The tangent of the bearing of VaPCs Is 



530 -U50 _ 2.330 

 <-, '20 -3,640 480 



and 



log 2,880 8.876fc8 1<>- 2,3808.8 



log 480 = 2.68124 !.. sin 7s :w =- 9.89186 



log tan 78 36' = O.K. log 2,428= 3.38523 



giving for VaPCs the bearing and length of 

 N 78 36' E, 2,428 ft. 



We now scale the map for a rough check on 

 these computed bearings and distances. 



4. COMPUTE CENTRAL ANGLES OF 

 CURVES. It is evident from the map that the 

 central angle for curve No. 1 is 



Ai - PjVl + v,V s = 63 39 ' + 45 58/ - 

 and the central angle for curve No. 2 is 



Aa - ^Vg + ^V 2 PC 8 = 45 58 ' + 78 36< = 124 34'. 



We roughly check these values by scaling the 

 map with the protractor. 



5. COMPUTE TANGENT - DISTANCES OF 

 CURVES. Using a "table of tangent-distances 

 for a 1 curve" we find the tangent-distance for 

 curve No. 1 is 



Ti = 1,355; 

 and for curve No. 2, 



Ta = 1,364. 



We obtain a rough check on these by scaling. 

 If in any case the value of A is beyond the 

 limits of the table, of course the tangent-distance 

 must be computed from the formula: 

 T = E tan A/2. 



6. COMPUTE CURVE LENGTHS. Length of 

 curve is equal to one hundred times the ratio of 

 central angle to degree of curve. 



Length of curve No. 1 is 



LI 100 Ai / DI 100 (1090 37)/6 100 (109.62) / 6 

 = 1,827. 



8 



