spindla
Ri= Radius at Tip
Rz- Radius at Root
Q-' Set Back at Tip
Pi = Pitch at Tip
P2= Pitch at Root
Fig. 724.
The slopes of the various helices will not be altered by giving the
blade a set back and the construction given in Fig. 723 will therefore
still apply in finding these slopes but the plans of the helices on the
face of the blade will not now pass through the point o but through
points determined as explained in connection with Fig. 722.
382
PRACTICAL GEOMETRY
A convenient construction for finding at the same time the slopes
and the positions of the plans of the various helices on the face of the
blade is clearly shown in Fig. 724 and needs no further description.
Still another modification of the design of the face of the blade of
Leading Edge
A 2. 3
Ri = Radius at Tip
R2= Radius at Root
Pi = PiUh of Leading Edge
P3= Pitch of ToUoimg Edge
ij'iG. 725.
a screw propeller has to be mentioned. The part of the face of the
blade towards the leading edge may have a less pitch than the remain-
ing part towards the following edge. The face of the blade is then
said to have an axially increasing pitch. The necessary modification in
Fig.
^^^Radias at Tip
^-^Radius at Root
Q^ Set Back at Tip
, - Pitch of Leading Edge at Tip
?i= Pitch of Leading Edge at Root
?i=-Pitch of Following Edge at Tip
Pa'^ Pitch (^ Following Edge at Root
726.
the construction of the plan of the blade is shown in Figs. 725 and
726. In Fig. 725 the face of the blade has an axially increasing pitch
but a radially constant pitch and no set back. In Fig. 726 there is
an axially increasing pitch, a radially increasing pitch, and also a set
back.
HELICES AND SCREWS 383
Exercises XXV
1. A right-handed helix of 2 inches pitch, and a left-handed helix of 1 inch
pitch are traced on a vertical cylinder 2 inches in diameter. Draw the elevation
of two turns of the first helix and four turns of the second. The lower ends of
the two helices to be at opposite extremities of a diameter of the cylinder which
is perpendicular to the ground line.
2. Two helices are traced on the surface of a vertical cylinder 2 inches in
diameter. The helices are at right angles to one another at their intersection
and one of them, which is right-handed, has a pitch of 4 inches. Show as much
of the two helices as is contained in 4 inches length of cylinder. Draw the
elevations of the tangents to these helices at points which are 1 inch, 2 inches,
and 3 inches above the lower end of the cylinder.
3. Show one turn of a helix of 3 inches pitch on a vertical cylinder 1 inch in
diameter. Draw the locus of the horizontal trace of a moving tangent to the
helix, and determine the true form of a vertical section of the surface described
by the moving tangent, the plane of section to be at a distance of 1 inch from the
axis of the cylinder.
4. A cylinder 1-5 inches in diameter and 3 inches long has three helices of
3 inches pitch traced on its surface at equal distances apart. Draw the plan of
the cylinder and helices when the axis of the cylinder is inclined at 45^^ to the
horizontal plane.
5. Work the example on the helix of increasing pitch illustrated by Fig. 701,
p. 364, having given, — diameter of cylinder 2 inches, height of cylinder 3 inches.
On the development draw the curve whose ordinates are equal to the pitch at
each point of the helix of increasing pitch.
6. A straight line 1-5 inches long is at right angles to a fixed vertical axis and
has one end on that axis. This straight line describes a screw surface of 3 inches
pitch. Draw the elevation of one complete turn of this screw surface, showing
on it the generating line in positions whose plans are at intervals of 15^. Show
also the helices described by points on the generating line which divide it into
three equal parts in addition to the helix described by the outer end.
7. Same as the preceding exercise except that the generating line is 2*4 inches
long and is inclined at 45° to the axis.
8. Eeferring to the example illustrated by Fig. 703, p. 365, draw the elevation
of one complete convolution of the surface generated by the quadrant of a circle,
having given, — radius of quadrant 2 inches, axial advance per revolution 3 inches.
Show the generating arc in positions whose plans are at intervals of 15'^. Also,
in addition to the helix described by the lower end of the generating arc, show
the helices described by three intermediate points. Lastly draw the elevation of
a section of the surface by a plane parallel to the vertical plane of projection and
0-5 inch in front of the axis.
9. An equilateral triangle of 2*5 inches side moves with one side on a vertical
axis so that the other two sides describe screw surfaces of 1*5 inches pitch. Draw
the elevation of the surfaces described by four revolutions of the triangle. On
the lower half of the elevation show the generating lines in positions whose plans
are at intervals of 15*^. Show the helix which is the intersection of the two
screw surfaces.
10. Same as the preceding exercise except that the pitch is 2*5 inches instead
of 1-5 inches. [In this case the two screw surfaces do not intersect.]
11. Full size axial sections of various screw threads are shown in Fig. 727.
In each the diameter over the threads is 3 inches, the screws are all right-
handed and single threaded. Draw for each a projection on a plane parallel to
the axis of the screw, showing as much of the screw as falls within a length of
4 inches measured parallel to the axis. Show also for each a section of the nut
by a plane containing the axis. Height of nut in each case 3 inches.
12. A right-handed treble threaded worm 3*5 inches external diameter and
384
PRACTICAL GEOMETRY
4 inches long has threads of the form and dimensions shown by the axial section
{d) Fig. 727. Draw a projection of this worm on a plane parallel to its axis.
(a)
'^^iii^^
Section on A A
13. The following particulars relate to three helical springs. — No. 1. Axial
section a rectangle 0*5 inch x 0-125 inch with the longer sides parallel to the
axis. External diameter 2 inches. Pitch 1 inch. No. 2. Axial section a rect-
angle 0-5 inch x 0-125 inch with the longer side perpendicular to the axis.
External diameter 3 inches. Pitch 1 inch. No, 3, Normal section a circle 0*5
inch diameter. External diameter 3-5 inches. Pitch 1-75
inches.
14. A cylinder 3 inches in diameter and 4 inches long
has a left-handed helical groove cut in it. A section of the
groove by a plane at right angles to the axis of the cylinder
is 1 inch wide and 1 inch deep, the sides being parallel to
and the bottom at right angles to the mid-radial line of the
section. Draw a projection of the cylinder and groove on
a plane parallel to the axis of the cylinder and determine
the true forms of axial and normal sections of the groove,
15. A portion of a 1| inch twist drill is shown in Fig,
728, consisting of a cylinder with a conical end, cut with
two spiral grooves each of 12 inches pitch and of the form
defined by the given sectional elevation. Complete the
plan, showing the curve BB, and the helical' grooves cor-
rectly projected for a distance of 6 inches from the line AA.
[B,E.]
16. A circle 1^ inches in diameter, whose plane
is vertical, revolves with uniform angular velocity
about a vertical axis in its plane, describing an annulus
or anchor ring whose mean radius is 1^ inches. While
the circle is revolving a point P on its circumference
moves round the circle with uniform velocity. The
point P travels sixteen times round the circle while
the circle revolves three times about the vertical
axis. Draw the plan of the curvd traced by the point
P. The result is shown to a reduced scale in Fig. 729.
Note that the curve is endless.
17. ABO is a right angled triangle. The sides AC
and BC which contain the right angle are 3*5 inches
and 1-5 inches respectively. The triangle revolves
with uniform angular velocity about AC which is ver-
tical. A point P starting at A moves along AB with uniform velocity and
HELICES AND SCREWS
385
reaches B while AB makes two revolutions. Draw the plan and elevation of
the path of P. Draw also the development of the surface described by AB and
show on it the curve traced by P.
18. The radius of the base of a right cone is 1-25 inches, the axis, which is
4-5 inches long, being inclined at 40^ to the H.P. The apex is above the base
and pointing to the right. From the highest point in the base, a cord 0-25 inch
in diameter is coiled round the cone in a right-handed spiral. The clear distance
between the coils, measured in a straight line parallel to the surface of the cone
is 1-25 inches. Draw the plan of the cone showing two turns of the cord, [b.e.]
19. The centre line CDDO of a portion of a handrail is shown in plan and
elevation in Fig. 730, the parts CD, CD being straight, and the part DD being an
elliptic arc. Draw the elevation of the centre line on X'Y'. Find the true shape
of the figure CDDC. [b.e.]
20. Referring to Fig. 730, draw the development of the surface of which cddc
is the plan and show on it the centre line CDDC.
r
Fig. 730. Fig. 731. Fig. 732.
The above Fig's, are to be reproduced double size.
21. The plan and part bt an elevation of the centre line ABCD of a portion of
a handrail are shown in Fig. 731. Assuming that the straight and curved parts
are in the same plane complete the elevation and draw another elevation on X'Y'.
Show also the true form of the centre line ABCD.
22. Fig. 732 shows the plan and part of an elevation of the centre line ABCDE
of a portion of a handrail. The parts AB and DE are straight. ST is the tangent
to the centre line at and this tangent meets AB produced at S and ED produced
at T. Assuming that BC is in the plane containing AS and SC and that CD is in
the plane containing CT and TE, complete the elevation of the centre line and
draw another elevation on X'Y'. Draw also the development of the vertical
surface containing the centre line ABCDE showing on it that centre line.
23. Fig. 733 gives the plan and part of the elevation of a portion of a handrail
of rectangular cross-section (as prepared for moulding) ; complete the elevation of
the rail. Draw a second elevation of the rail on X'Y'. Set out the " face mould "
for this rail. [b.e.J
2c
386
PRACTICAL GEOMETRY
24. Fig. 734 gives the plan and part of the elevation of a portion of a handrail
of rectangular cross-section (as prepared for moulding) ; complete the elevation of
the rail. Set out the "face mould" for this rail. Draw also the " falling mould "
for the central vertical section CDDC of the rail. [b.e.]
Fig. 733. Fig. 734. Fig. 735.
The above Figs, are to be reproduced double size.
25. Stair rail (Fig. 735). To be made in two parts, cut from planks and con- *
nected at C by a joint perpendicular to FG. A plan and part elevation are
given. Complete the elevation of the centre line ABODE and draw a second
elevation of this line on X'Y'. Show the plan and two elevations of the joint at
C. Draw the "face mould" for the wreath ABC, and set out the "bevels" for
its ends.
Plot the " falling moulds " for the central section and the inner and outer
surfaces of the wreath ABC. Complete the elevation of the wreath ABC. [b.e.]
26. In Fig. 736 are given the flange and the maximum thickness section of
one blade of a three-bladed propeller ; the dotted curve is the developed face of
the blade. The face of the blade is a true right-handed screw surface whose pitch
is one and a quarter times the diameter of the propeller. Draw the figure double
size and then construct a plan and two elevations together with sections of the
blade as shown in Fig. 721, p. 378.
27. Same as the preceding exercise except that the blade is to have a set back
which, measured at the tip, is 0*06 of the diameter of the propeller ; also the
screw is to be left-handed instead of right-handed.
28. The flange and the maximum thickness section of one of the four de-
tachable blades of a screw propeller are shown in Fig. 737. The dotted curve is
the developed face of the blade. The diameter of the propeller is 19 ft 6 in. at
the tip and 5 ft. 2 in. at the root. The pitch of the face towards the leading edge
increases from 19 ft. at the root to 21 ft. at the tip, and the pitch of the face
towards the following edge increases from 19 ft. at the root to 23 feet at the tip.
HELICES AND SCREWS
387
Draw the plan and two elevations of this blade to a scale of one inch to a foot,
and determine the developed sections of the blade by cylindrical surfaces of 3A,
44, 5i, 6i, 7^ and 8.^ ft. radii.
t*- 18 — ^
1 "^^
J
A
1 \
/
1 \
1 tu
! ^
■^ i
1 u.
1 .'
\
\
\
1 /
1 /
\^
x/
\
1/
AXIS OF
1
SHAFT i
DIMENSIONS
IN INCHES
T""1 1
Fig. 736.
AXIS OF SHAFT
Fig. 737.
CHAPTER XXVI
INTERSECTION OF SURFACES
320. General Method. — The general method of finding the
intersection of two surfaces is as follows. Let A and B denote two
given surfaces whose intersection with one another is required. Cut
the surfaces A and B by a third surface C, the latter surface being so
chosen and employed that the projections of its intersections with A
and B are lines, such as straight lines or circles, which can easily be
drawn. Let A' denote the line of intersection of C and A, and let B'
denote the line of intersection of C and B. Let the lines A' and
B' meet at a point P. Then the point P lying on the intersection
of C and A, lies on A. Also since the point P lies on the intersection
of C and B, it lies on B. The point P therefore lies on A and B, and
therefore it must be a point on the intersection of A and B, By
moving the surface C into diflPerent positions, or by cutting A and B
by other surfaces similar to C, any number of points on the intersection
of A and B may be determined.
On the intersection of two surfaces there are generally certain
important points which should be first determined. For example if
the projection of the intersection meets a boundary line of the pro-
jection of one of the surfaces, the meeting point will be an important
one to determine and in the case of a curved surface this will be a
tangent point.
In moving the cutting surface in one direction from a position
which gives points on the required intersection to a position which
gives no such points, there will be an intermediate position which
gives the last of the points obtained by moving the surface in that
direction and these points are usually important.
As a general rule after the important points have been determined
only a few others are necessary to fix the required line of intersection.
It is a good plan to number or letter the cutting surfaces in order,
and the points determined by them, each point having the same number
or letter as the cutting surface on which it is situated.
321. Intersection of Two Cylinders. — The general method
of finding the intersection of the surfaces of two cylinders is to cut the
surfaces by planes parallel to their axes or to their generating lines.
These planes will intersect the surfaces of the cylinders in straight
INTERSECTION OF SURFACES
389
f? -s^^^
lines, the intersection of which with one another will determine points
on the intersection required.
Example 1. — The horizontal trace of a vertical cylinder is an
ellipse, 32 inches X 2-4 inches, the major axis being parallel to XY.
A horizontal cylinder, 1 '6 inches in diameter has its axis parallel to
the vertical plane and 0'3 inch in front of the axis of the vertical
cylinder. It is required to show the elevation of the intersection of
the cylinders.
There will be two identical curves in this case and in Fig. 738
only enough of each of the cylinders is shown for determining one of
these curves.
Cutting planes parallel to the axes of both cylinders will in this
case be vertical planes parallel to the vertical plane of projection.
HT is the horizontal trace of one of
these planes. This plane will cut the
vertical cylinder in two vertical straight
lines, one of which will have the point a
for its plan, and a perpendicular to XY
from a for its elevation. This same
plane will cut the horizontal cylinder in
two horizontal lines of which ah will be
the plan. To find the elevations of
these lines, take an elevation of the
horizontal cylinder on a plane perpen-
dicular to its axis, this elevation will be
a circle, but only half of it need be
drawn, as shown. The cutting plane
now being considered will have a trace
on this new vertical plane which will
coincide with its horizontal trace. The
point 6/ where this trace cuts the circle,
or semicircle, just mentioned will be the
end elevation of AB one of the lines in Fiq, 733.
which the horizontal cylinder is cut by
the cutting plane HT, and the length hhi will be the vertical distance
of these lines above and below the horizontal plane containing the axis
of that cylinder. Hence the required elevations a'6', a'b' will be at
distances equal to 66/ above and below the elevation of the axis of the
horizontal cylinder.
The intersections of a'h' and a'b' with the vertical line through a
determine points on the intersection required.
In like manner, by taking other planes, other points can be found.
