New Road Safety Measures

[Christophe]

Why doesn't it work? Obviously this is approximate because cars will dissipate impact energy better than the wall ...

In any case at the level of the stored kinetic energy it is however quite fair ... yes they must have the same weight ...

[Did67]

I agree with Chatelot: each will dissipate "its" kinetic energy in the deformation of its body.

If they are two identical cars, in fact, going into a wall at 90 km / h or entering into a collision with the same car opposite also driving at 90 km / h, it's kif-kif ...

On the other hand, if the cars are not identical, things get worse: the heavier and the fastest will "pass" its kinetic energy to the lighter / slower ... And there, it was the little one who trinuqe: it dissipates its kinetic energy + part of the other!

[remember the little toy with 6 suspended steel balls, we release the first one, and the last one "jumps"]

[Christophe]

I agree with Chatelot: each will dissipate "its" kinetic energy in the deformation of its body.

If they are two identical cars, in fact, going into a wall at 90 km / h or entering into a collision with the same car opposite also driving at 90 km / h, it's kif-kif ...

Well no it's not kif kif unless one of the 2 cars is stopped ... it's the relative speed of the impact that counts ... so 2 vehicles at 90 km / h have a lot more energy to dissipate only one at 90 km / h against a wall ...

After yes, on the physical details: the masses and the respective deformation of the bodies and the fact that the 2 cars are not anchored in the ground during the impact will mean that the damage will be "a priori" less important than against a wall ... but the a priori should be taken with a grain of salt because there are other obstacles on the road ....

Finally, what should be remembered is that the results of the crash test at 190km / h do not happen until you are driving at 190km / h.

The frontal impacts at 190km / h against a wall must be extremely rare but the wrecks resembling that of the video quite common ... that's all I wanted to say ...

[Flytox]

Anyway, even "that" at 90 km / h there will only be meat left on a wall ....

[Macro]

I don’t explain to you how I freak out at the wheel of my soapbox which is paddling at 75..When I see the panzers tumbling in front and behind me at twice my speed ... They send it to me in orbit the Chernomobile , if they hit me ..

[Gaston]

If they are two identical cars, in fact, going into a wall at 90 km / h or entering into a collision with the same car opposite also driving at 90 km / h, it's kif-kif ...

Well no it's not kif kif unless one of the 2 cars is stopped ...

I agree with Did67 ... it's kif-kif.

it's the relative speed of the impact that counts ... so 2 vehicles at 90 km / h have a lot more energy to dissipate than one at 90 km / h against a wall ...

This story of relative speed is a mistake.
Two vehicles at 90 km / h have exactly two people times the kinetic energy ofun vehicle at 90 km / h, not the energy ofun vehicle at 180 km / h which it is four times higher.

As this energy is to be dissipated in two people identical bodies, each therefore dissipates its own kinetic energy, as if it had entered a wall (which dissipates nothing ...).

[Did67]

This story of relative speed is a mistake.
Two vehicles at 90 km / h have exactly two people times the kinetic energy ofun vehicle at 90 km / h, not the energy ofun vehicle at 180 km / h which it is four times higher.

As this energy is to be dissipated in two people identical bodies, each therefore dissipates its own kinetic energy, as if it had entered a wall (which dissipates nothing ...).

That's what I was trying to say. But it is much better said !!!!

So I agree with chatelot, gaston and ... myself.

[Christophe]

Ah yes now I understand

Actually I got stuck ... I just had to do a corner calculation that I didn't do.

Since 1/2 mv² = kinetic energy; 2 same cars at 90km / h have half the energy of the same car at 180km / h (2 times 90 km / h): 2 * 1/2 * M * 90² = 8100 M = 1/2 * 1/2 * M * 180²

Ok for the remark on the deformation of the 2 bodies ...

So we can say that the body a car at 180 km / h against a wall must absorb 4 times more energy (coef 2 of kinetic energy times coef 2 of the 2 bodies) than in the case of a frontal impact at 90 km / h with an identical car ...... The fact remains that relative impact speed of the frontal collision is indeed 180 km / h... therefore why it does not intervene in the energy calculation of the shock?

It's this thing that bothers me from the start ... why is this a Gaston error? In this case, the energy coefficient to be dissipated compared to the wall at 180 km / h would only be 2 (2 bodywork instead of one) ...

[Gaston]

The fact remains that relative impact speed of the frontal collision is indeed 180 km / h... therefore why it does not intervene in the energy calculation of the shock?

It is not that it does not intervene, it is that it is necessary to make the complete calculation:

If you measure the speed of one car in relation to the other, it means that you are placing yourself in a benchmark which goes to the speed of the second car (the latter therefore has a zero speed in this benchmark before the accident).

In this benchmark, the kinetic energy before the shock is reduced to that of the first car and is worth 1/2 * m * (2 * V) ².

It's this thing that bothers me from the start ... why is this a Gaston error?

For the calculations to be correct, this benchmark must have a constant speed. So it should not stop at the moment of impact, but continue to move away at 90 km / h.

In this benchmark, the two damaged cars move at 90 km / h, the kinetic energy after the impact is not zero and is worth 1/2 * (2 * m) * V²

The energy to be dissipated is therefore
Energy before shock - Energy after shock
= 1/2 * m * (2 * V) ² - 1/2 * (2 * m) * V²
= 2 * m * V² - m * V²
= m * V²
= 2 * (1/2 * m * V²)

And we find the same amount of energy as when placing ourselves in a fixed frame of reference.

We can do the calculation with any benchmark moving at any speed (not relativistic) compared to the two cars, the difference between the initial and final kinetic energy is always the same

[Fakir]

For me, there are 3 situations:
Taking the hypothesis of stopping after colision:
1) 90 km / h on a wall: Energy to dissipate = 312m,
2) 90 km / h on a car at 90 km / h: Energy to dissipate = 625m,
1) 180 km on a wall: Energy to dissipate = 1250m.


180 km on a wall = 2 X 90 km / h on a car at 90 km / h = 4 x 90 km / h on a wall

And that in any of the repositories.

On the fixed one. Vi1 = -Vi2 = 90 and Vf1 = Vf2 = 0
In which is mobile Vi1 = 0 Vi2 = 180 and Vf1 = Vf2 = 90.
Energy to be dissipated in the latter case = 1250m - 625m = 625m