Is the NHTSA's Method for Testing Frontal Collisions Accurate?

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SUMMARY

The National Highway Traffic Safety Administration (NHTSA) conducts frontal collision tests by crashing vehicles into a fixed barrier at 35 mph, simulating a head-on collision between two vehicles of similar weight. The discussion clarifies that the forces experienced by vehicles in both scenarios are equivalent, as the reaction forces from a barrier mimic those from another vehicle. The key takeaway is that the NHTSA's testing method accurately reflects real-world collision dynamics, as the energy and damage are consistent whether one vehicle hits a barrier or another vehicle. Understanding this symmetry is crucial for interpreting crash test results.

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amazinghl
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Am I right or is NHTSA right?

In http://www.nhtsa.com/cars/testing/ncap/Info.html#iq8 they stated,

For testing frontal collisions, crash-test dummies are placed in driver and front passenger seats and secured with the vehicle's seat belts. Vehicles are crashed into a fixed barrier at 35 miles per hour (mph), which is equivalent to a head-on collision between two similar vehicles each moving at 35 mph. Since the test reflects a crash between two similar vehicles, make sure you compare vehicles from the same weight class, ± 250 lbs., when looking at frontal crash protection ratings.

I thought the difference of head-on with a car and crashing to the wall is

Car A 100N ---> <--- 100N Car B
Then you have the reaction force coming back at you.
Car A 100N <--- ---> 100N Car B

Thus if I'm in car A, I will have the 100N from car B AND the reactional force of 100N from my own car, which adds up to a total of 200N for EACH car.

VS.

| <--- 100N Vehicle
| ---> 100N Wall

I just have the 100N from the reactional force of the wall.

Which make that "Vehicles are crashed into a fixed barrier at 35 miles per hour (mph), which is equivalent to a head-on collision between two similar vehicles each moving at 35 mph." wrong.

Am I right or are they right?

Thanks for your time.
 
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Am I right or are they right?
They're right. You need nothing more than a symmetry argument. Two cars head-on at same speed, suffer the ame damage. So there is twice as much total damage compared with one car on a barrier. But there is also twice as much initial energy since there are two cars. So each car is damaged the same as in the car-barrier case.

Or look at it like this: As the two cars are colliding head-on, the interface between the two is fixed. Therefore if you imagine an infinitely thin but infinitely strong barrier between the cars, nothing would change. Now since the barrier is infinitely strong, imagine taking one of the cars away. Does not affect the remaining car.
 
Wow, this is an interesting thread, and after some thought, I think I see what krab is saying.
A fixed barrier hit by a car at 35 mph is the same as a head on collision between 2 cars going 35 mph because the fixed barrier imposes a reaction force when hit of, guess what? 35 mph!
Fascinating! But makes sense to me.
The important thing to realize, though, is that this is not the case where one of the cars in a head on collision is going faster than the other.
If I am geting this right, to simulate a head on collision between car "A" going at 35 mph and car "B" going at 45 mph, one merely needs to have car "A" indeed traveling that 35 mph speed, but the barrier needs to move towards it at only 10 mph if the mass of the moving barrier is 4.5 times more(or something like that) than the car it replaces.
I may not have my math right, but I get the point.
 

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