100
ENGINEER AND MACHINIST’S DRAWING-BOOK.
ject, H K the picture line, and O the station point, then
the vanishing point of A G and all its parallels will be
found by drawing O a parallel to it to intersect the pic-
ture line produced, when a is the vanishing point. The
vanishing point of A B and its parallels will be b. The j
vanishing point of F E and all other lines perpendicular
to the picture will be the point of sight e; and the lines
G F, B C, D E, being parallel to the picture, their per-
spectives will also be parallel to it.
These rules, which are thus established, enable us to
abridge, in many instances, the operations of drawing per-
spectives, as may be thus illustrated.
Let A B C D (Fig. 195) be the plan of a square, O the
place of the spectator, e f the line of projection, and
0 C the central plane.
spectator or point of distance, the line b E and the inter-
section of the lines at A will be the perspective of the
point required.
{ng. 196.) jLet it be required to
draw the perspective of
a square, situated in the
original plane, at any
angle whatever to the
plane of the picture. To
solve this problem it is
necessary only to find
the perspective of a single point, which may be done in
various ways.
Let A B (Fig. 197) be the plan of the square, O the place
of the spectator, E K the line of projection, and 0 c the
(Fig. 195.)
O
Draw ef to represent the vertical projection of the base
of the picture, G K the horizontal line, o the point of
sight, and O o the vertical plane. Then to draw the per-
spective of the square, transfer the side C B to oh (see
Prob. I. Rule 1), and draw b a, which is the perspective
of the side B A of the square produced to its vanishing
point (see, Prob. I. Rule 3). Then, as the diagonals of
the square form an angle of 45° with the picture, from o,
set off on the horizontal line o G, o K, each equal to the dis-
tance O C (see Prob. I. Rule 4), and draw h G, o K, the
perspective of the diagonals, produced to their vanishing
points, and join the points ad where these intersect the
perspectives of the sides by the line d a, parallel to the
base of the picture (see Prob. I. Rule 2), and o d a b will
be the perspective of the square.
If the perspective of the point A alone had to be sought,
the operation would be simply to draw b o and to inter-
sect it by o K.
Let it be required, for example, to find the perspective
of an original point A, situated 35 feet from the central
plane of the picture, and 63 feet from the base.
Set off from o (Fig. 196) to/, 35 feet from any conve-
nient scale, and drawf o, c being the point of sight; set off
from / to b, 63 feet, and draw to E the distance of the
(Fig. 197.)
central plane. We see that the point A is on the horizon-
tal trace of the central plane, and at the distance A c from
the picture, and that, therefore, the perspective of A is to
be found on the central line, in the vertical projection 0 e.
Set off the distance A c from c to g, and draw from g to
the point of distance h the line g h, and its intersection
with the central line at a is the perspective of A. Thus
two points in the perspective are obtained, as B, being on
the base of the picture, will have for its perspective b.
The vanishing points of the sides of the square are
found to be K and m, by drawing parallels to these sides
through O to meet the picture. Transfer the distances,
therefore, K and m, to the horizon line ef in np, and to
these, from b and a, draw the perspective representations
of the sides of the square. The operation is shown re-
peated above the horizon, the same parts being indicated
by the same letters accented.
In the examples hitherto, we have operated by methods
ENGINEER AND MACHINIST’S DRAWING-BOOK.
ject, H K the picture line, and O the station point, then
the vanishing point of A G and all its parallels will be
found by drawing O a parallel to it to intersect the pic-
ture line produced, when a is the vanishing point. The
vanishing point of A B and its parallels will be b. The j
vanishing point of F E and all other lines perpendicular
to the picture will be the point of sight e; and the lines
G F, B C, D E, being parallel to the picture, their per-
spectives will also be parallel to it.
These rules, which are thus established, enable us to
abridge, in many instances, the operations of drawing per-
spectives, as may be thus illustrated.
Let A B C D (Fig. 195) be the plan of a square, O the
place of the spectator, e f the line of projection, and
0 C the central plane.
spectator or point of distance, the line b E and the inter-
section of the lines at A will be the perspective of the
point required.
{ng. 196.) jLet it be required to
draw the perspective of
a square, situated in the
original plane, at any
angle whatever to the
plane of the picture. To
solve this problem it is
necessary only to find
the perspective of a single point, which may be done in
various ways.
Let A B (Fig. 197) be the plan of the square, O the place
of the spectator, E K the line of projection, and 0 c the
(Fig. 195.)
O
Draw ef to represent the vertical projection of the base
of the picture, G K the horizontal line, o the point of
sight, and O o the vertical plane. Then to draw the per-
spective of the square, transfer the side C B to oh (see
Prob. I. Rule 1), and draw b a, which is the perspective
of the side B A of the square produced to its vanishing
point (see, Prob. I. Rule 3). Then, as the diagonals of
the square form an angle of 45° with the picture, from o,
set off on the horizontal line o G, o K, each equal to the dis-
tance O C (see Prob. I. Rule 4), and draw h G, o K, the
perspective of the diagonals, produced to their vanishing
points, and join the points ad where these intersect the
perspectives of the sides by the line d a, parallel to the
base of the picture (see Prob. I. Rule 2), and o d a b will
be the perspective of the square.
If the perspective of the point A alone had to be sought,
the operation would be simply to draw b o and to inter-
sect it by o K.
Let it be required, for example, to find the perspective
of an original point A, situated 35 feet from the central
plane of the picture, and 63 feet from the base.
Set off from o (Fig. 196) to/, 35 feet from any conve-
nient scale, and drawf o, c being the point of sight; set off
from / to b, 63 feet, and draw to E the distance of the
(Fig. 197.)
central plane. We see that the point A is on the horizon-
tal trace of the central plane, and at the distance A c from
the picture, and that, therefore, the perspective of A is to
be found on the central line, in the vertical projection 0 e.
Set off the distance A c from c to g, and draw from g to
the point of distance h the line g h, and its intersection
with the central line at a is the perspective of A. Thus
two points in the perspective are obtained, as B, being on
the base of the picture, will have for its perspective b.
The vanishing points of the sides of the square are
found to be K and m, by drawing parallels to these sides
through O to meet the picture. Transfer the distances,
therefore, K and m, to the horizon line ef in np, and to
these, from b and a, draw the perspective representations
of the sides of the square. The operation is shown re-
peated above the horizon, the same parts being indicated
by the same letters accented.
In the examples hitherto, we have operated by methods