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Thread: winch motor
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25th Jan 2021, 11:49 AM #1Most Valued Member
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winch motor
Hi fella's, free running (no load) this little 12v ATV winch motor pulls 17.5 amps in working/load direction and 12 amps in the feed the cable out direction?
The brushes are 90 deg apart and there is no brake.
I want to run it in the reverse direction to take advantage of the lower amperage as I only have 30 amps at the plug on the back of my car, also I am using a 2:1 pulley reduction for the cable so it will be only lifting under 100kgs.
It has an epicyclic gear box, all spur gears so no probs there.
Why the amps are differant from 1 direction to the other has me confused and I am wondering if the low amp direction might be less efficient and actually pull more amps under load than the other direction?
Has anyone got any clues about this?
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25th Jan 2021, 03:58 PM #2Senior Member
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What happens when you reverse the polarity? Could be a timing issue.
Last edited by waxen; 25th Jan 2021 at 03:59 PM. Reason: forgot stuff
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25th Jan 2021, 04:21 PM #3Most Valued Member
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25th Jan 2021, 05:59 PM #4Member
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Have you measured the motor current under load? It may be less than 30 amps in either direction.
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25th Jan 2021, 06:10 PM #5
Load (Amps) increases as motor is under load (mechanical). Up until the point it stalls and draws many, many amps, and melts cables/blows fuses.
Irrespective of the way it's wired, your circuit (cabling / fuse) would need to cater for more than what you observe under no load. I believe you can calculate stall load based off motor HP (as a general guide only).
Your mileage may vary, but I'd consider wiring to at least 16mm^2 cable for power and earth side, and adjust cable size UP accordingly for voltage drop based off distance from the power source (battery). Any connection would need to cater for the Amp draw you are expecting. You may be able to spec-down such connections, but keep in mind that will then be your weak-link in the circuit, and your duty cycle/trigger finger will need to be adjusted accordingly.
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25th Jan 2021, 06:33 PM #6
It is quite common for brushed DC motors to have the timing advanced for the nominal direction of rotation that the motor is expected to run in. This increases the torque available and the power that can be consumed. I believe that this is the case in your situation. If you were to attempt to operate the unit in the reverse direction while under a pulling load, it would be less efficient and draw more current than it would pulling the same load in the nominal pulling direction. i.e, operating the motor under load in reverse direction draws more than the same load in the normal direction would.
A potential solution would be to invest in a V8 or diesel capable jump starter and power the winch from that, or install an extra battery and charge controller in the boot to avoid needing to pull a lot of current through the existing cabling.I used to be an engineer, I'm not an engineer any more, but on the really good days I can remember when I was.
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25th Jan 2021, 07:25 PM #7
Reading again, might add the following:
The device that is a load in the circuit draws the current - Amps are not 'available' to use, they are drawn by the device requiring power.
While the device free-spools at a low amp draw, this changes when mechanical load on the motor is applied. Having a gearbox (torque multiplier) minimizes the load the motor 'sees'.
Now I'm not 100% up to speed with how the brush orientation, air gap, and magnetic field affect torque, but I am led to believe this has something to do with it
TE_edit.jpg
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25th Jan 2021, 08:15 PM #8Most Valued Member
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25th Jan 2021, 08:17 PM #9Most Valued Member
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25th Jan 2021, 08:18 PM #10Most Valued Member
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ok some clarification
The spec's for the winch are from what I can ascertain...
12 volt 1 or 1.2 kw motor
Gear ratio 153:1
Load rating 3000 pounds = 1360kgs
1200w / 12v = 100amps
As I only have 30amps at the back of the car I am trying to make this winch operate at less than 30 amps as I don't want to have to upgrade the wiring unless I have to.
So where I am at is I have fitted a cable/pulley reduction of 2:1, so now the motor thinks it has a 306:1 gearbox to make its life a bit easier.
My dodgy estimated calcs,
full load of 1360kgs pulls 100amps from the motor
1360kg / 100kg = 13.6 (lifting load)
100amps / 13.6 = 7.35amps
No load amps + dodgy calculated amps 17.5 + 7.35 = 24.85 amps .......
Time will tell
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25th Jan 2021, 10:25 PM #11
Hi John,
The motor current draw when spooling out will always be less than the current when pulling a load ! You can't change that !
I my opinion you have gone in the right direction by increasing the gear ratio. A two to one reduction should reduce the motor loading by the same amount, however you might find that the motor current draw doesn't reduce in the same ratio.
A secondary battery and short heavy cables will help tremendously. I would take pains to actually measure the current draw of the motor under the actual load that you want it to move. A temporary fully charged battery and heavy cables using a clamp amp will assist in getting the parameters right.
Bear also in mind that you will have to charge a secondary battery and keep it charged otherwise you may find that the vehicle wiring cannot cope with the motor load. So you will need to provide suitable fusing to the secondary battery.
HTH.Best Regards:
Baron J.
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25th Jan 2021, 11:13 PM #12Most Valued Member
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26th Jan 2021, 12:15 AM #13
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26th Jan 2021, 08:21 AM #14Senior Member
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[QUOTE=shedhappens;1980424
I want to run it in the reverse direction to take advantage of the lower amperage as I only have 30 amps at the plug on the back of my car, also I am using a 2:1 pulley reduction for the cable so it will be only lifting under 100kgs.
It has an epicyclic gear box, all spur gears so no probs there.
Has anyone got any clues about this?[/QUOTE]
A secondary gearbox with a built-in 2:1 reduction and tumbler reverse between the drum and the epicyclic gearbox output would do the job.
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26th Jan 2021, 02:20 PM #15Most Valued Member
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