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

[haruspex]

the time errors of the cine-cameras have never been in question.

It's not errors in the timing of the frames. It's looking for an event in frames 1/24s apart and trying to deduce exactly when the event occurred. But if you are actually using frames much closer in time then I withdraw my concern.

 

[alex33]

From front wheels reaching top of dip to rear wheels arriving at dip. Duration Δt, approximately 1/6s.

Okay. After the physical modeling I will do a more accurate check on the frames to confirm if the times are right.

In this phase, there is still a normal force on the rear wheels, but the front wheels are in the air.

Yes, it's believable. What we see is that the front wheels, before leaving the ground, at the exit of the dip, still rotate, up to the 18th frame after their entry (where the wheel is seen again as complete on the ground). But we know that this is why the suspensions are involved and we are interested in the chassis instead.

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).

For the moment I have read superficially ... everything seems understandable to me. It's 4:30 am here and I need to get some sleep. Tomorrow compatibly with my work, I will study all the adjustments.

It's not errors in the timing of the frames. It's looking for an event in frames 1/24s apart and trying to deduce exactly when the event occurred. But if you are actually using frames much closer in time then I withdraw my concern.

I have a lot of evidence of the validity of time references. Keep in mind that I have analyzed many other scenes before this one. I mainly studied free fall motions, with elementary physics (given my limited skills) but after analyzing the data collected with software such as Origin pro, I always had confirmation of the applied models. In verifying the results and in the way of procuring the data, I have already had support from physicists and technicians in the past. It's the first time I've studied a more complex motion from a physical point of view, and I'm very grateful to you for the help you're giving me. Good night to me and good day to you!

 

[pbuk]

I'm a bit late to this discussion and I haven't read all the posts, but it looks like the main reason the front wheels leave the ground is suspension (including tyre) rebound. Look at the compression of particularly the RH front suspension in frames 0-16: that compression stores energy which is released in the following frames, accelerating the front of the vehicle upwards. This appears to be the dominant effect with torque and rotational momentum being negligable.

Or has this already been covered?

 

[alex33]

I'm a bit late to this discussion and I haven't read all the posts, but it looks like the main reason the front wheels leave the ground is suspension (including tyre) rebound. Look at the compression of particularly the RH front suspension in frames 0-16: that compression stores energy which is released in the following frames, accelerating the front of the vehicle upwards. This appears to be the dominant effect with torque and rotational momentum being negligable.

Or has this already been covered?

Dear pbuk thanks for joining this thread.
At the moment all those who have expressed an opinion agreed in considering the suspensions a marginal factor, certainly not the factor that triggers the jump. In fact I think the role of the suspension is to make the motion smoother without the vehicle constantly jumping.
I imagine this was the precise object of study for the Boeing Company that built the vehicle. I chose these frames in the famous sequence of the LRV Grand Prix, because it is one of the biggest jumps that the vehicle makes, even though the lunar ground is very bumpy.

It's not errors in the timing of the frames. It's looking for an event in frames 1/24s apart and trying to deduce exactly when the event occurred. But if you are actually using frames much closer in time then I withdraw my concern.

Returning to this topic... We have no more frames than were taken. So the frames are 1/24 s apart, but the event they capture depends on how long the shutter is open. The Maurer Data Acquisition Camera (DAC) had these exposure times: 1/60 s, 1/125 s, 1/250 s, 1/500 s, 1/1000 s. In the Apollo 16 mission data logs there is no information on what the shutter aperture was for this sequence (at least I haven't found it yet). One clue is the aperture used for the diaphragm: f/8 (this has been officially stated). With this aperture the correct exposure would be 1/60 s, but since the time was set manually, for now it's just a guess. I propose that the study of the error is done at the end of the modeling.

 

[pbuk]

At the moment all those who have expressed an opinion agreed in considering the suspensions a marginal factor, certainly not the factor that triggers the jump.

As far as I can tell the comments about suspension were before (1) you revealed that you were talking about a vehicle in 1/6 of Earth gravity; (2) you revealed that you were talking about a Lunar Rover which has very different suspension design constraints to a normal car; (3) you revealed the video which clearly shows extensive compression and rebound.