The most important points on the curves of intersection in this
case are those determined by the following cutting planes. (1) The
two planes which touch the horizontal cylinder, these determine the
points 1' and 5' on the elevation. (2) The plane which contains the
axis of the horizontal cylinder, this determines the points marked 3'.
(3) The plane which contains the axis of the vertical cylinder, this
determines the points marked 4'.
390
PRACTICAL GEOMETRY
The foregoing example may also be worked by using horizontal
cutting planes, and the student would do well to work the problem in
this way.
Example 2. — A vertical tube, external diameter 2-75 inches,
internal diameter 1-5 inches, has a horizontal cylindrical hole bored
through it 1-25 inches in diameter. The axis of the hole is 0-25 inch
from the axis of the tube. It is required to draw an elevation on a
vertical plane inclined at 35° to the axis of the hole.
First determine the intersection of the boring cylinder with the
outside of the tube exactly, as in the preceding example. The fact of
the horizontal cylinder being inclined to
the vertical plane will make no difference
in the working, as the heights of all the
lines in which the cutting planes inter-
sect the horizontal cylinder will remain
the same whatever be its inclination to
the vertical plane.
Next in like manner the intersection
of the boring cylinder with the interior
surface of the tube is determined. It will
be found that in this example this
interior intersection consists of one line
only.
The cutting planes which give the
most important points on the curves of
intersection in this example are the same
as those in example 1, with the addition
of the plane which touches the inner
surface of the tube.
The result of this example is shown
in Fig. 739, but the construction lines
have been omitted.
This example, like the preceding one, may also be worked by using
horizontal cutting planes.
Example 3 (Fig. 740).— a6 a'h' is the axis of a cylinder whose
horizontal trace is a circle 2*25 inches in diameter, cd c'd' is the axis
of a cylinder whose horizontal trace is an ellipse (major axis 3 inches,
minor axis 2 inches), whose minor axis is parallel to XY. It is required
to show, in plan and elevation, the intersection of the surfaces.
In this example the planes which will intersect the surfaces of both
cylinders in straight lines will be inclined to both planes of projection.
To find the directions of the traces of the cutting planes take a
point j>p' in ab a'h'. Through this point draw pq p'^ parallel to cd c'd'.
The line aq which joins the horizontal traces of ah a'h' and pq p'^ will
be the horizontal trace of a plane which contains the axis of one
cylinder and is parallel to the axis of the other, and this plane, and
planes parallel to it will, if they cut the cylinders at all, cut them in
straight lines parallel to their respective axes. In this particular
example the horizontal traces of these planes are parallel to XY.
Fig. 739.
INTERSECTION OF SURFACES
391
Seven planes whose horizontal traces are shown, and numbered
from 1 to 7, will determine all the important points, and no other
planes need be taken.
Fig. 740.
Consider plane number 3, This plane cuts each cylinder in two
straight lines whose plans are parallel to the plans of the axes of the
respective cylinders, and whose horizontal traces are at the points
where the horizontal trace of the plane cuts the horizontal traces of
the cylinders. The elevations of these lines will be parallel to the
392 PRACTICAL GEOMETRY
elevations of the axes of the respective cylinders upon which they lie.
These four lines intersect at four points in plan and in elevation and
determine the four points on the intersection of the cylinders which
have the number 3 attached to them.
It will be observed that plane number 1 touches both cylinders.
As already stated the horizontal traces of all the necessary cutting
planes are shown, but the remaining construction lines are only shown
for planes 1, 3, and 6, for the sake of making the figure clearer.
The portion of the curve which is seen in plan, and which must
therefore be put in as a full line, is found from the intersection of those
lines which lie on the upper surface of one cylinder with those which
lie on the upper surface of the other, that is, from the intersection of
the lines whose horizontal traces lie on the part fgh of the ellipse with
those whose horizontal traces lie on the part Imn of the circle. The
remainder of the curve in plan will be hidden by one or other or both
of the cylinders, and will therefore be put in as a dotted line.
In like manner the part of the curve which is seen in the elevation
is found from the intersection of the lines which lie on the front of one
cylinder with those which lie on the front of the other, that is from
the intersection of the lines whose horizontal traces lie on the halves
of the horizontal traces of the cylinders which are furthest from XY.
Instead of using the horizontal traces of the cylinders and the
horizontal traces of the cutting planes, the vertical traces may be
employed if the vertical traces of the cylinders come within a con-
venient distance on the paper. When one cylinder has a vertical trace
but no horizontal trace within a convenient distance, and the other
has a horizontal trace but no vertical trace within a convenient
distance, it is necessary to use both the horizontal and vertical traces
of the cutting planes.
322. Intersection of Cylinder and Cone. — The general
method of finding the intersection of a cylinder and cone is to cut the
surfaces by planes parallel to the axis of the cylinder and passing
through the vertex of the cone. These planes will cut the surface of
the cylinder in straight lines parallel to its axis, and the surface of the
cone in straight lines passing through its vertex. The intersection of
the lines on the cylinder with the lines on the cone determine points
on the intersection required.
Example 1 (Eig. 741). A right cone, base 2*9 inches diameter, axis
3-8 inches long, has its base horizontal. A cylinder 2 inches in
diameter has its axis horizontal and 1*2 inches above the base, and 02
inch from the axis of the cone. It is required to draw the plan of the
intersection of the surfaces and an elevation on a vertical plane in-
clined at 30° to the axis of the cylinder.
The general method which has just been described can easily be
applied to this example, but in this case the cutting planes may be
horizontal, because horizontal sections of the cone are circles and hori-
zontal sections of the cylinder are straight lines. In Fig. 741, hori-
zontal cutting planes have been taken.
Draw an elevation of the cylinder and cone on a plane perpendicular
INTERSECTION OF SURFACES
393
to the axis of the cylinder. In Fig. 741 this elevation is shown brought
round into the plane of the other elevation.
Consider the cutting plane of which L'M' is the vertical trace.
This plane intersects the cone in a circle of which the diameter is equal
to a/6/. The plan of this circle is a circle concentric with the plan of
the cone. This same cutting plane intersects the curved surface of the
cylinder in two straight lines of which the points 6/ and c/ are the end
elevations. The plans of these lines are parallel to the plan of the
axis of the cylinder, and at distances from it equal to half of t/c/. The
points h and e where these lines on the plan meet the circle already
mentioned, are the plans of points on the curve of intersection, and
Fig. 741.
perpendiculars from b and c to XY to meet L'M' at h' and c' determine
the elevations of these points. In like manner any number of points
on the curve of intersection may be found.
In Fig. 741 eight cutting planes are shown numbered from 1 to 8,
and these should be sufficient.
The most important points are on the planes 1, 3, 4, 5, 7, and 8.
The position of the plane number 5 is obtained as follows, jp/^/ is the
vertical trace of a plane which touches the cylinder and passes through
the vertex of the cone, p/ being the end elevation of the line of contact
of this plane and the cylinder, o^p^ is of course perpendicular to
2?/^/. The vertical trace of plane number 5 passes through p^'. The
I
394
PRACTICAL GEOMETRY
lines in which the plane p^ql cuts the cone are also shown and as
these lines are tangents to the curve of intersection they assist in the
correct drawing of the curve at important points.
Example 2 (Fig. 742). vv' is the vertex and va v'a' the axis of a
cone whose vertical angle is 30°. he h'c is the axis of a cylinder 2 inches
in diameter. The dimensions which fix the positions of the axes of
the cone and cylindei' are given at the top right hand corner of
Fig. 742.
Fig. 742. It is required to show, in plan and elevation, the inter-
section of the surfaces of the cone and cylinder.
This example is worked by using cutting planes passing through
the vertex of the cone and parallel to the axis of the cylinder which,
as already stated, is the general method of finding the intersection of
the surfaces of a cone and cylinder.
The first step is to determine the horizontal traces of the cylinder
and cone by Arts. 219 and 226 respectively. The construction lines for
this are omitted in Fig. 742.
All planes which are parallel to the axis of the cylinder and pass
1
INTERSECTION OF SURlFACES
395
through the vertex of the cone will contain the line vt v't' drawn
through v>v' parallel to he h'c . Hence the horizontal traces of all the
cutting planes will pass through t, the horizontal trace of vt v't'.
There are eight planes required to determine all the important
points but only three of these planes, numbered 1, 4, and 8, are shown
in Fig. 742. Plane number 1 touches the cylinder and* cuts the cone,
and plane number 8 touches the cone and cuts the cylinder ; each of
these planes will therefore determine two points only on the inter-
section required. Plane number 4 cuts both surfaces and will
determine four points on the intersection required. The omitted
planes 2, 3, 5, 6, and 7 will each determine four points.
323. Intersection of Two Cones. — The general method in this
Fig. 743.
•ase is to use cutting planes passing through the vertex of each cone.
These planes will cut the surface of each cone iu straight lines, and
396
PRACTICAL GEOMETRY
they will all contain the line joining the vertices of the cones. Hence
the horizontal traces of all the cutting planes will pass through the
horizontal trace of the line joining the vertices of the cones. Also, the
vertical traces of the cutting planes will pass through the vertical
trace of the line joining the vertices of the cones.
Example (Fig. 743). v{ol is the vertex of a cone whose horizontal
trace is a circle 3 inches in diameter, v^v.^ is the vertex of a cone
whose vertical trace is a circle 2*5 inches in diameter. The other
dimensions are given on the figure. It is required to show, in plan
and elevation, the intersection of the surfaces of the two cones.
Seven cutting planes passing through the vertex of each cone
will determine all the important points in this example, but in Fig. 743
only four of these planes, numbered 1, 2, 3, and 4, are shown. Both
traces of each cutting plane are employed. All the horizontal traces
of the cutting planes pass through v^, and all the vertical traces pass
through v/.
Consider plane number 1. This plane touches one cone and cuts
the other. The horizontal trace of the plane is drawn first, and if the
point where this trace meets the ground line comes within the paper
the vertical trace is obtained by joining «?/ to this point. If however
the point where the horizontal trace meets the ground line is off
the paper, then the construction given in Art. 11, p. 9, may be used
for drawing the vertical trace. The intersections of the line in which
plane number 1 touches the one cone with the lines in which it cuts
the surface of the other cone determine the points numbered 1 on
the intersection required. In like manner the other points are
determined.
324. Intersections of Cylinders and Cones enveloping
the same Sphere. — If two cylinders envelop the same sphere their
Fig. 744.
Fig. 745.
Fig. 746.
intersection will be plane sections of both and will therefore be
ellipses. Fig. 744 shows a projection of two such cylinders on a plane
parallel to their axes. The straight lines ah and cd form the projection
of the intersection of the cylinders.
The same remarks apply to the cone and cylinder (Fig. 745), and
to two cones (Fig. 746). Referring to Fig. 745, if the vertex of the
INTERSECTION OF SURFACES
397
cone is placed on the curved surface of the cylinder, one of the
ellipses will become a straight line. Referring to Fig. 746, it is
obvious that if the cone v.2ad be turned round so as to make V2d more
nearly parallel to v^a, the straight line cd will become more nearly
parallel to v^a and v^, and when v./l is parallel to v^a, cd will also be
parallel to v^a and v.^, and the intersection cd will then be a parabolic
section of each cone. Continuing the motion of the cone v.^ad in the
same direction, the intersection cd will become a hyperbolic section of
each cone.
325. Intersection of Cylinder and Sphere. — Since the sur-
face of a sphere is a particular form of a surface of revolution this
problem is a particular case of that discussed in Art. 326, p. 398. If
\
3^*diam.
I
Fig. 747.
the cylinder is also a surface of revolution, that is, a right circular
cylinder, then this problem is also a particular case of that considered
in Art. 328, p. 401. But instead of using the method described in
Art. 326, or the method of Art. 328 cutting planes parallel to the axis
of the cylinder and perpendicular to one of the planes of projection
may be employed, because such planes will cut the sphere in circles
and the cylinder in straight lines.
If the axis of the cylinder is inclined to both planes of projection
then an auxiliary elevation on a vertical plane parallel to the axis of
398 PRACTICAL GEOMETRY
the cylinder should be drawn, and the cutting planes being vertical
and parallel to the axis of the cylinder they will cut the sphere in
circles which will appear as circles in the auxiliary elevation.
Example (Fig. 747). The plan and elevation of a cylinder and
sphere are given, the dimensions being marked on the figure. It is
required to show, in plan and elevation, the intersection of the surfaces
of the cylinder and sphere.
The method used here is that of cutting the surfaces by horizontal
planes.
The upper right hand portion of Fig. 747 is a projection of the
cylinder and sphere on a plane perpendicular to the axis of the cylinder
and it is from this projection that the positions of the cutting planes
which give the important points on the required intersection arc
determined.
326. Intersection of Cylinder and Surface of Revolution. —
In all the problems hitherto considered on the intersection of surfaces,
the auxiliary cutting surfaces which have been used, in order to find
points on the required intersection, have been planes. The only simple
plane sections of a surface of revolution are, in general, sections at right
angles to its axis, which are always circles. Now it is evident that
only in very particular cases would a plane which cuts the surface of
revolution in a circle cut the surface of the cylinder in a circle or in
straight lines. For instance, a plane which cuts the surface of revolu-
tion in a circle may cut the cylinder in an ellipse, and it would clearly
be a laborious process to construct an ellipse for each cutting plane.
This objection may however be got over in the present problem by using
a tracing of the section of the cylinder by a plane which cuts the surface
of revolution in a circle, in a manner to be explained at the end of this
article.
Instead of cutting the given surfaces by planes they may be cut by
the surfaces of cylinders and points on the required intersection obtained
by means of circles and straight lines. Let a circle which is a section
of the surface of revolution be taken and let a straight line move in
contact with this circle and also remain parallel to the axis of the given
cylinder. This moving line will describe the surface of a cylinder which
intersects the surface of revolution in a circle and the surface of the
given cylinder in straight lines, and the points in which these straight
lines cut the circle will be points on the intersection required.
In order that the projections of the circular sections of the surface
of revolution may be circles and straight lines the axis of revolution
must be arranged perpendicular to one of the planes of projection.
Referring to Fig. 748, mn, m'n' is the axis of a cylinder which
intersects a surface of revolution whose axis is vertical. The con-
struction for the outline of the elevation of the surface of revolution is
shown at (c).
The ellipse which is the horizontal trace of the cylinder is deter-
mined as explained in Art. 219, p. 253.
ab, a'b' is a circular «ection of the surf acfe of revolution, oo' being the
centre of this circle. o<, o't' is a line through oo' parallel to mn, m'n'.
INTERSECTION OF SURFACES
399
ot, o't' is the axis of an auxiliary cylinder of which the circle a&, a'h' is
one horizontal section, and all horizontal sections of this cylinder will
be circles of the same diameter, it' being the horizontal trace of the
axis of this auxiliary cylinder the horizontal trace of its surface will be
a circle whose centre is t and radius equal to oa. The horizontal trace
f t
■7
Fig. 748.
of this auxiliary cylinder intersects the horizontal trace of the given
cylinder at s and r. The auxiliary cylinder intersects the given
cylinder in straight lines sjp, s'p' and r^, r'q' which are parallel to wm,
m'n'. pp' and qq the intersection of the lines 8p, s'p' and rq, r'q' with
400 PRACTICAL GEOMETRY
the circle ah, a'b' are points on the line of intersection of the given
cylinder and surface of revolution. By taking other auxiliary cylinders,
any number of points on the required intersection may be found.
