532 



SEISMIC METHODS 



[Chap. 9 



a:i3u = 1006 ml Xisd = 1085 m' 



X23u = 1325 m> from i4 X2Sd = 1890 m/- from B 



Xi2u = 890 mj xiu = 525 m^ 



In the calculation, steps 1 to 3 are carried out as before and give: 

 ^ = 3°; i = 36°50'; Vj = 3000 m-sec"'; Huc = 250 m; hu = 119 m. 



(4) 



(5) 



— = sin (|S - ^) = ^^^ = 0.412 = sin 24°20'; )3 = 27°20' 

 V3„ 4370 



- = sin (a + v?) = ^ = 0.383 = sin 22''30'; a = 19''30' 

 V3d 4700 



^^ = sin fe + (iA - <p)] = ^:^ = 0.557 = sin 33*50' 



sinti 

 sin jS 



0.6 



^r = sin [ii - (^ - ,p)] = "^ = 0.764 = sin 49*50' 

 sm ii 0.6 



,/, - ^ = -8*; i2 = 41°50' 



(6) ^ = -5= 



/»\ Vj 3000 . -rtrt — 1 



(7) V3 = ^-^ = „-^rT7-„ = 4500m-sec 



(8a) //2« = 



sin ii 0.666 



a;i3i/ [1 - sin (^ - ,p)] - Hi [cos (a + ^) + cos (/J - ^)] 



2 sin ii cos 12 cos ^ 

 2 sin ti cos 12 cos ^ = 0.891 



cos (a + <p) + cos (^ — (f) = 1.835 

 1 



H2U — 



(86) H2„ = 



0.891 



[1006.0.588 - 458] = 150 m 



X23a [sin {ii — if>) — sin (jS — ^)] - /? i« [cos {a-\- ^) -\- cos 03 - (p)] 



2 sin t'l cos t2 cos ^ 



1 

 0891 



+ Hiu • 2 cos t'l cos ip 

 2 sin ii cos t2 cos ^ 



[1325-0.145 - 458 + 250.2.0.8.0.997] = 149 m. 



