DOI Seite / Zitierlink:
https://doi.org/10.11588/diglit.25888#0057
DRAWING OF MACHINERY BY ORDINARY GEOMETRICAL PROJECTION.
41
Figs. 7 and 8.—To find the horizontal projection of the
transverse section of a regular five-sided pyramid, cut
by a plane perpendicular to the vertical, but inclined to
the horizontal plane; and let one edge of the pyramid
be in a plane perpendicular to both planes of projection.
This example, though, on the whole, very similar to
the preceding, is introduced as illustrating a method
which we have not hitherto had occasion to point out.
The projections of the pyramid are constructed by
describing from the centre S' a circle circumscribing the
base, and cutting the axis S S' at B'; starting from this
point, the circumference is to be divided into five equal
parts, and the contiguous points thus obtained to be joined
by straight lines, forming the polygon A' B' C' D' E', each
of whose angles, being joined to the centre S', show the
projections of the edges of the pyramid; the vertical
projection is then derived from the plan in the manner
already pointed out. Then, following the method ex-
plained in the preceding example, we obtain the horizon-
tal projection of the section made by the plane a c. But
it will be remarked that that method will not suffice
for the determination of the point b', because the perpendi-
cular let fall from the corresponding point b, in the eleva-
tion, coincides with the projection of the edge B S. It is
therefore necessary to find some other mode of finding the
distance of the point b' from the axis. Let the pyramid
be supposed to be turned a quarter of a revolution round
its axis ; the line B' S' will then have assumed the posi-
tion S' b2. Project the point b2 to 63, and join S 63. Then,
since the required point must also be conceived to have
described a quarter of a circle in a plane parallel to the
horizontal plane, and that its new position must be in the
line S 63, it is obvious that its vertical projection is the
point bi, the intersection of a horizontal line drawn
through b, with that line. The distance b 64, then,
being transferred from S' to b' determines the position
of the latter point in the plan; or, following a more
methodical process, by projecting the point 64 to A, and
describing a circle from the centre S' passing through A;
its intersection with B' S' is the point sought.
Projections of a Prism.—Plate II.
Figs. 1 and 2.—Required to represent a regular six-
sided Prism, in an upright position, in vertical and
horizontal projections.
Previously to laying down these figures, the ground
line, centre lines, and border, should be drawn as in the
preceding Plate. Then, from the centre S', with such a
radius as will circumscribe the base of the required prism,
draw a circle, and inscribe the hexagon as already directed.
Project the base thus delineated, by perpendiculars to the
ground line from each of its angular points; and, since
the prism is upright, it is obvious that these angular
points themselves represent the horizontal projections of
all its edges, and that their elevations coincide with the
perpendiculars A' G, B' H, &c. Set off from G to A the
height of the prism, and through A draw A D, parallel to
the ground line. This will be the vertical projection of
the upper surface. The edges, being all parallel to the
vertical plane, are, of course, seen in their actual length.
Figs. 3 and 4.—To form the projections of the same
prism, supposing it to have been moved round the point
G, in a plane parallel to the vertical plane.
Since, with the exception of the inclination, the vertical
projection of the prism is exactly the same as formerly,
all that is required is to copy Fig. 1. The method of pro-
ceeding in this case is so entirely analogous to that ex-
plained in reference to Fig. 3 of the preceding Plate, that
we consider it unnecessary to recapitulate it. This figure
being constructed, determines the horizontal projection of
the prism, which, being supposed to have moved in a
plane parallel to the vertical plane of projection, will still
have as its centre line in the plan, the line A' K'; and
consequently the lines F' E' and B' C', Fig. 2, produced,
will represent the outline of the sides parallel to the ver-
tical plane. Now, by letting fall perpendiculars from all
the angles in the elevation, and joining the contiguous
points of intersection with the horizontal lines appro-
priate to these points respectively, we obtain the polygon
A' B' C' D' E' F' as the projection of the upper surface,
and G' H' I' K' L' M' as that of the base of the prism.
Finally, it will be observed that all the edges are repre-
sented, in the horizontal projection, by equal straight
lines, as D' K', A' G', &c., and that the sides A' B', G' H',
&c., remain still parallel to each other, which will afford
the means of verifying the accuracy of the drawings.
Figs. 5 and 6.—Required the projections of the same
prism set into a position inclined to both planes of pro-
jection.
Assuming that the inclination of the prism upon the
horizontal plane is the same as in the preceding figures,
which, for the sake of simplifying the operation, we shall
suppose to be already constructed, the first process is to
copy Fig. 4, which may be done by drawing a centre line
X X, so as to form the required angle of the prism with
the vertical plane; then, having set off upon this line a
distance equal to A' K', Fig. 4, transfer the distances A' G'
and D' K' also to Fig. 6 ; and, in order to find the remain-
ing angular points, make A' a equal to the corresponding
distance in Fig. 4, and through a draw B' F' perpendicular
to the centre line, and transfer the distances a B', a F'.
Through the points B' and F', draw straight lines parallel
to A' K', and join A' B', A' F'; and, since we have already
seen that all the other sides must be parallel to these, the
figure is completed by drawing through the points G', D ,
and K', straight lines parallel to A' B and A' F' respectively.
Now, since the prism has been supposed to have pre-
served its former inclination to the horizontal plane, it is
obvious that every point in it, such as A, has, in assuming
its new position, simply moved in a horizontal plane, and
will, therefore, be in the line A A parallel to the ground
line, and since the same point has been projected to A,
Fig. 6, it will also be in the perpendicular A' A; the
point of intersection A, Fig. 5, is, therefore, its projection
in the elevation. The refraining angular points, in this
view, are all determined in the same manner by the aid
41
Figs. 7 and 8.—To find the horizontal projection of the
transverse section of a regular five-sided pyramid, cut
by a plane perpendicular to the vertical, but inclined to
the horizontal plane; and let one edge of the pyramid
be in a plane perpendicular to both planes of projection.
