ON PADDLE WHEELS. 149 



The general expression of the moment of resistance due to one of the floats is therefore 



P = ^ Rn3wb { rg pa + 3 6 ft 02] a _ r- 39 jR2 a + 6 a s] sin a 

 648000 x 2ffl l 



-[9R 3 + 12 Ra* + 24 R ak~\ /3 + [32 R* a + 7 R 1 k+I2 a* k 8ak*+2 k 3 ] sin. j, 



and the number of horse powers exerted through the medium of the two wheels with m floats 

 on each, 



H. P. = -^-^^A- ( [9 /? 3 + 36 /{ a 2 ] - [39 J? a + 6 a 3 ] sin. a 

 178200000 x 2ff\ L 



The effective pressure being in this wheel equal to the total pressure on the floats, or p, 

 and the moment of effective resistance being equal to the product of this quantity by v or 



^-?, the moment of effective resistance due to one of the floats in any given position, its 



upper edge not being immersed, is equal to 



Tr 3 a n 3 w b 



27000 x 2ff 

 the mean value of which is equal to 



[R cos. <j> a] 3 , 



a 

 1 a n 3 w b 



27000 x 2ff< 

 o 



/[TZcos. <f> a] 3 .d<j>, 



or 



! n 3 w b 



162000 x 2ff 



The quantity to be deducted for the over-immersion of the float is 



ft 



/[R cos. < a] 2 [_R cos. < A] . d <f>, 



a w 3 w b 



27000 x 2 ff, 

 o 



or 



Tr 2 a n 3 w b 



162000 x 2ff 

 so that the general expression of the moment of effective resistance due to one of the floats is 



E.P.= ~- n 3 wb_f - [9 fl 2 a + 6 a 3 ] + [4 E 3 + 11 jRa 2 ] sin. a 

 162000 x 2ff \ 



sin./9J; 



and the number of horse powers effectively employed, or the propelling effect of the two 

 wheels, 



H. P. E. = ** n * W b m ( - [9 R* a + 6 a 3 ] a + [4 R 3 + 11 R a 2 ] sin. 

 44550000 x2^l 



[4 R 3 + 6Ra? + 6Ra k-RP~\ sin. . . (2). 

 To find the formula for the centre of pressure, let x be the distance of that point from the 



