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reedcrr
Apr 5, 2004, 9:27 AM
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Those damn euro's have confussed our american minds again! In repeated threads I have read on RC.com there are many post's regarding gear failure thresholds and testing parameters and so on. I am not convinced that most climbers understand what vendors are talking about when the claim a locking biner has, "a max. "long axis load" of 28Kn before possible failure".
So I will try and explain it in layman's terms to those who are interested...if your not continue on to another thread! Stick with me this is easy to understand once you know the math!
Possible failure of any piece of equipment has many complex factors but let's deal only with the basics that can be understood quickly.
In our test piece we will be simulating a test on a Petzl, Williams Locking carabiner. ( Legal crap...We are not actually going to test this item and the numbers given are for education purposes only, refer to your owners manual that came with your biner for actual factors!) We will be using an 11mm Dynamic rope and a 165 lb. lead climber with a belayer that weighs 165lbs. and is not anchored and does not know how to do a running belay. Whew what a mouthful! :D
First we need to convert english to metric...so
1 lb. = 0.45 kilograms so... 165 lbs. = 74.8 Kg (in mass).
We will then have our climber take a 5 meter fall while attempting to clip. (He missied the clip and came off, poor guy!! :cry: ) This is a common fall distance in sport and trad climbing.
1 meter = 3.28 Ft
5 meters = 16.4 Ft or roughly 8.25 ft above the last piece of protection when he came off the rock.
But we said he was going for the clip when he came off so lets add 1.5 meters of slack in the line (from the belayer) so the total drop distance from start to finish is now 21 Feet, 8.25 Feet above the bolt and 12.75 ft below the bolt. Let's hope this climber is up high enough to fall that far safely! :shock:
The Williams biner has three measurements associated with it for load testing (so does every other biner made) and these are:
Load across the long axis : 28 Kn (this is the normal load on the biner if used correctly)
Gate open : 8 Kn (This is forgetting to close the screwgate and the gate comes open under the load of the falling climber)
Short axis : 7 Kn (This is cross-loading, common load failure of an incorrect belay setup or a possible failure due to clipping the lead climber in rather then tying in. Biner tends to want to "walk" while climbing and may get stuck in this cross-loaded position.
Now we need to figure out the applied force of our falling climber when he comes to the bottom of the fall.
1 Kn = 224.8 Lbs./force (this is a force measurement, not weight)
Again.. the Williams Locking carabiner possible failure limits:
28 Kn = 6295 Lbs./force (Applied force in pounds before possible failure)
8 Kn = 1799 Lbs./force (Applied force before possible failure)
7 Kn = 1574 Lbs./force (Applied force before possible failure)
Big difference! With the gate locked and closed and oriented in the correct position the biner is 4 times stronger then the same biner in a cross-loaded situation. So what does that mean to our climber in the above test? Can we even get close to 1574 lbs./force with our 165 lb. climber? Sounds like a lot of force...but just think if you jump off of a two foot chair on to a beer can you are applying about 2 Kn of applied force on to the top of that can with your foot! Or roughly 450 lbs./force...(Again, assuming you weigh in at 165 lbs.) If your going to try the beer can test don't do it barefoot!
Well our climber coming to a stop after falling 21 Feet is going to generate around 1230 Dekanewtons (DaN) of force.
1 DaN = 1.01 Kilograms/force or 2.25 lbs/force
1230 DaN = 2765 Lbs./force (Or 12.29 Kn!)
This means that in this senario this 165 Lb. climber, falling 21 Feet (again remember 8.25 ft. above the protection to 12.25 ft. below the protection) is placing 12.29 Kn of force on the belayers locking carabiner... or if the leader is clipped in then 12.29 Kn on their own locking carabiner.
Under normal circumstance with the load going along the long axis of the locking carabiner the belay would hold and the climber would be safe, but if the biner gate is open or the biner is cross-loaded the possibility that the biner will fail is fairly high and a severe risk is being taken.
The biner is normally the weakest link in most climbing setups because most trad protection properly placed will hold 28 - 30+ Kn. and a 3/8 in. x 4in. Stainless rawl bolt properly placed will also hold around 30 Kn before shearing under load. (Depending on the PSI of the rock)
I know this was a long post but hopefully it is informative as to what those numbers stamped on the side of equipment really means in a way we can all understand it. Take it seriously and do not forget to lock the biner and do not put the biner in a "walking" situation where it can get caught and cross-loaded!
Have fun and climb hard!
*edit for typo*
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wvredline
Apr 5, 2004, 12:16 PM
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You lost me when you got to dekanewtons. I get 165 pounds * 21 feet * 4.45 Newtons per pound foot = 15.4 kNewtons. What am I missing?
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geezergecko
Apr 5, 2004, 1:12 PM
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1 deca Newton = 10 Newtons
1 kilo Newton = 1000 Newtons
Therefore 1 kN = 100 daN
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icekubes7
Apr 5, 2004, 1:28 PM
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Thanks man, very informative. What about if you take a whipper on just a normal non-locking quickdraw carabiner? Would it possibly come open during a fall, and therefore severely weaken it?
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quickclips
Apr 5, 2004, 1:35 PM
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Thats the force applied to the belayers biner, but the draw is a different story. It has the force from climber to the biner and from the biner to the belayer, so it is (if everything is aligned) twice that load.
Very nice explination.
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lqdslvr
Apr 5, 2004, 1:43 PM
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Couple of follow-up questions. I didn't see the dynamic rope enter into the calculations. Perhaps that's factored into the ultimate number of 12.3 Kn, but it wasn't apparent. Also, isn't a crucial factor in making this caculation how much rope is out? This same 21 foot fall is going to be much different with 30' of rope out than with 100'. Finally, I also always understood that the friction and pulley effect with the biner on the piece catching the load would create about 1/3 less force on the belayer than on the leader. In other words, the leader may take 12.3 Kn, but the belayer will only feel 8.2 (and these two numbers must be added to calculate theforce applied to the biner onthe piece catching the fall, e.g, the top draw). I apologize if these are stupid questions; I'm just trying to understand the theory.
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scottd
Apr 5, 2004, 1:53 PM
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I didn't see anywhere in the post, the length of rope from the belayer to the climber. This will greatly affect the force applied to the biner... Or is the total distance 21 feet?
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oldsalt
Apr 5, 2004, 2:05 PM
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I understood that 12 kn is likely a fatal strain on the human body. As noted in a previous post, you could be describing the effects of a static rope. Dynamic rope stretch makes all the difference.
Good intent in the post, clarification needed.
