Measuring the Distance to a Thunderstorm using your Ears

As a child I often heard the rule of thumb:

"When you want to know how far a storm is you should count the seconds between lightning and thunder and then divide by 3 to obtain the distance in kilometres."

My goal today is to check if this is really true.

The lightspeed is determined to be $c = 3 \cdot 10^{8}\ m/s$. The speed of sound in dry air at 20° C is $v_{s} = 343\ m/s\ = 0,343\ km/s$. As a result, the following equations hold where $\Delta t_{l}$ and $\Delta t_{s}$ are the respective times that the light needs to reach us: $$c = \frac{\Delta x}{\Delta t_{l}},\ v_{s} = \frac{\Delta x}{\Delta t_{s}}$$ As we do not have any idea of where the storm is, we can only measure the time $\Delta t_{m}$ from the occurrence of the lightning until the moment we hear the thunder. $$\Delta t_{m} = \Delta t_{s} - \Delta t_{l}$$ As an approximation, we assume that the time the lightning needs to reach us is much smaller than the time the sound needs to reach us. $$\Delta t_{m} \approx \Delta t_{s}$$ Then we can state that the storm is approximately a distance of $\Delta x$ away (expressed in kilometres: sorry Americans 🦅 but you can also easily derive a similar rule of thumb for your metric system). After approximating $0,343$ by $1/3$, the following estimate is obtained. $$\Delta x \approx v_{s}\Delta t_{m} = 0,343\ km/s\ \ \Delta t_{m} \approx \frac{\Delta t_{m}}{3}\ km\ $$ So far so good, it turns out that the old wisdom makes sense. If you want to be more correct then you can easily change the sound speed for different conditions (temperature, humidity, ...).

What if we increase the distance such that the light does need to travel further? Then the time between sound and light will be the following. $$\Delta t_{m} = \Delta x\left( \frac{1}{v_{s}} - \frac{1}{c} \right) = \frac{\Delta x}{v_{s}} - \frac{\Delta x}{c}$$ However note that we earlier choose $\Delta t_{m} \approx \Delta t_{s}$, but in this case for a larger distance the approximation error in time $\Delta t_{E}$ made due to neglecting the light leads will increase. $$\Delta t_{E} = \Delta t_{s} - \left( \Delta t_{s} - \Delta t_{l} \right) = \Delta t_{l} = \frac{\Delta x}{c}$$ When the storm is nearby you, the approximation error will be zero. However, I don't think you will need to use this trick to know that the storm is close to you as you already might have come to that conclusion ☔.

But if we use our approximation for when the storm is further away and you still want to know its distance, the result might be inexact. Without any further approximations of the speed of sound, the error in distance is the following: $$\Delta x_{E} = \Delta x - v_{s}\Delta t_{m} = \Delta x - v_{s}\left( \frac{\Delta x}{v_{s}} - \frac{\Delta x}{c} \right) = \frac{v_{s}}{c}\Delta x \approx 1,1433\cdots\ \cdot 10^{- 6}\ \Delta x$$ The constant in front of the distance shows that when the speed of sound increases the error goes up. This makes sense as we previously assumed that the speed is much smaller than the speed of sound. If it goes up then the difference in speed between sound and light goes down, however this does not pose huge problems as $v_{s} \ll c$. In other words, this trick usually works as the speed of sound $v_{s}$ is much much smaller than the speed of light $c$.

The conclusion here is that, unless you want to know highly reliably estimate distance to the storm, you usually will obtain good estimates for the distance to the storm.