Articles in English

Drop test physics I

Tuomas Pöysti 2021

I don’t have a drop test tower. I don’t even have a proper lab with solid anchors. That’s ok, because I don’t have a real need for professional test setups, either. I play with my load cells to learn. Sometimes the sweetest bits of learning take place when figuring out how to cope with the suboptimal lab settings!

This article series is a little something I have learned recently. The first article is actually not that recent stuff for me, it has been a looong time since I was in high school. The following ones will get into the more recent stuff, and you can skip the intro if this is all clear to you.

Falling onto a cowstail

The thing I was interested in the beginning was the physics of cowstails misused as fall protection. What happens when a fall is arrested by a cowstail, where does the energy go and in what manner and so on.

This DMM’s article about slings seems to have been an active topic among many climbers for a decade now. It’s not (completely) the article’s fault, but sometimes it gets interpreted as “if you fall onto your static cowstail, you may rip a bolt off”. I don’t know if it is possible for that to actually happen, but I think the video does not suggest anything like that. Even about generating forces high enough to break the cowstail.

Force and acceleration

There will be some basic Newtonian physics introduced here. Maybe the most well-known piece of it is

F = ma


a = F/m

where F is force, m is mass and a is acceleration. This states that if force F is applied to an object at rest, the acceleration of the object will become the ratio of the force and the object’s mass.

If an object has acceleration of 3 m/s^2, every second it’s velocity will increase by three meters per second. Acceleration is the chance in velocity divided by the time the change took.

a = dv/dt

Acceleration means any change in velocity – if you ask a physician, deceleration is acceleration with a minus sign. This means that if an object with mass m and velocity v has to be stopped in t seconds, force F is needed, where F is

F = ma = m * dv/dt = m*v/t

Force and energy

Breaking a cowstail or anything always needs energy. Force is a mechanical way of transmitting energy. Mechanical energy is often called work. Their relation is roughly:

When a force moves an object, the energy involved is the product of the force and the disrtance the object moves, or displacement.

E = Fx

For example, if a winch applies a constant force of 5 kN to a car, pulling the car upwards a steady hill for a distance of 4 meters, the work done by the winch is 20 kJ. N and J are the SI units for force (newton, N) and energy or work (joule, J), and k is for kilo and it means simply 1000. 5 kN * 4 m = 20 kJ.

On the other hand, the car has done the same work to resist the efforts of the winch, to “brake” it. The work is done only once and energy involved is 20 kJ, but depending on the point of view, it can be seen in different ways.

Energy conservation

Energy is a very important way of measuring things in physics, because it tends to be conserved. For example, if a roller coaster car is pushed uphill, as we learned, energy F * x is stored somewhere, if force F pushed and the hill was x meters long. The energy is stored as gravitational potential energy of the roller coaster car: the car now has potential to roll the same or another hill downwards pulled by gravity.

The nice thing is that even if the energy changes it’s form, the amount is conserved. If the roller coaster car is let roll downhill, the potential energy is transformed into kinetic energy. That is, we can precisely calculate the kinetic energy and thus velocity of the car if we know how much downwards it has rolled from the starting point.

Energy losses

Or we could, if we knew all the complex ways of energy transformations during the experiment. The potential energy won’t simply transfer into kinetic energy, but there are also losses in form of friction in the wheels and axles of the car, air friction etc. These all follow the law of energy conservation, but we cannot keep track on them. In general, such losses transform energy into heat (the form of energy that tends to make material warmer).

Because of energy losses, the roller coaster is not a conservative system in the sense we’d like to: the car will not ever run on top of equally high next hill without some kind of extra energy – you know the clickety-click-thing that gives the extra push. On the other hand, it is quite convenient that brakes can be used to stop the cars and the energy simply vanishes into air, which just gets a bit warmer.


In physics, spring is a device that can be used to store potential energy (work). In real life, everything is a spring. Some springs are stiffer than other (a steel bar vs. rubber band) and some do not bend much before breaking (stone vs wood), but in their own league, they are all springs.

The most important point is that nothing is absolutely rigid; everything bends – the greater the bending force the more, but some amount anyway.

