Let it be required to calculate the ore reserves in a mine opened up on a vein with a mean cross section of 6 feet; a cubic foot of the vein matter in place weighing 150 lb. The ore stopes are generally very irregular. In this case, however, it may be supposed that the stope faces are 11 feet apart and 8 feet high. There is an inclined shaft, 10 feet by 6 feet, following the dip of the vein, and six levels, each 7 feet by 6 feet, 100 feet apart. The lengths of the levels are—

I. 200 feet west 150 feet east.
II. 160 feet west 100 feet east.
III. 120 feet west 400 feet east.
IV. 100 feet west 140 feet east.
V. 165 feet west 180 feet east.
VI. 350 feet west 150 feet east.

The longest level west is 350 feet, and the shortest 100 feet.

Assuming the bounding line of the area of available ore to be at a distance west of the shaft—

If the longest level east is 400 feet, and the shortest 100 feet, the bounding line in this direction, calculated in a similar way, will be at a distance of 250 feet from the shaft.

The inclined shaft has opened up the vein for 670 feet. Deducting, say, 15 feet for the irregularity of the surface, the quantity of ore in sight will be a rectangular block 655 feet deep, 225 + 250, or 475 feet long and 6 feet wide, that is 1,866,750 cubic feet.

From this quantity, however, must be deducted the quantity of ore extracted, namely:—

Cubic Feet
Inclinedshaft 665x10x6=39,900
LevelI. 350x7x6=14,700
LevelII. 260x7x6=10,920
LevelIII. 520x7x6=21,840
LevelIV. 240x7x6=10,080
LevelV. 345x7x6=14,490
LevelVI. 500x7x6=21,000
LevelI. Stoppedeast(roughestimate) 3,400
LevelI. ” ”west 6,500
LevelII. ” ”west 7,000
LevelIII. ” ”east 20,000
LevelVI. ” ”west 12,000
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