240 ANNUAL REPORT 
Meyer’s Method. —There are also other methods of calculating 
cement mixtures. Meyer thus advises the use of the formula: 
x((3CaO) SiO,) +y((nCaO) SiO,R,O,), 
in which the value of n=3 to 4. This is highly commended by some Euro- 
pean chemists. This formula has also been stated as follows: aMO-+- 
bAI,O,+1S10,, in which MO is the sum of the equivalents of lime, 
magnesia, iron oxide and alkalies, a@ is equal to the minimum value, 3, 
and the maximum value 3+-b, b of course is the equivalent of alumina. 
Illustrating the application of this formula by an example* and assuming 
that we are dealing with two materials of the following composition: 
Limestone. Marl clay. 
ROJO UK CEs GPs RCE een eas rant Haniel Pon Ts 2 42.0 
PANU Daaluals ppeMen cme a Meme, Manama xc ahi4 Oe 0.5 LOE 2 
Calciumioxid eae ee ean ree 50.4 22.4 
Magiesias: Zapb ae kace. pias Barer eee eae 1.6 Soters 
HELrric OXI Ci. pernee Aaa Sea ee 4.0 
ILGSS CHM, WANTON coocccoscocncsse0d 46.3 21.4 
we obtain the equivalents by dividing each percentage by its molecular 
weight. 
If now x is the number of parts by weight for one part of clay, 
the mixture, raw or burnt, contains: 
Lime plus fluxes (=MO)o0.94 x+0.95 
Alumina 0.005.1-++0.10 
Silica 0.02 +7-+0.70 
For a cement of the lower lime limit (CaO=3510,) we have 
_-  0.947-+-0.45=3(0.02"-10.70). 
For the upper lime limit (CaO=AI,O,+3Si0,) we obtain: 
0.944-+-0.45=0.0054--0.10--0.024-0.70. 
For a medium limit we have: 
0.947-+0.45—=% (0.005.7-+0.10) +3 (0.0217-+0.70). 
Solving for 1, we find it equal to 1.94. Hence for one part of clay there 
must be used 1.94 parts of limestone. The cement contains, then: | 
2.27MO-+0.1097A1,0;-+0.7388Si0,. 
Letting SiO,=1, the formula will be: 
3.075MO-+0.15Al,0, +1Si0,, 
which is the desired composition. 
*Taken from the Tonindustrie Zeitung. 
