Engineers? Lifting Keel Winding Torque?

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Inselaffe

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Hello,

A question for any engineers, or others, out there..

I am trying to calculate the torque required to lift a keel of 500kg. I know that 320 turns lifts it 0.8m.

Is the answer 12.3Nm ignoring any friction?
Or have I forgot a 2(PI)?

Edit: Thought some more:
1 turn = 0.8/320 m
Assume handle 1m long
Work Done = F*Dist
WD 1 turn handle = Handle Force *2(PI) *1
WD 1 turn lift = 500*9.81*0.8/320
Ignoring friction, WD handle = WD lift
giving handle Force = 1.95N
Since handle 1m long torque = 1.95Nm
Is this correct?

Also, the mechanism is a big stainless bar with a very coarse square thread (like in a steering box on and old car) on which a nylon threaded bush travels. At the top there are two crown wheels to turn the rotation through 90 degrees. All liberally smeared with graphite grease. The keel box is sloped forwards at an angle of maybe 15degrees giving more friction between keel and box.

Presumably if I double my answer above (or the correct one if I am wrong) to allow for friction then this would be a very conservative answer.

Presumably someone will suggest a torque wrench, which I will try, but I am interested in the calculation too!

Ta V much.
 
I seem to recall reading many years ago that if (and only if) a gear train has an efficiency of over 50% can it be run in reverse, i.e. moving the output will make the input move. Now the screw pitch on your keel lift is far too shallow for the keel to drop down of its own volition, spinning the screw as it does so - even with pounding from wave action applied to "unstick" things. So I would think that the efficiency of the lifting system is guaranteed to be well below 50%.

Approaching the problem from the other end, I used to use a ratchet handle of about 0.25 m length to turn my Etap 22's lifting screw. For the (undamaged) majority of its length I would estimate that I needed to apply about 10 kgf to the handle which implies a torque of about 25 Nm. Much less force was needed to lower the keel - but the fact that any force at all was required confirms the low efficiency.

I had no gears to turn the drive direction horizontal, however: I operated directly on the screw shaft end, i.e. about 10 degrees from vertical. Are your turning gears 1:1 ratio or do they give mechanical advantage?
 
That's interesting, and makes sense.
I know that my battery drill doesn't quite have enough torque, and its rated at 20Nm.
If my calculation is correct then the efficiency is below 10%. Thinking about it I guess this is reasonable since I am effectively dragging the keel up the sloping leading edge of the keel box.
I am pretty sure the two crown wheels are 1:1, but I took the overall transmission ratio anyway from keel to handle. These cogs must further reduce the efficiency though.

I want to buy a new drill, but don't know if I can get away with 30Nm or should go to 40Nm (I know 40Nm works, I borrowed a drill), price is the problem...
I guess I will have to just wait until I'm on the boat and use a torque wrench, but I'm a bit of a geek and these kind of calcs interest me!
 
Part of your problem is "Stiction"
It will take a higher torque to start the screw turning than to maintain its turn.
if you know the pitch of the screw you can calculate the the torque required to turn the mechanism once you have it going, but th e"breakaway" torque is probably best measured - as you say - with a torque wrench.

Mike
 
Get a spring balance and attached it to the handle used to lift it. If the distance from the pivot point of the handle is 1 foot and you need 30 lbs to shift it then the torque required is 30 lb-ft. If it's 2 feet from the point and 30lbs then the torque required is 60 lb-ft and so on.

If you're a young whipper-snapper or from distant lands then use kilograms and meters and call it Newton-Meters. Don't forget to keep the scales vertical or you'll probably get a duff readling.

Geoff
 
I would have thought that you would have to look at the gearing of the line to the drum drive and the worm gear. Then of course there's friction....

Your answer suggests that the weight of the handle would be more than sufficient to lift the keel ( for a third of a turn at least)
 
Think in terms of mechanical advantage: one turn of handle will raise the keel 0.8/320 metres. If that handle is 1 m. long then its end will move through 2 pi metres so your mechanical advantage will be 2 pi*1*320/0.8 or 800 pi or approx. 2500. Thus to lift a weight of 500 Kg. would require a force of 0.2 Kg at the end of the handle. In practice friction at the screw threads will dominate things and the above result will be way too low.
 
That's the answer I got (1.95Nm), but using a simpler method.

I am amazed at how much friction there is, since I know that the actual torque needed is between 20 & 40Nm.

Probably not so surprising though given that the keel is dragged up the leading edge of the keel box

I guess better that way or I would break my wrist lowering the keel /forums/images/graemlins/smile.gif
 
ETAP22_KEELSmall.jpg


This is the Mark I without the two crown wheels at the top.
 
Sorry, I just read my post again and it sounded very abrupt. Wouldn't all the friction and gearing and so on all be reflected in the effort needed at the handle? I may have mis-understood the original question of course..
 
That's a really useful picture! It shows what a properly engineered bit of kit the Etap's keel is. When you compare that with, say, the E-boat keel with its variously shaped chunks of metal to hold it in place and a twanging wire cable pulled up by a dinghy launching winch or maybe a Tirfor...

Shows how the ageing memory plays tricks, however - I see that the shaft angle from the vertical is actually 25 not 10 degrees! /forums/images/graemlins/blush.gif The arrangement at the top is the same as mine was, though, direct drive on to the shaft.

(Deleted query re. gears having re-read your last sentence!)
 
Very true. And if your screw had a kink in it (no comments, please) because the keel had at some time been bashed hard, it was even harder work. Used to take me 40 minutes or so to lift the thing but usually only once a year. The idea that "she's a lifter so it doesn't matter if we touch, we'll just wheech the keel up" wasn't really on!
 
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