Difference in peak forces between a human body and steel

Bob Mehew

Well-known member
I have been asked a question which is proving somewhat difficult to answer with confidence.  So can people help with some references?

The background is in dynamic rope testing one usually drops a steel mass of 80 or 100kg and see what peak force is reached by a rope (semi static or dynamic) in arresting the mass's fall.  Clearly steel is more ridged than a human body, so the question arises how much would the peak force be reduced by if it were a human body rather than a steel mass?

Years ago I was told 50% was a guide.  But I have just come across https://m.petzl.com/GB/en/Sport/Fall-comparison-with-rigid-human-mass?ActivityName=Rock-climbing which (making a few assumptions over the probable poor translation) implies the reduction is between 50 & 70%.

I also found http://www.itrsonline.org/PapersFolder/2009/Holden-May-Farnham2009_ITRSPaper.pdf which suggests 80%.

I have found a You Tube video at https://www.youtube.com/watch?v=eqZQnCGl24A&feature=youtu.be but can't get any numerical value out of the commentary.

And most intriguingly a comment at https://www.mountainproject.com/edit/forum-message/0?replyToId=114310951&quoteId=114321524 which implies a value of 80%.

To put this in context, Crawford, see http://www.hse.gov.uk/research/hsl_pdf/2003/hsl03-09.pdf suggests that 6kN (as used in EN standards) as a maximum survivable force for the type of harnesses used in caving.

So has any one got any other data on the topic of relating the impact of falls between steel masses and human bodies?
 

nickwilliams

Well-known member
It might be worth contacting HSL in Buxton to ask if they can help answer this. They have done a lot of work on the effectivness of fall restraint protection systems, including research into how well the CE marking standards (which mostly use a falling weight) represent how a FPS actually works when the weight is replaced with a body.

http://lmgtfy.com/?q=HSE+research+report+fall+protection will provide some links to reports which may provide some insight, and will certainly give you some names to try talking to.
 

Mark Wright

Active member
I don?t think there is any question that the forces applied to a real body or a soft dummy type test weight would be less than when using a rigid test weight.
The main problem is that if you tested dropping a soft dummy 10 times you would probably get 10 different test results.
I question the reason for asking the question in the first place. Is it so the questioner can put themselves in a more dangerous situation and possibly get away with it?
Mark
 

Badlad

Administrator
Staff member
Yes, an impossible question to answer.  The human body is so very different in all manner of ways associated with a fall compared to a test weight.  Plus using human bodies in fall testing is generally discouraged these days.  I believe Paul Seddon tried a few tests at Troll many many years ago. To try to place a figure on it with any sort of authority though would be asking for trouble.  As Mark suggests someone is just trying to get away with hammering a round peg into a square hole - or using something in a way they shouldn't - like using a jammer for lifelining when there are far better methods around.
 

Chocolate fireguard

Active member
The last 2 posts puzzle me.
There have been several attempts to relate the forces produced in arresting a steel mass to those when arresting a human body, or a mannequin. Results have varied widely.
It's not desperately important that someone comes up with an answer, and as Mark says there likely won't be a definitive one.
But it's interesting, and if some people with the time, resources, ability and inquiring minds want to have a look at previous work where's the harm?
We don't all have all 4 of the above, but no intelligent person should feel the need to rubbish such an attempt.
Even if (perhaps especially if) they earn a living from caving.
 

Bob Mehew

Well-known member
Mark Wright said:
I question the reason for asking the question in the first place. Is it so the questioner can put themselves in a more dangerous situation and possibly get away with it?
My reason is because I would like to be able to relate peak force on a steel mass to that on a human in an argument I wish to put forward in a few weeks time.  I can see counter arguments being raised if I assume 100%.  They will say that is unjustifiably conservative.  Given previously I had thought 50%, that value would reduce a peak force of say 8kN to 4kN which is worth taking into account.  It would mean that in such a fall predicting a difference between some one being harmed as opposed to walking away from the fall.  75 to 80% seems to make that argument much weaker.

The mountain project link implies that in developing various EN standards it was concluded that that a 80kg value for a steel mass would reflect the behavour of a 100kg human body.  (I had previously thought all climbers were skinny unlike many cavers.)