What was your motivation for not revealing this vital information at the start of the thread? Did you not think it was important?

 

[pbuk]

Returning to this topic... We have no more frames than were taken. So the frames are 1/24 s apart, but the event they capture depends on how long the shutter is open.

You are missing the point. If you are trying to determine at what time, for example, the tyres first left the ground and you have one frame with the tyres touching the ground and one 1/24 s later with them not touching then it doesn't matter how fast the shutter speed is, all you know is that the tyres actually left the ground somewhere in that 1/24 s window.

 

[haruspex]

I'm a bit late to this discussion and I haven't read all the posts, but it looks like the main reason the front wheels leave the ground is suspension (including tyre) rebound. Look at the compression of particularly the RH front suspension in frames 0-16: that compression stores energy which is released in the following frames, accelerating the front of the vehicle upwards. This appears to be the dominant effect with torque and rotational momentum being negligable.

That is an interesting possibility, but I would think a lunar rover would have suspension that is particularly well damped.
I calculate that when the front wheels reach the top the mass centre is rising at 0.14m/s and the whole is pitching up at 0.27 rad/s. That would account for the front axle rising 10cm above its normal position shortly after, but whether that is enough to see daylight under the tyre I am unsure.

 

[pbuk]

That is an interesting possibility, but I would think a lunar rover would have suspension that is particularly well damped.

Doesn't look like it, particularly in the original video

 

[alex33]

You are missing the point. If you are trying to determine at what time, for example, the tyres first left the ground and you have one frame with the tyres touching the ground and one 1/24 s later with them not touching then it doesn't matter how fast the shutter speed is, all you know is that the tyres actually left the ground somewhere in that 1/24 s window.

What we know for sure is that 1/23.976 s elapses between shots. The camera imprinted on the film an image within this time window spending a time of no more than 1/60 s. The regularity of these intervals is extremely important for the shooting process. A minimal misalignment would cause problems in the editing and post-production processes. In what part of the time window did the events we see occur? If the shutter is well synchronized, most likely in the center of the window, but in any case always at the same time distance from each other. Not "somewhere". Okay?

What we need to model is not what happens between event A and event B, but how much time has passed between A and B. Right?

 

[haruspex]

What we know for sure is that 1/23.976 s elapses between shots. The camera imprinted on the film an image within this time window spending a time of no more than 1/60 s. The regularity of these intervals is extremely important for the shooting process. A minimal misalignment would cause problems in the editing and post-production processes. In what part of the time window did the events we see occur? If the shutter is well synchronized, most likely in the center of the window, but in any case always at the same time distance from each other. Not "somewhere". Okay?

What we need to model is not what happens between event A and event B, but how much time has passed between A and B. Right?

@pbuk is raising the same concern I had. It does not matter how precise the shutter rate is, you are trying to judge exactly where between two frames the event occurred. That is limited by the magnitude of the shutter interval, not its precision.
Maybe I misunderstood, but I thought you responded that the real shutter rate is much faster and you have access to that.

 

[haruspex]

Doesn't look like it, particularly in the original video

The original video is somehow protected from my prying eyes. Don't forget this is lunar gravity, so it will look pretty bouncy even with well damped suspension.

 

[alex33]

@pbuk is raising the same concern I had. It does not matter how precise the shutter rate is, you are trying to judge exactly where between two frames the event occurred. That is limited by the magnitude of the shutter interval, not its precision.
Maybe I misunderstood, but I thought you responded that the real shutter rate is much faster and you have access to that.

It is necessary to distinguish between framerate and shutter opening time. The frame corresponds to a film sector that will be impressed. There are 24 photos (24 sectors of film) that will be captured on the film every second. If the shutter stays closed during one of those shots, I get a black frame. The framerate and shutter opening time are synchronized. The camera carriage will advance the film one sector every 1/23.976 s. In the central moment of this interval the shutter will open and let the light imprint the image on the sensor, in a time which is not less than 1/60 s. The result we see in the image is the overlap of everything happens in that 1/60 s. So is it clearer?

 

[pbuk]

Not "somewhere". Okay?

