Engineer wanted , power/load ratio

[sailbleu]

I intend to install this winch as an electric means of hoisting the dinghy into the davits.
I already have one , waterprioof protected , to get the outboard in . Works great . Well it's to say , all the hardware is in place ( pole windgenerator) ,it's electrically connected and operational , but not yet fully tested by pulling the outboard up . A yamaha 6 HP that weighs 28 kg . I expect no problems though.
And because of that apparent success , I want a similar winch for the dinghy that has about double the weight of the yamaha.

According to the specs this winch will draw 120 amps ( 12 volt) whilst pulling 900 kg.
What I would like to know is , what would the amps be when pulling up , lets say , 50 kg . Or better yet 90 kg ( dinghy + outboard) which is as you know 1/10 of 900 kg .
Would it be safe to say that the power consumption is also 1/10 of 120 amps ?
Or is there no linear increase/decrease of power draw in ratio with the weight/load ?
All has to do with the required wiring , thickness of the supply leads you see .

Thanks for helping me out here.

Regards

[Don Reaves]

I suppose you have already asked the manufacturer this question? That would be my first choice.

I did a Google search for "motor load vs current" and found a lot of interesting stuff, in case you don't get a technical response here.

Don

[sailbleu]

Jan 8, 2017 12:14:52 GMT Don Reaves said:

I suppose you have already asked the manufacturer this question? That would be my first choice.

I did a Google search for "motor load vs current" and found a lot of interesting stuff, in case you don't get a technical response here.

Don

I don't speak Chinese Don

Regards

[minnewaska]

I don't know the official engineering/physics answer to this. However, it stands to reason that the power draw is not linear. If you had zero load on the running motor, it wouldn't be zero draw. I'll throw a WAG out there to say it draws two thirds of its rating, even when not under load.

[rxc]

It also depends on the lifting rig that you use. Since the power that you are need is directly proportional to the force you are applying, a 90 kg load should theoretically require 1/10 the power of a 900 kg load, if they are lifted at the same rate. But then you have losses in the blocks to consider, for which there are probably no easily available data. A straight lift with no mechanical advantage (i.e., a simple "whip" rig) would represent this situation. On my boat, I have 2x6-part falls on the dinghy, and 1x4-part falls for the engine, when it is hauled separately from the dinghy. You can do the multiplcation for the multipart falls, but then the losses in the blocks needs to be accounted for. And, as the lines age and get stiffer, the losses thru the falls gets worse.

I think I could easily say that with 6-part falls, you would need a lot less power (lower starting and running current), but you will end up doing more total work/expend more energy (total amp-hours) because of the losses thru the blocks.

My criteria for stuff like this is whether it is easy enough so that my wife can be useful. She can lift the bow of the dinghy when the engine is still mounted on it, but not the stern. For hauling the dinghy onto the foredeck, she runs the anchor windlass while I manhandle the dinghy into position. She also runs the anchor windlass to haul me up the mast. For some reason the phrase "Rich widow" runs thru my head constantly while I am up there...

Edited with some correction to the terms. Power is energy used per unit of time. So, lifting 900 kg 1 meter requires 10 times the energy as lifting 90 kg. Lifting 900 kg 1 meter in 1 minute requires 10 times the power as lifting 90 kg 1 meter in 1 minute. If you take 10 minutes to lift 900 kg 1 meter, the amount of power required is the same as lifting 90 kg 1 meter in 1 minute. The amount of energy required to lift 900 kg 1 meter in 10 minutes is still 10 time the amount of energy to lift 90 kg 1 meter in 1 minute. Hopefully this explains the basic relationship. It does not account for losses in the rigging or the motor. (same concept applies when using English units)

[vasko]

why you just not try it and measure it  

[sailbleu]

Look what I found on ebay.de :

www.ebay.de/itm/elektrische-Elektro-Seilwinde-Elektrowinde-12-Volt-1-PS-900-kg-15-m-Vor-Rucklauf-/311601883698?hash=item488ceb6a32:g:UvsAAOSwtO5XInkF


1. Seillage Zugkraft 907 kg, Seillänge 2 m
2. Seillage Zugkraft 740 kg, Seillänge 4 m
3. Seillage Zugkraft 620 kg, Seillänge 7 m
4. Seillage Zugkraft 540 kg, Seillänge 10 m
5. Seillage Zugkraft 470 kg, Seillänge 14 m
6. Seillage Zugkraft 420 kg, Seillänge 15 m


