Car physics - Page 5

geckoGT · 19-12-2009, 10:12 PM

I found a very interesting article on motor vehicle crash physics. Read the section on 2 car crashes, it explains a hell of a lot.

Physics article

Essentially they are saying that two cars of similar characteristics traveling at a similar speed in opposite direction colliding is equivalent to hitting a solid, immovable object at the same velocity. It goes on to say that most modern cars are survivable in a crash at 35 (56 km/h) mph into a solid object, therefore two cars traveling at 35 mph (70 mph or 112 km/h closing speed is survivable).

There are some limitations to this theory, the most import one for our dilemma is that of structural integrity of the vehicles involved. All vehicles have sections of the front that have more longitudinal strength than other areas. For this above theory to be true, strong sections must strike strong sections on the other car in order to maximise use of the crumple zones. If the stronger areas are offset to each other, one vehicle will take a higher proportion of the load from the impact.

Now lets look at a simple truth that we all know, true head on with another vehicle is very rare. I have been to a number of head ons and I have never seen one that was a direct head on where structural members of the two cars would be even closely aligned. I also asked around work and I could not find another paramedic that had been to a direct head on, even those that were traveling in true opposite directions were offset from each other. That was asking every paramedic I ran into at station and hospital, more than 300 years combined experience talking there. Additionally all stated physics aside, the prangs they have been to, the head ons were always worse than static object collisions at speeds of 100 and over.

I also asked each one to answer one more question.

If they were driving at 100 km/h and two cars came at them in the opposite direction at the same speed, across both lanes, leaving them the choice of hitting a car or a very large tree. Would they take the car or the tree?

Most answered the tree.

So, in summary. If you are ever faced with this choice you can take either as long as you can guarantee the other car is traveling at the same speed (more is bad, less is good), it is the same mass (not a loaded falcon ute etc), has the same fuel and passenger load, does not have illegal mods altering its frontal rigidity and will line up perfectly with the front of your car. If you can not guarantee all these factors will be equal, take the wall or tree.

XCXDBABF · 20-12-2009, 08:07 AM

Flappist, I owe you an apology. I'm sorry.

I thought I could prove my assertions from studies I'd done quite some time ago. Whether I'm now missing something in my workings, I can not produce the results I expected, and as such obviously can't verify my assertions.

I should have taken more time considering the arguments placed by you and others.

flappist · 20-12-2009, 10:11 AM

Quote:

Originally Posted by XCXDBABF

Flappist, I owe you an apology. I'm sorry.

I thought I could prove my assertions from studies I'd done quite some time ago. Whether I'm now missing something in my workings, I can not produce the results I expected, and as such obviously can't verify my assertions.

I should have taken more time considering the arguments placed by you and others.

No apology necessary at all but thank you for offering it.

Forums are for debating and everyone is right or wrong sometimes.

Life goes on......

The Stylist · 20-12-2009, 10:20 AM

Certainly an enlightening read this thread has been, I'm glad people have brought these questions up as I remember seeing an advert years ago stating that two cars crashing into each other head-on at 100km/h would have equivalent impact forces of crashing at 200Km/h, and I've always wondered if this was true or not.

4Vman · 20-12-2009, 10:52 AM

Quote:

Originally Posted by Wally

NHTSA test:

Thankyou Wally, looks like this one is solved now..

05MkIIFutura · 20-12-2009, 11:25 AM

Quote:

Originally Posted by 4Vman

Correct, equal and opposing forces will cancel each other out, the variables are weight, angle of collision and speed of the 2 objects, 2 cars of equal weight (inertia) hitting squarely head at 100kpm each will have the effect of hitting a padded (compensate for crumple of second vehicle) brick wall at 200kph...

Except force (inertia) isn't linear. If you go twice the speed (hit the wall at 200km/h), you would need more than twice the crumple zone.

Similar concept to brake performance. Double the speed, you triple the required braking distance.

xtremerus · 20-12-2009, 05:36 PM

Love to see a real tennis match between XCXDBABF and Flappist

:

geckoGT · 20-12-2009, 06:52 PM

Quote:

Originally Posted by 05MkIIFutura

Except force (inertia) isn't linear. If you go twice the speed (hit the wall at 200km/h), you would need more than twice the crumple zone.

