Does 2:1 halyard lessen load on mast head sheave?

[savoir]

Originally Posted by ronbird

Ha ragione.
I certainly admit that it seems to defy "common sense", but then again so does Quantum Physics and anyone telling you otherwise is not being honest with you.
The Wikipedia link I referenced is worth a read and I am sure there are many other links. In Physics, we call it Statics and devote much of a semester to such discussions. So it is not trivial stuff, but LOTS of fun.

I took this diagram from that discussion. As you can see, there is that troublesome 150 again. You could make it 100 by not pulling down (red arrow). Bring it up, maintaining tension, and tie it off at the top block.

This diagram follows the same thinking as the previous one but both are still incorrect.

The total system shown is holding a force of 100 newtons. You can't fabricate another 50n from nowhere and claim that the system now holds 150n in total. The system is 2:1 and the diagram correctly shows 50n load on the two ropes on the right. If you could ignore the upper block and insert a strain gauge into the line on the right and hang the gauge from somewhere else then it would show 50n.

Perhaps the confusion arises because the upper block is just a take off block and not a purchase adding block. The 50n shown on the left is the same 50n as that shown on the right (just in a different direction) rather than additional.

The system shown could be made 3:1 by making the bottom block fixed and the top one moveable in which case the load on each section of rope would be 33n.

The only way to increase the upper load would be to raise the weight. That would vary with speed but applying 51n of force to the rope on the left would overcome gravity and start to raise the weight. The upper load would be 101n or more depending on the speed of raising.

[Snowpetrel]

Originally Posted by savoir

You do not explain how a weight of 100kg can create a total 150kg load.

Telling everyone " The diagram is correct " doesn't cut it.

The diagram is correct (provided we ignore the very minor friction effects) as previously explained the extra 50 kg comes from the effect of the halyard running down the mast to the base. Without the 2:1 the sheave load would be doubled and the total force 200kg.

Stretch in the halyard will be approximately half as well due to the reduced load.

[a64pilot]

A lot is over my head, but a two to one or three to one etc does not lessen the load on the masthead or the sail, it does of course decrease the force required on the line by increasing the length of pull.
So the forces on the line are lessened.
Reason I'd like a 2 to 1 is to get the same forces but half the required pull on the line by doubling the length of pull. Easier on the winch and clutch.

[Adelie]

Originally Posted by savoir

The diagram is not correct. The yellow fixed point above B will have a load of 50kg assuming no intervening factors. A weight of 100kg cannot create a load of 150kg shared between two fixed points. Whatever load each point may have the two combined cannot exceed 100kg.



The number above B should be 50kg.

The hand is pulling down with 50kg. So all down forces sum to 150kg and all up forces sum to 150kg.

Technically this should be 1500N but people understand kg better.

[rhubstuff]

Originally Posted by Kiwi.

Remember there is only one halyard going over the sheave,; the multipart is all on the sail side. So the full load of the sail is on the sheave, what else is up there to take the load?

Answer: the tang is up there, and its function is to take half the load. If the tang takes no load, are you saying that it can just be removed? Electricity is not linear force and cannot be equated.

[Privilege]

On a slightly different note, I happened to email my mast manufacturers (U.S Spars) the other day in relation to my 2:1 halyard setup. Here is what i asked;

I wonder if you might be able to offer me a quick bit of advice on my mast/running rigging setup.

I have a 58’ Z1001 mast on my 44’ catamaran. The mast head is fitted as standard that came with the mast a few years ago. The main has a 52’ luff. I currently have a 2:1 main halyard for raising the sail but I’m getting a lot of twisting issues which I’m trying to rectify. I have a battcar system, so raising the main with a 1:1 halyard is not really a problem for me and would alleviate the twisting issues. The question is, would it present a problem for the masthead? Is this going to put too much strain on the block or would it be better to stick with the 2:1? I don’t race the boat or sail hard.

I’d appreciate any input.

David,

This was their reply today;

The sheave in the masthead is able to handle the load and the other option would be to change out the rope for one that doesn’t twist.

[savoir]

Originally Posted by Adelie

The hand is pulling down with 50kg. So all down forces sum to 150kg and all up forces sum to 150kg.

Technically this should be 1500N but people understand kg better.

If your proposition was correct then the weight would be rising because 150n worth of force is being applied to a pulley system carrying 100n weight. Nothing else is possible.

[estarzinger]

Originally Posted by Privilege

(U.S Spars) . . . the other option would be to change out the rope for one that doesn’t twist.

If you follow up with them, I would love to hear their specific suggestion/recommendation for a specific brand/model of rope which will not twist.

I tried a couple (T900 and Endura Braid and a NZ brand dyneema cored double braid) and they all twisted.

I found the best things to do were #1 'lock' the head block so it cannot swivel and #2 move the masthead dead end as far from the sheave as practical to create separation which causes the rope to 'untwist' when it gets up there, and #3 make sure you don't put any twists in while coiling (eg always figure 8 coil) and #4 just take the halyard out of the mast every once and a while and trail it behind the boat to make it straight.

