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Old April 23rd, 2008, 00:48   #1
gunscythe
 
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Power Measurement

I had my inline power meter out today and tried out a few full-power readings.

First off, my non-airsoft.

Stryker high-speed plane - 250Watts
Typhoon aerobatic plane - 150Watts

Now, on to airsoft.

ICS MP5 on full auto.

7.2V 3300MaH Sub C battery - 72 Watts @ approx 10 amps
8.4V 3700MaH Sub C battery - 96 Watts @ approx 11 amps
9.6V 1700MaH RIS battery - 92 Watts @ approx 9.5 amps

What I found very interesting was that the 7 cell battery outperformed the 8 cell battery. I knew it was possible - I just wasn't expecting it.

Using these figures, I could fire full-auto on the 7.2V battery for 20 minutes solid.

I tried again on my ICS M16 and got the same result. My CA M4 wasn't too much different, so I quit comparing. Robert Goulet's are 10Watts.

Food for thought.

(I use an Eflite power meter that shows Volts, Amps, and Watts. I have it setup with inline Dean's connectors. This handy device gives me instant readings for my R/C stuff so I can make easy tweaks powered on.

http://www.horizonhobby.com/Products...ProdID=EFLA110

)
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Old April 23rd, 2008, 01:07   #2
Flatlander
 
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Power = Voltage x Current

Multiply the rated voltage of the batteries by the amps you measured that they put out and you get the power output/consumed in Watts (or very close to).

This is why higher Mah rated batteries (lower voltage) can out perfom (ie. higher ROF) batteries with a higher voltage.
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Old April 23rd, 2008, 01:10   #3
mcguyver
 
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If you want a true measurement, use the voltage delivered to the motor while under load and multiply that by the current you measure. The voltage on some packs will drop below rated voltage during this test.

I'm not surprised that a sub-C pack outperformed a "AA" pack. They usually do.
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Old April 23rd, 2008, 01:12   #4
gunscythe
 
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Quote:
Originally Posted by Flatlander View Post
Power = Voltage x Current

Multiply the rated voltage of the batteries by the amps you measured that they put out and you get the power output/consumed in Watts (or very close to).

This is why higher Mah rated batteries (lower voltage) can outperfom (ie. higher ROF) batteries with a higher voltage.
The amps were kind of guesstimates. I looked at one, but never switched the meter over for the others...it'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.
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Old April 24th, 2008, 17:53   #5
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Quote:
Originally Posted by gunscythe View Post
The amps were kind of guesstimates. I looked at one, but never switched the meter over for the others...it'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)...now 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)?

Shorts:

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 20:55..
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