craniology, part 1.5: the force of a falling object

Thanks to Amanda’s super Sunday Blog Roundup, I found this really interesting article from Eventing Nation about what a helmet will and won’t do to protect you in a fall.  It was incredibly fascinating and well worth the read, but left me with a couple of questions:

First, what are the forces at which the helmets in question were tested at when they failed the impact tests?

Second, what are the forces that are frequently experienced by an equestrian falling off of their horse?

gravity can be a real bitch

I couldn’t find any answers to the first question, which is an important one to me.  If the authors were testing helmets at 10,000 gs (G-forces) and they fail to protect the head against two impacts at that force well… that’s unrealistic.  But if the authors were testing helmets at 350 gs (a force easily acquired by an object falling from 3.5 meters with a stopping distance of only 1 cm), then that’s a much more reasonable test.  Fear not, I’ve emailed the main author of the study to get a little more information about how they conducted their research.  I’ll update you when I find out.  (I did discover that they use a really funny machine to test these things!)

To the second question, the answer is: it turns out it depends.  The basic calculation of g-forces involves the rate of deceleration and the force of gravity.  Rate of deceleration can be calculated by using velocity and stopping distance.  And therein lies the rub: stopping distance is highly variable on difference surfaces.  For example, when you drop your phone on asphalt, the stopping distance is basically just the crunch factor of your phone — as little as 1-2 mm in some cases.

The other complication is that there are multiple force vectors at work when you fall off a horse.  Not only do you fall from some height to the ground, but you also stop your forward motion — motion that can vary from 300 mpm (slow canter) to 600 mpm (Rolex gallop) depending on the speed you’re moving.

lets not forget all the physics lessons Murray has given me

I took physics (twice), though, and was pretty confident that I could figure out the force vectors, so set about determining the g forces likely experienced by helmets and riders falling at different speeds.  G forces are a convenient unit to calculate this in, since they are the same no matter the mass of the falling object — and that’s the unit in which the “brain trauma” numbers were reported.

The set up

You can use this handy-dandy fall force calculator from VaultCanada if you’d like to calculate g-forces of objects falling straight down.  It doesn’t add it in horizontal movement, but it is a bit of fun to play around with.

Then there are the basic equations.  I didn’t actually consult a physics text book, but I did use the VaultCanada site (linked above), the physics classroom, and the engineering toolbox.  Pertinent equations are below.

 

Pythagoras will help me with that last one

There are a couple of other caveats here.  Because I was using physics equations, they aren’t taking into account any of those pesky real-life forces like friction (ground or air), the oddity of the human/horse shape, what part of the body hits the ground first, etc. etc.  We are also not taking into account the fact that different parts of the body are likely to encounter different amounts of force by landing at different times.  Just assume that the hypothetical human in these equations always lands directly on their head.  So we’re working with a frictionless, lawn-darting, spherical humanoid riding in a vacuum.  You know.  A really realistic one.

Also, I could be wrong. There aren’t any internet resources that confirmed my plan to add these forces using the Pythagorean theorem.  I bulled ahead regardless.

Also, force vectors for helmet calculation will be pointing in toward the object.  Because what we are interested in is the force the surface is exerting on the object (one’s head, for example) during the stop*, so the force vectors should be reversed from the above image.  It doesn’t affect the calculations.

*You know that old joke about cliff jumping.  It’s not the long drop that kills you; it’s the sudden stop at the end.

With that information, let’s answer a few of my burning questions about forces and riding!

We’ll start with an easy one.  If you drop the helmet on the ground, are you really voiding the protective abilities of your helmet?

Let’s say you’re about my height (1.5 meters), and drop your helmet from your hands (about 1 meter high) on to arena sand, which compresses about 2cm when the helmet hits the ground.

100 cm / 2 cm = 50 gs of force

If the stopping distance decreases to 0.5 cm (5 mm)

100 cm / 0.5 cm = 200 gs of force

I don’t actually know how much force the outside of a helmet can withstand.  But 200 gs is well below what a helmet is supposed to protect you against (300 gs), and likely below what the helmets are tested at.  I’m not going to make any declarations about the safety of a helmet after being dropped on anything but arena sand — which I don’t think is the end of the world.  (When some butterfingers drops her helmet repeatedly on the barn aisle though….)

More pertinent to humans, if someone falls off of a horse onto arena sand, what are the forces experienced by their helmet?  And by their head?