1. Lose the attitude.
2. You are still missing the point. If we have an image at time 42 s with the wheels on the ground and an image at time 42 10/240 s with the wheels 10cm off the ground then the wheels could have left the ground at, say 42 2/240 s or 42 6/240 s. In the first case the average vertical velocity (while in vertical motion) is $\frac{0.1}{\frac{8}{240}} = 2 \text{ m/s}$ and in the second case $\frac{0.1}{\frac{4}{240}} = 4 \text{ m/s}$.

 

[alex33]

The original video is somehow protected from my prying eyes.

It is not possible to see it in the forum, you have to click on the link...

https://www.facebook.com/apolloflightjournal/videos/315560545801914

 

[pbuk]

So is it clearer?

1. Really lose the attitude, how old are you?
2. Stop trying to tell us how a camera works, we know that. Start listening to us when we tell you what the implications of that are on accuracy of measurement.

 

[alex33]

1. Lose the attitude.
2. You are still missing the point. If we have an image at time 42 s with the wheels on the ground and an image at time 42 10/240 s with the wheels 10cm off the ground then the wheels could have left the ground at, say 42 2/240 s or 42 6/240 s. In the first case the average vertical velocity is $\frac{0.1}{\frac{8}{240}} = 2 \text{ m/s}$ and in the second case $\frac{0.1}{\frac{4}{240}} = 4 \text{ m/s}$.

1) Sorry if I gave the impression of having a certain attitude... I don't have any. My English is very basic and I don't express myself well. I'm glad you pose questions. Questions and different points of view are very welcome here. I have already acknowledged my poor understanding of physics many times. For this I am grateful for the help that everyone can give. With haruspex we have been doing well up to now.

Stop trying to tell us how a camera works, we know that.

I insisted on the technical aspects of the camera because haruspex asked me to clarify whether or not I had access to more frames per second. I explained that the maximum number of frames we have is the one that was actually taken by astronauts (24 fps), but that the uncertainty about where to place the event captured by the camera in the time line is linked to the exposure time and not to the framerate.

2) We can't know what happened between event A and event B, that's true. But we know what happened in time 42 + 1/24 s (+/- 1/120 s). It cannot have happened at +2/240 s because the shutter was still closed. It also doesn't make sense to calculate a speed based on two frames. It's not what we'll do. We have to track the entire motion and feed the data into a system with their error range. From this system we will receive an evaluation in statistical terms of which is the reliability of the model. Please see the research by Mihaly Horanyi (Colorado Center for Lunar Dust and Atmospheric Studies, NASA Lunar Science Institute) and Hsiang-Wen Hsu (Laboratory of Atmospheric and Space Physics, University of Colorado Boulder), https://www.researchgate.net/publication/258468670 They expose their error ranges.

 

[pbuk]

the uncertainty about where to place the event captured by the camera in the time line is linked to the exposure time and not to the framerate.

But the event "the wheel leaves the ground" is not captured by the camera. Instead of using words to justify your own argument, look at my calculations to see the quantitative implications of this. And note that the length of the exposure does not feature in this calculation, only frame rate.

2) We can't know what happened between event A and event B, that's true. But we know what happened in time 42 + 1/24 s (+/- 1/60 s). It cannot have happened at +2/240 s because the shutter was still closed.

The wheel cannot have left the ground because the camera shutter was still closed? This does not make sense.

It also doesn't make sense to calculate a speed based on two frames. It's not what we'll do. We have to track the entire motion and feed the data into a system with their error range. From this system we will receive an evaluation in statistical terms of which is the reliability of the model.

What model? What system? What are we actually trying to do here? Are you trying to reverse engineer a physical model of a LRV from a few frames of film on an unknown terrain? Why when you can more easily construct such a model from published NASA data?

Let's go back to your original questions:

- Why do the front wheels leave the ground?

Because the suspension (including the tyres) is compressed by the forward momentum of the vehicle as it decelerates due to hitting the front of the dip. The stored energy is then released, accelerating the front of the vehicle upwards.

- Why are the rear wheels only slightly leave the ground?

Because the rear of the car is not subjected to as great vertical forces.

- Is the weight of the vehicle equally distributed on the 4 wheels?

