DEVELOPMENT 79 
by which a table can be drawn up for the developer used. An 
example will make this clear. Suppose that we have a 
developer which we know from practical trial gives just the 
negative which we want with a factor of 10. Suppose that 
we make the above-described trial with a developer at 90° C. 
(48° F.) and 20° C. (68° F.), and with the lower tempera- 
ture it took 50 seconds for the first appearance of the image, 
and at the higher temperature 22 seconds; then from a table 
of logarithms we find that the log. of 50 is 1.699 and that for 
22 is 1.342, then: 
log. 50 = 1.699 
logs 22 — 1342 
0.357 
Now the difference between the two temperatures was 
20 —9 = 11, therefore 0.357 — 11 — 0.0324, which is the 
logarithm of the difference in time of development for 1° C. 
or the log. factor. As we know that the developer has a 
factor of 10, therefore, if the time of appearance at 9° C. was 
50 seconds, the total time of development will be 50 * 10 = 
500 seconds. Then, if we want to find the time of develop- 
ment for 10° C., we subtract the log. factor 0.0324 from the 
log. of 50, and multiply by the factor to find the correct time 
of development. Thus log. of 50 = 1.699 — 0.0324 — 1.6666; 
from a table of logs. we find that this is the log. of 46.4, and, 
multiplying by the factor 10, we have 464 seconds as the 
result, instead of 500 at 9 degrees. If the temperature is 
lower, then we add the log. factor; thus, assuming that the 
temperature has dropped 2 degrees to 7° C., then log. 50 = 
1.699 + (0.0324 & 2) = 1.699 + 0.0688 = 1.7638; from 
our table we find this to be the log. of 58.05, and again, using 
our factor, we have 58.05 & 10 — 580.5 seconds as the correct 
duration of development at the lower temperature. The 
