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I asked this in the physics stack but hopefully will get more interest here. So I understand that when driving on a curved path, the inner wheel must travel a shorter distance than the outer wheel. But why?

As the car begins to turn, both wheels want to continue rolling at the current speed in linear motion due to conservation of angular momentum. If the wheels are being driven by the engine, then the torque is equal to rolling resistance on both wheels so both maintain a constant speed. If the car is coasting, then both wheels are braked by rolling resistance by equal amounts. In order to change the speed of the inner and outer wheels, there must be additional torque(s), which must be different for the two wheels. What is this torque, and where does it come from? Assume that both wheels are rolling without slipping.

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    $\begingroup$ This is a classical mechanics problem. It is absolutely an engineering and physics question because this is about dynamics. $\endgroup$
    – Reese
    Jan 9 at 9:32
  • $\begingroup$ If so, why was it not answered on the Physics Stack - they definitely understand all the concepts... $\endgroup$
    – Solar Mike
    Jan 9 at 9:42
  • $\begingroup$ I only asked recently, there may be an answer in the next few days, or it may simply not garner any interest. Maybe some people think the question is beneath them and not worth answering. However, it is absolutely an engineering question, and a branch of engineering that falls under classical mechanics. The open differential is, after all, a fundamental feature of the vast majority of automobiles. $\endgroup$
    – Reese
    Jan 9 at 9:45
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    $\begingroup$ "What is this [differential] torque [causing angular acceleration of wheels], and where does it come from?" -- same source as if you put the car in neurtal, wheels straight, and push on the car from outside. From the contract point on the ground. Exactly per NMech's answer. $\endgroup$
    – Pete W
    Jan 9 at 15:49
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    $\begingroup$ This is one of those thought experiments that might need some practical. Consider getting a toy car with ackerman steering and test your theories. Often it helps to get hands-on. $\endgroup$
    – Criggie
    Jan 10 at 21:56

6 Answers 6

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In order to change the speed of the inner and outer wheels, there must be additional torque(s), which must be different for the two wheels. What is this torque, and where does it come from?

If the car is turning, there must be a torque relative to the center of mass of the car (note that torque is always relative to some axis; the torque on the car as a whole is different from the torque on the wheels). Suppose the car is turning to the left. Then the rotation involves the left front wheel moving backwards relative to the COM of the car, and the right front wheel moving forward. This then means there are torques relative to the axles of the wheels (there's one shared axle, but the differential means that to some extent it acts like two axles).

Angling front tires -> sideways force on front wheels -> torque relative to car COM -> rotation of car -> lateral movement of wheels -> torque relative to axles.

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  • $\begingroup$ Thank you, this explains everything clearly, elegantly, and above all, correctly. Everything checks out. Thank you again! $\endgroup$
    – Reese
    Jan 10 at 19:25
  • $\begingroup$ @Acccumulation I am still trying to understand what this question was about, maybe since you got it answered you can help. I am fine with all the steps up to "Angling front tires -> sideways force on front wheels -> torque relative to car COM -> rotation of car -> lateral movement of wheels ", however I am having trouble visualising the last one i.e. lateral movement of wheels -> torque relative to axles. (cont'd) $\endgroup$
    – NMech
    Jan 11 at 6:58
  • $\begingroup$ @acccumulation For simplicity in the ideal case RR=0, the lateral movement of the wheels would create forces that are along the axis of the axle, so the moment with respect to the axle that those forces create would be perpendicular to the axle (thus causing bending, instead of acceleration/deceleration). Could you elaborate on that? Maybe I haven't understood correctly the lateral movement? or is it something more fundamental than that? $\endgroup$
    – NMech
    Jan 11 at 6:58
  • $\begingroup$ @NMech I'm not clear on what you mean. The forces are going to be at the point of contact between the road and the tire, so being perpendicular would mean they're vertical. There are vertical forces (normal force opposing the weight of the car), but those aren't special to turning. When the inner wheel moves backward, that opposes the rotation of the wheel, and so is a "negative" torque. $\endgroup$ Jan 11 at 23:27
  • $\begingroup$ @Acccumulation I think now I understand what you were writing on what is causing the acceleration. You are saying that the turning moment on the car, is making the wheels pivot and that relative motion "activates" the friction force(which I consider as fraction). $\endgroup$
    – NMech
    Jan 12 at 7:25
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TL;DR: The kinematic constraint imposes the redistribution of forces on the four wheels. This redistribution is permitted due to the car differential.

