ON PADDLE WHEELS. 



151 



6-25 



We have R = 9-25 ft., a = 3'75, cos. = ^ , k = 6-25, cos. /3 = - r , and, since p = a, 



9-25 



we have n = 30-439. 

 Log. 



55 



R = 0-96614 

 a = 0-57403" 

 a = 0-06197 

 sin. a =1-96101 



Substituting these values, we find, 



Log. 4 R 3 sin. a = 3-46149 

 11 Ra?sm. a= 3-11660 

 6 R? a /3 = 3-20297 

 3 R 2 k /3 = 3-12379 

 6 a 2 k /3 = 2-64060 

 fl/Psin. /3 = 2-42548 



Log. 9 # 2 a a = 3-52252 



6 a 3 a = 2-56221 



4 I? 3 sin. /3 = 3-36806 



6 R a 2 sin. /3 = 2*75993 



6 Raksin./3= 2-98178 



9-25 



Log. A = 0-79588 



/3 =-1-91851 



sin. ,8 =-1-86758 



4 fl 3 sin. a 

 1 1 .R a 2 sin. a. 

 6.R 2 a/3 

 3/Z a /t/3 

 6 a 2 k 13 

 R /t 2 sin. /3 



2893-9 



1308-0 



1595-8 



1329-8 



437-1 



266-4 



Sum of positive terms = 7831-0 



9 R* aa. 



6 a 3 a 

 4 R 3 sin. /3 

 6 1? a 2 sin. /3 

 6 R a k sin. /3 



Sum of negative terms 

 Sum of positive terms 



3330-6 

 364-9 



2333-8 

 575-4 



958-8 



7563-5 

 7831-0 



267-5 



b -1-94611 



b = 0-8833 ft. = 10-6 inches. 



The other question to be solved is, what is the power necessary to produce the required 

 effect. This is found by substituting, in formula (1), the value of b just obtained, as well as 

 those of all the other quantities which enter into it. This operation is in substance as 

 follows : 



