



brunoschull
Feb 6, 2011, 1:27 PM
Post #1 of 15
(5252 views)
Shortcut
Registered: Nov 10, 2004
Posts: 47

Hi folks, I explored past posts about force calculations looking for a good basic equation to calculate forces on ropes, climbers, protection and so on. rgold's formula is great. Thanks! My question involves the difference between calculating forces on climbers vs. protection. I believe the answer is related to the pulley effect and friction, but I'm not sure of the details. I hope somebody can help me see this clearly. rgold's formula provides the force on the climber. To calculate the force on the protection, various posts suggest multiplying by 1.6. For example: "The impact force calculated from this equation is the force on the climber. Multiply that by 1.6 to get the impact force on the piece of pro that held the fall." or "While we're on the subject, if you take account of friction between the rope and the top piece, then tension in the belayer's side of the rope is about 60% of the tension in the climber's side. The sum of these two forces is the force on the top piece. So, the top piece feels about 1.6 times the climber's weight, in a static situation. This ignores any other friction in the system, such as between the rope and intermediate protection or between the rope and the rock." Can somebody explain the number 1.6 to me in more detail? Is this number based on an estimation of the friction between ropes and carabinersor is it mathematically more precise than that? Also, I can understand that because of friction the forces on the climber's and belayer's side are different, however, I would think that friction would add force to the protection, in the same way that friction adds force to an anchor in a hauling system, and you would have to multiply by more than 2 instead of 1.6. But I am probably not thinking about this correctly. Can anybody help? Many thanks, Bruno





jt512
Feb 6, 2011, 2:03 PM
Post #2 of 15
(5233 views)
Shortcut
Registered: Apr 11, 2001
Posts: 21904

brunoschull wrote: "While we're on the subject, if you take account of friction between the rope and the top piece, then tension in the belayer's side of the rope is about 60% of the tension in the climber's side. The sum of these two forces is the force on the top piece. So, the top piece feels about 1.6 times the climber's weight, in a static situation. This ignores any other friction in the system, such as between the rope and intermediate protection or between the rope and the rock." Can somebody explain the number 1.6 to me in more detail? Is this number based on an estimation of the friction between ropes and carabinersor is it mathematically more precise than that? My understanding is that it is an empirical result from experiments in which the tension in the belayer's side of the rope was found to be around 0.6 (or 0.67) times the tension in the climber's side. Since the total downward force on the anchor is the sum of these two tensions, the total force on the anchor is 1.6 (or 1.67) times the tension in the belayer's side of the rope (or, equivalently, the impact force on the climber). Jay





blondgecko
Moderator
Feb 6, 2011, 3:12 PM
Post #3 of 15
(5214 views)
Shortcut
Registered: Jul 2, 2004
Posts: 7666

jt512 wrote: brunoschull wrote: "While we're on the subject, if you take account of friction between the rope and the top piece, then tension in the belayer's side of the rope is about 60% of the tension in the climber's side. The sum of these two forces is the force on the top piece. So, the top piece feels about 1.6 times the climber's weight, in a static situation. This ignores any other friction in the system, such as between the rope and intermediate protection or between the rope and the rock." Can somebody explain the number 1.6 to me in more detail? Is this number based on an estimation of the friction between ropes and carabinersor is it mathematically more precise than that? My understanding is that it is an empirical result from experiments in which the tension in the belayer's side of the rope was found to be around 0.6 (or 0.67) times the tension in the climber's side. Since the total downward force on the anchor is the sum of these two tensions, the total force on the anchor is 1.6 (or 1.67) times the tension in the belayer's side of the rope (or, equivalently, the impact force on the climber). Jay This is the limiting case where the climber and belayer's ends of the rope are both parallel. As the angle deviates from this, the contribution of the belayer's side to the tension in the top piece will reduce. Counteracting this, however, is the fact that reduced contact area between rope and carabiner leads to reduced friction, increasing the tension in the belayer's side. So... it's complicated (read: more or less impossible) to calculate from first principles, and can only be estimated based on experiments.





