A vehicle drives in a circle on a track at constant speed at with radius of curvature $\rho$. The vehicle's acceleration is $$a = \upsilon' T + \kappa (\upsilon)^2 N \\ = \kappa (\upsilon)^2 N.$$

The track is banked at angle $$\theta = \arctan \frac {\kappa \upsilon^2}{g}$$ so that all the force required is provided by the normal force.

Question: Does the vehicle's driver need to turn the steering wheel?

If yes: Why? Isn't all the force needed to turn provided by the bank? What additional force is needed from turned wheels?

If no: How does the vehicle roll (and not skid) if its wheels are pointed in the wrong direction?

Perhaps I am misunderstanding how turned wheels work: my assumption is that turned wheels do not skid, as long as the vehicle is moving in that direction, which would imply that for any angle that wheels turn, there is exactly one trajectory that the vehicle can take without skidding. Is that correct? How exactly do turning wheels function?

  • $\begingroup$ All the force needed to turn is indeed provided by the bank. The car will slide around the bank, but, absent friction, will rotate up the bank. This apparent rotation is of course just inertia, but in the frame of the car it seems like there is some force rotating the car. This force is a fictitious force caused by the fact that the road is curved. It seems like you are studying differential geometry. Keep studying and you'll learn about parallel transport, christoffel symbols, etc. That is the language that will help you think more precisely about why the driver needs to turn the wheels. $\endgroup$ Aug 18 at 6:03
  • $\begingroup$ Intuition tells me that a bank any less than 90 is a yes. But that you'd have to include gravity, and the (true, including down force) weight of the car, in any equation. And what kind of tires, and how hot they are. And track conditions. And... air resistance. Which means take it to the track and find out. Verify in field. $\endgroup$
    – Mazura
    Aug 19 at 18:53
  • $\begingroup$ there is exactly one trajectory that the vehicle can take without skidding - depends on how full the tires are. When my truck tires are low I can feel them crumple over themselves in a fast turn. There's a +/- for that one trajectory before it actually skids. $\endgroup$
    – Mazura
    Aug 19 at 19:06
  • $\begingroup$ Not if it's a 90-degree bank. (which seems unlikely because -- gravity) $\endgroup$ Aug 20 at 18:10

6 Answers 6



Your question is why we need to turn on a banked road if the normal force is already sending us around the curve. It's instructive, then, to consider what happens if there really is no turning force on the car. You can imagine the road is icy -- so icy that there is no friction force between the wheels and the road.

We can think of our car, then, as a vector $\vec{w}$ that points from the back of the car to the front of the car. This vector is sent around the road at a fixed height set by the normal force, as you say. To borrow video game terminology, if the car had no collision, we would expect $\frac{d\vec{w}}{dt} = 0$ as the car goes around the curve. This is, if you like, conservation of angular momentum: the car doesn't rotate unless a torque makes it rotate, but there is no torque because the road is icy.

There's only one problem: the car does have collision. As much as angular momentum conservation wants $\frac{d\vec{w}}{dt} = 0$, the road has other ideas. The normal force from the road will act as a torque which keeps the four wheels of the car on the road instead of passing through the road. In other words, there will be a torque from the road which makes sure that $\vec{w}$ remains tangent to the road. But within that tangent plane to the road, $\vec{w}$ is free to rotate.

How, then, will $\vec{w}$ rotate in the plane of the road. A first instinct might be that it doesn't rotate. If we're a bit more careful, we'll ask "doesn't rotate with respect to what?" The frame of the car is clearly not inertial, so that probably shouldn't be the frame in which the car isn't rotating. Instead, the car will not rotate in the ground frame, which, perhaps unintuitively, will look like a rotation in the car frame. The point is this: we can't have $\frac{d\vec{w}}{dt} = 0$ because the road is solid and forces $\vec{w}$ to be tangent to it, but we can have the next best thing $\frac{d\vec{w}}{dt} = \lambda(t) \vec{n}$, where $\vec{n}$ is the normal vector to the road. In other words, all of the change in the orientation of the car happens perpendicular to the road, but there is no change in orientation in the plane of the road, since, as we have said, the road is icy and has no way of forcing that to happen.

