vals, will represent the mass from which it was taken. Taking a heap of ore, A, and selecting one out of every twenty spade-, bag-, barrow-, or wagon-fuls, according to the quantity of stuff in the heap, there is obtained a second heap, B, containing one-twentieth of the stuff of the heap A. If we crush the stuff in B until this heap contains approximately the same number of stones as A did--which means, crushing every stone in B into about twenty pieces--B will become the counterpart of A. Selecting in the same manner 5 per cent. of B, there is got a third heap, C. This alternate reduction and pulverising must be carried on until a sample of suitable size is obtained. This may be expressed very clearly thus:--
A = 1000 tons of rocks and lumpy ore. B = 50 " " rough stones, 1/20th of A. C = 2.5 " " small stones, 1/20th of B. D = 0.125 " " coarse powder, 1/20th of C.
[Illustration: FIG. 1.
PARTLY REDUCED CONE
PLAN OF FRUSTRUM DIVIDED.
ELEVATION OF FRUSTRUM DIVIDED.
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