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HighGravity
Trad climber
Southern California
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Topic Author's Original Post - Apr 30, 2009 - 08:14pm PT
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Does anyone have a graph or scale of how much force is applied when a static rope or cable is shock loaded from different heights or weights attached? I need it as close to everyday terminology, because I'm trying to explain the forces involved to not the brightest crowd. Any help out there would be great.
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Yvergenhauf
Gym climber
UT
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Apr 30, 2009 - 08:48pm PT
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Here is one estimate. Not guaranteed for accuracy. For a climber of around 165 pounds falling around five feet, the force would be around 1200 newtons.
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TradIsGood
Chalkless climber
the Gunks end of the country
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Apr 30, 2009 - 09:02pm PT
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Forget what he just said!
The force changes as the rope is stretched. It is a function of how long the rope is, how long the fall is, and how heavy the climber is and a host of other factors.
Maybe just play with rubber bands cut to different lengths. Attach to a piece of string of a fixed length. Let a weight fall that length and see that the short rubber bands break and the longer ones don't.
If you want a "device", the cheapest most accessible thing that I can think of is your standard fish scale.
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tolman_paul
Trad climber
Anchorage, AK
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Apr 30, 2009 - 09:37pm PT
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First of all, there is no such thing as a static rope. Static means unmoving, and is trying to convey a rope that does not stretch, but that is not true, as all ropes stretch. What we refer to as static ropes are low stretch ropes.
What happens in the arrest of a fall is fairly basic Newtonian physics. You are taking a body that is falling and adding a force in the opposite direction to accelerate the body until the downward motion stops.
F=ma, where F is the force, m is the mass, and a is the acceleration. To figure out a, you need to know the velocity of the body, and the distance over which the body is accelerated to reduce the velocity to zero. If you had a truly unyielding system, then the acceleration would be inifinite, as it would take an infinite force to instantaneously stop the body.
The best way to find the force is not to calculate it, but to measure it with an accelerometer.
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Ed Hartouni
Trad climber
Livermore, CA
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as far as I can tell, shock loading in this context is just the force of stopping a falling object, where that object has time to free fall.
That being the case, the force is the change in momentum of the object and the time it takes to change that momentum:
F = dp/dt
but this is generally not useful for calculating the impact force on a rope. What is usually done is to integrate the work done stretching a cord for objects that have fallen a certain distance, this results in the "impact force" calculated with fall factors which you can find on various websites, try:
http://www.bealplanet.com/portail-2006/index.php?page=force_choc&lang=us
That is also not obvious...
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Ghost
climber
A long way from where I started
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First of all, there is no such thing as a static rope. Static means unmoving, and is trying to convey a rope that does not stretch, but that is not true, as all ropes stretch. What we refer to as static ropes are low stretch ropes.
Right. So instead of your spine snapping instantly as it would do if your fall was stopped by a steel cable, it actually takes 1.472 milliseconds for your spine to snap when your fall is stopped by a static rope.
I think the op is looking for a simple way of describing a real-world event, not for another dissertation on the physics of rope stretch.
Dynamic rope = bungee cord
Static rope = steel cable.
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jenren
Trad climber
Sac, CA
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Ghoulwe J would know..he knows all about that..has done all the testing etc...
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Dr.Sprock
Boulder climber
Sprocketville
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The very term "Impact Force" is somewhat of a handicap to the climbing world.
There are so many things wrong with it, that if you were to use the term in a college physics class, you would probably get the boot.
Confusing terms means that a climber probably coined the term Impact Force".
There is such a thing as in Impulse, and such a thing as a Force.
They are strict in their definition.
I = F (net) * Time Interval of Event
See that the force must be constant, which is never the case in rope stretch, so we integrate.
Impulse Units: kg m/s = N·s)
Definition of Force?
Simple and Eloquent
F = ma.
Newtons basic laws.
Math can be applied.
If you do a search on Impact Force, you will not find a classical derivation from Newton's 2nd Law.
You will just find a bunch of nonsense.
Use the conservation of energy, and a accelertion meter, like one of the posters said.
You need a graph of the event.
X axis will be time.
Y axis will be force.
You will have recoil, which means signs change, net forces increase, all kinds of cool calculus with the rope, note that the middle of the rope does not stretch one bit.
So do not feel bad if you do not understand impact force.
there really is no such thing.
it is a concept, not really based on scientific method.
it is a feeble attempt to quantify a complex event.
If you tell a propeller head , "We need to go to the Moon.
