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

In summary: 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.
  • #1
Tabaristiio
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2
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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  • #2
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
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
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?
 
  • #5
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
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
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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  • #8
jim hardy said:
And you'll have to allow for friction.
You certainly will! :smile:
 
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  • #9
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.
 

1. How do I determine the weight of the object I want to lift?

To determine the weight of an object, you can use a scale or other measuring device specifically designed for weighing objects. Alternatively, you can calculate the weight by multiplying the object's mass (in kilograms) by the acceleration due to gravity (9.8 meters per second squared).

2. What is the relationship between watts and amps when lifting a weight?

Watts and amps are both units of measurement for electrical power. The relationship between them is that watts are equal to amps multiplied by voltage. In the context of lifting a weight, this means that the amount of watts needed will depend on the amps required to lift the weight and the voltage of the power source.

3. How do I calculate the watts needed to lift a weight?

To calculate the watts needed to lift a weight, you will need to know the weight of the object, the distance it needs to be lifted, and the time it takes to lift it. The formula for calculating watts is watts = (mass x distance) / time. This will give you the amount of power (in watts) needed to lift the weight within the specified time frame.

4. Can I use the same formula for determining watts and amps for any weightlifting scenario?

The formula for calculating watts and amps for lifting a weight can vary depending on the specific scenario. For example, if you are using a hydraulic system to lift the weight, the formula will be different compared to using an electric motor. It is important to consider the type of lifting mechanism being used when determining the appropriate formula for calculating watts and amps.

5. What other factors should I consider when determining the amount of watts and amps needed for lifting a weight?

In addition to the weight of the object and the type of lifting mechanism, other factors that can affect the amount of watts and amps needed include the distance the weight needs to be lifted, the speed at which it needs to be lifted, and the frequency at which it needs to be lifted. It is important to take all of these factors into consideration when determining the appropriate amount of power needed for lifting a weight.

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