Reeving a Tackle to Advantage -- What am I Missing?

[2Hulls]

Originally Posted by TeddyDiver

Sorry but your math is sound and angle makes a difference..

OK, then show how it works and what difference it makes.

Dave

[Seaworthy Lass]

Originally Posted by 2Hulls

OK, I'll try once more, using conachair's very artistic drawing.

Premise: the angle of pull on the running end has no effect on the MA nor for the force to move the load.

To show that the premise may be true, let's assume it is false and test what happens if it is false, e..g., test that the angle makes a difference.

If the angle in the drawing was 0*, i.e., vertically up parallel to the rest of the rig, the load in this 3:1 system would be 1/3 the weight of the load, right?

I assume those arguing that the angle makes a difference must be using the logic that any angle from vertical must still apply the same vertical component as if the angle was vertical or 0*. So, if you postulate that the force on the rope holding the load has to increase as the angle of the force departs from vertical, i.e., increases from 0*, then to maintain the same vertical component you would have to divide the force by the cosine of the angle. This is the only trigonometry that can apply.

To keep it simple, let's neglect friction and assume a large weight that will not "swing" significantly when pulling it sideways. If this bothers you, we could modify the drawing and make the mechanical advantage 100:1 or something else very large such that the side pull was very small. But for the math we'll use the drawing MA of 3:1.

So, starting back at 0* angle, the force on the rope, F:

F = 1/3 X weight / cos 0* Cosine 0* = 1, so the force = .33 X weight. We all agreed to this a minute ago. No angle, no increase in F is needed to hold the weight. No surprise.

Now let's consider the 30* angle case of the drawing:

F = 1/3 X weight / cos 30* Cosine 30* = .866, so doing the math F = .38 X weight. A small increase. Stopping here, this appears to validate that more F is required to support the weight as you increase the angle from vertical.

But let's keep going. Let's consider a 45* angle.

F = 1/3 X weight / cos 45* Cosine 45* = .707, so doing the math F = .43 X weight, and sure enough, more F is required to hold the weight as the angle increases.

For 60*, similarly doing the math, F = .67 X weight. Still more F.

But any body getting worried yet?

Now let's consider the special case of 70.529* whose cos = .333333333

Wow, doing the math, F = 1 X weight. All of the mechanical advantage of using the block and tackle has disappeared! How can that be?

OK, consider an 89* angle. cos 89 = .01745. More wow, F = 19.1 X weight. Who believes that?

Finally, disproving that the angle makes any difference means that when it finally reaches 90*, cos 90 = 0 and division by 0 is impossible. The Limit of F as the angle approaches 0 = infinity X weight.

QED

For those arguing that the angle still makes a difference and don't agree with my proof that it can't, please pick an angle and do the math to arrive at what the force will be. Then do the math with the same logic at the extremes of the angles.

Dave

Dave, I think your formula is wrong, but perhaps someone with a mechanical engineering background can step in.

As I see it, the "tension" on the tail
= total load / sum of the cosines of the angle each "supporting line" is off the direction the object will move if you pull on the line
(A "supporting line" is one acting directly on the object)

Once the tail reaches 90° it is no longer one of these "supporting lines" and its cos is not included.

So in Conachair's example if there is a situation where there is no sidewards movement, if you hold the tail at 0° ie vertical and parallel to the other two lines, then the tension on the tail
= total load / (cos0 + cos0 + cos0)
= 1/3 of the total load

If the tail is at 30°, then the tension on the tail
= total load / (cos0 + cos0 +cos30)
= 1/2.866 of the total load

If the tail is at 60°, tension on the tail is 1/2.5 of the total load

If the tail is at 90°, tension on the tail is 1/2 of the total load

Beyond 90 the tail is no longer acting directly on the direction the load needs to be moved, so it remains 1/2

[TeddyDiver]

Originally Posted by 2Hulls

OK, then show how it works and what difference it makes.

Dave

SWL made a valid point about the load moving to abroad where the load is dealt with two lines so 1/3 of the load for the tail. The angle never reaches 90deg just becouse the load would be infinite as you noted..

[2Hulls]

Originally Posted by Seaworthy Lass

Dave, I think your formula is wrong, but perhaps someone with a mechanical engineering background can step in.

I already did.

But block and tackles is no engineer's area of expertise and I am not so arrogant to believe I didn't overlook something.

To wit:

Let's continue my "proof" to its logical conclusion and take the angles beyond 90*, i.e., past horizontal. By the time you get to 180*, and ignoring any direction changes through the last sheave, you end up with a 2:1 system instead of a 3:1 system. So through the arc of 0 to 180*, you've lost the last purchase of MA.

Beer?

That's my story and I'm sticking to it.......

Dave

[Seaworthy Lass]

Originally Posted by 2Hulls

I already did.

But block and tackles is no engineer's area of expertise and I am not so arrogant to believe I didn't overlook something.

To wit:

Let's continue my "proof" to its logical conclusion and take the angles beyond 90*, i.e., past horizontal. By the time you get to 180*, and ignoring any direction changes through the last sheave, you end up with a 2:1 system instead of a 3:1 system. So through the arc of 0 to 180*, you've lost the last purchase of MA.

Beer?

That's my story and I'm sticking to it.......

Dave

Dave, maybe it will help if you simplify the situation and think of a 2:1 system, not 3:1 (ie half the load if cos = 0).

