# Form of Buckingham Pi Function

Per https://en.wikipedia.org/wiki/Buckingham_%CF%80_theorem#Statement, the conclusion of the Buckingham Pi Theorem involves a function depending only on the so-called dimensionless "pi terms": $$F(\pi_1,\pi_2,...,\pi_p)=0$$

Question 1: Does the theorem say anything about the form of $F$? For example, if there is only one pi term, $\pi_1$, can $F$ be a complicated function like $F(\pi_1)=\sin(\pi_1)+\log(\pi_1)$?

Question 2: For a simple application of the Buckingham Pi theorem, an example using the relationship between speed, distance, and time is shown here: https://en.wikipedia.org/wiki/Buckingham_%CF%80_theorem#Speed. $pi$ is determined to be $TV/D$. Toward the end, it is concluded the form of $F(\pi)$ is $F(\pi)=C\pi=CTV/D$, or $T=CD/V$, where C is a constant. How does one know from the Buckingham Pi Theorem that $F(\pi)$ is in such a simple form?

The theorem does not say anything about the function F. It comes from an outside physical phenomenon . Buckingham $\pi$ theorem only serves to verify it.

Again it is one cited example only. By this process $F$ is not determined, but fulfills a necessary condition.
For instance an $F$ form $F( g , \omega, v, r )$ =0 is mathematically and physically permissible in:
$$\dfrac {g}{\omega ^2 } - \frac{v^2}{2 g }= 2 r$$
• Thanks Narasimham. I have two followup question. 1. For the cited example of speed, how was the author able to say $F(\pi)=C\pi$, if the Buckingham theorem does not say anything about the form? 2. Does the "dynamic equilibrium of forces" refer to Newton's 2nd law? Is Buckingham useless because there is only 1 unit (forces)? – FreshAir Jun 26 '15 at 19:55