It is actually even hard to imagine what an absolutely rigid object might be like, or especially two of them. When two real life objects are hit together, they act like springs. In addition, they might also break in pieces, but at least they store some of the kinetic energy into a special kind of potential energy, potential energy of a spring. Then, a bit later, the energy may turn back to kinetic energy, throwing the objects or their pieces back where they came from. Cue balls do exactly this, even though their deformation is impossible to observe by naked eye.


Keeping this very brief: Damping is a feature of oscillating systems which reduces the oscillating motion. Whereas a spring stores work into potential energy, a damper turns work into heat.

The most important point is, again, that all springs are also partly dampers. No spring is absolutely conservative, but have some kind of internal friction that dissipates part of the work.

Work done by a fall arrester

Now we are armed with basic physics to do a little calculation. A climber who weighs 82 kg (me) is lifted one meter upwards, so he has a (gravitational) potential energy of

E = 82 kg * 9.81 m/s^2 * 1 m = 804 J.

Actually, the climber does not have to be lifted in the beginning, as long as it is possible for him to fall one meter – it is up to us to decide the “zero level”.

The thing we cannot avoid is: if the climber moves one meter down, this amount of energy has to go somewhere. No matter if he falls or slides or climbs or takes an elevator. If he falls one meter, he will have a kinetic energy of (a little less than) 804 J. If he climbs down, his muscles need to deliver “braking” work equal to about the same. If he falls and is then stopped or if he uses a descender to start with, the fall arrest or braking device must put the 804 J somewhere else.

By the way, 804 J is quite a large amount of heat. Joule equals to Watt * second

J = W * s

and thus 804 J equals to a very bright 15 W led light shining for almost a minute. It’s easy to miss this point, since friction eats these kinds energy amounts all the time, usually just making life a bit harder.

Fall arrester forces

Usually we tend to start with forces when talking about dynamic ropes, fall factors and so on. What’s even worse, we stay there. But as I said earlier, to break something (a lanyard or a human spine, for example), energy is always needed. Plus, all objects are springs, some worse than others – and springs gain potential energy when stretched.

The thing is, when a fall arrest lanyard stops a fall, it (or something else) has to stretch or deform in general. If it does not,

  1. The kinetic energy cannot go anywhere
  2. The acceleration of the object is infinite

The latter one follows from the fact that if there’s no deformation, there’s zero amount of time between the points in time when the fall arrest point starts to decelerate the object and when it is halted.

Luckily every real life object is a spring, and besides that, also some kind of a damper. For example, an EN 892 climbing rope is very stretchy, but even more “dampy”. Its viscoelastic behavior keeps the falling leader from bouncing back after a fall and oscillating up and down until lowered. Imagine what it would be like to have a trampoline as bouldering mat!

Now that the rope is not a spring that temporarily stores the kinetic energy as potential energy and then returns it, where does the energy go? Probably mostly to a slight increase of the rope’s temperature.

How about an actual fall arrest lanyard, then? They come with shock absorbers. Everything is a spring, but shock absorbers are especially bad springs, since they are designed to be dampers. As the load exceeds certain limit, the shock absorber starts to tear open and extend, without any intent to bounce back to it’s original lenght as a decent spring would do. The energy is transferred into heat, probably by friction where the stitches tear open.

If the 804 J kinetic energy from an earlier example is to be absorbed, and the shock absorber tears at constant force of 3 kN, we can calculate that the required extension is

E = Fx => x = E/F = 804J/3kN = 0.27m.


Just one last thing so that I can go straight into business in the following texts. Impulse and momentum are important parts of Newtonian physics.

Momentum p

p = mv

is the product of object’s mass and velocity. It is a bit similar to kinetic energy, but completely different, and useful in different situations.

Impulse J

J = Ft

is, at it’s simplest form, the product of a force F and the time the force acts t.

One of the useful aspects of these is that if force F acts on an object with mass m for t seconds, that is, giving the object impulse J, we know that the change in the objects momentum will be equal to J.

For example, if we happened to know that the brakes of a roller coaster applied a decelerating force of 1 kN to a car, but had no idea what distance the car moved during that, we’d be clueless with the energy method.

Instead, if we had a stopwatch and could measure that the force acted for five seconds, we could calculate the impulse:

J = 5s * 1kN = 5kNs

which is exactly the change in the car’s momentum. If we knew the car’s weight is 500 kg, we could calculate the change in it’s velocity:

dv = dp/m = 5kNs/500kg = 10m/s.

In part two, we actually use these things to do something!