sp. g. = (4+5)/(0.53+1.25) = 9/1.78 = 5.06
It is necessary in this case to calculate the volumes of the galena and of the blende, which is done by dividing the weights by the sp. gravities: thus, 4 divided by 7.5 gives 0.53 and 5 divided by 4 gives 1.25.
The converse problem is a little more difficult. Given the sp. g. of a mixture and of each of the two ingredients, the percentage by weight of the heavier ingredient may be ascertained by the following rule, which is best expressed as a formula. There are three sp. gravities given; if the highest be written H, the lowest L and that of the mixture M, then:
Percentage of heavier mineral = (100×H×(M-L))/(M×(H-L))
Suppose a sample of tailings has a sp. g. of 3.0, and is made up of quartz (sp. g. 2.6) and pyrites (sp. g. 5.1): then the percentage of pyrites is 27:
(100×5.1×(3-2.6))/(3×(5.1-2.6)) = (510×0.4)/(3×2.5) = 204/7.5 = 27.2
The same problem could be solved with the help of a little algebra by the rule already given, as thus: the sp. g. of a mixture equals the sum of the weights of the constituents divided by the sum of the volumes. Then 100 grams of the tailings with x per cent. of pyrites contain 100-x per cent. of quartz. The sum of the weights is 100. The volume of the pyrites is x/5.1 and of the quartz (100-x)/2.6.
Then we have by the rule
3 = 100/((x/5.1)+(100-x)/2.6)
3 = 1326/(510-2.5x)
204 = 7.5x
and x = 27.2
If the percentage (P) and sp. g. (H) of one constituent and the sp. g. (M) of the mixture are known, the sp. g. of the other constituent may be calculated by the following formula, in which x is the required sp. g.: