Fall Force Generated

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Fall Force Generated

Postby Hann » Fri Jun 01, 2007 10:04 am

Help me if I'm wrong. Tell me if I'm not.

1N = (kg x m) / 1m per second acceleration.

Safe Working Load of a rope with breaking strain of 22kN = 2kN (10%)

Climber of 70kg (me carrying a trad rack, after a big breakfast)
Falling from 2m above his ‘pro

(70kg x 4m) / 9.8m.s.s
= 28.57N

= 0.2857kN
This can’t be right, can it?

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The Jimmy
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Postby The Jimmy » Fri Jun 01, 2007 10:57 am

mmm no it doesn't look right. I'm putting on my nerdy glasses and putting fresh batteries into my calculator as we speak. Get back to you soon.


Russell Warren
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Postby Russell Warren » Fri Jun 01, 2007 11:01 am

Newtons are a measure of Force
Force = mass x acceleration

In the case of a fall the force generated is a combination of gravitational force i.e. 9.81 x Mass + the deceleration force (which is a function of the stretch in the rope and how dynamic the belay is) x Mass.

To work out the force generated would be quite difficult as you would have to time exactly how long it takes to stop from the moment resistance to falling is felt i.e. more or less from the moment the rope starts to pull tight to when you come to the lowest part of your fall. The problem with this simplistic approach is that you then assume that the deccelaration is a constant i.e. average over the duration of the force applied by the rope. This is not entirely correct for several reasons:

Slippage on the belay device as most belay devices allow slippage during the initial stages of the take. As soon as the belayer locks off the entire system applies more force i.e. the belayers body is lifted off the floor so the entire mass x (gravity + upward acceleration) of the belayer is applied to the belayers end of the rope. The force on the other end exceeds the force require to lift the belayer as there is friction in the remainder of the system.

That said in my opinion the timing method will be close enough to get a good idea of how much force is being applied.

The other thing one needs to know is how far did the person fall before any rope resistance was felt and then one can work out his starting velocity from the equations of motion. v^2=U^2+2*a*s where v = final velocity, U = initial velocity and a = acceleration and S = distance fallen. For this equation a = g =9.81 m/s^2. Therefore V is calculated to be used in the next formula which is V=U + a*t. In this formula t is the time measured as discussed and U = V from the last calculation and V = 0m/s as you are stopped by the rope. From that you can solve for a. After that it is easy. Add the calculated a to 9.81 giving both the same sign to get a value greater than 9.81 and multiply by your mass to get the force exerted on your top piece of gear.

As you can see I was didn't have much to do at work. My apologies to those who I have bored to tears.

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Postby Jonah » Fri Jun 01, 2007 1:02 pm

Hey Russel help me with conceptualising what you just said.. something confuses me :?

The one contributing force is the gravitational force (9.81 x Mass): A rope would experience this force or tension if one had to simply hang on the rope.

Then the other force is the deceleration force: A mass with initial velocity (V) is brought to final velocity (U) with a deceleration dependant on time or distance between V and U.

It makes sense that you would add these two forces, because the total force is obviously greater than just the gravitational force. But I'm struggling to conceptualise the force system from fudamental principles..

The gravitational force is acting downwards, right?
Surely the deceleration force acts upwards, because its deceleration, in which case it should cancel out some of the gravitational force...(which it shouldn't)..?

Then if I look at it another way it makes sense that the only contributing force is the deceleration force: The rope simply brings a mass with initial velocity to zero velocity in time t. At (t-1) the rope had zero tension and at (t+1) the rope has only the gravitational force acting on it. But it seems like this gravitational force comes into effect after the deceleration force, when the rope is hanging still, as if the force got changed from deceleration to gravitation over time. I mean when you're considering the force required to make a rope break, it makes sense that you would only consider the deceleration force, since thats the force the rope experiences during the fall when the rope is likely to break (during t).

But as I said it makes sense and I would presume that you add the two forces. Could you help me understand exactly how the forces are added?

That would ease a lot of tension from my mind!!

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Postby BAbycoat » Fri Jun 01, 2007 1:26 pm

Think of it this way:

- gravity pulls you down the whole time
- the rope counter-acts for only part of the time
- therefore, the rope has to work extra-hard when it is working.

