
enjoimx
Feb 22, 2004, 7:49 PM
Post #1 of 24
(9258 views)
Shortcut
Registered: Feb 22, 2004
Posts: 378

I was wondering how you calculate the force acting on various aspects of your setup (biners, pro etc.) I know from physics that the force is equal to the mass times the acceleration, which are both constant when you take a fall. How then, does the length of the fall matter? Does it have to do with Work, which is equal to force times distance? Thanks





hosh
Feb 22, 2004, 7:53 PM
Post #2 of 24
(9258 views)
Shortcut
Registered: Dec 15, 2003
Posts: 1662

I think "freedom of the hills" hasa something about that. Can't remember though...





curt
Feb 22, 2004, 8:02 PM
Post #3 of 24
(9258 views)
Shortcut
Registered: Aug 26, 2002
Posts: 18272

In reply to: I was wondering how you calculate the force acting on various aspects of your setup (biners, pro etc.) I know from physics that the force is equal to the mass times the acceleration, which are both constant when you take a fall. How then, does the length of the fall matter? It doesn't. This is why the concept of "fall factor" was created. Curt





nurocks
Feb 22, 2004, 8:06 PM
Post #4 of 24
(9258 views)
Shortcut
Registered: Jul 18, 2003
Posts: 788

It's been a few years since physics, but I am fairly confident that the following is true. Mass in Kilograms*distance in Meters*(the gravitationaly constant here on earth) 9.8= a force in Newtons. Knock off three zeros and you have a total in Kilo Newtons. 1. I hope I'm right. 2. I hope that helps.





mulligan
Feb 22, 2004, 8:20 PM
Post #5 of 24
(9258 views)
Shortcut
Registered: Nov 10, 2003
Posts: 30

Yes it has to do with work and the conservation of energy. Work = Force * Distance. You are also going to have to use Kinetic energy which is: KE = .5(Mass)(Change in Velocity). So in the case of a fall it would be you total velocity since you are starting at rest. After solving for KE which would also = Work you could solve for the force. Granted this would be very general and would neglect teh fact that you are using a dynamic rope but this should give you a basic answer.





andypro
Feb 22, 2004, 8:21 PM
Post #6 of 24
(9258 views)
Shortcut
Registered: Aug 23, 2003
Posts: 1077

In reply to: I know from physics that the force is equal to the mass times the acceleration, which are both constant when you take a fall. The equation is right, your view of it is somehwat wrong. Acceleration is not a constant in this case. You need to do two things to use it. 1) Use the gravitational constant and the distance of the fall to determine the speed at which the climber begins slowing down. (theoretical vacuum so we can avoid fluid dynamics calculus). 2) Use the speed found in step one, and the time the climber took to come to a stop, as the acceleration figure for the equation. This will give you a vaguely accurate measure of the forces taking place. Theres alot more involved for any real accuracy, most of which involves calculus I couldn't remember to save my life.





curt
Feb 22, 2004, 8:24 PM
Post #7 of 24
(9258 views)
Shortcut
Registered: Aug 26, 2002
Posts: 18272

In reply to: Yes it has to do with work and the conservation of energy. Work = Force * Distance. You are also going to have to use Kinetic energy which is: KE = .5(Mass)(Change in Velocity). So in the case of a fall it would be you total velocity since you are starting at rest. After solving for KE which would also = Work you could solve for the force. Granted this would be very general and would neglect teh fact that you are using a dynamic rope but this should give you a basic answer. In a real world situation, you can not neglect the fact that you are using a dynamic rope. That is why the fall factor (length of fall taken divided by length of rope out) is important. And, it does (in terms of impact force on the gear and climber) make the total distance fallen irrelevant. Curt





curt
Feb 22, 2004, 8:29 PM
Post #8 of 24
(9258 views)
Shortcut
Registered: Aug 26, 2002
Posts: 18272

