Wheelie of a car coming out of a ditch: what is the correct model?

[alex33]

Rear wheel angular velocity retrieval
$mgh=\frac {1}{2}Jω^2$
I make the model for a parallelepiped of height h, width w, length l:
$J=\frac {1}{12}m(h^2+w^2)$
$ω=\sqrt{\frac {24gh}{h^2+w^2}}$

Here again the basic motion data:
Centre of the arc AB (the dip arc): (X=0; Z=0)
Length of the arc: AB = 2,082 m
Car wheelbase Length L: 2,286 m
Car Length L: 2,286 m (for the moment in order to simplify)
Radius of the arc: r = 2,875 m (reviewed)
Angle subtended by the arc AB: β = 2α = 41,5° (reviewed)
Angle between ground and arc tangent at A and B: α = 20,75° (reviewed)
Initial speed of the car at the entrance to the dip: V₀= 2,61 m/s (reviewed)
Car Centre of Mass = 1/2 L (for the moment in order to simplify)
g = 1,62 m/s²
X₀= r*SEN(α)
Z₀= r*COS(α)
V_x0= V₀*COS(α)
V_z0= V0*SEN(α)

The equations of motion of the rear wheels:
$Z(t) = Z_0 + V_z0 t + \frac {1}{2} gt^2 - \frac {1}{2}L * SEN(ωt)$
$X(t) = X_0 + V_x0 t + (\frac {1}{2}L - \frac {1}{2}L * COS(ωt))$

If they're right, I'll do the spreadsheet tomorrow.
THANK YOU

 

[haruspex]

Rear wheel angular velocity retrieval
$mgh=\frac {1}{2}Jω^2$

How do you get that? The right hand side looks like energy, but I don't see why it should equal the term on the left.

As I wrote, you need to start by finding the velocity components at the front wheels just as they reach the top of the dip. If the rear wheels have velocity v (all horizontal), what is the horizontal velocity of the front wheels? Given that the front wheels are still going up the last bit of the ramp, what is their vertical velocity?
What, then, are:
- the vertical and horizontal velocities of the mass centre
- the angular velocity of the chassis?
This much is just kinematics; no need to consider forces, torques, momentum or energy.

 

[alex33]

How do you get that?

I thought that a key to solving the problem related to the falling motion of the rear wheels could be the conservation of energy. The potential energy mg x height of the fall h should be conserved with the kinetic energy as a function of the moment of inertia and the angular velocity... so isn't ?

just as they reach the top of the dip. If the rear wheels have velocity v (all horizontal), what is the horizontal velocity of the front wheels?

...in that moment the car is launched with the same speed value V and direction according to the tangent of the arc just crossed, so I think:
V_x0 = V₀*COS(α)
...when the rear wheels are in free fall also, a rotational motion will be added with angular speed ω which on the X axis will be:
$ω_x = \frac{ω}{\sin(ωt)}$

What, then, are:
- the vertical and horizontal velocities of the mass centre
- the angular velocity of the chassis?

During the free fall the centre of mass have:
V_x0 = V₀*COS(α)
V_z0 = V₀*SEN(α)

Angular velocity of the chassis:
$ω = \frac{2V}{L}$

 

[haruspex]

I thought that a key to solving the problem related to the falling motion of the rear wheels could be the conservation of energy

We need to find the angular velocity when the rear wheels reach the dip. They have not descended yet. Besides, most the potential energy will be turned into vertical KE. That will then turn into elastic potential energy in the suspension. Suspensions are not designed to be highly elastic; after going over a bump you don't want to be bouncing up and down for the next 3 km. They are damped systems, converting the energy to heat.
Depending on the exact characteristics, the bounce might help the vehicle get up the other side of the dip, but it depends on the timing.

...in that moment the car is launched with the same speed value V and direction according to the tangent of the arc just crossed, so I think:
V_x0 = V₀*COS(α)

No. The car has a fixed length, and the rear is moving horizontally, directly towards the front wheels. If the front wheels have a different speed in that direction the car must be getting shorter or longer.
So the horizontal component of the front wheels' velocity is also v.
However, they are going up a ramp at angle $\alpha$, so what is their vertical component?

 

[alex33]

What you are saying is gospel to me!

However, they are going up a ramp at angle α, so what is their vertical component?

$V_z = V_x * \tan (α)$

 

[haruspex]

What you are saying is gospel to me!$V_z = V_x * \tan (α)$

Right.
So, can you figure out the angular velocity of the car?

 

[alex33]

However, this implies that the total speed of the front wheels is higher than the rear ones. It's correct?

 

[haruspex]

However, this implies that the total speed of the front wheels is higher than the rear ones. It's correct?

Yes, well spotted. It's $v\sec(\alpha)$.

