Documentation:FIB book/problems/Injury Scaling and Crush

A Ford Mondeo vears off a 50 km/hr road and crashes into a tree. An accident reconstructionist is asked to determine if the driver was speeding and if the accident could have been prevented. She has gathered the following information from various sources.

The car had a kerb mass of 1578 kg and the driver weighed 89 kg.
Crush Measurments
- Total crush width: 1.89 m
- C1: 0.07 m, C2: 0.18 m, C3: 0.54 m, C4: 0.64 m, C5: 0.19 m, C6: 0.12 m
- A: 879.948 N/cm, B: 130.013 N/cm2
Estimated drag factor of 0.59
AIS code: 450202.2
Crash simulation output:

Question 1

What section of the body is the driver's injury and how severe was it? How could an injury like this be caused to this area of the body?


SOLUTION (expand to show)
The first digit in the AIS code shows which area of the body the injury relates to; the thorax (4). The final digit in the AIS code indicates the severity of the injury; moderate (2). This could have been caused by contact with the seatbelt if the passenger was restrained. It may also have been caused by the passenger contacting the airbag or the steering wheel.

Question 2

Calculate the velocity of the vehicle at the moment of impact. State any assumption you have made. (Hint: Although C1 and C6 are different, assume $\theta$ is zero).


SOLUTION (expand to show)
Assume the velocity of the vehicle at the moment of impact in this situation is equal to the Barrier Equivalent Velocity (BEV). $G={\frac {A^{2}}{2B}}={\frac {87994.8^{2}}{21300130}}=2977.81J$ Assume $\theta$ is zero (in this case the impact of $\theta$ on the calculated BEV is negligible). $E={\frac {W}{5}}{\Biggl (}5G+{\frac {A}{2}}(C1+2C2+2C3+2C4+2C5+C6)+{\frac {B}{6}}(C1^{2}+2C2^{2}+2C3^{2}+2C4^{2}+2C5^{2}+C6^{2}+C1C2+C2C3+C3C4+C4C5+C5C6){\Biggr )}$ $E=237474.65J=237.47kJ$ $BEV={\sqrt {\frac {2gE}{mg}}}={\sqrt {\frac {2237474}{1667}}}=16.88m/s=60.77km/hr$

Question 3

What was the velocity of the car before it started skidding?


SOLUTION (expand to show)
Assuming total energy before and after skid are equal: ${\frac {1}{2}}mv_{i}^{2}-fwd={\frac {1}{2}}mv_{f}^{2}$ where $v_{i}$ is the velocity of the car before the skid (initial velocity). $f$ is the drag force between the tyers and the road. $d$ is the skid distance. $v_{f}$ is the velocity of the car just before crashing after skid has stopped (final velocity). Rearrange this formula to calculate the initial velocity. $v_{i}={\sqrt {v_{f}^{2}+2fgd}}=19.28m/s=69.40km/hr$

Question 4

Using information given in the crash simulation diagram, if the car was going at the speed limit and applied the breaks at the same point in the road would the car still have crashed?


SOLUTION (expand to show)
Again, assuming total energy before and after the skid is equal: ${\frac {1}{2}}mv_{i}^{2}-fwd={\frac {1}{2}}mv_{f}^{2}$ Rearrange to calculate skid distance. $d={\frac {v_{i}^{2}-v_{f}^{2}}{2fg}}$ Assume final velocity is 0 m/s as the skid would have stopped the car moving. $d=16.6m$ As this distance is greater than the distance between where the skid begins and the tree, the car still would have crashed if it was going at the speed at the speed limit.