The student should notice that each line of intersection of an
auxiliary cylinder and the given cylinder intersects the corresponding
circle on the surface of revolution in one point only although its plan
may cut the plan of the circle in two points.
A convenient and practical method of solving the problem which
has just been considered is to take horizontal sections of both the given
surfaces. The sections of the cylinder are ellipses but these ellipses are
all of the same size and if one of them be drawn and a tracing of it
made, it only remains to draw a sufficient number of circular sections
of the surface of revolution and apply the tracing of the ellipse to each
to find points in the required intersection. The position of the centre
of the ellipse corresponding to a particular circular section of the surface
of revolution is where the plane of that section cuts the axis of the
cylinder. For the circular section afe, a'b' (Fig. 748) e is the position
of the centre of the ellipse in the plan and the tracing of the ellipse is
placed so that the centre is at e and the major axis on mn. The points
where the ellipse cuts the circle ah are points on the plan of the inter-
section required and these points may be pricked through. The eleva-
tions of the points are of course perpendicularly over their plans and on
the elevation of the corresponding circular section.
327. Intersection of Cone and Surface of Revolution. —
Placing the surface of revolution so that its axis is vertical, horizontal
sections of it will be circles, but except in the special case where the
cone is a right circular cone and its axis is vertical, horizontal sections
of the cone will not be circles or straight lines and all the horizontal
sections will be different. The tracing paper method which is applicable
to the intersection of a cylinder and a surface of revolution and which
was described in the latter part of the preceding Art., is therefore not
suitable in the case of the intersection of a cone and a surface of
revolution.
Auxiliary cones are taken which have their vertices at the vertex
of the given cone and for their directrices they have circular sections
of the surface of revolution. These auxiliary cones will intersect the
given cone in straight lines and the intersection of these straight lines
with the corresponding circles on the surface of revolution will deter-
mine points on the intersection required.
Referring to Fig. 749, vn, v'n' is the axis of a cone whose vertical
angle is 30° and which intersects a surface of revolution whose axis is
vertical. The dimensions are given on the figure.
The horizontal trace of the cone is determined as explained in Art.
226, p. 259. ah, a'h' is a circular section of the surface of revolution,
oo' being the centre of this circle, vt, v't' is a line passing through oo',
the centre of the circle, and the vertex of the cone, vt, v't' is the axis
of an auxiliary cone of which the circle ah, a!h' is one horizontal section.
All horizontal sections of this auxiliary cone will be circles but they
will be of different diameters. The point U' being the horizontal trace
INTERSECTION OF SURFACES
401
of the avxis of the auxiliary cone it will be the centre of the circle which
is the horizontal trace of that cone. The radius of this circle is equal
to /'W where m' is found by joining v' to a' and producing it to meet XY
as shown. The horizontal trace of the auxiliary cone and the horizontal
trace of the given cone intersect at s and r, and the auxiliary cone
intersects the given cone in straight lines vr, v'r' and t's, v's'. pp and
Fig. 749.
qcl the points of intersection of the lines vr, v'r' and vs, v's' with the
circle ah, a'h' are points on the intersection of the given cone and
surface of revolution. By taking other auxiliary cones any number of
points on the required intersection may be found.
328. Intersection of Two Surfaces of Revolution whose
Axes are Parallel. — Since all sections of a surface of revolution by
planes perpendicular to its axis are circles, it follows, that if the axes
of two surfaces of revolution are parallel, a plane which is perpendicular
2d
402
PRACTICAL GEOMETRY
to the axis of one will be perpendicular to the axis of the other, and
this plane, if it cuts both surfaces, will cut them in circles the inter-
section of which with one
another determines points
on the intersection required.
The surfaces of revolu-
tion should be arranged so
that their axes are perpen-
dicular to one of the planes
of projection. The projec-
tions of the circles men-
tioned above will then be
circles and straight lines
which are easily drawn.
An example is shown in
Fig. 750. An " anchor ring "
whose surface is described
by the revolution of a vertical
circle about a vertical axis is
shown penetrated by a cone
of revolution whose axis is
also vertical. One cutting
plane is shown intersecting
the anchor ring in two cir-
cles ah, a'b' and cd, c'd'.
This same plane intersects -p^^^ rj^
the cone in the circle mn,
m'n'. From the plan it is seen that the circle mn on the cone only inter-
sects the circle ah on the anchor ring determining the two points p
on the plan of the intersection of the cone and anchor ring. It will
be seen that the cone touches the surface of the anchor ring at the
point qq'.
329. Intersection of Two Surfaces of Revolution whose
Axes Intersect. — Place the surfaces so that the axis of one of them
is vertical and the axis of the other is parallel to the vertical plane of
projection. Referring to Fig. 751, vn v'n' is the axis of a cone of
revolution which intersects a surface described by the revolution of an
arc of a circle about a vertical axis as shown. The axes of the two
surfaces intersect at oo'. h'a'd', the elevation of a sphere is shown, the
centre of this sphere being at the intersection of the axes of the given
surfaces of revolution. This sphere intersects the surface whose axis is
vertical in a circle whose elevation is the horizontal line a'h' and whose
plan is the circle ah having its centre at o. This sphere also intersects
the other given surface of revolution in a circle whose elevation is the
straight line c'd'. These two circles lie on the sphere and one being on
one of the given surfaces of revolution and the other on the other, their
points of intersection are points on the intersection required. ^>', the
point of intersection of a'h' and c'd', is the elevation of the points of
intersection of the two circles and their plans are determined by a
INTERSECTION OF SURFACES
403
projector to intersect the
circle ah. Observe that the
plan of the circle of which
c'd' is the elevation, and
which would be an ellipse,
need not be drawn. Taking
other spheres with their
centres at oo any number
of points on the inter-
section of the given sur-
faces may be found.
330. Intersection of
Two Spheres.— Since the
surface of a sphere is a
surface of revolution the
methods described in the
two preceding articles for
determining the intersec-
tion of two surfaces of
revolution are applicable
to the case of two spheres.
But, since the intersection
of two spheres is a circle,
the projections of this cir-
cle may be found by a
simpler method.
Referring to Fig. 752,
o^ and 0.2 are the plans of
the centres of two spheres
whose radii are r^ and n
respectively. An elevation
is drawn on a ground line
XY parallel to o^o^. The
common chord a'h' of the
circles which are the ele-
vations of the two spheres
on the ground line XY is
an elevation of the circle
which is the intersection
of the spheres. The plan
of this circle is an ellipse
whose minor axis is ab and
whose major axis is equal
to a'h'. From the plan and
elevation shown an eleva-
tion on any other ground
line is easily determined.
The intersection of the
surfaces of the two spheres
Fig. 752.
404 PRACTICAL GEOMETRY
referred to above is the locus of a point whose distances from the points
Oy and O^ are 7\ and r.> respectively.
In like manner the intersection of a third sphere whose centre is
0;{ and radius r.^ with the second sphere whose centre is O2 and radius
r.j is the locus of a point whose distances from Oo and O.. are n and r..,
respectively. The plan of this third sphere and the ellipse which is
the plan of its intersection with the second sphere are shown in Fig.
752. The second ellipse is found in the same manner as the first from
an elevation on a ground line parallel to 0.^0.^. The circle which is the
intersection of the first and second spheres intersects the circle which
is the intersection of the second and third spheres at points whose plans
are p and q. The points P and Q are evidently points whose
distances from 0^, Og and O3 are r^, ^2 and r^ respectively.
If a'h' is inclined to XY at an angle which is nearly a right angle
the positions of p' and ' on the surface of a ver-
tical cylinder. The con-
struction in this case is Fig. 766.
obvious.
At (c), oc> is the centre of a sphere upon the surface of which the
point pp casts the shadow p^p^. The vertical plane containing rr' is
taken. This plane intersects the sphere in a circle. Turning this
plane with the ray and circle in it about the vertical axis of the sphere
until the plane is parallel to the vertical plane of projection, the inter-
section of the ray and circle in the new position is found. Turning
this plane with the ray and circle and their point of intersection back
to its original position, p^p^ is determined as shown.
335. Cast Shadow of a Line. — If the line is a straight line,
then, whether the rays of light are parallel or proceed from a point, it
is evident that all the rays which meet the line are in the same plane.
Hence the cast shadow of the line on any surface will be a portion of the
intersection of this plane with the surface. The extremities of the cast
shadow of the line will be the cast shadows of its extremities. The
cast shadow of a straight line on a single plane is the straight line
joining the cast shadows of the extremities of the line.
Fig. 767 shows the shadow cast by the straight line «&, dV on the
plane L'MN and on the horizontal plane of projection, the rays of light
being parallel to rr' . The traces of the plane containing the line ah a'h'
and parallel to rr' are first determined. The required cast shadow is
made up of a part of the horizontal trace of this plane and a part of
the intersection of this plane with the plane L'MN as shown.
If the line whose cast shadow is required is a curved line, and the
rays of light are parallel, all the rays which meet the line will lie on a
PROJECTION OF SHADOWS
413
cylindrical surface ; but if the rays all proceed from a point those which
meet the curved line will lie on a conical surface. The intersection of
the fore-mentioned cylindrical or conical surface with the surface upon
which the shadow is cast is the cast shadow required.
Generally when the line is curved its cast shadow is determined by
first finding, by Art. 334, the cast shadows of a sufficient number of
points in the line and then drawing a fair curve through these. Fig.
768 shows the cast shadow of a curved line ah a'b' the rays of light
being parallel to rr'. The shadow is cast partly on the surface of a
horizontal cylinder and partly on the horizontal plane of projection.
V. ^
Fig. 767
Fig. 769.
Fig. 770.
A simple case of importance is that of the cast shadow of a circle
on a plane parallel to the plane of the circle. The cast shadow in this
case is a circle whose centre is at the cast shadow of the centre of the
original circle. If the rays of light are parallel (Fig. 769) the diameter
of the cast shadow is equal to that of the original circle. If the rays
of light proceed from a point (Fig. 770) the diameter of the cast
shadow is greater than that of the original circle and is found by an
obvious construction.
Two other simple cases of importance relate to the cast shadows of
parallel lines on a plane.
If the rays of light are
parallel then the cast
shadows of parallel lines
on a plane are themselves
parallel. If the rays of
light proceed from a
point then the cast
shadows of parallel lines
on a plane will, produced
if necessary, meet at a
point. The latter case is
illustrated by Fig. 771
where ah a'h' and cd c'd' Fig. 771.
are parallel lines which
cast their shadows on the inclined plane L'MN, the light coming from
414
PRACTICAL GEOMETRY
the point ss. The point W to which the cast shadows converge is the
trace on the plane L'MN of the line sf s't' which is parallel to the lines
ah a'b' and cd c'd'.
One other case may be mentioned. If a straight line casts a shadow,
partly on one plane and partly on another plane parallel to the first,
the shadow on the second plane is parallel to the shadow on the first.
336. Shadow of a Solid having Plane Faces. — When a solid
having plane faces is placed in a beam of light which is large enough
to embrace the whole of the solid, it is evident that one part of a face
of the solid cannot be in light and another part in shade, unless the
solid has re-entrant angles, in which case one part of the solid may
cast a shadow on another part. Hence the shade line must be made
up of edges of the solid. Those edges which make up the shade line
can generally be determined by inspection. A particular edge is
part of the shade line if a line representing a ray of light meeting a
point on the edge in question does not enter the solid at that point.
Fig. 772.
Fig. 773.
Having determined the shade line, its cast shadow, which is the
outline of the cast shadow of the solid, may be determined by the con-
structions of the two preceding Arts.
Instead of first determining the shade line, the cast shadow of each
edge of the solid may be found. The resulting figure is either a
parallel or a conical projection of the solid according as the rays of
light are parallel or proceed from a point. The boundary line of this
projection is the cast shadow of the solid.
Eig. 772 shows the shadow cast by a short chimney on a roof and
also the shadow cast by the coping on the chimney itself. The rays of
light are parallel to rr. All the necessary construction lines are
shown.
Eig. 773 shows the shadows cast by a hexagonal pyramid and a tri-
angular prism on the horizontal plane and also the shadow cast by the
pyramid on the prism. The rays of light are parallel to rr'. All the
necessary construction lines are shown.
PROJECTION OF SHADOWS
415
337. Shadow of a Cylinder. — A cylinder which casts a shadow
i^^enerally has its shade line made up of two straight lines and two
curved lines. The straight lines are the lines of contact of the two
tangent planes to the cylinder, these tangent planes are parallel to the
rays of light or pass through the luminous point. Conceive a plane to
contain these straight lines. This plane will divide each end of the
cylinder into two segments, and the curved boundary line of one of
these segments on each end will constitute the curved parts of the
shade line. Those segments of the ends which must be taken in get-
ting the curved parts of the shade line will be evident from inspection.
In the particular case where the rays of light are parallel to the
axis of the cylinder the surface of the cylinder is itself the shadow
surface, and in the particular case where the rays of light diverge from
a point within the cylinder produced, the outline of that end of the
cylinder which is nearest to the luminous point is the shade line.
Fig. 774.
Having determined the shade line, its cast shadow, which is the
outline of the cast shadow of the cylinder, may be determined by the
constructions of Arts. 334 and 335.
Fig. 774 shows the shadow cast by a right circular cylinder on the
horizontal plane. The axis of the cylinder is horizontal, and the rays
of light are parallel to rr'. All the necessary construction lines are
shown.
Fig. 775 shows the cast shadow of a right circular cylinder whose
axis is vertical. The light proceeds from the point sr\ and the shadow
is cast partly on the horizontal plane and partly on the inclined plane
L'MN. All the necessary construction lines are shown.
338. Shadow of a Cone. — A cone which casts a shadow gene-
rally has its shade line made up of two straight lines and one curved
line. The straight lines are the lines of contact of two tangent planes
to the cone, these tangent planes are parallel to the rays of light or
416
PRACTICAL GEOMETRY
pass through the luminous point. A plane containing these sti-aight
lines will divide the base of the cone into two segments and the curved
boundary line of one of these will be the curved part of the shade line.
The segment of the base which must be taken in getting the curved
part of the shade line will be evident from inspection.
In the case where the line which represents the ray of light
through the vertex of the cone falls inside the cone, the shade line will
consist simply of the whole outline of the base of the cone.
Figs. 776 and 777 show two cases of the shadow cast on the hori-
zontal plane by a right circular cone, the axis of the cone being per-
pendicular to the vertical plane of projection. In Fig. 776 the rays
of light are parallel to rr, while in Fig. 777 they diverge from the
point ss. it' is the trace on the plane of the base of a line through the
Fig. 776.
Fig. 777.
vertex of the cone parallel to rr' in Fig. 776 and through ss' in Fig. 777.
Lines drawn from the vertex of the cone to the points of contact of
the tangents from W to the base of the cone are the straight portions
of the shade line. The remainder of the construction is obvious.
339. Shadow of a Sphere. — A sphere which casts a shadow,
when the rays of light are parallel, will have for its shade line the
circle of contact of the enveloping cylinder whose axis is parallel to the
rays of light. This cylinder and its line of contact with the sphere
are determined by the construction of Art. 221, p. 256.