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aklimerguy
Apr 5, 2004, 2:21 PM
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Reedcrr said :
"The biner is normally the weakest link in most climbing setups because most trad protection properly placed will hold 28 - 30+ Kn. and a 3/8 in. x 4in. Stainless rawl bolt properly placed will also hold around 30 Kn before shearing under load. "
Trad pieces range between 8 and 15 kN and your rope at the knot, 12 to 14 kN. Which would make the biner the strongest link, except maybe if you're clipped onto a bolt in granite rock. Or if the biner gate opens at the moment of the fall. Hail to wiregates!
You also forgot to talk about the fall factor (length of the fall/length of the rope between leader and belayer) which is in my point of view, the most important point to consider.
You did however do a great job explaining what a kiloNewton was, which I think was your main objective.
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ben87
Apr 5, 2004, 2:33 PM
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Interesting. I think the basic calculations of force above are all correct -- the effect of a dynamic rope is very important, obviously. I'd also like to see a close examination of the forces on the top piece, especially the carabiner -- which in many leading situations is a non-locking biner, even a wiregate. it seems to me that this is the point in the system where the most force is generated....?
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aklimerguy
Apr 5, 2004, 2:37 PM
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Ben wrote :
"I'd also like to see a close examination of the forces on the top piece, especially the carabiner -- which in many leading situations is a non-locking biner, even a wiregate. it seems to me that this is the point in the system where the most force is generated....?"
I agree, totally!
Does someone out there know of some "field" tests which would of been conducted?
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rockprodigy
Apr 5, 2004, 3:18 PM
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Just goes to show that anyone can post anything they want on the inernet.
Whether it's true or not!
I don't have time to correct this entire post, but I would like to mention that a pound is not a measure of mass, it is a measure of force. A "slug" is a measure of mass.
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quickclips
Apr 5, 2004, 6:10 PM
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I haven't checked the numbers, but I think these calcs didn't take in to consideration as most people have pointed out the dynamic properties of the rope. Being that a rope stretches say 7%, it increased the total distance the force is being applied for, thereby lowering acceleration and reducing the force. So how much rope that is out is a huge factor to the fall.
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fargoan
Apr 5, 2004, 6:24 PM
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In reply to: Ben wrote : "I'd also like to see a close examination of the forces on the top piece, especially the carabiner -- which in many leading situations is a non-locking biner, even a wiregate. it seems to me that this is the point in the system where the most force is generated....?" I agree, totally! Does someone out there know of some "field" tests which would of been conducted?
hey guys, good timing! check out last month's rock & ice (with the taquitz photo on the cover). they are doing a 3-issue series on fall forces using actual falls to get some "field data."
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robmcc
Apr 5, 2004, 6:29 PM
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Jeebus. That's not a good post at all, it's total misinformation and misunderstanding. That's not describing the impact force on a static rope, dynamic rope, or anything at all. It's just completely wrong. You simply cannot determine the force generated in the fall without taking into account the deceleration time, which is absolutely ignored in the post.
In reply to: 1 Kn = 224.8 Lbs./force (this is a force measurement, not weight)
What is lbs/force, exactly?!?
In reply to: Well our climber coming to a stop after falling 21 Feet is going to generate around 1230 Dekanewtons (DaN) of force.
Explain this, please. You have no idea how much force will be required to arrest the climber, you only know how much kinetic energy he possesses.
In reply to: 1230 DaN = 2765 Lbs./force (Or 12.29 Kn!)
It should be a warning sign to you that your calculations are coming up with numbers comparable to or greater than those generated in UIAA 1.8whatever fall factor tests, and your FF is apparently less than 1.
In reply to: The biner is normally the weakest link in most climbing setups because most trad protection properly placed will hold 28 - 30+ Kn. and a 3/8 in. x 4in. Stainless rawl bolt properly placed will also hold around 30 Kn before shearing under load. (Depending on the PSI of the rock)
Source, please? I don't think I own any pro rated to 30Kn. Harnesses, for that matter, are typically rated to 15 or so. I'm sure as hell not rated for 30Kn. I don't even think my hexes are rated to 30kn.
In reply to: I know this was a long post but hopefully it is informative...
Nope, quite misinformative, in fact. Thou shalt understand physics before trying to explan it to others.
Rob
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curt
Apr 5, 2004, 7:14 PM
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robmcc already pointed out a few good points. In addition to robmcc's comments, reedcrr failed to point out that the maximum force felt by the climber will be equal to the peak tension in the rope, during the time that the climber's fall is being arrested. The highest force on the top piece will be twice this figure, minus friction. The greatest negative acceleration experienced by the falling climber is also inversely proportional to the distance over which the falling climber is brought to a stop. This is why knowing something about the fall factor and rope elasticity is required to properly calculate the actual force experienced. rgold has posted elsewhere how to calculate this.
Curt
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drunkencabanaboy
Apr 5, 2004, 7:26 PM
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ROFL OMG - this thread is a cluster... a moderator should just delete it ROFL... its really pointless.
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billcoe_
Apr 5, 2004, 7:37 PM
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Well, I was going to thank him for his efforts to share with us. Awwww, I'm gong to do it anyway. Point is, you can break a cross-loaded or open biner with a leader fall. The pull on the belayer should be less than on the lead biner due to the friction and runig through the lead biner should it not?
Thanks anyway, not that I understand the math:
Bill
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andypro
Apr 5, 2004, 8:31 PM
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In reply to: The highest force on the top piece will be twice this figure, minus friction.
According to testing done by Petzl (I read it in a two page blurb kinda thing in climbing magazine a long time ago...I could porbably dig through my old ones and find it to be sure if anyones interested) Says that friction is responsible for up to 40% loss in the actual force on the topmost piece. So instead of 200% your lookin at more like 160% or so. They also went on to describe the same effects with a static rope, and then a via feratta setup. I dont remember those numbers though, I do remember that the via feratta one was rediculously high and almost certainly lethal. :shock:
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robmcc
Apr 5, 2004, 9:00 PM
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In reply to: You lost me when you got to dekanewtons. I get 165 pounds * 21 feet * 4.45 Newtons per pound foot = 15.4 kNewtons.
It might be worth saying again that foot pounds is a measure of energy while newtons is a measure of force. You can't convert one to the other any more than you can convert pounds to seconds. They measure different things.
Rob
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reedcrr
Apr 5, 2004, 10:23 PM
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I quoted:
"The biner is normally the weakest link in most climbing setups because most trad protection properly placed will hold 28 - 30+ Kn. and a 3/8 in. x 4in. Stainless rawl bolt properly placed will also hold around 30 Kn before shearing under load. (Depending on the PSI of the rock)"
I retract that statement, most of you spotted this and were correct in the fact that most pro will not hold that much, my mistake in not checking what I was writing. And the calc on the bolt was for a 1/2 x 4 bolt. A 3/8 X 4 bolt will shear at 4500 PSI not 6900 PSI...Sorry for this incorrect information. I did however go over the math used in the calc and it is correct.
