280 MR J. CLERK MAXWELL ON COLOUR, 



tional to the several components of the colour placed at their respective angular 

 points, and taking the centre of gravity of the three masses. In this way, each 

 colour will indicate by its position the proportions of the elements of which it is 

 composed. The total intensity of the colour is to be measured by the whole 

 number of divisions of V, U, and EG, of which it is composed. This may be 

 indicated by a number or coefficient appended to the name of the colour, by 

 which the number of divisions it occupies must be multiplied to obtain its mass 

 in calculating the results of new combinations. 



This will be best explained by an example on the diagram (No. 1). We have, 

 by experiment (1), 



•37 V + -27 U + -36 EG - -28 SW + -72 Bk 



To find the position of the resultant neutral tint, we must conceive a mass 

 of -37 at V, of -27 at U, and of 36 at EG, and find the centre of gravity. This 

 may be done by taking the line UV, and dividing it in the proportion of -37 to -27 

 at the point a, where 



a V : a U : : 27 : 37 



Then, joining a with EG, divide the joining line in W in the proportion of -36 to 

 (•37 + -27), W will be the position of the neutral tint required, which is not white, 

 but 028 of white, diluted with 072 of black, which has hardly any effect what- 

 ever, except in decreasing the amount of the other colour. The total intensity of 

 our white paper will be represented by ^=357 ; so that, whenever white enters 

 into an equation, the number of divisions must be multiplied by the coefficient 

 357 before any true results can be obtained. 



We may take, as the next example, the method of representing the relation 

 of pale chrome to the standard colours on our diagram, by making use of experi- 

 ment (2), in which pale chrome, ultramarine, and emerald green, produced a neu- 

 tral gray. The resulting equation was 



•33PC + -55U + -12EG = -37SW + -63Bk .... (2). 



In order to obtain the total intensity of white, we must multiply the number 

 of divisions, 37, by the proper coefficient, which is 357. The result is 1*32, which 

 therefore measures the total intensity on both sides of the equation. 



Subtracting the intensity of 55 U+12 EG, or -67 from 132, we obtain -65 as 

 the corrected value of "32 PC. It will be convenient to use these corrected values 

 of the different colours, taking care to distinguish them by small initials instead 

 of capitals. 



Equation (2) then becomes 



•65pc + -55U + -12EG = l-32w 



