462 Mr. J. J. Sylvester's Second Note on 



order, we may write its equation under the form 



where a is the radius of the base-circles, </> is zero at the apse, i.e. 

 the point which divides the curve into two equal and symmetrical 

 branches. Hence to find the nearest double tangent* we have 



<£ 2 -y + (<£ + 7t) 2 - 7 =0. 

 Putting <£ + — =^, this equation becomes 



IT 2 



The tailoi the secondary cyclode (say 7) is obviously the distance of 

 the apse (in respect of the centre) from the node of the parent first 



cyclode; and its length is ^ (7 — 2), the distance of the apse from 



the centre being -7. As 7 increases towards 2, ty decreases; 



that is, either double tangent tends more and more towards the 

 horizontal ; the Moorish arch therefore sinks, and the adjoining 



Tj-2 



haunches rise until when <j> = 0, i. e. 7= — , the two double tan- 



gents are in direct opposition and merge into a triple tangent 

 touching the arch at its centre. 



As y continues to decrease by <£ becoming negative, the 

 central arc sinks below the level of the adjoining branches, and 

 the double tangents slope more and more towards its extremities, 

 until at length they pass through the cusps; when this takes 

 place the tangent to the second cyclode becomes perpendicular 

 to the parent cyclode, and consequently touches the originating 

 circle so thatjo=— a, 

 a 





z I - 

 . 2 tt 2 4 a (it 2\ 2 



As 7 goes on decreasing, the double tangents quit the Moorish 

 arch altogether and connect the two infinite branches, which 

 turn their protuberances towards each other more and more, 

 until finally they touch and the double tangents coincide. 



TT IT 



This happens when </> = — — , i. e. 7 = -r* 



* For of course we may write in general 



(0 3 -7H((f +S+1 7r) s -y)e0, 

 i being any integer, and (/> will give the direction of a double tangent. 



