5 ms·

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 fr

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.

deleted 2y ago
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eigenket 2y ago
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.
- ValentinA23 2y ago
  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).
  - eigenket 2y ago
    Sure, if you change either of the units you can always change the other one to fix the equation again.
    - koolala 2y ago
      But does it work when you use the right Imperial technique?
      - ValentinA23 2y ago
        If I come up with my own measuring unit, let's call it the sneezle (whatever the actual length I assign to it) I will be able to also define a duration unit (say, the snifflebeat) based on the time it takes for a pendulum one sneezle long to complete a full oscillation, and vice versa I can define the sneezle by adjusting the length of a pendulum so that it oscillates in two snifflebeats. Here are the maths: T = 2π√(l/g) T/2π = √(l/g) (T/2π)^2=l/g g = l/(T/2π)^2 g = l/(T^2/4π^2) = 4π^2xl/T^2 Now substitue T with 2 and l with 1 an you get g = 4π^2x1/2^2 = π^2 It doesn't matter what the pair of units assigned to T and l are. However, they'll be interrelated. There is nothing arbitrary, and no coincidences behind g =~ π^2. It just requires to do some history of metrology and some basic maths/physics. If you want to discuss coincidences, may I suggest you to comment on this remark I made and which hasn't received any attention yet ? https://news.ycombinator.com/item?id=41209612 https://news.ycombinator.com/item?id=41209612
        koolala 2y ago
        Different planets, Different Sneezle length, Same Sneezle Beat I'll try to see these units you linked thanks
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 ago
  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.
  - koolala 2y ago
    Sounds universal. Get a different value on the Moon? Of course... pi squares differently on the moon :)
    - theginger 2y ago
      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.
      - koolala 2y ago
        I think I understand. Distance scales with Planet Size? That actually makes a lot of sense.
- phkahler 2y ago
  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.
  - SAI_Peregrinus 2y ago
    π^2 ≈ 32 to you?
mannykannot 2y ago
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.

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eigenket 2y ago

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.

ValentinA23 2y ago
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).
- eigenket 2y ago
  Sure, if you change either of the units you can always change the other one to fix the equation again.
  - koolala 2y ago
    But does it work when you use the right Imperial technique?
    - ValentinA23 2y ago
      If I come up with my own measuring unit, let's call it the sneezle (whatever the actual length I assign to it) I will be able to also define a duration unit (say, the snifflebeat) based on the time it takes for a pendulum one sneezle long to complete a full oscillation, and vice versa I can define the sneezle by adjusting the length of a pendulum so that it oscillates in two snifflebeats. Here are the maths: T = 2π√(l/g) T/2π = √(l/g) (T/2π)^2=l/g g = l/(T/2π)^2 g = l/(T^2/4π^2) = 4π^2xl/T^2 Now substitue T with 2 and l with 1 an you get g = 4π^2x1/2^2 = π^2 It doesn't matter what the pair of units assigned to T and l are. However, they'll be interrelated. There is nothing arbitrary, and no coincidences behind g =~ π^2. It just requires to do some history of metrology and some basic maths/physics. If you want to discuss coincidences, may I suggest you to comment on this remark I made and which hasn't received any attention yet ? https://news.ycombinator.com/item?id=41209612 https://news.ycombinator.com/item?id=41209612
      - koolala 2y ago
        Different planets, Different Sneezle length, Same Sneezle Beat I'll try to see these units you linked thanks

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 ago
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.
- koolala 2y ago
  Sounds universal. Get a different value on the Moon? Of course... pi squares differently on the moon :)
  - theginger 2y ago
    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.
    - koolala 2y ago
      I think I understand. Distance scales with Planet Size? That actually makes a lot of sense.
phkahler 2y ago
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.
- SAI_Peregrinus 2y ago
  π^2 ≈ 32 to you?

mannykannot 2y ago

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.