Similar concept to brake performance. Double the speed, you triple the required braking distance.

I think you need to read the article I posted in my last post. Even the man responsible for designing car safety systems for VW says the forces are the same, with a few caveats.

shyun · 20-12-2009, 07:51 PM

The main factor in the collision is the momentum of the objects involved. When two identical cars are travelling at each other in opposite direction at the same speed, the net momentum is zero. After they collide the the net momentum is still zero. This means that after the collision both will move away at the same speed. There is no way to predict what the speed may be, but it depends on how much energy is lost in the collision. Energy is used to deform the cars, as heat and as sound, and whatever is left over is retained in the cars.

Someone already brought it up but an 'immovable' object is realy immovable it is just attached to the earth. Whatever momentum a car colliding with an 'immovable' object loses is transfered to the angular momentum of the earth. Saying two cars colliding is like having a wall in between is different than a car colliding with a wall because of this. The car colliding with the wall transfers its momentum to the wall/earth. The imaginary wall between the two cars would gain no momentum because each car exerts a force on it from opposite directions, balancing out.

With regards which is worse, it's easier to use the fact that

Force*Time=Mass*Velocity=momentum

The force is the determining factor of the severity of the injuries (realy the acceleration), and the change of momentum is determined by how much the velocity of the car changes. The crumples zones increase the time of the collision ,reducing the force.
So hitting a wall only has 1 crumple zone and hitting another car has 2. 2 cars colliding at 100km/h in opposite directions is the same as 1 car colliding with a stationary car at 200km/h (2 crumple zones here)-Case 1

So effectively you have to compare this with 1 car hitting a stationary wall at 100km/h (1 crumple zone here).-Case 2

In the first case the change of momentum is double that of the second, but only 2 crumple zone.
In the second case the change of momentum is half the first, but with 1 crumple zones
So as far as i can see, 2 crumple zones will double the time the collision is happenning for. ie F*t=m*v and F*(2t)=2*(m*v), therefore the force in both situations is the same, and since they weigh the same so is the acceleration.
That's my take on the physics.

bob^ · 20-12-2009, 08:12 PM

What about if the car was on a large treadmill that would speed up to match the speed of the car. Would it take off? ;)

scoupedy · 21-12-2009, 07:01 AM

Quote:

Originally Posted by talk2tiny

I realise that there's no reason to trust my credibility any more than anyone else who's posted here but if you're interested, I have degrees in mathematics and theoretical physics and have almost completed a PhD in theoretical quantum physics. I can assure you that from a purely mathematical standpoint, two solid objects travelling at 100km/h towards each other results in the exact same forces as a single object travelling at 200km/h hitting a stationary object.

It's all to do with relative velocities but I won't go into the details. Those who don't believe me yet won't be swayed by a proof either. Obviously I can't speak for the effects of crumple zones and glancing collisions etc but to answer the physics part of the question, definitely aim for the wall rather than the car.

Mate you should maybe have a talk to a first year physics lecturer / student OR Dynamics lecturer. The only thing that matters here is Momentum MV2 ( mass x Velocity(squared)) If you double the speed you Quadruple the energy if you remove the second car then all of the energy is absorbed by the single car, two cars mean two objects that absorb the energy.
If the wall does not move absorb heat or flex in any way then it is sending ALL of the energy back into the car. If you really do look into this you will see that you are wrong

Energy in crash example
m1= 1000kg
m2= 1000kg
v1= 2m/s
v2= 4m/s

what you are saying that two cars at 100kmph is same as one at 200kmph
lets explore that statement:

2(mv2)= mv2 is what you are saying - two cars same weight equals one car same weight travelling twice the speed.

