be delineated. Then, beginning with the plane through the last course, conceive the other planes to be revolved, in order, each about its intersection with the preceding one, to coincide with it, and so on till all are brought into coincidence with the fixed one. The lines of the traverse will then be situated in one plane, and a plot of them, in this position, will be the profile required. The distances from the traverse to the floor and roof of the tunnel, at different points, enable us to complete the profile.

SECTION IV.

PRACTICAL APPLICATIONS.

401. To determine the position and depth of a mine shaft, CB, Fig. 188, which is to connect with a tunuel AB.

B

500'

FIG. 188.

If the course of the tunnel is not a straight line it must be traversed. Assuming it to be a straight line, the bearing is obtained with the transit. The length of the tunnel is then accurately measured with a steel tape or a rod. The surveyor now stakes out the course on the surface, then beginning at the

400

mouth of the tunnel the first peg is driven on a level with the floor, and the measurement of the tunnel, say 1000 feet, is laid off on the course established on the surface. If the slope rod is used, the sum of the vertical measurements or readings will be the vertical distance of C above A. Should the tunnel at B be higher or lower than at A, the difference subtracted from or added to the sum of the vertical readings, will be the required depth of the shaft. If the slope-rod is not used, a line of levels must be run to find the difference of level between 4 and C.

402. To find the distance necessary to connect the main shaft GF and the tunnel EF (Fig. 189).

400'

200'

Determine the depth of the shaft DE, by suspending a chain from the surface or by a steel tape with a weight attached. Measure the length of the tunnel EH, already completed, and, with the slope rod, or otherwise, obtain the horizontal distance DI, and the difference of level between the points D and G. This difference of level added to the depth of the shaft DE the total depth of the shaft GF. Find the depth of the shaft GJ already completed. Then GF-GJ = JF, and EF— EH — HF, the required distances. In this case the tunnel is assumed to be

straight and the position of the shaft determined as in the preceding problem.

403. To find the depth of a mine shaft AB, Fig. 190, the distance from the outcrop at C and the dip of the vein CB

Draw to any suitable scale the distance CD 600 feet; from C draw the line CD, making an angle of 60° with the line CD; from the point D draw a line perpendicular to CD, until it intersects CB at B; produce it upward to A, until distance from D to A difference of elevation between C and A. Measure the line AB on the scale, which gives the required depth. Measure also CB on the scale, which is the distance from the outcrop to the point of intersection.

404. Given the depth of the shaft DE 588.5 feet, Fig. 191, and the horizontal distance FD from the outcrop to the ton

of shaft 500 feet, and the angle GFD 60°, to find the

angle, we have (Davies' Leg., Trig., Art. 37),

588:5'

If a depression or elevation from the outcrop occurs at the point, it must be subtracted from or added to the depth of the shaft; for example, if there is a depression of 50 ft. it will only be necessary to sink 538.5 ft. to the point of intersection. The horizontal distance so found, subtracted from the real distance of the shaft from the outcrop = length of cross-cut 160.22, and

FG = √DE2 + (FD — GE)2 = 679.5.

Draw the line FD = 500 feet to any convenient scale. F draw the line FG, making the angle DFG = 60°. DE = 588.5 ft., and from E draw EG parallel to FD. tances can now be taken from the scale.

From

Draw

The dis

Required

405. Given the angle LJK 60°. The distance JL 700 ft., and the difference of level between J and H = 100 ft. the depth of a shaft from H that will intersect a cross-cut of 350 ft. from the point K, Fig. 192.

From the point K draw to JL a line parallel to HI; it will intersect JL 700-350 350 ft. from the point J. This line is the perpendicular of a right-angled triangle, and we have it 350 tan 60°. Add to this the difference of elevation, which is the required depth.

Draw the line JL=700 ft., and from J draw JK, making the angle LJK = 60°. From K set off the distance KI, parallel to JL, and make it 350 ft. From I raise the perpendicular IL and extend it to H, making the distance LH = 100 ft. of the scale. The distance IH is now taken from the scale.

Problems of this kind occur only where there are inclined

shafts.