As far as a pound not being a measurement of mass? Come on...
And to the question what is Lbf. or pound-force it is this:
It is a unit of force equal to the mass of 1 pound with an acceceration of free fall. (Or 32 ft/sec/sec) Unrestrained mass falling with gravity pulling.
I doubt very seriously that any of you are going to figure out fall factors before tying in and climbing. Also I doubt that before coming off the wall you are going to run calculations on how many Kn you are going to exert on the equipment you have. That is not the point of this thread...
In my first post I ommitted all of the calculations that would send most people in to tailspins, such as the dynamic elongation of the rope and the calculation getting to the final number of 12.29 Kn. Omitted on purpose... Yes a static rope will apply more energy into the equation and a belayer jumping at the right time colud lessen the impact by absorbing energy...so many factors to considered... so for the point of this thread I stuck with a basic pulley system and known forces applied within that system.
As a climber you must understand that there are many factors relating to the possible failure of your equipment, such as weight, force, heat, acceleration and deacceleration and so on and the main reason they print that warning on the side of most equipment is to give you an indication as to what that peice can take before possible failure.
In my post I used a 165 lb. person with a common fall from 8.25 feet above the last piece of protection and the final fall of 21 feet, again fairly common. With this simple example it shows that an average person taking an average fall is applying enough force on a biner to possibly cause serious failure if the screwgate is not locked or the biner has been cross-loaded, simple as that.
If you want to contest the applied physics PM and we will have fun...but the calcs used in my first post are close enough not to warrent a retraction on my part...except the paragraph that I have already retracted above.
So to all of my naysayers out there... pick it apart. The main point was to explain what a Kn is and I think I have done so...
By the way foot-pound is a measurement against gravity and pound-force is a measurement with gravity...just a note...
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jt512
Apr 5, 2004, 10:58 PM
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In reply to: I quoted: "The biner is normally the weakest link in most climbing setups because most trad protection properly placed will hold 28 - 30+ Kn. and a 3/8 in. x 4in. Stainless rawl bolt properly placed will also hold around 30 Kn before shearing under load. (Depending on the PSI of the rock)" I retract that statement, most of you spotted this and were correct...
Your first post was such a mess, I didn't know where to start correcting it. Fortunately for me, a few other knowledgeable people posted up.
In reply to: As far as a pound not being a measurement of mass? Come on...
The pound is both a unit of mass and a unit of force, though the latter is the more common usage.
In reply to: I doubt very seriously that any of you are going to figure out fall factors before tying in and climbing.
You damn well better understand the basic principle behind fall factors before you tie in. If you don't you're a double fatality waiting to happen.
In reply to: Also I doubt that before coming off the wall you are going to run calculations on how many Kn you are going to exert on the equipment you have. That is not the point of this thread...
What the point of your first post was was not clear. You stated that it was about understanding what a kilonewton is. I can explain what kN is in one sentence: 1 kN is equal to 225 pounds of force. What the rest of your post had to do with what a kilonewton is, I have no idea.
In reply to: In my first post I ommitted all of the calculations that would send most people in to tailspins, such as the dynamic elongation of the rope and the calculation getting to the final number of 12.29 Kn.
You should have realized that your result was too high, and rechecked your calculations. You did something wrong!
In reply to: The main point was to explain what a Kn is and I think I have done so...
1 kN = 225 lb.
So did I, using half a line.
-Jay
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trenchdigger
Apr 5, 2004, 11:09 PM
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In reply to: By the way foot-pound is a measurement against gravity and pound-force is a measurement with gravity...just a note...
Foot-Pound is a unit or torque in the english system.
The Pound (or Pound-Force if you like) is a measure of force. Toward gravity, against gravity, in a weightless environ - it doesn't matter. A Pound is a unit of force, period.
I appreciate your point, but I beg to differ on your application of pyhsics. I'm not a professor though and don't have the time or inclination to try to explain.
Relating kilonewtons and pounds is simple since both are units of force.
1 kilonewton = 225 pounds.
Nuf said.
The force exerted by a fall on a dynamic rope is complex and can be roughly calculated by a formula you should be able to find by sifting through posts on this site about that subject. However it's hardly and exact science. See the attached link for more info about pounds, kilonewtons, and gear strength.
http://home.vicnet.net.au/...%205%20Equipment.pdf
~Adam~
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jason1
Apr 5, 2004, 11:12 PM
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Force(N)= Mass(Kg)xAcceleration(M/s2) or F=ma
N=kg(m/s2)
kilo- prefix meaing 1000
acceleration of gravity=9.81 m/s2
Impulse= force/time (N/s) impulse is a force applied over a time... to soften a fall you add more time in dynamics to reduce the impulse... or a large force applied over a long time can be just as hard as a small force applied over short time...
i believe the slug is the english unit of mass
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jt512
Apr 5, 2004, 11:23 PM
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In reply to: In reply to: By the way foot-pound is a measurement against gravity and pound-force is a measurement with gravity...just a note... Huh? Foot-Pound is a unit or torque in the english system.
The foot-pound is commonly confused with foot-mouth-pound, which measures the amount of force required to remove one's foot from one's mouth, a high figure in reedcrr's case, apparently.
-Jay
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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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bigwallgumbie
Apr 6, 2004, 5:55 AM
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No akclimber, I don't think there are ANY smart people posting in this thread. None.
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ryanhos
Apr 6, 2004, 6:06 AM
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In reply to: No. You are confused. The "weight" you are dropping weighs 225 lb, or 1 kN, regardless of how high it is dropped from.
In reply to: 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. |
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andypro
Apr 6, 2004, 6:13 AM
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In reply to: total weight on the anchor is 2(climbers + gear) + 1 rope. (multiply by gravity to get force...) Friction does not apply here.
actually...weight is already a multiplication of gravity. This si a case of either/or, not both.
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jason1
Apr 6, 2004, 6:55 AM
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In reply to: 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
Correct... so the force applied to the top piece is going to be around twice that if you don't account for friction... if you do the common factor is around 1/3... i think petzel uses 2/5 but there again you'd have to figure how much rope is actually running to determine how much friction is really a factor... in a factor 2 fall the amount of force reduction caused by friction is nil.... probally only slightly better in cases of short (16-20') falls or in falls which the dynamic properties of the rope is limited to such a lenght...
