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This is a bit of childish question. When a bus goes around a corner, does the person sitting at the back travel further distance than the person sitting at the front? My thought is no because the bus is connected and every point moves along the same trajectory. My friend thinks yes because the trajectory of two endpoints can be visualized as traveling along two different arcs of a concentric circle. My problem with his argument is that the center is not static, nor is at the center.

I want to know the true answer though.

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    $\begingroup$ It's not a childish question! It's surprisingly complicated because calculating the trajectories of the front and back of a vehicle is not trivial. $\endgroup$ Mar 4 at 6:56
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    $\begingroup$ Not in the question, but it's also worth noting that the left and right sides of the vehicle travel different distances. Which is greater depends on the direction of the turn. (The outside edge must travel farther than the inside, to be clear). $\endgroup$ Mar 4 at 20:37
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    $\begingroup$ Is drifting allowed? Because I think it makes a big difference. $\endgroup$ Mar 4 at 21:00
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    $\begingroup$ Clarify please - are the two people sitting over the axles/ground contact points or are they equallly distant from the axle, or some combination? Busses like this exist: img.freepik.com/premium-vector/… $\endgroup$
    – Criggie
    Mar 4 at 21:01
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    $\begingroup$ The extreme case of trajectory was Jackie Chan in "Police Story", hanging off the outside rear top of a Hong Kong double-decker bus by an umbrella [youtube.com/watch?v=nFo_NjgwJbQ] $\endgroup$
    – smci
    Mar 6 at 0:45

5 Answers 5

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The center of the front axle must travel along a larger arc than the center of the rear axle or any point between those points in order for the vehicle to avoid whatever obstacle the vehicle is navigating around with the turn (i.e. the curb, the edge of the lane, a lamp post...). This is because the fixed wheels of the rear axle constrain the rear axle to move towards the front axle (as long as they are rolling without slipping), while the front axle is free to move in any direction the steerable front wheels are pointed. Being in front, for whatever arc the front axle is following, the front axle is that much farther along that arc than the rear axle, so the rear axle, always moving towards the front axle, will take the "inside track".

If you're a driver, you've experienced this even in a small personal vehicle when easing forwards out of very tight parking spaces or parallel parking: you have to wait until the rear of the wheel base has passed the obstacle before initiating your turn, otherwise you'll run the side of your vehicle into the obstacle.

Points forward of the front axle travel farther than the front axle, and points rearward of the rear axle travel farther than the rear axle. There's no limit to how much farther: if you had a mile long weightless extension on the back of your bus with a weightless passenger sitting on the end of it, she would traverse an arc of about $\pi/2$ miles when the bus took a 90 degree turn. For real vehicles, the farthest-turning point is almost always the front bumper.

Adding articulation reduces the difference in the size of the arcs of the front-most and rear-most axles by adding intermediate axles which all chase the axle directly in front, instead of the rearmost axle chasing the frontmost axle.

The rear axle(s) can travel wider arcs than the front axle(s) if the rear wheels are not constrained to move towards the front axle. This is "spinning out" or "fishtailing".

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    $\begingroup$ I love your explanation of additional articulations "chasing" their axles directly in front - I wonder if it will help me back up a car with a trailer attached. =) $\endgroup$ Mar 4 at 23:18
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    $\begingroup$ With a bicycle, you can actually see the traces of the tires after riding through a puddle. Indeed, the back wheel follows a smaller circle than the front wheel. That is most obvious in the extreme case when the front wheel is turned 90 degrees! $\endgroup$ Mar 5 at 16:23
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    $\begingroup$ There are many school buses in the Unites States in which the seats at the rear are far behind the rear axle. See this image: th.bing.com/th/id/… Is it a probably correct to say that for such a bus, the rearmost seats do travel farther than the other seats? $\endgroup$
    – Eddified
    Mar 5 at 23:13
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    $\begingroup$ "you have to wait until the rear of the wheel base has passed the obstacle before initiating your turn". My car has a nice gash to remind me of this, in case I forget a second time. $\endgroup$
    – pipe
    Mar 6 at 6:20
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    $\begingroup$ Here's a diagram that explains your words: tse2.mm.bing.net/… $\endgroup$ Mar 7 at 12:06
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The opposite is true, the person with the longest track is always the driver.

The point is that all busses that I know of have the front wheels steered. The remaining axle(s) simply follow the front axle. And because those following axles are always moving in the direction where the steered axle currently is (location only, the angle of the front wheels is totally irrelevant), each trailing axle cuts the corner a bit more than the one before it.

