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Old April 24th, 2008, 16:53   #5
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Originally Posted by gunscythe View Post
The amps were kind of guesstimates. I looked at one, but never switched the meter over for the's the wattage that I cared about.

The high voltage battery SHOULD have had higher amperage. It didn't, due to a higher internal resistance (of the battery). The bigger batteries (sub C) were able to dump more current. I'll have to try a 8 Cell Sub C pack and measure it.

Maybe some of you sparky's can elighten me on a few things. I've taken an AC/DC circuits course and can throw out all sorts of useless formulas and play with imaginary numbers and all the good stuff...just need some help understanding what they mean!

Motor draw:

From what I understand, the motors in our guns are not controlled or regulated (not sure of the correct terminology), therefore they will draw as much current as possible from the batteries:

I = V/R

Where the voltage of the battery is constant and the resistance (motor, wiring, internal, etc) of the system is assumed constant. This is why your 8.4v had a higher current draw than the 7.2v (sub C's).

This is why you see an increase in ROF with higher voltage batteries - they can simply 'push' more current resulting in more power being delivered.

P = VI
P = I^2*R
p = V^2/R

Higher mah rated batteries can also boost the ROF on a gun slightly (assuming same voltage) is this because:

1) They have less internal resistance in the cells AND/OR
2) The smaller mah rated batteries are simply 'tapped out' in their current delivering capabilities and the larger ones can 'pick up some slack'

Or maybe 2) directly correlates to 1)?


When the system has a short circuit the resistance significantly decreases (close to zero), causes the current to increase to ~infinity - bad things can ensue!

Mechbox Lockup:

This is where I'm confused. When the mechbox locks up, I see this as an increase of resistance on the system (motor). Yet this results in fuses blowing from too much current draw (or wires getting VERY hot if no fuse). Or does this somehow create some kind of short in the motor?

I hear the term 'it draws what it needs'...if this were the case, a higher voltage battery would see a drop in current, no??? But this makes sense in the case of a lockup - it draws more current to try and turnover the motor.

My brain hurts now...
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Last edited by Flatlander; April 25th, 2008 at 19:55..
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