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Y-hang

ChrisB

Well-known member
Apologies if the points below appear pedantic, but people read threads like this and take away the exact wording as their understanding, so the words need to be as accurate as possible even if those who understand the physics know what was meant without them. So I'm happy to be corrected if my words aren't perfect.
that more krab is in contact with the rope, so friction increases
Dry, static friction is dependent only on the force between the surfaces and the coefficient of friction, contact area isn't significant, nor is rope deformation. I would agree, however, that more bend in the rope is associated with more force, and therefore more friction.
The main anchor load is only 100% in a frictionless system
If the deviation cord bisects the angle of the bend in the rope, it's effectively totally frictionless. The tension in the rope, in that situation, is symmetrical relative to the krab, and the deviation cord is inline with the resultant force and can't carry anything but axial load, so there's no path for any friction force to be transferred. It would be accurate to say that "the main anchor load is 100% in a frictionless system or one with the same geometry". In practice, the cord won't be positioned to exactly bisect the angle, so there will be a friction force, which could be up or down.
 

Chocolate fireguard

Active member
Ah yes. I was forgetting. In my original post on the subject, #98, I wrongly used the reduction in contact area (factor of 3) to estimate the reduction in friction, whereas it should have been a factor of 2. That is going from a redirect angle of 180 degrees (force between surfaces of twice caver weight) to 60 degrees (force equal to caver weight).

So the figure for extra load on main belay, using my method, would not be about 15% of caver weight but about 25%.
 

mikem

Well-known member
Picture & text from alpinesavvy.com (link posted by phizz4):
directional+toprope+forces.jpg

A clever engineer friend of mine calculated that the force on each anchor is just a little over the climber’s body weight. (To be more specific, the right anchor takes about 1.25x the climber’s weight, and the left anchor takes about 0.9x the climber’s weight.)
So what is this system if it isn't a fixed rope with 2 (abrupt) deviations? If the load on the belayer was equal to the weight of the climber, then the loads on the anchors would be the same as each other, but they aren't & that is due to the friction...
 

Cantclimbtom

Well-known member
There was a thought a while back of placing climbing style anchors at the top of pitch heads of popular pull through routes (the kind with a single loop connected by chains to each anchor to remove the death triangle situation, does anyone know if this has been employed anywhere?
I know of a couple of mine through trips where there are 2 bolts joined by chain for pull through, which make a chain Y hang when rope threaded as intended. In neither mine do they get enough traffic that chain wear is a concern so there's no Maillon, just threading a link of chain
 

Chocolate fireguard

Active member
Picture & text from alpinesavvy.com (link posted by phizz4):
directional+toprope+forces.jpg


So what is this system if it isn't a fixed rope with 2 (abrupt) deviations? If the load on the belayer was equal to the weight of the climber, then the loads on the anchors would be the same as each other, but they aren't & that is due to the friction...
It's necessary to work out the directions of the forces applied to the rope by the deviations.
It's easy enough using the numbers given.

For the LH one it's to the left and 37 degrees above the horizontal. So pulling more on the horizontal rope.
For the RH one it's to the right and 53 degrees above the horizontal. So pulling more on the vertical rope.

So taking the bottom guy as a caver on a rope below 2 deviations, both of them are increasing the tension in the rope above them, because they are acting as deviations that have been pushed to higher positions and are therefore pulling "down" on the rope. The top guy is the main belay and the load on that is greater than the weight of the caver.

Reversing the jobs of the 2 guys, the RH one is on a rope where the 2 deviations "above" him are in their lower positions pulling "up" on the rope and so helping to support his weight, reducing the load on the main belay.

With no friction (back to the diagram on post #90) the angles would be 45 degrees, with the deviations producing no change in the rope tension and the main belay would see the weight of the caver.

Sorry if I got any of the LH/RH stuff wrong - it's getting late.
 

mikem

Well-known member
In which case, when I hang a 4kg load on a rope from my luggage scales (photo looking horizontally) & then use a karabiner to "deviate" it to 45 degrees (photo looking down at an angle), why does it now show only 3kg? & drops even quicker when I use my finger instead (more friction)...
RIMG8930.JPG
RIMG8929.JPG
 

Fjell

Well-known member
A krab used as a pulley has a friction factor of about 0.5 for a 180 deg bend. At zero degrees it would obviously be a factor of zero. I would just do linear interpolation as a first pass. A 45deg bend is thus 0.125 factor. So a 80kg load would be reduced to 70kg in principle. But this might be an underestimate given the complexity of a soft rope conforming to a bar giving higher frictional area.

If you used a ball-bearing pulley the loss would be negligible at 45deg.
 

