[*Math: (\frac{\pi d^2}{4})]

multiplied by the velocity, therefore the velocity in feet per
second = 4/(pi d^2) x 2497/60 = 2497/(15 pi d^2) and the
formula then becomes

2497/(15 pi d^2) = 124 x * 3rd_root(d^2)/3rd_root(4^3*) x
sqrt(1)/sqrt(300)

or d^2 x 3rd_root(d^2) = 3rd_root(d^6) = (2497 x 3rd_root(16) x
sqrt(300)) / (124 x 15 x 3.14159*)

or (8 x log d)/3 = log 2497 + (1/3 x log 16) + (* x log 300) -
log 124 - log 15 - log 3.14159;

or log d = 3/8 (3.397419 + 0.401373 + 1.238561 - 2.093422 -
1.176091 - 0.497150) = 3/8 (1.270690) = 0.476509.

* d = 2.9958* feet = 35.9496, say 36 inches.

As it happens, this could have been obtained direct from the
tables where the discharge of a 36 in pipe at a gradient of 1
in 300 = 2,506 cubic feet per minute, as against 2,497 cubic
feet required, but the above shows the method of working when
the figures in the tables do not agree with those relating to
the particular case in hand.

This result differs somewhat from the one previously obtained,
but there remains a third method, which we can now make trial