This example, though, on the whole, very similar to
the preceding, is introduced as illustrating a method
which we have not hitherto had occasion to point out.
The projections of the pyramid are constructed by
describing from the centre S' a circle circumscribing the
base, and cutting the axis S S' at B'; starting from this
point, the circumference is to be divided into five equal
parts, and the contiguous points thus obtained to be joined
by straight lines, forming the polygon A' B' C' D' E', each
of whose angles, being joined to the centre S', show the
projections of the edges of the pyramid; the vertical
projection is then derived from the plan in the manner
already pointed out. Then, following the method ex-
plained in the preceding example, we obtain the horizon-
tal projection of the section made by the plane a c. But
it will be remarked that that method will not suffice
for the determination of the point b', because the perpendi-
cular let fall from the corresponding point b, in the eleva-
tion, coincides with the projection of the edge B S. It is
therefore necessary to find some other mode of finding the
distance of the point b' from the axis. Let the pyramid
be supposed to be turned a quarter of a revolution round
its axis ; the line B' S' will then have assumed the posi-
tion S' b2. Project the point b2 to 63, and join S 63. Then,
since the required point must also be conceived to have
described a quarter of a circle in a plane parallel to the
horizontal plane, and that its new position must be in the
line S 63, it is obvious that its vertical projection is the
point bi, the intersection of a horizontal line drawn
through b, with that line. The distance b 64, then,
being transferred from S' to b' determines the position
of the latter point in the plan; or, following a more
methodical process, by projecting the point 64 to A, and
describing a circle from the centre S' passing through A;
its intersection with B' S' is the point sought.
Projections of a Prism.—Plate II.
Figs. 1 and 2.—Required to represent a regular six-
sided Prism, in an upright position, in vertical and
horizontal projections.
Previously to laying down these figures, the ground
line, centre lines, and border, should be drawn as in the
preceding Plate. Then, from the centre S', with such a
radius as will circumscribe the base of the required prism,
draw a circle, and inscribe the hexagon as already directed.
Project the base thus delineated, by perpendiculars to the
ground line from each of its angular points; and, since
the prism is upright, it is obvious that these angular
points themselves represent the horizontal projections of
all its edges, and that their elevations coincide with the
perpendiculars A' G, B' H, &c. Set off from G to A the
height of the prism, and through A draw A D, parallel to
the ground line. This will be the vertical projection of
the upper surface. The edges, being all parallel to the
vertical plane, are, of course, seen in their actual length.
Figs. 3 and 4.—To form the projections of the same
prism, supposing it to have been moved round the point
G, in a plane parallel to the vertical plane.
Since, with the exception of the inclination, the vertical
projection of the prism is exactly the same as formerly,
all that is required is to copy Fig. 1. The method of pro-
ceeding in this case is so entirely analogous to that ex-
plained in reference to Fig. 3 of the preceding Plate, that
we consider it unnecessary to recapitulate it. This figure
being constructed, determines the horizontal projection of
the prism, which, being supposed to have moved in a
plane parallel to the vertical plane of projection, will still
have as its centre line in the plan, the line A' K'; and
consequently the lines F' E' and B' C', Fig. 2, produced,
will represent the outline of the sides parallel to the ver-
tical plane. Now, by letting fall perpendiculars from all
the angles in the elevation, and joining the contiguous
points of intersection with the horizontal lines appro-
priate to these points respectively, we obtain the polygon
A' B' C' D' E' F' as the projection of the upper surface,
and G' H' I' K' L' M' as that of the base of the prism.
Finally, it will be observed that all the edges are repre-
sented, in the horizontal projection, by equal straight
lines, as D' K', A' G', &c., and that the sides A' B', G' H',
&c., remain still parallel to each other, which will afford
the means of verifying the accuracy of the drawings.
Figs. 5 and 6.—Required the projections of the same
prism set into a position inclined to both planes of pro-
jection.
Assuming that the inclination of the prism upon the
horizontal plane is the same as in the preceding figures,
which, for the sake of simplifying the operation, we shall
suppose to be already constructed, the first process is to
copy Fig. 4, which may be done by drawing a centre line
X X, so as to form the required angle of the prism with
the vertical plane; then, having set off upon this line a
distance equal to A' K', Fig. 4, transfer the distances A' G'
and D' K' also to Fig. 6 ; and, in order to find the remain-
ing angular points, make A' a equal to the corresponding
distance in Fig. 4, and through a draw B' F' perpendicular
to the centre line, and transfer the distances a B', a F'.
Through the points B' and F', draw straight lines parallel
to A' K', and join A' B', A' F'; and, since we have already
seen that all the other sides must be parallel to these, the
figure is completed by drawing through the points G', D ,
and K', straight lines parallel to A' B and A' F' respectively.
Now, since the prism has been supposed to have pre-
served its former inclination to the horizontal plane, it is
obvious that every point in it, such as A, has, in assuming
its new position, simply moved in a horizontal plane, and
will, therefore, be in the line A A parallel to the ground
line, and since the same point has been projected to A,
Fig. 6, it will also be in the perpendicular A' A; the
point of intersection A, Fig. 5, is, therefore, its projection
in the elevation. The refraining angular points, in this
view, are all determined in the same manner by the aid