If the rays of light proceed from a point, the sphere which casts a
shadow will have for its shade line the circle of contact of the envelop-
ing cone whose vertex is at the luminous point. This cone and its
circle of contact w4th the sphere are determined by the construction
of Art. 227, p. 263.
The intersection of the surface of the enveloping cylinder or
PROJECTIOlSr OF SHADOWS
4r
enveloping cone with any given surface will be the cast shadow of the
sphere on that surface.
Two examples on the shadow cast by a sphere are illustrated by
Figs. 778 and 779. The results only are shown, all the construction
lines being omitted. The student should work out these examples to
the dimensions given, which are in inches.
Fig. 778 shows the shadow cast by a sphere partly on the horizontal
plane and partly on a concave cylindrical surface. The part of the
Fig. 778.
Fig. 779.
sphere in shade is also indicated. The rays of light are parallel
to rr'.
Fig. 779 shows the shadow cast by a sphere partly on the horizontal
plane, partly on the vertical plane of projection and partly on the
vertical plane L'MN. The part of the sphere in shade is also indicated.
The rays of light proceed from the point ss'.
340. Shadow of a Solid of Revolution. — If the rays of light
are parallel determine, by Art. 292, p. 327, a cylinder to envelop the
surface of the solid of revolution, the generatrices of the cylinder being
parallel to the rays of light. The trace of this cylinder on the surface
upon which the shadow is to be cast will determine the required cast
shadow.
If the surface of the enveloping cylinder does not intersect the
surface of the solid (it may touch at one part but cut at another) then
the line of contact of the surface of the solid and the enveloping
cylinder will be the line of separation between light and shade on the
surface of the solid. If the surface of the enveloping cylinder also
intersects the surface of the solid, then this intersection must be found
in order to complete the shadow on the solid itself.
2b
418
PRACTICAL GEOMETRY
The solution of an example, which the student should work out
carefully, is shown in Fig. 780. The solid, which is in the form of a
vase, has its axis vertical. The rays of light are parallel to rr\ and the
shadow of the vase is cast on the horizontal plane of projection.
The chief difficulty is with the line of separation on the upper or
concave part of the vase. a!h'dd' and e'/'gltJIi! are the elevations of the
lines of contact of the cylinder or cylinders enveloping the upper part
of the solid. A part of the line CD casts a shadow MG on the neck of
the vase, A part of CD and a part of the solid in the neighbourhood
of H cast a shadow on a part of the solid below H and the outline of
this shadow is the curve HN. The curves MG and HN are obtained
by the construction of Art. 334, p. 411.
If the rays of light proceed from a point the problem may be solved
by the application of the construction of Art. 293, p. 328.
PROJECTION OF SHADOWS
419
Exercises XXVII
1. The projections of a rectangle ABCD and a square MNOP are given in Fig.
781. Determine the shadow cast by the square on the rectangle and the shadows
cast by both figures on the planes of projection. The rays of light are parallel
to rr'.
2. Determine the shadows cast by the triangle ahc a'h'd (Fig. 782) on the
planes of projection. The rays of light proceed from the point ss'.
»' Ex.l. i
^
dC
\
...
c-
H
d
K
y
y
n
Y
^
,
1
f-
c
ft
^
h-
1
n
n
P
\ 1
s'
. Ex.2.
d.
r
/
/
1
/
*Y
a
Jf
c
I
A
1'
J
T—
EX.3.1
>v
r
\
r'
\
1
a
\
1
J
V
J
/'
^
K
/
r,
/
\
1
'
V
]
EX.4.
^i
r
f^>\
^
r
k
/
"1
X
M/
\
1
Y
1
/
^
■^
N^
1
1/
\
Vi
(a
S-
r ^
k
\
h-
\
/
In
V
^
^
y
Fig. 781.
Fig. 782.
Fig. 783.
Fig. 784.
In reproducing the above diagrams take the small sqtiares as of half inch side.
3. The semicircles abc and a'b'c' (Fig. 783) are the plan and elevation re-
spectively of a certain curve. Show the shadows cast by this curve on the planes
of projection when the rays of light are parallel to rr'.
4. The plan ab and elevation a'b' of a horizontal circle are given in Fig. 784.
An oblique plane L'MN is also given. Determine the shadows cast by the circle
on the oblique plane and on the horizontal plane when the rays of light are
parallel to rr'.
5. ab, the plan of a straight line makes 45° with XY. a and b are ^ inch 1^
inches respectively below XY. A and B are 2 inches and 1 inch respectively
above the H.P. The shadow of AB on the V.P. is a horizontal line 3 inches long.
(1) Determine the directions, in plan and elevation, of parallel rays of light which
would cast this shadow. (2) The shadow being cast by light from a luminous
point which is in a plane perpendicular to XY and which contains the point A,
determine the plan and elevation of the luminous point. Show the shadow in
each case.
6. A cube of 1^ inches edge has its base horizontal and 1 inch above the H.P.
Determine the shadow cast by the cube on the H.P. when the rays of light are
parallel to a diagonal of the solid.
7. A solid of the form of the letter T stands on the floor in the angle of two ,
vertical walls as shown in Fig. 785. Determine the shadows cast by the solid on
the floor and on the walls, the rays of light being parallel to rr'.
8. Fig. 786 represents a square pyramid penetrating a square prism. Determine
the shadows cast by these solids on one another and on the planes of projection,
the rays of light being parallel to rr'.
9. Determine the shadow cast by the object shown in Fig. 787 on itself and on
the horizontal plane. The rays of light are parallel to rr'.
10. Determine the shadow cast by the object shown in Fig. 788 on itself and
on the horizontal plane. The rays of light are parallel to rr'.
11. o is the centre of a circle 2 inches in diameter, oab is a straight line
cutting the circle at c. oa = 0*5 inch, o6 = 2 inches. The circle is the plan of a
right circular cylinder standing on the H.P. Height of cylinder, 2 inches, ab is
the plan of a straight line which touches the upper end of the cylinder at G and
has the end B on the H.P. Show in plan and elevation the shadow cast by tha
420
PRACTICAL GEOMETRY
line on the cylinder and on the H.P. The ground line for the elevation makes
75° with ab. The plans of the rays of light make 45° with the ground line and
30° with ab, and the elevations of the rays are perpendicular to their plans.
\
EX.7 1
\
X
Y
y//. ■////.
W
'?W
Y/<
■/,
H
,
V
\
''(,
//
s
\ A EX.8
m
■ \.
7 I
XL 3y
^
^-5^
2w^
^ ! \A\ r
/"[">
^ /N^^-^
^N-r-
-*.
"^
1 1 EX.10 1
v!
\
X
Y
N^
^
•^
^-.
\
r\
^
^^
J
^
-
Fig. 785. Fig. 786. Fig. 787. Fig. 788.
In reproducing the above diagrams take the small squares as of half inch side.
12. The solid shown in Fig. 789 is illuminated by rays of light which are
parallel to rr'. Determine the parts of the surface of the solid which are in
shadow.
13. A solid made up of two cylinders is shown in Fig. 790. Determine the
shadow cast by this solid on the V.P. when the rays of light proceed from the
point ss'. Also, indicate the parts of the surface of the solid which are in shadow.
14. A cone and cylinder are shown in ?'ig. 791. Determine the shadows cast
by these solids on the H.P., also the shadow cast by the cone on the cylinder.
Indicate the parts of the surfaces of the solids which are not illuminated. The
rays of light are parallel to rr'.
15. A right cone, base 3 inches diameter, axis 3*5 inches long, lies on the
ground, one of the generating lines of the cone being the line of contact. The
rays of light being parallel, inclined at 45° to the ground, and their plans inclined
^Z\ *^
uv lyy*
L-^-\
^5^
(^t2j^ ^
I t^
|5^-=/2y ^
^ ^ :^
V
XL
:,, 3S yy
y
----- i
Ex.l2^
■~, ^ ...
S"
t
£ 1
W\ \W'^ }\k
vi ^r' W
^^ X /0(
y ^^q^-^rA:^^^ V
/'^■-5>v
Vx 1/ V
S EX.16 Z ZS
i^zm^i
VZ Z 2
^^2 ^2^
^^r:-?^
Fig. 789. Fig. 790. Fig. 791. Fig. 792.
In reproducing the above diagrams take the small squares as of half inch side.
at 35° to the plan of the axis of the cone, and the base of the cone being in shadow,
determine the shadow cast by the cone on the ground.
16. Fig. 792 shows a hemispherical cup with a cross bar at the top. Show on
the plan the shadow cast on the inside surface of the cup. The rays of light are
parallel to rr'.
17. The curved surface of a bowl is a zone of a sphere. Larger diameter of
zone, 2-75 inches. Smaller diameter, 1*75 inches. Height, 1 inch. The bowl
stands with its base on the ground and is illuminated from a point which is 2-25
inches above the ground and whose plan is 2 inches from the centre of the plan
PROJECTION OF SHADOWS
421
of the bowl. Determine the shadow of the bowl on the ground and the portions
of the bowl in shade, both in plan and elevation. The ground line for the elevation
makes 30° with the line joining the plan of the source of light and the centre of
the plan of the bowl. The thickness of the bowl is to be neglected.
18. a is the centre of a circle 3"25 inches in diameter, b is the centre of a
circle 2 inches in diameter, ab = 1*75 inches, ca is a line making the angle
cab = 10°. The larger circle is the plan of a hemispherical hole in the ground.
The smaller circle is the plan of a sphere resting on the ground. Determine the
plan of the shadows cast in the hole and on the sphere by parallel rays of light.
The plans of the rays of light are parallel to ca and the rays are inclined at 35°
to the ground.
19. Front and side elevations of a corbel projecting from a vertical wall are
given in Pig. 793. Show on the front elevation the shadow cast by the corbel on
the wall, also the part of the corbel in shade. The direction of the parallel rays
of light is given.
^
ff
A
^
^
::^
^
^
A
*^
^
f
\
^
\
k
^
y
J
EX.19
~^
1
/y
^
>
"^
\
K
y
j^
\
<
^
^
^
je
\
}
Is
\
/
\
/
; N
^
?
\
\
.Si. -
EX.20
\
^"
■"
EX.21
*<
V
^
"^
^
^
ijXii
{
^
\
ii
Fig. 798.
Fig. 794.
Fig. 795.
In reproducing the above diagrams take the small squares as of half inch side.
20. Two elevations of a wall bracket and a hanging conical lamp shade are
given in Fig. 794. Show on the left hand elevation the shadow cast by the
bracket and shade on the wall. One of the parallel rays of light is shown.
21. Fig. 795 shows a suspended sphere and a niche in a vertical wall. The
Fig. 796.
Fig. 798.
TJiese diagrams are to be reproduced twice this size.
422
PRACTICAL GEOMETRY
surface of the niche is cylindrical. Determine the shadows cast in the niche
and the part of the sphere in shade. The rays of light are parallel to rr' .
22. The circular arc ABC (Fig. 796) rotates about its chord AG. Draw the
elevation and plan of the figure generated. Determine the shadow cast on
the horizontal plane by parallel rays of light, one of which K is given. Show the
margin of light and shade on the surface of the solid. [b,e.]
23. The elevation and half plan of a solid of revolution are given in Fig. 797,
the axis being vertical. Find the projections of the limits of light and shade on
the solid, and the complete outline of the shadow thrown by the solid on the
horizontal plane. The arrows indicate the direction of the parallel rays of
light. [B.E.]
24. The line LL (Fig. 798) rotates about the axis AA. Draw the plan and
elevation of the figure generated. Determine the shadow cast on the horizontal
plane by parallel rays of light, one of which R is given. [b.e.]
25. An annulus or anchor ring is cut in two by an axial plane. One of the
halves is shown in plan (Fig. 799), with its section ends resting on the horizontal
plane. Draw the elevation of this semi-annulus on XY. Determine also the
shadow cast on the horizontal plane by rays of light, parallel to the vertical
plane, and inclined at 30° to the horizontal plane, one of which is shown in plan
at r. And show the projections of the limits of light and shade on the surface of
the solid. [b.e.]
Fig. 799.
Fig. 800.
26. Two elevations of a pipe bend, with two circular flanges, are shown in
Fig. 800. Reproduce these elevations three times the size given, and determine
the projections of the limits of light and shade on the solid when it is illuminated
by parallel rays of light inclined as shown. Show also on the right hand eleva-
tion the shadow cast by the whole solid on the vertical plane containing the face
of. the vertical flange. To avoid the interference of the left hand elevation with
the shadow cast on the vertical plane the two elevations may be placed further
apart.
27. The plan of an anchor ring resting on the horizontal plane is two con-
centric circles, the larger being 3'1 inches in diameter, and the smaller 1-1 inches
in diameter. The smaller circle is also the plan of a vertical right cylinder
(height 1*6 inches) standing on the horizontal plane. Determine the shadow
cast on the horizontal plane, and the unilluminated portion of the anchor ring,
taking a point as a source of light which is 2-7 inches above the horizontal plane
and whose plan is 2-2 inches from the centre of the plan of the ring.
CHAPTEE XXVIII
MISCELLANEOUS PKOBLEMS IN SOLID GEOMETRY
341. The Regular Dodecahedron. — The regular dodecahedron
is one of the five regular solids and has twelve faces all equal and
regular pentagons. This solid is shown in its simplest position, in
relation to the planes of projection, in Fig. 801. One face ABODE is
on the horizontal plane and AB,
one edge of that face, is at right
angles to XY. Let o be the centre
of the circumscribing circle of the
regular pentagon abcde. The sides
of the pentagon ahcde are the
horizontal traces of five of the
faces of the dodecahedron, and
these five faces meet in straight
lines whose plans pass through the
angular points a, h, c, cZ, and e,
and if produced these plans pass
through o.
Consider the two faces whose
horizontal traces are ah and fee;
these faces meet in a line whose
plan gh, when produced, passes
through 0. To find the point g,
imagine the face ABG to revolve
about AB in the clockwise direction
until it is in the horizontal plane.
It will then evidently coincide with
the pentagon ahcde and the plan
of the point G will travel in the
line gc perpendicular to ah Hence
a line through c perpendicular to
ah to meet a radial line ohg at g
determines the point g.
The plan may now be completed as follows. With centre o and
radius og describe a circle. Divide the circumference of this circle into
ten equal parts at the points /, g, h, etc. Draw the circumscribing
circle of the pentagon ahcde. Draw the radial lines fm, hn, etc. and
join the various points thus found, as shown.
Fig. 801.
424
PRACTICAL GEOMETRY
To draw the elevation : with centre a' and radius a'd! describe the
arc d'f to cut the projector from / at f. Join a'f. The line a'f is the
elevation of the face whose horizontal trace is ah. A projector from g
to meet a'f at g' determines the elevation of the point G. Points
whose plans are the alternate angular points of the outer polygon,
beginning with g, have their elevations at the same level as g', and the
points whose plans are the remaining angular points of this outer poly-
gon have their elevations at the same level as /'. Points such as M
and N" on the top face are at a height above F equal to the height of G
above the horizontal plane.
From the plan and elevation thus determined other projections may
be drawn in the usual way. For example, if a ground line be taken
perpendicular to m'd\ the elevation of one of the axes of the solid, a
plan of the solid, when that axis is vertical, may be projected from the
elevation already drawn. But the student is recommended to try and
draw directly the plan and an elevation of the dodecahedron when its axis
is vertical, without first drawing it with one face on the horizontal plane.
