80 Mr HOLDITCH, ON ROLLING CURVES. 



that if nO = 0, r = J, 



4\/2. ab 



IT 



12' '" «.(2v/2 + VS + 1) + ft. (2^2 - v^ - 1)' 



2tt 4aft 



/* = 



12 ' ~ 2.(a + ft) + (o — ft) Vs' 

 2\/2.«ft 



>7T 



r = 



12 ' ' « . (i + v'a) + ft . (\/2 - l) ' 



4nr 4aft 



r = 



12 ' 3« + ft' 



5tt 4\/2.aft 



r = 



12' " 0.(2^2 + \/3 - 1) + ft. (2\/2 + 1 -VV 



6tt 2aft 



/• = 



12 ' « + ft' 



7tt 4\/2.aft 



r = 



12' " o.(2^/2 + 1 - y§) + ft.(2\/2 + \/3 - 1)' 



8tt 4aft 



12 ' r " a + 3ft' 



9ir 2\/2.aft 



r = 



12' " «.(V2- 1) + ft.(\/2 + 1)' 



IOtt 4>ab 



r = 



12 ' " 2. (a + ft)- (a-ft)-\/3' 



11 7T 4\/2.aft 



r = 



12 ' " a . (2 -v/2 - ^3 - 1) + ft . (2 y/2 + Vs + 1) ' 



7r, r = a. 



Hence the following rule : Describe the circle whose radius is the 

 minor distance, and divide it into n equal parts, each of which will 

 form the base of a lobe ; divide half the base into twelve equal parts, 

 and draw straight lines from the centre, through the points of division, 

 respectively equal to the above values: and the curve drawn through 

 their extremities will be the outline of half a lobe (fig. 20). 



