friction diffrence

[Fairfield]

Is the friction loss the same in these two examples:

1. One Petzl P50 Rescue pulley where the angel of contact with the rope is 180 degrees
2. Two P50’s where the angel of contact is 90 degrees each

I understand that the answer is yes when it comes to karabiner, bollards, rock, figure of eight etc, ref. the capstan equation. But is the eqn. true when it comes to pulleys which have rolling friction?


This is a question posted on another blog site that I would like to get all your views on.

 

[Blinky]

EDIT:
Never mind, I misunderstood the question.

 

[moray]

What a great problem! My first guess was that the pulleys would have behaved like rings or biners or what-have-you, following the capstan friction formula. But the axle friction isn't quite the same thing--there is no tensioned cord or belt dragging around the surface. I made the following drawing to explain my analysis.

Petzl describes the P50 as having 90% efficiency. What this means is shown by the single pulley on the right. It takes 100 units of downward force to raise 90 units on the other leg. I have measured a P50 in the past and my result was almost precisely 90%. Note that Petzl doesn't specify the amount of force, just the ratio of the two legs. In the diagram I have indicated the total force experienced by the pulley axle, 190 units pointing straight down. It takes 10 units to turn the pulley when the axle is loaded with 190 units, so we can conclude 1/19, or just over 5%, of the load is required to turn the pulley.

Pulleys A and B show the 2-pulley setup. Again we lift 90 units at the left. The arrows from A and B show the respective axle loads on those pulleys. Multiplying each of those by 1/19 gives the force needed to turn the pulley. You need 6.96 units to turn A, so the tension in the horizontal rope is 96.96 units. In the same way, the two rope legs pulling on B give a total axle force of 142.53 units, which tells us it will take 7.5 units to turn wheel B.

I hope someone will check my math, but the result seems very clear: two pulleys are worse than one. The overall efficiency has dropped from 90% to 86%.

 

[TLHamel]

Those are the approximate numbers I get while calculating raising the load. The opposite is true while lowering the load. Also with slightly less force on the anchors.

 

[matdand]

[ QUOTE ]
The arrows from A and B show the respective axle loads on those pulleys.

[/ QUOTE ]

How did you guys get to those numbers?

 

[moray]

I know I didn't really explain how to do the math in the previous post; I was trying to explain what I think is going on. The load on a pulley axle is a simple vector addition problem. Here's an example:

We have two forces at right angles to one another acting on pulley P. Their magnitudes, as shown, are 106 and 62. If you finish drawing the rectangle, the diagonal from P to R represents both the direction and the magnitude of the resultant force. That force of 123 units is what P feels--it cannot tell the difference between the single force of 123 and the combined forces of 62 and 106.
If the two forces 62 and 106 were not at 90 degrees, your completed diagram would be a parallelogram, not a rectangle, and the diagonal would still represent the correct sum of forces.

The larger problem, which is an algebra problem, makes use of the vector sum procedure just described. I won't irritate anyone by writing down the equations (which is not how I did it anyway), but I'll state the problem. Take the left pulley in the upper diagram. The problem is to find the exact tension in the horizontal rope (96.96) that combines with the downward load of 90 to produce a resultant (132.29) such that 1/19 X 132.29 = 96.96 - 90. The difference between the two forces, 6.96, is the friction we are after, the force needed to turn the pulley. Isn't math fun.

 

[Ron]

Generally two pulleys will always be less efficient than a single pulley. Two pulleys introduce two losses, the magnitude of the loss depends on the specific configuration.

If two pulleys are configured in a MA, i.e. 3:1 etc. and there is a difference in pulley efficiency, the more efficient pulley should be located nearest the user's pull line.

 

[moray]

Yesterday I was showing an arborist friend how to write a tiny computer program to solve the pulley problem in this thread. It occurred to me it would be interesting to extend the problem to more pulleys. To get several pulleys to share a load more or less equally, they need to be arranged along a big vertically oriented semicircle. This is hard to do in real life, but easy in the virtual world of a computer program. The burning question is: if we keep adding more and more pulleys to the problem, does the extra friction just keep growing and growing? The answer is actually yes and no. Every extra pulley adds to the frictional load, but there is a limit, and the limit is quite low. This is possible because the added friction from each added pulley becomes extremely small.

Here are some calculated results showing the force needed to lift a 90-lb load using 90% efficient pulleys:

Pulleys-----Force
---1---------100
---2---------104.464
---3---------105.415
-180--------106.2064
5000--------106.2066

Now for real pulleys that have axle friction even when the axle is not loaded, these results are completely bogus for large numbers of pulleys. But for any practical rigging problem where the rigger wants to use several pulleys instead of just one, the extra pulleys will share the load very nicely with very little impact on overall efficiency. This same cost-free load sharing applies to setting an SRT line over a number of branches in the canopy--the branches really do share the load, and the load each branch feels is (roughly) proportional to the degrees of bend the rope makes around that branch.