Modeling vehicle lift in uphill front end collision

[Heidi Henkel]

Homework Statement

This is a video of the phenomenon. Start at 5:08 in this video and watch the next two crashes. That first one, where the car hits a wall, straight on, is the most helpful. The phenomenon is that when a car going uphill crashes, it has an upward speed/momentum vector from its path before it crashes (there is a horizontal vector and a vertical vector, in going uphill). Crashing into the wall stops the car from going forward, but doesn't stop the vertical vector, especially in the rear of the car. (The car is long enough that the movement is different in the front and rear. It's not like, say, a marble, where the mass is concentrated in one place. The mass is spread out the length of the car.) In the video, it seems that it doesn't take a huge amount of speed for a car going uphill, to lift off the ground on impact.

The questions is, if you know how far you have to jack up the rear of the vehicle to get the rear wheels to leave the ground, and the grade, can you find the minimum speed needed for the vehicle to leave the ground? If the vehicle was braking before impact, the rear wheels would have already started to elevate due to braking. ( See https://en.wikipedia.org/wiki/Weight_transfer and http://www.welltall.com/ymc/discovery/car/wt_xfer.html )[/B]

If a car has been braking and hasn't slowed down much, and then hits a wall, can we mathematically model what the minimum speed is they would have to carry, for the rear wheels to lift off the ground? For all 4 wheels to lift off the ground, like in the video? We are primarily looking for an estimated mathematical model with accuracy within about 10% source of error using precalculus math, though a more precise model using more advanced math would also be welcome.

The grade progressed steadily from 10% to 1% grade during braking, then the car hit the wall and the rear wheels lifted off and the front wheels just barely lifted off.

The distance the car needs to be jacked up, from normal ride height to wheel lift off is 0.3m.
The coefficient of friction of tires on road surface is 0.3.

Homework Equations

basic trig to make vehicle's speed up the hill into horizontal and vertical vectors
d= .5at^2+V_0t[/B]

The Attempt at a Solution

One thing we need to look at is the grade of the road. However, the grade of a road is not usually constant. Would we look at average grade over the course of a certain amount of time back from the collision? Average grade during the time spent braking, before the collision? Grade right at the moment of collision? Grade at start of braking? It seems that the rear wheel lift while braking would be related to the grade throughout the braking process. I am not sure what grade to use, to get the vertical and horizontal vectors. Does what grade to use, depend partly on the friction in the braking process? Better friction means the grade at the start of braking is more important, and worse friction means the grade toward the end of braking is more important?[/B]

It seems the impact moment, itself, is fairly simple if we understand what happened before that (How much the rear wheels have already lifted before impact, and what the momentum vectors are in the rear of the car before impact...which might have more to do with the grade some distance back, than the current grade).

At moment of impact I am using d= -.5at^2+V_0t
This is just vertical motion in relation to gravity.
V_0 would be the vertical component of the velocity at impact.
V_0 would equal delta V because this equation is just looking at from first moment of impact until the car reaches its maximum height, when V=0. So I can substitute at for V_0 on the first round and solve for t, then use the same equation again, plugging in my t, and solve for V_0. I am not sure that's right, though, because when I do this process with higher values for d (which is how far the car lifts off the ground) I seem to get lower values for V_0, which makes no sense.

If I can get a model for the collision moment, I can go back and experiment with what to use for grade, and just figure out how much difference it even makes, how I account for grade. It's possible that the choice of how to account for grade makes a small enough difference that it doesn't matter?

Are there other dynamics at work, that I am not taking into account?

 

[haruspex]

I cannot see that this would be a result of the car's uphill movement. The critical question is how high the barrier is in relation to the car's mass centre. If the mass centre is higher, the impact will create torque tending to lift the rear of the car. Indeed, the worst case would be going downhill.

 

[Heidi Henkel]

Watch the video of the racing car crash compilation I put in the introduction. The upward moving car crashes into a straight wall that is taller than the car. This makes it clear that the barrier is not causing the upward lift. The car's momentum is causing the upward lift. We are talking about the rear of the car lifting, not the front of the car lifting. If we were talking about the front of the car lifting, you would be right- the height of the barrier would be important and if the car's center of mass is higher, it could cause the car's front to lift. We are talking about the rear of the car lifting and this is demonstrated in a video where the barrier is much taller than the car. The car's upward vector lifts the car when it collides. However, the rear end of the car can already be starting to lift due to braking, as in this video if you start at 1:38, the braking immediately after that.
The car has an upward velocity vector due to its upward travel, and its rear end also is going upward due to braking.

