HP Required to move a robot 10meters at 1.5m/sec

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Jeremy Sawatzky
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TL;DR
I need to move a 4000lb robot along linear rails. I have my time constraints but I dont have my gearmotor selected
I need some assistance with calculating hp requirements for moving a load. - The 4000lb load is resting on ball bearing rails so friction is effectively zero and will be covered by my added power contingencies.

Load: 4000lbs
Distance to travel: 10 meters.
Time to Travel: 7.5 seconds
Need to accelerate the load from a stop to a nominal speed then decelerate coming to a stop.

My power delivery method will be a gearmotor driving a gear rack. - I suspect the pinion gear to be about 3-4in in diameter. So the gearmotor will need to run around 300rpm.

From experience, I would ballpark the hp requirement to be 4-5hp. but I would like the math to back it up.

Thank you!
 
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Perfectly horizontal movement?
Geared rails?
How much acceleration you need?
How will the mass be stopped?
 
Jeremy Sawatzky said:
TL;DR Summary: I need to move a 4000lb robot along linear rails. I have my time constraints but I dont have my gearmotor selected

From experience, I would ballpark the hp requirement to be 4-5hp. but I would like the math to back it up
So, you're not asking for HELP with the math, you are asking us to do it for you. How about you make some effort yourself first?
 
Jeremy Sawatzky said:
Need to accelerate the load from a stop to a nominal speed then decelerate coming to a stop.
Assume maximum motor torque is proportional to maximum rated motor current. Then maximum forward motor current will flow for the first 3.75 seconds, reaching the 5 metre mark. Then maximum reverse current will flow, braking the load during the last 3.75 seconds, reaching and stopping at the 10 metre mark.

First find the acceleration needed to get halfway in half the time.
s = ½ · a · t²
a = 2 · s / t²
s = 5 metre; t = 3.75 sec.
a = 2 * 5 / ( 3.75 * 3.75 )
a = 0.7111 m/s/s

Next, find the force that must be applied to the mass.
4000 lbs = 1814.37 kg.
F = m · a
F = 1815. * 0.7111
F = 1291. newton, over a distance of 5 metres.
Work done in 3.75 sec is force times distance 1291 * 5 = 6455 joule.
Work done per second is 6455 / 3.75 = 1721.3 joule/sec = 1721.3 watt.
So the motor needs to be rated at about 1.75 kW.

Next go back and check my memory for equations, and my word processor's arithmetic.
 
Jeremy Sawatzky said:
TL;DR Summary: I need to move a 4000lb robot along linear rails. I have my time constraints but I dont have my gearmotor selected

From experience, I would ballpark the hp requirement to be 4-5hp. but I would like the math to back it up.
You are correct. For moving 4000 lbs, you need some math. Here is how I solve this type of problem.
Step 1: Sketch an acceleration - velocity - position diagram that includes all available information. A pencil sketch is good enough at this stage. Your diagram would start as follows:
Accel Vel Pos.webp

You know the total move time and distance, but do not know acceleration, deceleration, or peak speed. It would require infinite acceleration and deceleration to move at constant speed for the entire distance, so that motion profile is clearly impossible. Another limiting case is constant acceleration for halfway, then decelerate at the same rate to bring it to a stop.

The math is as follows:
1) Velocity equals the integral of acceleration up to that time, AKA the area under the acceleration curve.
2) Position equals the integral of velocity up to that time, AKA the area under the velocity curve.
3) Sketch Accel - Vel - Pos curves for acceleration halfway, and equal deceleration for the last half, and convince yourself that the calculations in Post #4 are correct.
4) Now calculate the acceleration force and peak velocity.
5) For a 4000 pound mass, you cannot assume that friction will be small enough to ignore. You need data. Possibly pull a similar system using a spring scale.
6) The gearmotor must deliver enough torque to meet the sum of the acceleration torque plus friction torque. It must be geared to deliver the peak speed.

Now you can iterate. Reducing the portion of the move for acceleration and deceleration will increase the peak torque requirement and decrease the peak speed. Use the diagram method to help you figure out the equations.

