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

[Training Wheels]

Originally Posted by Kiwi.

I just thought of another way of explaining it. The force on the sail side of the sheave is whatever the force that the sail is applying, otherwise where is that force going? So the other side of the sheave must have an equal force, otherwise the halyard must be moving.

Exactly right. Each side must have the same load. If the sheave has 100%, then the tang would have 100%, hence the head would be moving towards the masthead. The sheave would have half the load.

[savoir]

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?

The tang

[atmartin]

Line (rope, whatever) can only hold tension. Ignoring trig and the fact that the load is dynamic and not straight down, the line is under half the tension of the sail load; however, so Kiwi is right, the sheave reacts to the load from the halyard winch and half the load of the sail (so the full load).

Picture works better here:


Sorry for my knee-jerk, incorrect response.

[Training Wheels]

True, but if it were single part, the sheave in the example would be seeing twice the load, or 200 kg. So ,2 parted, the sheave is seeing half the load of the single part load.

[atmartin]

Originally Posted by Training Wheels

True, but if it were single part, the sheave in the example would be seeing twice the load, or 200 kg. So ,2 parted, the sheave is seeing half the load of the single part load.

An unfortunate use of the term "load" on my part. If "load" is defined as the force on the sail, this set-up doesn't experience half the "load", but it does experience half the FORCE of a single sheave set-up.

Boy, what a "LOAD"

[savoir]

Originally Posted by atmartin

Line (rope, whatever) can only hold tension. Ignoring trig and the fact that the load is dynamic and not straight down, the line is under half the tension of the sail load; however, so Kiwi is right, the sheave reacts to the load from the halyard winch and half the load of the sail (so the full load).

Picture works better here:


Sorry for my knee-jerk, incorrect response.

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.

[atmartin]

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 diagram is correct. There are different ways to think about this:

1: Tension in a line is constant throughout the line, in this case the line is under 50 kg tension. Solve for reaction forces and you get the diagram.

2: There is a weight of 100 kg from the 'load' and 50 kg from the person holding the load, for a total of 150 kg pulling 'down', the sheaves pull 'up' to equalize this.

As Training Wheels mentioned, a single sheave setup would yield a 200 kg force on the sheave. 100 kg 'load' + 100 kg from the winch/deck-ape whatever is pulling on the other side.

[ronbird]

I would need a diagram showing loads because I am not very rig savvy. As stated it is really just "simple" high school physics. Simple is in quotes as I always found physics a bit challenging and I have a PhD in the subject.
But the answer is in this link for all that are more nautically inclined than me. Simple once it can be properly diagrammed-something I cant do.

https://en.wikipedia.org/wiki/Block_and_tackle

[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.

[ronbird]

Originally Posted by atmartin

The diagram is correct. There are different ways to think about this:

1: Tension in a line is constant throughout the line, in this case the line is under 50 kg tension. Solve for reaction forces and you get the diagram.

2: There is a weight of 100 kg from the 'load' and 50 kg from the person holding the load, for a total of 150 kg pulling 'down', the sheaves pull 'up' to equalize this.

As Training Wheels mentioned, a single sheave setup would yield a 200 kg force on the sheave. 100 kg 'load' + 100 kg from the winch/deck-ape whatever is pulling on the other side.

I think you are indeed correct as pertains to that diagram. Must be a mechanical engineer(:>).

[atmartin]

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.

Well, I took a shot at explaining it, but I guess it didn't work. Here is a crude modification of one of wiki's diagrams:



It might help to think of the winch or 'working end' of the tackle as another weight, like in the drawing. Note the balance of forces. Otherwise, the wikipedia article linked by ronbird does a better job explaining than I can.

[ronbird]

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.

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.

[Training Wheels]

The 150kg. Was the approximate amount of compression force on the mast

[UNCIVILIZED]

What's the story with using knots to attach the ends of the halyards at the masthead? Not an ideal method for high modulus cordage.

[mikereed100]

What kind of furler do you have on your screacher? If it is a continuous line furler, the 2:1 halyard may be there to prevent twisting when you furl, not to make it easier to hoist.