jt512
Feb 6, 2011, 3:27 PM
Post #4 of 15
(5209 views)
Shortcut
Registered: Apr 11, 2001
Posts: 21904

blondgecko wrote: jt512 wrote: brunoschull wrote: "While we're on the subject, if you take account of friction between the rope and the top piece, then tension in the belayer's side of the rope is about 60% of the tension in the climber's side. The sum of these two forces is the force on the top piece. So, the top piece feels about 1.6 times the climber's weight, in a static situation. This ignores any other friction in the system, such as between the rope and intermediate protection or between the rope and the rock." Can somebody explain the number 1.6 to me in more detail? Is this number based on an estimation of the friction between ropes and carabinersor is it mathematically more precise than that? My understanding is that it is an empirical result from experiments in which the tension in the belayer's side of the rope was found to be around 0.6 (or 0.67) times the tension in the climber's side. Since the total downward force on the anchor is the sum of these two tensions, the total force on the anchor is 1.6 (or 1.67) times the tension in the belayer's side of the rope (or, equivalently, the impact force on the climber). Jay This is the limiting case where the climber and belayer's ends of the rope are both parallel. As the angle deviates from this, the contribution of the belayer's side to the tension in the top piece will reduce. Counteracting this, however, is the fact that reduced contact area between rope and carabiner leads to reduced friction, increasing the tension in the belayer's side. So... it's complicated (read: more or less impossible) to calculate from first principles, and can only be estimated based on experiments. The angle issue is minor enough that I think we can concentrate on a model in which both ropes are vertical. If you're going to get picky about rope angles, then you shouldn't consider the "parallel" case to be "limiting." The positioning of the belayer relative to the top anchor can result in a negative angle between the ropes at the top anchor (ie, the ropes cross) as easily as a positive angle. Jay





blondgecko
Moderator
Feb 6, 2011, 4:16 PM
Post #5 of 15
(5190 views)
Shortcut
Registered: Jul 2, 2004
Posts: 7666

jt512 wrote: blondgecko wrote: jt512 wrote: brunoschull wrote: "While we're on the subject, if you take account of friction between the rope and the top piece, then tension in the belayer's side of the rope is about 60% of the tension in the climber's side. The sum of these two forces is the force on the top piece. So, the top piece feels about 1.6 times the climber's weight, in a static situation. This ignores any other friction in the system, such as between the rope and intermediate protection or between the rope and the rock." Can somebody explain the number 1.6 to me in more detail? Is this number based on an estimation of the friction between ropes and carabinersor is it mathematically more precise than that? My understanding is that it is an empirical result from experiments in which the tension in the belayer's side of the rope was found to be around 0.6 (or 0.67) times the tension in the climber's side. Since the total downward force on the anchor is the sum of these two tensions, the total force on the anchor is 1.6 (or 1.67) times the tension in the belayer's side of the rope (or, equivalently, the impact force on the climber). Jay This is the limiting case where the climber and belayer's ends of the rope are both parallel. As the angle deviates from this, the contribution of the belayer's side to the tension in the top piece will reduce. Counteracting this, however, is the fact that reduced contact area between rope and carabiner leads to reduced friction, increasing the tension in the belayer's side. So... it's complicated (read: more or less impossible) to calculate from first principles, and can only be estimated based on experiments. The angle issue is minor enough that I think we can concentrate on a model in which both ropes are vertical. If you're going to get picky about rope angles, then you shouldn't consider the "parallel" case to be "limiting." The positioning of the belayer relative to the top anchor can result in a negative angle between the ropes at the top anchor (ie, the ropes cross) as easily as a positive angle. Jay Guess I wasn't clear enough. A bit more detail: In the below force diagram, F is the force on the climber, and f is the frictional force through the carabiner, so that tension on the belayer side is F  f. Then as the angle changes, the vertical (Fv) and horizontal (Fh) components of the forces on the top carabiner are as shown. The total force on the top piece (Ft) is then calculated by good old Pythagoras. Plugging in some numbers (setting F arbitrarily to 1, and f to 0.33 so that F  f = 0.67), we get the following for Ft vs. angle: . That's what I was getting at, anyway  that maximum force will be seen where belayer and climber are pulling in the same direction.

