Tesla Crash in Indy

[wdolson]

No, this is not correct.

There are two factors at work here:

1) Total energy that must be dissipated. This is 0.5 * m * v^2 for each moving object.
2) Available mass to deform and dissipate that energy once the collision occurs.

Let's do actual calculations with reasonable numbers:

Car #1 (Telsa): m1: 2000 kg
Car #2 (2nd Tesla): m2: 2000 kg
Wall: Immovable, infinitely strong, i.e. no deformation or ability to dissipate energy.

Now we'll look at 5 different scenarios, in order of crash severity (using the dissipation factor as the crash severity indicator):

A. Car #1 @ 100 km/hr vs. Car #2 stationary:
Total Energy = 0.5 * m1 * v1^2 = 0.5 * 2000 kg * 100 km/hr ^ 2 = 772 kJ.
Dissipation factor = Total Energy / Total Mass = 772 kJ / 4000 kg = 0.193 J/g.

B. Car #1 @ 100 km/hr vs. Car #2 @ 100 km/hr:
Total Energy = 0.5 * m1 * v1^2 + 0.5 * m2 * v2^2 = 0.5 * 2000 kg * 100 km/hr ^ 2 + 0.5 * 2000 kg * 100 km/hr ^ 2= 1543 kJ.
Dissipation factor = Total Energy / Total Mass = 1543 kJ / 4000 kg = 0.386 J/g.

C. Car #1 @ 100 km/hr vs Wall:
Total Energy = 0.5 * m1 * v1^2 = 0.5 * 2000 kg * 100 km/hr ^ 2 = 772 kJ.
Dissipation factor = Total Energy / Total Mass = 772 kJ / 2000 kg = 0.386 J/g.

D. Car #1 @ 200 km/hr vs. Car #2 stationary:
Total Energy = 0.5 * m1 * v1^2 = 0.5 * 2000 kg * 200 km/hr ^ 2 = 3086 kJ.
Dissipation factor = Total Energy / Total Mass = 3086 kJ / 4000 kg = 0.772 J/g.

E. Car #1 @ 200 km/hr vs. Wall:
Total Energy = 0.5 * m1 * v1^2 = 0.5 * 2000 kg * 200 km/hr ^ 2 = 3086 kJ.
Dissipation factor = Total Energy / Total Mass = 3086 kJ / 2000 kg = 1.543 J/g.


You can see how much the speed affects the outcome. Physics doesn't lie, and that squared velocity will always catch up to you. Compare scenarios B and D: Same closing speed of 200 km/hr, but with all the speed in one car, the crash is twice as severe as the same collision with the speed evenly split between the two cars.

But the available mass to dissipate the crash energy is the other forgotten factor. Compare scenarios D and E: Same closing speed of 200 km/hr, but without the extra 2000 kg of the other car to dissipate the crash energy, the collision with the wall is twice as severe as the collision with another car.

I was comparing scenario B vs D. With relativistic velocities, the math gets different, but we're nowhere near the speed of light here, so plain Newtonian Physics is fine. When defining a system in Newtonian system, velocities are relative to the frame of reference. Walking 5 mph on a train moving at 50 mph can be described as 45 or 55 mph depending on the direction your walking on the train if the frame of reference is from outside the train. But as far as your body is concerned you are only traveling 5 mph as long as the train doesn't change direction or speed.

The math gets a little trickier when dealing with kinetic energy because the velocity is squared and you also have to ask the question whether you're looking at the entire energy of the collision or the energy absorbed by just one car in the head on collision. This explains the math:
http://mathforum.org/library/drmath/view/60747.html

The math is more straightforward when the two cars are the same weight and the collision is just right to make the math easy, which doesn't happen in the real world. But ultimately I didn't do the greatest job framing the problem in my head before trying to describe it.

That's true, but misleading. There is a significant difference between crashing into an immovable wall and a stationary car.

I was not comparing the situation when two cars hit one another head on vs a car traveling twice as fast into a wall. The results are going to be different because of the difference in characteristics of the wall.

