- #1

xXNightEagleXx

- 3

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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 - ∑ F

_{resistance}= ma

which is pretty straightforward.

However speaking of tires in specific the driving tires receives this Drive torque from the shaft sometimes called

Tq

_{wheel}= T

_{e}* xg * xd * η

which is translate into traction force

F

_{t}= T

_{wheel}/ 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 F

_{z}to obtain F

_{x}(which in my understanding is the traction force F

_{t})

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

F

_{t}= μ(SR) * F

_{z}

However this lead me to a math problem, to something that seems to be a contradiction. First of all, first equation says that F

_{t}is directly related to Tq

_{wheel}, but the other says that it is related to the friction coefficient and the load, in other words the friction force. Moreover, at t

_{0}with both V and Ω equal to 0, if i apply some Tq

_{wheel}that rotates the tire at t

_{1}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 F

_{fs}

F

_{fs}= 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 F

_{fs}equation above would mean again instantaneous force...

This video says that

F

_{fs}= 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 Tq

_{wheel}according to throttle pedal

- compute Tq

_{resistance}(braking, rolling)

- compute F

_{t}with dynamic weights using Tq

_{total}(1st doubt here)

- compute F

_{lat}= Coefficient of stiffness * Fz

- limit the total force to the amount of grip on the tire

(front)

- compute Tq

_{friction}with dynamic weights

- compute Tq

_{resistance}(braking, rolling)

- compute F

_{t}using Tq

_{total}

- compute F

_{lat}= 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