## Brain Stretching of a Spinning Skater

So I was wasting time on reddit the other day (happens all too often) where I came across the following video of a girl practicing spinning for figure skating using an off-ice apparatus:

You can see the post I read here, there were a lot of comments about how spinning that fast can’t possibly be good for the girl’s brain. My initial response was that since her head is at the center of the axis of rotation there really isn’t any force on it, but technically that’s not true: if you have something soft and squishy and you spin it really fast, it will bulge out in the middle and compress slightly at the top and bottom along the axis of rotation:

Note: thanks to thingiverse for the brain model, and blender for the rendering.

It probably doesn’t stretch out this much. But can we estimate how much it actually does stretch out when someone’s head is spinning around this fast? How would we estimate this? There are three pieces of information we need:

1. How fast the girl is spinning
2. Physical parameters of the brain
3. A model or equation to calculate/estimate the deformation of the brain while its spinning

## Speed of Spinning Girl

How can we figure out how quickly the girl is spinning? This should be fairly straightforward with some video analysis. Analyzing the video the way we need to though, requires us to be able to scan the video one frame at a time, and have a timestamp for each frame. Then we can start at the point in the video where the skater is spinning the fastest, walk through several rotations, and count the time elapsed by comparing timestamps. Then the rotation speed is simply # of rotations / time elapsed.

I don’t have any expensive video editing/analysis software, but a quick google search for a free open source software for basic video editing led me to Avidemux.

The software lets me step through one frame at a time with the Left and Right keys, and gives me a timestamp at the bottom. More than sufficient for my needs. Starting at 4.733s and ending at 7.833s, she goes through 14 rotations. That comes out to 4.52 revolutions/s.

## Mechanical Properties of the Brain

Doing a quick google search for the mechanical properties of the brain almost gave me too much information, there were hundreds of academic papers discussing it. I ended up using this paper: Biomechanical Modeling of Brain Soft Tissues for Medical Applications. It gives values across a range of values from other studies, but I ended up going with a lower-end value (since that will give a larger deformation) of Young’s modulus E = 2 kPa, and Poisson’s ratio of ν = 0.45. For density we’ll just assume it’s about the same as water, since it’s soft tissue and our body is after all mostly water, so that’s a density of 1000 kg/m3.

## Deformation Model

This was actually a bit harder for me to find. Searching for “deformation of rotating elastic sphere” and other similar search terms got me lots of papers and pages showing how the earth (or some other planet) bulges at the center due to its spin. However in that case the restoring force trying to return it back to being spherical is gravity, while in our case for the spinning brain it’s simply the object’s own elasticity.

Eventually I was able to find an online solid mechanics textbook that had the derivation for the elastic deformation of a spinning circular plate. Sure, it’s not quite the same as a sphere, but for our intents and purposes we can approximate the brain as a cylinder as much as we can a sphere. For the displacement of the cylinder/disc it gives the following equation:

$$\mathbf{u}=(1-\nu) \frac{\rho \omega^2}{8E} \{ (3+\nu)a^2 r – (1+\nu)r^3 \}\mathbf{e}_r – \nu \frac{\rho \omega^2 z}{8E} \{ 2(3+\nu)a^2 – (3\nu +2)r^2 \}\mathbf{e}_z$$

That’s given in vector form, so since we’re only interested in deformation in the r direction at at the very edge where $r=a$, we can simplify it to the following:

$$u_r(a) = \frac{\rho \omega^2 a^3}{4E} (1-\nu)(3-\nu)$$

The only parameter we’re missing is the radius of our cylinder. A quick google search tells me that the average human brain has a volume of 1400cm3. If we approximate the brain as a cylinder with equal height and diameter, we can calculate the radius using the following derivation:

$$V = \pi r^2 h = \pi r^2 (2r) = 2 \pi r^3$$

$$r = \left( \frac{V}{2 \pi} \right)^{1/3} = \left( \frac{1400 \text{cm}^3}{2 \pi} \right)^{1/3} = 6.0 \text{cm}$$

We also need to convert our rotation speed to radians/s, which is just multiplying rotations/s by $2 \pi$ which gives us $\omega = 28.38 \text{/s}$.

