Posted by 440roadrunner on April 29, 2009 at 08:38:34 from (98.145.76.31):
In Reply to: Electric math problem posted by MarkB_MI on April 29, 2009 at 02:55:38:
MarkB_MI said: (quoted from post at 02:55:38 04/29/09)
......for low-current loads you'll probably get very close the the rated amp-hours. But for a high current load, the voltage will quickly drop off causing the current demand to increase and a further drop in voltage......
The result is that you'll be discharging the battery much faster than the rate at which it was rated, so you can't expect to get the full rated amp-hours. I'd suggest you assume you can get fifty percent of the rated amp-hours. You might do a little better, but you probably won't do any worse.
Read that again. It's important, and is the point I was trying to make. Amp hour ratings are not an exact science, and depend on how the battery was rated, the honesty of the published ratings, and the load that you are applying.
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