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reedcrr
Apr 6, 2004, 8:21 AM
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Wow so many bright young indiviuals in this post...
Let me see where do I begin....
Let's just sum this up....
At 28 Kn it is going to take 6295 pounds of force (Lbf. (either pushing or pulling) to possibly cause a failure in the Williams locking carabiner. Want an easier way to see this visually? Take a 1/2 x 4 stainless steel rawl bolt commonly used to set up anchors...drill a hole into solid rock and place the bolt properly. Once tightened and in place it will take the combined weight of 41 men of average weight (150 lbs.), and one small child, to break the bolt in to two pieces or at the very least bend it to the point of breaking. No this does not count jumping on the bolt to add more force!
At 7 Kn it will only take 10 men and a golden retriever.... but get this the bolt will hold! However the Williams locking carabiner will not! Bottom line don't cross-load the carabiner or leave the gate unlocked where it can open on it's own!
As stated in a previous post.... 1Kn = 225 Lbf (although you incorrectly stated pounds...not pounds of force!) or my little brother Jim sitting on my chest trying to get me to say uncle! Both Jim and gravity are keeping him on my chest and the amount of force I would need to get him off of me would have to equal his weight, provided that he did not move!...unless of course he just gives up! :)
Hope that clears the intent of my post up clearly...if not go cry in your beer and go back to school and take more tests!
Peace to all the Elvi out there!
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davidgduval
Apr 6, 2004, 9:49 AM
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In reply to: Well our climber coming to a stop after falling 21 Feet is going to generate around 1230 Dekanewtons (DaN) of force.
Could you please post the equation which you used to arrive at this figure?
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djjackson
Apr 6, 2004, 2:40 PM
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Does the group have any one who can take the original problem and attempt an explanation, to include type of rope fall factor, gravity ( force= mass x acceleration, if I recall basic physics)? Current issue of Rock and Ice has a fairly good article regarding dynamic and static belays, which touches on the problem in the thread. Also see website Rescue Dynamics for good articles regarding physics and actual trials of rope, carabiners,etc.
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rgold
Apr 6, 2004, 4:00 PM
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In reply to: This means that in this senario this 165 Lb. climber, falling 21 Feet (again remember 8.25 ft. above the protection to 12.25 ft. below the protection) is placing 12.29 Kn of force on the belayers locking carabiner.
A number of people have mentioned this makes no sense without specifying a fall factor. To that observation I'd add that these numbers are not very realistic in any case. I think it is pretty rare nowadays to find a dynamic rope with a UIAA impact force much more than 9 kN. Mammut, whose ropes tend to be the stiffest, makes an 11mm rope, (The Flex) with an impact rating of 9.2 kN. Bluewater's 11mm rope (the Enduro) is rated at 8.8 kN. These ratings are very close to the maximum rope tension obtainable, since they are obtained from a factor 1.78 fall. The post was describing a missed clip on a sport climb, and sport climbs are usually one pitch. This means that the fall factor will have to be less than 1 (or else the climber hits the ground, thereby sparing the cross-loaded carabiner). The impact rating are done for an 80 kg climber, whereas reedcrr uses a 75 kg climber, which would reduce the forces a little more. Now to this add the fact that after the climber's rope has passed through the top biner, the tension on the other side, because of rope friction on the biner, will be about 2/3 of the tension on the leader's side, so the belayer's locker will experience no more than 2/3 of the leader's impact. In reality, friction through other carabiners could reduce the load on the belayer much more.
Put all this together and you get a load on the belayer that is going to be way less than 12 kN, which is, by the way, the threshold set by the UIAA for bodily injury.
This doesn't alter any of the cautions about the weakness of cross-loaded biners. Those warnings could, however, have been made with numbers a little more relevant to climbing.
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mtnjunkie
Apr 6, 2004, 4:50 PM
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The only illustrative point made in this thread is the absolute need for the physics of climbing to be abstracted into fall factors.
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robmcc
Apr 6, 2004, 4:58 PM
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In reply to: The only illustrative point made in this thread is the absolute need for the physics of climbing to be abstracted into fall factors.
This may be the most insightful comment yet.
Rob
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robmcc
Apr 6, 2004, 5:07 PM
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In reply to: 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
Nope, absolutely wrong. 733 newtons is the force that climber places on the rope while hanging. Gravity is providing a downward acceleration of 9.8 m/s2 which is opposed precisely by the tension in the rope applying an upward acceleration of 9.8 m/s2. I'm hand waving a bit, but won't bother to explain unless someone feels like caling me on it. :)
As a fall is arrested, that upward acceleration is much higher. I remember in my biggest fall, which at around 35' wasn't that big, how strong the pull on the rope was. Say you decelerate at 5 gs or so, your equation becomes
74.8 kg X 49 m/s2 = 3665 newtons, or 3.7 kN
If instead you hit the ground and various parts of your body decelerate at an unhappy 50gs, you get a nice 37 kN impact force, hard parts break, soft parts tear, and you don't survive. Same fall, faster stop, radically different outcome. Soft catch or fatal impact, it all depends on how fast you stop. Anyone attempting to calculate this stuf without taking that into account is just plain wrong. Defend your erroneous calcs, original poster, or give it up.
Rob
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jason1
Apr 6, 2004, 8:45 PM
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i think you're confusing velocity and acceleration....
the acceleration due to gravity is constant...
velocity increases with time....
9.81m/s2(5s)= 49.05 m/s
9.81m/s2(6s)= 58.86 m/s
9.81m/s2(7s)= 68.67 m/s
you're right about crunching the math... it only gives guidelines... each fall is a case by case assessment and at best an estimate...
i wish we could hear what dr. kodos has to say on the subject... FREE KODOS!!!!!!
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robmcc
Apr 6, 2004, 8:50 PM
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In reply to: i think you're confusing velocity and acceleration.... the acceleration due to gravity is constant....
The key words there are "due to gravity". Gravity is not the only force acting on the climber. The rope exerts a force on the climber as well. Unlike gravity, that force is not constant.
Rob
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jason1
Apr 6, 2004, 9:05 PM
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you're absolutly right, i'm not arguing that gravity is the only force in this situation...
I am saying that the acceleration dosn't go up as you fall down... velocity dose...
acceleration would go down gradually as the friction of the system and the dynamic properties of the rope kick in... then if you have any screamers on the pieces count them in too... and any friction due to the rock...
if you want to find the acceleration it's the second derivitve of a position/time graph...
ROCK ME DR. KODOS!!!!! dr. kodos, dr. kodos
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robmcc
Apr 6, 2004, 9:21 PM
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In reply to: I am saying that the acceleration dosn't go up as you fall down... velocity dose...
