Battery drain time

Hi all, I have a question re battery run-down time. This is for a junior design project- I'm at OSU,ME dept. Basically a tracked/wheeled vehicle that carries a load.
I am using 2, 3.6 V motors (out of cordless screwdrivers),they have planetary gearboxes with a 68:1 ratio, 40 in-lbs Torque, 180 rpm. The mechanism will weigh between 4-6 lbs fully loaded (the uncertainty is large still, due to unfinished bin design issues). My competition run is 10 minutes long. I have rechargeable 1.2V NiCad batteries available. The asme rules allow us up to 30V. 9V is going in my RC controller, and there is also a motor on the rice bin to move a ball screw assembly to shift our Center of Mass to assist in the stair-climbing activity(the stairs are 4" high). I have probably overloaded you on information, but better too much than too little, I think. What I need to know is this- with 1.2 V batteries rated at 500 mAh, running these motors full-on for 10 min, how long can I expect them to last? I'd be satisfied with a WAY to calculate this, if no one feels like tackling it. thanks all, k wallace wallacek*at*engr*dot*orst*dot*edu
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using appropriate conversion factors, mAh * Volts = watts-hours. Test your motors under load to determine what wattage they draw. Watt-hours / watts hours.
This is a first order answer--you will be drawing power at a less than optimal point on the battery curve, so you'll need to derate your first-order answer--dividing by four or so is probably in the right ballpark.
There should be spec sheets available for your battery, which will give load curves, allowing for a more accurate answer.

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On Sun, 14 Nov 2004 18:39:48 -0800, k wallace

There are two crucial parameters mentioned above. 1) The batteries energy storage 0.5 Ah is extremely sensitive to discharge rate.
2) Running these motors 'full-on' Does this mean at rated speed or rated load, I wonder? It could be that the specified torque X speed is the rated power, but see below for the consequences of this assumption.
I am going to ignore these nuances....
Per motor: 3.6 volt using 3 X 1.2 volt 0.5 Ah cells = 3.6 X 0.5 amp X 3600 sec = 6480 joules storage capacity
40 in-lb at 180 rpm = 1 meter X 0.45 kg X 9.8 N/kg at 3 rotations/sec = 2 pi X 3 meters X 0.45 X 9.8 newtons/sec = 83 joules/sec 83 watts consumption rate at full power. (This would take at least 23 amps (!) and sounds quite unlikely )
In other words, you are probably quoting stall torque not rated power.
In ideal circumstances, the cells would last 6480/83 seconds = 78 seconds if one just used 3 in series as shown. But you specified 10 minutes or 600 seconds, which implies at the very least 24 cells in series parallel.
From the foregoing, I conclude that you don't expect to run the motors at the load you specified for very long, if at all. They would last ten minutes (about) if run at 0.5 amp X 3600 seconds / 600 seconds = 3 amps per triplet.
Do remember that I ignored derating battery capacity for fast discharge rates and I found your gear-motor rated power extremely unlikely....
Brian W
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