Something scary to think about!

[snoboy]

Check this out, It's from Will Gadd's ice pages:

"However significant damage was done to the ankle region. The belayer was unhurt but it is known for the belayer to be spun around if facing in the wrong directions. It should be noted that crampons do not have swivels between them and the boot. In the case of the guy coming off of Louise Falls a couple years ago. The leader takes 30 feet of air and walks away unscathed however the belayer was carted away with a tib fib spiral fracture."

 

Yikes!

[bolt_clipper]

Huh? How would that happen? The belayer stands twenty feet away from the ice, so he gets pulled in when the leader falls? Clarification?

[freeclimb9]

Always a mistake to use a rope attached to someone who falls.

[snoboy]

Huh? How would that happen? The belayer stands twenty feet away from the ice, so he gets pulled in when the leader falls? Clarification?

 

I think what he is talking about is a belayer who wasn't facing the direction of pull. His belay wasn't "aimed" properly, and when the leader hit the end of the rope, the belayer was twisted around to face him. His crampon didn't move, and his upper body did, leading to the spiral...

 

I have a similar experience. I slipped on some hard summer snow, and twisted my knee, but didn't realize. However, when I got to the bare ice of the glacier tongue, and put my crampon on, well... OUCH!

[bolt_clipper]

Dumbass. They should have taken a picture of the belayer, he could be featured in the next edition of How To Ice Climb.

[Alex]

I was one of the two people who responded to this accident, it happened in March 1995 on Louise falls. I was one of a party of two climbing directly below this party; I had just completed my WFR training a month before.

 

The leader of the party had just started out on the p3 pillar, but instead of taking the common line between the 2 pillars with some stemming, he came in from the bolted belay on the left and launched straight up the column. Half way up, with only 1 screw between himself and belayer, he started getting really shaky. Maurice and I were watching, because the leader was obviously in trouble. Then one of the leaders tools popped and he whipped, pulling the screw out of the ice. The leader hit the ledge below completely on his side, and kept going, tumbling down the route. He had lost one of his tools in the initial fall, it just ripped out of his hand, glove and all. He finally came to a stop as the belay held him, halfway down the 2nd pitch. In the meantime, the belayer casually announces, "ow. I just broke my leg." We helped the leader recover mentally, he was unhurt! Then maurice climbed up to the base of the third pitch and I rapidly followed, while a party below (doctor from Rocky Mountain House/Red Deer and a paramedic friend) helped leader dude down the route.

 

At the belay, it rapidly became apparent what had happened. The belayer had clipped into the belay with a pearabiner only on his swami. As the load from the fall came against the belay, his torso was whipped from facing the rock to facing the load of the fall, as the pearabiner slid around his waist. However, standing on ice, his crampons and boots didnt swivel like his torso, and he got a nice greenstick tib/fib right at the top of his Invernos. We used athletic tape to splint up his leg, then lowered him down the route. He got helo-evaced out ot there, while his buddy the leader had to walk out under his own power and drive to the hospital.

 

The moral is, always belay in such a manner so that if there is a fall directly against the belay, you are pulled toward the belay, not towards the fall. This could have been avoided with a quickdraw on the belay.

 

This accident made it into Accidents in NA, 1996 I believe.

 

Years later, I met the doctor and paramedics at Rampart, telling this story to some of their friends around the dinner table. It was only then that we were formally introduced

 

Alex

[CascadeClimber]

It is worth noting that running the rope through the belay can effectively double the impact force on the belay in such a situation. If the belay failed you would end up with a much bigger problem than a tib-fib fracture.

 

Sometimes I run the rope through the belay, and sometimes not. It depends on my opinion about the strength of the anchor system and the potential for a factor-2 fall.

 

-Loren

[Alex]

how does that work? the factor 2 load is directly against the belay, whether you feed the rope through the belay or not..? Isnt the load the same? Its not like your body can compensate for it, your harness is made of static webbing.

