498 Captain A. G. McKendrick. [Nov. 30, 
and equation (3) takes the form 
y(2—y-+ee) Cre 
or, finally, ¥(2—2)—(¥—-2) = 0. (8) 
6 \eg c 
At a point of diversion, where dz/dy = 0, we have the equations 
= CL, Y= C2) Whence, We C22 ue. (6,7, 97 
Further, when 2 is constant, dz/dy = 0 leads in a similar manner to 
equation (4) and the associated equations (6, 7, 9, 10). These last two 
results can easily be seen from the geometry alone of the figure. 
Equation (8) is the equilibrium condition of the theoretical equation (2), 
with the substitution of z in place of a and 2 in place of n. 
Equation (1) 
has now been arrived at. To prove that it is correct, it must agree with 
experimental results for all values of y, x, z, and ¢. I shall show from 
certain of these that it does so. 
If experiments be well planned, the constants ¢ and z can be obtained 
with ease. The simplest mode of procedure is to erect an experiment on 
a line « = constant, choosing a low value of w The concentration of y, 
which gives the greatest degree of lysis, fulfils the requirements of 
equations (6), (7), (9), (10), for y? = cx is the family curve of the points of 
diversion, and the tangent at the point of diversion is parallel to the y axis. 
If a small series of this sort be put up, the tube which shows the first 
flash of lysis may be taken as an approximation, and a. further series of 
tubes with concentrations differing only slightly from this may be put up 
forthwith. When the concentration of y which causes maximum lysis in 
the series has been determined, the equation y* = cx evaluates c. The 
maximum degree of lysis having been evaluated colorimetrically, the 
equation « = yz gives the value of z for that degree of lysis. 
In the experiment which I have described, another method had to be 
adopted. As I have shown, the asymptote for a constant value of 2 is 
ale—y+cz = 0, and its slope dxz/dy = z. | 
In the series (Chart 1) erected on the line «= y, complete lysis values 
are succeeded by incomplete in the upper half of the line. Thus, z for 
complete lysis is greater than 1. I chose its value at 1:25. 
In the series erected on the line « = 0-5y, partial lysis values are succeeded 
in the upper part of the line by a value “slight—partial.” Thus, z for 
partial lysis is slightly greater than 0°5. I chose its value as 0°55. 
