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

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Discussion Overview

The discussion revolves around determining the power requirements (in watts and amps) for electric motors to lift, push, or pull objects of specific weights. Participants explore the relationship between motor power, weight, and time, as well as the factors influencing these calculations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants suggest that the power required from an electric motor is dependent on the weight of the object and the time allowed for the lift, indicating that faster lifts require more powerful motors.
  • There is mention of the relationship between power, voltage, and current, with a formula provided: Power = Volts × Amps.
  • One participant notes that the torque developed in a motor is proportional to the square of the current, implying that higher currents are necessary for finite-time lifts.
  • Another participant emphasizes the need to compute work and power, factoring in friction losses, and suggests that there is no general answer to the question posed.
  • Discussion includes converting linear lifting force to torque and estimating motor efficiency based on motor type, indicating that efficiency can vary significantly.
  • Several participants reference the concept of horsepower and its historical context, discussing how it relates to the power output of motors and the calculations involved in lifting weights.
  • There are humorous exchanges about the concept of "Metric Horsepower" and its implications in marketing and advertising.

Areas of Agreement / Disagreement

Participants express multiple competing views on how to approach the calculations for motor power requirements, and the discussion remains unresolved regarding a definitive method or answer.

Contextual Notes

Limitations include the dependence on specific definitions of efficiency, the need for assumptions about friction, and the variability in motor types and their characteristics.

Tabaristiio
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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?

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!
 
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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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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.
 
NascentOxygen said:
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?
 
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.
 
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.
 
Tabaristiio said:
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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jim hardy said:
And you'll have to allow for friction.
You certainly will! :smile:
 
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At Last !
A Metric Horse !
upload_2017-8-1_14-28-3.png

https://en.wikipedia.org/wiki/Horsepower
 
  • #10
Do they call him SI
 
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  • #11
jim hardy said:
A Metric Horse !
Un cheval.
 
  • #12
jim hardy said:
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
jim hardy said:
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
Tom.G said:
and then generously rounded up
I was told that at School. I believe it too. Salesmen don't change.
 
  • #15
sophiecentaur said:
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
jim hardy said:
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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