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snoboy

Something scary to think about!

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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.

 

yay!

 

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.

 

Yes. and it is, different. If you want to get down to nitty gritty, the force on the leader will be almost identical if there was a draw or not. But who cares, thats not the point!!!

 

The points are this:

 

1) Every force has equal and opposite force. This addresses Forests argument that the force is multiplied when another component enters the system. You dont get systems where there is 1000 pounds on the anchor, 1000 pounds on the belayer, AND 1000 pounds on the leader... +1000+1000-1000 = 1000.....not 0. Thats not correct.

 

2) And if you are belaying someone, it only makes sense to put a draw on the belay if that someone is maybe gonna fall, to be pulled INTO the belay rather than away from it. The force on YOU (effectively the anchor) will be the same if you do or if you dont. Why not prevent getting whipped around and down, shockloading the anchor maybe, and just have the anchor take the load directly? This is what the point is.

 

...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).

 

Yes. but this is not what you said earlier. Lets revisit your first post:

 

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.

 

This contradicts what you are saying now, in agreement that summing all forces is zero. It does not doulble the impact force. The force is the same on the anchor, for the same fall.

 

There is only ONE thing in this entire system generating force. The falling leader. The force of the falling leader for the same fall (thats a fall of the same distance!!) in ALL systems is the same. Agree? Lets call it 1060. In the system where he falls directly against the achor, scenario 1, the force against the anchor and the Leader are equal and opposite. +1060 and -1060. Now the second scenario is almost the same, but there is a draw on the anchor, a few feet away from thebelayer. The fall is the same distance (maybe a few inches shorter because of the draw, but as we say, who cares?). The force on the draw, and thus the anchor is STILL 1060, or thereabouts. Agree? Technically, the force is now split between the belayer and the leader against the anchor, but realistically this is proportional to the amount of rope from the belayer to the draw, compared to the amount of rope from the leader to the draw. So the longer the fall, the more skewed to one side this is. The only thing different is the belayer is not getting whipped around into the direction of the fall, rather just getting slammed into the rock of ice by the belay.

 

Lets look at two complete equations:

 

Force T is the force applied to the anchor after joe has tied off to it and bungie jumped 100 feet with 100 feet of rope. OK? So our forces look like

 

Anchor: +T

joe: 100/100(-T)

 

+T = -100/100T = 0

 

Now second. Force T is for joe who just took a ~100 foot fall 50 feet above the belay. Jut Sally is belaying, 3 feet from the belay, with a draw:

 

Anchor: +T

joe: 97/100(-T)

sally: 3/100(-T)

 

but here we have +T = -97/100T - 3/100T = 0.

 

right? The forces here on sally are comparatively negligible, but sally isnt getting whipped into the direction of the fall. There is no Tib Fib. There is no "doubling" of the forces on the anchor.

 

Similarly, when you top rope someone from above, do you belay the second directly from your device? Or through a draw on the anchor? Anyone who has held a fall of a second knows that getting pulled in the direction of the climber is undesirealble. You want to be pulled in the direction of the anchor. The force calcs here in this situation are no different. The (body-weight)force is against the anchor.

 

Hope that helps.

 

Alex

Edited by Alex

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The force on the draw, and thus the anchor is STILL 1060, or thereabouts. Agree?

 

No. The force on the anchor is not the same. It can't be if the sum of all the forces equals 0.

 

The point I was trying to make, perhaps in an unclear manner, is that running the rope from the belayer through the belay anchor will result in the (near) doubling of the force on the belay anchor if the leader falls with no gear in above the belay (or if the gear fails), or if the second falls.

 

A fall of either type tranmits twice the force to the anchor, less frictional forces, as it does if the rope is not run from the belayer to the anchor to the climber.

 

So if I mistrust the belay anchor, I don't run the rope through it when I am bringing up a second or belaying a leader.

 

Maybe this is better settled over a beer...

Edited by CascadeClimber

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Alex you are wrong. The reason it is more comfortable getting pulled toward the anchor is because you don't have any weight on you. If you are not holding up this force (you're not even holding your own weight right?) then what is? It is the anchor.

 

When you do a slingshot belay the anchor is holding both of you fruit.giffruit.gif. When you belay directly then the anchor is possibly holding both of you eventually if the belayer goes for a ride, but often the belayer can do some holding of their own. HCL.gif

 

You should have a bombproof anchor or independent anchor when you do the slingshot belay. fruit.gif

 

I don't know what relevance this net force of zero thing has. I haven't read it very closely though I guess. Surely you're not saying that the net force on a slingshot belay anchor is zero? hahaha.gif

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Hmm, Loren is right here, assuming "ideal" conditions, that is, no friction, yadda yadda yadda...

