Need Help Calculating Torque Required For Project

  • Thread starter jgayetty
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Hello,

I am working on a home hobby project and I need a motor to move an approx 1000lb weight back and forth - similar motion to a pendulum - at a rate of about once every 4 seconds. The problem is, I don't know how powerful the motor has to be to accomplish this. And since gear motors are pretty expensive ($200-$800) I decided it may be best not to "guess" and be wrong on the motor size needed.

I have been looking at a gearmotor (12 rpm) made by Dayton that has 600 inch pounds of force. But again, I am guessing as to the size I need. I was hoping I would find a kind soul on this forum to help me narrow down how many inch pounds of force I need.

Here is a PDF drawing I made that shows what I am talking about.
http://www.yozod.com/doc.pdf

I appreciate any and all help!

Thanks!

Joe
 

Answers and Replies

  • #2
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The mass is not insignificant, so it is probably going to be something big, so its shape may be important. What is it? What are you going to use for the pendulum's arm? Its mass and shape may also be of importance.
 
  • #3
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Thanks for the reply.

A rocking bed that hangs from 4 foot arms on each corner. I came to the 1000 lb figure assuming a worse case scenario. The actual weight of 2 people and bed may be more like 600 lbs. I figured a simplistic explanation may help narrow it down to the bare physics issues, so I tried to simplify its explanation in my drawing and previous post. I suppose I can even now see that my drawing didn't explain it completely properly anyway.

I can see that I am trying to move 1000 lbs 16 inches back and forth, but of course the actual movement is like an arc from an ellipse...as the movement is not exactly horizontal (slightly pendulum-like). So there is a slight up and down issue involved here too.

If I knew that a gear-motor producing 600 inch pounds would do it, I suppose that would answer my question...but I guess I don't want extreme overkill either.

Thanks again...

Joe
 
  • #4
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Almost missed your last question: The arms will likely be common chain.
 
  • #5
CWatters
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If it behaves like a pendulum then the power required to run it might not be that great. Remember an ideal pendulum just converts energy back and forth between PE and KE and needs no power to run.

The power required to start it swinging is another matter.

Peak torque on the motor and gearbox might be more important that average power. The main issue might be impulse loads when people move or the bed has to stop suddenly. Not sure how to go about quantifying these issues. Measure them?

EDIT: Actually the first bit of this post is is nonsense. It only applies if the frequency is close to the natural frequency of the pendulum.
 
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  • #6
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How does that 4-second cycle look like? A free pendulum would not require any external work to swing, so the motor would just have to compensate friction. The timing might be wrong (depending on the size and shape of the mass), but this could be compensated by holding the position (left/right) for some time. In this case, maximal torque and the length of the arm would be more relevant than maximal power.
 
  • #7
CWatters
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I had another go.. Assume the bed moves with ..

displacement = A cos(ωt)

where

A = length of the arm
ω = 2πf

velocity = ω A Sin(ωt)
acceleration = ω2 A Cos(wt)................. (1)

The instantaneous torque on the shaft is

Torque = Horizontal Force * A Sin(wt)

Horizontal Force = mass of bed * acceleration

so

Torque = mass * acceleration * A Sin(wt)...............(2)

put (1) into (2)..

Torque = mass * ω2 * ACos(ωt) * ASin(ωt)

Cos(ωt) * Sin(ωt) = 1

so

Torque = mass * ω2 * A2

Mass = 1000bs = 453kg
ω = 2πf where f = 1/4 so ω = 1.6
A = 8" = 0.2m

Torque = 453 * 1.62 * 0.22
= 46NM

but perhaps someone check I've not made a mistake. I expected the torque to change sign as the arm rotates?
 
  • #8
CWatters
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Oops that's not quite right because

cos(x)sin(x) = 1/2 sin 2x

so

Torque = mass * ω2 * A2 0.5 sin(ωt)

So torque does vary as the arm rotates.

Max Torque occurs when sin(ωt) = 1 and gives..

