284 C. J. MERFIELD. 



Practical example. — Let it be required to find the constants for 

 a transition curve when 



x c = 8 xJR = 0-32 

 .# = 25 2 logR = 2-795880 



The argument of the table being x c /R we find opposite 0*32 the 

 following quantities with which are written the operations required 

 to reduce them to the values for R. 



<jf> = 9° 27'54-"4 

 Logm + 2 log i? = 9-734558 

 2 log R = 2-795880 

 logm = 6-938678 



x' = 0-164447 xR 



y c = 0-017783 xR 



h = 0-004169 xR 



s = 0-320886 xR. 



The value of y at any other point may be found from the equation 



y = mx 3 



I x 

 Or y = y c x[ 



x c 



Remarks. — It is not necessary that R be integral, any value 

 may be used, also any unit of measurement may be adopted for R, 

 feet, chains, metres, etc., etc. To avoid interpolation in general 

 cases, always take x c so that when divided by R the result will 

 give two significant figures, thus if it be desired to apply a tran- 

 sition when x equals about 5 say, and R equals 9, it will be better 

 to take 



x c = 4-95 or x c = 5-04 

 so that the argument will be either 0*55 or 0-56 respectively. 



