When you consider the energy generated (and not loads), then a fall of three metres is more dangerous, since you would have more kinetic energy at the end of the fall and that means that you can smash into the rock face at a higher speed! The fall is therefore just as "hard" However this only applies the force, that impacts on the rope. So a fall of 3 metres with 9 metres of payout rope has the same fall factor as a fall of one metre with three metres of payout rope. It is interesting to note that the fall factor does not change when the fall height and payout rope are changed in the same proportion. A shorter rope at the same fall height is more uncomfortable than a longer one. The same applies when shortening the payout rope. ![]() When the payout rope is four metres, a fall of two metres is more uncomfortable than a fall of one metre. This makes sense, when you think about the numbers. The fall factor is a measure of the "force" of the fall. You must minimise the fall factor, since the other values are set. If you want to minimise the stopping force, there is only one option. This in turn arises from the ratio of the fall height h to the length of the paid out rope L: Here m signifies the mass of the climber, g gravity, E the elasticity modulus (a value dependent on the material), A the rope diameter and f is the so-called fall factor. The maximum force which arises in a fall is: The lower the value, the lower the impact during fall arrest held and the smaller the force on the climber. ![]() This impact force is a quality feature of a dynamic rope. The force generated at the moment of maximum rope elongation, is known as impact force. When a fall is stopped, the body of the climber absorbs the energy that is generated from the rope being stretched and the movement of the belayer. In the following information, we want to clarify for you how these forces arise and what the numbers on the ropes mean. This is because, although ropes can withstand forces of several kilonewtons (kN) relatively problem free, people have substantial problems with these loads. ![]() But this does not mean, that you can't be injured in a fall. Modern ropes are so stable, that under normal conditions and with careful use they cannot break. Then you can solve this equation system to get t and a needed for the last part of calculations: F=m*a.On all ropes you will find information on impact force and rope elongation. Which one is correct ?Īnd here you assume that vk=0, s - displacement of the center of the mass after the crash. The problem is that they have different equation for the same thing:Īs you can see in the first one they assume that W=(1/2)*F*s and in the second one it is W=F*d (d is the same as s before). So I began searching and found 2 interesting articles. ![]() very complicated methods used in car crash mechanics, I don’t want to go so far with this very basic method (and results seem to be incorrect as it’s hard to assume the impact duration time) : F=m*a, a=v/t I’ve found some examples of such calculations in FEA applications but they can be divided into 2 groups: Let’s say you model car crash or some falling object impact and don’t want to use explicit dynamics analysis in which objects will actually crash. I would like to learn how to calculate impact force for use as a static load in FEA simulation.
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