0.5

1 cm

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