When an axis of the solid is vertical it should be noticed that the
planes of the three faces
meeting at an extremity of
that axis will be equally
inclined to the ground, and
being equally inclined to
one another, their lines of
intersection will in plan
make 1 20° with one another.
As an exercise in draw-
ing the projections of the
dodecahedron the edges of
the solid may be conveni-
ently taken, say, 1'25 inches
long.
342. The Regular
Icosahedron. — The regu-
lar icosahedron is another
one of the five regular solids
and has twenty faces all
equal equilateral triangles.
Fig. 802 shows a regular
icosahedron, in plan and
elevation, when one face
ABC is on the horizontal
plane and AB one edge of
that face is perpendicular
to XY. The plan of the
top face, which is also hori-
zontal, is an equilateral triangle inscribed in the same circle, centre o,
as the triangle ahc, the sides of the one being parallel to the sides of
the other, as shown.
Fig. 802.
MISCELLANEOUS PROBLEMS IN SOLID GEOMETRY 425
A property of the icosahedron, which leads to simple constructions
for drawing it, is, that any angular point of the solid is the vertex of
a right pentagonal pyramid, the five faces meeting at the vertex being
faces of the icosahedron. Thus the faces meeting at the point F are
the faces of a right pyramid whose base is the regular pentagon
ABEDE, and the sides of the pentagon are edges of the icosahedron.
To complete the plan proceed as follows. Draw on ah as base the
regular pentagon ah^J)^%. Through Ej draw E^e perpendicular to
ah to meet the radial line oae at e. With centre o and radius oe
describe a circle! A regular hexagon inscribed in this circle, one
angular point being at e, is the boundary line of the plan, which is
completed by joining the angular points of the hexagon to the angular
points of the two central equilateral triangles, as shown. The pentagon
at El Dj El is the rabatment, on the horizontal plane, of the pentagon
of which ahede is the plan. Keeping this in mind the construction of
the elevation easily follows.
The student should also draw directly a plan and an elevation of
the icosahedron when an axis of the solid is vertical. The extremities
of the axis in question will be the vertices of two of the right pen-
tagonal pyramids previously mentioned. The pentagonal bases of
these pyramids will be horizontal and their plans will be inscribed in
the same circle.
As an exercise in drawing the projections of the icosahedron the
edges may be conveniently taken, say, 1*75 inches long.
343. Solids Inscribed in the Sphere. — A solid is said to be
inscribed in a sphere when all its angular points are on the surface of
the sphere.
Fig. 803 shows the constructions
for finding the length of an edge of
each of the five regular solids when
inscribed in a sphere of given diameter.
AB is the diameter of the sphere.
On this a semicircle is described.
Take BD equal to one-third of AB.
Draw DE at right angles to AB to
meet the semicircle at E. Join AE
and BE. AE is the edge of the in-
scribed tetrahedron, and BE is the
edge of the inscribed cube.
C being the middle point of AB,
draw CF at right angles to AB to
meet the semicircle at F. Join AF.
octahedron.
Draw AG at right angles to AB and make AG equal to AB.
Draw CG cutting the semicircle at H. Join AH. AH is the edge of
the inscribed icosahedron.
On EA make EK equal to half of BE. Join BK. With centre K
and radius KE describe an arc to cut BK at L. BL is the edge of the
inscribed dodecahedron.
Fig. 803.
AF is the edge of the inscribed
I.
426 PRACTICAL GEOMETRY
The following facts should be kept in view when working problems
on the regular solids inscribed in the sphere.
The plane containing one edge of a tetrahedron and the centre of
the circumscribing sphere bisects the opposite edge at right angles.
The rectangle which has for its diagonals two of the diagonals of a
cube is inscribed in a great circle of the circumscribing sphere.
The octahedron can be divided into two square pyramids in three
different ways, and the square bases of all these pyramids will be
inscribed in great circles of the circumscribing sphere.
344. Solids Circumscribing the Sphere. — A solid is said to
circumscribe a sphere, or a sphere to be inscribed in a solid, when all
the faces of the solid are tangential to the surface of the sphere.
A plane bisecting the angle between any two faces of the solid
passes through the centre of the inscribed sphere.
In the case of any one of the five regular solids the centres of the
inscribed and circumscribing spheres coincide.
345. The Sphere, Cylinder and Cone in Contact. — If two
spheres touch one another, the point of contact and the centres of
the spheres are in the same straight line.
If a sphere touches a cylinder or cone it will do so at a point, and if
the cylinder or cone be a right circular cylinder or a right circular
cone, the centre of the sphere the point of contact, and the axis of the
cylinder or cone are in the same plane.
If two cylinders or two cones or a cylinder and a cone touch one
another, they will do so either along a straight line or at a point. If
the cylinders or cones are right circular cylinders or right circular
cones and they touch one another along a straight line, the line of
contact and the axes of the surfaces are in the same plane ; the axes
will therefore either be parallel or they will intersect.
When two right circular cylinders touch one another at a point
the common perpendicular to their axes passes through the point of
contact.
Two surfaces which touch one another have a common tangent
plane.
346. Projections of Four Spheres in Mutual Contact. —
Denote the spheres and their centres by A, B, C, and D, and let their
radii be r^, r^j rg, and r^ respectively. Assume that the spheres A, B,
and C are resting on the horizontal plane and that the line joining the
centres A and B is parallel to the vertical plane of projection. The
elevations of the spheres A and B (Fig. 804) are circles touching one
another and XY and after these are drawn the plans may be pro-
jected from them as shown. Next determine the horizontal distances
between the centres A and C, and B and C as shown on the elevation.
This determines the centre cc' and the plan and elevation of the sphere
C may then be drawn.
The centre D of the fourth sphere will be at a distance r^ + r^ from the
centre A of the first sphere, and at a distance ra + r^ from the centre B
of the second sphere. If the triangle ADB be conceived to rotate
about the side AB, the point D will describe a circle which will be
MISCELLANEOUS PROBLEMS IN SOLID GEOMETRY 427
the locus of the centre of a sphere of radius r^ which touches the spheres
A and B.
Remembering that ah, a'V is parallel to the vertical plane of pro-
jection, with centres a' and V and radii equal to r^ + r^ and r^, + r^^
respectively, describe arcs intersecting at d^. Draw d/e' to intersect a'h'
at right angles at e'. Then ^d{ is the radius of the circle described by
the point D when the triangle ADB is rotated about AB. The plan of
this circle is an ellipse whose semi-minor axis is ed^, the projection
Fig. 804.
of e'd^ on ah. The semi-major axis ef is at right angles to ah and equal
to e'd-l. A quarter of this ellipse is shown.
Again, the centre D of the fourth sphere will be at a distance
fi + r^ from the centre A of the first sphere, and at a distance r^ + ^4 from
the centre C of the third sphere. If the triangle ADC be conceived to
rotate about the side AC, the point D will describe a circle which will
be the locus of the centre of a sphere of radius r^ which touches the
spheres A and C. The circle whose centre is c^ is the elevation of the
sphere C when that sphere is brought round into the position in which
it touches the sphere A and the horizontal plane and has its centre in
a plane parallel to the vertical plane of projection and containing the
centre A of the first sphere. With centres a' and c^ and radii r^ -f r^
and r^ -f r^ respectively, describe arcs intersecting at d.2. Draw d.20.2 to
intersect aJc^ at right angles at O2'. Then Oa'^a' is the radius of the circle
428 PRACTICAL GEOMETRY
described by the point D when the triangle ADC is rotated about AC.
The plan of this circle is an ellipse whose semi-minor axis on is on ac
and is obtained as shown. The semi-major axis oh is at right angles to
ac and equal to o.Jd^. A quarter of this ellipse is shown.
The two circles which have been referred to as described by the
point D intersect at two points one of which has the point d for its
plan, d being on the two ellipses which are the plans of the circles.
The elevation d' is in e'd^' and in the projector from d. This determines
the centre of the fourth sphere and the plan and elevation of that
sphere may now be drawn.
The two circles referred to as described by the point D are on the
surface of a sphere whose centre is A and radius r^ -f- r^. The ellipses
which are the plans of these circles may intersect at four points, but
only two of these are plans of the points of intersection of the circles.
The student should have no difficulty in deciding on which of the points
of intersection of the ellipses are to be taken.
347. Spherical Roulettes. — If two cones be placed in line
contact with their vertices coinciding and if one cone be made to roll
on the other, which is fixed, any point carried by the rolling cone will
describe a spherical roulette. The describing point carried by the roll-
ing cone may be outside, or inside, or on the surface of that cone and
it will be at a constant distance from the common vertex of the two
cones ; hence the describing point will move on the surface of a fixed
sphere.
When the two cones are right circular cones, and the describing
point is on the surface of the rolling cone, the spherical roulette
becomes a spherical epicycloid or a spherical hypocycloid according as the
rolling cone rolls outside or inside the fixed cone.
Spherical roulettes are tortuous curves, and no single projection of
a tortuous curve can show its true form. A spherical roulette has
therefore to be represented by two projections.
Projections of a spherical epicycloid and a spherical hypocycloid are
shown in Fig. 805. The description which follows applies to either
curve, v'a'a' is the elevation of the fixed cone the axis of which is
vertical, and the semicircle aba is the half plan of this cone, v'a'e' is
the elevation of the rolling cone when its position is such that its axis
is parallel to the vertical plane of projection.
Let the rolling cone start from the position in which vb, v'V is the
line of contact and Iqt the point of contact hV of the bases of the cones
in this position be the initial position of the describing point. Next
suppose that the rolling cone rolls into the position in which the line of
contact has vm for its plan. Draw a circle mnr^ having a radius equal
to the radius of the base of the rolling cone and touching the plan of
the fixed cone at m. Consider this circle to be the rabatment of the
base of the rolling cone on the plane of the base of the fixed cone when
the rolling cone is in the position now being considered. The rabat-
ment of the describing point will be rj, the position of r^ being such
that the arc mnr^ is equal to the arc mh. Restoring the base of the
rolling cone to its inclined position the plan r of the describing point
MISCELLANEOUS PROBLEMS IN SOLID GEOMETRY 429
will be in the line through r^ parallel to mv. To fix the position of r
in the line r^r let the circle mnr^ be carried round about the centre v into
the position asiC^, the diameter ae.^ being in line with va and parallel to
the ground line, r^ will move to s^. Draw the projector s^s/ to meet
the ground line at s/. With centre a' and radius a's/ describe the arc
SjV to meet a'e' at s'. Draw the projector s's to meet s^s parallel to aci
at 8. With centre v and radius vs describe the arc sr to cut r^r at r.
i SPHERICAL
EPICYCLOID
SPHERICAL
HYPOCYCLOID
Fig. 805.
Draw the projector rr' to meet the horizontal line s'r' at r'. Then rr' is
a point on the spherical epicycloid or on the spherical hypocycloid. In
like manner any number of points maybe found. The length of curve
shown is that due to half a revolution of the rolling cone.
The points such as r^ lie on a curve, shown dotted, which is either
a plane epicycloid or a plane hypocycloid.
The tangent to the spherical epicycloid or spherical hypocycloid at
a point R lies on the tangent plane to the sphere whose centre is V
and radius YA or YR at the point R. It also lies on the tangent
plane to the sphere whose centre is M and radius MR at the point
430
PRACTICAL GEOMETRY
R. The tangent is therefore the line of intersection of these two
planes.
The spherical epicycloid and the spherical hypocycloid occur in
connection with the formation of the teeth of bevel wheels.
348. Reflections. — When a ray of light impinges on a polished
surface at a point Q, it is reflected so that the incident and reflected
rays and the normal to the reflecting surface at Q are in the same
plane ; also the incident and reflected rays are equally inclined to the
normal.
Referring to Figs. 806 and 807, let the plane of the paper be the
plane containing the incident ray PQ and the normal QT to the surface
whose section by the plane of the paper is AB, Then the reflected
ray QR will be in the plane of the paper and will make with QT an
angle 6 equal to the angle between PQ and QT.
Fig. 806.
Fig. 803.
If the reflecting surface is a plane (Figs. 807 and 808) it is obvious
that all incident rays from a point P will be reflected from the plane
as if they came direct from a point P^ on the other side of the plane and
in the normal PN to the plane, P^N being equal to PN. A knowledge
of this fundamental theorem makes the solution of problems on
reflections from plane mirrors quite simple. The point Pj is called the
image of the point P in the mirror AB. It should be noticed that
although the image of P is in PN produced it is not necessary that the
mirror should extend to the point N in order that the image may be
seen from the point R.
Fig. 809 shows how to trace the path of a ray of light which
passes from a fixed point P and is reflected in turn from a number of
plane mirrors AB, BC, CD, and DE and then passes through a fixed
point S, the mirrors being at right angles to the plane of the paper.
Pi is the image of P in AB, Pg is the image of Pi in BC, P3 is the
image of Pg in CD, and P4 is the image of P3 in DE. The ray which
is reflected from DE and passes through S must evidently be in the
line P4S. The incident ray on DE is the reflected ray from CD and
must therefore be in the straight line from Pg, and so on the ray is
traced back to the point P as shown.
Fig. 810 illustrates the general case of the problem : given the
incident ray and a plane from which it is reflected, to determine the
reflected ray. PQ is the given incident ray and H.T. and V.T. are the
MISCELLANEOUS PEOBLEMS IN SOLID GEOMETRY 431
traces of the given plane. Find Q the point of intersection of PQ and
the plane. From a point P in PQ draw the normal PNPj to the plane
and find N the point of intersection of this normal and the plane.
Make NP^ equal to NP. Join PiQ and produce it. QR the produced
part of this line is the reflected ray required.
p^r:
P3^-
Fig. 809.
Fig. 810.
If the given surface from which a given incident ray is Reflected is
a curved surface, the point of intersection of the incident ray and the
surface must be determined by the method for finding the intersection
of a straight line and the particular curved surface given. The normal
to the surface at the point where the incident ray strikes it must then
be found and the plane containing it and the incident ray determined.
It will then generally be necessary to obtain a projection of the
incident ray and the normal on a plane parallel to their plane, or to
obtain a rabatment of the incident ray and normal into one of the
planes of projection, in order to draw the reflected ray which must
make with the normal an angle equal to the angle between the incident
ray and the normal.
ELEVATION
-''
o
^,
-'
Of
UECT
PLA
IMAG
E
Fig. 811.
Fig. 812.
An example on the projection of an object and its image in a plane
mirror is illustrated by Figs. 811 and 812. The object and its image
and the mirror are shown in plan and elevation to the left in Fig. 811,
432 PRACTICAL GEOMETRY
the mirror being at right angles to the planes of projection. The same
Fig. shows to the right an elevation on a second ground line X^Y^
inclined to the first ground line XY. The pictorial projection in Fig.
812 illustrates further the relative positions of the object, mirror, and
Exercises XXVIII
1. Draw a circle, centre o and radius 1-75 inches. Take a point a 1 inch from
0. Draw a straight line ax making an angle of 60^ with oa. The circle is the
plan of a sphere and a is the plan of a point A on its upper surface. A is one
corner of a cube inscribed in the sphere and ax is the direction of the plan of
one edge containing the point A. Complete the plan of the cube.
2. Referring to the preceding exercise, make oa = 15 inches and the angle
oax = 60°. a is the plan of one angular point of a tetrahedron inscribed in the
sphere and ax is the direction of the plan of one edge containing the point A.