But it would be nice if there was a specific recommendation for a rope that was particularly twist resistant?

[Adelie]

The weight is pulling down with 100kg or 1000N. The hand is pulling down with 50kg or 500N. The sum of down forces is 150kg or 1500N.

This is the same as the sum of up forces.

[Pizzazz]

It may help if you imagine that there is a single block at the top of the mast and you have a line going from the mast base, through the top block and back into your hands. This will give you a 2:1 purchase for bending the mast.

If you understand this, then you must agree that in order to raise a 100 kg sail (pulling on one end of the block, you need to apply 100 kg (equivalent) of force on the other end. Thus the total bend force for the mast will be 200 kg.

Adding another block (as shown in the diagrams), halves the tension in the line from 100 kg to 50 kg, doubles the length and since you have three lines, the total force is 3x 50 kg, thus the mast bend moment is reduced by 1/4, the tension by half and the line length is doubled.

[sailjumanji]

OMG people. The sheave takes half the load, and the tang takes the other half. Remove the tang and dead-end the halyard to the head of the sail and the sheave takes 100% of the load.

[Pizzazz]

Originally Posted by sailjumanji

OMG people. The sheave takes half the load, and the tang takes the other half. Remove the tang and dead-end the halyard to the head of the sail and the sheave takes 100% of the load.

Not correct. The tension in the line is the same throughout the line. There is one line pulling on the tang, but there are two lines (one going to the sail, the other going to the winch) pulling on the sheave. Hence, the load on the sheave is 2x the load on the tang.

If you remove the tang and dead-end the line to the sail, the force on the sheave will be 2x the weight of the sail (split between the sail and the winch). If you dead end the winch end of the line to the sail as well, then the load will be 1x the weight of the sail but you will not be able to raise it. Think about it again. It is a little non-intuitive.

One more way, when using a 1:1 halyard, you have to pull hard when raising the sail. This bends the mast slightly. When using a 2:1 halyard, you pull less. Therefore, the force on the mast will be intuitively less. The only time the force on the mast will be equal to the weight of the sail is if you are actually sitting on top of the mast and pulling the sale up. Now may be it is clearer.

[sailjumanji]

Originally Posted by Pizzazz

Not correct. The tension in the line is the same throughout the line. There is one line pulling on the tang, but there are two lines (one going to the sail, the other going to the winch) pulling on the sheave. Hence, the load on the sheave is 2x the load on the tang.

If you remove the tang and dead-end the line to the sail, the force on the sheave will be 2x the weight of the sail (split between the sail and the winch). If you dead end the winch end of the line to the sail as well, then the load will be 1x the weight of the sail but you will not be able to raise it. Think about it again. It is a little non-intuitive.

One more way, when using a 1:1 halyard, you have to pull hard when raising the sail. This bends the mast slightly. When using a 2:1 halyard, you pull less. Therefore, the force on the mast will be intuitively less. The only time the force on the mast will be equal to the weight of the sail is if you are actually sitting on top of the mast and pulling the sale up. Now may be it is clearer.

Really, if you remove the tang and tie the line directly to the sail, the load on the sheave will be 2x the weight of the sail? Wrong. The load on the sheave would be 1x the weight of the sail.

So in your mind, if I had a line with each end dead-end into a separate tang, the load on each would be? Each 1x the weight of the sail? Wrong. Each would carry half the weight of the sail. Replacing one of the tangs doesn't increase the load on that point.

[Snowpetrel]

Originally Posted by savoir

If your proposition was correct then the weight would be rising because 150n worth of force is being applied to a pulley system carrying 100n weight. Nothing else is possible.

Originally Posted by sailjumanji

Really, if you remove the tang and tie the line directly to the sail, the load on the sheave will be 2x the weight of the sail? Wrong. The load on the sheave would be 1x the weight of the sail.

^^ All I can say is don't take up engineering.

Frankly I am astounded that some people cannot grasp some of the very basic realities of how blocks work. A 1:1 tackle rigged to disadvantage such as a masthead sheave ends up with near enough twice the force that is on the rope in the upper block.

Also the load on the system is far above just the weight of the sail. There are luff tensions, sheet loads and wind pressure loads which on a 40 footer can easily end up getting into tonnes of force, far far above the basic weight of the sail.

This is all VERY VERY basic mechanics. The diagrams are right. Wikipedia is right. The text books are right. It is not a conspiracy, just basic engineering.

Please do some research and a bit of critical thinking.

[charliehows]

bit late to chime in but...aside from the tortuous discussion on sheave loading the other reasonable consideration is tangling - i had a 2-1 halyard on my mizzen - it wasnt necessary on a small sail but i left it in place until the bloody thing jammed when i was trying to drop the mizzen in the middle of a very busy channel. If the extra tackle isnt necessary for strength get rid of it.