A more natural way to say it is that equality requires that the unit of length is the length of an arbitrary pendulum and the unit of time is the half-period of the same pendulum.

The pendulum is a device that relates pi to gravity.

Sounds universal. Get a different value on the Moon? Of course... pi squares differently on the moon :)

The arbitrary length pendulum with a period of 2 seconds which is your unit of length, (or 1 Catholic meter) is much shorter on the moon. In local Catholic meters gravity would be pi squared Catholic meters / second. As it would on any planet.

I think I understand. Distance scales with Planet Size? That actually makes a lot of sense.

The equation holds in imperial units as well. The length of the 2 second pendulum needs to be in feet AND the value of g in ft/sec2.

π^2 ≈ 32 to you?

Solving the equation for pi we get:

PI = sqrt(g/L)

g = 9.81. L=1

or

g = 32.174. L=3.174

Either way works to approximately pi. There is a particular length where it works out exactly to pi which is about 3.2 feet, or about 1 meter. My point was that equations like that remain true regardless of units.

The reason pi squared is approximately g is that the L required for a pendulum of 2 seconds period is approximately 1 meter.

Replace s in your calculation with imperial s instead of metric s and it isn't imperial feet per metric seconds.

Imperial seconds were very, very close to metric seconds.