Determine speed of DC motor

A tachometer (either photo-reflective or mechanical) can be used to measure the motor's speed.

The speed will be determined by the capabilities of the power supply (voltage & current available) and the load applied. You COULD run the motor on 12 or 24 volts for more speed, but a the cost of shortening the motor's life - possibly to a matter of seconds. Since the motor is specified, do the competition rules also limit the capacity of the power source?

In a Tug-of-War, the winner is determined by strength, not speed. You can gear down the motor to provide less speed and greater torque.

You need to specify what rules apply to power supply,, gearing, etc.

John

The power source is fixed, I'll have no control over that. My intention was to gear down for as much torque as I can get with the materials provided, but I want to ensure that the motor runs at it's maximum efficiency and I'm not sure how to determine that.

** Depends if YOUR vehicle is the one going forwards, is stationery or is going backwards. ** The vehicle with the most torque and weight at the drive wheels will dominate.

The competition rules that YOU stubbornly refuse to reveal ARE the key.

So piss off - TROLL.

..... Phil

is

key.

First of all, there are no competition rules. We're given a set of a materials and told to build a monorail. That's really all there is to it. I was the one asking for help, so wouldn't I disclose as much information as possible?

Don't take Phil's attitude personally. He seems to enjoy hanging around s.e.b and ranting at people who ask, well, basic questions -- although there are frequently good bits of advice amongst the flying spittle.

--
Rich Webb     Norfolk, VA

The speed depends upon the applied voltage and the load. If the voltage is fixed, then the speed will decrease as the load increases. The above figures state that, for an 8V supply, the speed will be 3500 RPM at no load, reducing to zero (stalled) at a load of 0.01Nm.

For tug-of-war, you don't need power, just torque, which means that you want the lowest possible gear ratio. IOW, this is a mechanical problem, not an electronic one.

OTOH, if you're allowed to introduce some electronics between the power source and the motor, then it would be an issue of finding the maximum current which the windings can handle and constructing a constant-current switching regulator to provide that current regardless of speed.

s

So what is the mathematical relationship between load and motor speed?

I want to be able to calculate the speed that the wheels will turn. My gear ratio has not been finalized.

There are really 2 limits for motors: a maximum speed above which things start to come apart & a maximum current above which the windings start to smoke.

Your motor is spec'ed at 3500 rpm, yet it's a small motor (small diam, small centrifugal forces) so it'll probably do more. More is better: power = speed x torque. If you have more than one of these, you could experiment with higher speeds. Just increase the voltage.

Torque is proportional to current. So maximum current is the current at maximum torque. Lock the rotor & adjust the current until maximum torque is reached. That is the maximum current. Motors are designed for a certain temperature rise. I.e., the maximum current is determined by the temperature limit. Excessive current => excessive temp => smoke. If your contest is short, you could run higher currents. Rated maximum current is for continuous running.

First you need to know how much torque you can put to use. I.e., before your driven wheel(s) start to slip. Then gear down to that torque. In use, control the voltage to give the maximum current.

I now see that speed hasn't come into this, so maybe I'm missing something. Oh, wait ... here's how it comes in: when your tractor is /moving/, the gearing-down for torque & the voltage used to get maximum current/torque could cause the motor to exceed it's speed rating. So when it moves, you would have to monitor speed.

HTH, Bob

The simplest theoretical model suggests it should be linear: speed is proportional to voltage, current is proportional to load. As the load increases, so does the current, so does the voltage lost due to winding resistance. IOW, draw a straight line from 3500RPM@0Nm to 0RPM@0.01Nm.

This assumes that either the impedance of the power supply is low (i.e. you still get 8V at stall current) or that any voltage drop has been accounted for in the quoted stall torque figure.

But there are enough other factors that you would be better off measuring it.

Nice response, esp the temperature limits. I suspect the OP is at the Power = torque x RPM level. He should just google dynamometer.

Thanks.

Yeah, I had a similar feeling - that most of what I said would go right by.

Bob

speed is proportional to the voltage across the motor without its resistance.

(treat the motor as a zero resistance ideal motor in series with a real resistor)

constant current will give you constant torque (if you can ignore friction and sources of drag)

you don't have sufficient information.

stall current (or DC resistance) would help.

also are you allowed to overdrive the motor?

what are the rules? is building a "crawler" (with extremely low gear ratio) all that's needed to win?