Windlass 12v wire - what size

pampas said:
The cooker cable is rated at 60 amps continuose, and if my voltmeter is correct the volt drop is so small that I could not measure it at the control panel, length of run about 18 feet.
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What current was flowing when you measured the voltage? Of course you would not be able to detect a voltage drop if all you were doing was measuring the static voltage.

No way with Cooker wire could you possibly get no voltage drop on 100 amps flow. You cannot defy physics. The recommended size at 50sq mm is very considerably larger than cooker wire. Go back and check it when the brute is running.
 
Onia said:
Yes we know thicker is better but I still don't see why you would want to have thick for 7metres followed by significantly thinner for two metres.
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Put thin cable in over a long run gives a large voltage drop. Instead of the lecy powering the winch it heats up the cable. Winch runs slowly and tries to take more lecy ....

What ever happens you end up with lower voltage at the winch which decreases it's life due to lower voltage.

The thin bits of cable just allow you to connect it easily. There is possibly nothing to stop them being shortened but I would not increase their length.
 
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pampas said:
The cooker cable is rated at 60 amps continuose, and if my voltmeter is correct the volt drop is so small that I could not measure it at the control panel, length of run about 18 feet.
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the voltage drop can only be measured under full load.
However, with a good meter set on its greatest sensitivity, you may detect an increase of resistance with thinner cables. ( resistance = ohms )
Greater the resistance, the greater the volt drop and as others have said you just warm the cables up and the motor has less power. ( power = Watts. Watts = Volts x Amps )
Back to Ohms law, Volts = Amps x Ohms

Reminds me of all those tedious formula I had to learn at school for Pythagorus' theory S=P/H C=B/H T=P/B
where S=Sine P=Perpendicular H=Hypotenuse C=Cosine B=Base T= Tangent and the way I used to remember it was with a rhyme, 'Some People Have Curley Black Hair Through Perpetual Brushing'.
Sad what old age does to one. :-)
 
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pampas said:
The cooker cable is rated at 60 amps continuose, and if my voltmeter is correct the volt drop is so small that I could not measure it at the control panel, length of run about 18 feet.
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How exactly are you measuring the drop? If you are just measuring at the panel end, you won't see anything except perhaps a slight drop in the voltage due to the internal resistance of the battery etc. It won't be the voltage drop on your cable runs.

The are a couple of accurate ways to measure voltage drop on a cable run. (assuming the voltmeter is sensitive enough)

The easiest way is to measure the voltage at the motor end of the cable tun and compare it to the voltage at the battery end. The difference is the voltage drop in the cable(s)

Another way is to get a long length of wire (it can be quite thin as the current feeding the meter is so small) and connect one end to the battery and the the other end to the SAME polarity at the other end. The meter will read zero when there is no load. As soon as you start to draw current, the meter will start reading. Put the meter on a low voltage scale and you will be reading the actual voltage drop on that cable. measure the positive and negative cables and you have the total voltage drop!

You need a digital multimeter, (or an old fashioned AVO or similar) One of those voltmeters for measuring car battery voltage won't do it. Uou are trying to measure voltages of less than half a volt (or differences in reading of less than half a volt in the first example).

If you had a one volt drop it would be 8% and that would not be good...
 
john_morris_uk said:
How exactly are you measuring the drop? If you are just measuring at the panel end, you won't see anything except perhaps a slight drop in the voltage due to the internal resistance of the battery etc. It won't be the voltage drop on your cable runs.

The are a couple of accurate ways to measure voltage drop on a cable run. (assuming the voltmeter is sensitive enough)

The easiest way is to measure the voltage at the motor end of the cable tun and compare it to the voltage at the battery end. The difference is the voltage drop in the cable(s)

Another way is to get a long length of wire (it can be quite thin as the current feeding the meter is so small) and connect one end to the battery and the the other end to the SAME polarity at the other end. The meter will read zero when there is no load. As soon as you start to draw current, the meter will start reading. Put the meter on a low voltage scale and you will be reading the actual voltage drop on that cable. measure the positive and negative cables and you have the total voltage drop!

You need a digital multimeter, (or an old fashioned AVO or similar) One of those voltmeters for measuring car battery voltage won't do it. Uou are trying to measure voltages of less than half a volt (or differences in reading of less than half a volt in the first example).

If you had a one volt drop it would be 8% and that would not be good...
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darned power of edit doesn't work out here on this slow connection. I meant to say use the thin wire to extend one of the meter leads and connect one end to the battery and the other end to the motor same polarity...
 
