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quietseas
Apr 5, 2004, 11:41 PM
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guys who are trying to figure out force on the top piece: It's the same as force on the climber who falls. Friction does not increase the force on the biner. If the rope is tied off at the bottom (belaying from a fixed anchor) then the friction comes from the rope giving, and the friction will make the rope give slower, thereby increasing the time for the climber to decelerate to zero, making LESS force. (Assuming heat is not a factor, but that would be REALLY complicated!) If the climber picks up the belayer, the force on the top biner then becomes constant at: 9.8m/s/s X Total weight of both climbers in Kg. Note that the unit of force called a Newton is defined as Kilogram meters per second per second. My 2 cents.
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vincent
Apr 5, 2004, 11:57 PM
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...from John Long's "More Climbing Anchors"...
"a bombproof anchor means it's good enough to withstand the greatest force that a climbing team can place on it, which virtually never exceeds 2,500 lbs
generating 2,700 pounds of forc on an anchor, bolt, piece whatever seems a little high for a 21 foot fall on a dynamic rope with a dynamic belay. i generally try to listen and learn from people who have proven records of safe climbing and safe technique. and the 2,500 lbs. that long is referring to is a direct FF 2 lead fall onto the anchor...
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jt512
Apr 5, 2004, 11:58 PM
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In reply to: guys who are trying to figure out force on the top piece: It's the same as force on the climber who falls.
Wrong. It is approximately 1.6 times the force on the climber.
In reply to: Friction does not increase the force on the biner.
We know. Friction decreases the force on the biner. That's why the force on the biner is 1.6 times the force on the climber, not double the force on the climber.
In reply to: If the rope is tied off at the bottom (belaying from a fixed anchor) then the friction comes from the rope giving, and the friction will make the rope give slower, thereby increasing the time for the climber to decelerate to zero, making LESS force. (Assuming heat is not a factor, but that would be REALLY complicated!)
Huh?
-Jay
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jason1
Apr 6, 2004, 12:05 AM
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friction can also reduce the amout of rope available to absorb a fall...
like when you have many points creating a lot of drag...
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quietseas
Apr 6, 2004, 3:27 AM
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Sorry i don't get this quoting stuff, but in response to the last one. it's impulse theory. If you can make the deceleration occur over a longer time, then it will produce less force, since the force is not a constant but an impulse.
For the (I think) first one, I didn't make clear what I had in my head as I was running off to class, but... assuming that the belayer was tied off, so that he could not give, run-out, or anything like that. The force on the top biner would never take the belayer's weight, it would only ever take the force of the falling climber. It will take the force he generates from decelerating as the rope stretches, which if the deceleration is uniform (an assumption, of course we know what that does) then it is an impulse equation.
Think of it this way: Force = Mass X Acceleration. The ground which the climber is tied off to, no matter how massive, is not accelerating so it exerts no force. The only force in the system is coming from the falling climber, so the maximum amount of force that can be placed on the top piece is the force of the falling climber.
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quietseas
Apr 6, 2004, 3:31 AM
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ugh, don't even want to think about that! Hmm... fall with 7% rope stretch and deceleration X for first 3 meters, deceleration y for next meter, deceleration z for next 2 meters.... AGHHH!! We're still arguing the most simple setup! If we had to go into that much detail, we'd need a whole new server! Hehe
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jumpingrock
Apr 6, 2004, 3:45 AM
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Heh funny stuff... if it wasn't so sad. Just climb safetly. Place protection early and often at the start. Learn how to place it and learn how to belay and tie a god damn figure 8. Then you don't have to worry about whether your biner is oriented properly (except the belayer but see note on learning to belay).
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kiwikt
Apr 6, 2004, 4:07 AM
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Hey, wicked to have someone post this in laymans terms. Getting the knowledge out there how important it is to ensure as far as possible your biners are going to be loded correctly is awesome! Especially when you present the figures to back in up :D
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andypro
Apr 6, 2004, 4:17 AM
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Hmm...one of the first things I learend when I started climbing is to not crossload a biner. I dont think people need a scientific (sort of) explenation as to why. Ii'ts jsut common sense. Look on the side of the biner...29 is higher than 7 or 8. Make sure it follows the arrows.
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albertonium
Apr 6, 2004, 4:27 AM
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Way to sum it up. Those pictures are there for a reason. Follow them. Lots of info though.
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curt
Apr 6, 2004, 4:27 AM
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In reply to: guys who are trying to figure out force on the top piece: It's the same as force on the climber who falls.
Really? Let's think about this by considering a static situation. Suppose you weigh 165lbs and you are hanging on a rope passing through a biner connected to a piece of gear. I am holding the other end of the rope as your belayer. Are you seriously saying that the force on the top piece of gear in question is not 330lbs minus friction? Please think before answering.