Attachments:

Force_diagram.gif
(7.33 KB)


Anchor_force_vs_angle.gif
(7.40 KB)





jt512
Feb 6, 2011, 4:27 PM
Post #6 of 15
(5182 views)
Shortcut
Registered: Apr 11, 2001
Posts: 21904

blondgecko wrote: jt512 wrote: blondgecko wrote: jt512 wrote: brunoschull wrote: "While we're on the subject, if you take account of friction between the rope and the top piece, then tension in the belayer's side of the rope is about 60% of the tension in the climber's side. The sum of these two forces is the force on the top piece. So, the top piece feels about 1.6 times the climber's weight, in a static situation. This ignores any other friction in the system, such as between the rope and intermediate protection or between the rope and the rock." Can somebody explain the number 1.6 to me in more detail? Is this number based on an estimation of the friction between ropes and carabinersor is it mathematically more precise than that? My understanding is that it is an empirical result from experiments in which the tension in the belayer's side of the rope was found to be around 0.6 (or 0.67) times the tension in the climber's side. Since the total downward force on the anchor is the sum of these two tensions, the total force on the anchor is 1.6 (or 1.67) times the tension in the belayer's side of the rope (or, equivalently, the impact force on the climber). Jay This is the limiting case where the climber and belayer's ends of the rope are both parallel. As the angle deviates from this, the contribution of the belayer's side to the tension in the top piece will reduce. Counteracting this, however, is the fact that reduced contact area between rope and carabiner leads to reduced friction, increasing the tension in the belayer's side. So... it's complicated (read: more or less impossible) to calculate from first principles, and can only be estimated based on experiments. The angle issue is minor enough that I think we can concentrate on a model in which both ropes are vertical. If you're going to get picky about rope angles, then you shouldn't consider the "parallel" case to be "limiting." The positioning of the belayer relative to the top anchor can result in a negative angle between the ropes at the top anchor (ie, the ropes cross) as easily as a positive angle. Jay Guess I wasn't clear enough. A bit more detail: In the below force diagram, F is the force on the climber, and f is the frictional force through the carabiner, so that tension on the belayer side is F  f. Then as the angle changes, the vertical (Fv) and horizontal (Fh) components of the forces on the top carabiner are as shown. The total force on the top piece (Ft) is then calculated by good old Pythagoras. Plugging in some numbers (setting F arbitrarily to 1, and f to 0.33 so that F  f = 0.67), we get the following for Ft vs. angle: . That's what I was getting at, anyway  that maximum force will be seen where belayer and climber are pulling in the same direction. My main point was just that we should ignore the affect of the angle, at least at this stage of answering the OP's question, and just assume that both ropes are vertical, or at whatever angle the experiments were conducted that came up with the 1/3 "friction factor." My much more picayune point was that if you are going to concern yourself with the effect of the angle on friction, then you shouldn't ignore cases where the angle is less than zero, and hence the friction is greater (I think) than where both ropes are plumb. And, BTW, wouldn't f itself be a function of theta, rather than a constant independent of theta? Jay
(This post was edited by jt512 on Feb 6, 2011, 4:29 PM)





blondgecko
Moderator
Feb 6, 2011, 4:42 PM
Post #7 of 15
(5172 views)
Shortcut
Registered: Jul 2, 2004
Posts: 7666