2(1000*(2*2)=(1000*(4*4)
8000 does not equal 16,000
twice the energy created where has it gone the wall didn't absorb it just vanished did it

shyun · 21-12-2009, 11:23 AM

Quote:

Originally Posted by scoupedy

Mate you should maybe have a talk to a first year physics lecturer / student OR Dynamics lecturer. The only thing that matters here is Momentum MV2 ( mass x Velocity(squared)) If you double the speed you Quadruple the energy if you remove the second car then all of the energy is absorbed by the single car, two cars mean two objects that absorb the energy.
If the wall does not move absorb heat or flex in any way then it is sending ALL of the energy back into the car. If you really do look into this you will see that you are wrong

Energy in crash example
m1= 1000kg
m2= 1000kg
v1= 2m/s
v2= 4m/s

what you are saying that two cars at 100kmph is same as one at 200kmph
lets explore that statement:

2(mv2)= mv2 is what you are saying - two cars same weight equals one car same weight travelling twice the speed.

2(1000*(2*2)=(1000*(4*4)
8000 does not equal 16,000
twice the energy created where has it gone the wall didn't absorb it just vanished did it

The amount of energy is relative to the frame of refernce. So if you change reference frames then so will the energy of the objects realtive to the frame you're dealing with. That's why it's better using momentum for calculations as this doesn't mess things up when you change refernce frames.

auxr · 21-12-2009, 11:56 AM

O'k, been away for a couple of days - I think a poll should be conducted to award the 2009 physics prize for the right answer to the original post.

:

scoupedy · 22-12-2009, 06:39 AM

^^^^^^^^^^^
shunya- Reference points are only really needed if there are angles involved and you have to resolve using vectors in this case you don't need to use vectors as they are opposite each other

EgoFG · 22-12-2009, 12:40 PM

The simple answer to the original post is actually a question. "Worse for Who?"

If we are not to take into account the end state of the other car, then there is no difference (assuming a uniform collision).

If we look at it from only the end state of the insurance companies, then the two car head-on is twice as bad as the "one car in the wall" (again assuming a uniform collision)

When I used to drive my 1974 Chrysler by Chrysler 360, I used to reason this way: "Those little cars will absorb all the energy of the collision in their new-fangled crumple zones, and protect me "

When I "grew up" I realised that there might be people who loved the
dat-zda-bish-ota-ndas that I would polish off my chrome work - so I started dodgin' 'em.

Airmon · 22-12-2009, 03:47 PM

Quote:

Originally Posted by scoupedy

Mate you should maybe have a talk to a first year physics lecturer / student OR Dynamics lecturer. The only thing that matters here is Momentum MV2 ( mass x Velocity(squared)) If you double the speed you Quadruple the energy if you remove the second car then all of the energy is absorbed by the single car, two cars mean two objects that absorb the energy.
If the wall does not move absorb heat or flex in any way then it is sending ALL of the energy back into the car. If you really do look into this you will see that you are wrong

Energy in crash example
m1= 1000kg
m2= 1000kg
v1= 2m/s
v2= 4m/s

what you are saying that two cars at 100kmph is same as one at 200kmph
lets explore that statement:

2(mv2)= mv2 is what you are saying - two cars same weight equals one car same weight travelling twice the speed.

2(1000*(2*2)=(1000*(4*4)
8000 does not equal 16,000
twice the energy created where has it gone the wall didn't absorb it just vanished did it

You're getting a little confused here with your forumula. Momentum = (mass x velocity). And is measured in kg.m/s.
You're thinking of Kinetic energy which equals 1/2 x mass x velocity^2. In which case the energy quadruples as the speed doubles.

His point mainly has to do with frame of reference.

And to be honest most of us have it wrong when we try to look at the momentum and energy involved. All that matters in the scenario are the forces, which as pointed out would be fairly equal in either case.

MAD · 22-12-2009, 03:52 PM

Quote:

Originally Posted by Airmon

You're getting a little confused here with your forumula. Momentum = (mass x velocity). And is measured in kg.m/s.
You're thinking of Kinetic energy which equals 1/2 x mass x velocity^2. In which case the energy quadruples as the speed doubles.

His point mainly has to do with frame of reference.

And to be honest most of us have it wrong when we try to look at the momentum and energy involved. All that matters in the scenario are the forces, which as pointed out would be fairly equal in either case.

