Why dimensional analysis is never off by more that $(2\pi)^{(\pm1)}$?

I've been reading about dimensions analysis and at one point it mentions that there could be constants that dimensional analysis fails to define and dimensional analysis is never off by more that $(2\pi)^{\pm1}$? why the factor $(2\pi)^{\pm1}$? how does this come about?

because people are saying this is untrue I thought I should link the resource: what I'm reading is here: http://www.physics.drexel.edu/~bob/Chapters/dimensional3.pdf

1 line above example 1b

• This is not true. Your source is probably writing about a specific problem/topic. – valerio Jan 20 '18 at 14:54
• That's totally wrong! You should throw that book in the trash. – knzhou Jan 20 '18 at 14:57
• Factor $2\pi$ is a difference between linear and angular frequency. Or between wavelength and inverse of wavenumber. Maybe thats where this came from. But yes, the statement is too bold. – A.V.S. Jan 20 '18 at 15:02
• The quote is specific to that example, where it's using a pendulum - which relates to the linear vs. angular frequency thing. But note that dimensional analysis normally just tells you dimensions and you're guessing the form based on that. The errors can be arbitrarily large in doing that if you guess the form completely wrong, even if your guess has the right dimensions. – StephenG Jan 20 '18 at 15:16

Counterexample: what is the surface area of a sphere? Dimensional analysis says $r^2$, which is off by $4\pi$.
• I thought dimensional analysis says it goes like $d^2$, which is only off by $\pi$. – JEB Jan 20 '18 at 18:35