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# Drop test physics II

Tuomas Pöysti 2021

This part is a bit of a warmup. If you need a refresh on physics for this, see the first part.

I am one of the “never fall onto your cowstail” type of instructors. Of course we all know you should not do that, but some of us tend to put more emphasis on that and tell horror stories about the dangers of inadequate shock absorption.

I’m not expecting to find anything to debunk them. But it is good to base your actions (and especially your teaching) on facts, and facts can be found using science. Confirming the current convention is not a bad scientific finding. By the way, if something else is to be found, it is not time to abandon the conventions, but to question them.

So, just to play with the idea, I hang my enForcer load cell on a semi rigid anchor, clipped into it using about 50 cm long CT cowstail and took some falls on it. I started with very small ones, measured the maximum force using 500Hz sampling rate, and went harder until the peak loads exceeded 3,5kN. It felt like a lead fall, but I thought it might be better to check tomorrow if my back still thought the same. I estimated the harderst drops were a bit over 20 cm.

## Force, not displacement

I recorded some of the tests to play with the data and see if I can learn something – or more probably figure out what I should study properly with a better setting. I might have had a stupid, absent-minded idea about calculating energies involved.

As we know, energy is (among other things) the product of force and displacement. A given amount of energy may result from different forces and how they apply. For example, the gray areas representing energy might have equal widths (displacement) and area (energy), yet their heights (the maximum force are very different.

I thought I might learn something about the “shape” of typical fall arrest event. Even without testing, it is easy to guess that in real life the force is not constant over the displacement. It is also easy to see that constant force would be the lowest possible one. An ideal tearing shock absorber might produce something like the left example in the picture above. The one on the right could represent the potential energy of a spring. Spring is notorious for having the force proportional to displacement. From basic geometry we know that the latter is twice the height (i.e. the twice the maximum force) of the former if E1=E2.

Of course I had to forget going directly to such calculations, since I did not have the data about displacement. No measurement of the location of my body’s center of gravity whatsoever. The enForcer data is, of course, just force vs. time.

We engineers tend to be conditioned to use the energy principle, which is a powerful tool and quickly develops into a first guess in every situation. It took me a while to shake that off and remember momentum and impulse. It is possible to calculate impulse using the data!

Be sure to remember this while reading ahead: the x axis is now time, not displacement, and the area below the curve is not energy but impulse!

But first I modified the data a bit: I subtracted my weight (hard to keep it a secret at this point) from the force. The total force is

F = ma + mg

where ma is the portion of the force that is generated by accelerating the mass of my body and mg is weight, the part that is generated by the Earth’s gravity acting on the mass of my body. g is the gravitational acceleration, which has standard value of 9.81m/s^2 and m is my mass 82 kg, so the weight is 804.4 N. This lefts only the mg part, and results in the final force nicely oscillating around x axis:

Now, I figured the first impulse that arrests the fall is the area highlighted in red. It starts when the force equivalent to my weight is accumulated and stops again when the same force is left. I have to admit, I cannot give better justification right now.

The fancy name for the area is integral. But in this case the calculation is far from fancy; it is the simplest version of numerical integration. The data is sampled at 500 Hz i.e. every 0.002 seconds, so we can view each sample representing a small slice of an impulse with value 0.002s * F, whatever the F happens to be in each case. We just need to sum these slices up, something a spreadsheet application manages perfectly.

The result was 0.2 kNs. That is, the system changed my momentum by this same amount. What does it mean? If we assume that I was stopped at the end of the event, what was my initial velocity? Let’s see. Momentum

p = mv => v = p/m = 0.2kNs/82kg = 2.4 m/s.

Sounds reasonable, but it is hard to be convinced, since velocity is not something you measure or even estimate in everyday life, in this kind of situations. What if we assumed this velocity is the result of a free fall (as it actually was), and calculate the corresponding kinetic energy?

E = 1/2 * mv^2 = 1/2 * 82kg * (2.4m/s)^2 = 236,2 J

Now we can calculate which height corresponds to this energy, if we assume it was originally gravitational potential energy:

E = mgh => h = E/mg = 236,2J/(82kg*9.81m/s^2) = 0.29 m.

That is a bit higher than I estimated, but on the other hand, we made an unjustified assumption on the way!

## To arrest means to stop

We assumed that the impulse represented by the red area just stopped my motion. But it didn’t; the force curve suggests my body made at least two oscillations (that is, I bounced up and down twice), although in a quickly decaying way. The “red” impulse did not just bring my velocity to zero, it actually went all the way to the positive side, sending me back upwards (the initial momentum was negative, downwards).

The “blue” impulse does exactly the same. After that, though, the oscillation is so damped that we can quite safely assume the oscillation has completely stopped.

Now, I’ll make another brave assumption to save us from the one before: Let’s assume my body (or it’s center of gravity to be exact) returned down at the same speed around point 3 that it was sent up with around point two – that is, no energy was lost while I took the little bump upwards.

This means that the blue impulse finally eats the rest of my momentum, the excess part from the red one. So if we subtract the blue area from the red one, we should end up with the impulse that was just enough to stop me.

Using the same numerical integration method, the blue impulse has value 0.034 kNs. Red minus blue: 0.2 kNs – 0.034 kNs = 0.166 kNs.

Repeating the velocity – kinetic energy – potential energy trick, we end up with free fall of 0.21 m. Nice, exactly what I estimated while doing the drops!

## Exploring further

This has been just a warmup excercise, and we did not yet achieve very tangible results. I think it is quite rewarding to be able to credibly(?) dig up secondary data such as free fall distance, using the simple data that contains just force values vs time.

If you like what I have done here at all, I’m quite sure you will love the next part!