That is actually one of the main points that bus/truck drivers need to learn. To go wide enough around corners that the following part of their vehicle won't flatten any lamp posts and/or pedestrians waiting at the traffic lights.

For a person in the back to go farther than the driver, the bus would need to have its rear axle before the midpoint of the bus. While possible in theory by placing the heavy engine right in the front, I don't know of any bus manufacturer who has been braindead enough to create such a design...


As an experiment, you can ride tight circles with your bike on some ground where your tires leave traces (I hope you have a bike): You will clearly see that the front wheel produces a trace that surrounds the trail of the rear wheel. In the extreme case of a 90° steering angle, your back wheel won't move at all while the front wheel circles around it.

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    $\begingroup$ What if there is a tour guide sitting next to the driver, equally close to the front screen, but in the opposite side? Then when the bus turns to one way, the driver has the longest track, but when the bus turns the other way, the tour guide has the longest track. $\endgroup$ Mar 4 at 16:30
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    $\begingroup$ @JeppeStigNielsen Yep, you found the hole in that statement. You have been awarded 1000 picky-perfection points. Your welcome ;-) $\endgroup$ Mar 4 at 18:58
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    $\begingroup$ As Criggie pointed out, there's another complication. Sometimes, there's a lot of distance behind the rear wheel and the back of the bus, meaning that the last passengers will swing out proportionally to the amount the front-wheels swing in (all relative to the rear wheels). $\endgroup$
    – jpaugh
    Mar 5 at 5:39
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    $\begingroup$ @jpaugh That's the point of the paragraph about the location of the rear axle: To have the back of the bus swing out further than the front, it must be farther away from the unsteered axle than the front of the bus (that's an argument of symmetry). And that means that the entire rear half of the bus must be in overhang. As I said, this is possible theoretically, but not done in practice for obvious reasons. $\endgroup$ Mar 5 at 10:12
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    $\begingroup$ @cmaster-reinstatemonica since picky points are on offer: *you're 🤓 $\endgroup$ Mar 5 at 22:12
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We can suppose that for intifinitisemaly short ammout of time $dt$ the bus rotates around one point. For this time we can expect every point in the bus to move along circles defined by the centre (defined by the rear axle, distance between axles and steering angle) and the point-centre distance. It is easy to see that points farther from the centre must travel longer distance to cover same angular motion.

Now let's simplify the bus in a line segment. Let's define three points there: $R$ as the rear axle, $S$ as the steering axle and $C$ as centre of the turn. Let's define vectors $\mathbf{r}$ and $\pmb{s}$ as vectors where the wheels are pointing and originating in their respective points $R$ and $S$.

It is easy to see several properties:

  • The centre $C$ lies on lines perpedicullar to vectors $\pmb{r}$ and $\pmb{s}$,
  • Points $R$, $C$ and $S$ form right triangle with right angle at $R$,
  • From Pythagorean theorem we can see $|SC|^2=|RC|^2+|SR|^2$.
  • The steering axle always travel around greater radius compared to the fixed axle.
  • We can also see the farther the axles are, the greater the difference is.

Let's quantify the difference in respect to the steering angle. If You draw the sketch, You'll find that the steering angle $\pmb{rs}$ is equal to the angle $\gamma$ at the corner $C$. Using the triangle notation for the sides and angles (Corners: $R$, $S$, $C$, angles: $\rho$, $\sigma$, $\gamma$, sides $r$, $s$, $c$) and using goniometric functions we san see:

$$\frac r s=\frac{1}{\cos\gamma},$$

where $\frac rs$ defines the ratio between the steering axle travel radius and fixed axle travel radius and $\gamma$ is the steering angle.

We can clearly see that the key factors for You (sitting by the driver) and Your friend (sitting in the rear) are the steernig angle $\gamma$ and your distances from the fixed axle $x_Y$ and $x_F$. Each of you have your own angle of attack $\xi_Y$ and $\xi_F$ and travel radius $r_Y$ and $r_F$.

For both of you we can define your angle of attack as $\tan\xi_n=\frac {x_n}s$ and compare it to the steering angle we can get:

$$\xi_n=\arctan{\frac{x_n\tan\gamma}{c}}$$

Using this in the travel ratio we got earlier:

$$\frac {r_n}s=\frac1{\cos\left({\arctan{\frac{x_n\tan\gamma}c}}\right)}.$$

To compare Your and Your friend's travel distance we just divide previous equations to get the ratio as follows:

$$\frac{r_F}{r_Y}=\frac{\cos\left({\arctan{\frac{x_Y\tan\gamma}c}}\right)}{\cos\left({\arctan{\frac{x_F\tan\gamma}c}}\right)}.$$

