Talk:Buckingham π theorem

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

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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)

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

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

Confusion in the example

I removed the example about the atomic bomb because it is not correct: Taylor did not use dimensional analysis to make his estimates. The example could be reinstated if it is rewritten to say that it is possible to arrive at a result similar to Taylor's using dimensional analysis.

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.

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.

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.

Examples don't fully explain

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

Its because $f(\pi _{0},\pi _{1})=0$. Because of this, you can say $\pi _{0}=g(\pi _{1})$. For example, if $f(\pi _{0},\pi _{1})=\pi _{1}^{2}-{\sqrt {\pi _{0}}}+3$ then, because $f(\pi _{0},\pi _{1})=0$ it follows that $\pi _{0}=(\pi _{1}^{2}+3)^{2}$ PAR 01:28, 1 February 2007 (UTC)

The g isn't a value, anyway, it symbolizes a function.

If this helps, choose a constant value for $\pi _{1}$ so then you can think of $f(\pi _{0},\pi _{1})=0$ (for all values of $\pi _{0}$) as defining a new function (which we may name $h_{\pi _{1}}$) such that $h_{\pi _{1}}(\pi _{0})=0$. Then simply invert it to see $\pi _{0}=h_{\pi _{1}}^{-1}(0)$. If you consider this proceedure for various values of $\pi _{1}$ you are basically saying that $\pi _{0}=g({\pi _{1}},0)$ or more importantly $\pi _{0}=g_{0}({\pi _{1}})$. An alternative approach is to define the function $g_{0}$ in the following way: for any single particular value of $\pi _{1}$, consider every possible value of $x$ and try inserting each of those values into $f(x,\pi _{1})$ until you find the one that gives you zero, then define "$g_{0}(\pi _{1})$" as that special value found for x. At least this ensures $f(g_{0}(\pi _{1}),\pi _{1})=0$. Provided that there is only exactly one possible value that works for $g_{0}(\pi _{1})$, this proves that in those circumstances $\pi _{0}$ would also share the same value. Cesiumfrog (talk) 23:54, 4 November 2010 (UTC)

I am inclined to agree with the original concerned poster...this aspect of the example is not as clear as it could be for the typical reader. It omits some intermediate steps that I think most won't immediately understand without some explanation. Without it, this example doesn't make sense to someone who simply wants to understand how to use the Buckingham $\pi$ methodology. This could be improved if a "qualified" person expands this part of the page. --Lacomj (talk) 23:22, 14 May 2011 (UTC)

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)

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)

"Proof" is terrible

The so-called "proof" is nothing but an explanation of how to treat dimensions as a vector space. It doesn't actually prove the theorem, even informally. Halberdo (talk) 09:15, 11 March 2009 (UTC)

Agreed. That "proof" is not a proof at all, most of all not formal. There is nothing said, for example, about the rank of the dimensional matrix. Besides, the theorem, that's at least my opinion, is wrongly stated. It's the something like the maximum number of linear independent dimensionless terms that the PI-Theorem makes a statement about, if I'm not mistaken. Added confusing template ManDay (talk) 14:13, 21 January 2011 (UTC)

Math notation

This article contained this:

(p=n-k)-dimensional

I changed it to this:

(p = n − k)-dimensional

Variables should be italicized; parentheses and digits should not. A minus sign differs from a hyphen. Proper spacing should be used.

See WP:MOSMATH. 138.192.56.24 (talk) 00:55, 6 May 2011 (UTC)