What size windlass battery and where?

Would you like to elaborate on that word of warning?

Most of the points have been covered already.

Personally I am not necessarily adverse to extra weight in the bows of many cruising yachts as it helps distribute the boats mass, but the main things to consider (without repeating too much what has already been said):

One should size the cabling from the battery to the alternator for the alternator output in any event. In most boats the alternator capacity will be around half the load current of the windlass motor (as a generalisation) so one is stuck with biggish cables in any event. Even so, a forward battery is likely to suffer through undercharging due to the voltage drop on the cables meaning that the battery will not get long runs at 14.whatever volts. One can play tricks to minimise the charge current so can use smaller cables but in the end they are counterproductive in that they emphasise the charging difficulty.

Small alternators (so limiting the charge current cable sizing to the forward battery) are usually found on smaller boats so the cable run from the engine battery is shorter anyway so may as well cable back to there (they also have lower draw windlasses).

Batteries are high maintenance items and even more so if not maintained fully charged. I wouldn't necessarily go along with the motion difficulty up in the bow as cranking batteries get plenty of motion in cars and off road vehicles and seem to survive ok.

Add to the comments made on draw from the alternator in that the alternator will provide high currents to the windlass even if the battery is in good condition unless a big bank is used due to the volt drop on the battery when under the windlass load. OK that can be limited but then one is putting the whole load on the battery - with the collapse of capacity of battery capacity under high load and their sensitivity to undercharging I can only see this as a bad scheme unless a lot of battery capacity is used (as may be with the very high loads of a bow thruster). While the actual AHr drawn even say recovering 75m of chain (say 5 minute run) is low it is at a comparatively high amperage compared to battery size so may be significant for a smallish battery (due to capacity collapse under high draw).

I cannot imagine anyone wanting to put a breaker in the cables to a forward battery to protect against the alternator charge capacity as has been suggested (a breaker should be used, obviously, but sized appropriately according to the alternator output).

Generally all this means that unless one has other very high forward current loads it is sensible to cable back to the cranking battery (or house bank if it is safely capable of the high current draw ie adequately CCA rated). Generally, anything different to this I have seen or heard of have been mainly associated with amateur adventures where it is considered challenging and exciting to add more relays, bulbs, batteries etc as gimmicks rather than driven by good sense. But not always, there are exceptions.

If it all becomes too hard due to high loads ie big boat, then 24v, hydraulic or 230 v ac off generator become the solutions (depending on boat size).

John
 
I ran a 1000w Lofrans on my last boat - with a 90 amp trip and no problems. I ran heavy duty electrical welding cable from the battery bank in the stern to the windlass - it was quite cheap. I always ran the engine when using the windlass.

A bare 12 v would give an 85 amp max draw at the windlass but since you are running the engine and if you havent stinted on the cable you will have more than 12v and therefore less than 85 amp.
 
[ QUOTE ]
A bare 12 v would give an 85 amp max draw at the windlass but since you are running the engine and if you havent stinted on the cable you will have more than 12v and therefore less than 85 amp.

[/ QUOTE ]You mean more than 85A, not less /forums/images/graemlins/smile.gif
 
Presumably that same would apply to Bow thrusters? The previous owner of my boat fitted a bow thruster and also put one huge battery in the bows, apart from the wieght it takes up a lotta space in the forcabin. I had wondered if it was neseccary as one only ever gives a blast on the things and again more than likely when the engines running. What say I get rid of the battery up front and either re home it elsewhere of get shot altogether?
 
You mean he's re-written Ohm's Law? Suppose anything is possible in TB Land. Glad I'm in Spain where V continues to be = IR /forums/images/graemlins/smile.gif
 
You could re-wire the system so that the battery was near to the other batteries, for example. Or even use the other batteries for the bowthruster and dispense with the other one altogether. However, you would need to check that the type and size of the existing batteries is suitable for such a large current (you need about 60A per horsepower of bowthruster). You would also need to upgrade the wires up to the bowthruster. People here have mentioned welding wire - I have never used it and don't know the spec. but it sounds like a good idea as it is flexible and said to be cheap. It is certainly readily available.
 
I believe that Tome and Birdseye are quite correct. Do your sums again on the assumption that POWER remains constant.

One assumes that changing the voltage to the windlass does not change the rate of energy expenditure needed to haul the anchor. So if voltage increases, current decreases - similar to other electric drives and is one reason why they may burn out at low voltage (current increases - I think you said much the same in an earlier post).

John
 
[ QUOTE ]
I believe that Tome and Birdseye are quite correct. Do your sums again on the assumption that POWER remains constant.......One assumes that changing the voltage to the windlass does not change the rate of energy expenditure needed to haul the anchor. So if voltage increases, current decreases

[/ QUOTE ] What sort of motor is in this weird windlass? Series, shunt, compound? I'm waiting with eager anticipation to hear about this windlass motor that actually draws less current as the voltage rises.

[ QUOTE ]
- similar to other electric drives and is one reason why they may burn out at low voltage (current increases - I think you said much the same in an earlier post).

[/ QUOTE ] No! No! No! That was an electronic inverter (or dc-dc converter) in which the current will rise as the input voltage falls. But not a motor - at least, not unless it has some kind of highly complex integral electronic controller.

Is this a leg pull?
 
Ohms law only applies to a resistive load - and a motor is anything but resistive.

As a motor spins it develops "back EMF" that is proportional to the speed of rotation - and that in turn reduces the current flow. So a windlass turning with no load is not using much power.

As the load on the windlass increases, its speed decreases and so the Back EMF decreases, causing an increase in current and therefore power consumption.

