130 C. J. MERFIELD. 



From Table I appended to this paper we may determine 

 x/B 2 with the argument <£, then h/B 2 can be found from 

 the table of Vol. xxxiv. 



The arc TC = [8*2418774] Bit* 



To fix the point "K" opposite T we have 

 TK = (B*-Ri) sin <£. x = TK + R x sin <£ = i? 2 sin <f>. 



Tables. 

 Table I gives the values of x/B with the argument </>. 

 This table has been prepared from the equation 5a. 



Table II contains <f> with the argument h/B. Equation 

 (2) has been used in its preparation. 



In both tables the numerical values of the derivatives & 

 F'(t) are tabulated and facilitate interpolation. We have 

 for a given value of h/B 



*n=* .+ n{o,Fl(t)} +-y |.o>Fi(*)-«J^(t)j 8 



or we may write 



<^=Vi-(l-w) [»>;<*)} + ( ' 1 ^^ {^K(fc)-o,F>)] ...9 



If we write h/B in place of </>, then the formulas apply 

 equally well for the interpolation from Table II, they are 

 general. 



The notation of these formulae will be understood from 

 the following example of their application. 



Example. 

 Find the angle <£ from Table II when h/B equals 0*0109428. 



Therefore n = 0*428 



a, F'(t) = + 7 17"*9. co F[(t) = + 7' 21"*5 



i{«Fi(t) - coF^(t)} = + r-s 



and 

 4> n = 17° 7 6"*7 + (0*428) x 7' 17 ,, *9 + (0*428) 2 x 1"*8. 

 = 17°14'26"*4- (0*572) X 7' 21"*5 + (0*572) 2 x 1"*8. 

 = 17° 10' 14 "'4 