If you're looking at each car the forces are identical in both cases, but if you're looking at the 'accident' there is twice as much force.

flappist · 22-12-2009, 04:11 PM

Quote:

Originally Posted by EgoFG

When I used to drive my 1974 Chrysler by Chrysler 360, I used to reason this way: "Those little cars will absorb all the energy of the collision in their new-fangled crumple zones, and protect me "

When I "grew up" I realised that there might be people who loved the
dat-zda-bish-ota-ndas that I would polish off my chrome work - so I started dodgin' 'em.

The other minor problem with your thinking is that while the difference in mass allowed you to push the impact back in the direction the little car was coming from there was still a significant amount ebsorbed by your car and while the driver of the bubble was being protected by the front of the bubble slowing down at a greater rate than the middle YOUR car is slowing down at the same rate in the front, middle and back so ALL of the force is demonstrated on the soft squidgy thing in the middle i.e.YOU.

So while the bubble car driver's pride and joy is smeared all over the front of your Valiant saving the driver, YOU are smeared all over the inside of your car saving IT.

Not such a good idea really........

Wally · 22-12-2009, 07:36 PM

OK, at great expense to me ($20 inc GST) I have cobbled together a test rig to see which camp is right. I'm not going to attest to the accuracy of said test rig, but I'm hoping it will provide a definitive result.

So what will do the most damage, the single vehicle or the headon?

I will be making a video for youtube.

Nows your chance to stick you neck out.

flappist · 22-12-2009, 08:00 PM

Quote:

Originally Posted by Wally

OK, at great expense to me ($20 inc GST) I have cobbled together a test rig to see which camp is right. I'm not going to attest to the accuracy of said test rig, but I'm hoping it will provide a definitive result.

So what will do the most damage, the single vehicle or the headon?

I will be making a video for youtube.

Nows your chance to stick you neck out.

Ok Sheldon, just make sure that there is not an electric can opener nearby......

Rodp · 22-12-2009, 08:22 PM

Quote:

Originally Posted by EgoFG

When I used to drive my 1974 Chrysler by Chrysler 360, I used to reason this way: "Those little cars will absorb all the energy of the collision in their new-fangled crumple zones, and protect me "

When I "grew up" I realised that there might be people who loved the
dat-zda-bish-ota-ndas that I would polish off my chrome work - so I started dodgin' 'em.

Doesn't quite work like that...

http://www.youtube.com/watch?v=_xwYBBpHg1I

EgoFG · 22-12-2009, 09:31 PM

Quote:

Originally Posted by flappist

The other minor problem with your thinking is that while the difference in mass allowed you to push the impact back in the direction the little car was coming from there was still a significant amount ebsorbed by your car and while the driver of the bubble was being protected by the front of the bubble slowing down at a greater rate than the middle YOUR car is slowing down at the same rate in the front, middle and back so ALL of the force is demonstrated on the soft squidgy thing in the middle i.e.YOU.

So while the bubble car driver's pride and joy is smeared all over the front of your Valiant saving the driver, YOU are smeared all over the inside of your car saving IT.

Not such a good idea really........

Nah, not the two tonne beast against a Half tonne car - it needed Kitten #2 - but I fixed the problem

:
BTW - the CbyC was sold 15 years ago - that vid is MUCH more recent than more humourous comment !

you say "Other" ???

flappist · 22-12-2009, 10:57 PM

Quote:

Originally Posted by EgoFG

Nah, not the two tonne beast against a Half tonne car - it needed Kitten #2 - but I fixed the problem

:
BTW - the CbyC was sold 15 years ago - that vid is MUCH more recent than more humourous comment !

you say "Other" ???

Can you give me a quick list of 500kg cars? I Can't seem to find too many. In 1994 all the how did you put it, " dat-zda-bish-ota-ndas" were a lot more than 500kg.

And why did you have 381kg of ballast in your 1619kg 1974 Chrysler by Chrysler?

Looks like your memory is as good as your grasp of engineering physics......

EgoFG · 23-12-2009, 10:39 AM

Quote:

Originally Posted by flappist

Looks like your memory is as good as your grasp of engineering physics......