If Joe falls with force T on a draw, and Sally is belaying him just below the anchor, she will also feel force T upwards, and the draw will have a force of 2T on it. So it definitely doubles.

 

Tie a string to your finger, pull down on it with a force of 10LBS. Your finger feels 10LBS. Now, instead, loop that string around your finger, and pull on one side with a force of 10LBS. In order for the string to remain in a fixed position (i.e. equivalent to Sally holding the climber in a static belay), someone else has to pull on the other side with 10LBS. So your finger will now feel a force of 20LBS.

 

Of course, in the real world there is biner friction, and the belay isn't static, so then things get really complicated, and the force probably isn't doubled...

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Sure. In Alex's example, Joe falls with force T when tied directly into the anchor, falling from height X above the anchor.

 

If sally belays him through a draw, and Joe falls from height X above the draw, he's not going to put force T on the draw, since the belay will slip a bit. But say he puts force 0.8T on the draw, Sally will feel an upward force of 0.8T, give a total force on the draw of 1.6T. Whoa! You're right bolt_clipper!

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You have all lost it. The best thing to do is (what i do) is put a piece in above anchor then the fall will be on it. I count this as part of my anchor.

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I don't see anyone really disagreeing with anyone else here, but confusion. You run your rope through a carabiner attached to pro and someone pitches, the pro experiences the force of the fall x the mech advantage of the simple pulley. In an ideal fbd this would be 2:1. As said before, there is flex in the rope, movement of the belayer, friction in the biner based on width etc. etc. to drop this force, but it is nevertheless multiplied. I believe Loren is saying if you use your belay anchor in this manner, as a directional, the belay anchor will be submitted to multiplied forces, just like anything else. There is simply no question in this.

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The article is excellent. But it doesn't discuss the situation in question here. The closest illustration is #3, where the force generated by the fall is 5.7kN, but the top draw experiences 8.9kN. This illustrates the additive effect I was pointing out. But what is doesn't do is show that the force on the belay without the gear would be equal to the force of the fall.

 

If the rope is run from the belayer to the belay anchor to the climber and the leader falls with no other protection in place (or the second falls at all) the resultant impact force on the belay anchor is significantly higher (double, ignoring friction) than if the rope is not run through the belay anchor.

 

So if I am belaying on a shitty swedgie anchor, or bad screws, I don't run the rope through the belay, inorder to minimize potential impact force on it. If the anchor is bomber, I do, because it is more comfortable and the force of a fall on me is less, due to friction. However, if the force is high enough, I could do a face-plant into the rock or ice which would make me even uglier than I already am!

 

-L

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If the rope doesn't go through the anchor can't we assume that some of the force is absorbed by the belayer and transfered into the ground, muscles, breaking bones, before it gets to the ancor. It seem like with all the details of newtonian physics some of the more intangible forces have been left out. Can be think of the belayer like a really big screamer? Such as when a person falls in a crevasse you can determine with physics how much the person has to hold, but that doesn't factor in the friction of the rope on the snow etc, which abosorbs quite a bit of the forces.

 

Newton was a hack.

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If the rope doesn't go through the anchor can't we assume that some of the force is absorbed by the belayer and transfered into the ground, muscles, breaking bones, before it gets to the ancor. It seem like with all the details of newtonian physics some of the more intangible forces have been left out.

 

I wish I had a nickel for every time a prof said, "But for now let's ignore those forces."

 

You are right. A dynamic rope absorbs energy in a fall, as does friction at each point where the rope runs over a surface, including the friction of the kern against the mantle as the kern stretches, and the friction of the individual kern fibers on eachother. It turns out that calculating some of these quantities is very difficult. The rope for example, would presumably act like a spring. But a spring resists pull/absorbs energy at a constant rate (the 'spring constant'), and ropes don't. The complication of the minutae is why they are often left out of calculations and why much of the data about impact forces, energy absorbtion, and ultimate strength of climbing gear are determined empirically- through testing. It is easier, faster, and more accurate to do it that way.

 

Can be think of the belayer like a really big screamer?

 

Yep. Especially if they end up with a compound spiral tib-fib fracture. That would make a screamer out of me.

 

Hey, this is the most I've used that dusty old ME degree in ten years!