Your formula is F = ma. it is Exact. Can you do it?"
Prop Head: "No Problema"
But give him F (impact) = cr-ou23r-9p2o540-[jhvn-024
and he will tell you that the missile might land on mars, or it might land on Jupiter, 180 degrees to the South.
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Paul Martzen
Trad climber
Fresno
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What is the context? Who are you trying to teach this to and why? Is it a climbing class, a science class?
For many situations, a hands on demonstration such as TradisGood suggests is much better than an explanation alone.
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HighGravity
Trad climber
Southern California
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Topic Author's Reply - May 1, 2009 - 04:34am PT
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We were working with a possible SAR situation on lowering from a helicopter. The fixed point we are lowering from has a quick release that will release the rope in case of an emergency. Well, someone decided it needed a backup safety release, which would not become tawt until the rope dropped about 6 feet. I was trying to explain to others that the force of the fall if the first "tie in point" failed would be so great it would cause the helicopter to lose its main axis point, thus causing a major emergency. During the discussion we determined the average weight of the person(s) being lowered would be right around 220 lbs.
It sound horrible, but my arguement was losing one person to a 30 foot fall would be much better then losing up to 5 people and a very expensive aircraft. So, I'm trying to show about how much weight the person falling could put onto the aircraft when it's shock loaded.
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rgold
Trad climber
Poughkeepsie, NY
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An appropriate industrial load-limiter back-up would completely solve this problem---basically a super screamer or battery of super-screamers. A little chat with Yates might be a good idea; he says on his site that he has made such things for NASA. With the appropriate load-limiter, you could view the steel cable as if it was totally rigid and forget about the loads it would develop if it had to absorb the fall energy. (But the weight of the cable has to be added to the weight of the hanging person and gear).
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kinnikinik
Trad climber
b.c.c
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a cool demo for this is; using 20# fishing line and a 2# weight and a 20# weight . Hang the 20# weight on a 2foot peice of fishing line (static rope) no break. now take the two# weight on two feet of line and drop it from the same height as the fixed upper end. The line will snap. The dynamic energy is clearly more than 10 times the static force.
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Double D
climber
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Tie a peice of thread around an uppeeled banana with about 3 feet of tail. Drop it and the banana will be cut in half. Do the same with a robber band attached at the end of thread...it won't cut. Pretty simple and effective.
The other day I was out climbing and some nubees were leading with a static rope. As much as I pleaded with them, they continued to go at it on stuff that was beyond their ability. Yikes. I wish I'd had my banana rig then because they just didn't get it.
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ontheedgeandscaredtodeath
Trad climber
San Francisco, Ca
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Just curious, but if someone releases the rope because of an emergency why would you want it caught again? Are you talking about accidental releases? I've seen helicopters dump their long lines and buckets on fires. Once where it got tangled in some trees and once where the pilot got into a turn he did not want to be in. I know they did not want the lines re-caught.
We had similar discussions when I was a smokejumper about what to do if someone's static line got caught and he was being towed behind the airplane. We spent countless hours on the subject, usually concluding that the hapless jumper must be cut loose and left to fend for himself with the reserve chute. It has never actually happened!
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Paul Martzen
Trad climber
Fresno
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I love the banana and fishing line experiments. With the fishing line, you could reduce the distance the 2# weight falls to find at what height it doesn't break the line. Then you could add an additional line to see how far of a drop with the 2# weight breaks 40# test.
HighGravity,
The additional info about the helicopter scenario really helps. Seems to me that a safety release on the rope is there exactly for situations where the helicopter and crew are in danger. If you have to drop the rope in order to save the helicopter, you don't want a back up system that prevents you from doing that.
If they think there is some real (rather than unrealistic) possibility that the release could be operated accidentally then a second equalized release could be installed or an additional safety on the present release could be utilized. If the release requires a couple steps to be operated, it is much harder to accidentally operate it. I suspect the engineers who designed the release have probably thought this all out in greater detail than the guys who want to add to it.
But I think it is a really good idea to show you teammates some easy ways to test and experience this effect themselves. Gets some different weights, such as plastic soda bottles filled with water. Have them hold the bottles in their hands. Easy. Then tie webbing around the bottle necks and have the guys hold the other ends of the webbing in their hands. Have them wear gloves. Start with the lightest weight and drop a bottle a short distance, so the guy gets a feel of it. Then increase the distance of the drop to find the maximum drop the guy can hold without the webbing ripping through his hands. It is not at all static since you can't hold your hand steady against the force and the webbing rips through your grip at pretty low loads, but it gives you a "hands on" perspective.