 

[stopcrazypp]

I was not comparing the situation when two cars hit one another head on vs a car traveling twice as fast into a wall. The results are going to be different because of the difference in characteristics of the wall.

I'm not talking about the smaller differences between wall and car crash structures, but rather about the larger kinetic energy difference that the wall would not be moving, while the two cars would still be moving immediately after the crash involving the stationary car. That is the major difference most people miss and why people continually are misled into the myth.

To illustrate my point, let's say for example there is a stationary car that is bolted to the ground, but otherwise is identical to the moving car.

Again KE = kinetic energy of car moving at 45mph.

2 cars at 45 mph head on: 2KE dissipated by 2 cars = 1KE per car
1 car at 90 mph into stationary car: 2KE by 2 cars (2 cars still moving at 45mph after crash) = 1KE per car
1 car at 90 mph into stationary car bolted to ground: 4KE by 2 cars = 2KE per car
1 car at 90 mph into wall: 4KE by 1 car = 4KE per car

Notice just by bolting the car to the ground, this doubles the amount of energy dissipated, because the cars will not be moving after the crash.

 

[SomeJoe7777]

I'm not talking about the smaller differences between wall and car crash structures, but rather about the larger kinetic energy difference that the wall would not be moving, while the two cars would still be moving immediately after the crash involving the stationary car. That is the major difference most people miss and why people continually are misled into the myth.

To illustrate my point, let's say for example there is a stationary car that is bolted to the ground, but otherwise is identical to the moving car.

Again KE = kinetic energy of car moving at 45mph.

2 cars at 45 mph head on: 2KE dissipated by 2 cars = 1KE per car
1 car at 90 mph into stationary car: 2KE by 2 cars (2 cars still moving at 45mph after crash) = 1KE per car
1 car at 90 mph into stationary car bolted to ground: 4KE by 2 cars = 2KE per car
1 car at 90 mph into wall: 4KE by 1 car = 4KE per car

Notice just by bolting the car to the ground, this doubles the amount of energy dissipated, because the cars will not be moving after the crash.

KE = 0.5 * m * v^2

Kinetic energy to be dissipated does not depend on whether an object is bolted to the ground or not.

1 car at 90 MPH = 4KE, period.

If the other car is not bolted and free to move, there are additional opportunities to dissipate energy -- the cars can travel after the collision, collide with the ground, with each other, with other objects. But one way or another, 4KE still has to be dissipated.

Please see my post earlier with this same analysis using real numbers. Total energy to be dissipated is dependent solely on speed and mass, just like the equation says.

 

[mr10012]

@stopcrazypp is correct... Think about the collision happening in space (or a surface with no friction). Then in the frame of reference where a 90mph car hits a stationary car, the post-crash 2-car wreckage would be moving at 45mph in the direction of the moving car (preservation of momentum), and thus the change in kinetic energy as a result of the crash would be 4KE - 2KE = 2KE. In the frame of reference where the 45mph cars hit each other, the wreckage would be stationary for a change in kinetic energy as a result of the crash of 2KE - 0 = 2KE. In both cases the same energy (and thus damage) is released at the crash.

Granted, however, that if the 90mph crash takes place on a real road, the additional 2KE of energy would need to be dissipated, most likely through friction with the pavement or additional collisions. That could result in additional post-crash damage, but almost certainly not comparable to that from hitting a wall (or a bolted car).

 

[SomeJoe7777]

@stopcrazypp is correct... Think about the collision happening in space (or a surface with no friction). Then in the frame of reference where a 90mph car hits a stationary car, the post-crash 2-car wreckage would be moving at 45mph in the direction of the moving car (preservation of momentum), and thus the change in kinetic energy as a result of the crash would be 4KE - 2KE = 2KE. In the frame of reference where the 45mph cars hit each other, the wreckage would be stationary for a change in kinetic energy as a result of the crash of 2KE - 0 = 2KE. In both cases the same energy (and thus damage) is released at the crash.

Your (incorrect) assumption here is that all energy in the system after the collision is still kinetic, i.e. held as motion of the objects. This is true only for a perfectly elastic collision.

If some of the energy of motion has already been converted to part deformation and/or fragmentation then the collision is inelastic, and final velocities of the objects will no longer behave in the way you describe.

