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

[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?

 

[alex33]

^* Unless you have data on the suspension and tyres?

Furthermore I finded:
Vertical Damping rate 17.3 LB-Sec^2/FT^2
Horizontal Suspension Rate 51,000 LB/FT
Horizontal Suspension Damping rate 2420 LB/(FT/SEC)
Wheel Radial Spring Rate 400 LB/FT (0-1.5 IN)
680 LB/FT (1-5.3 IN)
7300 LB/FT (3IN)

Wheel Damping Rate 2,5 LB/(FT/SEC)

Is that the horizontal speed?

yes. I have a doubt (maybe stupid) ... to calculate the horizontal speed in the dip, should I take the measurement of the arc or the one of the chord?

 

[haruspex]

View attachment 326801

Furthermore I finded:
Vertical Damping rate 17.3 LB-Sec^2/FT^2
Horizontal Suspension Rate 51,000 LB/FT
Horizontal Suspension Damping rate 2420 LB/(FT/SEC)
Wheel Radial Spring Rate 400 LB/FT (0-1.5 IN)
680 LB/FT (1-5.3 IN)
7300 LB/FT (3IN)

Wheel Damping Rate 2,5 LB/(FT/SEC)yes. I have a doubt (maybe stupid) ... to calculate the horizontal speed in the dip, should I take the measurement of the arc or the one of the chord?

I'll see what I can deduce from the damping rate.
For horizontal speed, yes, use the chord.

 

[alex33]

I'll see what I can deduce from the damping rate.
For horizontal speed, yes, use the chord.

SORRY... I had lost a data... there is one more...
VERTICAL SUSPENSION RATE 14 LB/IN (0-9 Inches)
500 LB/IN (< 0 or >9 IN)
SUSPENDED VEICLE MASS 1424 LBs
WHEEL MASS 24 LBs

 

[haruspex]

Wheel Radial Spring Rate 400 LB/FT (0-1.5 IN)
680 LB/FT (1-5.3 IN)
7300 LB/FT (3IN)

I need the spring 'constant' for the suspension. The above look ok, but imply it is not constant over the range, which is a nuisance. And I do not understand how the last number fits in. How can it average 680 over 1 to 5.3 inches but be 7800 at 3 inches? Seems like a typo or two.

 

[alex33]

Seems like a typo or two.

I try to check if I find the constant .... but I'll check better tomorrow, now I'm about to fall off my chair because I'm sleepy

...nothing at the moment... sorry

 

[haruspex]

View attachment 326803

I try to check if I find the constant .... but I'll check better tomorrow, now I'm about to fall off my chair because I'm sleepy

...nothing at the moment... sorry

Yes, typo in post #106:

1-5.3 IN

should read 1.5-3 IN.
Another puzzle is the vertical suspension rate. If it means what I think, I believe this should be more than the wheel spring rate, but it is only 170lb/ft. Anyway, it's the wheel spring rate that I need.

 

[alex33]

So, It's my fault in the trascription of the spring rate, due to late night work... Solved!
For vertical suspension rate I read also 500 LB/IN out of the range 0-9 IN.
Please tell me if I have to find others informations...

 

[haruspex]

So, It's my fault in the trascription of the spring rate, due to late night work... Solved!
For vertical suspension rate I read also 500 LB/IN out of the range 0-9 IN.
Please tell me if I have to find others informations...

No, that'll do for now. I just have to figure out how to use it without painful algebra.

 

[alex33]

Before starting the simulation of phase 2 (free fall), I ask you to confirm the starting data:

$V_0x = V = 2,52 m/s$ (Chord length / 18 frames ...I think it is the most reliable data)

$V_0z = \frac {1}{2}V* \tan(α)$

$ω_0 = \frac {1}{2L}V* \tan(α) - (\frac {3g*L}{h^2+4L^2}*4/24)$

$X_0c= \frac {chord-length}{2}-L$

$Z_0c = \frac {(ω_0 + \frac {1}{2L}V* \tan(α))}{2} * L * 4/24$

$θ_0 = \frac {V * \tan(α)}{2L}*4/24$

Thanks

 

[alex33]

Hi,

The position of the rear wheels is trickier.
xr=xc−Lcos⁡(θ)+hsin⁡(θ)
yr=yc−Lsin⁡(θ)−hcos⁡(θ)
Similarly the front wheels:
xf=xc+Lcos⁡(θ)+hsin⁡(θ)
yf=yc+Lsin⁡(θ)−hcos⁡(θ)

Setting the spreadsheet I realize that these expressions are a problem, I don't think they are correct: for example Yf gives a negative value, while Yc is positive. I think we should add h to Y_0c and for the positions Yr and Yf subtract or add to Yc LSin(θ) without adding the third term of the expression. Same thing should be done with X... It shouldn't be necessary to add or subtract the Cartesian components of h to the wheels coordinate if we have correctly defined X_0c and Y_0c.

PS: (I'm still using the letter Z instead of Y for the vertical axis).

 

[TonyStewart]

TL;DR just started to skim this thread.
I would follow the Italian method in#1 and ignore all the formulae.

It all depends on speed and shock compression with stored energy how much it springs back with upstrokes falling down much faster than the downstroke with wheels moving up.