Expect that a simple gearmotor with an induction motor cannot meet these requirements. You will need an industrial servomotor and drive system matched to an appropriate gear reducer. One manufacturer that I have had experience with is Allen-Bradley. They are not the only manufacturer that makes suitable motors, but an example of what an application like this needs. Link to their servo motor catalog: https://literature.rockwellautomation.com/idc/groups/literature/documents/td/knx-td001_-en-p.pdf. And a link to their catalog of drives needed to drive the motor: https://literature.rockwellautomation.com/idc/groups/literature/documents/sg/knx-sg001_-en-p.pdf. That's a total of 486 pages of technical information. Mechanical engineers typically find the page that lists the available motors, with their torques and speeds, make a preliminary motor selection, then call in a control engineer. The control engineer adjusts the motor selection, figures out how to drive it, and implements the motion profile.
 
Lnewqban said:
Perfectly horizontal movement?
Geared rails?
How much acceleration you need?
How will the mass be stopped?
Yes - horizontal, geared rail. - accel and decel will be as soft as possible within the time limit. - the motor will be doing the stopping.
 
Baluncore said:
Assume maximum motor torque is proportional to maximum rated motor current. Then maximum forward motor current will flow for the first 3.75 seconds, reaching the 5 metre mark. Then maximum reverse current will flow, braking the load during the last 3.75 seconds, reaching and stopping at the 10 metre mark.

First find the acceleration needed to get halfway in half the time.
s = ½ · a · t²
a = 2 · s / t²
s = 5 metre; t = 3.75 sec.
a = 2 * 5 / ( 3.75 * 3.75 )
a = 0.7111 m/s/s

Next, find the force that must be applied to the mass.
4000 lbs = 1814.37 kg.
F = m · a
F = 1815. * 0.7111
F = 1291. newton, over a distance of 5 metres.
Work done in 3.75 sec is force times distance 1291 * 5 = 6455 joule.
Work done per second is 6455 / 3.75 = 1721.3 joule/sec = 1721.3 watt.
So the motor needs to be rated at about 1.75 kW.

Next go back and check my memory for equations, and my word processor's arithmetic.
Thank you very much, that is immensely helpful. I would probably plan for some decent headroom then, putting likely a 2.5 -3kw motor to account for friction losses and the ability to go a little faster if we need to.
 
jrmichler said:
You are correct. For moving 4000 lbs, you need some math. Here is how I solve this type of problem.
Step 1: Sketch an acceleration - velocity - position diagram that includes all available information. A pencil sketch is good enough at this stage. Your diagram would start as follows:
View attachment 366227
You know the total move time and distance, but do not know acceleration, deceleration, or peak speed. It would require infinite acceleration and deceleration to move at constant speed for the entire distance, so that motion profile is clearly impossible. Another limiting case is constant acceleration for halfway, then decelerate at the same rate to bring it to a stop.

The math is as follows:
1) Velocity equals the integral of acceleration up to that time, AKA the area under the acceleration curve.
2) Position equals the integral of velocity up to that time, AKA the area under the velocity curve.
3) Sketch Accel - Vel - Pos curves for acceleration halfway, and equal deceleration for the last half, and convince yourself that the calculations in Post #4 are correct.
4) Now calculate the acceleration force and peak velocity.
5) For a 4000 pound mass, you cannot assume that friction will be small enough to ignore. You need data. Possibly pull a similar system using a spring scale.
6) The gearmotor must deliver enough torque to meet the sum of the acceleration torque plus friction torque. It must be geared to deliver the peak speed.

Now you can iterate. Reducing the portion of the move for acceleration and deceleration will increase the peak torque requirement and decrease the peak speed. Use the diagram method to help you figure out the equations.

Expect that a simple gearmotor with an induction motor cannot meet these requirements. You will need an industrial servomotor and drive system matched to an appropriate gear reducer. One manufacturer that I have had experience with is Allen-Bradley. They are not the only manufacturer that makes suitable motors, but an example of what an application like this needs. Link to their servo motor catalog: https://literature.rockwellautomation.com/idc/groups/literature/documents/td/knx-td001_-en-p.pdf. And a link to their catalog of drives needed to drive the motor: https://literature.rockwellautomation.com/idc/groups/literature/documents/sg/knx-sg001_-en-p.pdf. That's a total of 486 pages of technical information. Mechanical engineers typically find the page that lists the available motors, with their torques and speeds, make a preliminary motor selection, then call in a control engineer. The control engineer adjusts the motor selection, figures out how to drive it, and implements the motion profile.
Thank you very much. - Post 4 does have some solid math. Unfortunately it is actually not up to me to purchase a gearmotor or servo, I am just providing some ballpark numbers so that my customer can make an informed decision. I will work through your post, thanks again