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suicideneil
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Join Date: Feb 2007
Location: Devon, England
05.25.2011, 03:22 PM

Quote:
Originally Posted by captanabolic View Post
also would this be correct:
Divide the energy storage capacity of the battery (watthours) by the battery voltage (volts) to get Ah (Amphours). Multiply by 1000 to get mAh.

So the original battery:
300 wh x 29.6v
=10.135 x 1000
= 10,135mah

so i would need a 10000 mah 29.6v pack to equal the old power supply.

confusing lol
Unless Im trippin' that seems correct- you'd only need a couple 5,000mah 8s lipos in parallel it seems ( or 4x 4s 5000mah packs, wired up to get the same result ).

Depending on the brand and discharge ratings, you could build a suitable pack for as little as ~$400, though more like double that for a premium brand of battery say.

Would be interesting to have a watt-meter hooked up between the controller and batteyr to see what peak/ max current draw currently is...
   
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