Run the rope through the test stand (assume a polished surface on the redirect that simulates a biner) with a rope that passes at 12 kN.

Now, sand and scuff the redirect to simulate a manky biner that increases friction and what happens to the forces? Up right?

I think they are making the assumption that the rope must still pass the same 12 kN standard, but it either gets tested with a polished surface or a rough and passes and then it goes into the real world where circumstances change and the rope properties do not.

12kN is the maximum impact force allowable for a rope to pass the drop test.

The drop test is supposed to simulate an extremely hard fall; beyond what most people will experience. Most ropes have actually have an impact force in that test of 8 or 9kN which is quite a bit below the maximum.

Yes, but if you increase the friction in your test rig with an identical rope you won't get the same result.

What do you mean by "that passes"?

What forces go up? The frictional force does, obviously. But the frictional force is not part of the total force on the biner. The total force on the biner is the sum of the tensions in two sides of the rope. Let F_t, F_b, and F_c be the force on the top biner, belayer, and climber, respectively, and let µ be the frictional force. Then

F_t = F_b + F_c . But

F_c = F_b – µ . Therefore

F_t = 2F_b – µ .

Clearly, then, the greater the friction µ the less the total force on the top biner F_t.

Okay, I think I see where you're getting confused. In the standard model of impact force, friction has no effect on the tension in the rope on the belayer's side. If the maximum tension on the "belayer's" side of the rope in a UIAA test is 12 kN, it will be 12 kN regardless of how much friction there is across the top biner. This is true to a first approximation. In truth, friction will increase the force in the belayer's side of the rope slightly; however, this increase is more than offset by the reduction in force in the climber's side. So the net effect of friction is still to reduce the total force on the biner.

You apparently think that the peak force on the climber's side of the rope will be greater when the friction over the top biner is increased. Under the standard model of impact force, this is false. Why do you think that friction over the top biner would affect the force felt by the climber?

Jay

You are starting from the wrong end. The article is talking about the maximum possible values and in that sense the chart is perfectly valid. The 12kN force is alway sused as this is the maximum result allowed for a rope not over a karabiner. Changing the friction value changes the forces on the karabiner and the belayer when we assume the force is still 12kN.

With friction two things happen: the force on the climber is greater because you are isolating the section of rope between the climber and the pro which means that there is less rope in the system; the force on the gear is reduced because the force on the belay side of the biner becomes less.

If the length of rope on the belay side of the biner is small and there is a lot of friction then the force on the gear will be greatly reduced (up to nearly a factor of 2) while the force on the climber is nearly the same.

You apparently think that the peak force on the climber's side of the rope will be greater when the friction over the top biner is increased. Under the standard model of impact force, this is false. Why do you think that friction over the top biner would affect the force felt by the climber?

Jay

With friction two things happen: the force on the climber is greater because you are isolating the section of rope between the climber and the pro which means that there is less rope in the system; the force on the gear is reduced because the force on the belay side of the biner becomes less.

If the length of rope on the belay side of the biner is small and there is a lot of friction then the force on the gear will be greatly reduced (up to nearly a factor of 2) while the force on the climber is nearly the same.

Because of this:


I don't know that this is greater than the energy lost to friction or not, but it is the model that I have seen used and the frictional loss written off with wind resistance as insignificant.

Also, to clarify I am talking about NOT assuming that the climber's felt force stays the same. Change none of the conditions (including the elasticity of the rope) except the friction of the redirect.

Did you notice that in the document you linked to they have force on the runner dropping as friction in the pro biner increases? (table 1 p7) They seem to be assuming that the 12 kN in the UIAA drop test is a law of physics.

Okay, I think I see where you're getting confused. In the standard model of impact force, friction has no effect on the tension in the rope on the climber's side. If the maximum tension on the "climber's" side of the rope in a UIAA test is 12 kN, it will be 12 kN regardless of how much friction there is across the top biner. This is true to a first approximation. In truth, friction will increase the force in the climber's side of the rope slightly; however, this increase is more than offset by the reduction in force in the belayer's side. So the net effect of friction is still to reduce the total force on the biner.

Jay

You must have edited while I was responding.

Now this makes perfect sense. You are agreeing that the force felt by the anchor would go up from friction if you ignored the energy loss from that friction because you are effectively reducing the ropes ability to stretch (over part of it's length) and absorb energy over a longer time period, right?

The typical assumption is that the frictional force is one-third the force felt by the climber; that is, the "friction factor" u = 1/3. Assuming that this friction factor is correct for a UIAA test rig, then a standard UIAA drop test using a rope with an impact force rating of 12 kN will generate forces on the climber, belayer and anchor of 12, 8, and 20 kN, respectively. (These numbers agree with those of Petzl in Table 1 of the paper you referenced, suggesting that Petzl used u = 1/3 in their own calculation.) In comparison, if we do the UIAA test under identical conditions, except that we increase the friction factor to 1/2, then the forces on the climber, belayer, and anchor will be 12.11, 6.06, and 18.17 kN, respectively. Thus, for a 50% increase in friction over the top biner, the impact force on the climber has increased less than 1%. Moreover, owing to the reduction in force on the belayer, the impact force on the top anchor has decreased from 20 kN to just over 18 kN.

The methodology for these calculations is explained here.

Jay

That paragraph was in my original post, but I mixed up "climber" and "belayer," which I quickly corrected in an edit.

So their conclusion from the chart is correct, but aren't they actually using the incorrect model that I was using where friction is only figure into the proportion each leg takes of total force and frictional loss is ignored? That's what it seems from the numbers they get. They just assume that climber force is 12 kN and apearently that gives similar numbers to the correct (or closer to reality) model that you are using.

I'm still not really following what you're trying to get at.

The context of the chart is the force applied to the gear. The conclusion I get from the Maximum Force on a Running Belay section is that the frictionless scenario results in the highest impact force on the gear so that including friction isn't needed in the worst case scenario calculation and that 20kN holds as a good maximum impact force requirement.

That's why it just confused my poor little brain when I was trying to respond. (At least that's the excuse I'm gonna use!)