Placing the crank in the opposite position OB, we find by a construction precisely similar to the above, the radius of curvature E_z_ at the other extremity of the axis of the curve. It will at once be seen that E_z_ is less than D_x_, and that since the normal at P is vertical and infinite, the evolute of DPE will consist of two branches xN, zM, to which the vertical normal PL is a common asymptote. These two branches will not be similar, nor is the curve itself symmetrical with respect to PL or to any transverse line; all of which peculiarities characterize it as something quite different from the ellipse.
[Illustration: FIG. 23.]
[Illustration: FIG. 24.]
[Illustration: FIG. 25.]
Moreover, in Fig. 22, the locus of the instantaneous axis of the trammel bar (of which the part EH corresponds to the connecting rod, when a crank OH is added to the elliptograph there discussed) was found to be a circle. But in the present case this locus is very different. Beginning at A’, the instantaneous axis moves downward and to the right, as the crank travels from A in the direction of the arrow, until it becomes vertical, when the axis will be found upon C’R, at an infinite 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