How do I convert amps to pull force for an electric motor?

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Discussion Overview

The discussion revolves around converting electrical current (amps) to pull force for an electric motor, specifically in the context of modifying a motor that operates at 2.8 amps and 115 volts. Participants explore the relationship between electrical parameters and mechanical force, particularly in relation to neodymium magnets and the concept of horsepower.

Discussion Character

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

Main Points Raised

  • One participant seeks to convert amps to pull force, questioning how to determine the pull force from a known amperage.
  • Another participant asks if the force in question pertains to an electromagnet, indicating a need for clarification.
  • A participant mentions working with neodymium magnets and suggests that the strength of pull is rated similarly to pull force in pounds.
  • One participant states that amps measure electrical current and do not directly correlate to pull force, requesting more details about the intended application.
  • A participant provides a formula for calculating horsepower from amps and volts but expresses difficulty in converting horsepower to pull force.
  • Another participant suggests calculating torque instead of pull force, questioning the meaning of "pull force" in the context of the motor's operation.
  • Some participants express skepticism about the feasibility of creating a motor powered solely by magnets, indicating potential misconceptions about magnetic force and electrical components.
  • One participant discusses the relationship between power, pull force, and speed, emphasizing that power requirements change based on the speed of pulling and the force applied.
  • A participant provides a formula for calculating current in relation to voltage and angular velocity, highlighting the complexity of motor performance and the impact of modifications on motor constants.

Areas of Agreement / Disagreement

Participants express differing views on the feasibility of converting electrical parameters to mechanical pull force, with some questioning the validity of the original inquiry and others exploring various formulas and concepts without reaching a consensus.

Contextual Notes

Participants note that the relationship between electrical current, power, and mechanical force is complex and depends on various factors, including motor constants and operational conditions. There are unresolved assumptions regarding the definitions of pull force and the intended application of the motor modifications.

Makin_it_turn
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Coverting Amp to pull force??

I am trying to figure out a way to covert amps to pull force(lbs). I have a known amp that I would like to find out the pull force in lbs.

I have 2.8 amp @ 120v and would like to know the pull force that is needed to turn this type of amperage.
 
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Are you referring to the force from an electromagnet?
 


Yes and no...

No in the case that I am working with neodymium magnets...and yes that I believe that the strengh of pull are both rated the same (pull force lbs)
 


Amp is a measure of electrical current and does not directly relate to a pull force, so I don't understand your question. Are you trying to generate electrical power or something? Please give us some details of what you are trying to do.
 


I have an electric motor that I'm "modifying". The motor is rated at 2.8amp @ 115v. Given simplified version of calculating horsepower. (a * v = w)*.0013=HP, my motor is .43hp. Where I get stuck is converting HP to pull force, I think this would be the easiest way. I am trying to make a system to convert magnet pull force into a rated HP to replace said motor. In other words, replace electrically created magnetic force into just magnet force ie: electricless motor :) I think what I need is a formula to calculate two like forces (pull force and hp) ...I hope this helps! It's been hindering me for a week
 


Why don't you just find the torque generated by the motor and use that instead of "pull force". I'm not even sure what you mean by pull force. What is pulling what?
 


Makin_it_turn said:
I have an electric motor that I'm "modifying". The motor is rated at 2.8amp @ 115v. Given simplified version of calculating horsepower. (a * v = w)*.0013=HP, my motor is .43hp. Where I get stuck is converting HP to pull force, I think this would be the easiest way. I am trying to make a system to convert magnet pull force into a rated HP to replace said motor. In other words, replace electrically created magnetic force into just magnet force ie: electricless motor :) I think what I need is a formula to calculate two like forces (pull force and hp) ...I hope this helps! It's been hindering me for a week

Are you saying you are trying to create a motor powered by magnets only? If so I have some bad news for you.
 


mesa said:
Are you saying you are trying to create a motor powered by magnets only? If so I have some bad news for you.

Ah, I missed that part of his post I guess. You are correct, replacing the electrical components with permanent magnets will not work.
 


hp is without crank/arm of wrench to resist

resistance is proportional to the length of the arm the wrench
torqe = force * length of wrench
 
  • #10


possibly... Given my original amps and volts, is there a way to calculate the pull the the winding are generating?
 
  • #12


Power can't give you the pull force. It depends on how fast you are pulling something. To pull something twice as fast at the same pull force, you need twice as much power. And vice versa. To pull something with twice as much force at the same speed requires the same doubling of power.

Alternatively, but following the same logic as above, you can only get torque of an electric motor if you know at which RPM you are going to need that torque. Power divided by angular velocity will give you torque.

The current times voltage is, indeed, power consumed by the motor. You do have to subtract the power wasted to heat, however. So the full formula is IV - IR², where R is the resistance of the coil.

Finally, keep in mind that simply because the motor is rated at 2.8A and 115V, it doesn't mean it will be drawing 2.8A at 115V. The amount of current the motor draws will depend on the RPMs the motor is going at and applied voltage. The formula for current is I=(V-kω)/R, where R is aforementioned resistance, V is applied voltage, ω is angular velocity, and k is a constant unique to the motor. If you know the maximum RPM the motor reaches under no load and given voltage, you can estimate that constant by k=V/ωmax.

Replacing magnets in a motor will effectively alter the constant k. R will remain the same, unless you change the coils as well. And these two constants effectively determine performance of an electric motor under ideal conditions. Naturally, real world tends to be slightly more complicated, but it's a very good estimate to start off with.

Hope some of that helps.
 

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