You can find online that the mass of the LRV is about 200kg. I estimate a suited astronaut weighs about 150kg. Look at where he is sitting on the vehicle: what do you think?

- Will the parabola drawn by the front wheels at the exit of the valley respond to the hourly law Z(x)=V0t+12Gt2 ? Are there other elements to take into account in the equation of motion? If possible, please write the correct equation.

There are far to many parameters, many of which are unknown, affecting the motion of the front wheels to make this possible.

 

[alex33]

The wheel cannot have left the ground because the shutter was still closed? This does not make sense.

Dear Pbuk I did not say this. I said that the image/data we have can be placed in the timeline with an uncertainty given by the shutter opening time. Obviously, I repeat, we cannot know anything about what happened before and after.

There is really a lot of data available in the NASA documents. We are already using some of them. My shortcoming is in physical modeling and for this I asked for help. Your proposal to study suspension has been met with interest. We have to work on it.

You can find online that the mass of the LRV is about 200kg. I estimate a suited astronaut weighs about 150kg. Look at where he is sitting on the vehicle: what do you think?

Yes,
I also think that since there is only one astronaut on board, the weight cannot be well distributed as official data say. It is rather a Y-axis problem, which we have never included in the discussion yet. But the mass of the astronaut was less than what you proposed.

The technical data tell us that this was the situation:

LRV MASS FULL LOADED = 1.520,40 LBs
Chassis = 462 LBs

Load vehicle with 620 pounds distributed as follows:
245 lb. driver seat
245 lb. passenger seat
130 lb. equally distributed over rear

 

[alex33]

These lead to an angular acceleration ω˙=−3gLh2+4L2, where L is the half length of the car (near enough Rsin⁡(α)) and h is the height of the mass centre above the ground when on the flat.

Dear Haruspex,
Could you give me some more information about this formula? Is it derived from the principle of conservation of energy? (I had proposed a similar - but incorrect - formula in my post #36)

Can I ask you to keep the name of the variables other than those already adopted? Otherwise I'm afraid to get myself confused... until now we've talked about L to indicate the length of the Rover (actually it would only be the length of the wheelbase, the Rover is 80 cm longer). As for h, I had defined the height of the chassis (it's more or less the same thing... I can adapt).

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.

I agree: we don't know if the path inside the dip is really an arc of a circumference and how the dip is developed in 3 dimensions. But for the model it will be sufficient to approximate it to an elementary geometric shape.

To detect the end of the stage I tracked the distance from the rear wheels to the centre of curvature of the dip.

What is the right way to do this tracking? I had prepared a column in the spreadsheet in which for each frame I associated a Z coordinate deriving from the expressions we have found with an X coordinate on the arc of the circumference and I've compared it with the X coordinate of the motion of the rear wheels. C(x₀, Z₀) is the centre of the circle whose arc we are traversing.

Is this ok? Thanks

 

[haruspex]

Because the suspension (including the tyres) is compressed by the forward momentum of the vehicle as it decelerates due to hitting the front of the dip. The stored energy is then released, accelerating the front of the vehicle upwards.

That's just another way of looking at what we are already doing. Without suspension, the car will follow up the rising curve and leave the ground because of its linear and angular momenta. With suspension, that tracking of the ground curve is delayed.
The suspension cannot add any extra energy; indeed, it will absorb some. It will also make the effect less apparent because the wheels will drop back to the ground more rapidly than the chassis can follow.

 

[alex33]

Based on the low resolution video I posted (1 fps) I made a timeline of the events, arranging the zero event so that we can have Δt between T₁ and t₄. I hope it can be useful.

