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

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!!!

Look, whatever the purchase is, its not enough for the new carbon mast on the T, so we have now a system with more purchase, by making the floating block inside the boom a 3:1 system.

We then run either 4:1 or 3:1 on the small blocks at the back, the change is achieved by dropping the last block in or out of the system depending on what you want at the time.

Here you go mate...... I must spend too much time doing this $hit <img src="http://www.catsailor.com/forums/images/graemlins/shocked.gif" alt="" />

Dependents on which view point you have.

You can view it from a force points of view or from travelled distance point of view.

Force : 1 lbs effort => 9 lbs end result
distance : 9 units distances pulled => 1 unit distance travelled by boom

And there are many more, and several have to be called 9:1 and others 1:9

So I just the beach talk version that is fashion of here and that puts the 1 in front mostly because in Dutch it just flows betters. I think it does so too in English.


Wouter

Sorry Wouter gotta disagree here. In the UK, (IMO) people will always use the multiplier first.

Back to this multi purchase...

in the picture I agree that up to the last block on the boom, there is a an 8:1 system.

But I do not understand where "only" an extra 1 comes from.

The block is moving down, and there is a (at least a 8:1) moving over this block, BUT as this block is moving itself there MUST (I think) be some multiplyer on the rest of the system. I agree it cannot be 2 as there is not a 180 degree turn, but I cannot see how it can be 1 as the block on the boom is moving.

The "count the moving blocks" thing will only work on a standard system of blocks. The key issue here is that the black rope does not come back down to the beam. Instead, it feeds into a second system. If you removed that system and simply anchored it in the boom you'd have a 1:1 system.

I find these things easier to look at in terms of rope moved. What does it take to move the boom 1 unit down? Well, if you remove the final block and imagine yourself sitting on the boom, it's pretty obviously 8 units. Now sit back on the boat and use that final block. How much more rope do you have to pull to shift the boom that 1 unit? 1 unit, so you've pulled 9 units, so it's 9:1.

It's not going to be a multiplier because it's not a cascade. There's only one floating block in the system and that's the one that gives you the 2 x 4 = 8 : 1.

Paul

Although my first proposal on 3:1 and 5:1 was found incorrect, I cannot resist commenting on the "Macca-system".
The confusing point here is the 90 degree turn to put blocks inside the boom. If you straighten out the system, you find a 4:1 compund system and a 2:1 cascade. Total = 8:1 (Or 1:8 if you are the boom).
For cascades, you cannot simply count number of line segments.

Seriously:
May I propose that you publish mainsheet setups that you really like here. This would be most useful to all of us.

Macca has provided an exellent example.

Stein

Nope. If you straighten it out and anchor the "static" turning block (the one originally at the far right) to the boat not the boom (e.g. tie it to a cable from the top of the mast) then you'd be nearly right. In fact, you'd only get 7:1 because the black rope is tied off to the boom not the deck. Fix that and you do finally get to a standard 2 x 4 cascade.

Straightening out the system doesn't change it, and either way you can't get away from the fact that the cascade is anchored in the moving boom, which means you have to move an extra unit of rope.

Paul

Ok, here is the new system, you can change the purchase really easily by moving the red line from terminating on the boom to terminating at the bottom block on the bottom of the system. we call the min purchase 9:1 anf the max purchase 12:1 but I am sure you all will find 500 other ways to calc it out <img src="http://www.catsailor.com/forums/images/graemlins/smile.gif" alt="" />

Here is a simple geometric explanation:

With the sheet taught, lift the boom, pulling the 4:1 blocks 1" further apart. This pulls 4" of the red line into the 4:1 system, moving the floating block 4", and pulling 8" of the black line into the end of the boom. However, since the end of the boom moved 1" away, this pulls 9" of line through the main sheet cleat.

So, it's 9:1, with the "extra 1" coming from the boom movement as you sheet.

--

Here's another simple geometric explanation:

Imagine you had a conventional 8:1 rigged upside down like this:

     BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB Boom
       /O
      /4 4
     / 4 4
       4 4
        O

and then you take the line at the end and run it through the bottom set of blocks like this:

     BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB Boom
        O
       5 4
       5 4
        O______

Suddenly, it is 9:1. It's no different for the cascaded system. That's where the "extra 1" comes from.