Clearly people have worked in this area; Nick's suggestion has revealed another document written by A C Sulowski.  My enquiry was to see if anyone have come across other material.  I take it that neither of you are aware of such material.
 

andrewmcleod

Well-known member
Bob Mehew said:
My reason is because I would like to be able to relate peak force on a steel mass to that on a human in an argument I wish to put forward in a few weeks time.  I can see counter arguments being raised if I assume 100%.  They will say that is unjustifiably conservative.  Given previously I had thought 50%, that value would reduce a peak force of say 8kN to 4kN which is worth taking into account.  It would mean that in such a fall predicting a difference between some one being harmed as opposed to walking away from the fall.  75 to 80% seems to make that argument much weaker.

The other problem with this is - what is the 'safe' value of kN in a fall? There is, of course, no simple answer. There is the 6kN standard - but that is assuming that you are a 80kg steel mass. All the standard is saying is that if dropping a 80kg mass onto a system results in a peak load on that mass of less than 6kN, then the system is a safe attachment system for a person. I would argue that does NOT mean that 6kN is a safe force to subject the human body to via a standard sit/full body harness - or at the minimum it should be a worst-case scenario.

Have you seen the rest of the Petzl stuff?
https://www.petzl.com/GB/en/Sport/Forces-at-work-in-a-real-fall
When they were doing their FF0.7 tests, they initially didn't have redirect on the anchor. The force was 'only' 2kN, but they had to change the test protocol because the belayer was slammed painfully into the wall, while the 3kN on the climber in free space was acceptable.
Rope slippage is very important to reducing the load... I suspect that most of those large falls onto an ascender would have resulted in cut ropes (since there would be no slippage to reduce the load), particularly if the belaying was done directly (which increases the load further).
https://www.petzl.com/GB/en/Sport/Influence-of-the-belay-device?ActivityName=Rock-climbing

The non-rigidity of people is pretty much the only safety margin left for some of the stuff we do (like use of ascenders, should there be a rebelay failure).
 

Antwan

Member
Would using the formula F=MA work? Work out the difference in time it takes for a steel weight to 'stabilise' and the extra time for a body to 'stabilise' to give the reduced acceleration factor

I might be looking at it too simplistic but it looks like the solution to me at this time of the morning
 

Bob Mehew

Well-known member
Antwan said:
Would using the formula F=MA work?
Only if someone has directly measured A, the de-acceleration of the mass or body.  The Crawford link I gave provides some estimates for people falling in helicopters, parachutes and the alike but not directly relevant to the set up I have in mind.  Hence my asking the question.
2xw said:
Honest question - will fatter cavers therefore generate less force?
The expectation is they will because the fat layer is expected to behave differently from muscle.  Whether it is significant and thus measurable is another question I don't have a quick answer to.

The underlying physics is the mass 'm' of the person and the distance they fall 'h' determines the amount of energy 'E' which has to be absorbed and is given by E = m*g*h (where g is acceleration due to gravity).  Then given that energy is also equal to Force 'F' times distance 'd', all one has to do is having the distance over which a person de-accelerates to work out the force F.  Unfortunately we are talking about a dynamic set up so the force is changing with respect to time which makes it much more complicated.  But compare a length of wire rope to an elastic band and you can immediately appreciate which will generate the lower peak force. 
 

andrewmcleod

Well-known member
Antwan said:
Would using the formula F=MA work? Work out the difference in time it takes for a steel weight to 'stabilise' and the extra time for a body to 'stabilise' to give the reduced acceleration factor

Assuming you did know how long the time was for it to 'stabilise' (and how would you define the start and end points of that time?) it would indeed tell you the average force, but not the peak force (which is what is of interest). The force will vary, initially low and low at the end with a peak in the middle. The shape and magnitude of that peak will depend on the situation.

[quote author=Bob Mehew]
Then given that energy is also equal to Force 'F' times distance 'd', all one has to do is having the distance over which a person de-accelerates to work out the force F
[/quote]

That's true - for a totally rigid body. How do you measure the distance travelled by a non-rigid body that exhibits deformation? Some energy is dissipated in the deformation.

Actually you could probably get a pretty good estimate of the acceleration of a falling _rigid_ mass by videoing it and tracking a fixed point. But it's probably more accurate to just use a force meter and measure the force with time (not just the peak force).

Sometimes you can use simple formulae for complex situations. Sometimes you can't (not because the formulae are wrong, but because the inputs are complicated...)
 

Antwan

Member
Thinking it through I would propose that measuring the time between the force been applied (when it increases above the static force of the rope) to peak force would be the time to 'stabilise ' at peak force. Theorising that a lower peak force means the mass take longer to decelerate applying a lower force for longer.

