Talk:Buckingham π theorem

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I don't understand. Could you give an example? --non-mathematician

Surely the reduction is by the number of dependent variables? As it stands I don't think this explanation makes sense. David Martland 16:01 Dec 6, 2002 (UTC)

It's really a theorem about free abelian groups, though.

Charles Matthews 07:34, 17 Sep 2003 (UTC)

1 Confusion in the example
2 See Also: Use in Videogames
3 Any software to do Buckinghams Theorem?
4 Examples don't fully explain
5 Article could use better examples
6 Minor triviality: Is it π theorem or Π theorem?

[edit] Confusion in the example

The example about the period of a pendulum seems a little confused between G and g -- the universal gravitational constant of the known universe, or the acceleration due to gravity on the surface of the Earth. The period of a pendulum is independent of the mass of the pendulum, but not independent of the mass of the Earth or the radius from its center. The discussion is right except for giving the units of g: not "length cubed divided by time squared divided by mass", but "length divided by time squared".

I've gone ahead and made that change. Other minor readability things would still help here. Also sorry for forgetting a summary of the edit.

[edit] See Also: Use in Videogames

One would think such a simple yet powerful algorithm would have direct application to physics engines, despite the fact that it yields no solutions. Surprisingly, I can't seem to find any such uses. If anyone knows of such an implementation, please make the relevant changes.

[edit] Any software to do Buckinghams Theorem?

I wish there was some public-domain software available to use this theorem, as I think its wonderful to be able to magically generate a formula(s) for anything you like, but I'm not much of a mathematician.

Using the constraints of Levels of Measurement and Extensive and Intensive measurements could further constrain the formulas generated.

[edit] Examples don't fully explain

I've read through this about fifty times and I'm not clear on how $π 0$ changes to $g (π 1)$ . Shouldn't it be an addition of that $g$ value, and not a multiplication? --aciel 00:47, 1 February 2007 (UTC)

Its because

f (π 0,π 1) = 0

. Because of this, you can say

π 0 = g (π 1)

. For example, if $f(\pi_0,\pi_1)=\pi_1^2-\sqrt{\pi_0}+3$ then, because

f (π 0,π 1) = 0

it follows that $\pi_0=(\pi_1^2+3)^2$ PAR 01:28, 1 February 2007 (UTC)

[edit] Article could use better examples

Surely someone can right up a better example for the use of this theorem?

The best I can come up with off the top of my head would be matching parameters (Reynolds, Mach numbers, etc) for wind tunnel or other such scaled fluid flow modeling.

130.134.81.16 20:18, 10 July 2007 (UTC)

[edit] Minor triviality: Is it π theorem or Π theorem?

...as Π is usually for "product". Or does it matter? (to anyone?) Wikicat (temp-2k7) 14:10, 25 October 2007 (UTC)

I believe the name is taken from Buckinghams use of lower case π to represent the dimensionless parameters. So it would be lower case, not upper case. PAR 18:26, 25 October 2007 (UTC)