PT, respectively; and let O = angle SOT, O 1 = angle SOW, O 2 = angle TOW, m = tan jS 2 /
tan /J 1.- Express B 1 in terms of the known angles )SY P 2 and B
Then substitute numerical values to find the strike <f) of the vein. Thus
n w-cos0 //in,
tan' 1 = ^mT- (49)For this vein, m = tan ft/tan JS 1 = 130(180)/[220(142)] = 0.749040; 6 = 180° - (55°32' +
19°26') = 105°02'. Substituting gives tan B 1 = (0.749040 + 0.259381)70.965775; B 1 =
46
0
14'15";</>-55
0
32' + 46
0
14'15"-90
0
-ll
0
46'15";strikeofvein-Nll°46
v
15"E.- Compute the dip of the vein
Use Eq. 48, considering PS as the line of known inclination. Thus, tan a = tan /B 1 IcOS B\
a = 48°45'25". - Verify these results
Apply Eq. 48, considering PT as the line of known inclination. Thus O 2 = 9- O 1 = 105°02'
- 46°14' 15" = 58°47'45"; tan a = tan ft/cos O 2 ; a = 48°45'25". This value agrees with the
earlier computed value.
DETERMINATION OF STRIKE, DIP, AND
THICKNESS FROM TWO SKEWBOREHOLES
In Fig. 27«, A and B represent points on the earth's surface through which skew boreholes
were sunk to penetrate a vein of ore. Point B is 110 ft (33.5 m) due south of A. The data
for these boreholes are as follows. Borehole through A: surface elevation = 870 ft (265.2
m); inclination = 49°; bearing of horizontal projection = N58°30'E. The hanging wall and
footwall (lower face of vein) were struck at distances of55ft(16.8m) and 205 ft (62.5
m), respectively, measured along the borehole. Borehole through B: surface elevation =
842 ft (256.6 m); inclination = 73°; bearing of horizontal projection = S44°50'E. The
hanging wall and footwall were struck at distances of 98 ft (29.9 m) and 182 ft (55.5 m),
respectively, measured along the borehole. Determine the strike, dip, and thickness of the
vein by both graphical construction and trigonometric calculations.Calculation Procedure:- Draw horizontal projections of the boreholes
Since the vein is assumed to have uniform thickness, the hanging wall and footwall are
parallel. Two straight lines determine a plane. In the present instance, two points on the
hanging wall and two points on the footwall are given, enabling one line to be drawn in
each of two parallel planes. These planes may be located by using these principles:
a. Consider a plane P and line L parallel to each other. If through any point on P a line is
drawn parallel to L, this line lies in plane P.
b. Lines that are parallel and equal in length appear to be parallel and equal in length in
all orthographic views.
These principles afford a means of locating a second line in the hanging wall or footwall.