My basic analysis of this is correct. I'm not asking about that. I'm asking about how to model it mathematically. Please watch the video.

 

[haruspex]

I was not as accurate as I should have been in my previous post. I should have said the height of the impact point in relation to the car's mass centre. No matter how high the wall is, the impact point may be quite low.

Edit: there is another possibility. If the impact point is somewhat higher than the car's mass centre then the torque will drive the rear of the car downwards. The tyres' elasticity might cause enough rebound to tip the car up.

 

[Heidi Henkel]

Yes, the impact point is low because the front of the car is already dipping down because of the front-loading effect of braking, as you can see in the more recent braking video I sent you. This is still a result of the car's own motion. The rear of the car also has an upward velocity vector because the car is traveling uphill. Both things, the front of the car dipping due to braking, making the impact point low, and the car's upward velocity due to going uphill, help the car's rear to go up on impact. See this video starting at 2:44. Both things- the tilting of the car forward during braking and the upward velocity component vector due to going uphill- contribute to the rear of the car lifting off the ground on impact. The front wheels may also lift off on impact, as in the first video. Different braking systems produce different degrees of forward tilt during braking. There are different philosophies of designing brakes. One philosophy is to overpower the front brakes in acceptance of the fact that the mass of the car shifts forward during braking. Another philosophy is to strengthen the rear brakes to help reduce this forward tilting. Notice that in that first video, the racing crash compilation, the car starting at 5:08 isn't tilting forward much during its braking prior to collision. So I guess there is a range of what happens in terms of forward tilt. Even with negligible forward tilt prior to impact, as in the first video, the car still lifts off the ground. In the first video, the lift seems to be primarily due to the upward vector of the car going uphill. I wonder if the rears of cars routinely lift in front end collisions on flat ground if they are already tilted forward before impact due to braking. It seems to me to be more of an uphill front end collision phenomenon, but if you can find any videos of the same phenomenon on level ground, I would be interested.

 

[Heidi Henkel]

Part of what I am trying to describe is the situation in which the front tires come just barely up off the road, also. In the braking on level ground models, the front of the car goes DOWN and the rear of the car goes UP. This does not look like a recipe, by itself, for both sets of wheels to lift on impact. The impact also including an overall upward vector of the car's travel, adds a significant reason for both sets of wheels to lift off the ground. If it's just braking and low point of impact due to the front of the car dipping down due to braking, I do not see how that would cause the front wheels to lift off the ground on impact. It seems that the upward momentum of the overall car has a significant role also.

Your "edit" is interesting, that if the first point of impact is high, the car's rear end might be pushed down then bounce up due to the elasticity of the tires, and, I would say, shocks and struts. This would take some time for all this elasticity to kick in, though. In that first video, it seems that the first impact is up; the upward lift is not due to a bounce from having first gone down on impact.

 

[Heidi Henkel]

I do think that without braking prior to impact, the impetus for the car to lift would be much less, and would require a higher overall velocity of the car. If the car is braking and the front end is already dipping and the rear is already rising, when impact happens, it takes very little overall velocity of car, for both sets of wheels to lift on impact. That first video is not at a very high speed. This is very interesting to me, how low a speed this car is moving at, yet it lifts. I think on a more slippery surface, the effect of braking in helping the car to lift would be less, but would still be much greater than not braking.

 

[haruspex]

In the braking on level ground models, the front of the car goes DOWN and the rear of the car goes UP.

Yes, this makes more sense. Braking creates a torque tipping the car forwards, lowering the point of impact.

The impact also including an overall upward vector of the car's travel, adds a significant reason for both sets of wheels to lift off the ground. If it's just braking and low point of impact due to the front of the car dipping down due to braking, I do not see how that would cause the front wheels to lift off the ground on impact. It seems that the upward momentum of the overall car has a significant role also.

No, this simply does not work. Draw a diagram of the car on a level road striking a barrier at below the height of the car's centre of mass. There is no torque, so the car does not flip. Now rotate the whole picture so that the car is going uphill at the time. The point of impact is still in the line of movement of the car's centre of mass, so there is still no flipping torque.
Start again, on the level, now with a lower impact point, so there is a flipping torque. Rotating this, there is exactly the same flipping torque from the impact. All that changes is that the car has to rotate further before gravity stops opposing the flip, so it is less likely to flip right over.

Where your intuition goes wrong is that a) the upward momentum applies to all parts of the car, so does not lead in itself to a rotation, and b) the impact force acts to oppose the car's motion, i.e. if the car is going uphill then the impact force on it acts downhill.

