When a bus goes around a corner, does the person sitting at the back travel further distance than the person sitting at the front?

Question

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.

Answer 1

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".

Answer 2

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...

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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.

Answer 3

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.

Answer 4

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.