Also, while conservation of momentum does apply in the general sense, the cars alone are not the isolated system where the momentum is conserved. The atmosphere, ground, and other objects also have momentum transferred to them.

 

[RogerHScott]

one way or another, 4KE still has to be dissipated

Sure, but any car travelling at this speed has that same KE, and they all need to dissipate it, yet, miraculously, 99.999997% of them
manage to do it without any crash or damage. That's because it is possible for a car to lose 2KE harmlessly by normal slowing down.
So the interesting question here is not what the total change in KE is but, rather, what the additional change in KE implied by the crash
is, above and beyond what would have been (safely and harmlessly) dissipated without a crash.

If some of the energy of motion has already been converted to part deformation and/or fragmentation then the collision is inelastic, and final velocities of the objects will no longer behave in the way you describe.

This, of course, is exactly the stuff you can't ignore when you're comparing the vehicle-vehicle case to the vehicle-wall case, unless you
are fortunate enough to be running into a rubber (or crushable) wall.

 

[stopcrazypp]

KE = 0.5 * m * v^2

Kinetic energy to be dissipated does not depend on whether an object is bolted to the ground or not.

1 car at 90 MPH = 4KE, period.

If the other car is not bolted and free to move, there are additional opportunities to dissipate energy -- the cars can travel after the collision, collide with the ground, with each other, with other objects. But one way or another, 4KE still has to be dissipated.

Please see my post earlier with this same analysis using real numbers. Total energy to be dissipated is dependent solely on speed and mass, just like the equation says.

To be clear, I'm talking solely about the energy dissipated immediately in the crash itself as damage to the cars. I thought that was obvious, but I guess it wasn't.

I'm not concerned about energy that will be dissipated anyways just from friction/air resistance etc in the normal course of travel (for example if the car let off the throttle and it never crashed, it would eventually come to a stop anyways, with no damage to the car). Yes, the left over 2KE will also be dissipated eventually by those forces, but that is not the concern here (only the parts that are attributable to the damage to the vehicles).

As per the other points about inelasticity etc, yes I'm aware those obviously will change things, but the conservation of momentum above contributes a much larger/obvious proportion from a macro level.

 

[SomeJoe7777]

Sure, but any car travelling at this speed has that same KE, and they all need to dissipate it, yet, miraculously, 99.999997% of them manage to do it without any crash or damage. That's because it is possible for a car to lose 2KE harmlessly by normal slowing down. So the interesting question here is not what the total change in KE is but, rather, what the additional change in KE implied by the crash is, above and beyond what would have been (safely and harmlessly) dissipated without a crash.

I'm not concerned about energy that will be dissipated anyways just from friction/air resistance etc in the normal course of travel (for example if the car let off the throttle and it never crashed, it would eventually come to a stop anyways, with no damage to the car). Yes, the left over 2KE will also be dissipated eventually by those forces, but that is not the concern here (only the parts that are attributable to the damage to the vehicles).

Ah, I believe we're actually on the same page here with the physics analysis. Yes, the objects moving have an energy and momentum, and yes, the energy and momentum must be conserved.

What we're talking about here is when the energy, no matter how much it is, gets dissipated, what proportion of it gets dissipated in a preferable way vs. a non-preferable way.

Preferable ways of dissipating energy would be regen braking, friction brakes, and then maybe tire skids.

Non-preferable ways are collisions involving part deformation and fragmentation.

So, in the above examples, a vehicle with 4KE vs. a bolted-down stationary vehicle dissipates more of the 4KE in non-preferable ways than a vehicle with 4KE vs. a non-bolted down stationary vehicle.

This concept can be extended to a 4KE vehicle vs. a wall, where nearly all of the 4KE will be dissipated in a non-preferable way.


My one remaining objection to your analysis is the estimate of the kinetic energy left in the vehicles after the 4KE car vs. stationary non-bolted car immediately after the collision, which you estimated at 2KE. While we can say for sure that the total kinetic energy left in the masses of both cars immediately after the collision is less than 4KE, we do not know exactly how much less. That completely depends on the exact collision parameters, amount of deformation, amount of inelasticity, etc. Some of that remaining kinetic energy after the collision may then be able to be dissipated in more preferable ways, such as skidding along the ground. This is also what I was saying earlier in that there are additional opportunities to dissipate energy.

