Is it possible to estimate the speed of a passing vehicle using a musical ear and the doppler effect?

Let us consider that you are at rest and the car, which emits at frequency $f_0$, approaches you with speed $v$. The frequency you receive increases to $$f_1=f_0\frac{c}{c-v},$$ where $c$ is the speed of sound. When the car get passed you the perceived frequency is reduced to $$f_2=f_0\frac{c}{c+v}.$$ The ratio is $$\frac{f_1}{f_2}=\frac{c+v}{c-v}.$$ Now solve this equation for $v$, $$v=\frac{r-1}{r+1}c,$$ where $r=f_1/f_2$.

Edit

Let us consider some examples. If the ratio corresponds to an octave (2:1), $r=2$, the car speed is $c/3\approx400\, km/h$, and that should be a Bugatti Veyron. If you notice a fifth (3:2), $r=3/2$ and $v\approx 240\, km/h$, which may be a nice sport car. A minor third (6:5), $r=6/5$, corresponds to $v\approx 110\, km/h$ which can even be a bus. For a difference in frequency corresponding to a semitone, $r\approx 1.06$, the speed is about $36\, km/h$ and for a tone, $r\approx 1.12$, the result is $v\approx 70\, km/h$. In all examples the speed of sound was taken as $c\approx 1240\, km/h$.

You can do such estimations! It even turns out that you don't need perfect pitch. As a sanity check, picture a train going past you, blowing its horn. Its horn consists of three notes forming a chord (incidentally, the chord was chosen because it was annoying). Now let the train go by you. It's still the same chord, just with a lower root. If the doppler equations you are looking for depended on perfect pitch, it would mean the effect of the changing velocity would affect different pitches differently. Because you observe the same chord as it passes, just shifting downward as a whole, that tells you that the equations you are looking for are frequency independent. It's just the interval that matters!

The equation for doppler shift is

$$f=f_0\frac{c+v_r}{c+v_s}$$

Where $f_0$ is the emitted frequency, $c$ is the speed of the wave (aka speed of sound), $v_r$ is the velocity of the receiver and $v_s$ is the velocity of the source.

Now if you're moving with respect to the air, you'd need to know your speed. Perhaps you could determine this by listening to an object on the ground (like listening to the ding of the railway crossing bells as you go past them). But for simplicity, let's assume we're holding still. $v_r=0$. We can rearrange a bit to get:

$$v_s = c\left(\frac{f_0}{f}-1\right)$$

Now we can also look at the pitch in the opposite direction, which will be $f+\Delta$, where $\Delta$ is the difference in pitch between when its coming to you and away from you. $$ -v_s = c\left(\frac{f_0}{f+\Delta}-1\right)$$

We can combine these equations to get:

$$ \frac{f_0}{f}-1 = -\left(\frac{f_0}{f+\Delta}-1\right)$$ $$ \frac{f_0}{f} = -\frac{f_0}{f+\Delta}+2$$ $$ f_0f+f_0\Delta=-f_0f+2f(f+\Delta)$$ $$ f_0 = f\left(1+\frac{1}{2\frac{f}{\Delta}+1}\right)$$

Substituting in to the earlier equation:

$$ v_s=c\left(1+\frac{1}{2\frac{f}{\Delta}+1}\right)$$

Now the neat thing about this is, if I did my math correctly, the equations for velocity only depend on $\frac{f}{\Delta}$, which is information you'd pick up from the interval alone. No perfect pitch required!

While not explicitly an answer, there is a nice historical connection to your question. In effect, the first public experiment that decisively illustrated the Doppler effect was almost exactly what you're describing.

In 1845, Christrophe Ballot placed one group of trumpet players on a moving train and another group at a station. Having tuned everyone up beforehand, he had both groups play and hold the same note as the train passed the station and observed the effects. Nothing like the classiness of using a group of musicians in a scientific experiment!

Further reading:

• http://www.weirdexperiments.com/steamtrain.htm (this article includes the tidbit that Doppler calculated a speed difference of $68ft/s$ ($70 km/h$) for one semitone difference in pitch)
• http://io9.gizmodo.com/5974423/the-first-proof-of-the-doppler-effect