finite distance
below AB', the locus for this quarter of the revolution being a curve
A'G, to which C'R is an asymptote. After the crank pin passes C, the axis
will be found above AB' and to the right of C'R, moving in a curve HB',
which is the locus for the second quadrant. Since the path of P is
symmetrical with respect to DE, the completion of the revolution will
result in the formation of two other curves, continuous and symmetrical
with those above described, the whole appearing as in Fig. 24, the
vertical line through C' being a common asymptote.

In order to find the radius of curvature at any point on the generated
curve, it is necessary to find not only the location of the instantaneous
axis, but its motion. This is done as shown in Fig. 25. P being the given
point, CD is the corresponding position of the connecting rod, OC that of
the crank. Draw through D a perpendicular to OD, produce OC to cut it in
E, the instantaneous axis. Assume C A perpendicular to OC, as the motion
of the crank. Then the point E in OC produced will have the motion EF
perpendicular to OE, of a magnitude determined by producing OA to cut
this perpendicular in F. But since the _intersection_ E of the crank
produced is to be with a vertical line through the other end of the rod,
the instantaneous axis has a motion which, so far as it depends upon the
movement of C only, is in the direction DE. Therefore EF is a component,
whose resultant EG is found by drawing FG perpendicular to EF. Now D is
moving to the left with a velocity which may be determined either by
drawing through A a perpendicular to CD, and through C a horizontal line
to cut this perpendicular in H, or by making the angle DEI equal to the
angle CEA, giving on DO the distance DI, equal to CH. Make EK = DI or
CH, complete the rectangle KEGL, and its diagonal ES is, finally, the
motion of the instantaneous axis.

EP is the normal, and the actual motion of P is PM, perpendicular to EP,
the angle PEM being made equal to CEA. Find now the component EN of the
motion ES, which is perpendicular to EP. Draw NM and produce it to cut EP
produced in R the center of curvature at P.

This point evidently lies upon the branch _z_M of the evolute in Fig. 23.
The process of finding one upon the other branch _x_N is shown in the
lower part of the diagram, Fig. 25. The operations being exactly like
those above described, will be readily traced by the reader without
further explanation.