Curt
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jumpingrock
Apr 6, 2004, 4:51 AM
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In reply to: Please think
Thinking hurts though :cry:
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ryanhos
Apr 6, 2004, 4:57 AM
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In reply to: In reply to: guys who are trying to figure out force on the top piece: It's the same as force on the climber who falls. Really? Let's think about this by considering a static situation. Suppose you weigh 165lbs and you are hanging on a rope passing through a biner connected to a piece of gear. I am holding the other end of the rope as your belayer. Are you seriously saying that the force on the top piece of gear in question is not 330lbs minus friction? Please think before answering. Curt
Curt: Are you standing or hanging? It makes a difference.
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jt512
Apr 6, 2004, 5:01 AM
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In reply to: Think of it this way: Force = Mass X Acceleration. The ground which the climber is tied off to, no matter how massive, is not accelerating so it exerts no force. The only force in the system is coming from the falling climber, so the maximum amount of force that can be placed on the top piece is the force of the falling climber.
What are you talking about? The earth exerts no force? Where does the tension in the rope come from?
-Jay
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curt
Apr 6, 2004, 5:02 AM
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In reply to: In reply to: In reply to: guys who are trying to figure out force on the top piece: It's the same as force on the climber who falls. Really? Let's think about this by considering a static situation. Suppose you weigh 165lbs and you are hanging on a rope passing through a biner connected to a piece of gear. I am holding the other end of the rope as your belayer. Are you seriously saying that the force on the top piece of gear in question is not 330lbs minus friction? Please think before answering. Curt Curt: Are you standing or hanging? It makes a difference.
I think it makes no difference as far as I can see. Why would it?
Curt
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andypro
Apr 6, 2004, 5:04 AM
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In reply to: The only force in the system is coming from the falling climber, so the maximum amount of force that can be placed on the top piece is the force of the falling climber.
In order for the belayer to stop the climber, they have to provide an equal and opposite force. Since the top piece is in effect a pulley, it changes the direction of force, so it may not appear to be opposite, but it is. If the falling climber generates 500 pounds of force, the belay system needs to generate 500 pounds of force to stop the climber from accelerating. 500+500=1000. Figuring the friction as stated numerous times previously, 500 x 1.6 is 800 pounds of force on the top piece, so not quite twice as much. In reality, it's much more involved, but this is a basic example.
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curt
Apr 6, 2004, 5:05 AM
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In reply to: In reply to: Think of it this way: Force = Mass X Acceleration. The ground which the climber is tied off to, no matter how massive, is not accelerating so it exerts no force. The only force in the system is coming from the falling climber, so the maximum amount of force that can be placed on the top piece is the force of the falling climber. What are you talking about? The earth exerts no force? Where does the tension in the rope come from? -Jay
Jay,
You are truly bad. Just really really bad. Don't think that I do not appreciate it. Hahahahahaha.
Curt
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ryanhos
Apr 6, 2004, 5:16 AM
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In reply to: 1 kN = 225 lb. -Jay
Would you please stop saying that!
I'll prove it's incorrect. Imagine yourself laying on the floor. You have two choices. I can drop a 225lb weight on your torso from 6 inches, or I can drop a 225 lb weight on your torso from 6 feet. You choose. By your logic, since 225lbs = 1kN, it shouldn't matter from what height I drop it, because it will exert the same force on your body when it hits. Math, and common sense, tells us that this is incorrect.
No physics book ever printed has published an equation to convert lbs or kg to kN. It just doesn't make sense.
You do however make some good points in your post. The original author needs to recheck his math and reasoning. I actually PMed him asking him to retract his post in order to save us from the clusterfsck. (and now I'm just adding to it...*sigh*)
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darelparker
Apr 6, 2004, 5:23 AM
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Stupid math...my head hurts.
I'm just going to stop falling.
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curt
Apr 6, 2004, 5:31 AM
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In reply to: In reply to: 1 kN = 225 lb. -Jay Would you please stop saying that! I'll prove it's incorrect. Imagine yourself laying on the floor. You have two choices. I can drop a 225lb weight on your torso from 6 inches, or I can drop a 225 lb weight on your torso from 6 feet. You choose. By your logic, since 225lbs = 1kN, it shouldn't matter from what height I drop it, because it will exert the same force on your body when it hits. Math, and common sense, tells us that this is incorrect. No physics book ever printed has published an equation to convert lbs or kg to kN. It just doesn't make sense. You do however make some good points in your post. The original author needs to recheck his math and reasoning. I actually PMed him asking him to retract his post in order to save us from the clusterfsck. (and now I'm just adding to it...*sigh*)
You have only proved that you didn't ever pass physics 101.
Curt
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ryanhos
Apr 6, 2004, 5:43 AM
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In reply to: I think it makes no difference as far as I can see. Why would it? Curt
It does indeed. *hrm* How to explain this.....