If you read anywhere in anything that I've written that I think gravitational acceleration is not constant, you've misread. This thread started out with a completely incorrect description of the forces in a fall. Correcting those errors seemed worthwhile. Correcting your misreading of my posts doesn't.
Rob
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drunkencabanaboy
Apr 6, 2004, 9:22 PM
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In reply to: i think you're confusing velocity and acceleration.... the acceleration due to gravity is constant... velocity increases with time.... 9.81m/s2(5s)= 49.05 m/s 9.81m/s2(6s)= 58.86 m/s 9.81m/s2(7s)= 68.67 m/s you're right about crunching the math... it only gives guidelines... each fall is a case by case assessment and at best an estimate... i wish we could hear what dr. kodos has to say on the subject... FREE KODOS!!!!!!
Oh yah - but what about Terminal Velocity!?
ROFL ROFL
how friggin ridiculous is all of this? No one is getting any safer. LOL
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jason1
Apr 6, 2004, 9:40 PM
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74.8 kg X 49 m/s2 = 3665 newtons, or 3.7 kN
sorry if i misread...
terminal velocity is irrelavant when dealing in terms of acceleration...
like i said guys it's an estimate... the goal of putting up newton's 2nd law was to explain what a KN is, and give a general sense of how it is derived...
and as i said before... free kodos
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jason1
Apr 6, 2004, 9:51 PM
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If you read anywhere in anything that I've written that I think gravitational acceleration is not constant, you've misread. This thread started out with a completely incorrect description of the forces in a fall. Correcting those errors seemed worthwhile. Correcting your misreading of my posts doesn't.
Rob
jeeze, dood... ain't no need to go all kodos on me and $H1T... i'm just putting it out there...
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robmcc
Apr 6, 2004, 9:52 PM
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49 m/s2 is a sample acceleration imposed on the climber AWAY from the earth by the tension in the rope. As I said, gravity is only one of the forces acting on the climber.
Rob
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bigwallgumbie
Apr 6, 2004, 9:53 PM
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In reply to: In reply to: Oh yah - but what about Terminal Velocity!? terminal velocity is irrelavant when dealing in terms of acceleration...
Score 5 points for drunkencabanaboy....
This is quickly becoming my alltime favorite thread on rc.com
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jason1
Apr 6, 2004, 9:56 PM
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i'm sorry, i don't understand...
can you explain how you came up with 49m/s2?
mea culpa, mea culpa, mea moxie, mea culpa...
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drunkencabanaboy
Apr 6, 2004, 10:34 PM
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In reply to: terminal velocity is irrelavant when dealing in terms of acceleration...
au contraire!
http://hypertextbook.com/facts/JianHuang.shtml
:lol:
This thread is about as useless as the guy with the really loud and obnoxious leaf blower outside. He's prob getting paid $20/hr to blow around dust... psh. My rent money at work.
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ipsofacto
Apr 6, 2004, 11:13 PM
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23456789
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jason1
Apr 7, 2004, 3:03 AM
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ok,you're right about terminal velocity being another factor that we have to account for with rope stretch and friction of every molecule that we brush up against and whats lost for heat and sound and did i hear a butterfly flapping its wings...
and you gotta admit that's a long fall if you're playing with terminal velocity....
write your congressman, FREE KODOS!!!
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drunkencabanaboy
Apr 7, 2004, 2:38 PM
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In reply to: ok,you're right about terminal velocity being another factor that we have to account for with rope stretch and friction of every molecule that we brush up against and whats lost for heat and sound and did i hear a butterfly flapping its wings... and you gotta admit that's a long fall if you're playing with terminal velocity.... write your congressman, FREE KODOS!!!
NOW we are getting to the important stuff! :lol:
LOL
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hosh
Apr 7, 2004, 4:17 PM
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... :?:
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shank
Apr 7, 2004, 4:46 PM
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Force=Mass x Acceleration
But in figuring out the force on gear that is stopping you,
Force=Mass x DEcceleration
So you have to determine how fast you are falling just when the slack come out of the rope, then figure in rope stretch and all that good stuff to determine how much time it will take for you to come to a complete stop. Figure your DEcceleration from that. I'll work out an example and post it later if anyone is interested.
Steve
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drunkencabanaboy
Apr 7, 2004, 4:49 PM
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In reply to: Force=Mass x Acceleration But in figuring out the force on gear that is stopping you, Force=Mass x DEcceleration So you have to determine how fast you are falling just when the slack come out of the rope, then figure in rope stretch and all that good stuff to determine how much time it will take for you to come to a complete stop. Figure your DEcceleration from that. I'll work out an example and post it later if anyone is interested. Steve
LOL - no one is interested :D HAHA
sorry - I'm just being a jerk because this whole thread is a laugh and i need to "stop watching" it.
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jt512
Apr 7, 2004, 4:50 PM
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In reply to: As stated in a previous post.... 1Kn = 225 Lbf (although you incorrectly stated pounds...not pounds of force!)
The pound is a unit of force. "Pounds of force" is, therefore, redundant.
-Jay
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andypro
Apr 7, 2004, 4:51 PM
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Edit: I posted too slow. This was meant for shank.
Eesh. That will require calculus that I've voluntarily forgot! But hey...if you up to it, lets see force over time graphs :twisted: That'll fix this mess...
P.S.-it wasn't the puff of smoke from the grassy noll, it was friggin Jackie O!!!!
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shank
Apr 7, 2004, 7:03 PM
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Ok. Someone correct me if I'm wrong here:
1 kg (kilogram)=2.2046 lbs (pounds)
1 m (meter)=3.28083 ft (feet)
All conversions are rounded from here on.
29.4 m (96 ft) of rope out
9.8 m (32 ft) lead fall
7% elongation (rope stretch)
I'm gonna use sport climbing as my example here.
You have a mass of 80 kg (176 lbs.), you are 29.4 m (96 ft) off the deck, and 4.9 m (16 ft) above the last bolt with an anchored belayer and no slack or friction in the system at all. You fall. This is a factor .3 fall (fall factor= (2 x distance above the bolt)/length of rope between you and belayer=(2x4.9)/29.4=9.8/29.4=.3)
When you fall you accelerate downward at 9.8m/s/s for 1s before the rope is weighted at all. with a static rope this is where your spine breaks, but with a dynamic rope it starts to strech and slow you down (decellerate). At 7% elongation it will stretch 2m and has to slow you down from 9.8m/s to 0 in that 2m.
So at 9.8m/s it will take you around .2s to go that 2m which is a decceleration of 49m/s/s roughly.
