Energy transformed in a collision

[ToniG]

When reconstructing traffic accidents to obtain the collition velocities it is needed to obtain the "crush energy" which is the amount of kinetic energy transformed into permanent damage to the cars involved. From that kinetic energy we get the Energy Barrier Speed which is used with the momentum equations to determine the speed of the cars involved . The crush energy is based on measurements of the deformation of the cars involved in the collision.

When in a two car collision only one of the two cars "Crash energy" is known, the total Crash energy of the two cars can be obtained if the relationship between the masses of the two cars is known. That formula is shown in "SAE Technical Paper 920604: Wood, D., Collision Speed Estimation Using a Single Normalised Crush Depth-Impact Speed Characteristic".
I attach a snapshot of the paper showing the formula.

Et = E1*(1+ M1/M2)
Where Et is the total crush energy, E1 is the Crush energy of car 1, M1 is mass of car 1 and M2 is mass of car 2.

It is not intuitive to me how can the relationship of the energy dissipated into each body only depend on the relationship of masses. I thought that the "stiffness" of each car, or other properties should be involved but it seems not according to the paper.

Can you help understand where the above formula is derived from?

 

[Frabjous]

Set the magnitude of two impulses equal to each other
ebs2= m1ebs1/m2

add the crush energies
e1=0.5m1ebs1^2
e2=0.5m2ebs2^2=e1(m1/m2)

 

[ToniG]

Thank you so much

 

[ToniG]

Even though the formulas are clear to me now, when I try to picture its implications in a car collision there is something that I don't fully understand. I am sure that I may be missing something, anyway please let me explain my reasoning:

Scenario 1: two equal cars, car1 and car2 collide when traveling at the same speed and opposite direction. The kinetic energy transformed into crush energy will be the same on both cars resulting in the same amount of damage / body deformation. Everything is clear so far.

Second 2: we increase the rigidity of car2 and keep the rest of variables as in the previous scenario (same mass for the two cars, same speeds). Since we increased the rigidity car2 the amount of energy absorbed by car 2 must be lower. According to the formula, the energy absorbed by car 1 will be lower in the same amount (since the masses of the two cars are still the same even though they have different rigidity). So we would end up with greater post collision speed for both cars and a lower deformation for both cars (less kinetic energy transformed).

What keeps me thinking is this:
How can be possible that strengthening the structure of one of the two cars results in a reduction of damage to the structure of the weaker car?
Would then a rigid barrier cause less damage than a deformable barrier in a collision?
Am I wrong when assuming that an increase in rigidity means less absorbed energy?

Sorry for asking this again even though the formula derivation is fully clear. But I needed to know if there is something missing or it is simply that what seems logic to me it is not necessarily what happens in real life.

 

[PeroK]

I'm struggling to make sense of what you are assuming and what you are asking.

That said, I'm not familiar with EBS.

 

[Frabjous]

I do not know this field.

The formula was derived given that one only measured the damage to ONE car. If you had the other car, you could measure its “ebs” directly and if the vehicle had significantly different structural behavior, its “ebs” would be different than from the one derived above. It is not clear by how much. In addition to this, there is a lot of physics that is absorbed into the “ebs” variable.

The field uses this approximate formula, so I assume that it has some validity.

In general, vehicles are generally designed to protect occupants, not minimize damage. An elastic car would be much more lethal to its occupants, so there are some limitations on how much stiffer a car could be.

Given that a rigid barrier does not move, a vehicle hitting one will decelerate in a shorter distance and consequently take more damage.