How to determine how many watts / amps for lifting a weight

  • #1
Is there a way to determine how many amps / watts is required from an electric motor in order to move (lift, push or pull) an object weighing a specific amount?

Conversely, is there a way to determine the maximum amount of weight an electric motor can move (lift, push or pull) based on its output amps / watts?

Or are there other factors involved? If yes, please specify!

Thanks in advanced!
 

Answers and Replies

  • #2
NascentOxygen
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Using an electric motor to lift something usually has an associated time requirement. If you are prepared to allow it a year to happen, you might get away with using a motor out of an electric clock to hoist a car for inspection. The faster you need something lifted, the more powerful the motor required.

For an electric motor, the power it develops = Volts × Amps
 
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  • #3
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The electromagnetic torque developed in a motor is proportional to the square of the current through the windings. If you will accept a quasi-static lift (and described by NacentOxygen), then you can get the minimum current from this idea. If you want to lift the weight in finite time, you will have to accelerate it, and that requires more force and hence more torque, so a higher current.
 
  • #4
Using an electric motor to lift something usually has an associated time requirement. If you are prepared to allow it a year to happen, you might get away with using a motor out of an electric clock to hoist a car for inspection. The faster you need something lifted, the more powerful the motor required.

For an electric motor, the power it develops = Volts × Amps

The amount of time to move / lift the object would be the amount of time an average human takes to lift a standard object they can lift easily, such as a rack sack. So say, in less than 5 seconds.

If a motor were to lift an object weighing 100 pounds, how much watts / amps would it need to lift it 5 feet off the ground in 1 second?
 
  • #5
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I think you will need to compute the work you want to do, the power required, and then select a motor based on the required power (allowing for friction losses). Then find the current requirements for your specific motor. I don't think this question has a general answer.
 
  • #6
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Starting with the force needed, convert to a torque through converting the linear (lifting) to rotational (shaft) motion. I would then apply a "good estimate" of motor efficiency depending on the motor type used. ( DC, Stepper, BLDC/Servo, or induction) - this is like 75 to 95%...

From this you can get power required, etc. (Watts)

Then also - the efficiency of the control / driver if used.

It is not difficult but there is no ONE answer to this based on the way you have phrased it.
 
  • #7
jim hardy
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Is there a way to determine how many amps / watts is required from an electric motor in order to move (lift, push or pull) an object weighing a specific amount?

Go back to your basics.
Power is rate of doing work.
Work is force X distance.
A horsepower is 550 foot-pounds per second. Somebody studied horses and figured out a good one could lift a 550 pound weight against gravity at that rate, a foot per second.

A horsepower is 746 Watts.
A Watt is a Joule per second.
A Joule is a Newton(force) X a Meter(distance)
From above you should be able to figure out the answer your question in SI units. But as @NascentOxygen observed you'll have to decide at what rate do you wish to lift the object.

And you'll have to allow for friction.
 
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  • #8
sophiecentaur
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And you'll have to allow for friction.
You certainly will!!! :smile:
 
  • #9
jim hardy
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At Last !
A Metric Horse !!!
upload_2017-8-1_14-28-3.png

https://en.wikipedia.org/wiki/Horsepower
 
  • #10
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Do they call him SI
 
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A horsepower is 550 foot-pounds per second. Somebody studied horses and figured out a good one could lift a 550 pound weight against gravity at that rate, a foot per second.

Hahaha I always wondered how they came on that figure
 
  • #13
Tom.G
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Somebody studied horses and figured out a good one could lift a 550 pound weight against gravity at that rate, a foot per second.
I think James Watt, the steam engine inventor, was invovlved in that. The story goes something like this:
He needed a way to tell potential customers how much work his steam engine could accomplish. Since the Prime Mover in those days was the horse, he decided the unit of measure to be the Horsepower. He surveyed, or had surveyed, the amount of work a horse could accomplish and then generously rounded up. That was his "safety factor." If he touted his engine as delivering the power of 10 horses and his customers could eliminate 12 horses, he was golden.

(Another successful enterprise based on astute marketing?)
 
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  • #14
sophiecentaur
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and then generously rounded up
I was told that at School. I believe it too. Salesmen don't change.
 
  • #15
jim hardy
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Salesmen don't change.
Yeah, wait'll advertisers get wind of that Metric Horsepower . It's 735/746 of a SAE horsepower. All metric engines get a free 1.5% hp boost .
 
  • #16
sophiecentaur
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Yeah, wait'll advertisers get wind of that Metric Horsepower . It's 735/746 of a SAE horsepower. All metric engines get a free 1.5% hp boost .
We tend to talk kW these days. No one understands those things but it doesn't matter. The added Gismos are far more important to your average car buyer; Sat Nav, DVD in rear seats. Engine? IS there an engine under there? I've never looked.
 

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