Dude, dont get into a flap. The comment about my 360 was obviously meant in jest (you dont need a high cut polish on Chrome).
Oh, and the extra kg - 95 was me, the rest was my hair and my ego (and two jerry cans of fuel to get me to the nearest servo if I ran out).

My physics is sound - per my comment on the original post.
The most successful way I have ever been able to demonstrate the head on vss 'immovable object' is with pool balls on a pool table.

the damage to the car is analogous to how far the balls bounce back.

If you get the balls to hit directly at the same speed, both will bounce back say x cm.

If you placed an 'immovable object' at the point of impact, and only rolled one ball - it would bounce back the same distance - x cm

To that one ball the impact of the collision is the same either way.

The fun thing about this is that it also shows how unlikely a uniform collision is - try to get those two pool balls to have one !
PS: dont put a house brick on a new pool table !

YOOT · 23-12-2009, 06:59 PM

If it hasn't been covered Fun = MA

Force (undefined) = Mass x Accelaration

F = 1500kg x 55.5m/s (Falcon doing 200kph)

F = 83100N

F = 3000kg x 27.7m/s (Two Falcons doing 100kph)

F = 83100N

In theory, exactly the same, but in reality there are other factors. In Theory Falcons should sell better than Commodores.

Airmon · 23-12-2009, 07:32 PM

Quote:

Originally Posted by YOOT

If it hasn't been covered Fun = MA

Force (undefined) = Mass x Accelaration

F = 1500kg x 55.5m/s (Falcon doing 200kph)

F = 83100N

F = 3000kg x 27.7m/s (Two Falcons doing 100kph)

F = 83100N

In theory, exactly the same, but in reality there are other factors. In Theory Falcons should sell better than Commodores.

In the scenario acceleration is 0 as the velocity is constant.

geckoGT · 23-12-2009, 08:52 PM

Quote:

Originally Posted by YOOT

If it hasn't been covered Fun = MA

Force (undefined) = Mass x Accelaration

F = 1500kg x 55.5m/s (Falcon doing 200kph)

F = 83100N

F = 3000kg x 27.7m/s (Two Falcons doing 100kph)

F = 83100N

In theory, exactly the same, but in reality there are other factors. In Theory Falcons should sell better than Commodores.

In the time this thread has been going the original question appears to have been lost.

The original question was which is worse, a car into a immovable object at 100 or two cars head, both traveling at 100?

EgoFG · 23-12-2009, 10:00 PM

Quote:

Originally Posted by geckoGT

In the time this thread has been going the original question appears to have been lost.

The original question was which is worse, a car into a immovable object at 100 or two cars head, both traveling at 100?

I think I answered it.

Quote:

Originally Posted by EgoFG

The simple answer to the original post is actually a question. "Worse for Who?"

If we are not to take into account the end state of the other car, then there is no difference (assuming a uniform collision).

If we look at it from only the end state of the insurance companies, then the two car head-on is twice as bad as the "one car in the wall" (again assuming a uniform collision)

Perhaps I should have said "Both Cars" rather than being flip and referring to "the insurance companies"

geckoGT · 23-12-2009, 11:52 PM

Quote:

Originally Posted by EgoFG

I think I answered it.

Perhaps I should have said "Both Cars" rather than being flip and referring to "the insurance companies"

I was actually referring to his use of speeds in his comment, he stated 200 into a wall in his calculations, the original question was 100.

EgoFG · 24-12-2009, 12:26 AM

Quote:

Originally Posted by geckoGT

I was actually referring to his use of speeds in his comment, he stated 200 into a wall in his calculations, the original question was 100.

there is actually no 200kmph in the OP.

There is this closing speed concept - but in a uniform collision this is irrelevant- it is the same as two 100kmph wall hits, but the OP is compareing one 100kmph wall hit to a 100kmph both ways head-on.

My comment stands
If you only consider one of the cars, the hit is the same.

If you do not care about the rest of humanity, then take the head-on every time - you have a better chance of the dice rolling your way - there are no dice on a wall hit.