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 87 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 
@V-s O27 3-37 
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 0:28 of white, diluted with 0°72 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 a= 3'57 ; so that, whenever white enters 
into an equation, the number of divisions must be multiplied by the coefficient 
3°57 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 
°33 PC +°55U+'12 EG="37 SW+°63 Bk. : . «(ene 
In order to obtain the total intensity of white, we must multiply the number 
of divisions, ‘37, by the proper coefficient, which is 3°57. 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 1:32, 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 
65 pe+'55 U+'12 EG=1:32 w 