Into the Mathematical Weeds

Alright, let's roll up our sleeves and say what it means to have $\frac{d\vec{w}}{dt} = \lambda(t) \vec{n}$. First we need to describe the road. I'll take the road to be a section of a cone. This gives a parametrization $$ \vec{x}(\rho,\phi) = (\rho\cos\phi, \rho\sin\phi,\rho\tan\theta) $$ where $\rho$ is the distance from the axis of the bank, $\phi$ is the angle around the turn, and $\theta$ is the angle of incline of the bank. At each point on the road we have a natural basis: \begin{align} \vec{x}_\rho &\equiv \frac{\partial \vec{x}}{\partial \rho} = (\cos\phi, \sin\phi, \tan\theta) \quad \vec{x}_\phi \equiv \frac{\partial \vec{x}}{\partial \phi} = (-\rho\sin\phi, \rho\cos\phi, 0) \\ \vec{N} &= \vec{x}_\rho \times \vec{x}_\phi = (-\rho\cos\phi\tan\theta, -\rho\sin\phi\tan\theta, \rho) \end{align} $\vec{x}_\rho$ is the vector that points outward and up along the bank, $\vec{x}_\phi$ is the vector that points forward on the road, and $\vec{N}$ is a vector perpendicular to the plane of the road. Note that these vectors are not normalized, but they still make a perfectly good basis.

Our car takes a path $\gamma(t) = (\rho_0 \cos(\omega t), \rho_0 \sin(\omega t), \rho_0 \tan\theta)$ around the bank at some fixed distance $\rho_0$ from the axis of the bank. Since $\vec{w}$ is tangent to the road, we'll have \begin{align} \vec{w}(t) &= w_\rho (t) \vec{x}_\rho(\gamma(t)) + w_\phi(t) \vec{x}_\phi(\gamma(t)) \\&= w_\rho(t) (\cos\omega t ,\sin\omega t, \tan\theta) + w_\phi (t)(-\rho_0\sin\omega t, \rho_0 \cos\omega t, 0) \end{align} We calculate \begin{align} \frac{d\vec{w}}{dt} &= w_\rho ' \vec{x}_\rho + w_\rho\omega(-\sin\omega t, \cos\omega t, 0) + w_\phi ' \vec{x}_\phi + w_\phi \omega (-\rho_0 \cos\omega t, -\rho_0\sin\omega t, 0) \\&=w_\rho ' \vec{x}_\rho + w_\rho\frac{\omega}{\rho_0} \vec{x}_\phi(\gamma(t)) + w_\phi ' \vec{x}_\phi + w_\phi \omega\frac{\tan\theta \vec{N} - \rho_0 \vec{x}_\rho(\gamma(t))}{\sec^2\theta} \\&= \left[w_\rho' - \frac{w_\phi \omega\rho_0}{\sec^2\theta}\right]\vec{x}_\rho + \left[\frac{w_\rho \omega}{\rho_0} + w_\phi '\right]\vec{x}_\phi + w_\phi \omega \sin\theta\cos\theta\vec{N} \end{align} Now remember that we said above we should have $\frac{d\vec{w}}{dt} = \lambda(t)\vec{n}$ where $\vec{n}$ is the normal vector to the plane. This is just the statement that all but the $\vec{N}$ components of $\frac{d\vec{w}}{dt}$ must vanish. We immediately obtain the system of differential equations \begin{align} w_\rho ' - \frac{w_\phi\omega \rho_0}{\sec^2\theta} &= 0 \\ \frac{w_\rho \omega}{\rho_0} + w_\phi ' &= 0 \end{align} Rearranging \begin{align} w_\rho '' &= \frac{\omega\rho_0}{\sec^2 \theta} w_\phi ' = - \frac{\omega^2}{\sec^2\theta} w_\rho \\ w_\phi'' &= -\frac{\omega}{\rho_0} w_\rho ' = - \frac{\omega^2}{\sec^2\theta} w_\phi \end{align} Thus $\vec{w}$ rotates in the tangent plane to the road at an angular velocity of $\frac{\omega}{\sec\theta} = \omega \cos\theta$.


