Assuming the bounding line of the area of available ore to be at a distance west of the shaft-- 100 + (350-100) = 225 feet --------- 2 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. Inclined shaft 665 × 10 × 6 = 39,900 Level I. 350 × 7 × 6 = 14,700 ” II. 260 × 7 × 6 = 10,920 ” III. 520 × 7 × 6 = 21,840 ” IV. 240 × 7 × 6 = 10,080 ” V. 345 × 7 × 6 = 14,490 ” VI. 500 × 7 × 6 = 21,000 ” I. Stoped east (rough estimate) 3,400 ” I. ” west 6,500 ” II. ” west 7,000 ” III. ” east 20,000 ” VI. ” west 12,000 -------- Total 181,530 Or in round numbers 182,000 This quantity, deducted from 1,866,750 cubic feet, leave 1,684,750 cubic feet. Divided by 13½, the number of cubic feet required for a ton, this gives 124,797 tons of ore in sight. The quantity of ore discovered in a mine may be estimated from its specific gravity and the average size of the vein. The specific gravity of the ore, with that of water taken at 1000 for standard is equal to the number of ounces in a cubic foot. Great caution is necessary to determine the proportion of the vein which may be considered solid ore. A vein 6 feet square and 1 inch thick, contains 3 cubic feet, therefore, in order to find the number of cubic feet per square fathom of a vein, it is merely necessary to multiply the thickness in inches by three.