Update 2016-03-19: Mia Culpa! I’ve been doing some revision of my physics since it’s been a good 12 years since I looked at this closely. The text I’ve been reading through for this is Physics for Scientists and Engineers with Modern Physics, Serway and Jewett, 6th Edition (International student edition), ISBN 0-534-40949-0 .

Seems in that time my physics has gone a bit rusty. The actual position with respect to time is given by:

$h(t)={1 \over 2} at^2 + v_0t + h_0$ So the results below are not quite correct. I shall re-do the calculations shortly.

Well, as a starting point, I figured I’d look at what happens in the current tests that are performed. I’ll have to dig up the relevant Australian Standards to see how they do things, but thanks to the Bicycle Helmet Safety Institute, we can have a look at the rig used in American CPSC labs.

They also make some interesting remarks about MIPS .

I’ll put some diagrams up, but for now bear with me. The typical test apparatus basically tries to measure the acceleration of the “headform” as it strikes a shaped anvil from some fixed height. In AS/NZS 2068, this height is 1.5m.

There’s a couple of different headforms they use, they’re basically a head-shaped block of wood, metal or plastic with an embedded accelerometer, with a known fixed mass and fixed dimensions. They strap the helmet under test to the headform, raise it to a fixed height (1.5m) and let it drop.

So let’s model this.

The following are our initial constants:

 Variable Symbol used Value Notes Free-fall height $h_0$ 1.5 m From AS/NZS:2068 Free-fall acceleration $a$ 9.8 m/s² Gravitational acceleration constant Headform+helmet mass $m$ 5 kg Educated guess here. Initial Velocity $v_0$ 0 m/s

We’ll start by trying to figure out the flight time, or time to impact. We’ll ignore wind resistance and the test apparatus, the height of the headform at any given time prior to impact is given by the equation:

$h(t) = at^2 + v_0t + h_0$ We simply solve this for $h(t_I)=0$ .

Most of the terms disappear, since we’re starting at rest and have a known starting height. We wind up with:

$0 = -9.8t_I^2 + 0t_I + 1.5$ We re-arrange this to find that the time to impact was 391.230 msec.

$t_I=\sqrt{-1.5 \over -9.8}=0.391230$

We can also determine that during this time, the headform accelerated to a velocity of -3.834 m/s. $v_I=at_I=-9.8\times0.391230 = -3.834$ This is at the point when helmet (or headform) meets anvil. The intention of the helmet is to absorb as much of the momentum as possible, so the worst thing that could happen here is a perfectly elastic collision.

The momentum at impact is given by the equation:

$p_I=mv_I=5 \times -3.834=-19.170$ The worst case is all of this momentum is reflected back to the headform itself. Let’s assume that happened over the course of 1 msec. So momentum after impact:

$p_A=19.170$ and the change in momentum:

$\Delta p = 19.170 - (-19.170) = 38.340$ which over 1msec, gives us a force of:

$F_A={\Delta p \over \Delta t}={38.340 \over 10^-3}=38340$ So 38.34kN, and what about the acceleration?

$F_A=ma_A$ $a_A={38340\over5}=7668$

7668m/s² is 782g. Our cyclist would be dead.

Suppose the helmet did its job, and over 3 msec, managed to attenuate that to the 200g as specified in AS/NZS 2068. This equates to -1960m/s² acceleration, or a downward force of -9.8kN. What would the change in momentum need to be?

$-9800 = {p_A - p_I} \over {3\times 10^-3}$ Re-arranging, we get an after-collision momentum of 1.534 Ns. So to meet the standard, the helmet has to attenuate that momentum. So what happens to the brain inside all this? Suppose we had a headform that modelled this.

The human brain is around 1.5kg on average, and has an approximate volume of 1130 cm³. It resides in the cranial vault, which has an approximate volume of 1170 cm³. These differ between males and females, and can vary wildly from this. For simplicity’s sake, let’s assume both are spherical. We can work out how much space there is around the brain using the following equation to calculate the radius of brain and vault:

$V={4 \over 3}\pi r^3$ Plugging these values in, we get a brain that has a radius of 6.461 cm, and a cranial vault of 6.537 cm. This leaves a gap of about 700um around the brain in which it can move. This is less than I expected, but let’s see what happens.

We know the headform was travelling at -3.834 m/s just before striking the anvil, and at this point, the brain is still moving at about that speed. We know it’ll continue to move forward that 700um before it hits the cranial vault, but how long do we have? About 183 microseconds.

Given the such small gap and time window involved, we could possibly consider the cranial vault in a simulated headform as being a gel with similar properties to the brain. It’ll deform as it hits the vault walls and “bounce” back, possibly causing it to ricochet into the opposing wall.

If we’re to have any hope in preventing this, we need to start speed reduction much earlier.