So the force on you is
49m/s/s x 80kg=3920N (newton) or 3.92 KN (kilonewton)
Also the belay has to exert this much force to stop you so the force on him is the same, but the force on the top gear is twice this amount, which is 7.84 KN.
Any corrections or questions?
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dirtineye
Apr 7, 2004, 7:17 PM
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YOU left out the friction over the biner. I believe Rgold and someone else covered the fact taht friction over the top biner reduces the force felt by the belayer by about 2/3.
Notice that if this friction over the top biner affects the force felt by the belayer, it must also affect the fall factor, and so using the theoretical fall factor is incorrect.
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robmcc
Apr 7, 2004, 7:22 PM
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Isn't this about the point where we start with some simplifying assumptions? I propose the following:
1) We replace the biner with a frictionless pulley. Clearly incorrect, but it puffs up our numbers, and as everyone knows, Size Matters!
2) We replace the climber with a spherical cow of uniform density. Just because. No real preference for what kind of cow, but in deference to the fact that this thread has been milked for all its worth, I'd stick to dairy cows.
Rob
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drunkencabanaboy
Apr 7, 2004, 7:24 PM
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In reply to: YOU left out the friction over the biner. I believe Rgold and someone else covered the fact taht friction over the top biner reduces the force felt by the belayer by about 2/3. Notice that if this friction over the top biner affects the force felt by the belayer, it must also affect the fall factor, and so using the theoretical fall factor is incorrect.
And don't forget about air resistance! ROFL - and the pointless thread starts all over again just as it began.
the person who commented about fall factors was right.
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drunkencabanaboy
Apr 7, 2004, 7:26 PM
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In reply to: 2) We replace the climber with a spherical cow of uniform density. Just because. No real preference for what kind of cow, but in deference to the fact that this thread has been milked for all its worth, I'd stick to dairy cows.
this is - in my opinion - the best idea yet.
Maybe the cows we get chocolate milk from would be even better?
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drunkencabanaboy
Apr 7, 2004, 7:31 PM
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In reply to: So the force on you is 49m/s/s x 80kg=3920N (newton) or 3.92 KN (kilonewton) Also the belay has to exert this much force to stop you so the force on him is the same, but the force on the top gear is twice this amount, which is 7.84 KN. Any corrections or questions?
actually - i do have a real comment - this is incorrect i believe. For this simple reason:
the force from the dynamic rope is not constant.
There is a peak force that is higher than the # you quote.
Blah - this thread will go on forever. Next we will be talking about the quantum possibility that all the molecules in the rope will simultaneously phase through the biner.
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robmcc
Apr 7, 2004, 7:34 PM
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Yep, I think you're right, DCB. That's probably the average force, not peak.
I think I'm to the point where I'd rather do something more fun than play in this thread. Like get a big mouth full of lemon juice. Right after sticking my tongue in a blender.
Rob
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drunkencabanaboy
Apr 7, 2004, 7:36 PM
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In reply to: Yep, I think you're right, DCB. That's probably the average force, not peak.
you DO agree with my chocolate milk cow statement too right? That it central to my whole thesis
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robmcc
Apr 7, 2004, 7:39 PM
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In reply to: In reply to: Yep, I think you're right, DCB. That's probably the average force, not peak. you DO agree with my chocolate milk cow statement too right? That it central to my whole thesis
Of course. That idea was so clearly an improvement over my suggestion that I was too embarassed to address it directly. Chocolate cow it is. Just make sure it's a cow, not a bull. If it ain't a cow, it ain't milk...
Rob
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bigwallgumbie
Apr 7, 2004, 7:48 PM
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Shank, good attempt, but no cigar:
In reply to: 1 kg (kilogram)=2.2046 lbs (pounds)
In reply to: When you fall you accelerate downward at 9.8m/s/s for 1s before the rope is weighted at all
No, the climber falls for 9.8m. If he was moving at 9.8 m/s then yes, it would take (9.8 m)/(9.8 m/s) = 1 s but the climber starts with a velocity of 0 m/s. So the climbers velocity after falling 9.8 m is:
V = sqrt(2*a*x) = sqrt (2*9.8m/s*9.8m )
V = 13.9 m/s
And the fall takes:
V = a*t
t = V/a = (13.9m/s) / (9.8m/s/s)
t = 1.4 s
And to doulbe check:
x = (1/2) *a*t^2 = (0.5)*(9.8m/s/s)*(1.4s)^2
x = 9.6 m (not 9.8 but i did some rounding so... )
In reply to: At 7% elongation it will stretch 2m and has to slow you down from 9.8m/s to 0 in that 2m.
Problem: I'm not sure where you got that 7% stretch, but I'm assuming its in relation to either a static loading of the rope or the strech during a UIAA test fall, either way, the amount the rope streches is going to vary on the the imact on the climber which will be related to the fall factor and friction between the rope and biner(s). So, nice idea, but using that 7% is a no go.
So i didn't really help figure out the load, but thought i could help clean up the physics. If i f*cked anything up somebody correct me, its been a good year and a lot of brain cells since i got outa school.
Mike
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drunkencabanaboy
Apr 7, 2004, 7:54 PM
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Mike is oh-so-right about the t=1.4 s - not the 1s previously quoted - but he forgot the most important remaining question -
what type of harness would you use for our chocolate milk cow? BD Alpine Bob maybe? I Think such a diaper-style harness would fit around her utters.
Also, to make the physics easier, treat the rope like a spring. Spring physics are taught in HS physics classes.
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bigwallgumbie
Apr 7, 2004, 8:05 PM
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Drunken texan (W?):
The diaper harness sounds like a good idea (I know home much it hurts to get a nad up under a leg loop, never mind an udder) but I'm thinking she might like a chest harness to keep her upright, them cows don't have really big hips ya know. Shit, we're talking about a spherical cow ain't we? Hmmm... Haulbag? Being as it is a brown cow, how do you tell if shes get s the "milk squeezed outta her" or if its the shit-scared-outta-her (esp if said cow just got back from the potrero....) Could be a nasty suprise when you go for that glass of free chocolate milk...
Extra ghetto climbing physics: If a factor 2 fall is like a 2000lb load on the climber, is a factor .3 fall a 2000/(2/.3) = 333 lb load on the climber? Kinda breaks down on real small fall factors though...
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robmcc
Apr 7, 2004, 8:10 PM
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Maybe we should go with a rigid, spherical, chocolate cow of uniform density. Then we could just bolt it. Otherwise, I'd go with the haulbag idea, or maybe a cargo net.
