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.
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