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Author Topic: Motor torque selection  (Read 784 times)

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Offline madoTopic starter

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Motor torque selection
« on: May 14, 2013, 12:53:38 AM »
I have a carriage  that moves on 4 wheels and weight  65kg I made design  for the motor based on  http://www.societyofrobots.com/mechanics_dynamics.shtml  but the resulted torque was very high
Velocity = 3.14 ft/s
Acceleration = 1.5 ft/s
Mass = 143 lbs or 65kg
wheel diameter= .5ft
RMF = 143*  3.14*1.4 /2*pi = 107 .2 lbs ft rps
If I choosed rps to be 2 rps the torque will be 53 lbs ft and this is very high so where is the error here?

Offline jwatte

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Re: Motor torque selection
« Reply #1 on: May 14, 2013, 10:41:04 AM »
65 kg is a lot of payload. Accelerating at 1.5 ft/s/s is a fair bit of thrust (evel though it's "only" 1/20th of a g, if I can do unit conversions in my head.) You will need strong motors to achieve that level of acceleration.
Also note that typical motors run in the thousands of rpm, and don't have a lot of torque, but use gear boxes to trade rpm for torque. Thus, a 12V motor with 1,000 kV constant and 2 ozin of torque would run at 12,000 rpm at 12 V; use a 500:1 gearbox and it gets to 24 rpm and has 1,000 ozin of torque! (That motor would be too small for your design case, but that's a pretty small RC car motor I described...)

Offline madoTopic starter

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Re: Motor torque selection
« Reply #2 on: May 14, 2013, 12:18:09 PM »
if i used two motors of 2 oz of torque , will it be enough to move the load ?
i need to know how to get enough torque .
how to calculate the required torque?
« Last Edit: May 14, 2013, 12:26:22 PM by mado »

Offline jwatte

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Re: Motor torque selection
« Reply #3 on: May 14, 2013, 04:51:51 PM »
"2 oz" makes no sense. For torque, you need force times lever length. Do you mean 2 ozin?

No, that by itself is unlikely to overcome the losses in the system. However, a 2 ozin motor with a big enough gearbox (say, 1000:1) might be able to start moving the load. It will likely not have the degree of acceleration you need, though.

You already calculated the necessary parameters in post 1.

Offline madoTopic starter

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Re: Motor torque selection
« Reply #4 on: May 15, 2013, 01:09:39 AM »
so referring to your talk i got that the my acceleration need to be bigger and the problem here is to get gear box with 1000:1 ratio , did i hit?

Offline newInRobotics

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Re: Motor torque selection
« Reply #5 on: May 15, 2013, 01:51:03 AM »
how to calculate the required torque?
m = 65kg
a = 1.5 ft/s2
v = 3.14 ft/s
D = 0.5 ft
r = D / 2 = 0.25 ft

F = m * a = 65kg * 1.5ft/s2 = 29.718N
T = F * r = 29.718N * 0.25 ft = 2.2645116 Nm

C = pi * D = 3.14 * 0.5 ft = 47.8536 cm
f = v / C = 3.14 ft/s / 47.8536 cm = 2 Hz
RPM = f * 60 = 2 Hz * 60 = 120

Theoretically, to achieve desired velocity and desired acceleration You need 1 motor with RPM = 120 and T = 2.2645116 Nm, however in real world You have friction, errors and the like, so add at least 25% to compensate for that, so what You are looking for is RPM = 150 and T = 2.8306395 Nm. If You have 2 driving motors, then divide torque value by 2 for each motor, if You have 4 driving motors, then divide torque value by 4 for each motor.

P.S. Derived torque value only applies to acceleration on flat ground, not a slope. If You want to find out required torque to accelerate on a slope You have to include gravity into equation.
« Last Edit: May 15, 2013, 01:54:54 AM by newInRobotics »
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