I imagine a half moon pothole big enough to drop the wheels and the body depend greatly on the horizontal and vertical speed, spring time constant, momentum, and a little wheel acceleration during the extra circumference travelled in the hole while the vehicle is at constant speed. That speed could make the difference from a wheelie to a damaged front tire or front end strut from the increase downward momentum and sharp edge while storing spring energy to increase the height of the wheelie.

To me it would look not much different than driving over a curb in terms of tire and front end impact starting from the bottom of the pothole. It would look like a step function input rather than a pulse of a curb so the angular momentum of the body is different. But the damage to the front end could be similar at different speeds. The step will reveal the damping factor of the suspension and % of overshoot aggravated by the upward thrust of the quarter pipe exit. .

I'll let you guys go back to your $V_ot +\frac 1 2 at^2$ where a=0 which doesn't begin to describe the angular momentum stored and released and compression on impact before the wheelie.

 

[haruspex]

Yf gives a negative value, while Yc is positive

Yc should start at h, so I don't see how yf would be negative immediately. Are you saying it goes negative later?

 

[alex33]

Yc should start at h, so I don't see how yf would be negative immediately. Are you saying it goes negative later?

In fact it shouldn't be negative. In the formula that returns Z_0c we have not taken into account h, so the point is a few centimeters above the ground line. In yf=yc+Lsin⁡(θ)−hcos⁡(θ) (as example) the third element of the expression is greater than the sum of the first 2, so already at t₀ I get a negative value. I think we should add h to Z_0c and leave only the first 2 terms in the wheel coordinates.

 

[alex33]

TL;DR just started to skim this thread.
I would follow the Italian method in#1 and ignore all the formulae.

It all depends on speed and shock compression with stored energy how much it springs back with upstrokes falling down much faster than the downstroke with wheels moving up.

I imagine a half moon pothole big enough to drop the wheels and the body depend greatly on the horizontal and vertical speed, spring time constant, momentum, and a little wheel acceleration during the extra circumference travelled in the hole while the vehicle is at constant speed. That speed could make the difference from a wheelie to a damaged front tire or front end strut from the increase downward momentum and sharp edge while storing spring energy to increase the height of the wheelie.

To me it would look not much different than driving over a curb in terms of tire and front end impact starting from the bottom of the pothole. It would look like a step function input rather than a pulse of a curb so the angular momentum of the body is different. But the damage to the front end could be similar at different speeds. The step will reveal the damping factor of the suspension and % of overshoot aggravated by the upward thrust of the quarter pipe exit. .

I'll let you guys go back to your $V_ot +\frac 1 2 at^2$ where a=0 which doesn't begin to describe the angular momentum stored and released and compression on impact before the wheelie.

Thanks Tony !
we went from theory to numbers just to verify the first proposed solution to the problem, with respect to a real event, but identifying the correct physical model is always the main objective of the question I asked in this thread. So thank you for your comments on which I and the other friends who have made themselves available will reflect for sure.

 

[haruspex]

In the formula that returns Z_0c we have not taken into account h,

Ah, yes, I see my mistake. In post #92 I omitted an "h+" at the start of the formula. It's in my spreadsheet, but I missed it.

 

[alex33]

Ah, yes, I see my mistake. In post #92 I omitted an "h+" at the start of the formula. It's in my spreadsheet, but I missed it.

So you confirm that the 3rd element in the wheels coordinate expressions is neessary. I'll reflect a bit more on this.

My expressions in #114 are ok?

 

[haruspex]

Before starting the simulation of phase 2 (free fall), I ask you to confirm the starting data:

$V_0x = V = 2,52 m/s$ (Chord length / 18 frames ...I think it is the most reliable data)

$V_0z = \frac {1}{2}V* \tan(α)$

$ω_0 = \frac {1}{2L}V* \tan(α) - (\frac {3g*L}{h^2+4L^2}*4/24)$

$X_0c= \frac {chord-length}{2}-L$

$Z_0c = \frac {(ω_0 + \frac {1}{2L}V* \tan(α))}{2} * L * 4/24$

$θ_0 = \frac {V * \tan(α)}{2L}*4/24$

Thanks

$V_0z$ should be $=ω_0 L$.
$Z_0c$ needs "h+", as discussed.
For $θ_0$ I use the average rotation rate during del t:
$(\frac {V * \tan(α)}{2L}+ω_0)/2*4/24$

 

[TonyStewart]

Thanks Tony !
we went from theory to numbers just to verify the first proposed solution to the problem, with respect to a real event, but identifying the correct physical model is always the main objective of the question I asked in this thread. So thank you for your comments on which I and the other friends who have made themselves available will reflect for sure.

The first model is the absorption of sharp edges of the tires, then the spring mass shock model. Otherwise this buggycar will drive anyone buggy without more assumptions being defined or become just a wooden car with no suspension and solid rubber tires with an extremely rough ride.

 

[alex33]

$V_0z$ should be $=ω_0 L$.
$Z_0c$ needs "h+", as discussed.
For $θ_0$ I use the average rotation rate during del t:
$(\frac {V * \tan(α)}{2L}+ω_0)/2*4/24$

Dear Haruspex, do you check the values I send you? Please take a look at last calculations we shared. Thanks

 

[haruspex]

Dear Haruspex, do you check the values I send you? Please take a look at last calculations we shared. Thanks

I will do, but did not get a chance today.

 

[alex33]

Ok, thanks... don't worry