4 ms·
No, the equality requires the length of a 2 second period pendulum be g / pi^2. Change your definition of length - that no longer holds true. g in imperial u
by usaar333 2y ago
No, the equality requires the length of a 2 second period pendulum be g / pi^2. Change your definition of length - that no longer holds true.
g in imperial units is 32 after all. g has units; pi does not
- lupire 2y agoA 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.
- koolala 2y agoSounds universal. Get a different value on the Moon? Of course... pi squares differently on the moon :)
- theginger 2y agoThe 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.
- koolala 2y agoI think I understand. Distance scales with Planet Size? That actually makes a lot of sense.
- phkahler 2y agoThe 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.
- SAI_Peregrinus 2y agoπ^2 ≈ 32 to you?
- koolala 2y agoReplace s in your calculation with imperial s instead of metric s and it isn't imperial feet per metric seconds.
- SAI_Peregrinus 2y agoImperial seconds were very, very close to metric seconds.
- phkahler 2y agoSolving 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.