In reply to: In reply to: I know from physics that the force is equal to the mass times the acceleration, which are both constant when you take a fall. The equation is right, your view of it is somehwat wrong. Acceleration is not a constant in this case. You need to do two things to use it. 1) Use the gravitational constant and the distance of the fall to determine the speed at which the climber begins slowing down. (theoretical vacuum so we can avoid fluid dynamics calculus). 2) Use the speed found in step one, and the time the climber took to come to a stop, as the acceleration figure for the equation. This will give you a vaguely accurate measure of the forces taking place. Theres alot more involved for any real accuracy, most of which involves calculus I couldn't remember to save my life. Andypro, The gravitational constant has nothing to do with F=ma in calculating the force on the gear and falling climber. The acceleration that is relevant here is the negative acceleration related to the distance over which the falling climber is brought to rest. Curt





muncher
Feb 22, 2004, 8:33 PM
Post #9 of 24
(9258 views)
Shortcut
Registered: May 4, 2003
Posts: 454

It all depends on how long it takes you to stop. It starts to get interesting the moment you begin to weight the rope and start to decelarate. The more time it takes you to stop the smaller the decelaration thus the smaller the force. As was mentioned earlier it is as much a function of how much rope is out as how long the fall is. The more rope out, the more the rope can stretch, the longer it will then take you to stop thus lesssening the force on your gear. If you take a similar length fall with much less rope out the force on yourself and the grea will be much greater as the rope cannot elongate as much thus the time it takes you to come to rest is much shorter. Then you have to take into account friction on gear, dynamic/static belay, slippage through the belay device etc.





andypro
Feb 22, 2004, 8:42 PM
Post #10 of 24
(9258 views)
Shortcut
Registered: Aug 23, 2003
Posts: 1077

In reply to: Andypro, The gravitational constant has nothing to do with F=ma in calculating the force on the gear and falling climber. The acceleration that is relevant here is the negative acceleration related to the distance over which the falling climber is brought to rest. Curt There is no negative acceleration. Acceleration is simply a change in speed or direction. To get the acceleration for a change in speed, you would use the absolute value of the difference of the two speeds over time (which is where meters per second per second comes from, or meters per second squared). To get these numbers, you need ot use the gravitational constant and the length of the fall before the climber begins to slow down to get the starting speed to plug into the equation. All of this is inconsequential in the end, however, due to the fact that none of it takes into account the dynamic stretch of the rope dependant on forces it takes, the friction in the chain of protection, the fact that a climber doens't immediately come to a complete stop (more of a jagged sinewave of force transmitted), air resistance, or any one of a multitude of other factors.





trenchdigger
Feb 22, 2004, 8:54 PM
Post #11 of 24
(9258 views)
Shortcut
Registered: Mar 8, 2003
Posts: 1447

andypro wrote:
In reply to: There is no negative acceleration. Acceleration is generally considered to be a vector (having a magnitude and direction) defined as dV/dt. If velocity is positive in the chosen coordinate system and the change in velocity (dV) is negative, acceleration will be negative. ~Adam~





curt
Feb 22, 2004, 9:07 PM
Post #12 of 24
(9258 views)
Shortcut
Registered: Aug 26, 2002
Posts: 18272

In reply to: In reply to: Andypro, The gravitational constant has nothing to do with F=ma in calculating the force on the gear and falling climber. The acceleration that is relevant here is the negative acceleration related to the distance over which the falling climber is brought to rest. Curt There is no negative acceleration. Acceleration is simply a change in speed or direction. Acceleration is the change in velocityperiod.
In reply to: All of this is inconsequential in the end, however, due to the fact that none of it takes into account the dynamic stretch of the rope dependant on forces it takes, the friction in the chain of protection, the fact that a climber doens't immediately come to a complete stop (more of a jagged sinewave of force transmitted), air resistance, or any one of a multitude of other factors. Is there any chance you read my previous post? The things you are talking about regarding dynamic ropes is taken into account by the fall factor. Curt





trenchdigger
Feb 22, 2004, 9:19 PM
Post #13 of 24
(9258 views)
Shortcut
Registered: Mar 8, 2003
Posts: 1447