-17​	Start
-16​	Front Wheels on the flat
-15​	Front Wheels down in the dip
-14​	Front Wheels down in the dip
-13​	Front Wheels down in the dip
-12​	Front Wheels down in the dip
-11​	Front Wheels down in the dip
-10​	Front Wheels down in the dip
-9​	Front Wheels down in the dip
-8​	Front Wheels down in the dip
-7​	Front Wheels down in the dip
-6​	Front Wheels up in the dip
-5​	Front Wheels up in the dip
-4​	Front Wheels up in the dip
-3​	Front Wheels up in the dip
-2​	Front Wheels up in the dip
-1​	Front Wheels up in the dip
0​	Front Wheels up in the dip
1​	Front Wheels rotate coming out from dip
2​	Front Wheels rotate coming out from dip
3​	Front Wheels in free fall
4​	Front Wheels in free fall
5​	Rear Wheels in the dip
6​	Rear Wheels in the dip
7​	Rear Wheels in the dip
8​	Rear Wheels in the dip
9​	Rear Wheels in the dip
10​	Rear Wheels in the dip
11​	Rear Wheels in the dip
12​	Rear Wheels in the dip
13​	Rear Wheels in the dip
14​	Rear Wheels in the dip
15​	Rear Wheels in the dip
16​	Rear Wheels in the dip
17​	Rear Wheels in the dip
18​	Rear Wheels in the dip
19​	Rear Wheels in the dip
20​	Rear Wheels in the dip
21​	Rear Wheels in the dip
22​	Rear Wheels in the dip
23​	Front Wheels and Rear Weels in free fall
24​	Front Wheels and Rear Weels in free fall
25​	Front Wheels and Rear Weels in free fall
26​	Front Wheels and Rear Weels in free fall
27​	Left Front Wheel landing
28​	…
29​	…
30​	…
31​	…
32​	…
33​	…
34​	…
35​	…

 

[haruspex]

Can I ask you to keep the name of the variables other than those already adopted? Otherwise I'm afraid to get myself confused... until now we've talked about L to indicate the length of the Rover (actually it would only be the length of the wheelbase, the Rover is 80 cm longer). As for h, I had defined the height of the chassis

As a matter of course, I generally define length variables as half lengths of symmetric objects, otherwise fractions keep appearing in the algebra.

I start with these constants:

r	2.982245324	L	1.019987973	I/m	0.394925155
v	2.6	omega	0.4638891017	J/m	0.68
alf	0.3490658504	del t	0.1666666667	s
h	0.38
g	1.65

Note that my L (half car length) is $r\sin(\alpha)$. I know now it should be a bit longer, but I don't think it changes much.
I/m is the moment of inertia of the car per unit mass: $\frac 13(h^2+L^2)$.
J/m is the impulse on the rear axle on landing, per unit mass. The impulse is normal to the ground.
del t (Δt) is the time between the front wheels exiting the dip and the rear wheels reaching it.
Omega is the angular velocity when the front wheels reach the dip. As discussed, the vertical velocity of the mass centre then is $\frac 12v\tan(\alpha)$, giving a rotation rate of $\frac 1{2L}v\tan(\alpha)$.

As origin, I use the point halfway across the dip but at the surrounding ground level.
Velocities and displacements are positive to the left and up; rotations are clockwise from horizontal.

The simulation during a free fall tracks the following:
t: time since start of phase
xc, yc: coordinates of mass centre
xf, yf: coordinates of front wheels
xr, yr: etc.
theta: chassis orientation

v0x

v0y

w0

x0c

y0c

theta0

These are the initial values of the velocity and position of the mass centre and the orientation at the start of the current free fall phase. During such a phase, we can compute the current values from those and the time since the start of the phase.

I also track total mechanical energy, E= kinetic+potential as a check on the equations and to ensure J is not excessive.

During del t, the rear wheels are still on the ground but the front wheels are airborne.
Taking moments about the rear wheels, gravity on the mass centre exerts a torque -mgL. The moment of inertia about that axis is $\frac 13(h^2+4L^2)m$, so the angular acceleration during del t is $-3\frac{gL}{(h^2+4L^2)}$. (Its magnitude will reduce slightly during del t, but not much.) Hence, when the rear wheels reach the dip the rotation rate is $\frac 1{2L}v\tan(\alpha)-3\frac{gL}{(h^2+4L^2)}\Delta t$. This is $w_0$ for the first free fall phase.

From that we can deduce $v_{0c}=w_0*L$ and for $y_{0c}$ I use the average rotation rate during del t: $y_{0c}=(w_0+\frac 1{2L}v\tan(\alpha))/2*L*\Delta t$. (Yes, these are approximations based on theta still being quite small.)