jt512 wrote: blondgecko wrote: jt512 wrote: blondgecko wrote: jt512 wrote: brunoschull wrote: "While we're on the subject, if you take account of friction between the rope and the top piece, then tension in the belayer's side of the rope is about 60% of the tension in the climber's side. The sum of these two forces is the force on the top piece. So, the top piece feels about 1.6 times the climber's weight, in a static situation. This ignores any other friction in the system, such as between the rope and intermediate protection or between the rope and the rock." Can somebody explain the number 1.6 to me in more detail? Is this number based on an estimation of the friction between ropes and carabinersor is it mathematically more precise than that? My understanding is that it is an empirical result from experiments in which the tension in the belayer's side of the rope was found to be around 0.6 (or 0.67) times the tension in the climber's side. Since the total downward force on the anchor is the sum of these two tensions, the total force on the anchor is 1.6 (or 1.67) times the tension in the belayer's side of the rope (or, equivalently, the impact force on the climber). Jay This is the limiting case where the climber and belayer's ends of the rope are both parallel. As the angle deviates from this, the contribution of the belayer's side to the tension in the top piece will reduce. Counteracting this, however, is the fact that reduced contact area between rope and carabiner leads to reduced friction, increasing the tension in the belayer's side. So... it's complicated (read: more or less impossible) to calculate from first principles, and can only be estimated based on experiments. The angle issue is minor enough that I think we can concentrate on a model in which both ropes are vertical. If you're going to get picky about rope angles, then you shouldn't consider the "parallel" case to be "limiting." The positioning of the belayer relative to the top anchor can result in a negative angle between the ropes at the top anchor (ie, the ropes cross) as easily as a positive angle. Jay Guess I wasn't clear enough. A bit more detail: In the below force diagram, F is the force on the climber, and f is the frictional force through the carabiner, so that tension on the belayer side is F  f. Then as the angle changes, the vertical (Fv) and horizontal (Fh) components of the forces on the top carabiner are as shown. The total force on the top piece (Ft) is then calculated by good old Pythagoras. Plugging in some numbers (setting F arbitrarily to 1, and f to 0.33 so that F  f = 0.67), we get the following for Ft vs. angle: . That's what I was getting at, anyway  that maximum force will be seen where belayer and climber are pulling in the same direction. My main point was just that we should ignore the affect of the angle, at least at this stage of answering the OP's question, and just assume that both ropes are vertical, or at whatever angle the experiments were conducted that came up with the 1/3 "friction factor." My much more picayune point was that if you are going to concern yourself with the effect of the angle on friction, then you shouldn't ignore cases where the angle is less than zero, and hence the friction is greater (I think) than where both ropes are plumb. And, BTW, wouldn't f itself be a function of theta, rather than a constant independent of theta? Jay Indeed it would be  which is why, as I said, the only way to get good estimates of forces is by experiments with various configurations. I'm just splitting hairs, really.





jt512
Feb 6, 2011, 5:04 PM
Post #8 of 15
(5162 views)
Shortcut
Registered: Apr 11, 2001
Posts: 21904

blondgecko wrote: I'm just splitting hairs, really. Me too. Jay
(This post was edited by jt512 on Feb 6, 2011, 5:05 PM)





JimTitt
Feb 6, 2011, 11:59 PM
Post #9 of 15
(5098 views)
Shortcut
Registered: Aug 7, 2008
Posts: 1002

The value of 1.6 is roughly what you get on the top runner if you perform a static belay drop tests (in other words tie a knot in the rope) but through a belay plate you get rather different results. For something like a Grigri you might be getting up this high but for the rest of the time the factor will be considerably lower. As jt says the force on the top piece is the sum of the two forces BUT this is not the sum of the two maximum forces, rather the sum of the two forces at the same time. (The belayer side force lags considerably behind that of the faller because of frictional hysterises and the top runner force is inbetween.) If you look at the force graphs for a typical device you will see that at the point of maximum force for the faller (3.08kN) the belayer side force is 0.8kN and the top runner force is 3.9kN which all adds up nicely. However the belayer force continues to rise to a value of 1.5kN while the faller force drops to 1.5kN and the measured top runner force is down to 3.1kN which also adds up nicely. All well and good. However if you are going to use the maximum forces and give a factor to convert then it all goes a bit haywire, here are the measured forces for two common plates, one a soft one and one more powerful. 2.4kN (faller) 2,8kN (runner) Factor 1.17 3,1kN (faller) 3,9kN (runner) Factor 1.26 The ratio depends a bit on the actual impact forces (and in turn on the power of the device) and a reasonable value would probably be around 1.25 for a typical plate. And now Iīm off to another trade show to look at yet more belay devices! Jim





ptlong2
Feb 7, 2011, 12:09 PM
Post #10 of 15
(5034 views)
Shortcut
Registered: Aug 10, 2010
Posts: 102

blondgecko wrote: So... it's complicated (read: more or less impossible) to calculate from first principles..... the only way to get good estimates of forces is by experiments with various configurations. This isn't true if by first principles you mean starting with the same assumptions you did. It is relatively straightforward to calculate Ft while including f as a function of theta. As theta varies up to a pretty large angle the difference in force is small enough to be considered noise, as you've already admitted. And of course Jim Titt pointed out there are much larger real effects that are being ignored by this "first principles" approach. But that doesn't mean that real forces cannot be estimated with models.





