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kurios


Dec 25, 2008, 4:15 PM
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DMM Revolver
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Has anyone lab tested the claims of DMM that the Revolver carabiner lessens the impact force at the last (or top) piece of gear - ie. dropping a suitable weight as in a fall -several fall factors- and measuring force at the anchor?

I'm not 100% convinced by their claims, but my physics is a little rusty these days.

I realise that decreasing friction at the last carabiner would help spread the load on both sides of the carabiner- belay side and climber side - so that there is less force felt by the climber (though more by the belayer) than with a conventional carabiner (without a pulley). However, surely this would increase the force at the anchor (last/top carabiner).

My logic is: Fa = F1 + F2
Fa is the force at the anchor.
F1 is the force on the climber side
F2 is the force on the belayer side.

With 100% friction (no slippage) on the anchor Fa=F1 as F2 is 0. This is like tying off the climber at the last carabiner.

With 0% friction (perfect pulley scenario) F1 = F2 and so Fa = F1 + F2 or Fa = 2F1.
So more force at the anchor.

Obviously there is a continuum from 100% to 0% frction.

Since the revolver lessens the friction the situation shifts to more like the 0% friction end of the continuum so F1 becomes more = to F2 and so increases the force at the anchor.


johnwesely


Dec 25, 2008, 4:31 PM
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Re: [kurios] DMM Revolver [In reply to]
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I imagine that the difference is negligible. I think that they work by allowing the rope to stretch more.


keep_it_real


Dec 25, 2008, 4:41 PM
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Re: [kurios] DMM Revolver [In reply to]
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If the climber is tied off at the anchor (100% friction) and they're leading so they're above the anchor, they will experience a force 2 fall. That's the worst fall you can have because of the huge deceleration. Basically with more friction, the falling climber will decelerate faster creating more force that the anchor needs to hold.

remember f=ma

in any almost any lead fall, both the belayer and the climber will be supported by the anchor so m (mass) in the equation remains the same for most falls. a (acceleration) is what the revolver changes. The pulley slows down acceleration making the forces smaller.

example
acceleration caused by regular biner = 25 m/s/s
acceleration caused by revolver = 20 m/s/s
(I don't know what kind of acceleration you really experience in a fall but it's probably more than that)
mass of cllimber and belayer = 150 kg

regular f=25*150= 3750 N
revolver f=20*150= 3000 N

revolver creates a lower force.


(This post was edited by keep_it_real on Dec 25, 2008, 4:45 PM)


dj69


Dec 25, 2008, 8:46 PM
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Re: [keep_it_real] DMM Revolver [In reply to]
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keep_it_real wrote:
example
acceleration caused by regular biner = 25 m/s/s
acceleration caused by revolver = 20 m/s/s
(I don't know what kind of acceleration you really experience in a fall but it's probably more than that)
mass of cllimber and belayer = 150 kg

revolver creates a lower force.

Doesn't gravity have a constant acceleration of 9.8 m/s/s

I am pretty sure the revolver just increase the time interval in which the fall is being absorbed, so if the same force is being absorbed over a longer period of time, the peak load, would be reduced. (isn't the same concept as jumping, during a lead fall.)

This sounds right to me, but then again my physics is alittle rusty.Crazy


keep_it_real


Dec 25, 2008, 9:27 PM
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Re: [dj69] DMM Revolver [In reply to]
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what I was referring to was the deceleration when you stop not the acceleration of gravity.

and yes the revolver does increase the time it takes to stop which means the acceleration is less as I said earlier.

I'm confusing myself over all this acceleration/deceleration. They're equal and opposite.


(This post was edited by keep_it_real on Dec 25, 2008, 9:29 PM)


kurios


Dec 26, 2008, 1:51 AM
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Re: [keep_it_real] DMM Revolver [In reply to]
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OK but there is a trade off happening here. With less friction there is effectively more rope to stretch and so less deceleration to lesson the force on the climber (F1).

However, at the same time there is more force being spread to the belayer (F2).

And since the force at the anchor is the sum of the forces on the climber and on the belayer (Fa = F1 + F2), I'm wondering whether the lessening of force F1 is more than the increasing of force F2. (Remember my question was about the force on the anchor (top piece) which seems more critical than that on the climber because gear placed is usually the weakest link -unless a bolt/hanger).

I suspect it is not and that the increase in force F2 > than the decrease in force F1.

Wouldn't mind some equations or some data from a lab test to check this out.

Anyone done the work on either the physics or in the lab?


joeforte


Dec 26, 2008, 5:06 AM
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Re: [kurios] DMM Revolver [In reply to]
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Kurios, "Keep it real" is right. The deceleration is numerically less with a revolver, due to the lower friction. This allows the rope on the belayers side to see more of the force, which means more rope stretch, a longer fall, and more force on the belayer. This can be a good or a bad thing, depending on the situation. The revolver is a very useful tool, when used at the RIGHT TIME. I keep one on the back of my harness, on my only screamer, for critical, yet marginal gear placements. Your belayer should also know what to expect (a ride) if you fall on one!


brenta


Dec 26, 2008, 8:29 AM
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Re: [kurios] DMM Revolver [In reply to]
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I'm not sure how to read the DMM blurb. I see no explicit mention of the Revolver being at the top anchor. If you place a Revolver at one of the intermediate anchors, especially one where the rope makes a sharp turn, then it's definitely going to reduce the force on the top anchor. Given that I know of no one climbing with a full rack of Revolvers, this seems the most likely scenario.

If the Revolver is on the top anchor, then it depends on how much rope is on either side of the anchor. If there is much more rope on the leader's side than on the belayer's side, the Revolver will increase the force on the anchor relative to a standard biner. In the opposite situation, it will help. Computing the exact break-even point is a non-trivial exercise.

As for the accelerations during the arrest of a fall, we know that the impact force on an 80 kg climber in non-pathological situations is at most 12 kN. The net upward force on our climber is bounded by 11.2 kN (subtract gravity). Therefore, the acceleration is at most 14g (divide by climber's mass). If the impact force is 4 kN, the acceleration is 4g, and so on.


adatesman


Dec 26, 2008, 8:40 AM
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brenta


Dec 26, 2008, 8:56 AM
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Re: [adatesman] DMM Revolver [In reply to]
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adatesman wrote:
I seem to recall reading in a thread shortly after they came out that the pulley only functions under low load situations.
This is a very good point. The pulley may in fact be permanently damaged by loads in excess of 11 kN, according to DMM. However, during the arrest of a fall, the pulley would progressively get stuck and hence allow some more rope to slip than a regular biner.


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