Cars overtaking

While cycling during one of the unseasonably sunny October days, I passed lots of cars on the opposite side of the road, but none seemed to be overtaking me. I wondered: how many more I should expect on one side compared to the other?

Making some unrealistic assumptions: I ride my bicycle at a constant speed of b mph in traffic that is constant, with evenly spaced cars (a poor assumption) travelling at a constant speed c mph (not slowing when they overtake—a realistic assumption, in my experience), the answer is easily calculated.

The unslowing cars approach from behind at a merciless relative speed of c-b mph, and those safely across the central double line zoom past at c+b mph. So, for every car I pass on the opposite side of the road, I would expect to be overtaken \frac{c-b}{c+b} times. 

Diagram of cars overtaking a bike

A pictorial calculation: cars in the green block start ahead of me; those in the red and yellow have gone by after one hour.

For example, if I cycle at 10mph in traffic of 30mph, I should be expected to be overtaken by half as many cars as zoom by in the other direction. (Quick sanity check: the formula also works if I am cycling at the same speed or faster than the cars; or if we change the units.)

Of course, that was all too easy: another implicit assumption I made, that was reasonable, is that we’re all travelling at velocities nowhere near the speed of light.  Otherwise, pointlessly taking special relativity into account, relative to a fixed observer standing on the side of the road, if I travel at u \, \mathrm{ms}^{-1} , and the cars are travel at \pm v \, \mathrm{ms}^{-1} , then, relative to me, they will be approaching from behind at \frac{v-u}{1-\frac{uv}{c^2}} and towards me from the front at \frac{v+u}{1+\frac{uv}{c^2}} , where c is now the speed of light (about 3 \times 10^{8} \mathrm{ms}^{-1}) giving the new ratio:

\frac{v-u}{v+u} \frac{c^2 +uv}{c^2 - uv} .

As a check, suppose instead of cars overtaking, we think of light beams (ie. v=c ). Then the two sides of the fraction will cancel to 1, no matter what u is: the speed of light is constant for any velocity |u| < c , so I’ll still have equal amounts passing in either direction (though red and blue-shifted, I believe).

For instance, if  u=\frac{v}{3}=\frac{c}{6}, then instead of 1:2 the ratio of overtaking to oncoming cars (or now, possibly, red to blue cars, if I keep looking forward) will narrow slightly to 13:22. Not a massive change there then.


Leave a comment

Filed under Accessible, Maths in Life

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s