I'm a mechanical engineer and it's been years since I've worried about electrical engineering so please forgive the ignorance. I'm interested in using a winch driven by an electric motor (Harbor freight item number 61840) with a duty cycle of 5% (45 sec at max rated load of 2500lbs and 14 min 15 sec of rest). However, the max load I am lifting is 250 lbs...can I use a ratio to calculate the duty cycle for this load or is the duty cycle a constant? Also, please let me know if there is any literature that can help. Thanks in advance!
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Duty Cycle
- Thread starter akpalican
- Start date
Skogsgurra
Electrical
That duty cycle is what you need to stick to if you are lifting 2500 lbs.
For lighter lifts, you can (to make things easier) increase the duty cycle according to the ratio Rated Load/Actual Load. That would be an increase from 5% to 50%.
And, this is a rather conservative calculation. You will probably be able to run that winch continuously because heat in a motor is usually proportional to load squared. And now, the nice thing with a winch is that you cannot run it continuously - if you don't have an endless lifting wire. So you are fine. No worries.
Gunnar Englund
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.
For lighter lifts, you can (to make things easier) increase the duty cycle according to the ratio Rated Load/Actual Load. That would be an increase from 5% to 50%.
And, this is a rather conservative calculation. You will probably be able to run that winch continuously because heat in a motor is usually proportional to load squared. And now, the nice thing with a winch is that you cannot run it continuously - if you don't have an endless lifting wire. So you are fine. No worries.
Gunnar Englund
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.
BrianPetersen
Mechanical
Reality ought to be somewhere between a constant (i.e. 5%) and proportional (i.e. ten times that, 50%, considering that you are only operating it at one-tenth capacity). The problem is that unless the manufacturer provides this information, you have no way to know where in this enormous gulf between the two, that the reality lies.
If you can measure the motor current under the two conditions (full load, and your expected 10% of full load), you could make a "swag" that the motor heating is in proportion to the current. The current at one-tenth load will be more than one-tenth the full-load current (and the current at no load will be non-zero) because of the magnetizing current, plus frictional losses in the mechanism. But by how much, there's no calculation that can be done without more information.
How much is at risk if you are wrong in the absence of solid data ... determines what assumption you can use.
If you can measure the motor current under the two conditions (full load, and your expected 10% of full load), you could make a "swag" that the motor heating is in proportion to the current. The current at one-tenth load will be more than one-tenth the full-load current (and the current at no load will be non-zero) because of the magnetizing current, plus frictional losses in the mechanism. But by how much, there's no calculation that can be done without more information.
How much is at risk if you are wrong in the absence of solid data ... determines what assumption you can use.
- Thread starter
- #5
Thank you all for your help!
Brian-
The table in the spec sheet for the motor said that the amp draw at 12V was 132 at 2500lb line pull. Using a ratio, I approximated that the amp draw is 13.2 for 250 lb line pull. I also just found an equation for estimating the duty cycle: duty cycle = [(rated current at req'd duty cycle) / (max current at req'd duty cycle)^2] * rated duty cycle
This wasn't the formula you were talking about, was it?
Brian-
The table in the spec sheet for the motor said that the amp draw at 12V was 132 at 2500lb line pull. Using a ratio, I approximated that the amp draw is 13.2 for 250 lb line pull. I also just found an equation for estimating the duty cycle: duty cycle = [(rated current at req'd duty cycle) / (max current at req'd duty cycle)^2] * rated duty cycle
This wasn't the formula you were talking about, was it?
BrianPetersen
Mechanical
Scotty is right about I-squared-R losses, but that's not the only thing going on, and I would not jump to the conclusion that the current draw is in proportion to the load that the winch is pulling. There will be some inherent friction in the mechanism and the motor will require a certain amount of magnetizing current, even at no load.
Go with the formula you found for duty cycle but measure the actual running current - don't guess, and don't make the assumption that the running current is directly proportional to the load.
Go with the formula you found for duty cycle but measure the actual running current - don't guess, and don't make the assumption that the running current is directly proportional to the load.
Skogsgurra
Electrical
No, akpalican. That was me. (heat in a motor is usually proportional to load squared).
But we are all right, in different ways. The real problem is that we do not know what kind of motor you have got. If it is a PM DC motor (which I assumed), then the load^2 works well. But if it is an induction motor, it doesn't fit that well. For reasons mentioned by Brian.
And, of course, Bill and Scotty are also right. As always.
Bottom line: Don't worry - Just use it. And, if it is an essential part of your operation, keep a spare.
Gunnar Englund
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.
But we are all right, in different ways. The real problem is that we do not know what kind of motor you have got. If it is a PM DC motor (which I assumed), then the load^2 works well. But if it is an induction motor, it doesn't fit that well. For reasons mentioned by Brian.
And, of course, Bill and Scotty are also right. As always.
Bottom line: Don't worry - Just use it. And, if it is an essential part of your operation, keep a spare.
Gunnar Englund
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.
I googled the OP’s winch. First hit tells it all.
winch data
It also has a pdf data sheet (owner’s manual).
It is for a car. It uses a 12V supply. By interpolation a 250lb load draws 20A and runs at 12 feet per minute.
132A x 12V= 1584W.
20A x 12V = 240W.
(although when the manufacturer says “12V” they may mean more like the 13.8V you would expect when the engine is running.)
More to come …
winch data
It also has a pdf data sheet (owner’s manual).
It is for a car. It uses a 12V supply. By interpolation a 250lb load draws 20A and runs at 12 feet per minute.
132A x 12V= 1584W.
20A x 12V = 240W.
(although when the manufacturer says “12V” they may mean more like the 13.8V you would expect when the engine is running.)
More to come …
Quoting Scotty UK: Tell the wife.![[lol]](/data/assets/smilies/lol.gif "[lol] [lol]")
At the legacy steamship site where I volunteer, there is a sign on the side of the building where the WC's are located, indicating that the mens' facilities are to the left and the womens' are to the right. Beneath this is the reason: "Because women are always right."
CR
"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]
At the legacy steamship site where I volunteer, there is a sign on the side of the building where the WC's are located, indicating that the mens' facilities are to the left and the womens' are to the right. Beneath this is the reason: "Because women are always right."
CR
"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]
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