Automotive Normal load on the tires of car

[Yanna]

Hello there!

I am Yanna and I am building a racing car. I am now trying to calculate the tire drag of this car when cornering, but I'm stuck on something.

In order to find the tire drag, I need the normal load on each wheel. To find this, I made some equations:
1) the sum of the normal load over all four wheels equals the total gravity force;
2) the sum of the momenta around the Center of Gravity in pitch direction is zero;
3) the sum of the momenta around the Center of Gravity in roll direction is zero.
However, I now have 3 equations and 4 unknowns, so I cannot calculate the normal load on each wheel. What should the fourth equation be? I have some ideas:
a) The sum of the momenta around the Center of Gravity in yaw direction is zero;
b) Some equation regarding the torsion of the car being zero. (I would not know how to make this equation)

Is any of these ideas correct or is it another equation that I should use?

Thank you in advance for your help!

 

[Ranger Mike]

Race car suspension Class

above has much valuable info on this

 

[AZFIREBALL]

To be on the safe side why not just consider all the weight of the car on the two outside wheels in a turn?
It is clear from watching many races...there is very little car weight on the inside wheels in a tight turn.
BTW - are there any down forces applied to the car via aerodynamics?

 

[jrmichler]

Google is your friend: Search statically indeterminate problem. I found a section on statically indeterminate structures in my old undergrad strength of materials book. The key to solving this problem is that all four wheels are mounted on springs, and the ground keeps the four wheels aligned to a plane.

 

[jack action]

Let's look at the no springs case first. We can find the portion of the weight on the front axle ($W_f$) with the pitch moment equation:
$$W_f = \frac{d}{L}mg + \frac{h}{L}ma_x$$
Where $\frac{d}{L}m = m_f$, i.e. the portion of the vehicle's mass "acting" on the front. ($L$: wheelbase; $d$: distance from CG to rear axle; $h$: CG height; $m$: vehicle mass; $a_x$: longitudinal acceleration)

Now looking at the front axle only, we can look at the roll moment equation:
$$TW_{Lf} = \frac{T}{2}W_f + hm_fa_y$$
Or:
$$W_{Lf} = \frac{1}{2}\frac{d}{L}mg + \frac{1}{2}\frac{h}{L}ma_x + \frac{h}{T}\frac{d}{L}ma_y$$
($W_{Lf}$: load on left front wheel; $T$: track; $a_y$: lateral acceleration)

You can do the same for all other wheels, which will leave you with sign changes and/or $d$ replaced by $(L-d)$ as minor differences in the equations.

Now let's assume that there are springs and that the body rolls about a fixed roll center (the red dot in the following figure):

The lateral acceleration is now acting at $h_0$ instead of $h$. The moment $M$ - that transmits to the ground through the spring - is $(h-h_0)ma_y$. But that is for the whole car (front and rear). In this case, the red spring would represent the combination of both the front and rear springs. Because they are in parallel, the total spring (roll) stiffness is $K_f + K_r$. And the contribution of the front springs would be $\frac{K_f}{K_f + K_r}$, thus $M = \frac{K_f}{K_f + K_r}(h-h_0)ma_y$.

Therefore in our previous equation for $W_{Lf}$, we have to replace $hm_fa_y$ with $h_0m_fa_y + \frac{K_f}{K_f + K_r}(h-h_0)ma_y$, which leads to:
$$W_{Lf} = \frac{1}{2}\frac{d}{L}mg + \frac{1}{2}\frac{h}{L}ma_x + \frac{h_0\frac{d}{L}+(h-h_0)\frac{K_f}{K_f + K_r}}{T}ma_y$$

Just to be clear, $h_0$ is measured at the CG location (between the front & rear axles) where the roll axis passes. The roll axis is the line joining the front & rear roll centers. So it is the same value for the front or the rear axle.

Note that if $h_0=h$, then it is the same equation as with the no springs case (the entire moment is absorbed by the solid axle and the springs absorb nothing).

Note also that if the rear springs are infinitely stiff (i.e. solid bars) and $h_0 = 0$, then the rear axle absorb the entire roll moment and there is none on the front axle (if the rear springs are solid bars, then the body doesn't rotate and the front springs cannot absorb anything).

 

[Yanna]

Thank you! I found the normal load on each wheel now!