Re: UPS VA ratings (was: Re: [PLUG] equipment give-away on Wed evening)

["Edward M. Corrado" <ecorrado@athena.rider.edu>]

On Wed, 8 Jan 2003, Jeff Abrahamson wrote:

> Dividing by the square root of two is more than a good approximation:
> 
>    <http://whatis.techtarget.com/definition/0,,sid9_gci213722,00.html>
> 
> To compute the root mean square of a sinusoidal wave form, we want to
> compute the square root of the mean value of the square of the
> wave. In other words, square the sin, add it up (integrate), and
> divide by the length of what we integrated before taking a square
> root.
> 
> It suffices to find the root of the mean from 0 to pi, since sin
> squared is periodic with period pi:
> 
>     sqrt ( integral(0,pi)[sin^2 x dx] / pi )
> 
> In other words, compute the area under the curve of sin squared from 0
> to pi, divide by the length to get the mean, and then take the square
> root.
> 
> The integral of sin^2 x is 1/2 x - 1/2 sin x cos x + C. The second
> term is zero at both zero and pi, and the first term is zero at
> zero. Since the constant cancels out, the integral from 0 to pi is
> pi/2. Divide by pi and we get 1/2. Take the square root.
> 
> -Jeff

Hey, I'm finally using those Calculus classes I took when I earned my BA
in Mathematics! Who ever thought it would be to read a PLUG
posting! Anyway, just to clarify, It is the (approximate) square root of
1/2 that is the 0.707 as Jeff pointed out that you multiply by (instead of
the square root of 2, but my guess is that the number 2 was just a typo in
the original post.)

Ed C.

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