So our final calculation is:

$$u_r(a) = \frac{(1000 \text{kg}/\text{m}^3) (28.38 \text{/s})^2 (0.06 \text{m})^3}{4 (2000 \text{Pa})} (1-0.45)(3-0.45)=0.03 \text{m} = 3.0 \text{cm}$$

Well, that is a very large displacement, our brain is stretching from a 12cm diameter to 18cm! Now there are some problems that this result indicates:

• Our model assumed a simple perfectly elastic solid. In reality this assumption only holds for small deformations, after a certain amount of displacement it ceases to be elastic (i.e. if you relieve the stress then it returns to its original shape) and becomes plastic deformation (i.e. the shape permanently changes), etc.
• The brain is encased in the cranial cavity, and doesn’t have very much room to stretch. The gap between the brain and the cranium is 0.4mm to 7mm depending on the location.

Now if the maximum displacement had been < 0.1mm or so (which is what I assumed) it wouldn’t really make a difference, those assumptions would have been fine. But with this kind of a predicted large displacement we can’t use such a simplified model any more.

But wait! Maybe it isn’t the model itself that’s bad, but the boundary conditions! When solving for the equation for $u_r(a)$ above, we make the assumption that the radial stress is zero at $r=a$. However if the brain is constrained inside the cranial cavity so that it can’t expand, then a better boundary condition would be that the displacement $u_r(a)=0$. This won’t give us much (if any) displacement inside the brain, but we can still calculate the stress inside the brain material.

$$\frac{\partial}{\partial r}\left( \frac{1}{r} \frac{\partial}{\partial r} (ru) \right)=-\frac{1-\nu^2}{E}\rho \omega^2 r$$

Integrating twice gives us the displacement $u$ with the integration constants $A$ and $B$:

$$u=-\frac{1-\nu^2}{8E}\rho \omega^2 r^3 + Ar + \frac{B}{r}$$

The solution has to be finite at $r=0$, so that means $B=0$. To solve for $A$ we set the condition that the displacement has to zero at the edge, or $u(a)=0$. That gives us a formula for the displacement:

$$u(r)=\frac{1-\nu^2}{8E}\rho \omega^2 (a^2 r-r^3)$$

So our cylinder model of a brain is displacing a maximum of about 3mm around halfway between the center and the edge. A more general dimensionless version of the plot is here:

This is a more general version of the solution which is valid for any size cylinder with any known parameters of Young’s modulus $E$, density $\rho$, rotation speed $\omega$, radius $a$, and Poisson’s radio $\nu$.

Another thing we can calculate is how much stress there is inside our cylindrical brain as a function of radius.

$$\sigma_{rr}=\frac{E}{1-\nu^2}\left( \frac{\partial u}{\partial r} + \nu \frac{u}{r} \right)$$

Substituting in our derived function for $u$ we can simplify it to the following function:

$$\sigma_{rr}=\frac{\rho \omega^2}{8}\left( a^2(1+\nu) – r^2(3+\nu) \right)$$

This shows that the ‘brain’ is stretching in the center (positive stress), and being compressed at the edge (negative stress).

I tried to look up how much mechanical stress a human brain can withstand, but it’s almost impossible to filter out psychological stress from search results, and everything else is about sudden shocks to the brain: concussions, CTE, etc.

I found a few papers talking about different types of brain injury:

1. S. Kleiven, X. Li, A. Eriksson, and N. Lynøe, “Does High-Magnitude Centripetal Force and Abrupt Shift in Tangential Acceleration Explain High Risk of Subdural Hemorrhage?,” Neurotrauma Reports, vol. 3, no. 1, pp. 248–249, Jan. 2022, doi: 10.1089/neur.2022.0025.
2. B. D. Stemper et al., “Head Rotational Acceleration Characteristics Influence Behavioral and Diffusion Tensor Imaging Outcomes Following Concussion,” Ann Biomed Eng, vol. 43, no. 5, pp. 1071–1088, May 2015, doi: 10.1007/s10439-014-1171-9.
3. S. Kleiven, “Why Most Traumatic Brain Injuries are Not Caused by Linear Acceleration but Skull Fractures are,” Frontiers in Bioengineering and Biotechnology, vol. 1, 2013, Accessed: Mar. 05, 2023. [Online]. Available: https://www.frontiersin.org/articles/10.3389/fbioe.2013.00015

The second one is specifically related to rapid rotational acceleration, as opposed to impacts which is what almost everything else is. But it’s specifically about acceleration, where in our video the girl isn’t really accelerating that quickly in her rotational motion, even if her rotational velocity is quite high.