Complete the plan of the tetrahedron.
3. Same as exercise 2, except that the solid is an octahedron instead of a
tetrahedron.
4. The base of a pyramid is an equilateral triangle ABC of 2 5 inches side.
The vertex V of the pyramid is in a line through A at right angles to the plane
of ABC, and VA is 2*5 inches long. Placing the pyramid with its base on the
ground and AB parallel to XY, draw the projections of the Id scribed and circum-
scribing spheres.
5. A right circular cone, base 2*5 inches in diameter and altitude 2*5 inches,
stands with its base on the ground. A cylinder 2 inches in diameter lies on the
ground in contract with the cone, the axis of the cylinder being horizontal. A
sphere 1-25 inches in diameter rests on the ground in contact with the cone and
cylinder. Draw a plan of the group and an elevation on a vertical plane parallel
to the axis of the cylinder showing the projections of the points of contact of the
solids.
6. Three spheres, two of them 1-5 inches in diameter, and one 2 inches in
diameter, rest on the horizontal plane, each in contact with the other two. A
cylinder 2 inches in diameter rests, with its axis horizontal, on top of the spheres,
touching each of them. Draw the plan of the group.
7. Two straight lines ab and cd bisect one another at" right angles, ab = 4:
inches and cd = 3-5 inches. These lines are the plans of the axes of two cylinders
whose diameters are equal and which touch one another. The heights of the
points A, B, C, and D above the horizontal plane are 0-7 inch, 2*8 inches, 2*4
inches, and 4*5 inches respectively. Draw the plan of the cylinders and an
elevation on a ground line parallel to ab, showing the point of contact.
8. abc is a triangle, ab = 2 inches. Angle bac = angle abc = 30°. a, is the
plan of the centre of a sphere of 1*3 inches radius which rests on the horizontal
plane, b is the plan of the centre of a second sphere which rests on the horizontal
plane and touches the first sphere, c is the plan of a point on the upper surface
of the first sphere. Draw the plan of a third sphere which touches the other two,
c being its point of contact with the first sphere.
9. A cone of revolution, vertical angle 64°, of indefinite length, lies with its
curved surface on the ground ; draw ite plan. Determine a sphere of 0*6 inch
radius, which rests on the ground and touches the cone at a point 2-5 inches
from its vertex. Show the indexed plan of the point of contact, and determine
the common tangent plane at this point. [b.e.]
10. A right circular cone, base 2-5 inches in diameter and axis 3 inches long,
lies with its slant side on the ground. A sphere, 2 inches in diameter, moves in
contact with the ground and the surface of the cone. Draw the plan of the locus
of the point of contact.
11. vab is a triangle, va = 3*4 inches, vb = 2'4 inches, and ab = 2 inches, a
is the plan of the centre of a sphere of 2-8 inches diameter, A being 25 inches
MISCELLANEOUS PROBLEMS IN SOLID GEOMETRY 433
above the ground, h is the plan of the centre of a sphere of 1-75 inches diameter,
B being 1*5 inches above the ground, v is the plan of the vertex of a cone whose
semi-vertical angle is 20" and w^hich touches the two spheres, V being 2 inches
above the ground. Draw the plan of the group and an elevation on a ground line
parallel to av, showing the points of contact of the cone and spheres.
12. Same as preceding exercise except that a cylinder 1-5 inches in diameter
is to take the place of the cone, V being a point on the axis of the cylinder.
13. A cone, in elevation v'h'k' (Fig. 813) lies on the horizontal plane, its axis
VC parallel to the vertical plane. A cylinder of 1 inch radius, its axis parallel to
the line de, d'e', touches the cone at a point whose height is represented by the
horizontal g'h' . Draw the plan of the surfaces, and the projections of their point
of contact, showing the trace of the cylinder on the horizontal plane, and the
projections of its generator in contact with the cone. [b.e.]
60
50
^0°
dto
Ob -0-7"
^0\ .
Fig. 813.
-30
Fig. 815.
-.J
22
14. The straight line AB (Fig. 814) is tangential to a sphere of 1 inch radius
whose centre is on the straight line CD. Draw the figured plan of the lines and
sphere and show the point of contact of the sphere and the line AB. Unit for
indices O'l inch.
15. Two lines AB and CD are given in Fig. 815 by their figured plans, and a
plane is given by its scale of slope. Determine a sphere of 1*25 inches radius
touching the lines and plane. Unit 0*1 inch. The sphere is intended to be below
the plane. [b.e.]
16. v'a'o' (Fig. 816) is the half elevation of a fixed right circular cone (1 ) whose
axis VO is vertical and whose vertex is
V. v'a'e' is the elevation of a second
right circular cone (2) in line contact
with cone (1) and having its vertex at
V. The cone (2) rolls on the cone (1)
and a point on the circumference of
the base of the rolling cone describes
a spherical epicycloid of which a'c'h'
(above a'o') is the elevation, or a spheri-
cal hypocycloid of which a'c'b' (below
a'o') is the elevation. Draw the plan
and elevation of these spherical rou-
lettes.
Next take a right circular cone (3)
which has the base of the cone (1) for
a circular section and u' for the eleva-
tion of its vertex, the angle v'a'u' being
a right angle. Take the spherical
roulette, of which a'c'h' is the eleva-
tion, to be the directing curve of a
cone (4) with its vertex at V. Deter-
mine the curve of intersection (5) of the cones (3) and (4). Develop the surface
of the cone (3) with the curve (5) on it. a'CB is a sketch of this development.
2 F
Fig. 816.
434
PRACTICAL GEOMETRY
Lastly, draw on the circular arc a'B as base a plane epicycloid and a plane
hypocycloid with a rolling circle 1-8 inches in diameter (the diameter of the base
of the rolling cone) for comparison with the curves a'CB on opposite sides of
the arc a'B.
17. Taking the fixed and rolling cones (1) and (2) as given in the preceding
exercise and Fig. 816, draw the plan and elevation of the spherical epitrochoid
described by a point on a diameter of the base of the rolling cone produced, the
describing point to be 1-4 inches from the centre of the base of the rolling cone.
18. The ray of light rr' (Fig. 817) impinges on the horizontal plane and is
reflected on to the vertical plane of projection from which it is again reflected.
Show the path of the ray in plan and elevation.
19. A ray of light rr' (Fig. 818) impinges on the horizontal plane and is
reflected on to the vertical plane HTV from which it is again reflected. Draw
the plan and elevation of the path of the ray. Show also in plan and elevation the
path of a ray which after reflection from the horizontal plane and the plane HVT
passes through t-he point nn', the incident ray being parallel to rr'.
\
•2- SIT
\-W . :=
:ni 53n
^7
-' \ ^
^
^A% t-,<-
>,
J5
^i ^v
AI. ^Z _
/ \
?c
ISIy is- !S
_ii y X
. Sy «JL _
-L Jy
^
„ Ti
""r^^-
3 ^x
-«• -s-.^t
h>.
. V /
^^: /A\x \\\
/^
t -a-
y \h , r>j I fe»^i
^ A
^^mr .J,
Ml \A\\
I J
Fig. 817.
Fig. 818.
Fig. 819.
Fig. 820.
20. A polished right circular cylinder (Fig. 819) has its axis vertical. The
ray of light rr' strikes this cylinder and is reflected on to the vertical plane of
projection. Draw the plan and elevation of the path of the ray.
21. The surface of a shield (Fig. 820) is spherical. A projectile strikes this
shield and is deflected, rr' is the line of flight before impact. Show the line of
flight after impact, assuming that the projectile is deflected like a ray of light.
22. A right circular cone, diameter of base 2 inches, altitude 2 inches, stands
with its base on the ground. A ray of light is parallel to the ground line and
0"75 inch above the ground, and its plan is at a perpendicular distance of 0-25
inch from the centre of the plan of the cone. This ray of light impinges on the
surface of the cone and is reflected. Draw the plan and elevation of the reflected
ray.
23. An object and a mirror are given in Fig. 821. Draw an elevation of the
Fig. 821.
Fig. 822.
I
MISCELLANEOUS PROBLEMS IN SOLID GEOMETRY 435
object and its image in the mirror on a ground line inclined at 60° to XY.
Assume that the mirror is large enough to show the
whole of the image in the elevation asked for.
24. Fig. 822 represents a hanging lamp shade and a
tilted mirror. Draw the two elevations and the plan of
the image of the conical shade in the mirror. From the
plan project on a ground line inclined at 70° to XY the
elevation of the mirror, the shade, and as much of the
image as would be seen within the boundary of
the mirror. [b.e.]
25. The point cc' (Fig. 823) is the centre of a polished
sphere of 1 inch radius. A ray of light parallel to rr'
impinges on the sphere and is reflected, the reflected ray
passing through the point 2>p'. Draw the projections of
the path of the ray.
Fig. 823.
i
APPENDIX
MATHEMATICAL TABLES
The Mathematical Tables on the following five pages are those
supplied to Candidates at the Examinations of the Board of Education,
and are reprinted by permission of the Controller of H.M. Stationery
Office.
MATHEMATICAL TABLES
437
Angle. 1
Chord.
Sine.
Tangent.
Co-tangent.
Cosine.
De-
grees.
Radians.
0°
00
1
1^414
r5708
90"
89
88
87
86
1
2
3
4
•0175
•0349
•0524
•0698
■017
•035
•052
•070
•0175
•0349
•0523
•0698
•0175
•0349
•0524
•0699
57^2900
28^6363
19-0811
14-3007
•9998
-8994
-9986
-9976
r402
1^389
r377
1364
r5533
1-5359
1-5184
1-5010
5
•0873
•087
•0872
•0875
11^4301
•9962
1-351
1-4835
85
6
7
8
9
•1047
•1222
•1396
•1571
•105
•122
•140
•157
•1045
•1219
•1392
•1564
•1051
•1228
•1405
•1684
9^5144
8-1443
7-1154
6-3138
•9945
•9925
•9903
•9877
1338
1-325
1-312
1-299
1-4661
1-4486
1-4312
1-4137
81
83
82
81
10
•1745
•174
•1736
•1763
5-6713
•9848
1-286
1-3963
80
11
12
13
14
•1920
•2094
•2269
•2443
•192
•209
•226
•244
•1908
•2079
•2250
•2419
•1944
•2126
•2309
•2493
5-1446
4-7046
4-3315
4-0108
•9816
•9781
•9744
•9703
1-272
1-259
1-245
1-231
1-3788
1-3614
1-3439
1-3265
79
78
77
76
15
•2618
•261
•2588
•2679
3-7321
•9659
1-218
1-3090
75
16
17
18
19
•2793
•2967
•3142
■3316
•278
•296
•313
•330
•2756
•2924
•3090
•3256
•2867
•3057
•3249
•3443
34874
3-2709
3-0777
2-9042
•9613
•9563
•9511
•9455
1-204
1-190
1-176
1-161
1-2915
1-2741
1-2566
1-2392
74
73
72
71
20
•3491
•347
•3420
•3640
2-7475
•9397
1-147
1-2217
70
21
22
23
24
•3665
•3840
•4014
•4189
•364
•382
•399
•416
•3584
•3746
•3907
•4067
•3839
•4040
•4245
•4452
2-6051
2-4751
2-3559
2-2460
•9336
-9272
-9205
•9135
1-133
1-118
1-104
1-089
1-2043
1-1868
1-1694
1-1619
69
68
67
66
25
•4363
•433
•4226
•4663
2-1445
•9063
1-075
1-1345
65
26
27
28
29
•4538
•4712
•4887
•5061
•450
•467
•484
•501
•4384
•4540
•4695
•4848
•4877
•5095
•5317
•5543
2-0503
1-9626
1-8807
1-8040
•8988
•8910
•8829
■8746
1-060
1^045
1^030
1^015
1-1170
1-0996
1-0821
1-0647
64
63
62
61
30
•5236
•518
•5000
•5774
1-7321
•8660
1-000
1-0472
60
31
32
33
34
•5411
■5585
■5760
•5934
•534
•551
•668
■585
•5150
•5299
•5446
•6592
•6009
■6249
•6494
■6745
1-6643
1-6003
1-5399
1-4826
-8572
-8480
•8387
•8290
•985
•970
-954
-939
1-0297
1-0123
-9948
•9774
59
58
57
56
35
•6109
■601
■5736
•7002
1-4281
•8192
-923
-9599
55
36
37
3S
39
•6283
•6458
•6632
•6807
•618
•635
■651
•668
•5878
•6018
•6157
•6293
•7265
•7536
•7813
•8098
1-3764
1-3270
1-2799
1-2349
•8090
-7986
-7880
-7771
-908
-892
-877
-861
•9425
•9250
•9076
•8901
54
53
52
51
40
•6981
•684
•6428
•8391
1-1918
-7660
-845
•8727
50
49
48
47
46
45°
41
42
43
44
•7156
•7330
•7505
•7679
•700
•717
•733
•749
•6561
•6691
•6820
•6947
•8693
•9004
•9325
•9657
1-1504
1-1106
1-0724
1-0355
-7547
-7431
-7314
-7193
•829
•813
•797
•781
•8552
•8378
•8203
•8029
45°
•7854
•765 i '7071
1^0000
1-0000
•7071
•765
•7854
Cosine
Co-tangent
Tangent
Sine
Chord
Radians
Degrees
Angle 1
438
MATHEMATICAL TABLES
Logarithms
1
2
3
4
5
6
7
8
9
12 3 4
5
6 7 8 9
10
0000
0043
0086
0128
0170
0212
0253
0294
0334
0374
4 9 13 17
4 8 12 16
21
20
26 30 34 38
24 28 32 37
11
0414
0453
04S2
0531
0569
0607
0645
0682
0719
0755
4 8 12 15
4 7 11 15
19
19
23 27 31 35
22 26 30 33
12
0792
0828
0864
0899
0934
0969
1004
1038
1072
1106
3 7 11 14
3 7 10 14
18
17
21 25 28 32
20 24 27 31
13
14
1139
1173
1206
1239
1271
1303
1335
1367
1399
1430
3 7 10 13
3 7 10 12
16
16
20 23 26 30
19 22 25 29
1461
1492
1523
1553
1584
1614
1644
1673
1703
1732
3 6 9 12
3 6 9 12
15
15
18 21 24 28
17 20 23 26
15
1761
1790
1818
1847
1875
1903
1931
1959
1987
2014
3 6 9 11
3 5 8 11
14
14
17 20 23 26
16 19 22 25
16
2041
2068
2095
2122
2148
2175
2201
2227
2253
2279
3 5 8 11
3 5 8 10
14
13
16 19 22 24
15 18 21 23
17
2304
2330
2355
2380
2405
2430
2455
2480
2504
2529
3 5 8 10
2 5 7 10
13
12
15 18 20 23
15 17 19 22
18
2553
.2577
2601
2625
2648
2672
2695
2718
2742
2765
2 5 7 9
2 5 7 9
12
11
14 16 19 21
14 16 18 21
19
2788
2810
2833
2856
2878
2900
2923
2945
2967
2989
2 4 7 9
2 4 6 8
11
11
13 16 18 20
13 15 17 19
20
3010
3032
3054
3075
3096
3118
3139
3160
3181
3201
2 4 6 8
11
13 15 17 19
21
22
23
24
3222
3424
3617
3802
3243
3444
3636
3820
3263
3464
3655
3838
3284
3483
3674
3856
3304
3502
3692
3874
33^4
3522
3711
3892
3345
3541
3729
3909
3365
3560
3747
3927
3385
3579
3766
3945
3404
3598
3784
3962
2 4 6 8
2 4 6 8
2 4 6 7
2 4 5 7
10
10
9
9
12 14 16 18
12 14 15 17
11 13 15 17
11 12 14 16
25
3979
3997
4014
4031
4048
4065
4082
4099
4116
4133
2 3 5 7
9
10 12 14 15
26
27
28
29
4150
4314
4472
4624
4166
4330
4487
4639
4183
4346
4502
4654
4200
4362
4518
4669
4216
4378
4533
4683
4232
4393
4548
4698
4249
4409
4564
4713
4265
4425
4579
4728
4281
4440
4594
4742
4298
4456
4609
4757
2 3 5 7
2 3 5 6
2 3 5 6
13 4 6
8
8
?