 

[bak_phy]

Just for fun lets put these cars in space and with an onlooker between them (and a bit to the side),
To the onlooker you have 2 cars each with 1KE slamming into each other.
To the guy in car 1 it looks like he is just sitting there minding his own business when car 2 slams into him with 4KE of energy.
To the guy in car 2 it looks like he is just sitting there minding his own business when car 1 slams into him with 4KE of energy.

So.... After the collision there can only be 1 level of destruction but there are either 2KE or 4 KE of energy both of which are 100% correct.

The example with the wall is car one just minding his own business when a planet hits him with an effective infinite amount of KE.

(edited original for clarification)

 

[tga]

The Mythbusters episode mentioned up thread did a great job of explaining what happens when two cars hit each other vs hitting a stationary object. After reading this thread I had to go back and watch that episode, as I had always thought you added the speed of the two cars together when in fact that is not the case.

It's too bad they didn't add 100mph into a stationary car (probably ran out of budget for sacrificial cars) too show is not just closing speed (IE, @SomeJoe7777 's case B vs D)

 

[stopcrazypp]

Ah, I believe we're actually on the same page here with the physics analysis. Yes, the objects moving have an energy and momentum, and yes, the energy and momentum must be conserved.

What we're talking about here is when the energy, no matter how much it is, gets dissipated, what proportion of it gets dissipated in a preferable way vs. a non-preferable way.

Preferable ways of dissipating energy would be regen braking, friction brakes, and then maybe tire skids.

Non-preferable ways are collisions involving part deformation and fragmentation.

So, in the above examples, a vehicle with 4KE vs. a bolted-down stationary vehicle dissipates more of the 4KE in non-preferable ways than a vehicle with 4KE vs. a non-bolted down stationary vehicle.

This concept can be extended to a 4KE vehicle vs. a wall, where nearly all of the 4KE will be dissipated in a non-preferable way.


My one remaining objection to your analysis is the estimate of the kinetic energy left in the vehicles after the 4KE car vs. stationary non-bolted car immediately after the collision, which you estimated at 2KE. While we can say for sure that the total kinetic energy left in the masses of both cars immediately after the collision is less than 4KE, we do not know exactly how much less. That completely depends on the exact collision parameters, amount of deformation, amount of inelasticity, etc. Some of that remaining kinetic energy after the collision may then be able to be dissipated in more preferable ways, such as skidding along the ground. This is also what I was saying earlier in that there are additional opportunities to dissipate energy.

Well yes, obviously the 2KE number is just a very rough estimate, from the simple assumption that the two cars end up moving at 45mph after the crash. It certainly will be different in the real world once all the details are figured in.

 

[SomeJoe7777]

Just for fun lets put these cars in space and with an onlooker between them (and a bit to the side),
To the onlooker you have 2 cars each with 1KE slamming into each other.
To the guy in car 1 it looks like he is just sitting there minding his own business when car 2 slams into him with 4KE of energy.
To the guy in car 2 it looks like he is just sitting there minding his own business when car 1 slams into him with 4KE of energy.

So.... After the collision there can only be 1 level of destruction but there are either 2KE or 4 KE of energy both of which are 100% correct.

The example with the wall is car one just minding his own business when a planet hits him with an effective infinite amount of KE.

(edited original for clarification)

Yes, you're speaking of reference frames, which can change the apparent velocity, and therefore the apparent kinetic energy of each object.

However, the key here is that all objects after the collision must eventually come to rest within their reference frame (on Earth, at least), and at that point each object's v=0 and KE=0.

If each post-collision object (bulk of the car, pieces of the car, etc.) comes to rest (v=0) in the reference frame of the Earth/ground, then the kinetic energy that must be dissipated for each post-collision object is 0.5 * m * v^2, where v=initial velocity relative to the reference frame where the objects will come to rest.