The conclusion that the car rotates in the tangent plane to the road at an angular velocity of $\omega\cos\theta$ is a bit opaque. At this point it might aid interpretation to reintroduce some physical constants that describe the motion of the car around the bank. If $R$ is the radius of curvature of the bank, we have $$ \omega = \sqrt{\frac{g\tan\theta}{R}} $$ Thus $$ \omega\cos\theta = \sqrt{\frac{g\tan\theta}{R}} \cos\theta = \sqrt{\frac{g}{R}\sin\theta\cos\theta} = \sqrt{\frac{g}{2R} \sin(2\theta)} $$ Therefore, the car rotates in the plane of the road at an angular velocity of $\sqrt{\frac{g}{2R} \sin(2\theta)}$.

Note that this rotation is not caused by any force. After all, we supposed the road was icy. Rather, this rotation is purely due to us adopting the non-inertial frame $\vec{x}_\rho, \vec{x}_\phi$. From the ground, our car would not appear to be rotating it all. Rather, it would be maintaining, to the best of its ability, the orientation with which it entered into the turn. It just happens that maintaining your orientation on a banked turn requires rotating up the bank. This is very much like (in fact mathematically identical) to the situation with Foucault's pendulum. As the pendulum traces a curve around the sphere that is our planet, the path of the pendulum rotates from the perspective of observers on the ground near the pendulum. In fact, the pendulum is maintaining its orientation in space, but because we adopt the non-inertial frame of the pendulum, we see the pendulum rotate and knock over another domino for seemingly no reason.

So, to answer your question, we do need to turn. We need to turn into the bank to oppose this tendency of the car to rotate up the bank.

Further Reading

What we've really done here is parallel transport the vector of the car $\vec{w}$ around the curve $\gamma(t)$. Because the surface was curved in space, this parallel transport resulted in the vector $\vec{w}$ changing orientation as it moved along $\gamma(t)$.

All of this is covered in any textbook on differential geometry. My personal favorite is Frankel's Geometry of Physics. I don't know if it's a classic, but it absolutely should be. Everything you could want to know about the intersection of differential geometry and physics is in that text and is presented in a very clear way with helpful exercises.

  • $\begingroup$ Addendum: after a fairly lengthy calculation, which I can produce if you like, I found that the angular velocity of $\sqrt{\frac{g}{2R}\sin(2\theta)}$ does not depend on the profile of the bank. As long as a car takes a circular path around a bank with angle of elevation $\theta$ and radius of curvature $R$, it will tend to spin up the bank at angular velocity $\sqrt{\frac{g}{2R} \sin(2\theta)}$ in the absence of friction. $\endgroup$ Aug 19 at 2:03
  • $\begingroup$ By angle of elevation $\theta$, I mean the angle the vector $\vec{x}_\rho$ makes with the ground at the height of the car. Of course if the profile is not a straight line, this $\theta$ will change depending on the car's position higher or lower on the bank, but we will still have an angular velocity of $\sqrt{\frac{g}{2R}\sin(2\theta)}$. $\endgroup$ Aug 19 at 2:09

As to whether the driver needs to turn the steering wheel:
there is the following extreme case: a surface banked all the way to 90 degrees. In that extreme case the wheels of the car must be pointing straight, otherwise the the car would veer off.

So I expect that when a car is negotiating a banked turn the wheels will be pointing into the turn somewhat less than if the car would be going around a corner on a level road, but some amount of pointing of the wheels will be necessary.

  • $\begingroup$ Thanks. Can this be made more precise, by analyzing the forces? $\endgroup$ Aug 17 at 21:58
  • $\begingroup$ @SRobertJames I can try to make things precise, but it will take me a bit of time to write it up. I'll post my answer tomorrow if I get the chance. $\endgroup$ Aug 18 at 6:13
  • $\begingroup$ Don't we need to know the weight of the car? That would give us the speed range that w/o steering you neither fly off the track nor crash into ground. But not very interesting as that's probably like 12mph. At a 90; no (provided you're going fast enough to remain in orbit), +1. $\endgroup$
    – Mazura
    Aug 19 at 18:41

It depends on what you mean by the verb "turning."