Chris_Robb said:
What current was flowing when you measured the voltage? Of course you would not be able to detect a voltage drop if all you were doing was measuring the static voltage.

No way with Cooker wire could you possibly get no voltage drop on 100 amps flow. You cannot defy physics. The recommended size at 50sq mm is very considerably larger than cooker wire. Go back and check it when the brute is running.
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These winches are spur gears with a mag brake attached and only take 19 amps at full chat,unlike the rest that are worm and wheel with very high efficiencys lost in the drive. I aint stupid spent all my life in the electrical wourld so know where to measure for volt drops. it would appear that many on this post assume it takes 1000watts may I respectfully suggest a geek at Southern wince site would be in order.
 
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pampas said:
These winches are spur gears with a mag brake attached and only take 19 amps at full chat,unlike the rest that are worm and wheel with very high efficiencys lost in the drive. I aint stupid spent all my life in the electrical wourld so know where to measure for volt drops. it would appear that many on this post assume it takes 1000watts may I respectfully suggest a geek at Southern wince site would be in order.
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Can you post a link to these winches.

I wonder why any of us have been bothering with the conventional ones if you can operate one on 19 amps, which is less than the power taken by my dinghy blower upper.
 
Chris_Robb said:
Can you post a link to these winches.
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Here you go...

Note that the "1000" in the product name appears to allude to the power of equivalent worm drive windlasses, not the power of these units. They quote a current draw of 25A for the 12V "1000" (and recommend a 50A circuit breaker), so it's running at about 300W.

Came out very well in the Sailing Today test a few years ago, IIRC.

HTH

Andy
 
misterg said:
Here you go...

Note that the "1000" in the product name appears to allude to the power of equivalent worm drive windlasses, not the power of these units. They quote a current draw of 25A for the 12V "1000" (and recommend a 50A circuit breaker), so it's running at about 300W.

Came out very well in the Sailing Today test a few years ago, IIRC.

HTH

Andy
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Umm - is says 1100W input, 450W output - but claims 82% efficiency? and how do you get 1100W by drawing 25A from a 12V supply?

25A x 12V = 300W input. 300W input x 82% = <250W output.

The figures just don't add up....
 
bedouin said:
Umm - is says 1100W input, 450W output - but claims 82% efficiency? and how do you get 1100W by drawing 25A from a 12V supply?

25A x 12V = 300W input. 300W input x 82% = <250W output.

The figures just don't add up....
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Thats what I was scratching my head about. It would be wonderful if the fugures added up.....
 
Chris_Robb said:
Can you post a link to these winches.

I wonder why any of us have been bothering with the conventional ones if you can operate one on 19 amps, which is less than the power taken by my dinghy blower upper.
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Sorry but I googled it then found that Darglow engineering sold them but do not have their link to hand.

I bought a vertical one + capstan about 3 years ago, very satisfied with it Its only fractionally less powerful than conventional makes and trips at 20 amps does what it says on the box. When powering out it really goes a bit to to fast for, me but you get used to it, brilliant bit of kit cannot praise it enough!!!
 
In practice on the few occasions I have used an electric windlass, or seen one in use, it has not raised the anchor any faster than I can do by hand. I think for the sort of motion I use on my windlass I can probably output something like 200-250W (maybe a little less) so perhaps 250W output is actually realistic and these 1000W claims just vast exaggerations
 
bedouin said:
In practice on the few occasions I have used an electric windlass, or seen one in use, it has not raised the anchor any faster than I can do by hand. I think for the sort of motion I use on my windlass I can probably output something like 200-250W (maybe a little less) so perhaps 250W output is actually realistic and these 1000W claims just vast exaggerations
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When just pulling in chain they are hardly loaded. However, when we use the windlass to pull me up the mast, you know it must be working bloody hard - so probably near capacity of 1000W to raise 15 stone quickly. I suppose one could do the calculations about height weight and speed of lift..... and come up with the watts needed. (but I can't!).
 
I suppose one could do the calculations about height weight and speed of lift..... and come up with the watts needed. (but I can't!).
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Can't, why not? Went to school didn't you?

15 stones, call that 950 Newtons, multiply by the height of the mast in metres to get the energy required in Joules. A watt is 1 joule/second so if a minute (60 seconds) is quickly enough, for a 12 m mast I make that 950x12/60 = 190 watts. Not allowing for frictional losses of course or the less than 100% efficiency of the winch
 
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