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Seat belt force mismatch between test data and calculations

  1. Oct 16, 2012 #1
    Hello! I have a question about seat belt force. In a 55km/h collision maximum chest acceleration is 60 g, and if say torso weighs say 40kg, then max force should be 600 m/s^2 * 40kg, which equals a lot, but if you look at test data of seat belt force it is only 8KN which is strange, compared to 600*40 = 24KN. And such result are everywhere, i can't understand why belt forces are always 3-4 times smaller than the calculated force... Can someone help? :)
  2. jcsd
  3. Oct 16, 2012 #2
    The calculations do not model reality well. There is no way that the structure of the automobile can transmit 60 g to the person. It will start yielding and breaking up, which absorbs much energy and minimizes loads transmitted to the person. Anything over 40 g is not survivable under any conditions. In most cases it will take less than that to kill. That is why the USAF puts a 40 g survivability requirement on the crew station of all their aircraft.
    Last edited: Oct 16, 2012
  4. Oct 16, 2012 #3


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    Welcome to PF!

    The seat belt is connected to the car in three places. The angle of the belt at the connection won't be exactly in the direction the car is moving, so one would have to calculate the force each point can take, but I don't think it will be too far off from 8kn per connection.
  5. Oct 16, 2012 #4
    Pkruse, i agree that G forces are big, but the point is that those are dummies, and what i'm trying to understand what is it that i don't know, because, like i wrote it's is very strange that the force, energy disappears somewhere... yes, the car structure will collapse, but those numbers are from onboard data recorders, so yea if at one point chest experiences acceleration of 60 g's so using F=ma, and 40 kg it's 24 KiloNewtons and not 8 or 6 KN.

    russ_watters, you wrote about 3 points of seatbelt tensionning, but in such case, as then the whole body is restrained by the seat belt we get 80 kg * 600 m/s^2, which is 48KN. Concerning the angle i don't know if it matters if the whole weight of the driver is restrained by the belt. And as i understand belt force is measured by cutting the belt in two and connecting it again with a force meter (well not exactly, but you get the point).

    Again, any thoughts, maybe i'm missing something???
    Last edited: Oct 16, 2012
  6. Oct 16, 2012 #5
    Ok, i believe i solved it, see i didn't know much about pulleys, but yea it is that principle, plus there are also femur forces... So if say 40 g's for the body, then 80 kg * 400 = 32KN. Sum of femur left and right forces 9KN. Sum of shoulder and lap belt portions 11KN (numbers from test data) and we multiply by 2, because torso and pelvis portions of the belt are both anchored by two fixing points and there is torso and pelvis as weight subjected between those points, so 11 KN * 2 = 22 KN. And 22+9 = 31 KN ~ 32KN. Yay. Hopefully this logic is ok.

    Thank for help, of course if it is correct...

    Though i must say that if i look to another crash test data it is still that 10KN is missing..

    Well but if you think that head is stopped by the wheel -4.5 kg, also arms and lower legs also sort of stop by themselves, that leaves us with ~66kg, so maybe that's where 10 KN goes..
    Last edited: Oct 16, 2012
  7. Oct 16, 2012 #6


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    Don't forget that a crash dummy (just like a human) is not a rigid body. Some of the energy will be absorbed in deforming it.
  8. Oct 16, 2012 #7
    I doubt that it matters, because it is the final deceleration (I believe it is the final) after some absorbtion and deformation of the dummies body...
    Last edited: Oct 16, 2012
  9. Oct 16, 2012 #8


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    Staff: Mentor

    Well, I'm not sure. Because the arms, legs and head are free to move/pivot, they decelerate after the torso, so they may not affect the maximum force felt by the belt. It isn't a simple problem to model.
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