0
$\begingroup$

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?

$\endgroup$
5
  • 1
    $\begingroup$ 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. $\endgroup$
    – alephzero
    Nov 5, 2018 at 18:34
  • $\begingroup$ 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. $\endgroup$ Nov 5, 2018 at 18:42
  • $\begingroup$ See engineering.stackexchange.com/q/24485/10902 $\endgroup$
    – Solar Mike
    Nov 5, 2018 at 18:59
  • $\begingroup$ Cars have been completely crushed along with the driver between the back of one truck and the front of another... $\endgroup$
    – Solar Mike
    Nov 5, 2018 at 19:02
  • 1
    $\begingroup$ When collide, after the bumper of the 300 kg car (A) has deformed to the designed maximum, the bigger vehicle (B) will run it over with a slightly reduced speed (due to the damping effect from car A) and momentum. I suggest simplifying the scenario to what happens when two cars, with the same builds (mass) and speed, collide instead to make the prediction easier. $\endgroup$
    – r13
    Dec 20, 2021 at 18:40

2 Answers 2

1
$\begingroup$

"300-kg car going 10 m/s, and 1700-kg car going 20 m/s. Large car available crushing distance: 1.33. Small car crushing available crushing distance: ALTERNATIVE A: 2.367 m. ALTERNATIVE B: 0.1 m"

To keep the average acceleration of the small car to 15g which is probably survivable, you need 2.2m of crumple zone in the small car, the big one doesn't bend at all with a realistic crumple zone designed to give it a full frontal crash crush zone of 0.6m. Crush length for 50g sustained (which will kill you) is about 0.66m

Graph of small car crumple zone length vs average decel Incidentally all this talk of average g levels is a bit optimistic, here's a random graph from the interwebz some crash pulses

$\endgroup$
2
  • $\begingroup$ +1 for "the big one doesn't bend at all with a realistic crumple zone", although I have no idea whether you are correct. You seem to know what you are talking about, and you have spoken to an essential part of the question. Incidentally, we gave up on design for a two-car collision. We hope the occupant will be fine in a collision against a fixed barrier since the vehicle can't go faster than 20 miles per h. If interested, you can see what we're working on here: podthepeople.org/files/bids/rfq20210119/… . We'd love to have additional help. Thank you! $\endgroup$ Dec 17, 2022 at 22:42
  • $\begingroup$ It is approximately true that crumple zones are constant force, although there a lot of caveats. As such, if you want a 1700 kg car to stop from 19 m/s at 20g you should provide a crumple zone v^2/2/200 metres long, and of course the force would be 200*1700. So the big car will apply 200*1700 N to the small car before the crumple zone crumples. That's a lethal acceleration of 20g*1700/300. So the small car needs its own crumple zone, which as we've worked out is rather big. $\endgroup$ Dec 18, 2022 at 5:22
0
$\begingroup$

As per your statement if all the crash energy is absorbed in crash zone, and the two cars fuse together and move along direction of heavy car:

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

Where alpha is the excessive acceleration above your 50G. So, total G = 50.5

And the big car's crashing zone is going to experience a 14.5 G deceleration.

$\endgroup$
6
  • $\begingroup$ Would the force be 150 kN until my crash zone is consumed and then continue at 255 kN? Or would it oscillate around 50.5 G x 300 kg during the whole period of deceleration/acceleration? $\endgroup$ Nov 5, 2018 at 20:07
  • $\begingroup$ @JeffreyHood , We need more data to be able to measure that. We need the speed of the two cars, just as an example if the small car is going twice as fast then the crumple zone would run out before its kinetic energy runs out. $\endgroup$
    – kamran
    Nov 5, 2018 at 20:31
  • $\begingroup$ "And the big car's crashing zone is going to experience a 14.5 G deceleration." Why? The big car's crumple zone might a different depth from the small car's. Don't you think that might make a difference? $\endgroup$
    – alephzero
    Nov 5, 2018 at 20:41
  • $\begingroup$ Would you be interested in commenting on an example with more data? 300-kg car going 10 m/s, and 1700-kg car going 20 m/s. Large car available crushing distance: 1.33. Small car crushing available crushing distance: ALTERNATIVE A: 2.367 m. ALTERNATIVE B: 0.1 m $\endgroup$ Nov 5, 2018 at 21:15
  • $\begingroup$ The acceleration of a 300 kg car will always be 1700/300 times bigger than the 1700 kg car due to Newton 3 and 2. $\endgroup$ Dec 16, 2022 at 0:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.