Mainsheet system: advantage

I'm afraid you are incorrect. Each fall of line contributes to the overall mechanical advantage of the system regardless of whether it terminates on the

fixed

or the

moving

end (in fact, the two are moving relative to one another, so they are both moving). But regardless, a simple test: flip your blocks upside-down. You will find that you get exactly the same purcahse with the blocks in either orientation.

sm

Those blocks listed as 8:1 are 8:1. Do a load diagram...if you pull 8 inches and get 1 inch of movement than it is absolutely 8:1 - no way around it.

In hindsight, I put these block load arrows pointing the wrong way for a static diagram - but I think you get the point.

Just wait until you try to sort out a cascading system.

picture

You must both be right; your explantions seem highly plausible. It must be correct that advantage depends on ratio of (rope length pulled):(distance moving of boom).
My simple living-room experiment must be inaccuarate. And the general rule that only blocks on the object moved contribute to advantage, must be incorrect.

Thank you!

Stein

PS: I stick with my 10:1 setup for my Taipan 5.7, though.

Yeah - I think you are getting conturbulated thinking that the kinetic system (moving) is different than the static. All loads must be equal. To quicly determine the purchase of a (non-cascading) system, you can count the running lines between the two stationary points. 2 lines is 2:1, 3 lines is 3:1, 4....and so on.

If you have a cascading system, you separate the two systems and multiply them against each other. A lower block setup with 4 runs at 4:1 that attaches to a cascading system with 2 runs at 2:1 is a total of 8:1 purchase.

So just for practice, this one is 2 x 4 = 8, correct?

It looks like the black is a cascade doubler and the red on the left is 4:1.
The red-right is ineffective, turning only.

Assuming that the black and the red lines that just end at the boom are simply attached to the boom - yup. 8:1...the question is though....WHY all the complication?

I think is was Rolf that posted this as a Tornado setup. Why, you ask? Small diameter lines through the blocks, I'd guess. No taper needed. Maybe less total weight by using small blocks? Hidden mechanisms keeps the rookies staring? Easy to hide an illegal 10 or 12 setup? <img src=

alt=

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This drawing was done by Macca, I think.
It shows a low-friction, lightweight and very clean setup used on many top-notch Tornados.

There are no rules on how much purchase you can have on any sheet on the Tornado. Would be a silly rule in my opinion.

But this one has another line contributing to total boom movement...the black line on the extreme left. So won't is be 8:1 + 1:1 = 9:1?

Correct !

Wouter

Kind of like

Where's Waldo?

....I missed one.
The biggest, darkest line on the page!

9:1 is is!

I've been wanting to setup this exact system on my T...but I have not zero'd in on a double-block that is slim enough to fit into the Marstrom boom section. Plus, I have an outhaul system running inside already (that I quite like & use), so it would need to be removed/replaced with some other setup.

Right now I run a traditional 57mm harken carbomatic triple lower unit up to two carbo doubles (one with becket) hanging off the boom with spectra. I also put a 40mm single on the lower unit & run the tail back to the upper double's becket. Gives 9:1. Works pretty good.

The advantages are:
1. much lighter overall (blocks & 3mm vectran line)
2. Excellent easing (cascades are generally lower friction)
3. Lower cost can be achieved since blocks are small & cheap compared to 57 mm Harken Quads etc. (though I've seen several version of this running custom made blocks with carbon side plates...$$$)

Some disadvantages are:
1. can be a headache to work out
2. Can jam up inside boom if not setup right
3. Maintainance more difficult

Lots of top T guys run this setup, but it is not required to get to the top of fleet...Charlie Ogletree & Johonny Lovell run standard harken quads <img src=

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I have often thought this wold be a good way to go. I do not understand what turns the section where in the diagram the line goes from red to black... ??

some pics, someone can link them so they appear in the posts but i have never worked that part out <img src=

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This pic is the turning block at the front of the boom, the thin rope turns around this block and attaches to the floating block in the next pic.

this is the floating block, we have a new systems that allows us to change from 9:1 to 12:1 in seconds, but more on that when i put the boat back together and take pics. way to complex to explain here <img src=

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Many thanks for the photos as they will prove very useful to me. What I am having trouble understanding is how the larger sheet exits the boom at the sharp angle. is there a through block there?

Okay, have understood. : (

I make 8:1 + 1:1 = 9:2 [Same as 8]

My money is still on 8:1

Paul

right result; wrong maths

it's is actually a 8:1, with a 1:1 on it, so

8:1:1 is a 8:1; 9:2 is 4.5:1

Yes, definitely 9:1.

For anyone who doubts: 1 lb of tension on the sheet (black line) leads to 2lbs of tension on the red lines, which is obviously 9lbs of tension between the boom and main traveller, or 9:1.

Yes, and there are 4 red lines in the moving part of the system. (up and down from the boom to the back beam (well the wire to the back beam)

8:1

You have a 2:1 on the Black line system, the black line system is then pulling on a 4:1; 2:1:4:1 = 8:1

It is a 9:1 system guys.

No doubt about it.

Wouter

Agreed and I stand corrected - that is 9:1

I've had another look at this and think thus

Red system is a 4:1.

The black system is a 3:1 as the both blocks (2 and 3) on the boom are (in effect) moving in relation to the back beam.

As you pull the mainsheet in, the first block on the boom(2) is also moving down, as is the big block inside the boom(3), so the black system is a 3:1, making the total system a 12:1.

I cannot see how it can be a 9:1 as the internal cascade is 4:1 and so any multiplier must result in an even number unless there is just a

routing

block, which there are none as all the

big/black

blocks are moving except the back beam block.

Try this

Okay, let me explain how it works.

But first lets derive the tension in each line because we'll need that later.

Black line is held in hand and so all the black lines are under 1 unit tension; the pull of the skipper.

The red line is attached to the black line by an 1:2 purchase system so the tension in the red line is 2 units.

It is impossible for any line to have different tensions at different points. Think about this.

When a line makes a full 180 wrap around a block then it pulls 2 times the line tension downward, that is obvious. If a line makes a 90 wrap then the block is pulled down by 1 times the line tension. A termination point equal 1 times the line tension downward. All still pretty straight forward.

Now lets count the loads when looking only at the boom blocks, the others are unimportant.

Black line : about 1 partial wrap so roughly 1 unit down pull and nothing else

red line : 1 full wrap, 1 partial wrap and a termination point ; total = 4 times the line tension of 2 units = 8 units of downward pull

Add them all up and you get (by approximation) 9 units of downward pull for each single unit of pull by the skipper, ergo it is a 1:9 system.

And this is one of the few ways to do it right.

Wouter

Wouter,

You can't keep writing purchases a 1:x unless you are talking about an inverse purchase system. The standard nomenclature has the advantage shown FIRST.

8:1, 4:1, 6:1, 12:1 etc.

by writing 1:9 you are in effect saying that for every 1lb of effort applied at the load you need to apply 9lbs - usually bloody hard work!!!