So ultimately a bit inconclusive in this post, though I was able to learn a lot about solid mechanics and deformation.

## Cost/benefit analysis of martial arts training and self defense

I was thinking about my history of training martial arts and how it’s worth/benefit would be analyzed from a purely dispassionate/rationalist perspective. Many (possibly majority) of people start learning martial arts for self defense: i.e. to (hopefully) learn how to protect themselves if they are ever in a life or death violent situation. How well does learning martial arts actually do this? Can we analyze it from a quantitative point of view?

One simple way we could analyze would be to use the following equation:

$$[\text{Value}] = [\text{Benefits}]-[\text{Costs}]$$

Pretty simple. If our total benefits are greater than our total costs, it will be a positive value. So what are we exactly measuring here? Lets say time, as in lifespan. If someone attacks me and I die, then my total lifespan is shorter than if I successfully defend myself and am able to live longer. The associated cost in time would be the time spent training martial arts.

So how do we calculate or at least estimate the increased lifespan? I would formulate it like this:

$$[\text{Gained Lifespan}]=(Y-A)p$$

Where $Y$ is the average lifespan for your demographic (currently 79 years in the U.S.), $A$ is your current age (for sake of argument we’ll say 20 years old), and $p$ is probability that you encounter a situation where your martial arts training makes the difference between life and death or serious injury.

However, that equation isn’t quite right. It is the gained lifespan if I were attacked and am able to save myself right when I am 20 years old, in which case I would gain 59 years of life. But if I were attacked when I was 78 years old, on average I would only gain one additional year of life. So we need to take it from the average age I would be, which is just the half-way point between $Y=79$ years and $A=20$ years, so the corrected equation is:

$$[\text{Gained Lifespan}]=\frac{Y-A}{2}p$$

As for $p$, we currently have no idea what that probability could be, so let’s leave it for now.

The time cost is pretty easy to calculate. Let’s say you study martial arts for 10 hours/week for 10 years. The total time spent learning is then:

$$[\text{Time Cost}]=C=(10\; \text{hr}/\text{wk})(52\; \text{wk}/\text{yr})\left( \frac{1 \;\text{day}}{24\; \text{hr}} \right)\left( \frac{1\; \text{yr}}{365\; \text{day}} \right)(10\; \text{yr}) = 0.59\; \text{yr}$$

At this point we know every term in the equation to calculate $[\text{Value}]$ but the probability $p$. What we can do though is calculate what $p$ would have to be for us to break even, i.e. the point where the the expected or average gained lifespan is equal to the time we spent learning the martial arts in the first place. That’s simply setting $[\text{Value}]=0$ and solving for $p$. Doing so we get:

$$p=\frac{2C}{Y-A} = \frac{2 (0.59\;\text{yr})}{79 \;\text{yr} -20 \;\text{yr} }=0.02$$

So our final probability is $p=0.02$ or 2%. If our martial arts training is able to save our lives when we are in a life-threatening situation, then it’s worth it to spend 10 hrs/week training for 10 years if there is at least a 2% of us being killed by violence before we die of natural causes.

So what is the chance of us being killed? In the US, the CDC states that currently there are 5.8 homicides per 100,000 people per year. So assuming that homicides are completely random (which of course they aren’t, but for the sake of argument we’ll go with that simplification), then we have a $p=0.000058=5.8 \times 10^{-5}$ probability or a 0.0058% chance of being killed by homicide in any given year.

But we need to know the total probability of it happening over the course of $Y-A=59$ years, not just one! How do we calculate that? We can’t just add them all up 59 times and say the total probability is $P=59p$, since you can’t just add up probabilities like that. That’s like saying if you roll a 6-sided die six times that you’re guaranteed to roll 6, or if you flip a coin twice you’re guaranteed to get a heads. And we can’t multiply $p$ by itself 59 times either, $P=p^{59}$ is the probability of someone being killed ever year for 59 years! Of course you can only be killed once (that we know of), but even if you could be killed more than once it becomes an astronomically small chance.