Rob
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bigwallgumbie
Apr 7, 2004, 8:12 PM
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Any idea what the spring constant for a rope is? Shank, where'ld ya get that 7%? To be a little more accurate we could do a spring/damper thing but I ain't doing the math... I'm procrastinating hard, but not that hard
Wait, doesn't petzl have a program thing for all this stuff?
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jason1
Apr 7, 2004, 8:14 PM
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shouldn't someone be arguing that it's really an elastic collision...
really force diagram the on side of the top piece... then account for the other... then take away for friction determined by how much rope runs over the top biner(with rope stretch and distance belayer is pulled up)...
you guys should use one of those OSHA harnesses.... the big uncomfortable ones with the bungee on back...
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bigwallgumbie
Apr 7, 2004, 8:15 PM
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Mmm, solid cholacte cow, its like an easter bunny but 10000 times bigger!!! We'll need a lotta weed to eat that sucker.
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robmcc
Apr 7, 2004, 8:16 PM
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In reply to: Any idea what the spring constant for a rope is?
Well, there's the problem. It isn't constant for a climbing rope.
Rob
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robmcc
Apr 7, 2004, 8:19 PM
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I have to reluctantly concede that I was wrong about something very important. It's hard to admit, but I really have no choice.
This thread is actually MORE fun than a mouth full of lemon juice after sucking on the business end of a blender.
Rob
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drunkencabanaboy
Apr 7, 2004, 8:30 PM
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As you can see from my photo - chest harnesses are not needed - and they don't release their milk any easier than you do. Let's not be ridiculous people.
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drunkencabanaboy
Apr 7, 2004, 8:35 PM
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In reply to: In reply to: Any idea what the spring constant for a rope is? Well, there's the problem. It isn't constant for a climbing rope. Rob
BTW - it _is_ constant. If you don't think so - you don't know what a spring constant is.
http://www.vishay.com/brands/measurements_group/guide/glossary/spr_con.htm
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robmcc
Apr 7, 2004, 8:35 PM
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I don't think ANYONE has EVER considered the Wedgie Factor, either. In the transition from unweighted harness to weighted one, the harness becomes quite intimately well, intimate with the wearer. You're especially familiar with that if you've ever climbed sans underwear. What about that extra 3-6" of give as climber compresses into the harness on a hard fall? That's 6-12" of extra squish when you count both climber and belayer. Even more if we're talking about cows, as long as they're non-rigid cows.
Rob
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drunkencabanaboy
Apr 7, 2004, 8:36 PM
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In reply to: I don't think ANYONE has EVER considered the Wedgie Factor, either. In the transition from unweighted harness to weighted one, the harness becomes quite intimately well, intimate with the wearer. You're especially familiar with that if you've ever climbed sans underwear. What about that extra 3-6" of give as climber compresses into the harness on a hard fall? That's 6-12" of extra squish when you count both climber and belayer. Even more if we're talking about cows, as long as they're non-rigid cows. Rob
EXCELLENT POINT! You are a true scholar of which the likes of us climbers have never seen before. Except maybe for JT - HA - j/k - about JT that is.
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drunkencabanaboy
Apr 7, 2004, 8:43 PM
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In reply to: Drunken texan (W?): Extra ghetto climbing physics: If a factor 2 fall is like a 2000lb load on the climber, is a factor .3 fall a 2000/(2/.3) = 333 lb load on the climber? Kinda breaks down on real small fall factors though...
I think you are referring to the loads on the anchor - in which case - that number may not be AS FAR off. (think: load to climber ~ 160 lbs + load to belayer ~ 160 lbs)
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robmcc
Apr 7, 2004, 8:49 PM
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In reply to: BTW - it _is_ constant. If you don't think so - you don't know what a spring constant is. http://www.vishay.com/brands/measurements_group/guide/glossary/spr_con.htm
I know what it is, thanks. :)
I read something yesterday that made the interesting point that ropes typically stretch 7% or so under an 80 kg load. How much should they stretch under an 800kg load, as in a severe fall? 70%? Doesn't happen. Or at least ISTR that max stretch is in the neighborhood of 40%.
Anywho, I'm finding stuff now that contradicts that claim. Some sources say ropes _do_ act like springs, but I've found at least one that says they don't. Until such time as I can do some testing of my own, I'll withdraw the assertion that they don't.
Rob
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drunkencabanaboy
Apr 7, 2004, 8:52 PM
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In reply to: I'll withdraw the assertion that they don't.
Great.
What do you think of my climbing cows photo?
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jason1
Apr 7, 2004, 8:54 PM
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now that's some funney $(-)!T....
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bigwallgumbie
Apr 7, 2004, 8:56 PM
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Bitchen... are those indusrial strength paper clips load reducing at high forces??
'Ma! I sah besty a rock clamberin!'
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drunkencabanaboy
Apr 7, 2004, 8:58 PM
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In reply to: b----... are those indusrial strength paper clips load reducing at high forces?? 'Ma! I sah besty a rock clamberin!'
Yes
Yes they are.
I hear they work as well as keychain biners.
And those a pure-bread texan chocolate milk cows.
YE-HAW.
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bigwallgumbie
Apr 7, 2004, 9:15 PM
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From http://www.bealplanet.com/produits/anglais/produit1.html
For the beal 10.2 Flyer II rope:
Imapct force: 7.4 kN
Extension during first fall: 37 %
Static elongation: <= 10% (less than or equal)
So making an ass of u and me:
spring constant for static case:
F = 80 kg * 9.81 m/s^2 = 0.78 kN
k = F/x = 0.78kN/0.1 = 7.8 (kN/%)
for first fall:
k = 7.4kn/.37 = 20 (kn/%)
so the spring constant is 7.8 for the static and 20 for the first fall, so the spring constant does change??
Ghetto physics disclamers: k in kN/%??? a little weird but i think it'll work, also the first fall should be less than 7.4 kN, but that just makes the spring constant bigger, also the elongation for the static case is less than or equal to 10% but whatever, close a fucking nuff. That and bluewaters figures are pretty similar (plus the 2004 BW catalog has a pic of sharma with the reddest eyes I've EVER seen, friggin rad)
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robmcc
Apr 7, 2004, 9:15 PM
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In reply to: What do you think of my climbing cows photo?
Classic. I do have to point out, though, that your cows are not spherical.
