Fig. 63 



Hence AB cos v = sr. In other words, the interval read must be 

 multiplied by the cos <v to get the rod reading from which FiM may 

 be deduced. Then, 



CM = FiM + (c + f) = (c + f) + K (AB cos v) 

 CCi = (c + f ) cos\ and Ci-H. I. = Fi M cos v; but 

 FiM = K ( AB cos v), therefore 

 C-H.I. = (c + f) cos v + K(AB cos v) cos v 



= (c + f) cos v + K (AB cos 2 v) 

 Let I = actual reading, or interval AB; then, 

 H=C - H. I. = K I res- v 4- (c + f ) cos v. 



P'or the vertical component the distance H. I.-M is required. 

 H. I.-M = CM sinv, and substituting the value of CM as above, 

 H. I.-M = [K (AB cos v) + (c + f)] sin v 



= K ( AB. cos v. sin v ) -f- (c -f- f ) sin v, but 

 cos v. sin v ]/2 sin 2v, .'. 

 V H.I.-M =- K. I l / 2 sin 2v + (c + f ) sin v 



110 



