66 ANALYSIS OF MILK. 



multiple of tlie number of grammes per 100 c.c. ; for if we suppose 

 that the substance A is water, Sa will equal 1, and equation (2) 

 can then be written — 



S = 1 + BS X Kb. 



which is practically equation (3). 



Formulae for Calculations. — In order to deduce a formula 

 expressing the relation between specific gravity and percentage 

 by weight of fat and solids not fat, let us call the specific gravity 

 (for convenience) 1 + S, the percentage by weight of fat F, and 

 of solids not fat N. 



Then the number of grammes of fat per 100 c.c. will be 

 F X (1 + S) and of solids not fat N x (1 + S). 



The weight of the water in 100 c.c. is then 



100 X (1 + S) - N X a + S) - F X (1 + S) grammes ; 

 and its volume 100 x. (1 + S) - N x (1 + S) - F x (1 + S) c.c. 



The volume of fat and solids not fat in 100 c.c. is therefore 



100 - (100 X (1 + S) - N" X a + S) - F X (1 + S) ) CO. 



which equals 



N X (1 + S) i- F X (1 + S) - 100 S . (4) 



Let us assume that the specific gravity of fit is / and of solids 

 not fat n. 



Then the volume of fiit in 100 c.c. is ^ ' and of solids 



^ , N X (1 + S) , 



not lat : therefore 



11 



Nx (1 + S) + F X (1 +S) - 100S = gjii^-tJl + N X (1 +S) 



/ n 



or 100 S = N X (1 + S) - -^:^1-+S) + J, , ,1 + s) - ^-^l±^ 



or 100S = Nx(l + S) (^)+Fx(l + S)('^y 

 Now, as n and / are constant, we may write fo: 

 and for (^^\ h. 



Then the equation stands 



100 8 ^^ , ^ 



l-fS""" N + 6 F (5) 



It is usual, however, to estimate total solids (T) and fat in 

 an analysis. 



T = iSr + F, and, therefore, N = T - F. 