First, what height are they falling from?  Let’s say we have a 16.2 hand horse, and the human’s head is 2.5 feet above that height — like a medium-height person sitting on a relatively tall horse.  That’s 241.2 centimeters high.  Let’s stick with that 2cm stopping distance again.  Without any horizontal motion (i.e. horse is standing still and our spherical rider just falls off the side):

241.2 cm / 2 cm = 120.6 gs of force — a not insubstantial amount of force, but not enough to for sure cause a serious brain injury

If the helmet foam compresses by 2mm during the impact, then the head in question experiences a slightly different amount of force

241.2 cm / 2.2cm = 109.6 gs of force — a pretty awesome decrease, thanks Charles Owen!

What if you fall off at the peak of a fence?  My head was about 292 cm in the air at the peak of that charming 2’6″ fence above (I used a really specific measuring system called PowerPoint), so if we go with our previously calculated stopping distances

292 cm / 2 cm = 146 gs
292 cm / 4 cm = 73 gs

Okay, let’s talk horses in motion.

In stadium, posted speeds vary between 300 and 400 mpm — I think? I’ve never really paid attention except when writing up programs — which is well within the range of a canter.  Let’s say that rider and pony have an unfortunate parting of ways at 350 mpm involving a dirty stop, and our lawn-darting, spherical rider goes flying over pony’s head and on to the ground, where she skids for 30 cm (1 foot) in arena sand that compresses to 4 cm (a little more cushy than the aforementioned arena sand).

vertical forces = 241.2 cm / 4 cm = 60.3 gs
horizontal forces = (5.8 mps)^2 / (2 * 0.3 m * 9.8 m/s/s) = 5.7 gs
Pythagoras says: combined forces = 60.57 gs

That’s really a fairly gentle fall, all things considered — barely more significant than falling off your stationary horse.

What if we jack up the speed to 520 mpm?

vertical forces = 60.3 gs (same as above)
horizontal forces = (8.76 mps)^2 / (2 * 0.3 m * 9.8 m/s/s) = 38 gs
Pythagoras says: combined forces = 71.44 gs

All in all, it seems like falling in soft, squishy footing in stadium (or dressage, I guess) isn’t all that awful.  4 cm isn’t even that deep of an impact crater.

Cross country is where things get sketchy.

Cross country footing is nowhere near as squishy as stadium footing, so I’m conservatively estimating a 1 cm vertical stopping distance.  This drastically increases the vertical forces right off the bat.

vertical forces = 241.2 cm / 1 cm = 241.2 gs

If our spherical rider falls off of her horse traveling at 400 mpm, and stops in a distance of 10cm (skidding), then

horizontal forces = (6.67 mps)^2 / (2 * 0.1 m * 9.8 m/s/s) = 22 gs
Pythagoras says: combined forces = 242.26 gs

If said rider is traveling at 600 mpm and has the same stopping distance (10 cm), it looks worse

vertical forces = 241.2 gs
horizontal forces = (10 mps)^2 / (2 * 0.1 m * 9.8 m/s/s) = 51 gs
Pythagoras says: combined forces = 247 gs

Let’s say that the helmet our hypothetical rider is wearing compresses by 2mm (0.2 cm) during the impact, as above.  Then we get

vertical forces = 241.2 / 1.2 = 201 gs
horizontal forces = (10 mps)^2 / (2 * 0.102 m * 9.8 m/s/s) = 38 gs
Pythagoras says: combined forces = 208 gs

Once again, that’s an appreciable decrease in force caused by only a 2mm compression within the helmet.  Helmets are cool, man.

I ran out of fall-adjacent media

Just for shits, here’s what happens if someone traveling at Rolex speeds falls and has a stopping distance of only 1cm (maybe they hit a fence, I dunno)

vertical forces = 214.2 gs
horizontal forces = (10 m/s)^2 / (2 * 0.01 m * 9.8 m/s/s) = 510 gs
Pythagoras says: combined forces = 563 gs

Helmet compression of 2mm decreases this impact force to 395 gs, which is a really appreciable decrease.  As if I needed more reason to wear a helmet — gotta protect that money maker.

These estimates are probably not all that realistic, given the many caveats I had to list above.  I’m also not taking into account multiple concussions, bouncing, being kicked by horses, hitting other objects… Really, there’s a lot more that could go in to this model to make it more accurate (though as we know, the more predictors you add to a model the less predictive it actually becomes — statistics is a cruel mistress).  But it was an interesting exercise in practicing physics and geometry again, and gave me some idea of the serious forces we put our heads under.  Thank you, helmet, for protecting my melon so often!

(Now I have other questions. What forces are needed to break certain bones, for example?!)