STATE GEOLOGIST. 239 
| Tensile strength 
Formula employed. 1:3, 28 days. Boiling test. 
Pounds per sq. in. 
@.OCXOVSO ; GCHOWN Oyo ccsacosssaease: 290 Good. 
(@.SCxONSO, (ACHOWN Ons.csoceccnacconne 360 Good. 
(OQ FCHOVSO,, COHOVM Ons dscnecosssdene: 225 Good, off glass 
Five briquettes were used in each test. This work was repeated 
with practically the same results by two students. 
Examples of Cement Batch Calculation.—To illustrate the method 
of calculating the composition of a cement mixture, the following example 
is worked out: ; 
Given the clay mentioned above: 
68.82 % silica, 
19.20 % alumina and iron, 
2.69 % calcium oxide, 
2.46 % magnesium oxide. 
and the limestone: 
91.80 % calcium carbonate, 
3.52 % silica, 
3.51 % alumina and iron. 
The formula (2.8CaO)Si0,, (2CaO)AI,O, requires for every part 
by weight of silica 4.66 parts of calcium carbonate, and for every part 
of alumina two parts of the carbonate. The clay requires, therefore, 
68.82 x 4.66=320.7 parts, 
1O)2O X2.CO= Bowl, VAIS. 
359.1 parts calcium carbonate. 
But as the clay contains 2.69 per cent. calcium oxide, we must 
deduct this amount, as the carbonate, and we obtain: 359.1—( 2.69 X (100 
--56)=354.3 parts of calcium carbonate required. 
On the other hand, the limestone contains silica and alumina, which 
‘take up some of the calcium present, so that the amount available is 
91.8—(3.52X4.66+3.5xX2)=68.4% of calcium carbonate. The amount 
of limestone to be used for one part of the clay is therefore 354.3—68.4—= 
5.179 parts, or 100’parts of clay require 518 parts of limestone. In this 
calculation the ferric oxide has been taken together with the alumina, as the 
difference due to allowing for the iron is negligible. In practical work 
the calculation is simplified by the use of logarithms, curves, or tables 
analogous to the slag tables used in blast furnace work. 
