timing planes with sound

If you’re standing outside and a plane passes overhead, it sounds like it’s coming from a bit further back than it looks like it’s coming from. This is because light travels practically instantaneously (c = 299,792,458 m/s = 670,616,629 mph, so you’re 0.00004 light-seconds from a plane 12000m in the air) while sound takes a while (vs = 340 m/s = 761 mph, so you’re a whopping 35 sound-seconds from the same plane).

You can use this to estimate the speed of a plane. In particular if you know

  • \(v_s\) — the speed of sound, and
  • \(\theta\) — the angle between the visual and acoustic images of the plane,

you can find the speed of the plane: it’s just \(v_s\theta\)! (Assuming you measure \(\theta\) in radians.) I should really make a sketch, but here’s the idea:

Suppose it takes time \(T\) for the sound from the plane to reach the ground. Then, calling the height of the plane \(h\), we have:

\(v_s T = h\).

In this time, the plane moves an angle \(\theta\), as seen from the ground. This angle corresponds to an actual distance of \(h \theta\) (if we measure \(\theta\) in radians). So:

\(v_p T = h\theta\).

If we divide these two formulas, woah, \(T\) and \(h\) cancel out! All you’re left with is

\(v_p/v_s = \theta\),


\(v_p = v_s\theta\).

Nifty. There’s probably a neat practical “rule-of-thumb” version of this people could use (with their thumbs?) to quickly time planes without having to use radians or anything like that. I am, as of yet, too lazy to come up with it.

[Note: Speed of sound varies with altitude, going down to about 295 m/s at cruising altitude.]