PROBLEMS 271 



the full rise FC, ^d and h the difference in depth of the 

 deepest hole A and the shallowest C: 



h = j—^ r tan /S = CiF tan 5 = ^-^ ^ tan a 



cos {^d — ^2) cos (iSi — /Sd) 



Dividing we get 



Whence 



cos ((3d — 132) = I, ^ tan /? ^ J^ 

 cos (iSi — ^d) tan a m 



, ^ cos 1^2 — A; cos j8i .^p.v 



tan 13d = — —. — ^ J — -. — ^ (60) 



sm (32 — k sm jSi 



Which gives the full rise bearing from which /3d + 180 deg. 

 is the full dip bearing and the amount of dip can be got from 

 the fundamental formula (51), thus 



, , tan (8 



tan d = 



(61) 



cos {(3d — (32) 

 or 



tan a 



cos (,81 - ^d) 



First Graphical Solution. — Assume the surface survey 

 reduced to level is as shown in Fig. 180 and the stratum 

 is 800 ft. deep at A, 550 at B, and 200 at C. 



Plot triangle ACiBi from field notes and erect perpendicu- 

 lar CiC on BiCi at Ci and equal to the difference in elevation 

 between C and A. On this line measure off CE equal 

 to the difference in elevation of C and B. Draw EB 

 parallel and equal to CiBi. Connect C and B and produce 

 to meet CiBi produced in D. Join AD and draw CiF 

 perpendicular to AD; and on CiB lay off CiM = CiF. 

 Join C and M. Now CiF is the direction of dip and CMCi 

 its amount. 



Second Graphical Solution. — Lay off the surface level 

 triangle ABC or the original triangle (Fig. 181) leveled 

 to a given datum. Set off at the shallowest and deepest 

 holes, A and C, their respective depths AAi, 800 ft., and 



