Talk:Buckingham π theorem

message: 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)
thread_title: Untitled

message: 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.    — Preceding unsigned comment added by 82.83.58.235 (talk) 19:11, 1 December 2011 (UTC)
thread_title:  Confusion in the example 

message: 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)
replies: text: 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)

text: 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)
thread_title:  Examples don't fully explain 

message: 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)
thread_title: Article could use better examples

message: ...as Π is usually for "product".  Or does it matter? (to anyone?) Wikicat (temp-2k7) 14:10, 25 October 2007 (UTC)
thread_title:  Minor triviality: Is it π theorem or Π theorem? 

message: 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)
replies: text: Agreed. The "proof" is only a statement about dimensions. there's not even a hint of why this matters at all, or why this could help with scaling. 2803:9800:9504:7B33:D973:7310:557A:AC54 (talk) 02:27, 17 May 2022 (UTC)
thread_title:  "Proof" is terrible 

message: This article contained this:
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: (p=n-k)-dimensional
I changed it to this:
: (p = n − k)-dimensional
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Variables should be italicized; parentheses and digits should not.  A minus sign differs from a hyphen.  Proper spacing should be used.
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See WP:MOSMATH. 138.192.56.24 (talk) 00:55, 6 May 2011 (UTC)
thread_title:  Math notation