# Sizing crumple zone for multi-vehicle collisions

I need to size bumper beams for the crumple zone of a very lightweight vehicle. Together with the driver, the vehicle will weigh 300kg.

Sizing the beams for a single-vehicle collision against a fixed object is simple. To prevent brain injury, I need to keep the average deceleration under 50 G, so the average crushing force must be no more than ~ 150 kN. Dividing the kinetic energy by the force gives me the needed crush distance.

However, if the vehicle is in a collision with another (heavier) vehicle, I am unclear about when and how the two crumple zones will collapse. Looking at NHTSA crash test data, it seems the crumple zones in most passenger cars are designed for a deceleration of about 15 G against a rigid barrier. For a 1,700-kg car, this would mean an average crushing force of 255 kN.

If my vehicle with beams designed for a crushing force of 150 kN crashes into a vehicle designed with a crushing force of 255 kN, what happens?

a) Assuming my crush zone can absorb all the excess kinetic energy in the collision, what will the deceleration of my vehicle be?

b) Can I rely on the crush zone of the other vehicle absorbing some of the energy before the force on my vehicle over a significant period of time rises above 150 kN? If so, how much?

• You are probably attempting the impossible here (which is why the regulations call for tests against a fixed barrier). If your lightweight vehicle picks a fight with a 30 ton truck, with a head-on relative velocity of say 120 MPH, you are going to lose, however much design work you do on your beams. – alephzero Nov 5 '18 at 18:34
• I'm trying to limit to 50 G with a collision of 20 mph for the 300-kg vehicle against 40 mph for the 1700-kg vehicle. Ambitious? Yes. Crazy? Maybe. Impossible? No. – Jeffrey Hood Nov 5 '18 at 18:42
• – Solar Mike Nov 5 '18 at 18:59
• Cars have been completely crushed along with the driver between the back of one truck and the front of another... – Solar Mike Nov 5 '18 at 19:02

$$\alpha = (255-150)kN / (1700 +300) \ 105000 /(2000) \times 9.8 ~ 5m/s, \ \ or \ \ g/2$$