We are working on a computer-controlled electric vehicle project which is based on an ATV chasis.

While I was looking for a DC motor to move this vehicle (which is estimated to be 350kg), I found an electric wheelchair motor. It is 500 W 108 rpm 24V DC motor and has its own differential. The salesman told me that it can move up to 540 kg. I don't know which fact this number is based on.

I also have a 500 W 1500 rpm 24 V DC motor. If I decrease rpm and increase torque with the same ratio of electric wheelchair motor via reductors, can it move up to 540 kg? Or does the max weight that a motor can move depend on another factors? How can I measure this?

Thanks in advance.

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    \$\begingroup\$ You don't measure it before you calculate it. It's basic physics. You have an object with a certain weight, to change it's speed you will need an amount of energy. There's also friction to overcome to keep it at a certain speed. The energy needed comes from the motor. Motors with more power can apply the same amount of energy in less time. So: study physics and make calculations. \$\endgroup\$ – Bimpelrekkie Jan 28 '16 at 11:21
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    \$\begingroup\$ Oh, and: never trust a salesman. Of course the motor can move 540 kg, how much it can move does not depend on the motor but on the friction the weight has to it's surroundings. In space, you can move 540 kg with your finger. \$\endgroup\$ – Bimpelrekkie Jan 28 '16 at 11:25
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    \$\begingroup\$ The motor that spins the rotating restaurant on the top of the Space Needle in Seattle is on the order of a one-HP motor. I assure you the restaurant weighs more than 540 kg. Your question is not answerable without knowing how fast you want the thing to spin and what forces are working to slow it back down. \$\endgroup\$ – Eric Lippert Jan 28 '16 at 20:22
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    \$\begingroup\$ To put things in perspective: A Tesla Model S has 270000 W of power, more than 500 times what you have. \$\endgroup\$ – MSalters Jan 28 '16 at 20:52
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    \$\begingroup\$ Give me a lever and a place to stand and I'll move the world. The same can be said about gearing. \$\endgroup\$ – Steve Jan 28 '16 at 21:26

"The salesman told me ..." LOL! If physics was his thing, he'd probably not be a salesman. In any case his job is to make a sale, whether you get your device working or not.

The first thing you need to decide is how much power the motor needs to have. You can then worry about torque and speed later. Those can be traded off against each other, but you can't cheat the physics that requires a certain power for certain tasks.

There are two end-user criteria you need to look at to decide the power. These are how fast you want to be able to accelerate, and what minimum speed you want to be able to go up a decent hill. I'll use the hill criterion as example.

Let's say you want to be able to go 20 MPH up a 20% grade. 20% grade means you go up 1 part for every 5 forward. Since the only physics work being done is going up, the problem reduces to raising 350 kg straight up at 4 MPH. 4 MPH is 1.8 m/s, and here on earth 350 kg weighs 3.43 kN. The power expended is therefore:

  (1.8 m/s)(3.43 kN) = 6.13 kW

Of course there will be some friction to overcome, so you'd want about 10 kW in this example. Since 500 W isn't even close, you either have to specify much lower performance or get a much bigger motor (and the power source to feed it).

Let's flip this around and see what 500 W can do.

  (500 W)/(3.43 kN) = 146 mm/s

That's how fast 500 W can lift the whole unit straight up. Applying that to a 20% grade, for example, it can move along at 5x that, or 730 mm/s, or 1.63 MPH. In reality there will be friction and other losses, so probably not more than 500 mm/s = 1.1 MPH.

Added about torque

You should start with power as described above. Once you have decided how much power the motor must put out, you face the torque/speed tradeoff. You can figure out what torque/speed you need at the wheels, but that will usually be much too slow and too high torque for a reasonable electric motor to produce directly. As a result, there will be some gearing between the wheel shaft and the motor shaft. Since gearing is in there anyway, pick a good motor and then design the gear ratio accordingly, not the other way around.

To put this in perspective, let's look at the torque and speed required go up a 20% grade at 20 MPH as described above. Let's say the wheels are 500 mm in diameter, so 250 mm radius, so 1.57 m circumference. 20 MPH is 8.9 m/s, so the wheel must turn at 5.7 Hz. Your not likely to get a suitable motor with peak power and efficiency at 5.7 Hz (342 RPM). You'll probably end up with a 5x to 10x gear ratio, depending on the best available motor you find.

For example, let's say you've decided you need a 10 kW motor. That could come as 60 Hz (3600 RPM) and 26.5 Nm, 20 Hz and 80 Nm, or a variety of other combinations that all result in 10 kW. Suitable motors will only be available in limited combination, and the gearing will likely be custom designed anyway. Pick the motor, then let it dictate the gear ratio.

  • \$\begingroup\$ Thanks for your answer, but there is still some question marks in my mind: When is torque important ? Also, how does the calculations change for a straight horizontal way, is it same as with to lift straight up? Sorry for my lack of physics knowledge. \$\endgroup\$ – Kerem Zaman Jan 28 '16 at 15:09
  • \$\begingroup\$ With the right transmission you can change the torque however you want. You’ll need a certain wheel torque to go up a gradient or accelerate in a certain time (and to overcome friction and air resistance at higher speeds). For Olin Lathrops example the wheel torque would be 3.43kN*wheel radius. The rpm with 500W would be (146mm/wheel circumference)*60. So with a 30cm diameter wheel you’d need 514Nm wheel torque and you’d get 58.4rpm at the motor’s maximum power. Look up the datasheet of motor for the rpm/torque/power curve. \$\endgroup\$ – Michael Jan 28 '16 at 16:20
  • \$\begingroup\$ @Kerem: I have updated my answer to discuss torque, which is also basically what Michael said. \$\endgroup\$ – Olin Lathrop Jan 28 '16 at 16:25
  • \$\begingroup\$ Thanks for adding it. Silly me forgot to multiply my radius with 2pi for the circumference and I can’t edit anymore. \$\endgroup\$ – Michael Jan 28 '16 at 17:43
  • \$\begingroup\$ It might be worth mentioning that you are using mechanical output power in your equations, wheras the motors may be specced in terms of electrical (input) power. Real motors are far from 100% efficient, so a significant portion of the input electrical energy will be dissipated as heat due to resistive heating and friction. \$\endgroup\$ – Mels Jan 28 '16 at 18:40

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