IMHO it is easier understand this if you first understand the difference between kinematics and kinetics.

  • kinematics: are about the geometry of motion (kinematics don't take into account forces or work).
  • kinetics: this is how forces and moments affect the motion --- and vice versa (i.e. if you know the motion determine the forces).

small interlude/example

It is very common that first the kinematics is considered and then the kinetics are applied. Consider the following example of a hammer athlete:

enter image description here

Assuming:

  • the person and hammer is rotating at a constant angular speed $\omega$ and
  • the distance between the axis of rotation and the hammer is R
  • the string has no weight

then through the kinematics I know that the angular velocity is $\omega$, and the angular acceleration is $0$. And I can calculate at any time the angular position if I know the angular position at t=0.

Given the above information I can find out the force that the athlete has to apply on the string that is required for this motion to be accomplished. This is done through the use of kinematics. I.e. knowing the motion I can calculate the forces requires to produce this motion.

My point here is that:

the motion constraints can impose/determine the required forces.


Car example

In the example you are providing, (the car during a turn without slipping), kinematics play the primary role. I.e. the rear inner wheel needs to travel a smaller distance over the same time compared to the outer wheel.

enter image description here

if the symbols in the above image are used, then for a turn of angle $\theta$ the distance travelled is:

  • rear inner: $=(R-\frac{w}{2})\theta$
  • rear outer: $=(R+\frac{w}{2})\theta$

(When there is no sliding) Kinematics also determines the velocity and acceleration for each of the wheels.

After that point, since the position, velocity and acceleration for each wheel are known, then you can use the kinetics (inverse) to see what forces are required to produce that motion.

In this instance, what happens (even if you consider a rigid chassis and infinite stiffness car suspension), what happens is that the turning redistributes the vertical forces on each wheel. That changes the magnitude of the friction on each wheel (and also affects the rolling resistance although rolling resistance is not as important). However the magic is in the differential.

enter image description here

The differential allows (at the same time) each wheel to spin at its own velocity and also allows them to "communicate" (for lack of a better word) and distribute the power to each wheel.

In the particular case of the open differential, when the car is jacked up, and the engine shaft is not rotating, then rotating one wheel has the effect of rotating the other wheel in the opposite way. So, the angular momentum and is preserved.

idling car during turn

I will focus on this case.

So, when an idling car is turning, each wheel is free to spin at its own rate, and at the same time if the outer is spinning faster, the open differential makes the inner wheel spin faster (see the video in this answer). At this point it is important to remind that the torque that slows down the wheels when the car is coasting is from the rolling resistance on each wheel (which is the longitudinal component of the friction). To simplify matters, we'll assume that the rolling resistance is zero (perfectly round wheel) and all the friction force between the wheel and the road is available as traction force.

The traction force

  • in the linear acceleration/decel is parallel to the acceleration.
  • In the case of uniform circular motion, the traction acts as the centripetal force (pulling the car inwards).

When the car is coasting, the force of the rolling resistance is usually small compared to the radial component (centripetal force). The moment caused by wheel traction (friction forces) in the radial direction (Centripetal) is what causes the rotation moment that turns the car. (See image below)

enter image description here

Also it is important to note that the contact of a wheel is not a single point but a contact patch that has a complex behavior.

enter image description here enter image description here
Forces on wheel system contact patch

car accelerating through the turn

When the car is using the engine power, then there is an additional factor that affects the kinetics (dynamics). I won't delve into this case because there are a number of factors that will also affect this outcome (namely FWD, RWD, 4WD etc). However bottom line is that there is added torque from the engine. In that case again the differential allows spliting of the torque (and redistribution of power) so that the kinematic constraints are satisfied.

