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impulse force on cams
11:02:32 AM

12:14:04 PM
how long is a piece of rope? :-)

12:49:28 PM
a few more things I think you need to know;

what's the fall factor;

what's tha angle between the contact point with the rock of the cam lobe, and the axis of rotation of the cam lobe? I guess since you said the cam was perfectly placed, the angle should be chosen such that the lowest impact force occurs - would this be 45 degrees?

Think this is one for myth busters...

1:02:21 PM
This is a difficult problem to supply more than a rough answer.

5m above the cam (I'll assume no slack in the system - very tight rope).
Fall distance 10m until rope begins to catch fall.
Velocity after falling 10m would be approximately 14m/s.
Impulse = mass by change in velocity = 1120kg.m/s
Lets assume the rope stretches by 10% because the maths is easy.
Further assume that the cam is 15m off the ground.
Total rope from belay to climber 20m = stretch 2m
Climber goes from 14m/s to 0m/s across a distance of 2m
Deceleration takes about 0.3s (linear deceleration)
Impulse = force by deceleration time
therefore average force = 3.7kN (once again assuming the deceleration is linear which it won't be)

The force acting on the cam will actually increase as the rope begins to catch the climber and reach a peak before finally settling at 0.8kN when everything stops moving. I've left the weight of the rope out of the calculation because its simpler that way. I've also assumed the belayer doesn't move and the rope doesn't slip through the belay device (both unrealistic). I've also assumed the cam doesn't move or rotate during the fall.
1:22:13 PM
My pissweak understanding of physics and dynamics of a fall leads me with just enough knowledge to query your logic but not enough to correct it.
Isn't 3.7kN the force exerted on the climber?
We want the force on the piece which I understand should be around 2x3.7kN (perhaps plus some allowance for friction).
Also, 10% stretch is very conservative, most ropes stretch 8% under body load (from memory), and up to 30% under maximum force. 20% might be a better approximation to work with.
More please!
1:41:06 PM
Don't forget about the give in the harness / body interface. Would be quite significant when you include both the belayer and the climber's harnesses.

Without the harnes factor and with a rope modulus of 19.46kn, 10m fall with 20m rope out, I get 7.92kn force on top anchor, rope stretch close to 25% 4.9m. About 6g's.

Assume that there is no friction at the top piece and that the belay is static, eg. the rope does not slip when the climber falls. In real life the rope slips a little decreasing the impact force further, unless you tied it to a tree or something.
3:27:44 PM
Easy, dangerously close to the limits of a black alien.

3:36:31 PM
I wanted to simplify the problem. There are forces acting on the cam during the first 10m of the fall but these would be difficult to calculate as they depend on the tragectory of the climber and rope. I've simplified the problem by only considering the forces once the rope starts to "catch the climber".
Final force of 0.8kN is one 80kg climber hanging under gravity with the assumption that the belayer is standing on the ground.
I made an arbitrary choice of 15m so I had a "fall factor" to work with. If you select a different fall factor you'll get a different answer.
The peak force is close to 8kN according to the Petzl Simulator.
Used: 80kg climber, 10.5mm single rope, belay 12mm boltsx2, first & second running belay = none, third running belay = friend at 15m, you fall 20m, gri-gri for belay device.
3:38:34 PM
If there is 20m of rope out then 10% stretch is 2m, not 1m. The whole thing is an idealisation anyway, so the fall of 10m is used to calculate how fast the climber is going when the rope starts to catch him/her. The deceleration won't begin until the force due to the rope catching the climber is equal to the climbers weight which would be sufficiently close to the point at which the rope came tight that it wouldn't make much difference to the outcome anyway, and would be much harder to calculate.

I would reccomend doing a numerical simulation using (Hookes) law for a spring (F=kx, where k is a constant and x is the displacement from equilibrium), and newtons second law (F=ma) and stepping though the situation with a small time step, if you want a more accurate number. You could use excel for this.
3:39:20 PM
yeah, what he said.
3:57:35 PM
Forget the spring shit, thats why you use the rope modulus.
4:02:25 PM
Check out that petzl simulator, it's not bad.

4:02:59 PM
This popped up on a quick google search.
Far more detailed than my response -
4:13:13 PM
If like me you cannot read equation 10 it should be this:

Equation 10

Force of impact = m.g {1 + square root of [ 1+ ( 2.y.m / m.g ) }

6:03:05 PM
Wow, this has finally proven it for me. Climbers are all geeks!!! = )

Shouldn't we be out climbing something instead of looking at this?
6:21:36 PM
Like you can talk engineer!

6:26:58 PM
See what I mean

11:01:39 PM
On 12/09/2005 ti wrote:
>Um, why did you need to presume that the cam was 15m off the ground? Is
>this so that you can factor in rope stretch? If so, would it be logical
>to say that the climber falls 11m (10m + 1m due to 10% rope stretch)? And
>from this calculate the time that the force is exerted on the cam?

Because when you are 5m above your last piece and it's 15m above your belayer, you have 20m of rope out. This means you'll fall 10m before stretch, and it'll be a ~Factor 0.5 fall. If the cam was right at your belayer, in which case it's a Factor 2 fall.. ouch (assuming it's a hanging belay so you don't deck out before the rope catches you!) Yet another reason why you should get something good in just off the ground.

Worry less about figuring out the actual force, and more about the fall factor - it's easier to calculate, and as meaningful.
10:34:08 AM
You need to use the rope modulus as it acurately describes the deceleration. The deceleration is not linear.

12:35:44 PM
why not just average the rated strength off a few cam brands and add your choice of factor of safety?

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There are 36 messages in this topic.


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