optimize order of purchase systems

If I have simple cascades 2:1 and 3:1 for a total of 6:1 does is it better to have the higher purchase closer to the load or away.

If the load is 6 units and 2:1 (1 shiv) then 3:1 (2 shivs and a becket) and the loss due to friction is 0.1 -

Option A) 2:1 to 3:1

load * friction on shiv
3 * .1 + 1 * .1 + 1 * .1


Opttion B) If you go 3:1 to 2:1

2 * .1 + 2 * .1 + 1 * .1

If there is a 'penalty' for more friction due to higher load then option A loses, you should go with higher purchase first
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It is not that simple. Not only is the friction of each sheave multiplicative (not additive i.e. total tension with three f=0.1 sheaves is 1.33 not 1.3) to the next while inhauling, it depends on how may ends are "dead" (i.e. not pulling between the things you want to pull on).

 

The different arrangements result in different available travel:
- 3:1 pulling on a 2:1 will allow travel of 1/2 the original length
-2:1 pulling on 3:1 only allows travel of 1/3 the original length

The maths of friction over the various blocks is more than I am keen to do right now, but I would not be at all surprised if it came out the same.

 

Welcome to the forum.

Looks like a textbook question to me.

Real world blocks don't always have 10% loss. I once came across 5/8 line forced thru a block designed for 3/8, resulting in greater than 95% loss.

Other constraints have a greater importance in the purchase systems I have designed.

 

If the equations under the friction section of Block and tackle - Wikipedia https://en.wikipedia.org/wiki/Block_and_tackle apply and are accurate, then the efficiency is independent of the load and both combinations of purchase order are the same.

There is efficiency * penalty * due to higher load.

Solving for total load

F(total) = L(total) / ( N (1-sheave)) * N ( 2 - sheave) * eff ( 1-sheave) * eff ( 2 - sheave)

 

The wiki that you quote is for only for single tackles and the word 'approximate" is thrown around a fair amount..

Ok, let us assume this is for a backstay.

To get a 6 to 1 tension advantage with a single tackle you need 6 sheaves (a three-fold purchase), with a compound tackle you only need 4 (two gun tackles backed). Assuming a 90% efficient sheave for the first case (a three-fold purchase) L= .9T+.81T+.73T+.66T+.59T+.53T => L=4.22 T . In the second case L =T1+.9T1+.81T1 where T1= .09T +. 81T => T1 = 1.71T => L=4.63 T. So the two backed gun tackles have a greater mechanical advantage (about 10%) even though both have the same 6 to 1 geometric advantage.​

 

As a simple thought experiment, no it doesn't matter provided friction is proportional to load and all blocks and lines are similar. But the real world doesn't work like that.

You would normally use different blocks in the two parts of the cascade, say 3" for the primary and 2.25" for the secondary. Now it comes down to load ratings and price and bearing options. Can you opt for lower load but less friction bearings, or do you have to use high load plain bushed sheaves?.

Line choice has a big effect. A lot of the friction comes from fitting the line into the groove. This is often more than the actual bearing losses. If you want the secondary to be easy on your hand, you can do better with an efficient high-strength 3:1 primary and a nice soft 2:1 secondary.

On the other hand, many becket blocks carry the same load rating as the single, which can make 3:1s a bit awkward to optimize. You don't actually gain any capacity by going with the 3:1. You can handle the same load with a 2:1, you just have to pull harder. This means that both arrangements have the same load capacity as each other, and as a 2:1 + 2:1. So the 2:1 +3:1 would be cheaper than the 3:1 +2:1 because of using more smaller parts. But if you don't use a becket block and instead dead-end the standing part separately, then you gain capacity with the 3:1 over the 2:1.

 

I find that the frictional loss is the same for both cascade arrangements. Working back from fixed T at the user's hand towards an unknown L at the load (a backwards way of thinking perhaps), and taking "a" to be the block's efficiency I get:

L = T(1+a)(1+a+a^2)

Which can be seen to give L = 6T for "a"= 1 (100% efficiency)

Note that in the better arrangement (user pulls the 3:1 directly) there are 2 blocks on the low tension line, and only 1 on the high tension line (but this line is in more tension than the other arrangement (L/2 rather than L/3)).

 

tlouth7, be careful in this case because you presume a "live" hauling end like for a on mast halyard tensioner. In my example I assumed a turning block to a cleat. Arrangement really drives some tackles because as Archimedes stated you need a place to stand.

 

So long as anything beyond the system in question is identical then this will make no difference to the fact that both configurations give the same frictional loss.

If you add 1 turning block between user and the system* then I get the same result as jehardiman.
*such that it is rigged to disadvantage, with a total of 4 sheaves.