just to get an idea of the kind of performance to expect from a

flooded conversion.

For this I used some basic equations from zeva.com.au (great website

by the way!)

So I tallied up the internal resistance (or ohms) in all the batteries

plus the wiring losses. I left out the losses in the contactors and

whatnot since I'm assuming it would be minuscule and I don't know

where to find that info.

So, as I'm sure many of you know, to find power loss you multiply

resistance by the square of current.

So I took the internal resistance, or ohms, of each battery and

multiplied it by how many batteries there would be. In this case, I

took 12 12 volt J-150 batteries, which have an internal resistance of

0.0036, as per evconvert.com. I assumed a use of about 20 feet of

copper, which came up to 0.00154 ohms. I know that basically the max.

amount of current you can safely pull out of flooded lead-acid is 500

amps, so I used that as my current. Here's the math!

12 * 0.0036 + 0.00154 = 0.04474 ohms.

P=500*500*0.044174

So the power loss is around 11 kilowatts. The amount of power

available, without resistance, is 72 kilowatts (144votls*500amps) So

the real amount of power is 61 kilowatts. Then you have to multiply

that by the controller and motor. So it comes down to: 61*.95*.90=

52kw or about 65-70hp.

I have come to the conclusion that you can either have cheap flooded

batteries with so-so performance, pricey AGMs with great performance,

or the lithium based batteries that cost a fortune but in the long run

are the better solution.

So anyway, were my numbers right?

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