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Yvergenhauf
Gym climber
UT
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Hey TIG,
I thought you said forget about that formula?
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midarockjock
climber
USA
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Helicopters are used quite frequently for moving static
mass greater than that and static ropes are ok. A helicopter
no differently than a crane should not be shock loaded.
I worked on a daily basis for years with shocks loads. A
1000 force is at times doubled and static ropes were used.
Also I usually generated a much greater than force of 2000 lbs.
with each shock load.
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Colt
climber
Midpines
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Here are some quick numbers assuming I understood the situation correctly and the rope and shock absorption properties are reasonable in the web calculator that I used. I assumed a 220lb man with a 6' static cord attatchment. The calcs were done using the web calculator here:
http://www.myoan.net/climbart/climbforcecal.html
Fall Distance (ft) Force (lbs)
1 419
2 838
3 1257
4 1676
5 2096
6 2515
Notably, the web calculator makes several assumptions about rope properties and other aboption components in the system. The reason a nice graph doesn't exist is because the forces can very drastically when the energy absoption properties are adjusted.
For example, if the same fall scenarios occured and energy was disipated through nothing but a .1 mm stetch in the cable the max forces would be exponentially higher. The following calcs assume decleratoin occuring linearly over a distance of .001m and no other mean of energy absoption exists.
Fall Distance (ft) Force (lbs)
1 66927
2 133854
3 200780
4 267707
5 334634
6 401561
This is why the physics geek always drag this issue into the more technically precise relm.
If deceleration occured over .1m (~4") the forces would be:
Fall Distance (ft) Force (lbs)
1 669
2 1339
3 2008
4 2677
5 3346
6 4016
Notably, I used this site to do the calcs:
http://hyperphysics.phy-astr.gsu.edu/hbase/flobi.html#c1
There are so many factors that effect the forces in a human fall scenario that sticking to the fall factor method and lab test numbers is best.
In the end the theoretical numbers always seem too high from what I have seen in the field. I took a 4' daisy whip when the bottom daisy was on a bomber bolt and a head blew as I was climbig up the upper ladder. It hurt...had to take a little time out...but I didn't experience and 1600lb back breaker that the math describes.
IMHO the math tends to show numbers that are somewhat higher that what will generally be experienced in the field b/c too many factors are left out. That being said any static falls should still be avioded at all costs. A few feet static fall will likely not break your back or tip a helicopter, but it will at least suck a great deal.
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the Fet
Supercaliyosemistic climber
Tu-Tok-A-Nu-La
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Use a climbing rope, then it's just a factor 1 fall.
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HighGravity
Trad climber
Southern California
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Topic Author's Reply - May 1, 2009 - 05:15pm PT
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Thanks for all the help, hopefully some of this will get through to them.
The reason some people want the backup is they were in the military and they always had a backup in case of accidental release. The problem is if you need a quick release due to a snag or some other emergency they time it would take to get to the back up would more then likely be too late for the crew and helicopter. Not to mention the "Oh $h!t" factor if someone were to shock load the cable or rope. I'd be more worried about staying in the helicopter then trying to hit the quick release.
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TradIsGood
Chalkless climber
the Gunks end of the country
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Y - I do not see anything about "formula" in any of my posts here. Not sure what you are talking about.
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Yvergenhauf
Gym climber
UT
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Your formula is roughly mgh. It is the same one I used to give the estimate that you were so quick to dis.
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noshoesnoshirt
climber
dangling off a wind turbine in a town near you
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As for the "industrial screamer", we use these every day on towers. Personal Fall Arrest Systems currently limit the loading experienced by the subject to an OSHA regulated max of 8kN, or Canadian std 4kN.
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Dr.Sprock
Boulder climber
Sprocketville
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The back up is used when not in critical situations, which is probably 90 percent of the time.
Then when things get weird, you un do the back up.
There is a really cool helicopter drop rope on ebay i was thinking of using for top roping.
it's bigger than a gym rope, huge braids of green poly.
60 feet weighs forty pounds, 250 dollah Buy It Now.
i noticed a steel cable running down about 5 feet down the rope, with a hook on the end.
this might be a back up system component, maybe not.
some special forces types could probably tell ya.
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TradIsGood
Chalkless climber
the Gunks end of the country
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Y - mgh is energy, not force.
Dissed your original because the correct answer is "it depends" (on a bunch of stuff and you made no attempt to discuss the assumptions that gave that answer)!
