ON PADDLE WHEELS. 135 



and find 0'561 and 0*562, of which we take the mean, 0-5615. The difference between 0-5455 

 and 0*5615, or 0'016, corresponding to a difference of 0*5 in the depth of the float,, we find 

 the difference corresponding to 0'38 by the proportion 0'5 : 0'016 : : 0*38 : 0*012, sufficiently 

 near for our purpose, and this added to 0*5455 gives 0'5575 for the proportion of the depth of 

 the float contained between its upper edge and the centre of pressure : the distance of this 

 point from the axis of the wheel, or y, is therefore equal to 9*3937 ft. 



Log. a = 0-6989700 

 p = 0-8813691 

 y 0-9728367 



6 = 0-0041126 



sin. = - 1-9276647 ' 

 Whence, 



Log. 2 y 3 6 = 3-2236527 2 y 3 6 = 1673-60 



(? y 6 = 2-7396875 p 2 y = 549-15 



ap 2 sin. 6 = 2-3893729 ap 2 sin. 6 = 245-12 



2467-87 

 ,,4py 2 sin. 6= 3-3567672 4pfs\n. 6= 2273-88 



S" = 193-99 



Log. S" = 2-2877771 



7T 2 = 0-9942998 



n 3 = 3-5282738 



b = 0-9420081 



/ 0-3979400 



w = 1-8070274 



m 1-2041200 



27 = - 2-1912490 



-148^000 = - 8-8282735 



Log. H.P. = 2*1809687 



H.P. = 151*69. 



The calculation of the effective power would be abridged in the same manner by means of 

 an analogous point, which we would call the centre of propelling effect. This point is so 

 situated, that, if the whole surface of the float were concentrated there, the propelling effect 

 would remain the same ; so that, calling z the distance of this point from the axis of the 

 wheel, and the inclination of the float when its centre of propelling effect enters and leaves 

 the water, the propelling effect of a pair of wheels may be found in the following manner : 

 We know the total pressure on one of the floats at the angle $ to be equal to 



7T 2 n 2 bfw I ^ 



900 x 2ff 



> 



