How high of a gear ratio can a motor have?

Question

I want to make a robot arm where a joint is powerful enough to lift 8 kg up at a distance of 1 meter.

This requires torque $tau = r*F = r*m*g$ = about 80 nm.

So now I am trying to find the requisite parts to put this together, i.e. motor + gears and then controller.

It seems that I will require a very high gear ratio to make this happen. For example this motor: http://www.robotshop.com/ca/en/banebots-rs-550-motor-12v-19300rpm.html has stats:

stall torque = 70.55 oz-in = 0.498 nm no load roation speed = 19300 rpm

To get my required torque, I need a gear ratio of 80/0.498 = about 161:1 (and the max speed drops to about 120 rpm).

My questions:

1) Do my calculations so far seem correct? It seems a bit suspect to me that an $8 motor with some gears can cause a 17.5lbs dumbbell to rotate about a circle of radius 1m twice a second (I'm barely that strong). This type of torque would be awesome for my application, but perhaps I'm missing something in the calculations and being too optimistic (e.g. efficiency).

2) Is it safe to operate a motor at such a high gear ratio? Gears are small, and I'm worried they'll easily crack/break/wear down quickly over time. Does anyone know of example gears that I should consider for this?

Thank you very much for any help, cheers.

Answer 1

Your calculation of about 80 N⋅m torque for lifting 8 kg with a 1 m lever arm is ok; more precisely, the number is 8 kg ⋅ 9.81 m/s² ⋅ 1 m = 78.48 N⋅m. As mentioned in other answers, you will need to scale up to account for gear inefficiency.

A simple calculation based on work shows that the Banebots RS-550 DC motor mentioned in the question is not powerful enough to spin the mass at 120 rpm in a vertical plane, geared down only 161:1. From $W = f\cdot s$, with $f = 78.48 N⋅m$ and $s=2 m$, we require 156.96 joules to raise the mass 2 m. Figures for most-efficient operation from the Banebots page show 120 watts (12 VDC ⋅ 10 A) power and 17250 rpm. Supposing efficiency is 85 to 90%, the motor will take from $156.96/(120\cdot 0.90) = 1.45$ to $156.96/(120\cdot 0.85) = 1.53$ seconds (ie, about 1.5 seconds) to deliver enough power to lift the mass 2 m, giving $60/(1.5⋅2) = 20$ rpm.

As noted in a previous answer, the RS-550's torque value of 0.498 N⋅m that is shown in the Banebots page is stall torque. Following is a typical small-DC-motor diagram showing current (red), speed (blue), power (yellow), and efficiency (green) versus torque.

From: JohnsonElectric.com PMDC Motors (permanent magnet DC brushed motors) webpage

You can see that speed is zero at highest torque, and since (as noted in wikipedia) mechanical power is given by $P = \tau\cdot\omega$, where $\tau$ is torque and $\omega$ is angular velocity in radians per second, ie $2\cdot\pi\cdot n$ at $n$ rpm, delivered power and efficiency also are zero at highest torque.

To compute actual values of torque, speed, current, power, and efficiency for this motor at various operating points, use the Basic DC motor equations given in wikipedia's Brushed DC electric motor article. Most of the parameters needed for the calculations are given in the Banebots RS-550 motor page, which apply when the motor is powered by 12VDC. Other required parameters can be calculated; for example, motor resistance is about $12V/85A = 0.14\Omega$, based on stall current.

If the 8 kg mass you mention always is attached to the arm, you probably should use a counterweight, a gas lift tube, or a spring to compensate for it, so that whatever motor you use will not have to do as much work. Or, if the 8 kg mass is an item being placed into a workcell, operated on, and then removed, you might use two arms, linked such that weight of an item being placed is offset by that of an item being removed.

Answer 2

It's quite possible for an $8 motor with enough gearing to move a mass as you describe. It's also possible to have gear ratios much more than a hundred. However:

1. Check that the gears you're using are rated for the torque you need.

2. What speed of rotation do you need? This defines the motor power you need (something people often forgot to think about). Power is usually the first thing you should consider when choosing a motor.

power(W) = force(N) x speed(m/s)

So, how quickly do you need the weight to travel?

3. The higher the gear ratio, the lower the efficiency of the gears. If the manufacturer doesn't specify the efficiency, then assume 50%, in which case, you'll need a motor of at least twice the power.

Answer 3

One problem I can see is, that you are using stall torque (at 0 rpm) and no load rotation speed (at 0 Nm). If you want the motor to still turn, it will have lower torque than this, and lower RPM.

Also, running the motor at (or close to) stall for long time will probably overheat it (it's gonna draw 85A (that is about 1KW) and dissipate it all as heat).

I don't have much experience with this, but for a first version I would guess you can get 50% of the stall torque and 50% of the no load rpm continually (or maybe 40%?). Then you could test it and go for different motor in second iteration.