That sets up the conditions at the start of this free fall phase. At time t into the phase we can calculate xc, yc and theta easily: $x_c=x_{0c}+v_{0x}*t$, $\theta=\theta_0+w_0*t$, $y_c=y_{0c}+v_{0y}*t-\frac 12gt^2$.
The position of the rear wheels is trickier.
$x_r=x_c-L\cos(\theta)+h\sin(\theta)$
$y_r=y_c-L\sin(\theta)-h\cos(\theta)$
Similarly the front wheels:
$x_f=x_c+L\cos(\theta)+h\sin(\theta)$
$y_f=y_c+L\sin(\theta)-h\cos(\theta)$

Finally, the "touch" column tracks whether the rear wheels have landed:
$r^2-x_r^2-(r\cos(\alpha)-y_r)^2$.
When that value goes negative, we've hit regolith.

For the second free fall phase, we have to set up a new set of initial conditions and start t at 0 again. For this we need a value for J/m. Something around 0.6 to 0.7 m/s seems to be reasonable. More on that later.

 

[alex33]

Dear Haruspex,
you are doing a tremendous job. I am very grateful to you.
Thanks for sharing all the formulas. During the day (which here in Europe started 4-5 hours ago), I will import everything into my excel.

With respect to the images, I wanted to tell you not to let yourself be influenced by the type of free fall curve we observe up to the impact of the front wheels, much less by the times recorded. There are some elements in play that I haven't told you about and will tell you about once we've defined the model. Let's go to define the model up to the actual free fall of the whole Rover (including the last bouncing of the rear wheels).

We have to build a credible and the simplest possible theoretical model.

Then I'll do a tracing on the high resolution images. I will first have to extract the frames from the official sources, correct the format, aberrations and perspective. Maybe it will take me a few days (compatibly with my job it could take 10 days). Eventually we will have about 120-140 pairs of coordinates for front wheels (51), center of mass (51)and (as much as possible) rear wheels (20-40). We will then test the model against a statistical analysis and look within what confidence intervals it is valid.

 

[haruspex]

Dear Haruspex,
you are doing a tremendous job. I am very grateful to you.
Thanks for sharing all the formulas. During the day (which here in Europe started 4-5 hours ago), I will import everything into my excel.

With respect to the images, I wanted to tell you not to let yourself be influenced by the type of free fall curve we observe up to the impact of the front wheels, much less by the times recorded. There are some elements in play that I haven't told you about and will tell you about once we've defined the model up to the actual free fall of the whole Rover (including the last bouncing of the rear wheels).

We have to build a credible and the simplest possible theoretical model.

Then I'll do a tracing on the high resolution images. I will first have to extract the frames from the official sources, correct the format, aberrations and perspective. Maybe it will take me a few days (compatibly with my job it could take 10 days). Eventually we will have about 120-140 pairs of coordinates for front wheels (51), center of mass (51)and (as much as possible) rear wheels (20-40). We will then test the model against a statistical analysis and look within what confidence intervals it is valid.

Are you able to create a video simulation from a spreadsheet? It would be great to see if it actually looks right.

 

[alex33]

Are you able to create a video simulation from a spreadsheet? It would be great to see if it actually looks right.

it should be possible

 

[jbriggs444]

There are some elements in play that I haven't told you about

Please tell us that it is not about a moon landing conspiracy theory.

 

[alex33]

Please tell us that it is not about a moon landing conspiracy theory.

Thank goodness I'm not a proponent of any conspiracy theory!

 

[berkeman]

With respect to the images, I wanted to tell you not to let yourself be influenced by the type of free fall curve we observe up to the impact of the front wheels, much less by the times recorded. There are some elements in play that I haven't told you about and will tell you about once we've defined the model.

Why do you think it is best to keep withholding information in this discussion? Is there a logical reason for doing this? The Mentors have been getting complaints about this behavior of yours in this thread.

 

[alex33]

Why do you think it is best to keep withholding information in this discussion? Is there a logical reason for doing this? The Mentors have been getting complaints about this behavior of yours in this thread.