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  • $\begingroup$ I thank you for the effort you put into this answer. There are still some unanswered questions, though. Firstly, you said that the diff allows the wheels to distribute the torque requirements, but all sources are unanimous that a differential always splits the torque evenly between the wheels, and only power can differ, due to different rotating speeds. Are you referring to the resistance torque? Do you mean that, in the case of a left turn, the right rear wheel experiences more friction, and this force opposes the forward rotation of the left wheel via the rotating pinion in the diff? $\endgroup$
    – Reese
    Jan 9 at 11:20
  • $\begingroup$ I was only wondering about the relevance of vertical forces. My train of thought is: In the diff, torque is delivered equally to both wheels and the reaction torque is the same from both axles in a straight line. In a left turn, the right wheel must accelerate forward and the left wheel accelerates in reverse. Acceleration in either direction requires a net force. The diff delivers the same torque (twisting force) to both axles in the same direction, yet the two axles accelerate in different directions. The only explanation for a change in the net force that the reaction force changes. $\endgroup$
    – Reese
    Jan 9 at 13:05
  • $\begingroup$ @reese I've modified my answer (although I think it will still require some more). I specifically decided to focus on the coasting case, so that there is no tangential acceleration/deceleration. The reason I did that was your comment that "the right wheel must accelerate forward and the left wheel accelerates in reverse" (somehow it implied that the engine offers this torque. $\endgroup$
    – NMech
    Jan 9 at 14:09
  • $\begingroup$ During coasting there is no additional torque from the the engine. The net sum is zero. When in straight line, the small rolling resistance forces generate opposite torque on each axle which balancce out. However when turning, the rolling resistance changes between inner and outer wheels. $\endgroup$
    – NMech
    Jan 9 at 14:11
  • $\begingroup$ I see. What's the cause of the change in the rolling resistance? $\endgroup$
    – Reese
    Jan 10 at 12:39
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This is a simple explanation/observation. A two-wheel dolly rotates along a circular path by a constant torque (T). The position of the wheels changed from point 1 to point 2 in one second (t = 1s), what you can say about the distance that each wheel has traveled, and what is the speed of each wheel at the turn?

enter image description here

ADD: I think it is imperative to understand what the "differential" in a car does. Here is an explanation I found that is easily understood by a non-automotive engineer or mechanical engineer:

What does a differential do inside car?

"The main purpose of this differential component is of 2-folds:

- Changes the direction of torque in a car that is heading to the drive wheels of the car

- Allows for the drive wheels in a car to turn at various speeds at any time

Normally, all of the torque of an engine usually goes into just one transaxle or transmission, and it sure needs to be directed at some point to the two wheels of the car. Now, the side gears of a differential are the ones that will mesh with the larger pinion gears to actually perform this function.

The spider gears are the ones that allow for one wheel of a car to turn faster than the rest, thereby preventing wheel scuff and binding whenever you are taking a corner or bend with your car." https://naijauto.com/car-maintenance/differential-in-car-3018#:~:text=Latest%20news%20articles%201%20What%20is%20a%20Differential%3F,...%203%20Simple%20signs%20of%20a%20bad%20differential

Another quote:

"Differential" - "As part of the front and/or rear axle assembly, the differential plays an integral role in how your car makes turns. The differential is designed to drive a pair of wheels while allowing them to rotate at different speeds. This function provides proportional RPMs between the left and right wheels. If the inside tire rotates 15 RPM less in a turn than going straight, then, the outside tire will rotate 15 RPM more than going straight." https://www.sunautoservice.com/what-is-a-differential-on-a-car/

Hope thess helps.

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  • $\begingroup$ Although I don't appreciate that you simply repeated the premise of my question, thanks for bringing up the turning torque, which was also brought up by kamran and accumulation. This would mean that the turning torque produced by the front wheels is transmitted to the rear wheels via the rigid body/chassis, and due to the different distances of the wheels from the axis of the torque, the tangential force on the outer wheel is larger than the inner wheel. This tangential force accelerates the outer wheel, and is therefore the source of the different wheel speeds. Is that correct? $\endgroup$
    – Reese
    Jan 10 at 13:21
  • $\begingroup$ Sorry, I'm not an automotive engineer, couldn't comment on your new question. $\endgroup$
    – r13
    Jan 10 at 15:46
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Let's imagine a car turns to the left 360 degrees, a complete clockwise circle.

If we look at the tracks left, we see two concentric circles (For the front wheels, and another pair for the rear wheels, as shown on the diagram ), inner wheels turn in the inner circle which is smaller and outer wheels turn in the outer and larger circle. As per the diagram.

  • Because the outer track's radius is larger its circumference is larger than the circumference of the inner circle.

  • This means both the front wheel and rear wheel on the outside circle, the right hand, have to cover a larger distance.

