Tuesday September 4, 2007

Bob Norman sez you have a better chance of surviving a head-on car crash if you *accelerate* into the crash then if you brake. Nonsense??

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Tuesday September 4, 2007

Bob Norman sez you have a better chance of surviving a head-on car crash if you *accelerate* into the crash then if you brake. Nonsense??

comments powered by Disqus

C L JahnTue Sep 4, 08:19 AM #Yeah, it’s nonsense.

Two cars colliding at 70mph is the same as running into a wall at 140mph.

Running into a wall at 90mph is much more deadly than hitting it at 40mph.

A truck doing 40mph will survive a head on collision with a Subaru doing 90mph. The Subaru won’t.

To survive a head on collision, you want to be in the bigger vehicle.

PulpWed Sep 5, 06:10 PM #CL Jahn, you completely miss the point, which is that running into another car is NOT THE SAME as running into a wall. Assuming equal weights, the car going the fastest will knock the other car back, causing more damage than it receives. It’s simple, really.

aleshWed Sep 5, 09:01 PM #I’m completely torn about this.

I aced physics in high school (AP, biotch), but my brain has since been wiped clear of that sort of practical knowledge. I tried to re-construct the knowledge yesterday, but it’s not so easy:

It seems that in physics there are two types of collisions: those between hard bodies and those between soft bodies.

Collisions between hard bodies (the example would be billiard balls) conserve momentum, which clearly is not true of head-on car collisions (what with crumple zones, etc). But nor are the collisions like those between soft bodies (which seem to be more like water balloons) . . .

As far as I can tell, when two cars collide, the most realistic way to simplify the math is to assume that they merge into one body, and you simply subtract the momentum of one from the momentum of the other. My back-of-envelope figuring the other day did in fact produce situations where a higher velocity (and therefore higher proportional momentum at the time of impact) for the vehicle of lesser mass did in fact produce a smaller change in velocity for that vehicle then a lower change in velocity.

Unfortunately, when I tried to put them into english, my calculations turned to mush.

Calling any high-school physics teacher (or competent student), who should be able to answer this in 1/100th of 60 minutes.

BUT!

I think that if you think it’s simple, you’re not really understanding the problem. An acceleration on the part of either vehicle obviously increases the force of the impact. It’s not enough to show that a greater proportion of the increase falls on the other vehicle — one needs to show that the impact for the accelerating vehicle actually decreases (that is, the change in it’s velocity on impact MORE THEN outweighs the increase in velocity before the impact).

More then anything, I would love to see an image of this problem worked out on a chalkboard (do they even have chalkboards in schools anymore?).

C L JahnSun Sep 9, 10:50 AM #Pulp, feel free to substitute “car” for “wall.” Two cars colliding head on with each doing 70mph will do the same amount of damage as one car doing 140mph impacting a parked car. The only thing that changes is where the two cars end up after the crash. Accelerating into a crash is a Bad Idea.

aleshSun Sep 9, 11:26 AM #CL~

You feel free to substitute “car” for “wall” only if you’ve forgotten that the “car” “moves.”

C L JahnSun Sep 9, 04:06 PM #Alesh-

So will a wall if you hit it hard enough.

aleshSun Sep 9, 08:16 PM #CL~ Ok… sorry if I was being hostile; I appreciate your input on all this stuff.

What I’m saying is that the MOMENTUM of the two objects that collide is the relevant thing here. Momentum is a product of mass and velocity. The wall is sitting still, so it’s velocity is 0, so it’s momentum is 0 too.

The moving car has some momentum

p(1) = x

YOUR car has momentum p(2):

p(2) = y

You can’t do anything about p(1), and you can’t do anything about you car’s mass. But you CAN change your car’s momentum by changing its velocity.

now when you get to

x – y

you’ve shifted the RATIO of how the impact is distributed. Of course if you’ve accelerated, you’ve also increased the overall force of the impact.

What we need is someone who really understands the physics to take this math to the next step and point us toward the correct equations that SHOW HOW THE FORCES GET DIVIDED between the two vehicles.

Again, I appreciate your input, but I’m not sure you’re the physics expert we’re looking for here.

AlexMon Sep 10, 12:36 AM #Pulp is wrong. In a head-to-head collision, no car can push another one back if it’s moving, no matter how fast is going (or what you see in the movies).

Momentum alone doesn’t explain why. Momentum is p=mv (mass times velocity). I’ll do it in metric units because that’s how the units work (and nevermind equal weight, let’s stack the chips in favor of one car). Say car A is a compact that weighs 1,000 kg and travels at 60kph; Car B is an SUV, weighs 3,000 kg and travels at 80 kph without braking as Pulp suggests (it doesn’t matter if the driver floors the accelerator because the effect will be negligible).

At the moment of collision, car A has a momentum of 60,000 kgm/s and car B has a momentum of 240,000 kgm/s. But here’s the rub: what matters is not their momentum but their

kinetic energyand if one is enough to alter the other —in other words: how much energy car B needs to generate in order to reverse the movement of car A and “knock it back”.The equation is Ek = momentum square divided by mass times 2. For car A is 1800 kjoules, for car B is 9600 kjoules. Because of Newton’s third law, we know that 1800 kjolues will act against car B, (so it has 7800 left) and the energy needed to move car A one meter can be calculated by a simple work equation (mass times gravity factor [9.81 m/s] times coeficient of friction). Some car accident investigator says in is site that the coeficient of friction of car tires (non-rolling, because the engine is still in gear) is more or less 0.47. The energy needed to move linearly a mass of 1,000 kg with that coeficient of friction is 4610.70 kjoules per meter.

So even if it were a perfect inelastic collision and the cars were strong enough to not deforme (imagine billiard balls that didn’t bounce) AND the cars collided perfectly square (to avoid deflection caused by angular momentum) Car B has enough force to move car A for just 1.6 meters.

In reality, we know the cars would crumple on impact and most of the kinetic energy will be absorbed by their bodies and converted into thermal energy, so there won’t be enough left to move either one.

Only if car B was much heavier (like a semi), travelling at a ridiculous speed —or if it could accelerate enough in the seconds before a crash— it would generate enough force to knock back car A. Such a car doesn’t exist and regardless, there’s only so much deceleration the human body can resist. The g-forces alone would kill the driver.

IMHO once you are in the path of an accident with no time to brake, the only thing you can do is steer.

C L JahnMon Sep 10, 09:03 PM #In other words, I was right aaaallll the way back in the first comment.

honzaTue Sep 11, 04:31 PM #Alex,

For an eye-popping account of just how many g-forces a human being can survive, read this Wikipedia account of F1 race car driver David Purley (next-to-last paragraph, 1977 British Grand Prix)