There are edge cases fo course, that are interesting:

  • For $\gamma=0$ the turning radius diverges to infinity, but the final equation return 1, meaning you are both travelling with the same velocity.
  • For $\gamma$ approaching right angle the equation is not defined. It is easy to guess the radius ratio end up eventually as (converges to): $$\lim_{\gamma\rightarrow\pi/2}\frac{r_F}{r_Y}=\frac{|x_F|}{|x_Y|}$$
  • Because $\cos x=\cos{-x}$, it doesn't matter if your friend sits behind the rear axle or in front of it, what really matters is his and Your distance from the rear axle.
  • Since all functions are "nice" (continuous, monotonous) for all values of $\gamma$ in the range from 0 to the right angle we can conclude that Your friend's radius is smaller then Yours when he is closer to the axle than you. $r_F\leq r_Y\iff x_F\leq x_Y$.

Last but not least we can translate radii into velocities and distances covered.

Since the bus behaves as a solid body (no defformations) we can expect that the angular velocity $\omega(t)$ is the same for all parts of the bus in every moment $(t)$. Then we can write: $$v_F(t)=\omega(t) r_F(t).$$

If we care for the distance we can write it as follows:

$$s_n=\int_0^T\left(\omega(t)r_n(t)\right)dt,$$ $$s_F-s_Y=\int_0^T\left(\omega(t)r_F(t)\right)dt-\int_0^T\left(\omega(t)r_Y(t)\right)dt=\int_0^T\left(\omega(t)\left(r_F(t)-r_Y(t)\right)\right)dt$$

  • It means we are adding up all little distances ($ds$) covered over little moments ($dt$).
  • If we consider the bus moves forward then the multiplication $\omega(t)r_n(t)>0$.
  • Then changing from left turn (say $\omega>0$) to right turn ($\omega<0$) switches the signs for radii too.
  • Then if we compare the sums with the respect to your positions in the bus, we can conclude that if your friend is closer to the fixed axle ($x_F\leq x_Y$) than you are he moves slower than you ($|v_F|\leq |v_Y|$) and as he is adding smaller or equal values every time.
  • For the difference you are then always adding zeros (when moving straight) or negative values there is no way you have no way to "compensate" for his loss.

So to judge your argument both claims of yours considering the bus has conventional dimensions, in other words the rearmost seat is closer to the fixed axle than the frontmost seat.

  1. We have proven that different points of the bus are travelling different trajecotries and distances, so your claim is false.
  2. Your friend's argument cannot be judged independently. If it follows my expectation of hte bus geometry. Then his claim is false too - You are travelling longer distance, but his explanation is true.
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There are several good answers already, so this is just a visualization. I, too, was at first tricked by the fact that the passengers are "affixed" to the bus, and so must "travel the same distance" as the bus. But then I thought of windshield wipers. If you turn them on, and put an odometer on the tip, what will it read compared to the pivot? Even though the wiper blades are connected to the vehicle, it should be pretty obvious that the tips travel further, by virtue of their additional motion.

Now, this is different from the passenger situation, where the passengers stay still w.r.t. the vehicle. But we can modify the scenario by rotating the wiper blades so that they are "sweeping the street". But also move them back so that the pivot is over the rear axle, and we see that the vehicle as a whole is very much like a wiper blade that can sweep side to side from a fixed rear pivot. This should make it fairly obvious why the passengers over a steering axle will always travel further, on average, than the passengers over a fixed axle.

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I believe so because it would be a larger turn for the person in the back then in the front because the turn would be larger in the back then in the front. Try it yourself: get a toy car and pencil and turn the car and draw the arc for the front of the car and the back of the car. The arc will be bigger in the back then the front, therefore being a longer distance travel. This distance though is at most a few feet and will not matter real world, but that example is static though so you could still be right, I'm just pitching in my thoughts.

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    $\begingroup$ You can maneuver a toy car in lots of ways that an actual car can't; and, you can maneuver an actual car in ways a bus can't, although to a lesser degree. I think that demonstration will prove whatever motion your hands want the toy car to make. $\endgroup$
    – jpaugh
    Mar 5 at 5:33
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    $\begingroup$ This needs some explanation - it runs counter to what I understand, and opposite to at least one other answer. $\endgroup$
    – MikeB
    Mar 6 at 10:28
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    $\begingroup$ @MikeB No, it does not need some explanation, it's just plain wrong. Sorry, Kellan. I've been doing this experiment over, and over again for years while commuting, and it's always the steered front wheels that move on the outside of the turn. It's best visible when you do the experiment with a bike or a kick scooter. $\endgroup$ Mar 6 at 10:45

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