 

[alex33]

... but for the Centre of Mass, what I'm guessing is that the vertical speed is lower than that of the front wheels? It's possible?

 

[haruspex]

... but for the Centre of Mass, what I'm guessing is that the vertical speed is lower than that of the front wheels? It's possible?

Yes. What should it be there, half way along the car?

 

[alex33]

more or less the centre of the dip

 

[haruspex]

more or less the centre of the dip

No, I mean what is the vertical velocity of the mass centre, in terms of $v, \alpha$?

 

[alex33]

this is what i don't understand... the tilt angle is the same (we are supporting the car on the rear wheels), since the relationship between Vx and Vz is the same i would be tempted to say that the vertical velocity of the center of mass is the same as the front wheels, but it is clear that it is not so... I don't understand what is missing...

 

[haruspex]

this is what i don't understand... the tilt angle is the same (we are supporting the car on the rear wheels), since the relationship between Vx and Vz is the same i would be tempted to say that the vertical velocity of the center of mass is the same as the front wheels, but it is clear that it is not so... I don't understand what is missing...

The front wheels have a vertical velocity $v\tan(\alpha)$, while the rear wheels have no vertical velocity. The mass centre, being directly above the midpoint of those, therefore has a vertical velocity $\frac 12v\tan(\alpha)$.
So now you have enough to write the expressions for the x and y coordinates of the mass centre at time t after 'lift off'.

What about the angular velocity?

 

[alex33]

What about the angular velocity?

To answer this question I would use the relation $ω = V * R$
Since the radius, in our case is $L$ (the whole chassis rotates)
and that the speed we've to use is the vertical one $V * \tan(α)$
...if I'm lucky, the angular speed for the front wheels is...
$ω = V * tan(α) * L$
on the centre of mass we'll have:
$ω = \frac{1}{2}V * tan(α) * L$
...the rear wheels are our pivot so there can't be an angular speed at the moment. We will therefore have a load on the suspension.

 

[haruspex]

To answer this question I would use the relation $ω = V * R$

No such relation; it would be dimensionally wrong for a start.
Try $ω = V / R$.
You may be thinking of angular momentum, mvr.

the angular speed for the front wheels is

Angular speed is a property of the whole rigid object.
E.g. considering how the front axle rotates about the rear axle, the angular speed is $v\tan(\alpha)/L$, or how the mass centre rotates about the rear axle, the angular speed is $v\frac 12\tan(\alpha)/(L/2)$, which is the same.

As with the horizontal velocity, there is no force acting on it while in free fall, so it remains constant. That allows you to write an expression for the tilt angle at time t.

 

[alex33]

Try ω=V/R.

what kind of bungler... I reversed the elements !!

Angular speed is a property of the whole rigid object.
E.g. considering how the front axle rotates about the rear axle, the angular speed is $v\tan(\alpha)/L$, or how the mass centre rotates about the rear axle, the angular speed is $v\frac 12\tan(\alpha)/(L/2)$, which is the same.

As with the horizontal velocity, there is no force acting on it while in free fall, so it remains constant. That allows you to write an expression for the tilt angle at time t.

OK

 

[alex33]

As with the horizontal velocity, there is no force acting on it while in free fall, so it remains constant. That allows you to write an expression for the tilt angle at time t.

Wait! we said we weren't in free fall yet: the rear wheels are still on the ground just before the dip. However we say that Vx remains constant because the driving force compensates the friction. Okay?

Unfortunately with α(t) i got stuck again. Please be patient... I would have:
$\frac{α}{t} = \frac{V* \tan(α)}{L}$ and so $\frac{α}{\tan(α)} = \frac{V * t}{L}$ but I have no idea how to get α if we have to divided it by its tangent.

 

[haruspex]

we said we weren't in free fall yet: the rear wheels are still on the ground just before the dip.

We have the velocity and rotation rate just as the rear wheels reach the dip. They are not going to change suddenly.

Unfortunately with α(t) i got stuck again. Please be patient... I would have:
$\frac{α}{t} = \frac{V* \tan(α)}{L}$ and so $\frac{α}{\tan(α)} = \frac{V * t}{L}$ but I have no idea how to get α if we have to divided it by its tangent.

$\alpha$ is in use as the half angle of the dip arc. Let's use $\omega$ for the rotation rate (taking the front rising as positive) and $\theta(t)$ for the angle reached (above horizontal).
$\theta=\omega t=\frac{V* \tan(α)}{L}t$.

 

[alex33]

Ok. I understand. θ(t) is the function for the rotation when the free fall start. α is the angle reached at the top of the dip and then fixed by sperimental data.

To stay close to the real event, I ask you to take into account the fact that from when the front wheels lift completely out of the hole, for 2/24 seconds the same front wheels travel without detaching from the ground (suspensions?) and at the same time the wheels rear they complete the last flat stretch before going down into the hole using the same 2/24 + other 2/24 seconds. This is proper event scanning.