Some of the others are right  because you're using a dynamic rope (well at least we hope you are!)  the foce exerted on the system by your fall is not as simple as just looking at how far you fall or how fast you're moving when the rope catches you. Fall factor and stretch of the rope must be considered. The actual max force on the rope should look something like this... F = ma*(1 + sqrt(1+ 2*f*k/mg )) where: F = force of fall k = spring constant of rope ( E*A/length of rope catching fall  note that the k and/or E*A value could fairly easily be found for a rope experimentally, but not just from the rope's specs) f = fall factor (fall length/length of rope catching the fall) for this purpose, length of rope catching the fall = the length of rope between the belayer and the climber So that'll give you the force of the fall on the rope. With no friction in the system, the force on the highest piece of gear in the system could be as much as 2X that force due to a pulley effect between the climber and belayer. Also note that friction of the rope against the anchor or rock between the belayer and falling climber will effectively increase the force on the system. Slippage of some rope through the belay device while catching the fall will decrease the force on the system. Once you have a good estimate of the force on the anchor or piece(s) of gear in question (quite an impressive feat!) simple geometry and physics will allow you to divide the load up between the pieces that make up the anchor. ~Adam~





enjoimx
Feb 22, 2004, 9:40 PM
Post #14 of 24
(9258 views)
Shortcut
Registered: Feb 22, 2004
Posts: 378

Thanks everyone!





jeffers_mz
Feb 22, 2004, 11:38 PM
Post #15 of 24
(9258 views)
Shortcut
Registered: Jul 11, 2002
Posts: 357

"k = spring constant of rope ( E*A/length of rope catching fall  note that the k and/or E*A value could fairly easily be found for a rope experimentally, but not just from the rope's specs) " Most (about 9 out of 10) ropes sold give figures for static elongation, useless for the purpose. A few also give dynamic elongation, but only for one specific fall, a "standard fall". Dynamic elongation is not a linear relationship between elongation and the energy absorbed, because the rope is an imperfect spring. Dynamic elongation approaches limits at zero velocity (static elongation) and infinity, when the rope breaks. In between it follows a curve.





ryanhos
Feb 22, 2004, 11:58 PM
Post #16 of 24
(9258 views)
Shortcut
Registered: Mar 8, 2003
Posts: 132

can't anybody use google anymore? http://www.google.com/search?hl=en&ie=UTF8&oe=UTF8&q=impact+force+rope&btnG=Google+Search and http://www.climbgeorgia.com/train/fall.html Long story short, it's capped at almost 2*(impact force of rope) for a top rope anchor and the (1*impact force of rope) when belaying right off the anchor.





rgold
Feb 23, 2004, 6:47 AM
Post #17 of 24
(9258 views)
Shortcut
Registered: Dec 3, 2002
Posts: 1804

Trenchdigger gives the correct equation (if a=g is taken to be acceleration due to gravity). Impact force calculations are usually conservation of energy calculations. In order for the falling climber to stop his or her fall energy has to be "absorbed." The fall energy is the total length of the fall (including distance travelled because of rope stretch). The mechanism of "absorbtion" is that conservation of energy dictates that the falling climber must do some kind of work in an amount equal to the fall energy. In the case of the belayed falling climber, the work is done by stretching the rope. The fact that the fall energy must equal the work done in stretching the rope allows one to calculate the maximum rope stretch. The rope is modeled as a spring (it is really a damped spring), which means that the tension in the rope is proportional to the relative stretch, (which can be taken as the percentage stretch, for example). A simple integration shows that the work done in stretching a rope of length L so that the elongation is s is W=(1/2)(K/L)s^2. This has to be equal to the total fall energy, which is mg(H+s) where m is the mass of the falling climber, g is the acceleration due to gravity, and H is the height of the fall. This equation can be solved for s, the maximum rope stretch, and the result multiplied by (K/L) to get the maximum tension in the rope. Trenchdigger's formula is the result. The fact that "the length of the fall doesn't matter" can be traced back to the fact that the tension in the rope is proportional to the relative stretch, and so what turns out to matter is the length of the fall relative to the amount of rope available for stretching. These calculations do not take into account the friction in the system caused by (even relatively small) bends around the carabiner. This friction has the effect of decreasing the amount of rope available for stretching, and thereby effectively increasing the H/L ratio. It is quite possible in practice to achieve impact loads corresponding to double the actual H/L ratio for the fall, and so the formula given by Trenchdigger must be viewed as a lower bound on the forces, a bound that can be exceeded considerably when there are multiple pieces of protection in the system.





jt512
Feb 23, 2004, 8:20 AM
Post #18 of 24
(9258 views)
Shortcut
Registered: Apr 11, 2001
Posts: 21904