Thanks for the question. It's not my intention to hide anything. I also have no more information than anyone searching the public data on this sequence can have. The problem is that I don't have the possibility to recover ALL the data NOW. Last and most importan, what I have raised is a purely theoretical problem (as I explained at the beginning): in my opinion it is incorrect that modeling is affected by data and measurements of weak value. This is an issue that has been debated on several occasions in this thread. The data obtained from the images are not completely reliable. Following the definition of the model we will try to do a statistical analysis to verify the model.

Anyone having problems with this process can suggest other solutions...

 

[alex33]

As a matter of course, I generally define length variables as half lengths of symmetric objects, otherwise fractions keep appearing in the algebra.

OK, I'll adapt my mind. But let's remember that in our case the half lengths are different because the centre of gravity is not "in the centre" of the axis.

J/m is the impulse on the rear axle on landing, per unit mass. The impulse is normal to the ground.

Sorry but how do you calculate the impulse? Shouldn't it be I/m * Δt ? What is the Δt you used?

These are the initial values of the velocity and position of the mass centre and the orientation at the start of the current free fall phase.

I was wondering... since one of the strongest data we have is the Rover's average speed BEFORE the dip enters, wouldn't it make more sense to start modeling from when the front wheels enter the hole? Is there an impact there too? How do it change the vertical speed? I think that only with these elements can we estimate the speed to use for the subsequent phases.

As origin, I use the point halfway across the dip but at the surrounding ground level.

Ok, I'll use the same origin... I suppose Z₀ will be the ground line.

 

[haruspex]

how do you calculate the impulse?

We cannot know the impulse. Its magnitude depends on the elasticity of tyres and suspension, while the direction depends on the angle of the floor at that point.

Shouldn't it be I/m * Δt ?

No, I define the impulse to be J (a vector, really, but I am assuming it acts normally to the presumed arc) , so the impulse per unit mass is J/m. That produces a velocity change $\Delta v=J/m$ at the mass centre, and a change to the angular momentum. We do not have to worry about how long it takes… take it as instantaneous.

wouldn't it make more sense to start modeling from when the front wheels enter the hole? Is there an impact there too? How do it change the vertical speed?

Do we know how the motor responds in the four wheel drive? Quite possibly the vehicle will accelerate as the front wheels descend, but would likely have returned to its original speed as the front wheels exit.

 

[alex33]

Do we know how the motor responds in the four wheel drive?

We have these graphs... but I don't know if that's what we need, I don't know how to interpret them

Is it essential to know how the 4 tractions respond? For the ascent from the dip we didn't ask ourselves...

No, I define the impulse to be J (a vector, really, but I am assuming it acts normally to the presumed arc)

Sorry, but it's not clear to me... You released a value of 0.68 for J/m, where does it come from? (bear with my shortsightedness please).

 

[haruspex]

Is it essential to know how the 4 tractions respond?

It's probably more a question of how the driver responds to descending into the dip. Eases off the accelerator? Brakes?
But assuming the power is constant, it would accelerate down into the dip then decelerate, more or less equivalently, on the way up. I see nothing better than to assume it is back to its speed on the flat.

You released a value of 0.68 for J/m, where does it come from?

As I wrote, we cannot know^* what the value is. We know it has to be enough to stop the wheels penetrating the ground (a 'deadcat bounce') but not so much that the car gains KE.
^* Unless you have data on the suspension and tyres?

 

[alex33]

We know it has to be enough to stop the wheels penetrating the ground (a 'deadcat bounce') but not so much that the car gains KE.

OK, so 0.68 is an estimate...

But assuming the power is constant, it would accelerate down into the dip then decelerate, more or less equivalently, on the way up.

If we take the frames as a reference to calculate the time: the entry speed, a few moments before reaching the hole, is 2.61 m/s. After we have 18 frames to run across the arc (2.52 m/s). If we take the ground as a reference to mark the position of the chassis, the dip is covered in 23 frames (we are at 2.38 m/s). If it is not necessary to define equations for this part of the sequence we use for the moment an averagee of the last 2 values.

 

[haruspex]

If we take the ground as a reference to mark the position of the chassis, the dip is covered in 23 frames (we are at 2.38 m/s).

Is that the horizontal speed?