[wdietsch]

maybe I am missing something here ..... sounds like the belay was not protected

[CascadeClimber]

Because to keep the system static, the anchor must absorb the force of the falling leader plus the equal force of the stationary belayer (less static and dynamic friction).

 

This is the best pic I could find quickly:

 

If you remove the rope from the quickdraw, the equal and opposite force on the anchor is only that of the falling leader.

 

This also applies to belaying a second through the anchor.

 

-Loren

[iain]

I think he means you effectively have a 2:1 pulley system by running the rope through the belay as a directional.

[iain]

whoops, now I am redundant

[Alex]

Loren, your picture makes sense to me, but your explaination does not.

 

From this picture, its obvious that the force on the anchor would be 1160 daN regardless of what the forces on the climbers are. By adding a draw, the force is now split between the climbers (good!), rather than being exerted completely on the leader(bad!) and the leader's belay tie in, as it was in the case described above on Louise Falls. Think about it. In your picture the belayer only has to hold 700 daN of force through the ATC, rather than 1160. This is good. The leader only takes 460 Newtons on his tie in. But the anchor will ALWAYS take 1160. There is no question. So by adding a draw you are lessening the forces on your own belay loop and belay device, and on the leader, but otherwise the forces are always going to be the same.

 

Alex

[Greg_W]

Thanks for that explanation, Alex. I have always had a question in my mind about loads on anchors and directionals (and I have an engineering degree ). That is one of the better explanations I've read.

[COL._Von_Spanker]

"...I was one of a party of two climbing directly below this party..."

 

?? Good thing he didn't take y'all with him, or drop other misc. shit on ya.

 

 

 

[forrest_m]

wait just a second, alex, I'm not sure you have it right. forget the diagram for a second and walk through this with me.

 

option a: falling straight onto the belay. when the climber hits the bottom, they put a force on the rope of, say, 1000 pounds. the anchor must therefore resist 1000 pounds to balance out the 1000 pounds of tension in the rope.

 

option b: falling on a quickdraw: same fall, rope from climber to QD had same 1000 pounds of tension; but to keep it from slipping, the rope from the QD to the belay must also have 1000 pounds in it, more or less in the same direction of pull (assuming both belayer and fallen climber are below the anchor). therefore, the QD has 2000 pounds of tension in it, because it is being pulled down by 2 lengths of rope, each with 1000 pounds. the anchor must therefore support double the weight.

 

this is how I've always understood it, but maybe I'm wrong?

 

btw, obviously the force won't be exactly double because the angle between the two rope strands > 0, friction, etc.

 

[CascadeClimber]

What Forrest said.

 

To look at it another way, it is a C-pulley system. 2-1 mechanical advantage, except it is a 2-1 mechanical disadvantage in this scenario.

 

-L

[Alex]

no

 

option a: falling straight onto the belay. when the climber hits the bottom, they put a force on the rope of, say, 1000 pounds. the anchor must therefore resist 1000 pounds to balance out the 1000 pounds of tension in the rope.

 

I absolutely agree.

 

option b: falling on a quickdraw: same fall, rope from climber to QD had same 1000 pounds of tension;

 

no, it doenst

 

but to keep it from slipping, the rope from the QD to the belay must also have 1000 pounds in it, more or less in the same direction of pull (assuming both belayer and fallen climber are below the anchor).

 

no, it doesnt.

 

therefore, the QD has 2000 pounds of tension in it, because it is being pulled down by 2 lengths of rope, each with 1000 pounds. the anchor must therefore support double the weight.

 

no, it doesnt. The anchor will always have 1000 pounds applied to it in this situation. It's the forces against the other components of the system you must now consider.

 

In the first scenario, there are 2 components to the system, the anchor and the leader (since our belayer is directly in line with the anchor->leader force vector and therefore doesnt have any role in the force calculations). We all agree on this.