This helps to explain just why voltage drop in the cables to a windlass is so important because, unlike a resistive load, the motor tries to compensate for the voltage drop by consuming more current, hence further increasing the drop.
 
For 'windlass' read bow-thruster - I'm sure the windlass will come before long!

My starter battery is dedicated and at present the thruster comes off the other, domestic, battery. It doesn't take much use (OK, digging a new channel into/out of the berth) to wipe out instruments e.g chartplotter.

I could easily transfer the thruster feed to that starter battery - and irrespective of your version of Ohm's Law I can grasp that if the engine is running, battery drain will be less - but to what advantage? Don't I run the risk, after a particularly heavy session crawling back home, that the starter motor would be getting low, at which point I moor up, switch off and go home. Which leaves me with a less than A1 battery with which to fire the engine next time.

Sorry, I'd prefer to cast off with a low domestic battery, wouldn't you? (No, I won't get lost without that little picture to show me the way - I'm the dinosaur with a paper chart, remember??)

And for the idealists, I am planning to access a void under the cockpit sole and install a third battery. Then thruster AND windlass would come off that.

Wodja think?

P.S Battery location isn't an issue: you raggies are bound to keep weight out of the ends, so no choice. I can out put it where I like, which is . . . next to the others of course.
 
A motor under load draws more current than a motor on no (or less) load but that has no bearing on this. What he said is that a motor with a fixed load (anchor and chain) will draw less current as the supply volts are decreased. That is untrue.
 
Consider a series wound DC driven windlass operating at max load (constant). We reduce the voltage and it slows (actually does less work) until it soon approaches stall (we are operating at max. load capacity for normal voltage) and then stalls. We started at max load so only a smallish drop in voltage will result in a stall.

As the motor appoaches stall and then stalls the current drawn will be very much higher than when operating at full load. So current has increased even though we have reduced the voltage.

Or are you claiming that if one operates a windlass under full load and stalls it by reducing the voltage below normal operating voltage then you are quite safe from burning it out at stall or close to stall because the current will be less than at full load?

John
 
John, I am not 'claiming' anything at all. I am telling you that if you reduce the input voltage to an anchor windlass (series wound motor) under load the current will fall. Fact. That is all we are talking about. End of story.
 
So you are saying that for a motor providing constant power if you reduce the voltage you reduce the current - umm - I don't think so!
 
Who ever said that this is a 'constant power' device? Where does this notion come from?

It does have a very constant speed for a set voltage, but not constant power (or speed) for varying supply voltages.
 
So, as I asked, you are saying that when you reduce the voltage of a loaded windlass from rated voltage to the point it is stalling then the stall current is less than the current at load at rated voltage? If you are saying that then you are saying that it is then safe to approach stall or stall the motor with no risk from current induced overheat.

If you are not saying that then it must be that if one reduces the voltage the current increases.

John
 
I don't think that there is a simple answer to that. The equations of a rotating machine are quite complicated - I haven't had to work with them since my undergraduate days and now that I am living aboard I don't have my old textbooks so I can't give chapter and verse. However, in estimating this stall point you would have to take the family of torque/speed curves for different drive voltages. You would need to make an estimate of the stiction as well as the dynamic friction and the load and equate that to the torque - you would need to do that graphically. The current would fall with voltage until stiction stopped the motor and then the current would rise. Not hugely, because the motor would already have been going quite slowly; you would just see a little blip.

Now, if you have the family of torque/speed curves for different voltages, you will see that the torque falls as the voltage falls. And the current falls with the voltage.

Maybe the easiest way for you to prove it to yourself is to ask yourself what would happen if you put a 12V dc supply onto a 24V dc windlass. Do you think that a huge current surge would result, with potential damage (which seems to be what your are saying) or that the windlass would just run very slowly with little power (which is what I am saying). Which?
 
Do you think that a huge current surge would result, with potential damage

Whether there is the potential for damage depends on at what voltage the motor would ordinarily have stalled at when fully loaded, as I think we are likely to agree that if one reduced the voltage to 1v current would be low and there is no likelihood of damage even though stalled.

At 12 v and rated load on a 24v loaded windlass I would assume that the motor is very likely already stalled - so, in the case that the motor would ordinarily have stalled at 18v, say, then the stalled current at 12 v may likely not be higher than or as high as the normal run current at rated voltage. But at 18v, taking that as the stall voltage for the load on the windlass, then I would expect it to be significantly higher than the run current and unless the motor is protected damage may result. Also, the fact that the voltage is 18 v may be not that we have artificially reduced the voltage to that for the sake of this example, but is the real life situation in that the 6v drop from 24v has come about through voltage drop on the cables and depression of the battery/charging voltage with the load.

Getting back to your suggested 12v and assuming that the motor actually stalls at 18v then the current will decrease as one lowers the voltage from 18v to 12 v, because we are now into the Ohm's Law territory that you originally suggested (there being no other influences from rotation), and the current at 12v may indeed (probably?) then be much less than the run current.

All along I have been careful to state the windlass to be loaded. I think it is obvious that if the windlass is lightly loaded or not loaded at all, then if we try to stall it by lowering the voltage the stall voltage will be very low indeed and the consequent locked rotor current also low and of little concern. However, that is not a real life scenario that occurs on the boat - we stall the motor at close to rated voltage.

I think what one believes gets down to whether one believes or not that under rated load in real life the windlass motor draws more current when approaching stall or stalled than when running, because at those points the voltage is at its lowest due to battery/charge system depression (obviously if that depression is exaggerated then no current problem in stall but maybe problem in running /forums/images/graemlins/smile.gif). The outcome of that belief (which is what I accept and is what happens in my experience because the breaker opens when the motor is slowed or stalled through overload) is that the motor draws more current when the voltage is reduced.

John
 
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