Chocolate fireguard

Active member
In which case, when I hang a 4kg load on a rope from my luggage scales (photo looking horizontally) & then use a karabiner to "deviate" it to 45 degrees (photo looking down at an angle), why does it now show only 3kg? & drops even quicker when I use my finger instead (more friction)...
View attachment 16903View attachment 16904
If you are going to use that setup to check it would be a good idea first to check that the scales read correctly at that angle.
Then to find out what happens when a deviation is at the extreme positions allowed by friction you should move the krab, or your finger, vertically as far as possible before it slides and see what happens to the reading.
 

mikem

Well-known member
Electronic scales may not give correct readings if held at an angle, but this is a physical spring, so I don't see why it shouldn't. I know it isn't accurate enough to give an exact reading, but I'm just looking for a proof of concept (& Fjell seems to agree with me).

This page suggests that 1 - friction loss = releasing force / load, so friction in above example would be 1 - (3/4) = 0.25 (which is in engineer's ball park of Fjell's 0.125). Where the hand is modelling the top anchor & the pulley an extreme deviation:
https://roperescuetraining.com/physics_friction_lowering.php
 

Chocolate fireguard

Active member
Electronic scales may not give correct readings if held at an angle, but this is a physical spring, so I don't see why it shouldn't. I know it isn't accurate enough to give an exact reading, but I'm just looking for a proof of concept (& Fjell seems to agree with me).

This page suggests that 1 - friction loss = releasing force / load, so friction in above example would be 1 - (3/4) = 0.25 (which is in engineer's ball park of Fjell's 0.125). Where the hand is modelling the top anchor & the pulley an extreme deviation:
https://roperescuetraining.com/physics_friction_lowering.php
Where the hand is modelling the top anchor.

So again you are choosing the top anchor to be at the low force end. Why must it always be so? Other than that it suits your argument :).

The direction of the frictional force at the pulley is anticlockwise - towards the hand.
In your scenario that corresponds to a deviation krab at a low position with the friction supporting some of the caver's weight. That's fine. The load on the main belay will be less than the caver's weight.

But if the krab is at a high position, with the frictional force on the rope downwards, it's necessary to have the hand modelling the caver because the frictional force adds to the load on the main belay.
 

ChrisB

Well-known member
I find it much easier to understand looking at the angle of the ropes, rather than the friction. Friction is part of the transfer of the force at the krab, and can act in either direction. It's not the same as friction in a haul system.

1. If the deviation bisects the angle of the rope, friction isn't involved in the force transfer and the load in the main rope doesn't change. In the diagram below, angles A and B are equal.

2. If the deviation is angled more steeply than the bisection line, angle A is smaller than B, the line of the deviation is closer to the rope to the caver, so it takes some of the caver's weight and the tension in the rope above the deviation decreases.

3. If the deviation is lower than the bisection line, angle B is smaller, the line of the deviation is pulling more on the belay than the caver, so the load on the belay increases, and the tension in the rope above the deviation increases.

In cases 2 and 3, the deviation angle can be moved away from the bisection line until there is no longer enough friction to transfer the load
devangle.jpg
 

Cantclimbtom

Well-known member
It's worth noting that the effects of friction on say a deviation krab aren't as significant as people think, because when someone bounces up and down on the rope below the deviation (ascending/descending) it will tend to creep a bit of rope past the krab and equalise most of any inequality caused by friction, so in MikeM's example, that "missing" 1 kg would probably return if there was something (to scale: a cat?) aggressively jumaring up the rope. (I expect to see that tested and photographed please Mike!)
 

mikem

Well-known member
My step ladder isn't high enough!

Okay, I now understand where you're coming from, but in reality there is normally more rope above that point than there is deviation cord, so the rope stretch will take the krab to a lower position, & when it goes slack (or especially when caver moves up past the krab) the deviation tends to drop to a lower point on the rope. It's only when a caver is descending that they are likely to clip it in higher up (or pushing it up when ascending), both of which generally involve less jugging than ascending the rest of the rope, as most real world deviations are low rather than high angle.

When the krab is higher on the rope it not only increases the load on main anchor, but also on the deviation anchor (analagous to shallow V rigging) & the forces in a deviation are somewhere between those of a frictionless system & a V rig of similar angle, as the friction means there is some grip on the rope, but not solidly held in position.
 

Fjell

Well-known member
I can see there are some exciting vector diagrams being made, but the basic thing going on is you have a force running along a cable that is attached above. In the absence of friction all that happens with the deviation is that the load on it varies with the angle. The load on the belay never changes without friction (and it can only reduce). If the load above and below the krab were different and the krab was frictionless, the rope would accelerate in accordance with Newtons rather persuasive First Law of Motion. Which would be interesting to experience no doubt.

So what happens in general is that the belay load doesn’t change, but the deviation load gets more extreme the higher the angle on the upper side. Best not to do that.
 
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ChrisB

Well-known member
The load on the belay never changes without friction (and it can only reduce)
I agree with everything except saying that it can only reduce. If the krab is pushed up above where it would be if it were frictionless, the deviation will pull down on the rope and increase the belay load.
 
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