HbG lies in the hanging wall, and this plane is therefore determined. The ensuing con-
struction parallels that in the previous calculation procedure.
- Establish a system of rectangular coordinate axes
Use A as the origin (Fig. 270). Make x the east-west axis, y the north-south axis, and jc the
vertical axis. - Apply the given data to obtain the coordinates
of the intersection points and point G
For example, with respect to F 09 y = 205 cos 49° cos 58°30'. The coordinates of G are ob-
tained by adding to the coordinates of Hb the differences between the coordinates of Fa
and Fb. The results are shown:
Point x, ft(m) y, ft (m) z,ft(m)
H 0 30.8 (9.39) 18.9 (5.76) -41.5 (-12.65)
Hb 20.2 (6.16) -130.3 (-39.72) -121.7 (-37.09)
Fa 114.7 (34.96) 70.3 (21.43) -154.7 (-47.15)
Fb 37.5 (11.43) -147.7 (-45.02) -202.0 (-61.57)
G 97.4 (29.69) 87.7 (26.73) -74.4 (-22.68)- For convenience, reproduce the plan of the intersection
points, and G
This is shown in Fig. 28a. - Locate the point S at the same elevation as G
In Fig. 28Z>, draw an elevation normal to H 0 H^ and locate the point S on this line at the
same elevation as G.
10. Establish the strike of the plane
Locate S in Fig. 28«, and draw the horizontal line SG. Since both S and G lie on the hang-
ing wall, the strike of this plane is now established.
11. Complete the graphical solution
In Fig. 28c, draw an elevation parallel to SG. The line through H 0 and Hb and that through
Fa and Fb should be parallel to each other. This drawing is an edge view of the vein, and it
presents the dip a and thickness t in their true magnitude. The graphical solution is now
completed.
12. Reproduce the plan view
For convenience, reproduce the plan of H 09 Hb, and G in Fig. 28J. Draw the horizontal
projection of the dip line, and label the angles as indicated.
13. Compute the lengths of lines HaHb and HaG
Compute these lengths as projected on each coordinate axis and as projected on a hori-
zontal plane. Use absolute values. Thus, line H 0 H 5 : Lx = 30.8 - 20.2 = 10.6 ft (3.23 m); Ly
= 18.9 - (-130.3) = 149.2 ft (45.48 m); L 2 = -41.5 - (-121.7) = 80.2 ft (24.44 m); Z,hor =
(10.6^2 + 149.2^2 )^0 -^5 = 149.6 ft (45.60 m). Line H 0 G: Lx = 97A - 30.8 = 66.6 ft (20.30 m); Ly
= 87.7 - 18.9 = 68.8 ft (20.97 m); L 2 = -41.5 - (-74.4) = 32.9 ft (10.03 m); Lhor = (66.6^2 +
68.8^2 )^0 -^5 = 95.8 ft (29.20 m).
14. Compute the bearing and inclination of lines HaHb and HaG
Let (^ 1 = bearing OfH 0 Hj,; $ 2 = bearing ofHaG; /B 1 = angle of inclination OfH 0 H 1 ,; j8 2 =
angle of inclination of H 0 G. Then tan ^ 1 = 10.6/149.2; ^ 1 = S4°04'W; tan c/> 2 = 66.6/68.8;
^ 2 = N44°04'E; tan Q 1 = 80.2/149.6; tan ft = 32.9/95.8.