 

-L

 

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It seems like the force absorbed by the belayer would be rather substantial, hence the broken bones. So this would not be a reccomended scenario to alleviate forces on the anchor, but a force that cannot be ignored in this situation.

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I don't think anyone disputes the fact that because there are so many pieces in a real world system, real forces won't equal the "worst case"physics situation, i.e. knots tightening, friction, belayer pulled up etc. I agree with paul, the best thing is to always protect the belay with an additional piece that I mentally consider part of the anchor. but...

 

what seems to be at issue is that alex is clinging to "common sense"over mathematics. he is insisting that since the fall is the same, it is impossible for greater forces to be generated. as evidence he cites that if there are more components in the system, the force felt on each must be less, but this is simply not true. the slingshot DOES add another component of force to the system, which if the pulley is perfect and the angles are perfect, is exactly the same as the original force.

 

maybe this will help explain. it seems we all agree on my scenario #1, a 1000 pound weight hanging off the QD puts 1000 pounds of force onto the anchor. I assume we would all also agree that if we tied a second 1000 pound weight on a second rope to the same QD that the anchor would now feel 2000 pounds? now, instead of tying the ropes to the QD, I tie them to each other, passing through the QD. Has the force on the anchor changed?

 

Loren, if I only had a nickel for every time my physics prof. answered questions like this by saying "Draw the Free-Body Diagram" !

 

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It depends entirely on the position of the belayer. In the best case scenario, which is not difficult to approximate in the field, almost 0 force/energy is applied to/absorbed by the belayer, it is simply passed through to the anchor.

 

All you have to do is get into a position that aligns the force of the fall with the anchor and remove slack in the system that connects the belayer to the anchor. I try to lean against the anchor and face perpendicular to the fall-force vector (facing 90 degrees from the direction of pull in the case of a fall).

 

So in Alex's example, the spiral fracture occurred because the belayer was looking up at the leader and was spun around by the fall. But if he was oriented differently, with no slack, the force would not have spun him and he would have felt much less force overall.

 

Running the rope through the anchor simply makes predicting the direction of the fall-force vector, and consequently the best body position, easier

 

-L

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Loren, if I only had a nickel for every time my physics prof. answered questions like this by saying "Draw the Free-Body Diagram" !

 

I was trying to do this last night. I gave up and watched The Daily Show instead. Maybe I'll break out my old notes and textbooks and try again. It gets complicated when you toss in the frictional forces...

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Looks like you guys need to sharpen up. Here you go, what are the mech. advantages of the following 2 systems? Please have your answers to me by classtime tomorrow. laugh.gif

 

fbd1.gif

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Alex,

This is from the article you linked

Why so much force on the top piece? Because the top piece experiences 2 forces:

  • The impace force of the climber's fall, and
  • The force that the belayer can apply with the belay device

 

If you run the rope through the anchor, that is the top piece. You will get more force on the anchor, but less on the belayer. That would have probably avoided the leg-breaking in the introductory anecdote. So in this case clipping it through your anchor with a quickdraw probably would have helped. However, if the belay anchor was shaky, there is a slight chance that it was strong enough to hold the force of the fall (which was attenuated by using up some energy snapping some bones) , but that it was not strong enough to hold 1.6 times the force. If that was the case, then running the rope through the anchor would have killed both those guys (and maybe you if you were in the fall line).

 

Paul makes a good point, which I alluded to in my previous post. You can make an independent anchor (i.e. first piece on climb, Pope's "frank nut"). That piece will experience the multiplied force, but if it blows you still have the anchor. That works fine and is the best option, IF you have another (good) placement nearby.

 

OK lets use this scenario:

 

You've just topped out off the Outer Space handcrack. You placed all your gear that would fit in the little crack (.5 Camalots +/-) above and are forced to belay your second anchored only to that spindly little, quickly dieing, tree. Do you

 

a) Run the rope up through the one sling around that tree, or

b) Snug yourself up to the "anchor", set your feet on some footholds, point yourself in the probable direction of the fall, and belay directly off your dee-vice?

 

My answer is b) because I am going to be able to hold part or all of the force of a fall with just my footholds and the friction of my butt wedged in the crack. I am being part of the anchor until I get totally pulled off my stance. If you choose a), then you have to trust that the tree can hold both you and the climber if he falls.

 

Once the second turns into the leader, you still should be in this "ready to catch, pointed down" position until the leader gets a piece in (the now cleaned .5 camalot).

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"Draw the Free-Body Diagram" !

AAAAAAHHHHHH!!! Make it stop, make it stop!!! The nightmares are returning!!!!

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