geometry of a car

The centripetal force can easily be calculated as: $F = (M*v^2)/R = (M*v^2)*sin(\delta)/L$. But how is this force distributed over the (front and rear) wheels? My initial thought was to just divide it by 4 for each wheel, but when you turn your front wheels 90 degrees, there will be no force over the rear wheels. So when simply dividing by 4 is wrong, then how is the distribution in reality?

Is it also safe to assume the forces on the front wheels are equal to each other, and also the same for the rear wheels?

  • $\begingroup$ Note that it's clear that the force on the inner wheels can drop to zero for real cars, since you can turn tightly enough that the inner wheels leave the road. $\endgroup$
    – user107153
    Nov 18, 2019 at 13:22
  • $\begingroup$ Be a little more diligent and decide which $v$ to use (different parts of the car move at different speeds), and also decided where centripetal force is applied. $\endgroup$ Jul 25, 2020 at 23:18

4 Answers 4


The two-wheel model shown in the question is good start for determining the loading on the tires. But this would only result with the total load on the front tires, and the total load on the back tires, without any further details on how these loads are distributed left-to-right. To get there, you also need to consider the height of the center of mass relative to the wheel axles.

In any case, I will proceed with the two-wheel model below.

The laws of motion provide with a balance of forces and well as a balance of torques for you to consider. Maybe doing a free body diagram first, might help. I hope you are familiar with high school physics, trigonometry, and the concept of vectors.


Here centerline of the front wheels is at A, the centerline of the rear wheels at B and the center of mass at C. The side forces at the rear are $\vec{F}_B$ directed towards the center of rotation O, the side forces at the front are $\vec{F}_A$ also directed towards O, driving force on the front is $\vec{P}_A$ (assume front wheel drive car), and the centrifugal force $\vec{F}_C$ is located at C pointing away from O.

Notice also that the radius of turn $R$ is located where the rear wheels are, describing an arc with radius equal to $$R = L \cot \delta$$ where $L$ is the distance $\overline{AB}$.

The rotational speed of the vehicle $$\Omega = \frac{v_B}{R} = \frac{v_B^2 \tan \delta}{L}$$ where $v_B$ is the speed on the rear wheels. Also the radius of the center of mass is $$R_C = \sqrt{R^2 + c^2} = \sqrt{\left( c^2 + L^2 \cot^2 \delta\right)} $$

The arc radius of the front wheels is needed also $R_A = \sqrt{L^2+R^2}$

So the centrifugal force magnitude is $$F_C = m \Omega^2 R_C = m \frac{v_B^2 \tan \delta}{L} \sqrt{\left( 1 + \frac{c^2 \tan^2 \delta}{L^2}\right)}$$

But the car might also accelerate or decelerate changing its rear-wheel speed with acceleration $a_B$ causing rotational acceleration $$\dot \Omega = \frac{a_B}{R}$$ and inertial force on the center of mass $$P_C = m \dot \Omega R_c = m a_B \sqrt{ \left( 1+ \frac{c^2 \tan^2 \delta}{L^2} \right)}$$

Additionally, the change in rotation requires a net torque $M_C$ to be applied to the center of mass that equals to $$M_C = I_C \dot \Omega = I_C \frac{a_B \tan \delta}{L}$$ where $I_C$ is the mass moment of inertia of the car about the center of mass. This can be estimated with $I_C \approx m c (L-c)$ by approximating the car as a lumped mass of $\tfrac{L-c}{L} m$ on the rear wheels and $\tfrac{c}{L}m$ on the front wheels.

And now for the balance of forces and torques (about the center of mass)

$$\begin{aligned} F_B + \tfrac{R}{R_A} F_A + \tfrac{c}{R_C} P_A - (m \dot \Omega R_C) \tfrac{c}{R_C} - (m \Omega^2 R_C) \tfrac{R}{R_A} & = 0 \\ P_A \tfrac{R}{R_C} - F_A \tfrac{L}{R_A} - (m \dot \Omega R_C) \tfrac{R}{R_C} + (m \Omega^2 R_C) \tfrac{L}{R_A} & = 0 \\ c \left( F_B \right) - (L-c) \left( F_A \tfrac{R}{R_A} + P_A \tfrac{L}{R_A} \right) - I_C \dot \Omega & = 0 \end{aligned}$$

Where front-wheel forces act along the $\delta = \tan^{-1} \left( \tfrac{L}{R} \right)$ angle and center-of-mass forces act along the $\delta_c = \tan^{-1} \left( \tfrac{c}{R} \right)$ angle. The above is to be solved for $F_A$, $F_B$ and $P_A$ given a specific rotational acceleration $\dot \Omega$.