Now I try to build the spreadsheet.

 

[haruspex]

when the front wheels lift completely out of the hole, for 2/24 seconds the same front wheels travel without detaching from the ground

Are you saying that the front wheels remain in contact with the ground for that time after exiting the dip but then lift off? Yes, that's just the suspension relaxing, but has negligible effect on what the chassis does. Once we have found the chassis path, it won't be hard to hang the wheels off it.

the same 2/24 + other 2/24 seconds

So the rear wheels reach the start of the dip 2/24 s after the front wheels leave it?

Now I try to build the spreadsheet.

Before you do that, please post the equations you have for:

• the x, y positions of the mass centre
• the attitude of the chassis, $\theta$
• the x, y positions of the rear wheels

as functions of time.

 

[alex33]

Are you saying that the front wheels remain in contact with the ground for that time after exiting the dip but then lift off?

Yes

So the rear wheels reach the start of the dip 2/24 s after the front wheels leave it?

No 4/24 after the front wheels leave the dip, 2/24 after the front wheels lift off from the ground

 

[alex33]

• the attitude of the chassis, $\theta$

...sorry the "attitude"?

 

[haruspex]

...sorry the "attitude"?

Angle to the horizontal.

 

[haruspex]

No 4/24 after the front wheels leave the dip, 2/24 after the front wheels lift off from the ground

That makes the car 0.43m longer than the dip. That will make a huge difference. The front wheels should be back on the ground before the rear wheels reach the dip.
No, it's ok. I made a mistake in how I adjusted the spreadsheet.

 

[alex33]

Before you do that, please post the equations you have for:

• the x, y positions of the mass centre
• the attitude of the chassis, $\theta$
• the x, y positions of the rear wheels

as functions of time.

Centre of Mass in the dip (from t₀ = 0/24 s to t₁₈ = 18/24 s)
$X(t) = X_0 + Vt$
$Z(t) = Z_0 + \frac {1}{2}V * \tan(α) * t$

Angle to the horizontal of the chassis
$θ = \frac{V*\tan(α)}{L} * t$

Rear Wheels in the dip (from t₂₂ = 22/24 s to t_? = ?? s dust not allow us to define)
$X(t) = X_0 + Vt$
$Z(t) = Z_0 + V * \tan(α) * t - \frac {1}{2} g * t^2$

 

[alex33]

it is a 1 fps version of the original sequence in 24 fps. Thus we can share the reliefs in time and space.

 

[haruspex]

Centre of Mass in the dip (from t₀ = 0/24 s to t₁₈ = 18/24 s)
$X(t) = X_0 + Vt$
$Z(t) = Z_0 + \frac {1}{2}V * \tan(α) * t$

The vertical velocity of the mass centre does not reach $\frac {1}{2}V * \tan(α)$ until the front wheels reach the top.
Why start the analysis any sooner?

 

[alex33]

That makes the car 0.43m longer than the dip.

The initial speed before the LRV enters the dip is $\frac{L * 24}{21} s = 2,61 m/s$
While for the sequence inside the dip it is $\frac{L * 24}{23} s = 2,38 m/s$
To calculate the difference between the chord of the arc and the length of the wheelbase I used the second one and the result is 0.40m (0,397 m)

 

[haruspex]

it is a 1 fps version of the original sequence in 24 fps. Thus we can share the reliefs in time and space.

If you deduced all the measurements from that video there must be considerable uncertainty in the values. In particular, how did you arrive at the angle at the edge of the dip?

 

[alex33]

The vertical velocity of the mass centre does not reach 12V∗tan⁡(α) until the front wheels reach the top.
Why start the analysis any sooner?

Sorry, I understood that this was the motion of the centre of mass throughout the dip. Is it valid only for α = 20°...? And before? Because I believe it has a vertical motion even before α = 20° ...

 

[haruspex]

Sorry, I understood that this was the motion of the centre of mass throughout the dip. Is it valid only for α = 20°...? And before? Because I believe it has a vertical motion even before α = 20° ...

Yes, it has vertical motion, down or up, for most of the time in the dip. But if we take the velocity of the rear wheels approaching the dip as fixed then we can deduce the vertical velocity of the front wheels when they reach the top. What vertical velocity they had prior to that is irrelevant. That's handy, since we have little idea what the shape of the dip is.

Your estimates are all based on a 24/s frame rate. That means that each time value has an inherent random error of $\pm 1/48s$. Taking the difference between two increases the rms value of that to $\sqrt 2/48s$. For your speed calculations that turns into an error of $\pm 0,1m/s$.

But the datum that most bothers me is $\alpha$. How did you arrive at a value for that?