blondgecko
Moderator
Feb 7, 2011, 12:23 PM
Post #11 of 15
(5026 views)
Shortcut
Registered: Jul 2, 2004
Posts: 7666

ptlong2 wrote: blondgecko wrote: So... it's complicated (read: more or less impossible) to calculate from first principles..... the only way to get good estimates of forces is by experiments with various configurations. This isn't true if by first principles you mean starting with the same assumptions you did. It is relatively straightforward to calculate Ft while including f as a function of theta. As theta varies up to a pretty large angle the difference in force is small enough to be considered noise, as you've already admitted. And of course Jim Titt pointed out there are much larger real effects that are being ignored by this "first principles" approach. But that doesn't mean that real forces cannot be estimated with models. The difficulty lies in predicting with any accuracy how f varies, and all the other unforeseen factors such as the (very interesting) hysteresis effects Jim pointed out. But that's my point, really. You could put together a model entirely from first principles, but it would be really quite difficult and most likely a fair way off from reality. Easier to collect a bunch of experimental data to feed into a semiempirical model.





ptlong2
Feb 7, 2011, 1:48 PM
Post #12 of 15
(5006 views)
Shortcut
Registered: Aug 10, 2010
Posts: 102

blondgecko wrote: The difficulty lies in predicting with any accuracy how f varies... But it isn't difficult.
In reply to: But that's my point, really. You could put together a model entirely from first principles, but it would be really quite difficult and most likely a fair way off from reality. Easier to collect a bunch of experimental data to feed into a semiempirical model. If that were really your original point it would have been better had you simply stated that in the first place. It would have led to a different and possibly less trivial discussion.





blondgecko
Moderator
Feb 7, 2011, 2:53 PM
Post #13 of 15
(4980 views)
Shortcut
Registered: Jul 2, 2004
Posts: 7666

ptlong2 wrote: blondgecko wrote: The difficulty lies in predicting with any accuracy how f varies... But it isn't difficult. It is when you start to consider factors such as rope age, stiffness and fuzziness, carabiner model, wetness, etc.
In reply to: In reply to: But that's my point, really. You could put together a model entirely from first principles, but it would be really quite difficult and most likely a fair way off from reality. Easier to collect a bunch of experimental data to feed into a semiempirical model. If that were really your original point it would have been better had you simply stated that in the first place. It would have led to a different and possibly less trivial discussion. Probably.





ptlong2
Feb 7, 2011, 4:20 PM
Post #14 of 15
(4963 views)
Shortcut
Registered: Aug 10, 2010
Posts: 102

blondgecko wrote: It is when you start to consider factors such as rope age, stiffness and fuzziness, carabiner model, wetness, etc. Now you've changed your claim, haven't you. You know we haven't got an elephant. I'm not prepared to simply sit here bickering. Take her away, Heinz!





JimTitt
Feb 8, 2011, 2:26 AM
Post #15 of 15
(4924 views)
Shortcut
Registered: Aug 7, 2008
Posts: 1002

The angle thing is sortof interesting but believe me to derive anything from first principles appears almost impossible, at some stage you have to plug some experimental data in somewhere and to get the data youīve effectively already done the tests and got the answers! The classic way to work out the friction around the karabiner is to plug the coefficient of friction and angle into Amontons capstan theory but this doesnīt work for ropes which have a significant bending resistance (stiffness) and capstan theory is a dead duck here. The frictional force varies with angle, rope stiffness and load and to calculate the rope stiffness is going to be a monster job with thousands of individual threads all moving around under load. And that is if the coefficient of friction is a constant, since however it varies with pressure and speed amongst other things this is another load of variables for each different case. Going to be a complicated formula! Obviously its going to be easier to just bend it around something and measure at various loads and angles which is what we wanted anyway. Iīve already done this for lower loads (that occur in the belay device) and itīs clear that after a wrap angle of 90° the increase in frictional force is levelling off and by 180° there is virtually no increase. (Technically it isnīt frictional force but pulley efficiency as friction is not the only effect). To test at the higher loads encountered in a fall is easy enough but practically testing at the speeds and loads together is going to be another matter. Since the normal drop tests (which are normally conducted at 150° to stop the dropweight landing on the belayer below) are about the worst case scenario it seems reasonable to use these results. Jim