In space, you will have to pick a reference frame, compute initial velocities using the speed relative to that reference frame, and also note that post-collision objects do not have to come to rest. Total energy will be conserved in any single reference frame. Picking a different reference frame may compute a different amount of initial kinetic energy, but that energy will also be conserved within that reference frame.

 

[bhzmark]

@SomeJoe7777 and @stopcrazypp and others who contributed,
As frustrating as this forum is sometimes, tangential discussions like above are what keeps me coming back. Well done and thank you for the educational reading.

 

[Max*]

As frustrating as this forum is sometimes, tangential discussions like above are what keeps me coming back. Well done and thank you for the educational reading.

I completely agree. I was just sitting on the sidelines reading.

 

[bak_phy]

@SomeJoe7777 Conservation of moment is conserved at all times and in all reference frames whether on earth or in space. This is regardless of any and all transfers which may happen between various forms of energy. E.g. All momentum is kinetic.
When all things appear to come to rest it's simply because any momentum which is not has been transferred the earth.
During the time in which a head on crash between two cars takes place this earth transfer is entirely meaningless as it initially takes place in the friction between the wheels and the road and then moments later between the friction of the other parts of the car and the road. E.g. After the destruction has taken place.
I do wish that the mythbusters episode had done the moving car hitting a stationary car experiment.
Depending on the elasticity of the collision, or how much energy was lost the stationary car would immediately after the impact be traveling backwards at anywhere from half the initial speed of the moving car to fully the initial speed.

 

[UberEV1]

How about adding one more variable to the discussion . . . the Smart Car . . . it is so small there are effectively no crush zones, thus it must somehow be "stiffer" to protect its occupants. This implies less crushing but greater deceleration, which is also not necessarily a good thing. Not sure how this part of the discussion will turn out, but I'm sure at its conclusion I will be glad to be driving the Model S.

 

[SomeJoe7777]

@SomeJoe7777 and @stopcrazypp and others who contributed,
As frustrating as this forum is sometimes, tangential discussions like above are what keeps me coming back. Well done and thank you for the educational reading.

Thanks, bhzmark! It's been a while since I've used some of my physics ... rattling a few cobwebs here and there ...

 

[RogerHScott]

it must somehow be "stiffer" to protect its occupants

Ummm.... how do you figure that any such "stiffness" protects the occupants? What you don't want is your flesh accelerating rapidly,
in the case where you're hit, or decelerating rapidly, in the case where you hit something. It is exactly that F(=MA)orce that does the
damage, and a stiff car will transfer it directly to your body.
This reminds me of one of the fallacies of superheroes (and other "enhanced" humans): they're shown using their flesh-and-bone bodies
to accelerate things (say, picking up a car and throwing it) that would involve forces that would shred their soft tissues.

 

[UberEV1]

Ummm.... how do you figure that any such "stiffness" protects the occupants? What you don't want is your flesh accelerating rapidly,
in the case where you're hit, or decelerating rapidly, in the case where you hit something. It is exactly that F(=MA)orce that does the
damage, and a stiff car will transfer it directly to your body.
This reminds me of one of the fallacies of superheroes (and other "enhanced" humans): they're shown using their flesh-and-bone bodies
to accelerate things (say, picking up a car and throwing it) that would involve forces that would shred their soft tissues.

I probably used the word "stiffness" too loosely, but in essence the idea is to protect the occupants of the Smart Car by leveraging the crush zones of the other car (as much as is possible for such a small car). In an excerpt HERE, the explanation reads: "The principle safety system is a high tensile steel safety cage M-B calls "the tridion safety cell." It protects its occupants like a walnut shell around the nut inside. The safety cell’s longitudinal and transverse rigidity actually activates the crumple zone design of the other vehicle involved in the accident to distribute the impact energy evenly over the little car's body." But as you point out, I would also be worried about the deceleration forces. In any case, a smaller car is (typically) no match for a larger car.

 

[SomeJoe7777]

in essence the idea is to protect the occupants of the Smart Car by leveraging the crush zones of the other car (as much as is possible for such a small car).

What happens if two Smart Cars hit each other?