132 1 XPERIMENTS ON COLOUR, AS PERCEIVED BY THE EYE. 



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 (l), 



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 



aV : aU :: '27 : '37. 



Then, joining a with EG, divide the joining line in W in the proportion of '3( 

 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 

 whatever, except in decreasing the amount of the other colour. The total in- 

 tensity of our white paper will be represented by ^ = 3*57; so that, whenever 

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

 the coefficient 3 '5 7 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 ex- 

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

 neutral 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 3'57. The result is 

 T32, which therefore measures the total intensity on both sides of the equation. 



Subtracting the intensity of '55 U + '12 EG, or '67 from T32, we obtain '65 

 as the corrected value of '3 3 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. 