All of this assumes the road surface is straight. If the car is cresting a hill then it already has a 'forward roll' angular momentum.
It also assumes the barrier height is the same, in terms of perpendicular distance from (the continuation of) the road surface. If you rotate the diagram in terms of the car and road but leave the barrier vertical the barrier might be effectively lower.

You mention all four wheels lifting on impact. I have not seen that in the videos. Please provide a timestamp for that.

 

[Heidi Henkel]

First paragraph is not on the topic. I am not discussing a car flipping. It also makes no sense even if it were on the topic. There are plenty of other reasons for the car to have the rear wheels or all 4 wheels lift off, besides this very narrow set of criteria.

Second paragraph: part a, the upward momentum applies to all parts of the car, and it does help lead to rotation because the front of the car gets temporarily pinned on the obstacle, preventing it from going up, but the rest of the car has nothing to prevent it from going up at any time. Part b is also incorrect. The impact force does not directly oppose the car's motion. The object the car crashes into is a vertical object and is effective in stopping the car's horizontal vector. It does nothing to change the car's vertical vector. Except the front of the car is pinned for an instant of collision while the rest of the car is still going up, initiating a rotation in the direction of forward roll.

3rd paragraph: Yes, cresting a hill, which is the topic I introduced, does cause the car to already have some forward roll angular momentum. Thanks. That is helpful.
The barrier goes from the ground to well above the height of the car, so barrier height is not a variable. Height of collision point would only be changed if the car's own motion causes the bumper of the car to be lower (such as due to braking prior to impact in certain designs of braking systems).

4th paragraph: I clearly stated this first in my description. The first video I posted, start at 5:08 and watch the collision of that car. It takes about 5 seconds. That car has all 4 wheels leave the ground.

 

[Heidi Henkel]

the instant when the front bumper is pinned on the obstacle during collision, is enough to initiate a forward roll rotation of the vehicle. Once that is started...an object that is in motion stays in motion unless a force changes that...it keeps going until a force stops it. That force is gravity.

 

[haruspex]

Second paragraph: part a, the upward momentum applies to all parts of the car, and it does help lead to rotation because the front of the car gets temporarily pinned on the obstacle, preventing it from going up, but the rest of the car has nothing to prevent it from going up at any time. Part b is also incorrect. The impact force does not directly oppose the car's motion. The object the car crashes into is a vertical object and is effective in stopping the car's horizontal vector. It does nothing to change the car's vertical vector. Except the front of the car is pinned for an instant of collision while the rest of the car is still going up, initiating a rotation in the direction of forward roll.

When two rigid bodies meet, there is a normal force and, potentially, a frictional force. The normal force acts to oppose the relative motion of the bodies perpendicular to the contact plane. The frictional force acts to prevent relative motion parallel to the contact plane.
If there is no slipping, the net force must be parallel to the relative motion of the bodies. The verticality of the barrier can only be relevant if the car can slide up it during impact.
When I mentioned flipping forwards, I did not mean to imply flipping right over necessarily, merely a tendency to pitch forwards.

4th paragraph: I clearly stated this first in my description. The first video I posted, start at 5:08 and watch the collision of that car. It takes about 5 seconds. That car has all 4 wheels leave the ground.

I watched that, and it does not show all four wheels lifting initially. Immediately after impact, the rear of the car lifts but the front does not. This is consistent with the fact that in that instance the impact point is low. Indeed, the front seems to bed right down.
On rebound from the barrier, the rear starts to fall, since it is airborne, while the front rises in response to the compressed suspension. Having rebounded, there is no forward pitching torque, but now a backward pitching torque.

 

[Heidi Henkel]

1st paragraph: If the car were a tiny particle, you would be right. The car is probably 9-10 feet long. While there is something preventing the front bumper from lifting up on impact, there is nothing preventing the rear of the car from lifting up on impact, hence the initiation of forward roll. Think about problems like forces on long I-beams in civil engineering. There is a lot of different stuff going on all along this long I-beam. When an object isn't a small particle, different things can be happening at different places and this can cause the object to do motions such as rotation. Another situation in which you see this with cars is when a car without ABS brakes tries to stop on a road surface where the right and left sides have different amounts of friction. The car spins. Because different things happen in different locations for the same object, because the object has dimensions and is not a small particle. Think about the kinodynamics of the rear of the car, independently from the front of the car.

That is an interesting analysis of the collision. There is a moment when all 4 wheels are simultaneously off the ground and the car rebounds backward a little bit- horizontally- while completely airborne.

