DOI Seite / Zitierlink:
https://doi.org/10.11588/diglit.25888#0061
DRAWING OF MACHINERY BY ORDINARY GEOMETRICAL PROJECTION.
45
the torus, and let the circle IT' /' L', and the rectangle
HIML be the analogous projections of the cylinder,
which passes perpendicularly through it. If now we con-
ceive, as before, a plane a b, Fig 13, to pass horizontally
through both solids, it will obviously cut the cylinder in
a circle which will be projected in the base H' /' L' itself,
and the ring in two other circles of which one only, part
of which is represented by the arc /' 63 b', will intersect
the cylinder at the points/' and A, which being projected
vertically to Fig 13, will give two points / and b2 in the
upper curve of penetration.
Another horizontal plane, taken at the same distance
below the centre line A B, as that marked a b is above it,
will evidently cut the ring in circles coinciding with those
already obtained ; consequently the points/' and 63 indi-
cate points in the lower, as well as in the upper curves
of penetration, and are projected vertically at d and e.
Thus we find that by laying down two planes at equal
distances on each side of A B, we obtain, by one operation,
four points in the curves required.
In order to determine the vertices to and n, we must
again have recourse to the method explained in the pre-
ceding problem; that is, to draw a plane O' n, passing
through the axis of the cylinder and the centre of the
ring, and to conceive this plane to be moved round the
point O', as on a hinge, until it has assumed the position
O' B', parallel to the vertical plane ; the point li, repre-
senting the extreme outline of the cylinder in plan, will
now be at r', and being projected vertically, that outline
will cut the ring in two points p and r, which would be
the limits of the curves of penetration in the supposed
relative position of the two solids ; and by drawing the
two horizontal lines r n and p to, and projecting the point
n' vertically, we obtain, by the intersections of these lines,
the two points to and n, which are the vertices of the
curves in the actual position of the penetrating bodies.
The points at which the curves are tangents to the out-
lines H I and L M of the cylinder, may readily be found
by describing arcs of circles from the centre O' through
the points H and L', which represent these lines in the
plan, and then proceeding, as above, to project the points
thus obtained upon the elevation. Lastly, to determine
the points, as j, z, &c., where the curves are tangents to
the horizontal outlines of the ring, draw a circle P' s' /
with a radius equal to that of the centre line of the ring,
namely P D ; the points of intersection z and j’ are the
horizontal projections of the points sought.
Required to represent the sections which would be
made in the ring now before us, by two planes, one of
which, N' T', is parcdlel to the vertical plane, while the
other T' E' is perpendicular to both planes of projection.
The section made by the last-named plane must obvi-
ously have its vertical projection in the line C D, which
indicates the position of the plane ;* but the former will
* We may here observe, that in drawing the details of machinery,
particularly if on a large scale or of the actual size, a mode of repre-
sentation similar to that exemplified in Fig. 13, is frequently resorted
to, and is attended with considerable economy of time and space.
be represented in its actual form and dimensions in the
elevation. To determine its outlines, let two horizontal
planes g q and i h, equidistant from the centre line A B,
be supposed to cut the ring; their lines of intersection
with it will have their horizontal projections in the two
circles g o' and li q which cut the given plane N T in o'
and f. These points being projected vertically to o, q, k,
&c., give four points in the curve required. The line
N' T' cutting the circle A' E' B' at N', the projection N
of this point is the extreme limit of the curve.
The circle P s' j', the centre line of the rim of the torus,
is cut by the planes N' T' at the point s', which being pro-
jected vertically upon the lines D P and G l, determines s
and l, the points of contact of the curve with the horizontal
outlines of the ring. Finally, the points, t and u are ob-
tained by drawing from the centre 0, a circle T' v tangent
to the given plane, and projecting the point of intersection
v to the points v and x, which are then to be replaced
upon C D by drawing the horizontals v t and x u.
Penetrations of Cylinders, Prisms, Spheres,
and Cones.—Plate VI.
Figs. 15 and 16.—Required to delineate the lines of
penetration of a sphere and a regular hexagonal prism
ivhose axis qicisses through the centre of the sphere.
The centres of the circles forming the two projections
of the sphere are, according to the terms of the problem,
upon the axis C C of the upright prism, which is projected
horizontally in the regular hexagon D' E' F' G' H' I'.
Hence it follows, that as all the lateral faces of the prism
are equidistant from the centre of the sphere, their lines
of intersection with it will necessarily be circles of equal
diameters. Now, the perpendicular face represented by
the line E' F' in the plan, will meet the surface of the
sphere in two circular arcs E F and L M, Fig. 15, described
from the centre C, with a radius equal to c b' or a' c.
And the intersections of the two oblique faces D' E' and
F G' will obviously be each projected in two arcs of an
ellipse whose major axis clg is equal to the diameter of
the circle a cb, and the minor axis is the vertical projec-
tion of that diameter, as represented at e' f, Fig. 16. But
as it is necessary to draw small portions only of these
curves, the following method may be employed.
It is sufficiently evident that the horizontal line D G
will pass through all the points where the edges of the
visible faces of the prism intersect the surface of the
sphere. Now, if we divide the portions E F and F G
respectively into the same number of equal parts, and,
drawing perpendiculars through the points of division,
set off from F G the distances from the corresponding
points in E F to the circular arc E' c F, we shall have as
many points in the elliptical arc required as we have
taken divisions upon the chords. The remaining ellip-
When the form of a piece is exactly symmetrical on either side of a
centre line, it is sufficient for all practical purposes to represent on
one side of the centre line, half the external elevation of the object,
and on the other, half the section or plan, according as its nature or
form may require.
