Shouldn't that be sqrt(10) times as bad?
The distance fallen under constant acceleration g for t seconds is 1/2 g t^2, so it takes sqrt(10) times as long to fall 10 times as far. The velocity after falling t seconds is g t, so falling sqrt(10) times as long gives a velocity of sqrt(10) times as much.
First consider that our kinetic energy is
K = mv^2/2
Additionally is the material has a spring constant k, then the force that results when the material deforms by x amount is
F(x) = kx
If we let d be the max distance the material, then we can integrate over the force to get the energy absorbed, which is equal to the kinetic energy.
int_0^d F(x) dx = K
kd^2/2 = mv^2/2
Now there are two ways the impact could work. Imagine we have a really thick crash pad, then we consider it to deform without limit and have a constant k, giving
d = sqrt(mv^2/k)
F_max = F(d) = v sqrt(mk)
But what about a helmet? It can only over it's thickness, at which point your head is basically in contact with the concrete. Thus, we have a constant deformation distance d, and get
k = mv^2/d^2
F(d) = mv^2/d
As we can see, depending on how we consider it we get either a force of sqrt(10) or 10 times the original amount. I probably failed to take into account something that someone who knew more about materials could point out.
tl;dr it's complicated because impact force does not necessarily scale linearly with velocity.
update: so after some conversation with others. It seems that head on concrete is better modeled by a constant d. So a drop of 10 times the height results in 10 times the force.
See also http://hyperphysics.phy-astr.gsu.edu/hbase/impulse.html
The square root in the velocity calculation goes away since
E_k = 1/2 MV^2
so the energy dispersed hitting the ground from a 10 inch fall is exactly 10x as bad as a 1 inch fall.
Of course the actual damage done by that energy transferred through your head will be some complex function due to bone breakage and cell tearing thresholds, etc. But as a first approximation of suck, I'd go with this simple linear model.
Still dropping your head an inch (painful) is a nice reminder of how much energy is involved from normal riding height.
Or a balloon. It recovers from lots of pressing, even repeatedly. But there's a pressure from which it's never going to recover and crossing that threshold has binary outcome.
I just bought a new helmet and the one with MIPS was among the cheapest options at the bike store I went to. There were many many other helmets without it that were vastly more expensive. I paid £50 for a Scott helmet that had MIPS, most other helmets were in the £80-150 range and most of them didn't have it. So I don't exactly feel like I paid extra for this technology, it just happened to be included with the helmet that was 1) within my budget 2) fit comfortably.
The MIPS and Wavecell seem to assume that helmets grab the road with great traction and then snap your skull producing a large rotational acceleration so great that your brain falls behind. In my experience helmets (which are at first approximation spherical) hit in the center, easily skid, and induce approximately zero rotation in your skull.
I do recall in the 1990s or so that as brands switched from a heavier plastic shell (like the ancient bell helmets) to the then common nylon sleeve over the styrofoam that one manufacturer picked a particularly high friction material which did cause additional injuries. Once that was fixed everyone was happy again.
So absorbing energy from the fall is great. Playing games with additional layers with low internal friction or fancy ways to collapse is useless. I'm not completely against the idea, but seems laughable to pay 3-6X for something unlikely to help. Bike helmets that meet the toughest impact standards are available for $50 or so.
I personally have fallen many times, had my helmet skip many times (the longest was more than 10 feet) and never noticed my helmet gripping the ground and trying to torque my head.