In reply to: Some of the others are right  because you're using a dynamic rope (well at least we hope you are!)  the foce exerted on the system by your fall is not as simple as just looking at how far you fall or how fast you're moving when the rope catches you. Fall factor and stretch of the rope must be considered. The actual max force on the rope should look something like this... F = ma*(1 + sqrt(1+ 2*f*k/mg )) where: F = force of fall k = spring constant of rope ( E*A/length of rope catching fall  note that the k and/or E*A value could fairly easily be found for a rope experimentally, but not just from the rope's specs) f = fall factor (fall length/length of rope catching the fall) for this purpose, length of rope catching the fall = the length of rope between the belayer and the climber I don't know what E and A are, but if I am interpreting your formula correctly, we don't even need the rope's specs to determine k. All we need to know are the specs of the UIAA drop test. Since F = ma, solving your equation for k, gives k = mg/2f For a UIAA test fall: m = 80 kg f = 1.77 So, we could solve for k. However, this implies that all ropes have the same k, which doesn't seem right. What did I do wrong? Jay





jt512
Feb 23, 2004, 11:39 AM
Post #20 of 24
(9258 views)
Shortcut
Registered: Apr 11, 2001
Posts: 21904

In reply to: In reply to: What did I do wrong? You didn't solve correctly for k. The solution is F(F2mg)  = k 2fmg with m=80, f=1.78, g=.0098, F=UIAA impact force in kN. I think I solved the equation that he wrote correctly, but according to your solution, his equation was wrong: a in his equation should have been g. At any rate, we can solve for k. For a Beal Booster F=7.3 kN, so k = 14.99. So, now we go make theoretical impact force calculations that don't take realworld factors into account. But it's a start. Jay





bukel
Feb 26, 2004, 10:02 PM
Post #21 of 24
(9258 views)
Shortcut
Registered: May 10, 2003
Posts: 42

It doesn't. This is why the concept of "fall factor" was created. Curt That is wrong Curt. The speed you aquire while falling is very important to calculate the real stress that the rope will hold. The fall factor is a simplification to make it easier to understand.





reprieve
Feb 26, 2004, 10:13 PM
Post #22 of 24
(9258 views)
Shortcut
Registered: Jan 24, 2004
Posts: 604

Yeow, I am a physics major...and I am not even going to jump into the fray on this one. Just one thing to add....all of these cases you are offering evaluations for are very idealized, and in reality it would depend on many more variables. Approximations can be useful as long as you take them with a grain of salt.





curt
Feb 28, 2004, 6:58 PM
Post #23 of 24
(9258 views)
Shortcut
Registered: Aug 26, 2002
Posts: 18272

In reply to: In reply to: It doesn't. This is why the concept of "fall factor" was created. Curt That is wrong Curt. The speed you aquire while falling is very important to calculate the real stress that the rope will hold. The fall factor is a simplification to make it easier to understand. Hey, thanks for posting something totally incorrect and nonsensical. Maybe you should take a physics class or two, learn that the correct term for "speed" is velocity and try to understand the subject matter about which you are posting. Curt





quietseas
Feb 28, 2004, 7:17 PM
Post #24 of 24
(9258 views)
Shortcut
Registered: Nov 2, 2002
Posts: 31

hey dude, not sure if anybody has answered this yet. look in your textbook under "impulse equation". figure how far you fall, get the velocity using 9.8 m/s/s for acceleration. then use the stretch of the rope (usually the mfgr. lists a % stretch or something like it) and figure how much distance you will take to stop. using initial velocity, and length to stop, you can figure out a new acceleration (the rate of DEceleration from the fall as the rope stops you). plug this into F=ma using your mass (in kilograms, 1kg=2.2 lbs) and you will get a force in Newtons (units will be Kg*m/s/s if done right). I do not remember the exact steps as I have done this all a long time ago. Good luck! It's really a fairly easy and fun thing to do on a rainy day if you don't mind a lot of numbers. BTW, on my rope, my mass, i remember that I could fall it's entire length at factor 2 and the harness would hold up quite nicely! The rope would have some issues, and I never even came close to stressing a sling or biner when using the dynamic rope in the equation. An interesting aside to this is that in theory, if using static rope, the impact force would be infinite. Not so considering that your body gives and slows itself, but this is a wonderful mathematic illustration of why you shouldn't EVER take a fall on a static rope. Besides the fact that even a two foot fall hurts like mad! (I was stupid once, and in the future will look back on today as being stupid as well!)