  • The front wheels are free, so the right front wheel turns more than the left front wheel.

  • The differential just unlocks the rear wheels, allowing the right rear wheel to turn more than the left wheel, or else the rear wheels would skid and could perhaps break the rear axel.

  • The differential does not impart any different force on the wheels, it just allows them to divide the transmission rotation, differently.

Edit

After OP.s comment.

Let's say the differential is removed for maintenance and let the rear wheel turn freely. If tow the car into a turn we see the outer wheels turn faster. the tighter the turn the greater the difference of the speed.

The differential just allows this. The steering pulls the front wheels to left with a force F to left. (in our example) $$F* D_{D=wheelbase}=Torque$$

This torque causes the rear wheels to turn at different speeds. Let's imagine the steering could turn left all the way to 90 degrees. Then the right wheel would accelerate forward and the left one backward.

.

differential

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  • $\begingroup$ How do the wheels divide the transmission rotation differently via the differential? $\endgroup$
    – Reese
    Jan 9 at 11:13
  • $\begingroup$ Ah, I think I understand. You mean to say that the rigid body (or chassis, at least) of the car itself is the medium through which the steering torque arrives at the wheels. The torque is generated from the front wheels due to the centripetal force that arises when they are directed to one side. If the car is towed, then the steering torque is delivered from the towing vehicle via the tension of the towing cable. Is this correct? $\endgroup$
    – Reese
    Jan 10 at 13:04
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This video went a long way to help me understand how differentials work and how resistance on one wheel translates to motion on the other.

Around The Corner - How Differential Steering Works (1937)

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  • $\begingroup$ Thank you but the concept in the video is not relevant to the question. In the scenario shown in the film, increased resistance on one wheel (grasping the wheel with your hand) causes the other wheel to speed up. But rolling resistance does not change when the car is making a turn, and not only that, but if it were to increase, it would increase in the outer wheel. In that case, the outer wheel would slow down and the inner wheel would speed up, which is clearly not what happens in reality. $\endgroup$
    – Reese
    Jan 10 at 15:02
  • $\begingroup$ This is the answer I was going to enter but was too slow. @Reese, why do you focus on rolling resistance? In your scenario wheels are controlled by position on road, steering wheel and engine. They do not slip you said. So wheels are forced by steering wheel to more through a certain trajectory on road. These trajectories have different length for the two wheels. Without slipping they will roll through these trajectories apparently for the same time but with the outer one with a higher rpm to compensate for the distance. Open differential allows for this speed difference. That's all. $\endgroup$ Jan 10 at 18:27
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Sorry for my bad English.

I believe that the confusion originates from a misunderstanding of the nature of rolling resistance. In the question the following is stated:

If the wheels are being driven by the engine, then the torque is equal to rolling resistance on both wheels so both maintain a constant.

That can only be stated if we assume the wheels are weightless, which in turn make the system undetermined meaning that from the known physical equations we cannot figure out the speed of the wheels, so additional assumptions like no slipping should be introduces. In such a case an explanation for the different rotation speed goes like this: since that wheels cannot slip and the outer radius is larger than the inner one we conclude that the outer wheel is rotating faster.

That is totally unsatisfactory if we want to understand the nature of rotation but allows us to ignore some aspects of the rolling resistance.

If we do not assume the weightlessness of the wheels there is no equilibrium between rolling resistance and the engine torque, which in the end results in different rotation speeds of outer and inner wheels.

The question of why rolling resistance of outer and inner wheels differ is far too complex to describe in depth but i hope that the following line of thought, although somewhat incorrect, would lead to understanding: when car starts to turn the outer wheel starts to cover more distance meaning more micro collisions with the road particles with more speed so the rolling resistance is more for the outer wheel.

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  • $\begingroup$ Yes, the assumption is no slip. Your answer is interesting, but I've read that rolling resistance itself is not dependent on speed, so I am unsure if this explanation makes sense. In fact, if there was no diff, like a horse-drawn carriage with independent L/R wheels, the larger resistance on the outer wheel would make it slow down in a turn! However, I really appreciate that you understood the point of the question, which is to ask why the wheels turn at different speeds, not simply repeating the obvious observation that the radii of the turning circles of the L/R wheels are different. $\endgroup$
    – Reese
    Jan 10 at 12:51

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