45
the torus, and let the circle IT' /' L', and the rectangle
HIML be the analogous projections of the cylinder,
which passes perpendicularly through it. If now we con-
ceive, as before, a plane a b, Fig 13, to pass horizontally
through both solids, it will obviously cut the cylinder in
a circle which will be projected in the base H' /' L' itself,
and the ring in two other circles of which one only, part
of which is represented by the arc /' 63 b', will intersect
the cylinder at the points/' and A, which being projected
vertically to Fig 13, will give two points / and b2 in the
upper curve of penetration.
Another horizontal plane, taken at the same distance
below the centre line A B, as that marked a b is above it,
will evidently cut the ring in circles coinciding with those
already obtained ; consequently the points/' and 63 indi-
cate points in the lower, as well as in the upper curves
of penetration, and are projected vertically at d and e.
Thus we find that by laying down two planes at equal
distances on each side of A B, we obtain, by one operation,
four points in the curves required.
In order to determine the vertices to and n, we must
again have recourse to the method explained in the pre-
ceding problem; that is, to draw a plane O' n, passing
through the axis of the cylinder and the centre of the
ring, and to conceive this plane to be moved round the
point O', as on a hinge, until it has assumed the position
O' B', parallel to the vertical plane ; the point li, repre-
senting the extreme outline of the cylinder in plan, will
now be at r', and being projected vertically, that outline
will cut the ring in two points p and r, which would be
the limits of the curves of penetration in the supposed
relative position of the two solids ; and by drawing the
two horizontal lines r n and p to, and projecting the point
n' vertically, we obtain, by the intersections of these lines,
the two points to and n, which are the vertices of the
curves in the actual position of the penetrating bodies.
The points at which the curves are tangents to the out-
lines H I and L M of the cylinder, may readily be found
by describing arcs of circles from the centre O' through
the points H and L', which represent these lines in the
plan, and then proceeding, as above, to project the points
thus obtained upon the elevation. Lastly, to determine
the points, as j, z, &c., where the curves are tangents to
the horizontal outlines of the ring, draw a circle P' s' /
with a radius equal to that of the centre line of the ring,
namely P D ; the points of intersection z and j’ are the
horizontal projections of the points sought.
Required to represent the sections which would be
made in the ring now before us, by two planes, one of
which, N' T', is parcdlel to the vertical plane, while the
other T' E' is perpendicular to both planes of projection.
The section made by the last-named plane must obvi-
ously have its vertical projection in the line C D, which
indicates the position of the plane ;* but the former will
* We may here observe, that in drawing the details of machinery,
particularly if on a large scale or of the actual size, a mode of repre-
sentation similar to that exemplified in Fig. 13, is frequently resorted
to, and is attended with considerable economy of time and space.
be represented in its actual form and dimensions in the
elevation. To determine its outlines, let two horizontal
planes g q and i h, equidistant from the centre line A B,
be supposed to cut the ring; their lines of intersection
with it will have their horizontal projections in the two
circles g o' and li q which cut the given plane N T in o'
and f. These points being projected vertically to o, q, k,
&c., give four points in the curve required. The line
N' T' cutting the circle A' E' B' at N', the projection N
of this point is the extreme limit of the curve.
The circle P s' j', the centre line of the rim of the torus,
is cut by the planes N' T' at the point s', which being pro-
jected vertically upon the lines D P and G l, determines s
and l, the points of contact of the curve with the horizontal
outlines of the ring. Finally, the points, t and u are ob-
tained by drawing from the centre 0, a circle T' v tangent
to the given plane, and projecting the point of intersection
v to the points v and x, which are then to be replaced
upon C D by drawing the horizontals v t and x u.
Penetrations of Cylinders, Prisms, Spheres,
and Cones.—Plate VI.
Figs. 15 and 16.—Required to delineate the lines of
penetration of a sphere and a regular hexagonal prism
ivhose axis qicisses through the centre of the sphere.
The centres of the circles forming the two projections
of the sphere are, according to the terms of the problem,
upon the axis C C of the upright prism, which is projected
horizontally in the regular hexagon D' E' F' G' H' I'.
Hence it follows, that as all the lateral faces of the prism
are equidistant from the centre of the sphere, their lines
of intersection with it will necessarily be circles of equal
diameters. Now, the perpendicular face represented by
the line E' F' in the plan, will meet the surface of the
sphere in two circular arcs E F and L M, Fig. 15, described
from the centre C, with a radius equal to c b' or a' c.
And the intersections of the two oblique faces D' E' and
F G' will obviously be each projected in two arcs of an
ellipse whose major axis clg is equal to the diameter of
the circle a cb, and the minor axis is the vertical projec-
tion of that diameter, as represented at e' f, Fig. 16. But
as it is necessary to draw small portions only of these
curves, the following method may be employed.
It is sufficiently evident that the horizontal line D G
will pass through all the points where the edges of the
visible faces of the prism intersect the surface of the
sphere. Now, if we divide the portions E F and F G
respectively into the same number of equal parts, and,
drawing perpendiculars through the points of division,
set off from F G the distances from the corresponding
points in E F to the circular arc E' c F, we shall have as
many points in the elliptical arc required as we have
taken divisions upon the chords. The remaining ellip-
When the form of a piece is exactly symmetrical on either side of a
centre line, it is sufficient for all practical purposes to represent on
one side of the centre line, half the external elevation of the object,
and on the other, half the section or plan, according as its nature or
form may require.