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

[Tom <tom.hornberger@verizon.net>]

Sorry,
Had a brain fart.  It's been a few you since tech school.  1/sqrt(2)*V (or A)  
is the right formula to find the RMS value of volts or amps.  The RMS value 
is the equivalent "DC" voltage (or current).  I.E. 120 VDC * 1 ADC = 120 W 
and 120 VAC (RMS) * 1 AAC(RMS) = 120 W in a pure resistive load.  The  actual 
usable power (Watts) drawn by an AC load is dependent on the phase angle 
(theta?) between the voltage and current.  In a pure resistive load theta (or 
power factor)=1, in any other load theta<1.  Actual AC power dissipated is V 
ac(RMS) * I ac(RMS) * Theta = W.  The effective power drawn is expressed in 
VoltAmps because the AC source has to supply the effective power whether the 
load actually dissipates that much power.  The electric company spends a lot 
of money putting capacitor banks at substations to try and raise the power 
factor to 1.0 to cut their generating costs.

Hope this wasn't too long winded.
Tom
On Wednesday 08 January 2003 09:30, Edward M. Corrado wrote:
> 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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