Physics 101

The beginnings of this I got from the Harken site.   I’m trying to design a mainsheet system for a sail of around 155 sq ft.   By my reckoning, in strong conditions, one needs up to 15 sq ft to one purchase [for comfort] but in light conditions, considerably less.

Therefore I was looking for a “two speed” system which didn’t involve rethreading the blocks as conditions changed.  I came up with this below and have some questions for you, if you’re not bad on physics [6th grade level or higher].

M and N are blocks on a traveller hawse running across the deck and fixing either side at each hull.   They have block A attached between them.   There is a double [or could be triple] purchase system hauling them port or starboard.   They are at deck level.

In the air, attached to the boom, are blocks B and C.   The sheet P runs through the blocks and wants to lead [through the air] back to each cockpit but it’s too short – maybe 15 feet.

Sheet J is an attachment sheet via snaphook H, which would join snaphook E.   The other end of sheet J runs to the cockpit.

Floating around on sheet J is a block L with becket [attachment point] K.   Letter I has no purpose – it was an error in the pic.  :)

F is a u-bolt attached to the hull.   G is a block which is open one side to allow a sheet to be clipped in – it is bolted to the hull as well.

OK, let’s begin.   By my reckoning, if D is attached firmly for the nonce, then sheet P is in a 3:1 purchase, if you were to haul directly on E.

Question 1:  is that so or is it 4:1?

In light conditions, assuming it’s 3:1, then that is about right for that sail and were I to attach H to E directly, then it does not alter the purchase.

However, if conditions become moderate, I need to increase my purchase.   So I clip H to F and block L to E.   By my reckoning, I’ve turned 4:1 into 6:1.

Question 2: is that so or is it some other purchase?

If conditions become bad, then I need 9:1.   I now remove H from F, run it through G [which is open and doesn't require any unthreading], go back up and attach H to becket K.   By my reckoning, I’ve just created 9:1.

Question 3: is that so or have I created 12:1?

If conditions become really bad,then I reduce sail.

Thanking you in anticipation.

6 Responses to “Physics 101”

  1. ivan October 9, 2012 at 22:46 Permalink

    It would help if you did a diagram showing how things were attached.

    Which line are you pulling on and what do you expect it to do?

    As you show it you have all single roller pulleys therefore there is no mechanical advantage gained at all.

    To get a 9:1 ratio it means you are pulling 9ft of rope to move the block 1ft.

    From my memory of the sailing werries on the broads most of their blocks were at least 4 rollers which gave a reasonable ratio for ease of lifting the sails.

    You could make a small model of your layout on a piece of plywood and use one of the old spring balances to check if you get any advantage with any setup,

    I will try and do some diagrams showing how pulley ratios are calculated – actually dig out the correct book and scan the necessary pages – and e-mail them to you

  2. Suzie October 10, 2012 at 00:34 Permalink

    James,

    (By my reckoning; it has been a while) I agree with the 4:1 ratio, which may be expressed as 3:1 depending on whether or not the 4 or 3 includes the useful force at point E. It’s a matter of preference.

    The important thing is that the force you are opposing is divided by four ‘fixed’ points – D, two at A and E (assuming you’re strong enough).

    The other point you’re assuming which isn’t necessarily valid is that the angles involved are all 180 degrees. Anything less than this and you can’t assume the load is shared – you’ll need to fetch your trigonometry books too. Frictional forces aside (how good are your pulleys?) if you can keep your angles tight, moving from question 1 to 2 should double your purchase to 8:1; moving from question 2 to 3, should double your purchase again. That’s in theory and isn’t consistent with the angles shown in the picture.

    Go build you boat old chap and try not to capsize it. I would so miss our sparring.

  3. James Higham October 10, 2012 at 00:48 Permalink

    Ivan, Suzie, I didn’t state it clearly enough above but B and C are on one boom up in the air, being drawn towards M, A and N on the deck. Everything over on the left is an attachment to E at one end and the hull at the other.

    To haul down vertically on E would use 9 feet of rope to move the boom 3 feet. That’s 3:1. As for separate blocks, it doesn’t matter if B and C are together or apart, as long as they are on the same boom and a few inches apart, the boom gets pulled down, which is done with mechanical advantage.

    The reason for keeping the blocks separate is for ease of running so they don’t get jammed and tangled.

    The question of whether the rope after the last block is ever part of the advantage or not has always puzzled me – this is moving on to other things now. For example, if E should feed into a block on the deck, it would be 4:1 but if it only goes to hand, then it would be less than that. How much less appears to depend on the angle.

    So if the line at E went 90 degrees to B, horizontally instead of vertically, that alters the advantage again. Or does it? Perhaps it doesn’t alter it at all. Is that vectors?

    Harken claim that a 150 sq ft sail only needs a 6:1 block but I think they mean in conjunction with a winch. I remember a jib of 50 sq ft with a 2:1 and that almost ripped the arms out in 15 knots of wind.

    From that I reasoned that 30 sq ft at 2:1 would be reasonable, hence 15 sq ft for one purchase.

  4. James Higham October 10, 2012 at 01:09 Permalink

    Yes, that all seems clear enough now. CAB are pretty well vertical, which is a definite 2:1 but the others, at angles, are not. Let’s say they’re 0.5 of a purchase each, making 3:1 all up – that would still be enough if it was doubled or tripled by the other system. The downside is the amount of rope that has to be hauled in for a tiny amount of movement of the spar but that’s MA.

    Thanks for the help.

  5. Robbo October 10, 2012 at 07:16 Permalink

    I’m puzzled by your calculations here. The system EBACD gives you 2:1 provided BA and AC are near vertical, and less if not.

    You can get a further 2:1 by attaching L to E and H to F and passing J round G as long as FB and GB are almost parallel. However, I think the distance EF needs to be four times the distance you want to move A, so this may be a constraint.

  6. James Higham October 10, 2012 at 08:11 Permalink

    Robbo, sure you’re right – EBACD certainly gives 2:1, if E goes to hand and D as well.

    However, if D is fixed, then ACD has a certain input, which, as you sy, “near vertical, and less if not”. As CD is clearly not vertical, then it can’t count as a full purchase.

    But let’s put D directly under C, beside A and it would count as a purchase. So, from a point above BC [hauling upwards], there’d be no advantage but below, parallel to CA and fixed to the same deck A is, there would be purchase.

    There must be a continuum in between. Whatever the MA is, as is now, HJLK will multiply it. As Ivan said, you’d need to see a diagram of the boat to work out the MA.

    In practical boating terms, the trick is to add a block above between B and C and below, beside A, which really would make it 4:1 plus a bit extra and using the add on at F/H-G/E-J[to cockpit], that would double that, which would be enough for that size sail.

    Those comments have been most helpful, thanks, more than you perhaps realize. It’s cleared away many of the preconceptions and shall be borne in mind when placing the blocks on the deck at strong points.

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