No, you don't really need to steer your vehicle to take that turn. You don't even need wheels for that. When thrown at the correct speed and angle, your car will slide through the turn without any problem. As you said, banking will provide all the forces for the turn. But it won't be a pleasant experience for the driver and the passengers, though.

We do not bank the roads to steer the car; we do it to ensure that we are in a good position to do so.

When we say "turning," we also silently assume that the car will align its direction with its trajectory automatically. This is not something to be taken for granted, and many factors are essential for this process to happen in a controlled fashion. We need some way to change the orientation of our vehicle, as well as its direction of motion.

Banking road geometry

Imagine a car on top of an imaginary 2-dimensional rectangular sheet of plastic. As long as you don't stretch or compress the sheet, we don't change the geometry. Thus, we can call two surfaces geometrically similar if one can be formed from the other without stretching or compressing. The steering of the wheels depends on the road's geometry. Even if we bend the road as shown, the person driving the car won't notice a difference as long as we do not change the local geometry. (Read more about intrinsic and extrinsic curvature)

You can now see that a 90-degree banked road is the same as a straight road for the car geometrically. You don't need to turn the car here.

Ideally, if all our roads were constructed like this, there wouldn't be any need for turning. But given our primitive public transport system and city planning, it is what it is for now. Coming back to reality, a 45-degree banked road is not the same as a rectangular straight road.

You will need to stretch some part of the plastic sheet in order to form this. If you make this banked road flat, that geometrically similar surface will look like this.

Here, you will need to turn your wheels like you would on a normal turn. Now how would turning the wheels make the car change its orientation?

Steering geometry

Steering is the process through which we change the direction of motion as well as the orientation of our vehicle. We make use of some properties of wheels to achieve this.

Wheels like to move precisely in the direction they are pointed towards. (This is a bit more complicated because real tires are tori, and the direction would depend on the point of contact.)

Now if you have a bike, we have two wheels trying to move like that. When you are moving straight, there is no problem because both wheels are moving on the same path. But once you turn the front wheel right, you have a situation where the front wheel wants to go right, but the back wheel wants to go straight. The only way this can be achieved is though the change of orientation of the bike. The bike moves in a circle so that both wheels are tangential to the circle.

This is how steering works. In addition to changing the direction of motion, notice how the vehicle's orientation is also changed.

Now consider a pair of this arrangement, as in a four-wheeler. For the same turning angle, you will immediately see that the left and right wheel pairs try to move in a separate circle. In order for this to happen, the chassis needs to break, or the wheels need to slip.

The only way to solve this problem is to make both wheels turn differently. If you are wondering, yes, the front wheels of your car are not parallel during turning. During a turn, the outer wheel turns less.

This is a well-known problem in mechanical engineering and was solved by a German carriage builder called Georg Lankensperger in 1816. You can read more about this here, for a start.

Image credits: Wikipedia

TL;DR: Your car is always moving along the arc of a circle formed by the point of contact of its four wheels.

If wheels do all the turning, why bank the road?

When you take a turn, your car is moving in a non-inertial frame of reference. The centrifugal force associated with this motion tries to throw the car away from the road. Usually, on flat roads, the friction between the tires and the road counteracts this centrifugal force. This friction has a limit to it, decided by the weight of the car and the coefficient of friction between tires and the road. But the centrifugal force can be arbitrarily high, depending on the curve's radius and the vehicle's speed. So there is a possibility that the centrifugal force overcomes a friction threshold, and your car flings out of the curve. This is why we have speed limits on curves.

Banking the road provides an another force in addition to the friction to counteract the centrifugal force: the normal force. On flat roads, normal forces are acting perpendicular to the centrifugal force, making it effectively useless. But on a banked road, the normal force is tilted, giving us a helpful component against the centrifugal force.

