Which brand of rope do you prefer?

[ynot]

The blue line in the graph is impact force on the last piece of protection measured at the runner?

[caribe]

My bad, I didn't mean to come off as a jerk, but you did put me down about the reading and not answering part, and even you tried to answer questions without much success, so you are just as guilty as I am.

_____ I am not being successful because you are not listening. Tell me when I lose you in the following text. You are talking about fall factor because you want to express severity of the fall. But fall factor is not the real issue, it is a component of the issue. If you want to talk about fall severity, aka stresses on the elements in the system, you want to talk about maximum impact force. Saying that talking about impact force is "changing the subject" is ridiculous.
_____ To wit Saxman's comment was transparent and spot on. I will translate Bram for you. "A high fall factor does not necessarily mean a severe fall." A high fall factor with negligible force in the system does not break gear or injury people. I love it when people take a 4 ft fall and worry about their ropes. It might be hard on the ankles, but its not so bad on the ropes.
_____ From the very start I said that fall factor is not part of rope stretch. Go back to page 3. The same fall on a dynamic and a static rope will have the same fall factor, but different stretch.

[Redpoint]

Sorry, I took your words to literally when you said:

"I don't think that the Fall factor accounts for stretch in rope."

and so I assumed you were only guessing and didn't know for sure. And so I was still determined to find out for %100 sure if it did or not.

Ya I know the fall factor is part of the equation for impact force, remember I am the one who posted all of the variables of the equation. I just couldn't post the equation itself unless I took a picture of it in my book, or wait I can make a picture of it in paint. I am just running on slow today, I was caving all day yesterday, drilling and banging hammers trying to get in to this virgin cave.

Here is the equation for anyone who wants to see it:



I = impact force

M = rope modulous in ft-lb/sec(squared). Material constant depending on cross-sectional area of rope, fiber, content, etc.

F = fall factor (distance fallen/amount of rope paid out)

m = mass

g = acceleration constant of gravity (32 ft/sec(squared)

The only thing that confused me is how my fall factor diagram for example shows someone being 3 feet above a bolt, falling and being 3 feet below a bolt, when they should have fallen farther than that with rope stretch. I also didn't understand why none of these authors bring up rope stretch when talking about fall factors.

But now when I think about how rope stretch is factored in when doing the impact forces equation, I see why they left it out of the fall forces equation; you don't want to factor rope stretch in twice(once for the fall factor part, and again for the impact forces equation.

Riddle solved.

Thanks for your help.

[caribe]

The blue line in the graph is impact force on the last piece of protection measured at the runner?

_____ Yes, that is correct. The dark blue is simulated on the computer with math models, the light blue is experimental--a real system. In an friction-less world with no give at the harnesses the force on the climbers harness would look very similar to the force as a function of time on the belayer's harness if they both weighed the same. In a frictionless world these two forces would play out in time and the sum of the two would always equal the force measured at the first runner.
_____ Doesn't this make sense intuitively? Think about two people hanging from one point at two ends of the same rope. The force due to gravity at the runner should be the sum of their weights; however each harness only has to support each climber's body weight. Each harness does not care how it is hanging, only that is is hanging and supporting its climber.
_____ At the beginning of the curve, the max impact force is roughly the sum of the force on the climber's and the belayer's harness, check it out. At the beginning of the curve the belayer is lifted and then lands again and the force on her harness again becomes zero.
_____ To the extent that the force at the two end points does not equal the force as a function of time on the runner is either friction or negative kinetic energy (mass moving upwards due to oscillation in the system instead of falling). The integral of the runner's force/time curve with respect to time minus analogous integrals of the belayer and the climber curves should quantify the force lost due to friction in the system throughout the duration of the fall.
_____ This is the extent to which I think that I understand the math. I could tell you more about the chemistry of polymer than I know about these mechanics. I am just an old country chemist (not engineer or physicist). I would love to learn more if there is anyone here geekier than me. Post-er up.
_____ No Redpoint I am not asking you to explain anything to me.

[caribe]

I just couldn't post the equation itself unless I took a picture of it in my book, or wait I can make a picture of it in paint.

You know that it gets a little raunchy on Red River climbing sometimes . . . are you old enough to be here?

[Redpoint]

I don't get it, what did that sentence have anything to do with me appearing to be young, was it the paint part? O did you think I meant finger paint lol

Don't worry I am 28.

I have been using Photoshop for so many years now, and ever since someone on here requested a picture in Paintbrush, I have been using Paint(Paintbrush-a free microsoft graphics program that comes with Windows.. just in case you are a confused mac user I thought this parenthesis sentence might be necessary) again for quick illustrations, and besides I missed Paintbrush, it's been too long.

[ynp1]

Redpoint, i also thought that. if you do not count the stretch in the rope then you would get fall factors over 2 for example. 100 feet off belay fall on to belay fall is 215 feet with stretch but rope is only 100 feet so the fall factor would be 2.15. i asked around and people told me that you count the stretch of the rope in with the length of the rope and that is why you cannot get a factor two fall with a climbing rope, static or dynamic. i could be wrong about this, but this is all i could figure out by asking around...

[caribe]

But now when I think about how rope stretch is factored in when doing the impact forces equation, I see why they left it out of the fall forces equation; you don't want to factor rope stretch in twice(once for the fall factor part, and again for the impact forces equation.

____ Yes! The square root in the expression that you posted comes from the same math that describes harmonic oscillation,
http://www.redriverclimbing.com/viewtop ... 486#199486
but I would have to play around to derive it. I am not sure I could in any reasonable time and I have other stuff to accomplish today.

[Redpoint]

Redpoint, i also thought that. if you do not count the stretch in the rope then you would get fall factors over 2 for example. 100 feet off belay fall on to belay fall is 215 feet with stretch but rope is only 100 feet so the fall factor would be 2.15. i asked around and people told me that you count the stretch of the rope in with the length of the rope and that is why you cannot get a factor two fall with a climbing rope, static or dynamic. i could be wrong about this, but this is all i could figure out by asking around...

Well I hope you are paying attention ynp1, caribe and I are both sure that the fall factor does not include rope stretch, that part comes later on in the impact forces equation.

Check out the post I just made with the impact forces equation, and you will see why the fall factor leaves out rope stretch. In the impact forces equation the rope stretch would be part of M.

[caribe]

I don't get it, what did that sentence have anything to do with me appearing to be young, was it the paint part? O did you think I meant finger paint lol

Try
I/mg = 1 + SQRT(1 + 2fM/mg)