Comment by phkahler
1 year ago
Doesn't the relationship hold if we change units? It seems like it must.
When I worked with electric water pumps I loved that power can be easily calculates from electrical, mechanical, and fluid measurements in the same way if you use the right units. VoltsAmps, torquerad/sec, pressure*flow_rate all give watts.
Nope, it completely vanishes in other units. If you do all your distance measurements in feet, for example, the value of pi is still about 3.14 but the acceleration due to gravity at the earth's surface is about 32 feet s^(-2). If you do your distance measurements in furlongs and your time measurements in hours then the acceleration due to gravity becomes about 630,000 furlongs per hour squared and pi (of course) doesn't change.
Only because you're using metric seconds instead of "imperial seconds" (the time it takes for a 1 foot long pendulum to complete a full oscillation).
Sure, if you change either of the units you can always change the other one to fix the equation again.
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This is not quite the same situation, as you are calculating a value having a dimension (that of power, or energy per second) three different ways using a single consistent system of units, and getting a result demonstrating / conforming to the conservation of energy. If you were to perform one of these calculations in British imperial units (such as from pressure in stones per square hand and rate of flow in slugs per fortnight) you would get a different numerical value (I think!) that nonetheless represents the same power expressed in different units. The article, however, is discussing a dimensionless ratio between a dimensionless constant and a physical measurement that is specific to one particular planet.
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
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 :)
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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?
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