I will look at the links when I get home from work and see how my theory based solely on this thread stacks up to accepted figures
 

andrewmcleod

Well-known member
Antwan said:
Thinking it through I would propose that measuring the time between the force been applied (when it increases above the static force of the rope)
What is the 'static force of the rope'?
to peak force would be the time to 'stabilise ' at peak force. Theorising that a lower peak force means the mass take longer to decelerate applying a lower force for longer.

This is _often_ true, but it is not required by the equation (or the laws of physics). The lowest possible peak force is the case where the force is constant throughout the deceleration - this is very unlikely. You could imagine a setup where a small dyneema cord backed up by dynamic rope is used. There will be a very high peak force until the dyneema snaps, followed by a slow deceleration at low force due to the dynamic rope. There were/are ropes made like this (with a superstatic core cord designed to snap in a big fall). A simpler setup with static rope might give a higher average force, and thus a quicker deceleration to zero, but with a lower peak force.

I will look at the links when I get home from work and see how my theory based solely on this thread stacks up to accepted figures

The general 1D equation, of which F=ma is just a simplification for constant values (or averages), is F(t) = m a(t), or F(t) = m dv(t)/dt, or F(t) = m d2x(t)/dt. All are time-dependent functions. Integrating once with respect to time over limits t=>0 to T and v=>-v0 to 0 (i.e. coming to a halt from a velocity in the opposite direction to the force), we find:

v0 = 1/m (0->T)? F(t) dt

which requires you to specify the form of F(t) (i.e. the force with time) to solve for T (by integrating F(t)).
F(t) could be almost anything - it could be a constant, it could be a extremely high peak at the start or the end, or whatever. All you can know is that the integral of the force with respect to time over the time period 0 to T where a body decelerates from v=v0 to v=0 is equal to its mass times its starting velocity.

All of this remains of little or no relevance because the real situation is complicated and not amenable to trivial modelling.
 

Antwan

Member
It would appear that what I thought would be happening in fact does not and therefore my method is completely meaningless. (A little knowledge is a dangerous thing!)

Plenty of analysis of hydraulic transients with and without shock absorbing measures show the time to peak force and back again to be the same, just lower. Assuming similarities between a solid steel weight and a bendy caver!
 

mikem

Well-known member
The problem with applying the climbing data is the second dynamic body of the belayer is providing quite a lot of that reduction in peak force.

Mike
 

Bob Mehew

Well-known member
mikem said:
The problem with applying the climbing data is the second dynamic body of the belayer is providing quite a lot of that reduction in peak force.
I am unsure about that.  Petzl has provided some useful information in their technical tips area, see

1 https://www.petzl.com/GB/en/Sport/Forces-at-work-in-a-real-fall?ActivityName=Rock-climbing

2 https://www.petzl.com/GB/en/Sport/Influence-of-the-belay-device?ActivityName=Rock-climbing

3 https://www.petzl.com/GB/en/Sport/Consequences-of-poor-rope-drag-management-in-a-common-fall?ActivityName=Rock-climbing

4 https://www.petzl.com/GB/en/Sport/Fall-comparison-with-rigid-human-mass?ActivityName=Rock-climbing

and

5 https://www.petzl.com/GB/en/Sport/Appendix-2--Detail-of-installation-on-two-ropes-with-two-ascenders?ActivityName=Rock-climbing

One bit of it does not make sense to me (see in 4 about the claim of 70%) but what is notable is the drag of crabs can much reduce the forces on a belayer (see in 3) but not the climber or top anchor.  Petzl suggest that the presence of the crabs effectively reduces the length of rope over which the energy of the fall is absorbed, hence raising the peak force on the climber and anchor but reduces the peak force (to zero in one case!) on the belayer. 


 

mikem

Well-known member
Tip 4 lists mainly reasons that are due to the belayer...

The differences in loading on the climber & belayer in Tip 1 are mostly due to the friction on the top krab, if you used a pulley it would be closer to a frictionless system & the loads would be more even. So yes, increased drag does reduce the effective length of rope that the load is spread over. Fortunately (for the maths), SRT situation should be easier to model than the climbing one, but the forces on the body will be greater (for the same length of fall).

& a 7O% increase from body to steel, is less than a 5O% decrease from steel to body...

US rope access standards allow for up to 8kN when using a harness - are they assuming larger workmen, or just factoring in tools carried as well? (Climbers & cavers both generally have a lot more kit hanging off them - in our case is it better to have the bag suspended from the central maillon or the harness?)
 
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