So if we start falling at t_1, the rope starts catching us at t_2, and the deceleration is complete at t_3, then the average force exerted throughout the deceleration is

Weight (in kN) x (t_3 - t_1) / (t_3 - t_2).

See why a dynamic belay helps?

(all this is modulo all Russell's caveats, and more).

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Postby ant » Fri Jun 01, 2007 3:21 pm

Without getting my textbooks out - yes Newtons are a measure of force.
1N is the force which 100g exerts in a static environment.
When you weigh yourself you're actually 'forcing' yourself, and calculating mass based on the assumption that gravity is a constant 9.81 or whatever.
(ie so when someone says 'May the Force be With You!' - they actually mean 'I hope you get fat and fall off a cliff)

<<insert lots of maths here >>
Force x time spent falling = Energy
The rope is designed to desipate this kinetic energy over enough distance (the 'boing') such that the force it exerts doesn't cause the rope to break or pop your legs out of the bottom of your harness.

the end result is that climbing ropes are designed to: (and this from memory so tell someone who cares if I'm slightly wrong)

be able to absorb the impact of a fall to within acceptable level of shock, using, at the minimum, half the length of rope as is the length of the fall.
(Hence the whole concern over fall factors, which I'm sure you can easily google)
re acceptable - I think acceptable is 11kN, but the built in breaking strain is still 22kN - (I stand to correction on this point)

Out of interest - the reason that 22kN was chosen as a standard:
When they were inventing climbing gear - the British Military had decided that 11kN was the maximum force that a paratrooper could withstand (when his parachute opened) without extreemly severe consequences on a par with hitting the ground. (ie your legs popping out of the bottom of your harness)
In climbing scenarios we often set up systems whereby the force is doubled due to a geared pully effect (ie anchor has weight of climber AND belayer) - therefore they decided that 2 x 11kN = 22kN would be a good benchmark without going over-board.

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Postby stillemans » Fri Jun 01, 2007 3:30 pm

Damn, this feels like a physics test.

If you assume that the rope is like a spring then the force that it applies on you (the tension in the rope) is proportional to the stretch multiplied by some spring constant:
F = k * d

The stretch will be a maximum when your velocity reaches zero (at the bottom of your fall) i.e. The maximum tension in the rope.

Before 4m your accelaration is 9.81m/s^2. Therefor your velocity at the point when rope starts to pull tight (at 4m) is 8.86m/s (if I calculated correctly)

After 4m your acceleration is 9.81 - (k/m)*d. Your starting velocity is 8.86m/s and your final velocity is zero so:
0 = (8.86)^2 + 2*d*(9.81 - (k/m)*d)

So lets assume the stretch in the rope is 0.5m then k=12363 kN/m

So the maximum force exerted on you by the rope would be 6.2kN

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The Jimmy
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Postby The Jimmy » Fri Jun 01, 2007 4:02 pm

Jonah you are correct in saying that the gravitational force acts downward and the deceleration force brought on by the rope acts upward, but his is when you use the CLIMBER as your point of reference.

In this case we want to determine the forces on the highest point of protection, in which case you will have a force acting downward on it brought on by the mass (in kg) of the falling climber x (g=9.81 + a=how russel calculated it)

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Postby Magnus » Fri Jun 01, 2007 5:10 pm

This is how I see it. Energy from fall gets put into the rope. That means:

mass x g x fall distance = avg tention in rope x rope stretch

The key factor is therefore the stretch, which depends on the amount of rope out and the \"stretch factor\" of your rope.

So, if you have 7.6% stretch and fall a full rope length (60m rope):
Tention in rope = (70kg x 9.81 x 60m) / (7.6% x 60m) = 9 kN = serious wedgie.

The amount of tention on the gear then depends on how much of the tention is absorbed by the belayer (Energy gets past along the system). If the belayer is static, or worse, if you have a lot of rope-drag, most of the tention goes into the gear. Another reason to have long runners when climbing on trad.

Bottom line, avoid rope drag and minimise the \"Fall factor\" by placing gear low down on a route.

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