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HighGravity
Trad climber
Southern California
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Topic Author's Reply - May 1, 2009 - 10:05pm PT
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Dr, that sounds like a fastrope used to insert troops in the military.
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Dr.Sprock
Boulder climber
Sprocketville
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yeah, it was beef.
2000 dollars new, but thats Govt. prices.
water proof, 3 million pound test, i thought it would be cool since everything i do is 60 foot sand stone.
just drape the mofo over the rock, with knots every 2 feet, and you can free climb around it without a harness and rigging.
just grab it when you pop.
did you know bill withers used to make airline toilets?
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Dr.Sprock
Boulder climber
Sprocketville
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found it, called the Fast Rope.
see here:
You are bidding on one roll of Fast Rope 60 foot in length Spliced end with 2 galvanized steel rings P/N 3336680. This rope was purchased through the US Government as surplus merchandise. The rope is unused and in new condition and still in the original manufactures packaging. Although this product was designed for air assault operations it could used in many different applications. Just let your imagination run wild (Deer stand egress might be fun). This rope sells retail for as much as $1200.00 per 60 foot coil. Through the buy it now function you can purchase this rope for $400.00 per coil plus shipping and handling (50 Pounds).
HELI-VAC™ FAST ROPE® SYSTEMS provide efficient and safe methods for infiltrating and evacuating special operation personnel to and from point targets.
A special low stretch, abrasion resistant fiber, Dacron M/P Type 77* and Pli-Moor® construction eliminate rope hockling or kinking, resist heat build up during
use, and afford fast, fully controlled descent and safe, sure assent.
The FAST ROPE® SYSTEMS were developed in the U.S. during the mid 1980’s in a coordinated effort with our nation’s military and law enforcement services.
In 1990, Columbian along with the U.S. Army Natick RD Center in Natick, MA authored and developed the only FAST ROPE® mil-spec which is still in use
today, Military Specification MIL-F-44422 Fiber Rope Assembly, Insertion.
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rgold
Trad climber
Poughkeepsie, NY
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The first site referenced by Colt is completely bogus. The calculator uses an incorrect formula for fall factor (and who know what other idiocy occurs after that). The second site is ok but not applicable, because it requires that you guess the rope stretch, and making that guess is equivalent to specifying the peak impact force. So all that is really happening there is guessing peak impact forces.
Unfortunately, you can't believe everything you read on the internet.
The equation in the Beal planet reference is good for dynamic ropes. But there is a constant that represents "system stiffness" that is hard to estimate, although it can be measured in drop tests. It is typical to replace system stiffness with the rope stiffness, the Hooke's Law spring constant for the rope and hope for a decent approximation. In the helicopter rescue scenario, the only additional contribution to system stiffness probably comes from knot tightening. But then there could be the downward displacement of the helicopter against lifting force, which might be significant.
There are two other flies in the ointment: first, static ropes are, I think, less well approximated by the equation for simple harmonic motion used for springs, apparently because there may not be a reliable spring constant for the full range of behavior. The other fly is not so serious: putting a human on the end of the rope will produce less impact than a solid object of the same weight. This is of interest, of course, to those taking falls on static material, but ignoring just gives you higher values, which you would want anyway as part of the safety margin.
Meanwhile, it is simple to calculate the design and deployment of "super-screamers" to keep the load at some predetermined maximum, so I'm having trouble seeing the need to estimate what happens if you entrust the energy absorbtion to the static rope.
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TradIsGood
Chalkless climber
the Gunks end of the country
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Actually, rgold, "You can't believe everything you read on the internet, but..."
Hey, BTW. I think I might have walked right past you and some taller guy on the carriage road last Friday.
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rgold
Trad climber
Poughkeepsie, NY
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TIG, you could have---I was there with John Bragg, who qualifies as a taller guy.
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TradIsGood
Chalkless climber
the Gunks end of the country
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Blondish guy?
It was one of those, "should I introduce myself to a total stranger moments."
Speaking of "shock loading", it was my first view of the Trapps this year! :-)\
Went to the Nears on Saturday to avoid the crowds. Felt like July!
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rgold
Trad climber
Poughkeepsie, NY
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Yeah---formerly blond...
Hey, feel free to introduce yourself, what's the worst that can happen? I'm old and pretty harmless.
A week ago, some guy walked past me, flashed a big grin, and said, "howya doin?" I didn't recognize him, but he obviously knew me, so rather than embarrass myself, I just said "great, how's it goin?"