The results are very complex, but they can be simplified further with some assumptions. For example, the constant speed assumption, $\dot \Omega = 0 $ has the following solution

$$ \begin{aligned} F_A & = (m \Omega^2) \frac{L^3 R_C-L^2 R_C c +L R_A c^2 + R^2 R_A c}{L(L^2-L c+R_A R_C)} \\ F_B & = (m \Omega^2) \frac{R (L-c)(L\,c+R^2)}{L(L^2-L c+R_A R_C)} \\ P_A & = (m \Omega^2) \frac{R\,R_C (c-L)}{L^2-L c+R_A R_C} \\ \end{aligned}$$

To see if this is even remotely correct, I created a small angle approximation for the general case (small steering angle $\delta$) to find that

$$ \begin{aligned} F_A & \approx m \frac{c v_B^2-(L^2-c^2) a_B}{L^2} \sin \delta \\ F_B & \approx m \frac{(L-c) v_B^2 + (L^2-c^2) a_B}{L^2} \sin \delta \\ P_A & \approx m a_B \cos \delta \\ \end{aligned}$$

which kind of makes intuitive sense, in terms of how speed $v_B$ and acceleration $a_B$ might affect the tire forces.

and for the non-accelerating case

$$ \begin{aligned} F_A & \approx m \frac{c\, v_B^2}{L^2} \sin \delta \\ F_B & \approx m \frac{(L-c) v_B^2 }{L^2} \sin \delta \\ P_A & \approx 0 \\ \end{aligned}$$

which is a result that matches the general rule of the best handling cars are the ones with $c = L/2$. Use this above to find $F_A \approx F_B$, which means the tires are equally loaded resulting in the best handling limit (This is the topic of different discussion).


This problem is statically indeterminate : with the limited information provided it is not possible to determine how the static friction force is distributed between the 4 tyres. The total friction force on the vehicle must be $Mv^2/R$, but it can be distributed between the 4 tyres in an infinite number of ways all of which are consistent with the laws of friction and Newtonian Mechanics.

The friction force on each tyre can have any value up to $\mu N$, where $\mu$ is the coefficient of limiting static friction between the tyres and the road and $N$ is the normal reaction at each wheel. Even if you know the normal forces on each tyre, this only tells you the maximum friction force which can be provided at each tyre : it does not tell you what the actual friction force is, unless you know that the tyre in question is on the point of slipping.

Neither can it be assumed that the friction forces are the same on the two inner or two outer tyres, nor on the two front or two rear tyres. Even if the mass of the vehicle were distributed symmetrically, like a uniform cuboid, and the coefficients of friction were identical at each tyre, this does not prevent there being different friction forces on each tyre.

In practice there is of course a definite value of the friction force on each tyre, which depends on the preceding motion as well as such things as weight distribution and coefficients of static friction.

For example, suppose the inner tyres encounter a patch of ice. Temporarily they cannot provide any friction force : the outer tyres must provide the whole friction force. After the patch of ice has been passed the inner tyres could again provide friction, but if the outer tyres continue to grip well there is no need for the friction force to be re-distributed.

See the following similar questions about the distribution of friction across the surface of a block or between blocks connected by a string :

The Parallelepiped on a Sloping Plane: A Reconsideration of an Old Friend
Will "A" experience max static friction if "A" is connected to "B" which is on the verge of slipping?
Determining the direction of friction

  • $\begingroup$ The problem isn't statically indeterminate exactly if you consider the height of the center of mass, and the moment balance on all directions. With the assumption of the left-right distribution of forces on the rear tires based on the lever rule. $\endgroup$ Jul 26, 2020 at 1:56

You cannot simply divide by 4, no. If you set up that formula for each wheel, then you'll have to take into account that the speeds and distances are different.

The one thing all four wheels have in common is angular velocity $\omega$. Even at a 90 degree turn, the rear and front wheels spin equally fast (degrees per second) around the rotation point. Otherwise the car would be breaking apart.

The angular velocity relation $$v=\omega r$$ helps you calculate the linear speed $v$ for each individual wheel, since you know the distances $r$. With this $v$ and $r$ per wheel you can calculate the centripetal acceleration for each wheel.

I would then divide the mass by 4. That would be a necessary assumption, namely that each wheel "carries" equally much mass. Then the centripetal force on each wheel can be calculated.

I have not done the calculations, but I would expect all four forces to be different. And varying differently with the turning angle of the front wheels (some will be cosine and some sine to the angle). For a 90 degree turn you will have a rotation point located at the first rear wheel. That wheel spins around, but doesn't move linearly. It has zero speed $v$ and distance $r$. According to your formula (if you plug in the relation $r=v/\omega$ in place of $r$), zero speed indeed shows zero centripetal force.

Is it also safe to assume the forces on the front wheels are equal to eachother, and also the same for the rear wheels?

If the turn is big, then differences in the distance $r$ become negligible. The differences in speeds will as well become negligible. And then all forces are more or less equal, yes.

  • 1
    $\begingroup$ But....wheels also have different camber, different static weight distributions (mostly F/B), then there are suspension (possibly active) dependent loading while turning at any radius, then you put a wing on the car and it goes way beyond distributing the mass of the car--the aero loads can be greater--you can pull 4+ lateral g's in a turn. Then put some electronic stability control and other assists that change power to each wheel. Bank the turn--assume there is a racing groove (position dependent static friction), finally add dirty air from the leader. this is a complex engineering question. $\endgroup$
    – JEB
    Apr 17, 2018 at 13:30

Try considering this

Since the car is driven by one engine let us assume all 4 wheels have same velocities at all time for simplicity.

And the weight of car is divided quite uniformly over the 4 wheels as well

Sonnow that we have made the m•(v^2) part of the equation same for all the wheels let us move on to the Raius of curvature part

And as for Radius os curvature it will be different for the wheels on the inner and on the outer (in case of the picture provided considering the car is turning clockwise the right and the left pairs of wheels respectively)

For the wheels on left "R" will be less thus they will get more of the centripetal force and will require more friction

For the wheels on right "R" will be more thus they will get less of the centripetal force and will require less friction

Thus i think the friction force should be divided not on the basis of front and back but on the basis of inner and outer


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