 

In the second scenario, there are 3 components to the sytem. The anchor, and the force on it, is unchanged. Its still the SAME FALL. The other components in the system change though. This is precisely what Lorens diagram above shows, actually. If you keep adding components, the force on EACH component will go down. Think about it. With 2 components, the force exerted on EACH is less than with 1, in the first scenario. How about if you now added 2 ropes to the belayer and 2 draws. The anchor STILL takes 1000 pounds, but now each Leader - draw takes half the force that it did before, and each belayer to draw also. There are now 4 components. Does that mean there are now 4000 pounds of force?? No, it means that each component now only has to deal with 200, 300 or so, whatever adds up to 1000.

 

This is because for every force there is an equal and opposite force. This is a fundamental law in Physics (Newtonian mechanics). If the force on your anchor is 1000 pounds as you state, the opposite force MUST BE 1000 pounds combined on the belayer and leader. Force vectors are addative here, the sum of all forces MUST be 0. Lorens diagram bears this out. This MIGHT transate to more on each component with some force vectors, which is why you tend to equalize anchors with less than 60 degree angles, as you hint at Forest, for example, but I dont usually do that much math to figure out how much more. However, you cannot magically conjure up another couple thousand pounds of force.

 

It makes intuitive sense. This is one reason you climb on twin ropes, and each rope takes some of the falls force. This is why its BETTER to have a piece between you and the anchor when you fall. And better to have a piece on the anchor, even if there is nothing else between you and leader.

 

Alex

Edited December 5, 2002 by Alex

[iain]

Wouldn't you agree if you replaced an ideal pulley for the quickdraw and you had a climber with force T falling and a secured belay on the other side that there would be a force of 2T on the pro? I'm not sure I see where you're going, whether you agree or disagree with this. If you just tied the rope to the draw, you would only have T on the pro.

[iain]

Here

 

There's 10N on each strand if the weight is to remain stationary but 20N on the anchor in this image. Agreed?

Edited December 5, 2002 by iain

[Alex]

iain, you are correct in both cases. But you left something out.

 

Wouldn't you agree if you replaced an ideal pulley for the quickdraw and you had a climber with force T falling and a secured belay on the other side that there would be a force of 2T on the pro?

 

so the force on the leader is +T, and the force on the anchor is -2T (right?) so whats the force on the belayer? +T. So the equation becomes +T+T-2T = 0

 

I'm not sure I see where you're going, whether you agree or disagree with this. If you just tied the rope to the draw, you would only have T on the pro.

 

yes, +T on the leader and -T on the pro. +T-T=0.

 

Alex

[Alex]

iain, yes, completely agree with the drawing. Its exactly what I have been saying. +10N +10N -20N = 0. Please read my post. All force has equal and opposite force. The sum of it all is 0.

 

Alex

[iain]

Okay so there's twice the force on the pro then (20N). I think that's all that was to be made clear.

[CascadeClimber]

Alex-

 

You are right, the sum of the forces in x, y, and z planes must be zero, or some part of the system must be in motion. Where you are mistaken is in your assertion that the force on the anchor is constant. It is not.

 

The constant in the system is the force related to the fall of the leader or second. Surely you can see that the y-vector downward force created by the fall cannot change as a result of the anchor system. F=MA, the mass of the climber times (in the case of a fall) the acceleration of gravity, 9.8m/s^2.

 

What you are saying is that the force of the falling leader is different if the rope runs through the anchor. The orientation of the rope has nothing to do with the mass of the climber and the acceleration of gravity.

 

It is counter-intuitive in this presentation, but if you think of it as a C-pulley, with the anchor as the victim, you know that the force required to hold the c-pulley in a static position is 1/2 the weight (in newtons or pounds-force) of the victim (ignoring friction).

 

I'll go dig through my Statics book to find a reference if this still seems wrong to you.

 

-L

[CascadeClimber]

Here, from the Beal web site:

 

In case of a fall, the anchor sustains a double load: that of the climber and that of the belayer. These two forces combine in a pulley effect. If the rope does not absorb enough energy, the load on the anchor point will be very high, with increased risk of failure… Rope with a low impact force is the key to limiting the consequences of the pulley effect.

 

 

-L