Rob
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drunkencabanaboy
Apr 7, 2004, 9:33 PM
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In reply to: (plus the 2004 BW catalog has a pic of sharma with the reddest eyes I've EVER seen, friggin rad)
:lol:
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jt512
Apr 7, 2004, 9:35 PM
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In reply to: From http://www.bealplanet.com/produits/anglais/produit1.html For the beal 10.2 Flyer II rope: Imapct force: 7.4 kN Extension during first fall: 37 % Static elongation: <= 10% (less than or equal) So making an ass of u and me: spring constant for static case: F = 80 kg * 9.81 m/s^2 = 0.78 kN k = F/x = 0.78kN/0.1 = 7.8 (kN/%) for first fall: k = 7.4kn/.37 = 20 (kn/%) so the spring constant is 7.8 for the static and 20 for the first fall, so the spring constant does change?? Ghetto physics disclamers: k in kN/%??? a little weird but i think it'll work, also the first fall should be less than 7.4 kN, but that just makes the spring constant bigger, also the elongation for the static case is less than or equal to 10% but whatever, close a f---ing nuff. That and bluewaters figures are pretty similar (plus the 2004 BW catalog has a pic of sharma with the reddest eyes I've EVER seen, friggin rad)
I calculated k = 14.99 for the Beal Booster II. See http://www.rockclimbing.com/...php?t=52735&start=19
-Jay
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shank
Apr 7, 2004, 9:38 PM
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In reply to: Shank, good attempt, but no cigar: In reply to: 1 kg (kilogram)=2.2046 lbs (pounds) Correct, but it should be lbs-m (pounds of mass)
I know and you know, but the people that didn't know would still know what I mean I thought.
In reply to: In reply to: When you fall you accelerate downward at 9.8m/s/s for 1s before the rope is weighted at all No, the climber falls for 9.8m. If he was moving at 9.8 m/s then yes, it would take (9.8 m)/(9.8 m/s) = 1 s but the climber starts with a velocity of 0 m/s. So the climbers velocity after falling 9.8 m is: V = sqrt(2*a*x) = sqrt (2*9.8m/s*9.8m ) V = 13.9 m/s And the fall takes: V = a*t t = V/a = (13.9m/s) / (9.8m/s/s) t = 1.4 s And to doulbe check: x = (1/2) *a*t^2 = (0.5)*(9.8m/s/s)*(1.4s)^2 x = 9.6 m (not 9.8 but i did some rounding so... ) In reply to: At 7% elongation it will stretch 2m and has to slow you down from 9.8m/s to 0 in that 2m. Problem: I'm not sure where you got that 7% stretch, but I'm assuming its in relation to either a static loading of the rope or the strech during a UIAA test fall, either way, the amount the rope streches is going to vary on the the imact on the climber which will be related to the fall factor and friction between the rope and biner(s). So, nice idea, but using that 7% is a no go. So i didn't really help figure out the load, but thought i could help clean up the physics. If i f*cked anything up somebody correct me, its been a good year and a lot of brain cells since i got outa school. Mike
Out of time, I'll get back for more later.
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robmcc
Apr 7, 2004, 9:39 PM
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In reply to: so the spring constant is 7.8 for the static and 20 for the first fall, so the spring constant does change??
Yeah, that's the kind of stuff I read. Given the static figures, a hard fall should stretch the rope a lot more than it actually does. Like 70-90%, which I think just doesn't happen ever. But whatever. I don't have conclusive data or sufficient credible information, so I'm not drawing a conclusion on that one yet.
Rob
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drunkencabanaboy
Apr 7, 2004, 9:45 PM
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In reply to: Classic. I do have to point out, though, that your cows are not spherical. Rob
Oh but they are
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drunkencabanaboy
Apr 7, 2004, 9:52 PM
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In reply to: In reply to: so the spring constant is 7.8 for the static and 20 for the first fall, so the spring constant does change?? Yeah, that's the kind of stuff I read. Given the static figures, a hard fall should stretch the rope a lot more than it actually does. Like 70-90%, which I think just doesn't happen ever. But whatever. I don't have conclusive data or sufficient credible information, so I'm not drawing a conclusion on that one yet. Rob
Appears you are quite right
http://bstorage.com/speleo/Pubs/rlenergy/Default.htm
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shorty
Apr 7, 2004, 10:33 PM
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In reply to: Let's not be ridiculous people.
I'm offended by the use of chocolate cows in this thread. Not only should PETA get involved, but using chocolate cows is racist and discriminates against the caramel, vanilla, and strawberry cows. I'm going to call the NAACB ("B" for bovine) as soon as I finish my Big Mac.
I'm offended by the potential injury inflicted on said cows. I would like to see a disclaimer on drunkencabanaboy's post which states "No heifers were injured in the production of this illustration." Next time, please use CG. Save our cattle for burgers! Or steaks, or roasts, or prime rib, or fillets....
I'm offended by the complex calculations which include kilonewtons, foot pounds, accelleration, decelleration, E=mc2, friction, rope stretch, lingerie stretch, wedgie factors, belay butt floss, and the loss of bowel control. We're trying to explain these concepts to cattle, fer chrissake! Cattle have only three things on their minds:
- When do I get to eat again?
- Where do I get to poop again?
- Wow, that new bull really does have big .....
Well, and maybe:
- Isn't that shorty warming up the grill again?
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jxl914
Apr 7, 2004, 11:14 PM
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I'm joining this topic a little late and was too lazy to review all 8 pages of stuff. I agree up to the point where you start talking about forces it takes to stop the falling climber. No offense, but I'd love to see how you figured that out (if there is 21 feet of rope and the climber stops after falling 21 feet, there is an infinte force generated). I'd imagine that the effective spring constant of the rope was not considered at all and would likely be non-linear and therefore, really hard to approximate. I also love the "rock psi" whatever that is (I'm guessing your talking about the ultimate tensile or compressive stress the rock can sustain before brittle fracture. I hate to say this but I am actually a "rocket scientist" and I don't feel qualified to takle the subject without alot more data. I think we should all call it quits here guys and gals. thanks for educating us though.
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bigwallgumbie
Apr 7, 2004, 11:29 PM
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Hmmm, sounds like jxl914 (WTF??) got aroused by the spherical cows and now feels embaressed.... It's alright man, cows need loving too. You can't fight who you are...
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easysteve
Apr 8, 2004, 1:03 AM
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Hey, Red, don't listen to most of these people, you did what you wanted.
Steve
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drunkencabanaboy
Apr 8, 2004, 1:32 AM
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In reply to: Hmmm, sounds like jxl914 (WTF??) got aroused by the spherical cows and now feels embaressed.... It's alright man, cows need loving too. You can't fight who you are...
Maybe (s)he is a said "strawberry cow" who was offended by our complicated physics equations. Notice it was her first post ever!
I was also wondering what happens as our cows approach (but - of course - does not reach) the speed of light on their travels down to earth? Would she perceive the belaying cow's utters shriveling up? While her utters remained pink and supple?
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