7
10 11 13 15
9 11 13 14
9 11 12 14
9 10 12 13
30
4771
4786
4800
4814
4829
4843
4857
4871
4886
4900
13 4 6
9 10 11 13
31
32
33
34
4914
5051
5185
5315
4928
5065
5198
5328
4942
5079
5211
5340
4955
5092
5224
5353
4969
5105
5237
5366
4983
5119
5250
5378
4997
5132
5263
5391
5011
5145
5276
5403
5024
5159
5289
5416
5038
5172
5302
5428
13 4 6
13 4 5
13 4 5
13 4 5
7
7
6
6
8 10 11 12
8 9 11 12
8 9 10 12
8 9 10 11
35
5441
5453
5465
5478
5490
5502
5514
5527
5539
5551
12 4 5
6
7 9 10 11
36
37
38
39
40
5563
5682
5798
5911
5575
5694
5809
5922
5587
5705
5821
5933
5599
5717
5832
5944
5611
5729
5843
5955
5623
5740
5855
5966
5635
5752
5866
5977
5647
5763
5877
5988
5658
5775
5888
5999
5670
5786
5899
6010
12 4 5
12 3 5
12 3 5
12 3 4
6
6
6
5
7 8 10 11
7 8 9 10
7 8 9 10
7 8 9 10
6021
6031
6042
6053
6064
6075
6085
6096
6107
6117
12 3 4
5
6 8 9 10
41
42
43
44
6128
6232
6335
6435
6138
6243
6345
6444
6149
6253
6355
6454
6160
6263
6365
6464
6170
6274
6375
6474
6180
6284
6385
6484
6191
6294
6395
6493
6201
6304
6405
6503
6212
6314
6415
6513
6222
6325
6425
6522
12 3 4
12 3 4
12 3 4
12 3 4
5
5
5
5
6 7 8 9
6 7 8 9
6 7 8 9
6 7 8 9
45
46
47
48
49
6532
6542
6551
6561
6571
6580
6590
6599
6609
6618
12 3 4
5
6 7 8 9
6628
6721
6812
6902
6637
6730
6821
6911
6646
6739
6830
6920
6656
6749
6839
6928
6665
6758
6848
6937
6675
6767
6857
6946
6684
6776
6866
6955
6693
6785
6875
6964
6702
6794
6884
6972
6712
6803
6893
6981
12 3 4
12 3 4
12 3 4
12 3 4
5
5
4
4
6 7 7 8
6 6 7 8
5 6 7 8
5 6 7 8
50
6990
6998
7007
7016
7024
7033
7042
7050
7059
7067
12 3 3
4
5 6 7 8
n^^^i^
MATHEMATICAL TABLES
Logarithms
439
1
2
3
4
5
6
7
8
9
1
2
3 4
5
6 7 8 9
51
52
53
54
7076
7160
7243
7324
7084
7168
7251
7332
7093
7177
7259
7340
7101
7185
7267
7348
7110
7193
7275
7356
7118
7202
7284
7364
7126
7210
7292
7372
7135
7218
7300
7380
7143
7226
7308
7388
7152
7235
7316
7396
2
2
2
2
3
2
2
2
3
3
3
3
4
4
4
4
5
5
5
6 7 8
6 7 7
6 6 7
6 6 7
55
7404
7412
7419
7427
7435
7443
7451
7459
7466
7474
2
2
3
4
5
5 6 7
56
57
58
59
7482
7559
7634
7709
7490
7566
7642
7716
7497
7574
7649
7723
7505
7582
7657
7731
7513
7589
7664
7738
7520
7597
7672
7745
7528
7604
7679
7752
7536
7612
7686
7760
7543
7619
7694
7767
7551
7627
7701
7774
2
2
2
2
2
2
3
3
3
3
4
4
4
4
5
5
5 6 7
5 6 7
5 6 7
5 6 7
60
7782
7789
7796
7803
7810
7818
7825
7832
7839
7846
2
3
4
5 6 6
61
62
63
64
7853
7924
7993
8062
7860
7931
8000
8069
7868
7938
8007
8075
7875
7945
8014
8082
7882
7952
8021
8089
7889
7959
8i)28
8096
7896
7966
8035
8102
7903
7973
8041
8109
7910
7980
8048
8116
7917
7987
8055
8122
2
2
2
2
3
3
3
3
4
3
3
3
5 6 6
5 6 6
5 5 6
5 5 6
65
8129
8136
8142
8149
8156
8162
8169
8176
8182
8189
2
3
3
5 5 6
66
67
68
69
8195
8261
8325
8388
8202
8267
8331
8395
8209
8274
8338
8401
8215
8280
8344
8407
8222
8287
b351
8414
8228
8293
8357
8420
8235
8299
8363
8426
8241
8306
8370
8432
8248
8312
8376
8439
8254
8319
8382
8445
2
2
2
3
3
3
2
3
3
3
3
5 5 6
5 5 6
4 5 6
4 5 6
70
8451
8457
8463
8470
8476
8482
848,s
8494
8500
8506
2
2
3
4 5 6
71
72
73
74
8513
8573
8633
8692
8519
8579
8639
8698
8525
8585
8645
8704
8531
8591
8651
8710
8537
8597
8657
8716
8543
8603
8663
8722
8549
8609
8669
8727
8555
8615
8675
8733
8561
8621
8681
8739
8567
h627
8686
8745
2
2
2
2
2
2
2
2
3
3
3
3
4 5 5
4 5 5
4 5 5
4 5 5
75
8751
8756
8762
8768
8774
8779
8785
8791
8797
8802
2
2
3
3
4 5 5
76
77
78
79
8808
8865
8921
8976
8814
8871
8927
8982
8820
8876
8932
8987
8825
8882
8938
8993
8831
8887
8943
8998
8837
8893
8949
9004
8842
8899
8954
9009
8848
8904
8960
9015
8854
8910
8965
9020
8859
8915
8971
9025
2
2
2
2
2
2
2
2
3
3
3
3
3
3
3
3
4 5 5
4 4 5
4 4 5
4 4 5
80
9031
9036
9042
9047
9053
9058
9063
9069
9074
9079
2
2
3
3
4 4 5
81
82
83
84
9085
9138
9191
9243
9090
9143
9196
9248
9096
9149
9201
9253
9101
9154
9206
9258
9106
9159
9212
9263
9112
9165
9217
9269
9117
9170
9222
9274
9122
9175
9227
9279
9128
9180
9232
9284
9133
9186
9238
9289
2
2
2
2
2
2
2
2
3
3
3
3
3
3
3
3
4 4 5
4 4 5
4 4 5
4 4 5
85
9294
9299
9304
9309
9315
9320
9325
9330
9335
9340
2
2
2
2
2
2
3
3
2
2
2
3
4 4 5
86
87
88
89
9345
9395
9445
9494
9350
9400
9450
9499
9355
9405
9455
9504
9360
9410
9460
9509
9365
9415
9465
9513
9370
9420
9469
9518
9375
9425
9474
9523
9380
9430
9479
9528
9385
9435
9484
9533
9390
9440
9489
9538
3
3
3
3
4 4 5
3 4 4
3 4 4
3 4 4
90
9542
9547
9552
9557
9562
9566
9571
9576
9581
9586
2
2
3
3 4 4
91
92
93
94
9590
9638
9685
9731
9595
9643
9689
9736
9600
9647
9694
9741
9605
9652
9699
9745
9609
9657
9703
9750
9614
9661
9708
9754
9619
9666
9713
9759
9624
9671
9717
9763
9628
9675
9722
9768
9633
9680
9727
9773
2
2
2
2
2
2
.2
2
3
3
3
3
3 4 4
3 4 4
3 4 4
3 4 4
95
96
97
98
99
9777
9782
9786
9791
9795
9800
9805
9809
9814
9818
2
2
3
3 4 4
9823
9868
9912
9956
9827
9872
9917
9961
9832
9877
9921
9965
98:'6
9881
9926
9969
9841
9886
99:0
9974
9845
9890
9934
9978
9850
9894
9939
9983
9854
9899
9943
9987
9859
9903
9948
9991
9863
9908
9952
9996
2
2
2
2
2
2
2
3
3
3
3
3 4 4
3 4 4
3 4 4
3 3 4
440
MATHEMATICAL TABLES
Antilogarithms
•00
♦01
•02
•03
•04
1
2
3
4
5
6
7
8
9
12 3 4
5
6 7 8 9
1000
1002
1005
1007
1009
1012
1014
1016
1019
1021
0011
1
1222
1023
1047
1072
1096
1026
1050
1074
1099
1028
1052
1076
1102
1030
1054
1079
1104
1033
1057
1081
1107
1035
1059
1084
1109
1038
1062
1086
1112
1040
1064
1089
1114
1042
1067
1091
1117
1045
1069
1094
1119
0011
0011
0011
0111
1
1
1
1
1222
1222
1222
2222
•05
1122
1125
1127
1130
1132
1135
1138
1140
1143
1146
0111
1
2222
•06
•07
•08
•09
1148
1175
1202
1230
1151
1178
1205
1233
1153
1180
1208
1236
1156
1183
1211
1239
1159
1186
1213
1242
1161
1189
1216
1245
1164
1191
1219
1247
1167
1194
1222
1250
1169
1197
1225
1253
1172
1199
1227
1256
0111
111
111
111
1
1
1
1
2222
2 2 2 2
222 3
2 2 2 3
•10
1259
1262
1265
1268
1271
1274
1276
1279
1282
1285
111
1
2 2 2 3
•11
•12
•13
•14
1288
1318
1349
1380
1291
1321
1352
1384
1294
1324
1355
1387
1297
1327
1358
1390
1300
1330
1361
1393
1303
1334
1365
1396
1306
1337
1368
1400
1309
1340
1371
1403
1312
1343
1374
1406
1315
1346
1377
1409
111
111
111
111
2
2
2
2
2 2 2 3
2 2 2 3
2 2 3 3
2 2 3 3
•15
1413
1416
1419
1422
1426
1429
1432
1435
1439
1442
111
2
2 2 3 3
•16
•17
•18
•19
1445
1479
1514
1549
1449
1483
1517
1652
1452
1486
1521
1556
1455
1489
1524
1560
1459
1493
1528
1563
1462
1496
1531
1567
1466
1500
1535
1570
1469
1503
1538
1574
1472
1507
1542
1578
1476
1510
1545
1581
111
111
111
111
2
2
2
2 2 3 3
2 2 3 3
2 2 3 3
2 3 3 3
•20
1585
1589
1592
1596
1600
1603
1607
1611
1614
1618
111
2
2 3 3 3
•21
•22
•33
•24
1622
1660
1698
1738
1626
1663
1702
1742
1629
1667
1706
1746
1633
1671
1710
1750
1637
1675
1714
1754
1641
1679
1718
1758
1644
1683
1722
1762
1648
1687
1726
1766
1652
1690
1730
1770
1656
1694
1734
1774
112
112
112
112
2
2
2
2
2 3 3 3
2 3 3 3
2 3 3 4
2 3 3 4
•25
1778
1782
1786
1791
1795
1799
1803
1807
1811
1816
112
2
2 3 3 4
•26
•27
•28
•29
1820
1862
1905
1950
1824
1866
1910
1954
1828
1871
1914
1959
1832
1875
1919
1963
1837
1879
1923
1968
1841
1884
1928
1972
1845
1888
1932
1977
1849
1892
1936
1982
1854
1897
1941
1986
1858
1901
1945
1991
112
112
112
112
2
2
2
2
3 3 3 4
3 3 3 4
3 3 4 4
3 3 4 4
•30
1995
2000
2004
2009
2014
2018
2023
2028
2032
2037
112
2
3 3 4 4
•31
•32
•33
•34
2042
2089
2138
2188
2046
2094
2143
2193
2051
2099
2148
2198
2056
2104
2153
2203
2061
2109
2158
2208
2065
2113
2163
2213
2070
2118
2168
2218
2075
2123
2173
2223
2080
2128
2178
2228
2084
2133
2183
2234
112
112
112
112 2
2
2
2
3
3 3 4 4
3 3 4 4
3 3 4 4
3 4 4 5
•35
2239
2244
2249
2254
2259
2265
2270
2275
2280
2286
112 2
3
3 4 4 5
•36
•37
•38
•39
2291
2344
2399
2455
2296
2350
2404
2460
2301
2355
2410
2466
2307
2360
2415
2472
2312
2366
2421
2477
2317
2371
2427
2483
2323
2377
2432
2489
2328
2382
2438
2495
2333
2388
2443
2500
2339
2393
2449
2506
112 2
112 2
112 2
112 2
3
3
3
3
3 4 4 5
3 4 4 5
3 4 4 5
3 4 5 5
•40
2512
2518
2523
2529
2535
2541
2547
2553
2559
2564
112 2
3
4 4 5 5
•41
•42
•43
•44
2570
2630
2692
2754
2576
2636
2698
2761
2582
2642
2704
2767
2588
2649
2710
2773
2594
2655
2716
2780
2600
2661
2723
2786
2606
2667
2729
2793
2612
2673
2735
2799
2618
2679
2742
2805
2624
2685
2748
2812
112 2
112 2
112 3
112 3
3
3
3
3
4 4 5 5
4 4 5 6
4 4 5 6
4 4 5 6
•45
2818
2825
2831
2838
2844
2851
2858
2864
2871
2877
112 3
3
4 5 5 6
•46
•47
•48
•49
2884
2951
3020
3090
2891
2958
3027
3097
2897
2965
3034
3105
2904
2972
3041
3112
2911
2979
3048
3119
2917
2985
3055
3126
2924
2992
3062
3133
2931
2999
3069
3141
2938
3006
3076
3148
2944
3013
3083
3155
112 3
112 3
112 3
112 3
3
3
4
4
4 6 5 6
4 5 5 6
4 5 6 6
4 5 6 6
MATHEMATICAL TABLES
Antilogarithms
441
•50
1
2
3
4
5
6
7
8
9
12 3 4
5
6 7 8 9
3162
3170
3177
3184
3192
3199
3206
3214
3221
3228
112 3
4
4 5 6 7
•51
•52
•63
•54
3236
3311
3388
3467
3243
3319
3396
3475
3251
3327
3404
3483
3258
3334
3412
3491
3266
3342
3420
3499
3273
3350
3428
3508
3281
3357
3436
3516
3289
3365
3443
3524
3296
3373
3451
3532
3304
3381
3459
3540
12 2 3
12 2 3
12 2 3
12 2 3
4
4
4
4
5 5 6 7
5 5 6 7
5 6 6 7
5 6 6 7
•55
3548
3556
3565
3573
3581
3589
3597
3606
3614
3622
12 2 3
4
5 6 77
•56
•57
•58
•59
3631
3715
3802
3890
3639
3724
3811
3899
3648
3733
3819
3908
3656
3741
3828
3917
3664
3750
3837
3926
3673
3758
3846
3936
3681
3767
3855
3945
3690
3776
3864
3954
3698
3784
3873
3963
3707
3793
3882
3972
12 3 3
12 3 3
12 3 4
12 3 4
4
4
4
5
5 6 7 8
5 6 7 8
5 6 7 8
5 6 7 8
•60
3981
3990
3999
4009
4018
4027
4036
4046
4055
4064
12 3 4
5
6 6 7 8
•61
•62
•63
•64
4074
4169
4266
4365
4083
4178
4276
4375
4093
4188
4285
4385
4102
4198
4295
4395
4111
4207
4305
4406
4121
4217
4315
4416
4130
4227
4325
4426
4140
4236
4335
4436
4150
4246
4345
4446
4159
4256
4355
4457
12 3 4
12 3 4
12 3 4
12 3 4
5
5
5
5
6 7 8 9
6 7 8 9
6 7 8 9
6 7 8 9
•65
4467
4477
4487
4498
4508
4519
4529
4539
4550
4560
12 3 4
5
6 7 8 9
•66
•67
•68
•69
•70
4571
4677
4786
4898
4581
4688
4797
4909
4592
4699
4808
4920
4603
4710
4819
4932
4613
4721.
