I Straight-Line Vehicle Motion problem

Summary
Impossible System?
Summary: Impossible System?

Hello, I am trying to simulate the dynamics of a vehicle accelerating from a standstill to top speed. The vehicle acceleration equation is:

Av = (Ft+Ff)/m

Where Ft = Traction Force, Ff = Friction Force (From Drag and Rolling Resistance), and m = Vehicle Mass

My problem here is that the traction force depends on vertical load on the tire, which also depends on the acceleration of the vehicle.

Wf,r,d = Wf,r,s+(h/L)*m*A

h = vehicle height, L = wheelbase, W(f,r,s) and W(f,r,d) is the static and dynamic weight (load) on the front and rear wheels.

The Traction force is calculated by multiplying the vertical load on the tire by the longitudinal tire force aka normalized longitudinal force. The latter is determined by using Pacejka's Magic Tire Formula. What I don't understand is how am I supposed to calculate the acceleration if the acceleration is needed to calculate...acceleration?

Here are some screenshots of the website I am referencing:
244608

244609
 
Last edited by a moderator:

jrmichler

Science Advisor
892
793
You need to understand the assumptions implicit in the equations.

1) If you apply a "small" torque to the wheel, the acceleration is determined by the torque after subtracting rolling friction.

2) If you apply a "large" torque to the wheel, it spins (lays rubber), and the acceleration is determined by the normal force and friction coefficient (maximum traction force).

The actual traction force is the lesser of the drive torque and the maximum traction force.

Slip ratio is a second order effect that can and should be ignored until after the large effects mentioned above are fully understood. Use search terms tire slip ratio for some good references.

And please edit your background. You have a degree in mechanical engineering, but are still in high school?:smile:
 
You need to understand the assumptions implicit in the equations.

1) If you apply a "small" torque to the wheel, the acceleration is determined by the torque after subtracting rolling friction.

2) If you apply a "large" torque to the wheel, it spins (lays rubber), and the acceleration is determined by the normal force and friction coefficient (maximum traction force).

The actual traction force is the lesser of the drive torque and the maximum traction force.

Slip ratio is a second order effect that can and should be ignored until after the large effects mentioned above are fully understood. Use search terms tire slip ratio for some good references.

And please edit your background. You have a degree in mechanical engineering, but are still in high school?:smile:
I am working towards a degree in mechanical engineering. I am graduating in a couple of weeks. Sorry for the misunderstanding. I will do some more research as to how the system works, but is the system even possible? I understand the effects but I want to simulate it in the most realistic way.
 
You need to understand the assumptions implicit in the equations.

If you apply a "large" torque to the wheel, it spins (lays rubber), and the acceleration is determined by the normal force and friction coefficient (maximum traction force).
The maximum traction force is the weight, or load, on the tire multiplied by the friction coefficient, yes. But the load on the rear wheels is ALSO dependent on acceleration.

Wf,r,d = Wf,r,s+(h/L)*m*A
 
663
250
The maximum traction force is the weight, or load, on the tire multiplied by the friction coefficient, yes. But the load on the rear wheels is ALSO dependent on acceleration.

Wf,r,d = Wf,r,s+(h/L)*m*A
I assume the load on the front wheels is similarly reduced by the (h/L) factor? I guess this is a rear wheel drive car so doesn't matter. If you put this into the longitudinal F=ma equation you just get an effective "mass" (that is reduced by (h/L))...yes?
 
I assume the load on the front wheels is similarly reduced by the (h/L) factor? I guess this is a rear wheel drive car so doesn't matter. If you put this into the longitudinal F=ma equation you just get an effective "mass" (that is reduced by (h/L))...yes?
Yes, this is true. Load will always shift from the front to the rear, but I would also like to simulate FWD and AWD systems as well. I mean, surely, there must be a way around this. Acceleration cannot depend on itself. Even if I make an if statement of the following:

IF(Load_Rear*Coeff_Friction < Force_Traction)
then(Load_Rear*Coeff_Friction = Force_Traction)
else (Force_Traction = Torque_Wheel/Radius_Wheel)

Load_Rear, or Load_Front, is always dependent on the vehicle acceleration, and if there is slip, it results in an error because the load transfer equation, which is dependent on vehicle acceleration, is being used in place of the traction force as the maximum traction force to calculate acceleration. You cannot calculate something that is dependent on itself. However, the website says otherwise:
Screen Shot 2019-06-05 at 11.29.42 AM.png

It says "after a brief moment, the amount of weight shifted"... does this mean I need to use the acceleration from the previous time step for the weight transfer?
 
663
250
Yes, this is true. Load will always shift from the front to the rear, but I would also like to simulate FWD and AWD systems as well. I mean, surely, there must be a way around this. Acceleration cannot depend on itself. Even if I make an if statement of the following:

IF(Load_Rear*Coeff_Friction < Force_Traction)
then(Load_Rear*Coeff_Friction = Force_Traction)
else (Force_Traction = Torque_Wheel/Radius_Wheel)

Load_Rear, or Load_Front, is always dependent on the vehicle acceleration, and if there is slip, it results in an error because the load transfer equation, which is dependent on vehicle acceleration, is being used in place of the traction force as the maximum traction force to calculate acceleration. You cannot calculate something that is dependent on itself. However, the website says otherwise:
View attachment 244620
It says "after a brief moment, the amount of weight shifted"... does this mean I need to use the acceleration from the previous time step for the weight transfer?
No it all happens at once. Suppose you wish to solve for the maximum acceleration you can achieve before slipping. What does that equation look like?
 
No it all happens at once. Suppose you wish to solve for the maximum acceleration you can achieve before slipping. What does that equation look like?
Well, I assume it would be the instantaneous tire load multiplied by the friction coefficient divided by the mass of the vehicle:

Am = Wm*Cf/m


Wouldn't this be it? Or is there another way?
 
663
250
Well, I assume it would be the instantaneous tire load multiplied by the friction coefficient divided by the mass of the vehicle:

Am = Wm*Cf/m


Wouldn't this be it? Or is there another way?
Yes and the instantaneous tire load implicitly contains A. Put that in explicitly and rearrange to again get A. This will give the max A without slipping but includes the dynamic loading.
 
Yes and the instantaneous tire load implicitly contains A. Put that in explicitly and rearrange to again get A. This will give the max A without slipping but includes the dynamic loading.
I will try to accomplish this, but if I fall short, could you break it down for me?
 

jrmichler

Science Advisor
892
793
I think I see the root cause of your confusion. You are trying to go straight to the equation(s), and skipping the free body diagram (FBD). An FBD is much more than a statics class homework problem. It shows that you have accounted for all of the forces, and it is the basis for deriving the necessary equations.

The engineering curriculum gives you the tools to solve a wide range of real world problems. Many of those problems do not have a readily available equation. Even if an equation was available, you still need to verify your solution, and you will need to justify your solution to others. Merely stating that you found an equation on the internet will not suffice.

This is not an impossible system, but a relatively simple and straightforward system. So make the FBD, post it, and let us critique it.
 
They say that C=1.0 I think and you have already done the W (see above).
Okay, I think I am starting to understand the system a bit better, however I don't know wether to use the instantaneous wheel load or the static load over the tires for the rolling resistance.
 
663
250
Okay, I think I am starting to understand the system a bit better, however I don't know wether to use the instantaneous wheel load or the static load over the tires for the rolling resistance.
Write down a solution attempt (complete) and we can proceed.
 

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