If the tail moves away from 0° it is going to become harder and harder to move the object, until when the tail is at 90°, you just have a 1:1 system. You can then move the tail 180° all the way to 270° and it will remain 1:1. When you move past 270, the load starts reducing again until you are back to half the load at 360° (You of course need to have a block that can spin ).

Have a think about it, or better still just set up a trial and see. The angle of the tail is critical in a "tackle rove to advantage" set up (it makes no difference in a "rove to disadvantage" situation - you can spin the tail 360° without affecting the purchase).

SWL

[2Hulls]

Originally Posted by Seaworthy Lass

Dave, maybe it will help if you simplify the situation and think of a 2:1 system, not 3:1 (ie half the load if cos = 0).

Well, that doesn't help me in that with a 2:1 you're either rove to advantage (for which we earlier agreed that angle doesn't matter) or you can't go beyond 90* because the weight would just run down the rope.

I have to stop now for todat becuase th rum it kickin in....

Dave

[TeddyDiver]

Originally Posted by Seaworthy Lass

If the tail moves away from 0° it is going to become harder and harder to move the object, until when the tail is at 90°, you just have a 1:1 system. You can then move the tail 180° all the way to 270° and it will remain 1:1. When you move past 270, the load starts reducing again until you are back to half the load at 360° (You of course need to have a block that can spin ).

When the tail comes (close to) 90 deg the bucket is up block to block and there's no more line left, job done

[Seaworthy Lass]

Originally Posted by 2Hulls

Well, that doesn't help me in that with a 2:1 you're either rove to advantage (for which we earlier agreed that angle doesn't matter) or you can't go beyond 90* because the weight would just run down the rope.

I have to stop now for todat becuase th rum it kickin in....

Dave

Dave I will draw a diagram to show the weight would not run anywhere.
Enjoy the sundowner .
Edited to add - laughing hard here - yes of course it would run down . I will just stick to drawing 0-90°

Originally Posted by TeddyDiver

When the tail comes (close to) 90 deg the bucket is up block to block and there's no more line left, job done

LOL, you should have started with a longer line or not pulled as hard or not pulled at all until you reached 90°

[TeddyDiver]

Originally Posted by Seaworthy Lass

LOL, you should have started with a longer line or not pulled as hard or not pulled at all until you reached 90°

Or just forget the purchase and fill the bucket with a single (simple) line

[Seaworthy Lass]

Originally Posted by TeddyDiver

Or just forget the purchase and fill the bucket with a single (simple) line

Yes, that would work too, but a bucket is not a good example - how about making the load an arm that can move freely up and down via a frictionless track? Then we don't have to worry about considering or hypothetically ignoring sidewards motion?

[Pete7]

Right back in the real world, we have now worked out the easy way to lift a 10kg bucket of tools up to the top of the mast.

So moving on what about the main sheet. Most seem to have the tail exit the lower block which is convenient for handling. Remind me which block the rope is attached to so I can check it next weekend down the boat. Does it really matter by the way?

[Seaworthy Lass]

Originally Posted by Pete7

Right back in the real world, we have now worked out the easy way to lift a 10kg bucket of tools up to the top of the mast.

So moving on what about the main sheet. Most seem to have the tail exit the lower block which is convenient for handling. Remind me which block the rope is attached to so I can check it next weekend down the boat. Does it really matter by the way?

Pete, because the tail exits a standing not moving block, this is a "reeved to disadvantage". The purchase would be greater if it was flipped, but as you said, ease of handling is more important (it needs to lead through a clutch and around a winch).

Preventers on the other hand, exit at the moving end. This is "reeved to advantage" if the jam cleat is low enough down so the line does not exit perpendicular to the block.

Interestingly I had a good peer at our preventer. The purchase would have increased by nearly 1 had the jam cleat been just a few cm lower. Bad design!

[jkishel]

Originally Posted by 2Hulls

OK, then show how it works and what difference it makes.

Dave

Here's the correct math:

With line tension T, the reaction at the top block, parallel to the two blocks is 2T.

With Tension T at 30 degrees, there is a horizontal force on the bottom block of Tsin30 or T/2 to the right.

The horizontal reaction at the top block must T/2 to the left and is equal to the reaction at the top block, 2T times sina where a is the angle the blocks make with the vertical. So T/2 = 2Tsina

Solving T/2 = 2Tsina for a: a = arcsin 1/4 or about 14.8 degrees

The sum of the vertical forces on the system must also be zero. For a 10kg weight:

Tcos30 + 2Tcos14.8 = 10

or T=3.6kg

As noted elsewhere, with T applied at an angle of 0, T would be one third of W or 3.33 kg

[monte]

Usually you can still pull in line with a preventer or boom vang. You lose the purchase a bit when you cleat it in the jammer after trimming. I've seen pics of some salty sailors holding the mainsheet exiting from the boom block, I guess they knew a thing or two...

[Seaworthy Lass]

Ok, here are the diagrams showing that the angle of the tail has a vital effect on the tension on the tail (and therefore the effective purchase).

To eliminate sidewards movement, rather than a bucket I have made the load an arm in a friction free track (all these discussions have involved friction free systems).

PS I am not sure how the term "purchase" is defined, but I think it can be applied like this - anyway it is just terminology.

If the photo can't be read easily the figures are:
A=0°, tension on tail = 1/2 x load (purchase = 2:1)
A=30°, tension on tail = 1/1.866 x load
A=60°, tension on tail = 1/1.5 x load
A=90°, tension on tail = 1 x load (purchase = 1:1)