Torque = 453 * 1.62 * 0.22 * 0.5
= 23NM = approx 17 foot lbs

The starting torque will be higher though.
 
  • #9
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It only applies if the frequency is close to the natural frequency of the pendulum.
Yes, a couple of you commented on this timing "issue".

Originally, I tried to design the propulsion of the bed so that it would synchronize with the natural frequency at which the bed would rock. Just like when I push my kids on a swing. I wait until they swing back naturally then I push again. In fact I initially hooked up a small gearmotor (a wiper motor) with a speed controller. I figured I could adjust the speed controller on the motor to match the timing of the bed...which I found is impossible to do with exactness. If the timing was off just 1/10th of a second, which would probably even be difficult to obtain, it just took a few strokes before the motor was fighting with the natural bed movement. And since that motor was very small, it lost the battle.

I realize a different mechanism "could" be designed to be triggered from the bed's cycle, etc etc, but I don't want to reinvent the wheel. So I decided a big strong motor would just move the bed back and forth without regard to the natural frequency of the bed. Simple and easy although maybe not the most efficient I agree. Efficiency isn't my primary concern.

I am going to go back and CWatters formulas a bit more.

Anyone think 600 inch pounds would be enough?

Thanks!
 
  • #10
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Oops that's not quite right because

cos(x)sin(x) = 1/2 sin 2x

so

Torque = mass * ω2 * A2 0.5 sin(ωt)

So torque does vary as the arm rotates.

Max Torque occurs when sin(ωt) = 1 and gives..

Torque = 453 * 1.62 * 0.22 * 0.5
= 23NM = approx 17 foot lbs

The starting torque will be higher though.
So that's about 200 inch pounds. Good so far (motor is 600 inch pounds)

How does one calculate starting torque required?

Thanks again!
 
  • #11
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The real issue here is that the period of free oscillations is unlikely to be 4 seconds. A simplistic calculation indicates it is going to be about 2 seconds. If the pendulum were rigid, then a counterbalance could be used to tune it to 4 seconds, then just about any motor would do.

Without this, the system can be modeled as a driven harmonic oscillator, and, assuming the kinematics of the system can be resolved into some nice force function, we should be able to estimate the maximum load on the motor. I guess we need an engineer who understands kinematics of mechanisms here :)
 
  • #12
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More questions. If that is a bed, what are its dimensions? Width, height, length? How is it going to be suspended? Top corners, bottom corners, some other way? Is it going to swing longitudinally or transversally?

Also, have you already built or at least designed the rocker mechanism? If so, then knowing its dimensions would be helpful.
 
  • #13
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May I reframe the issue a bit?

Instead of focusing on the "pendulum" aspect of the design, think of it as simply moving 600 pounds along the horizontal plane. Disregard friction as well. Just a basic horizontal movement of 600 lbs. In that case, how much torque would it take to get the movement started, and how much to maintain the movement?

I ask to reframe the issue/problem because I don't need an exact answer, just a ballpark. Here is why: if the above problem turns out to require 200 inch pounds, then I can be pretty much assured that 600 inch pounds will be more than enough...which is the motor I am considering. If the problem above requires 500 inch pounds, then I can either get more complex in the problem, or just buy a 1000 inch pound motor.

Can anyone help me with a formula to answer the above basic question?

Thanks everyone for your help so far. It IS appreciated!

Joe
 
  • #14
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As the pendulum frequency is not so far away from 4s, it is a relevant aspect.
If you neglect it, see CWatters' posts (#7 and #8)
 
  • #15
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As the pendulum frequency is not so far away from 4s, it is a relevant aspect.
If you neglect it, see CWatters' posts (#7 and #8)
Sorry, I thought I provided 99% of the relevant info and was hoping for an answer. Thank you to all who posted. I will just order a motor and try it...that seems to be the easiest way to know how much torque is required.

Here is a guy who built what I want to build. If anyone is interested...

http://www.youtube.com/watch?v=D_CKsKp0bwk

Thanks!

Joe
 

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