4831
4943
4624
4732
4842
4955
4634
4742
4853
4966
4645
4753
4864
4977
4656
4764
4875
4989
4667
4775
4887
5000
12 3 4
12 3 4
12 3 4
12 3 5
5
5
6
6
6 7 9 10
7 8 9 10
7 8 9 10
7 8 9 10
5012
5023
5035
5047
5058
5070
5082
5093
5105
5117
12 4 6
6
7 8 9 11
•71
•72
•73
•74
5129
5248
5370
5495
5623
5140
5260
5383
5508
5152
5272
5395
5521
5164
5284
5408
5534
5176
5297
5420
5546
5188
5309
5433
5559
5200
5321
5445
5572
5212
5333
5458
5585
5224
5346
5470
5598
5236
5358
5483
5610
12 4 5
12 4 5
13 4 5
13 4 5
6
6
6
6
7 8 10 11
7 9 10 11
8 9 10 11
8 9 10 12
•75
5636
5649
5662
5675
5689
5702
5715
5728
5741
13 4 5
7
8 9 10 12
•76
•77
•78
•79
5754
5888
6026
6166
5768
5902
6039
6180
5781
5916
6053
6194
5794
5929
6067
6209
5808
5943
6081
6223
5821
5957
6095
6237
5834
5970
6109
6252
5848
5984
6124
6266
5861
5998
6138
6281
5875
6012
6152
6295
13 4 5
13 4 5
13 4 6
13 4 6
7
1
7
7
8 9 11 12
8 10 11 12
8 10 11 13
9 10 11 13
•80
6310
6324
6339
6353
6368
6383
6397
6412
6427
6442
13 4 6
7
9 10 12 13
•81
•82
•83
•84
6457
6607
6761
6918
6471
6622
6776
6934
6486
6637
6792
6950
6501
6653
6808
6966
6516
6668
6823
6982
6531
6683
6839
6998
6546
6699
6855
7015
6561
6714
6871
7031
6577
6730
6887
7047
6592
6745
6902
7063
2 3 5 6
2 3 5 6
2 3 5 6
2 3 5 6
8
8
8
8
9 11 12 14
9 11 12 14
9 11' 13 14
10 11 13 15
•85
7079
7096
7112
7129
7145
7161
7178
7194
7211
7228
2 3 5 7
8
10 12 13 15
•86
•87
•88
•89
7244
7413
7586
7762
7261
7430
7603
7780
7278
7447
7621
7798
7295
7464
7638
7816
7311
7482
7656
7834
7328
7499
7674
7852
7345
7516
7691
7870
7362
7534
7709
7889
7379
7551
7727
7907
7396
7568
7745
7925
2 3 5 7
2 3 5 7
2 4 5 7
2 4 5 7
8
9
9
9
10 12 13 15
10 12 14 16
11 12 14 16
11 13 14 16
•90
7943
7962
7980
7998
8017
8035
8054
8072
8091
8110
2 4 6 7
9
11 13 15 17
•91
•92
•93
•94
8128
8318
8511
8710
8147
8337
8531
8730
8166
8356
8551
8750
8185
8375
8570
8770
8204
8395
8590
8790
8222
8414
8610
8810
8241
8433
8630
8831
8260
8453
8650
8851
8279
8472
8670
8872
8299
8492
8690
8892
2 4 6 8
2 4 6 8
2 4 6 8
2 4 6 8
9
10
10
10
•11 13 15 17
12 14 15 17
12 14 16 18
12 14 16 18
•95
8913
8933
8954
8974
8995
9016
9036
9057
9078
9099
2 4 6 8
10
12 15 17 19
•96
•97
•98
•99
9120
9333
9550
9772
9141
9354
9572
9795
9162
9376
9594
9817
9183
9397
9616
9840
9204
9419
9638
9863
9226
9441
9661
9886
9247
9462
9683
9908
9268
9484
9705
9931
9290
9506
9727
9954
9311
9528
9750
9977
2 4 6 8
2 4 7 9
2 4 7 9
2 5 7 9
11
11
11
11
13 15 17 19
13 15 17 20
13 16 18 20
14 16 18 20
INDEX
Numbers refer to pages
Abscissa, 128
Acceleration in harmonic motion, 156
Accuracy in drawing, 4
Acute angle, 2
Addition, 10
of vectors, 83, 84
Advance or lead, 151
Altitude of tetrahedron, 187
Amplitude, 149
Analysis, harmonic, 158
Anchor ring, 328
Angle between two planes, 200, 224
Angles, 1
', trigonometrical ratios of, 3
Antiparallel, 7
Approximate developments of undevelop-
able surfaces, 356
solutions, 61 et seq.
Arc of a circle, 1
Arch, linear, 123
, three-hinged, 121
Arched ribs, 122
Archimedean spiral, 140
Arithmetic, graphic, 10
Arithmetical mean, 11
Asymptotes, 13
Asymptotic cone of hyperboloid, 331, 339,
341
Auxiliary circle, 39, 149
projections, 191
of a point, 191
of solids, 193
Axial and normal sections of screw threads,
368
■ pitch of helix, 362
Axis of a conic, 30
, radical, 25
Axometric projection, 295
Bending moment diagrams,
Binormal, 343
Bow's notation, 92
Catenary as a roulette, 76
Centre of curvature, 13
of a conic, 47
of a roulette, 66
107
Centre of parallel forces, 95
of pressure, 98
of stress, 98
of vision, 301
, radical, 26
Centres of gravity* 96
■ of similarity, 14
of similitude, 14, 25
Centroid of a triangle, 7
Centroids, 96
Chord of a circle, 1
of a conic, 34
Chords, scale of, 3
Circle, 1, 16 et seq.
, equation to, 132
of curvature, 12
, properties of the, 16
Circles in contact, 21
Circular arcs, rectification of, 61
measure of an angle, 2
Circumference of a circle, 1
Circumscribed circle of spherical triangle.
237
Collar of hyperboloid, 331
Collinear points, 6
Complementary angles, 2
Composition of harmonic motions, 151,
154, 155
of vectors, 86
Concurrent straight lines, 6
Coney clic points, 6
Cone, 183
enveloping sphere, 263
surface of revolution, 328
, oblique, 266
, projections of, 187, 258
■ , sections of, 259
, tangent plane to, 317
Cones enveloping two spheres, 325
Conical projection, 167
Conic sections, 30 et seq. •
Conies, general properties of, 35
Conjugate axis, 38
Contour lines, 283
Contouring a surface from its equation,
284
444
INDEX
Co-ordinate geometry, plane, 128 c^ sea.
planes, 168
Coplanar forces, 91
Cosecant of an angle, 4
Cosine of an angle, 4
Cotangent of an angle, 4
Couples, 107
Co versed sine of an angle, 4
Cross-rolls, geometry of, 335
Cube, 182
Cubic equation, 144
Curtate cycloid, 69
Curvature, centre of, 13
, circle of, 12
, radius of, 13
Curve, harmonic, 149
y -a-\- bx», 137
y = ae'^"", 139
y = a logg bx, 140
y = bx'\ 136
y == x», 135
yx^ = c, 138
Curved surfaces and tangent planes, 313
et seq.
, sections of, 314
Curves, similar, 14
, tortuous, 343
Cusp, 13
Cycloid, 67
Cycloidal curve, 66
Cylinder, 182, 252
, elliptical, 255
enveloping sphere, 256
surface of revolution, 327
, oblique, 265
, projections of, 187, 253
, sections of, 253
Cylinders, intersection of, 388
Dams, masonry, 100
Definitions of solids, 181
Deflections of braced frames, 1 18
Descriptive geometry, 166
Developable surfaces, 314
Development of a surface, 349
of surface of cone, 353
of cylinder, 352
of prism, 349
of pyramid, 350
Developments, 349 et seq.
Dihedral angle, 200
Directrix, 313
of a conic, 30
Divided pitch of helices, 362
of screw, 367
Division, 11
Dodecahedron, 182, 423
Eccentricity of a conic, 30
Elevation, 168
, sectional, 241 ,
Ellipse, 30, 32, 38-45, 48, 49
, equation to, 133
Ellipsoid, 337
Elliptic paraboloid, 341
Elliptical cylinder, 255
Envelope, 13
glissette, 78
roulette, 77
Epicycloid, 70
, spherical, 428
Epitrochoid, 72
Equation, graph of, 129
Equations, graphic solution of, 142
Equiangular spiral, 141
Error in measuring, 4
Escribed circles of a ti
Evolute of a conic, 49
of a curve, 13
Exponential curve, 139
equation, 139
Face mould, 371, 374
Falling moulds, 373
Figured plans, 279
Focus of a conic, 30
Forces, coplanar, 91
, specification of, 91
Fourier series, 158
Frequency, 148
Funicular polygon, 92
Generating line, 313
Generation of surfaces, 313
Generatrix, 3i3
Geometrical mean, 11
Glissettes, 78
Gorge of hyperboloid, 331
Graph of equation, 129
Graphic arithmetic, 10
solution of equations, 142,
statics, 91 et seq.
Gravity, centres of, 96
Great circle of a sphere, 249
Ground line, 168, 301
Handrails, 369
Harmonic analysis, 158
curve, 119
motion, 148
motions, composition of, 161, 154,
155
Harmonical mean, li
Helical springs, 367
Helices and screws, 362 et seq.
Helix, 362
, axial pitch of, 362
, inclination of, 363
, left-handed, 363
, normal pitch of, 362
of increasing pitch, 363.
, pitch angle of, 363,
INDEX
445
Helix, ri^ht-handed, 363
Hip rafter, 245
Hipped roof, 245
Horizontal projection, 279 et seq.
Hyperbola, equation to, 134
, properties of, 45
, rectangular, 47
Hyperbolic paraboloid, 342
Hyperboloid of one sheet, 338
— — of revolution, 331
of two sheets, 340
Ilyperboloids of revolution in rolling con-
tact, 333
Hypocycloid, 73
— "■ — , spherical, 428
Hypotrochoid, 74
Icosahedron, 182, 424
Image, 9, 430
Inclination of a line to a plane, 223
of a plane, 280
of helix, 363
Inclined plane, 1 99
Indexed plan, 279
Inertia, moment of, 110
Inferior epitrochoid, 72
hypotrochoid, 74
trochoid, 69
Inflexion, point of, 13
Inscribed circle of spherical triangle, 237
of a triangle, 20
Intersection of cone and surface of revolu-
tion, 400
of cones, 395
of curved surface with prism or
pyramid, 404
of cylinder and cone, 392
of c'vlinder and sphere, 397
of cylinder and surface of revolution,
398
of cylinders, 388
of line and plane, 216, 283
of plane and contoured surface, 284
of i)risms and pyramids, 404
of spheres, 403
of straight line and cone, 265
of straight line and cylinder, 258
of straight line and curved surface,
317
• of straight line and sphere, 252
of surfaces, 388 et seq.
of surfaces of revolution, 401, 402
of two planes, 204, 282
Involute of a circle, 76
of a curve, 13
Involution, 12
Isometric projection, 291
scale, 292
Jack rafter, 245
Lag and lead, 150
Latus rectum of a conic, 34
Lead and lag, 150
Left-handed helix, 363
Level easing, 369
quarter, 369
Linear arch, 123
line of resistance, 101
Lines, I
Link polygon, 92
Locors, 82
Locus, 1
Logarithmic curve, 140
spiral, 141
Lune of a sphere, 249
Masonry dams, 100
Mean proportional, 11 •
Measuring point, 305
Medians of a triangle, 6
Meridian sections, 314
Method of sections, 1 1 7
Modulus figure, 99
Moment of a force, 104
of inertia, 110
Moments, principle of, 107
Motion, periodic, 148 et seq.
Moulding, raking, 244
Mouldings, 243
, sections of, 243
Multiple point, 13
Multiple -threaded screw, pitch and lead
of, 367
Multiplication, 11
Node, 13
Normal pitch of helix, 362
to a curve, 12
to a surface, 317
Notation, Bow's, 92
in projection, 170
Oblate spheroid, 329
Oblique cone, 266
co-ordinates, 128
cylinder, 265
parallel projection, 295
plane, 199
Obtuse angle, 2
Octahedron, 182
, projections of, 187
Orthocentre of a triangle, 7
Orthocentric triangle, 7
Orthogonal projection, 167
Orthographic projection, 167
Osculating plane, 343
Parabola, equation to, 133
, properties of, 36
Paraboloid, elliptic, 341
, hyperbolic, 342
446
INDEX
Paraboloid of revolution, 337
Parallel forces, centre of, 95
projection, 167
Parallelepiped, 182
Parameter, 37
Pedal triangle, 7
Perimeter of a triangle, 21
Periodic motion, 148 6-^ seq.
time, 148
Perpendicular plane, 199
projection, 167
Perspective projection, 167, 300 et seq.
Pictorial projections, 290 et seq.
Pitch and lead of multiple-threaded screw,
367
Pitch angle of helix, 363
Plan, 168
• , sectional, 241
Plane co-ordinate geometry, 128 et seq.
, inclined, 199
, oblique, 199
■ of projection, 166
, perpendicular, 199
section of a sphere, 249
, traces of, 198
Planes, representation of, 198
Point of distance, 302
of inflexion, 1 3
Polar co-ordiuates, 128
triangle, 234
Pole and polar, 24, 49
Polygon, funicular, 92
Polyhedron, 181
Pressure, centre of, 98
Principal elliptic section of hvperboloid,
338
normal, 343
section of oblique cone, 266
— of oblique cylinder, 265
Principle of moments, 107
Prism, 182
, projections of, 183
■ , section of, 241
Projecting surface of a line, 167
Projection, 166 tA WY^
LOAN
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U.C. BERKELEY LIBRARIES
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THE UNIVERSITY OF CALIFORNIA LIBRARY
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