 

[alex33]

If you deduced all the measurements from that video there must be considerable uncertainty in the values. In particular, how did you arrive at the angle at the edge of the dip?

This is a low resolution version. The NASA source is this:
https://www.facebook.com/apolloflightjournal/videos/315560545801914

I was able to extract every single frame and I must say that it has a very good resolution. The measurement error is very small indeed. The images were shot on 16mm film and in 2019 were rephotographed and digitized. Before starting I restored the images to their homographic dimensions and corrected the aberrations (it is a 10 mm lens). Sorry if on the experimental part I can't share all the information right away, but I will do soon.

For the half angle subtended by the arc, I initially used an approximation, comparing the measurements of the chord with those of the wheelbase. Then, since you helped me with the front wheel speed formula:
$V_z = V * \tan(α)$ I counted the frames and having the depth of the dip I determined V_z. V it was known and I obtained α which is 19.91° when with the other method I had 20°...

 

[alex33]

Yes, it has vertical motion, down or up, for most of the time in the dip. But if we take the velocity of the rear wheels approaching the dip as fixed then we can deduce the vertical velocity of the front wheels when they reach the top. What vertical velocity they had prior to that is irrelevant. That's handy, since we have little idea what the shape of the dip is.

So the expressions of Z(t) you helped me write are valid only with α = 20° ? For both the centre of mass and the front wheels?

Your estimates are all based on a 24/s frame rate. That means that each time value has an inherent random error of ±1/48s.

If the error were so great, photographic technologies would not lead to the results we know. There are equipment manuals, guidelines and good practices for photographic analysis, even in the forensic field... normally the spatial errors of the images (distortions and aberrations) are corrected, but although I have searched for the time errors of the cine-cameras have never been in question. Also consider that the times of the image imprinted on the film are not given so much by the framerate but by the shutter opening time, which is much much smaller than the framerate.

One thing I've glossed over at the moment is that the actual framerate isn't 24 fps but most likely 23,976 fps, which is almost always the case with this frame rate. The Maurer DAC camera that captured these images worked no differently.

When I'll finish this work, I will always have done something better than the Colorado University team in the 2012 study published in the AJF. (https://www.researchgate.net/publication/258468670) . They used the same technique with much poorer sources. They didn't insert any errors related to the framerate, but they got the framerate completely wrong: they analyzed a sequence shot at 24 fps believing it was a 29.97 fps... they managed to measure differences in the motion of the dust raised by the LRV even on the cloned frames during the conversion from 24 to 29.97 fps (American TV standard).

 

[haruspex]

This is a low resolution version. The NASA source is this:
https://www.facebook.com/apolloflightjournal/videos/315560545801914

I was able to extract every single frame and I must say that it has a very good resolution. The measurement error is very small indeed. The images were shot on 16mm film and in 2019 were rephotographed and digitized. Before starting I restored the images to their homographic dimensions and corrected the aberrations (it is a 10 mm lens). Sorry if on the experimental part I can't share all the information right away, but I will do soon.

For the half angle subtended by the arc, I initially used an approximation, comparing the measurements of the chord with those of the wheelbase. Then, since you helped me with the front wheel speed formula:
$V_z = V * \tan(α)$ I counted the frames and having the depth of the dip I determined V_z. V it was known and I obtained α which is 19.91° when with the other method I had 20°...

Ok.

Anyway, I made a mistake earlier; see post #60.
We now have three phases to deal with:

1. From front wheels reaching top of dip to rear wheels arriving at dip. Duration $\Delta t$, approximately 1/6s. In this phase, there is still a normal force on the rear wheels, but the front wheels are in the air. The easiest way to deal with this is to consider torques about the rear point of contact. These lead to an angular acceleration $\dot\omega=-\frac{3gL}{h^2+4L^2}$, where L is the half length of the car (near enough $R\sin(\alpha)$) and h is the height of the mass centre above the ground when on the flat. From that we get adjustments to $\omega$ of $\dot\omega\Delta t$, to $\theta$ of $\omega\Delta t$, and to the height of the mass centre of $\frac 12v\tan(\alpha)\Delta t$.
2. From the front wheels reaching the top of the dip to the rear wheels landing. This is the free fall stage. We can simulate this in a spreadsheet. I used hundredths of a second. To detect the end of the stage I tracked the distance from the rear wheels to the centre of curvature of the dip. When this exceeded the radius of curvature it had landed. Of course, we do not know that it is a simple arc of a circle. It may land a bit sooner.
3. When the rear wheels land, there is an impulse sufficient to prevent ground penetration (except transiently in the simulation), but not much more. Certainly not enough to increase the total KE. From that we can compute the new initial conditions for the next stage: from the rear wheels bouncing to whichever comes first of the front wheels landing or the rear wheels landing (which might or might not be still in the dip).