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

• alex33
In summary: So the front wheels would only leave the ground for a moment and then they would be back on the ground.
haruspex said:
* 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)

haruspex said:
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?

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alex33 said:
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.

haruspex said:
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

alex33 said:
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.

haruspex said:
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

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alex33 said:
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:
alex33 said:
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.

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...

alex33 said:
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.

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

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Hi,
haruspex said:
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 Y0c 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 X0c and Y0c.

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

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.

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alex33 said:
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?

haruspex said:
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 Z0c 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 Z0c and leave only the first 2 terms in the wheel coordinates.

TonyStewart said:
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.

alex33 said:
In the formula that returns Z0c 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.

haruspex said:
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?

alex33 said:
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##

alex33 said:
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.

haruspex said:
##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

alex33 said:
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.

Ok, thanks... don't worry

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