Understanding Tire Forces in Bicycle Simulation: Physics and Code Explained

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In summary, the author is trying to figure out how to calculate the traction force on a tire. He starts by reading a book and then finds a problem with the math. He tries to figure out what the problem is and ends up scratching his head.
  • #1
xXNightEagleXx
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I'm creating a game, simple 2d game but that relies on individual tires forces simulation in a bicycle model. There are code out there but I'm not the kind of guy that just copy and paste. I like to understand and i love physics.

I started by reading this web page

It gave me a good starting point . I then found a copy of the book Milliken & Milliken Race car vehicle dynamics. Everything was fine and easy to understand until i reached the tire forces calculation and application.

First of all, I've read that
Ft - ∑ Fresistance = ma

which is pretty straightforward.

However speaking of tires in specific the driving tires receives this Drive torque from the shaft sometimes called
Tqwheel = Te * xg * xd * η

which is translate into traction force
Ft = Twheel / r

So i can plugin this Ft in the first equation and i have a result

But both the link and book states that the longitudinal traction force in the tires is the result of a function of the slip ratio
SR = (ΩR/V) - 1

that gives us a normalized value that gets multiplied by Fz to obtain Fx (which in my understanding is the traction force Ft)

Since what push us forward is the friction force, like this video explain, then what the book says is not wrong (never doubted it)
Ft = μ(SR) * Fz

However this lead me to a math problem, to something that seems to be a contradiction. First of all, first equation says that Ft is directly related to Tqwheel, but the other says that it is related to the friction coefficient and the load, in other words the friction force. Moreover, at t0 with both V and Ω equal to 0, if i apply some Tqwheel that rotates the tire at t1 i'll still have SR = -1 (locked) because my V will still be 0...what am i missing here?

Trying to figure out what I'm missing, i thought that what happens here is that before the tire slips what matter is the Ffs
Ffs = mgμ (as static friction and not a function)

If that's the case then it means that initial instantaneous force is always maximum in proportion to the parameters, regardless you went full pedal or not...which seems something wrong...

Things get even stranger when i analyze front wheel (free wheel) where it creates a friction force in the opposite direction of the rear one. Using the Ffs equation above would mean again instantaneous force...:cry:

This video says that
Ffs = ma*1/2

That would make sense that the bigger the acceleration the bigger the friction force and the bigger the torque that rotates the front wheel but where does the coefficient of friction fit in this equation? I mean if i have 0 coefficient of friction then i have 0 torque.

As you can image at this point gets basically impossible to apply all of this.

In my head i thought that the code would be something like this (per tire)
(rear)
- compute Tqwheel according to throttle pedal
- compute Tqresistance (braking, rolling)
- compute Ft with dynamic weights using Tqtotal (1st doubt here)
- compute Flat = Coefficient of stiffness * Fz
- limit the total force to the amount of grip on the tire
(front)
- compute Tqfriction with dynamic weights
- compute Tqresistance (braking, rolling)
- compute Ft using Tqtotal
- compute Flat = Coefficient of stiffness * Fz
- limit the total force to the amount of grip on the tire
(finally)
- Apply forces for each wheel
- Calculate wheels acceleration
- Integrate the wheels

I'll end for here because just clarifying above points will be extremely helpful and probably even make the pseudo more clear

Thanks in advance
W
 
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  • #2
I guess i have found the root of my doubts... slip ratio at low speed (v0 equal or close to 0)... that stupid thing brought me to a whole world of problem, that simply confused me. Now think about waking up at 6am, overthinking everything up to the point to messed up what was logic and well known...that's what happened to me...desperation to solve a problem up to the point where you believe that you could be wrong from the beginning.

That being said, everything makes sense now, even how the traction force (rear tires) and the inverse friction force (front tires) acts and how the car moves due to traction force.
Problem is that once V = 0 or close, obviously, the ForceTraction = mu(slipratio) * N gets impossible with the "standard" definition of the slip ratio since it will be undefinited.

I'm trying to figure out the best way to deal with low speed situation, maybe even using pajecka magic numbers.

If anyone has a suggestion about how to deal at low speed i'll be glad to hear it.

Thanks for everyone
 
  • #3
While the driving force of the wheel is less than μ times the downward force on the wheel, the SR=0. That is the condition of normal street driving, i.e. gentle acceleration. The SR rises (the wheels spin) when the supplied engine torque exceeds μ times the downward force on the wheel (the sliding static friction).

See also this thread:
https://www.physicsforums.com/threads/why-do-tires-slip-easier-from-a-standstill-start.931807/
 

Related to Understanding Tire Forces in Bicycle Simulation: Physics and Code Explained

1. What are tire forces?

Tire forces are the forces that act on a tire when a vehicle is in motion. These forces include the normal force, lateral force, longitudinal force, and rolling resistance.

2. How do tire forces affect a vehicle's performance?

Tire forces play a crucial role in a vehicle's performance, as they determine the grip, handling, and stability of the vehicle. The amount and direction of these forces can impact acceleration, braking, and cornering abilities.

3. What factors influence tire forces?

There are several factors that can influence tire forces, such as tire tread, tire pressure, road conditions, and vehicle speed. The weight and design of the vehicle also play a significant role in the amount of force exerted on the tires.

4. How do tire forces differ in different driving conditions?

Tire forces can vary depending on the driving conditions. For example, driving on wet or icy roads can decrease the amount of grip and increase the risk of sliding. On the other hand, driving on a dry road with good traction can increase the amount of grip and improve the vehicle's performance.

5. How can tire forces be optimized for better performance?

To optimize tire forces, it is essential to regularly check and maintain tire pressure, rotate tires to ensure even wear, and choose tires with the appropriate tread pattern for the driving conditions. Additionally, driving at appropriate speeds for the road conditions can also help optimize tire forces and improve overall vehicle performance.

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