If you are hanging from the rope, there is no doubt that your ENTIRE weight is being supported by the rope. If it is supported by the rope, it is supported by the anchor. So if the belayer is hanging, total weight on the anchor is 2(climbers + gear) + 1 rope. (multiply by gravity to get force...) Friction does not apply here.
If you are standing on the ground, some of your weight is being supported by the ground. Therefore, it could not be supported by the rope or the anchor. Friction DOES apply here. When the rope stretched, it created tension between the belayer and the anchor. That tension removes ONLY SOME of your weight from the ground if you are still standing. Therefore, the (approx) 60% transmission of force from one side of the biner to the other does apply. Total force on the anchor will be less than scenario 1.
Before anyone posts a reply:
1.) This was in response to a comment about a STATIC situation.
2.) It is difficult to judge what happens at the power-point biner just as steady state is achived. This analysis assumes that the rope does not recoil through the biner at all once it has been pulled through by the falling climber. Depending on certain factors (stretch recovery rate, the bungee effect, number of previous, recent falls, etc.) this may be an incorrect assumption. However, if the assumption is incorrect, it does not weaken the argument. The fact remains that if you are physically standing on the ground, you are not transmitting 100% of your weight to the anchor and if you are hanging completely from the anchor, you are transmitting 100% of your weight to it.
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emjay
Apr 6, 2004, 5:44 AM
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In reply to: Force(N)= Mass(Kg)xAcceleration(M/s2) or F=ma N=kg(m/s2)
In other words, the length of the fall is irrelevant, correct? To use the original example, a 74.8 kg climber falling ANY distance generates
74.8 kg X 9.8 m/s2 = 733 newtons
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jt512
Apr 6, 2004, 5:47 AM
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In reply to: In reply to: 1 kN = 225 lb. -Jay Would you please stop saying that! I'll prove it's incorrect. Imagine yourself laying on the floor. You have two choices. I can drop a 225lb weight on your torso from 6 inches, or I can drop a 225 lb weight on your torso from 6 feet. You choose. By your logic, since 225lbs = 1kN, it shouldn't matter from what height I drop it, because it will exert the same force on your body when it hits. Math, and common sense, tells us that this is incorrect.
No. You are confused. The "weight" you are dropping weighs 225 lb, or 1 kN, regardless of how high it is dropped from. Both lb and kN are units of force, as commonly used scientifically. And weight is a force; specifically, the force produced by a mass as a result of gravity. However, all else equal, the force exerted as a result of dropping an object of a given weight will depend on the height from which it is dropped. This resulting force can still be measured in either pounds or kN. It will be greater than the weight of the object, but the resulting force in kN, when multiplied by 225, will still equal the resulting force in pounds.
In reply to: No physics book ever printed has published an equation to convert lbs or kg to kN. It just doesn't make sense.
Modern physics books consistently use SI units, in which kg are a unit of mass and kN a unit of force. However, on earth, all that is required to convert weight to mass is multiplication by a constant. Hence, on earth, or whenever one can assume a particular gravitaional acceleration, one unit implies the other. Thus, on earth, both force and mass can be expressed either in kg or kN.
-Jay
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akclimber
Apr 6, 2004, 5:52 AM
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all this smart people stuff is making my head hurt, shit no break and i got home in 1 piece=me happy...
:D :D :D :D :D
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curt
Apr 6, 2004, 5:54 AM
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In reply to: In reply to: I think it makes no difference as far as I can see. Why would it? Curt It does indeed. *hrm* How to explain this..... If you are hanging from the rope, there is no doubt that your ENTIRE weight is being supported by the rope. If it is supported by the rope, it is supported by the anchor. So if the belayer is hanging, total weight on the anchor is 2(climbers + gear) + 1 rope. Friction does not apply here. If you are standing on the ground, some of your weight is being supported by the ground. Therefore, it could not be supported by the rope or the anchor. Friction DOES apply here. When the rope stretched, it created tension between the belayer and the anchor. That tension removes ONLY SOME of your weight from the ground if you are still standing. Therefore, the (approx) 60% transmission of force from one side of the biner to the other does apply. Before anyone posts a reply: 1.) This was in response to a comment about a STATIC situation. 2.) It is difficult to judge what happens at the biner just as steady state is achived. This analysis assumes that the rope does not recoil through the biner at all once it has been pulled through by the falling climber. Depending on certain factors (stretch recovery rate, the bungee effect, number of falls previous, recent falls, etc.) this may be an incorrect assumption. However, if the assumption is incorrect, it does not weaken the argument. The fact remains that if you are physically standing on the ground, you are not transmitting 100% of your weight to the anchor and if you are hanging completely from the anchor, you are transmitting 100% of your weight to it.
I understand your position, but I think you are wrong. The climber (in a static situation) is exerting a force on the top piece of gear that is equal to his weight. For the climber to be held there, the force on the belayer's end of the rope must be equal to the climber's weight--minus friction over the gear. Do you disagree with this?
Curt
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