Last 907 kg, Geschwindigkeit 0,9 m/min, Strom 120 A
Last 680 kg, Geschwindigkeit 1,6 m/min, Strom 90 A
Last 450 kg, Geschwindigkeit 2,2 m/min, Strom 60 A
Last 220 kg, Geschwindigkeit 2,8 m/min, Strom 30 A
Leerlauf, Geschwindigkeit 3,2 m/min, Strom 8 A

Free run takes 8 amps , lifting 220 kg 30 A .
That means I can get away with 20 amps for 90 kg and in most case just 60 kg being the weight of the dinghy itself.
And it also means the existent wiring 3mm² for the outboard hoist will also work for the dinghy.I will never use both at the same time
Just in case anyone else is interested in this type of winch , please extend your search as this price is about the double of what >>>> I PAID <<<< , shipment to Sicily included !

[rene460]

Hi Sailbleu,

I am sorry that I switched off too early to see this yesterday.  It is made for me, you see, but even then the information in your last post is the part necessary to answer.

As rxc implies, it it the tension in the cable as it approaches the winch that is the important load, weight of your dinghy modified by the advantage of your pulley system and allowing for pulley and rope friction.  You can measure it with a luggage scale if within the range of the scale.  If you arrange a two to one advantage by pulleys, you will only require 404 kg plus the contribution of pulley and rope friction to lift your 907 kg.  Four to one should be around 230 plus friction.  You can't ignore friction as you increase the number of pulleys.  The centre board of my previous boat was 450 kg, I used 4:1 then put the line on a two speed winch.  I tried 6:1 but friction meant the load on the winch was insignificantly less but still the extra rope to wind in!
For a standard brushed DC motor you can assume the performance is close to linear.  Torque Reduces as speed increases from stall (zero speed) to free running (max speed), while current is proportional to torque, so no available torque at free running speed, max torque at slowest speed and max current.  At zero speed the very high current will quickly burn out the motor, if not the wiring, but preferably the fuse!  The second table is consistant with this if I interpret it as follows- 120 A, 0.9 m/min, for 907 kg.  60A, 2.2 m/min for 450 kg, 30A, 2.8 m/min for 220 kg, 8A, 3.2 m/min at no load.  Half the load gives half the current, and approx twice the speed.  I am not sure of the meaning of the number of meters in the first table, perhaps it is a specified max lift, proxy for time of run, which suggests a limited continuous lift at 120A?  This would be consistant with the high current, but your proposed load is well below this, fortunately.
So in theory your winch will lift your proposed load, as you have guessed, but I would also second vasko's suggestion, try it and measure the current (and the voltage at the winch).  Torque is also proportional to voltage, hence if you apply the same torque but at lower voltage you will require higher current, and tend to overheat the wiring and the motor.  The test run is the time honoured test of performance.

I hope this helps, but I will have another try if you need more help.  Apologies to those who can't read "Chinese".

rene460

[sailbleu]

Totally agree with all you have said Rene .
When I get back to the boat next week I will set up an experiment with my electric outboard battery to test the amps on different loads.
My biggest concern is indeed when the voltage drops (fairly long leads from batteries to davits) , the current will get out of control and burn the 25 amp fuse.
Although the run time to hoist the dinghy will be rater limited , so I guess there's no chance for the wiring to heat up too much .

But you can never be too careful .
So if in doubt , I'll run a thicker cable .
Not really looking forward to that.

Regards

[rene460]

Hi Sailbleu,

Unfortunately I think that you are correct and larger cables will be necessary.  Cables are rated on the basis of acceptable heating which gives you a fuse rating for cable protection but too high for most device protection, and gives too much voltage drop for most practical purposes.  Rational cable sizing requires a criteria based on voltage drop.  Most sources suggest three to five percent voltage drop max, and on this basis the overheating rating is only acceptable for about two meters of cable.  You have to add the length of positive and negative cable lengths as both direction suffer the voltage drop.  Barely enough for runs under the car bonnet, and clearly not enough for long runs on a boat.  A brief test for measurement purposes as you say should not have time to heat too much with you watching for any sign of trouble, and yields a lot of useful information.

I assume you will leave the existing wire in place, insulated and tagged at each end, it will be needed one day, and just run a new wire for your winch.  I totally agree that it is not a job to look forward to.  Sorry to be the bearer of bad news but you had already worked it out yourself.

rene460