Shock loading


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Trad climber
Southern California
Topic Author's Original Post - Apr 30, 2009 - 08:14pm PT
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.

Gym climber
Apr 30, 2009 - 08:48pm PT
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.

Chalkless climber
the Gunks end of the country
Apr 30, 2009 - 09:02pm PT
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.

Trad climber
Anchorage, AK
Apr 30, 2009 - 09:37pm PT
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.
Ed Hartouni

Trad climber
Livermore, CA
May 1, 2009 - 12:23am PT
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:

That is also not obvious...


A long way from where I started
May 1, 2009 - 12:50am PT
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.

Trad climber
Sac, CA
May 1, 2009 - 12:59am PT
Ghoulwe J would know..he knows all about that..has done all the testing etc...

Boulder climber
May 1, 2009 - 03:18am PT
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 = NĚs)

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.
Paul Martzen

Trad climber
May 1, 2009 - 03:42am PT
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.

Trad climber
Southern California
Topic Author's Reply - May 1, 2009 - 04:34am PT
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.

Trad climber
Poughkeepsie, NY
May 1, 2009 - 08:58am PT
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).

Trad climber
May 1, 2009 - 10:41am PT
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.
Double D

May 1, 2009 - 10:51am PT
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 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.


Chalkless climber
the Gunks end of the country
May 1, 2009 - 01:16pm PT
HighG - The maximum load in your example will be when then load starts at the base of the helicopter. You could test drop a 220 pound weight 6 feet and measure the maximum force at the top of the rope. The further down from the helicopter, the less the peak force.

To measure peak force you could drop it from a "seesaw" with known weight and calculate the leverage - adjust fulcrum to needs. (Subject to reduction due to bending of the seesaw.)

The energy dissipated will be mgh where g = 9.8 m/s/s h = 2 meters + stretch, m = 100 kg. The average force will be the energy dissipated divided by the stretch (approximately, not allowing for heat dissipitated in the rope).

 Torque, if the weight is not suspended from the aircraft CG.
 Wing loading.

Trad climber
San Francisco, Ca
May 1, 2009 - 01:44pm PT
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!
Paul Martzen

Trad climber
May 1, 2009 - 02:11pm PT
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.

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.

Gym climber
May 1, 2009 - 02:32pm PT
Hey TIG,
I thought you said forget about that formula?

May 1, 2009 - 02:45pm PT
Recommended reading,
order their block chart, it provided more than 180 degree factors.

For "working" arborist. Working = climbers.

Warning, if a arborist you should know more than rigging and start small.
del cross

May 1, 2009 - 02:52pm PT
Not the graph you were hoping for but here are a couple of sources with some numbers: drop test data on static lines for a few different fall factors. It appears that a 6 foot drop on 20 feet of rope with a 220 lb load would result in around 1000 lbs of force. I wonder what that would that do to a helicopter?

May 1, 2009 - 04:10pm PT
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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