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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.
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