What we do is instead calculate the probability that we aren’t killed in 1 year, multiply that by itself 59 times to get the probability we aren’t killed at all over 59 years, and then subtract that by 1 to get the probability that we would be killed over 59 years.

$$P = 1-\left( 1-p \right)^{Y-A}=1-\left( 1- 5.8 \times 10^{-5} \right)^{59} = 0.0034$$

Or a 0.34% chance of a person being killed over the course of 59 years. That’s pretty high (3 people in 1000), but it’s still much smaller than the 2% required for our martial arts training to be ‘worth it’ from a cost/benefit analysis perspective.

However all this analysis invites an obvious question: would this 10 hrs/week of training for 10 years actually give me the skills to protect myself if I were to be violently attacked by someone? Does it even make any difference?

In terms of a deadly assault where someone is attacking you with a knife or a gun, I’m afraid the answer is: probably not. Statistically it’s hard to evaluate this. Lots of martial arts claim to be able to teach people to defend themselves against an attacker with a knife or a gun (and I’ve trained some of them myself), but unfortunately this isn’t very realistic. If someone has a gun on you and they intend to kill you, it’s very difficult to survive. Similarly if someone has a knife on you and intends to slash and stab you, your chances of getting out without a serious or fatal injury is very difficult.

This second scenario is actually pretty easy to test in relative safety with a partner. Put on a fairly close-fitting shirt and pants (that can get stained), and the attacker has a large permanent marker to simulate a knife. The attacker’s goal is to mark your body with the magic marker as many times as they can, and the defender’s goal is to take the marker away from them without getting any marks on them. You’ll find it’s pretty much impossible, even for a skilled martial artist, to be able to successfully defend themselves even against a relative beginner without several marks on their clothing, often in very dangerous areas (torso, thigh, wrist, neck, etc.)

You can do the same type of thing with a paintball gun and some protective equipment, but it can get a lot more messy. The results are pretty much the same though: the odds of the defender not being killed or seriously injured/maimed are very small.

Some of the best realistic conversations on this I’ve seen on the internet are by a couple of guys that evaluate some of the least realistic fight scenes on film: Logan Lo and Chad Vasquez at the Scenic Fights YouTube channel. These guys have a lot of experience doing full-contact martial arts with full resistance and with weapons, and they have a lot of great commentary on what is or is not realistic in a combat situation, and the realistic portrayal of fighting someone with or without a weapon.

Whether you should fight back or not can depend on a lot of different factors. A drunk belligerent guy in a bar pushes you? De-escalate and walk away. Someone with a gun or a knife demands your wallet? Just give them your wallet. Several dudes with baseball bats start approaching you menacingly? Just run and hope you can get somewhere safe. But someone with a gun tries to get you into a car? Statistically the odds of you getting out alive aren’t good, you should probably resist and fight back. Someone takes a swing at you? Combat has already started, it’s hard to get out of it at that point and you need to defend yourself. So basically, context matters.

So how likely is an assault to become a murder? Is there any data that we can use to estimate this? A couple of searches got me to these two sites: number of aggravated assaults in the US in 2020, and number of homicides in the US in 2020.

Here is a python Jupyter notebook with the data and plots:

Assault and Murder Statistics 2020

You see that in general, there are 10x more assaults than there are homicides of any category, but the categories don’t all exactly match. Also some of the naming conventions are a little strange:

• Personal: means with hands, fists, kicking, etc. Basically assault without a weapon.
• Blunt: a blunt object of some kind
• Guns: the lists have lots of different classifications for guns, and the two lists don’t even agree.
• Asphyxiation vs strangulation: no explanation given
• None: what does that mean? No weapon used? In that case how is it different from Personal?
• How does one commit aggravated assault with drugs, fire, or explosives?

Most of the weird stuff is really small in numbers, so we can mostly ignore it on a statistical basis. I decided to do a bit of grouping and renaming in order to make the two lists agree a bit more and make more sense. From a self-defense perspective, we can simplify it to just six categories:

• Unarmed battery (punching, kicking, etc.)
• Firearm (all kinds, no need for 10 different categories here)
• Knife
• Blunt
• Strangle (also an unarmed attack, but trying to choke instead of punch/kick)
• Other (everything else)

By combining and sorting into those categories, we also need to add the assaults and the homicides together to get the total number of attacks for each category: presumably the homicides aren’t being double-counted as assaults as well.

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