1901 - 2 .] Prof. Chrystal on Trisector and Quartic Trisectrix. 19 
given eccentricity. In fig. 5, 0 and K are fixed centres. 0 A = 
LU = LM = 6; OL = AU = AM = £&; AP = c; K N = b‘ AN = OK 
= 2b. Hence the two contra-parallelograms 0 L M A, OANK, are 
similar, so that the angles A 0 K and A 0 L are equal. It follows 
that P traces the Limagon r=2c cosO + b, whose eccentricity is 
2c/6. We therefore get a hyperbolic Limagon, the Cardioid, or 
an elliptic Limagon, according as c> = <\b. The trisectrix is 
obtained by taking c = b. 
Other link-motions have been given for tracing the Limagon ; in 
particular, a five-bar one by Mr Hart ( Proc . Lond. Math. Soc., vol. 
vi. p. 138); hut they have not the advantage of being so readily 
adjustable as the above so as to give Limagons of varying eccen- 
tricity. A working model was shown to the Society. 
Finally, we may remark that the quartic trisectrix can be 
generated as an epitrochoid, the radius of the fixed and rolling 
circles being equal, and the distance of the tracing point from the 
centre of the rolling circle equal to the diameter. This property 
gives another simple mechanical process for tracing the curve. 
