Electric "power chair"

I suspect you're going to end up with a heavy 24 V DC power supply and as many soup-can capacitors as you can scrounge. :)

You should be able to Google up the battery part numbers to get a rough idea of what they can put out on a short-time basis.

For the low-dollar approach, All has a 100 A shunt for $13.50 and a

200 A shunt for $16. Use the voltmeter you already have to measure the drop across the shunt (0 to 50 mV). (Or, they sell the panel meters that go with these shunts.)

For the direct approach, VDO has car ammeters that might do what you want. Their "190 105" does +/- 100 A without a shunt and costs about $30 when ordered at your local O'Reilly Auto. Their "190 151" does

+/- 150 A but needs a shunt; the ammeter is $82 and the shunt is $56 at Jegs.

If the existing batteries are beat and you think they are limiting the current, perhaps use a couple of car batteries in series as a test. If you have two cars, you're set... if not, scrounge one from your car and hit up the junkyard for a used battery for $25 or $30. (It's worth $5 to $10 at the scrap yard when you get done shoving electrons with it.)

Possibly related: All is selling a DC-DC converter brick for $30 that puts out 24 V, 12.6 A. The usual suspects don't stock it but they will order one for $280 (!). The input range is conveniently exactly what you'd get if you slapped a bridge rectifier across the AC line. These can be paralleled for higher output.

Standard disclaimers apply: I don't get money or other consideration from any companies mentioned.

Matt Roberds

If its for indoor use on flat floors, you don't necessarily need to supply what the motors will take. Going into current limit reduces motor torque... at least as long as the controller keeps working.

Also no need to buy a shunt, a known bit of copper wire will do that. Calculate R from its dimensions.

NT

Yes. But, I imagine it will weigh *less* than a pair of (large) batteries AND will "work" after sitting in storage for an indefinite amount of time (whereas batteries would undoubtedly lose their charge and/or ability to *hold* a charge!)

I suspect the motors would be very tolerant of large amounts of ripple. OTOH, the translator/driver may have smarts in it that "forget what they were doing" between (poorly filtered) waveform peaks! :-/

I'm sure the chair uses much less than the batteries can deliver. E.g., the "translator" (motor driver?) has a 40A/motor limit. And, there's a 30A breaker in series with the battery pack.

What I was more interested in was shirt-cuff guesstimates as to the amount of "power" required to move a person-sized mass at the speeds and accelerations typically expected of a powered chair.

Or, first hand experience folks may have had repurposing such chairs.

E.g., the gear box seems very inefficient. But, putting a number on that (without empirical measurements) would just be a SWAG.

If you have line power available in storage, keep the batteries on a low-rate charger there. The charger that came with the chair (if available) may or may not be good at this. If the only charger you can get is "too big" (too much current output), you can cut the rate down a lot by low-frequency PWM with a $5 light timer from the hardware store.

Also, when the user slows or stops the chair, the motor controllers

*may* be attempting to regenerate into the battery. Many switching power supplies don't like this too much.

Google the breaker and see what its curves are like. A 30 A breaker might allow 60 A or more for a short time. This is easy for a battery and not as easy for most power supplies.

Assume 100 kg human, adjust as needed

Assume 50 kg chair, adjust as needed

Assume chair acceleratation of 1 m/s^2 (In other words, it takes one second to get from a stop to one meter per second, which is 2.2 mph. This is a pretty standard rate for electric *trains*. The actual rate of the chair is probably less.)

F = ma = (100 + 50) kg * 1 m/s^2 = 150 N

Assume two drive wheels, so 75 N required from each wheel

Assume each drive wheel has an 0.1 m radius, adjust as needed

Torque = r x F = 0.1 m * 75 N = 7.5 N m required from each wheel

Circumference of wheel = 2 * pi * 0.1 m = 0.628 m

Assume the chair is traveling at 1 m/s

1 m/s / 0.628 m/rev = 1.59 rev/sec

P = torque * angular velocity = 7.5 N m * 1.59 rev/sec = 11.9 J per wheel

or 23.8 J total

If this power is applied for 1 sec, this is 23.8 W delivered at the wheels, on level ground. Add more for losses in the drivetrain, motors, PWM controller, hills, etc.

How hot does it get?

Matt Roberds

I don't expect to use this often. Even on a float charge, there's no guarantee that it will be usable at any particular point in time.

While it's relatively easy to keep a spare set of batteries on hand for a flashlight, radio, etc., it's not practical for such large batterie$ as these (and, they too would die if not kept "floated").

By contrast, a lines powered supply has a much higher chance of being "available" at some random future time.

Good point. Though I was thinking of just a big linear... oversized XFMR to keep output impedance low, etc.

The FETs in the controller are rated at 40A so we're in the same ballpark.

Adjust how? Remove unnecessary appendages? :> ("New weight loss plan available in your area! Call now for an appointment!")

There's the rub. I assume the powertrain is, at best, 50% efficient. Maybe even less. Hence my question. The sorts of gearboxes, motor controllers, etc. used in industrial settings tend to be much higher quality/performance than this consumer kit. I was hoping someone (e.g., hobbyist) had a feel for actual numbers in an application like this.

Thanks for your effort, though! I'll file it away and see how it correlates with empirical data when I'm at that point.