This increases the threshold for centrifugal force to overcome. Hence we have larger speed limits on banked roads. Also, as pointed out by @Lawnmower Man in the comments, banked roads are much more comfortable for the passengers in the car. Humans inside the car are like inverted pendulums with those heavy heads. Banking helps reduce the torque acting on them, lowering their vulnerability to tip over. Same reason it is wise to crouch your heads on a merry-go-round (unlike these idiots below).

Banking the road ensures that we are in control. Banking the road ensures that all our geometry works properly.

I hope that this now answers your question. Have a nice day!

  • $\begingroup$ Great answer! I think the end would be improved by noting that the purpose of banking is to convert the centrifugal force into a surface normal force, which improves traction considerably by reducing/removing the lateral acceleration and converting it to "down force" relative to the car. This is also more comfortable for the passengers. $\endgroup$ Aug 20 at 6:40
  • $\begingroup$ @LawnmowerMan, yeah I got lazy there, should be better now. $\endgroup$
    – AlphaLife
    Aug 20 at 8:27

One reason you want to turn the wheels is to avoid introducing kinetic friction. You want the contact point between the tire and the road to be instantaneously at rest. Imagine you do not turn the wheel, and you are moving into a circular banked path towards the north direction and clockwise. Also imagine there is no friction in the road. After a quarter turn the CM of the car will be moving east, but the entire car will still be facing north, and at the contact point the wheels will be moving north/east (north due to the internal rotation and east because the entire translation of the car). If instead of ice you had pavement, then you would definitely be experiencing a lot of kinetic friction in a direction you dont want.


If the bank is part of a cones surface, it is flat metrically, meaning it can be cut radially and rolled out flat on a plane.

With all four wheels parallel, the car makes a parallel flat metric parallel transport on a straight line, touching the circle of movement on the cone at constant hight and leaving forward and backward over the outer rim.

Other profiles have other metrics with other geodesics of parallel transport, but a circle at constant height as a geodesic means by definition, that the two parallel circles of the inner and outer wheels have to have the same lengths up to a quadratic correction in their radial distance.

That happens for a vertical flat cylinder with a quadratic varying radius at its extremal radius, eg a barrel or a hyperboloid of revolution.


You ask

What additional force is needed from turned wheels?

Turned wheels don't create a force. If my car is parked and I turn the steering wheel, thus changing the angle of the front tires, that doesn't cause my car to start moving.

Likewise the car can keep going straight even if I turn the front wheels. Sometimes this is referred to as understeer, and I believe it's related to having the forces on the racetrack through the tires biased too much toward the back. You cure this by adjusting the forces towards the front wheels, for example by changing the angle of the front wings to apply more downforce to the front.

To be more serious, you are probably thinking about a situation where we are not close to the limit where the car starts to skid. The reason why the car doesn't skid is because that's how friction works. The lateral force at the point of contact between the tire and the tarmac is whatever is needed to keep the contact patch stationary. If the road is flat, the front tires are pointed straight and the car's alignment is perfect, then no lateral force is necessary. If the front wheels are turned, then they tend to make the car go in a circle, and that requires a lateral force at the point of contact between tires and the road. A lateral force is also applied to the rear wheels even though you don't change their angle relative to the car chassis. We know this because, back to my racecar example, if you overcorrect the understeer you end up with oversteer, where the rear wheels begin skidding first and the car spins but its center of mass doesn't actually turn, or turns less than is desired.

Now back to the situation where we are not close to skidding. When you are driving on a flat surface and turn the steering wheel, the car starts to travel in a circle. This means the inside tires are traveling a shorter distance than the outside tires.

If you don't turn the steering wheel, then the tires on both sides have to travel the same distance, otherwise at least two of the tires will be skidding. And we have assumed that nothing is skidding.

Therefore, you have to turn the steering wheel if the tires on one side of the car need to travel farther than the tires on the other.

And the answer to whether you have to turn the steering wheel on a banked turn is the same as the answer to whether the outside tires have to travel a farther distance than the inside tires.

If the track is banked at 90 degrees and your car has enough downforce that it can stay attached to the track, then you don't need to turn the wheels because the normal force provides all of the acceleration.


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