 

[haruspex]

1st paragraph: If the car were a tiny particle, you would be right. The car is probably 9-10 feet long. While there is something preventing the front bumper from lifting up on impact, there is nothing preventing the rear of the car from lifting up on impact, hence the initiation of forward roll. Think about problems like forces on long I-beams in civil engineering. There is a lot of different stuff going on all along this long I-beam. When an object isn't a small particle, different things can be happening at different places and this can cause the object to do motions such as rotation. Another situation in which you see this with cars is when a car without ABS brakes tries to stop on a road surface where the right and left sides have different amounts of friction. The car spins. Because different things happen in different locations for the same object, because the object has dimensions and is not a small particle. Think about the kinodynamics of the rear of the car, independently from the front of the car.

Suspension and crumpling apart, the car is a rigid body. It cannot start to rotate unless there is a torque acting on it. You have not explained how a car heading uphill into a cliff face, without braking, with the point of impact at or above the mass centre height, would experience a forward pitch torque. It is not enough to say, well, there's a lot going on. You have to find some credible explanation of what you think you have observed, or check again whether that is really what you see.
The spinning car with differential friction is another matter. That is a very clear and credible explanation.

 

[Heidi Henkel]

I think I have explained this many times and you have ignored me. The car's motion has a horizontal vector and a vertical vector. When the car hits the barrier, the front of the car is stopped in its horizontal and vertical motion. The rear of the car has nothing keeping it from going upward. The rest of the car keeps it from going horizontally but there is nothing preventing it from going up. So it goes up. Not only that, but there are several ways in which the car's forward pitch has already been initiated. One is the braking, which loads the front of the car and causes the front of the car to go down and the back of the car to go up. Another is (in my example, but not in the video) the cresting of the hill. All three of these mechanisms act on the car to cause the rear end to lift.

You also see this when a car has a front end collision at an angle. The front of the car is stopped and the rear of the car is not stopped and tries to continue to go forward, but is deflected into a slightly different path. The rear essentially tries to catch up to the front and this results in rotation.

When the front of the car and the rear of the car have two different force situations and do two different motions, this can cause the rotation. There doesn't need to be a rotational force to cause rotation.

Why the front wheels also lift is more of a question. It could be a rebound from the shocks and struts and tires, from having gone down due to braking, or it could be just an extension of the forward pitch.

 

[haruspex]

I think I have explained this many times and you have ignored me. The car's motion has a horizontal vector and a vertical vector. When the car hits the barrier, the front of the car is stopped in its horizontal and vertical motion. The rear of the car has nothing keeping it from going upward. The rest of the car keeps it from going horizontally but there is nothing preventing it from going up. So it goes up. Not only that, but there are several ways in which the car's forward pitch has already been initiated. One is the braking, which loads the front of the car and causes the front of the car to go down and the back of the car to go up. Another is (in my example, but not in the video) the cresting of the hill. All three of these mechanisms act on the car to cause the rear end to lift.

I'm not ignoring that point, I have answered it already, but you do not believe my answer.
Consider a barrier consisting of a horizontal rod. If the car strikes it horizontally, the impact on the car is horizontal. If the car strikes it while going up at an angle of 45 degrees, the impact from the rod is downward at an angle of 45 degrees.
If the barrier is a vertical plate and the car is going uphill, the result will be different if the car can slide up the plate, but even then that does not result in a forward pitch. Indeed, it would have the opposite consequence.
I already agreed that braking does cause a forward pitch, but striking the barrier can only add to this to the extent that the line of action of the impact force passes below the mass centre.

 

[Heidi Henkel]

So the cause of the lift off is the forward pitch caused by the reducing grade of the hill, and the impact point being below the center of mass due to the lowering of the front of the car during braking. And also the forward pitch movement caused by the braking (front end going down, back end going up). The fact that the car is going uphill doesn't add to it?

I am going to experiment with little Matchbox cars that are going up a constant grade incline and are not braking, to see if I can observe what you are saying. I am skeptical but I do understand what you are saying. If I push a little car up a straight ramp and let it coast and crash into something, it should not lift off if you are right.

 

[haruspex]

If I push a little car up a straight ramp and let it coast and crash into something, it should not lift off if you are right.

OK, but remember that you have to arrange that the impact point is not below the line of travel of the mass centre. That might not be simple with matchbox cars.

 

[Heidi Henkel]

I wonder if bumpers of some cars automatically place the impact point (the bumper, if slamming into a solid object) below the center of mass. Most of the engine, and most of the car, is usually above the bumper level.

 

[haruspex]

I wonder if bumpers of some cars automatically place the impact point (the bumper, if slamming into a solid object) below the center of mass. Most of the engine, and most of the car, is usually above the bumper level.

Many countries/states set maximum heights for bumpers.