I assume that you will have a DC/DC converter of some fashion, which
will only work over a certain input range.  Make the following
calculation using the delta V from full charge to dc/dc min.

energy (joules) = 1/2 C * V * V.

energy is watt seconds (IE 1J = 1W * 1S)

You ought to get in the ballpark with that.  Don't forget the
electrolytic's internal leakage!

Chris~

"J. Delmonico" wrote:
>
> I don't have my physics books handy - but someone
> asked me if there was a way to draw a rough
> equivalence between a battery of a certain mAh rating
> and a capacitance value.  I thought it was an
> interesting question!
>
> It is a given that capactors and batteries are
> different animals .  For instance, since the capactor
> stores charge, the discharge curve (volts at a certain
> current draw versus time) would be linear, in contrast
> to the battery whose chemical reations tend to try to
> hold the voltage up until the chemistry is nearly used
> up.
>
> So the interesting calculation is how many farads per
> mAh assuming we stop when the capacitor drops to
> (say)2/3 of it's fully charged voltage?  Multi farad
> caps are getting inexpensive and fairly small....
>
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