He said, "great," and walked on ahead (I don't move so fast any more...). But later on he stopped for a snack and I passed him. We smiled, and as I walked away, he shouted after me, "good seein' ya again Chuck!"
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Dr.Sprock
Boulder climber
Sprocketville
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You either have to know:
1) rope stretch
2) Delta T: The time it took from the onset of rope catch to the time when you stopped, with no recoil.
This will give you an exact answer, but unfortunately, the answer is Net Force.
Inside that Net Force will be a collection of different forces.
Force of Impact is a better term than Impact Force, if you want to take the S.I. Hayakawa approach, but I still get a nauseous feeling when working inexact formulae like that weird k and f deal.
Here is how I would approach this problem:
Climber Bob is a 100 kg person with all his crap on.
Thats about 220 lbs for all you wall climbers.
Bob reaches the anchor belay, and decides that Jeff, his partner, is taking too long to jug.
So Bob pops a King Cobra, and then pops off the rock.
He falls 10 meters. Or about 30+ feet for all you free climbers.
OK, so after the rope catches Bob, he travels another 1 meter. That's gotta hurt.
Well, lets find out:
1)Calculate Bob's velocity, v, when rope catches.
v = root (2* 9.8 m/s^2)(10 meters)
v = root 196
v = 14 meters per second.
2) Calculate the Impulse on Bob as if he landed on a ledge after his 10 Meter fall:
I = F * Delta t.
This is where our Impulse formula brakes down, when it tales Zero seconds to stop a freefall.
See that as the impulse time becomes close to zero, no mater what the mass and velocity of Bob, it would take an infinite force to stop him instantly.
But we can equate Impulse with a Change of Momentum to get around this.
Bob had 0 Momentum while drinking the King Cobra. The Delta P, or Total Change in Momentum will simply be his speed at impact times his mass.
p = mv
p = 100 kg * 14 m/s (from 1 above)
p = 1400
So I = p therefore,
I = 1400 N s.
3) Calculate the Average or Net Force needed to stop Bob over a distance of 1 meter:
In coming to a stop, Bob decelerates from 14 m/s to 0 m/s in a distance of 1 meter.
So his Average speed during this deceleration is simply
14/2 = 7 m/s.
Since we know Bob's Average Speed and braking Distance , we can get the amount of time he was decelerated in during that 1 Meter of rope stretch:
Time = Distance/Velocity
t = 1 Meter/7 m/s
t = 0.14 Seconds
So it only took Bob 140 milliseconds to stop after the rope caught.
Impulse = Average Force over the time of deceleration.
I = F * Delta t
Since we calculated Impulse in 2) above, we now have everything we need to calculate Bob's Average or Net Force during deceleration:
I = F * t
1400 N s = F * (0.14 s)
1400/0.14 = 10,000 Newtons, or
10 kN.
This is the Average Force experienced by Bob during that 1 Meter of deceleration.
Now you can adjust this for different rope droop, different fall factors, stuff like that, but you will be getting away from the exact Net Force little by little as you introduce uncertainties.
A Summary:
1)Get Velocity of Fall: v = root (2)(Gravity)(Distance) =
v = root 19.6 * Length of Fall
2) Get Impulse = I = Delta p = mv
3)Get Average Speed during Deceleration:
Avereage Speed = v/2
4)Get time of deceleration:
t = Distance/Velocity = D/v
5) Get the Net Force:
F = Impulse/Time = I/t
And there ya go...
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Dr.Sprock
Boulder climber
Sprocketville
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OK, i just combined all the steps from above, and it cancels into this easy to handle equation:
Fnet = mgh/stretch
eg:
Bob weighs 100 kg, falls 10 meters, is caught in 1 meter:
Fnet = (100 * 9.8 * 10) / 1
Fnet = 9800 = 9.8 kN, verifies to previous post.
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Dr.Sprock
Boulder climber
Sprocketville
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OK, HolyGravity, this is the answer to your question.
If we hold gravity constant, which hopefully it is, and pick a fixed height and weight for a climber, then the basic graph morphs into 1/x, observe:
so the bottom line is, as you approach a "cable like " scenario on your rope stretch, the force goes up dramatically.
But as soon as you get past a certain amount of stretch, it really does not matter from one static rope to the other, as long as you have enuff line out.
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| Messages 1 - 38 of total 38 in this topic |
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SuperTopo on the Web
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Torque, if the weight is not suspended from the aircraft CG.
Let us know!