Should I stay or should I go?

I was about to go out for a walk, but I read the weather forecast and there is an expected rainfall of 0.5mm over the next hour. That looks like not much, but what does 0.5mm actually mean?

1. Interpreting rainfall forecasts

0.5mm means if you put a recipient during the rainfall you would see the level of water rise by 0.5 millimetres. If your recipient were massive and had a surface of $1m^2$ (1 meter long, 1 meter wide) you would have stored¹

$$1m^2 · 0.5mm = 0.5dm^3.$$

This is non other than $0.5l$ ($l$ for litres), which is another common unit for providing these types of data; actually to keep units straight the actual figure is litres per squared metre, that is: $l/m^2$.

So it’s raining $0.5l/m^2$, thus we just need to know the surface of our body that’s exposed to the rain (in squared metres) and multiply it by this figure to see how wet we will get.

2. Best case scenario

To make a first easy calculation let’s assume rain falls fully vertically, i.e., there’s no wind and we are pretty much standing still.

In that case we are just concerned with the surface of our body as seen from above, which is covered mostly by our head and our shoulders (pregnant women and overweight people may have to make some adjustments).

Since I was already out when thinking about this, I decided whether I could get a reasonable figure from the top of my head. With my hand extended, my thumb to pinky distance is about $20cm$. My head could be $25\times15cm^2$, but let’s make it

$$20\times20cm^2 = 400cm^2$$

for ease. My shoulders are less than $20cm$ long and $10cm$ wide, so it’s something like

$$2·(20\times10cm^2)=400cm^2.$$

Thus we would have (we will round up to be conservative):

$$(200+400)cm^2 = 800cm^2 \simeq 1000cm^2 = 0.1m^2.$$

Thus we have $0.5l/m^2 · 0.1m^2 = 0.05l$, how much is that? A small bottle is $0.5l$, so $1/10$ of that. A glass is usually around $0.2l$, so $1/4$ of that. That is, being out there with a rainfall of $0.5mm$ would be like pouring a quarter of a glass over my head and shoulders spread out over the course of $1h$. Seems like a manageable amount, provided the rainfall is actually spread out during that time, as that’s key to give body heat time to evaporate the water and not get wet. If that amount falls in 5 minutes I’d rather have some kind of cover.

3. The real deal

Now anyone who has fought the rain knows that wind is the real issue.

Apart from drastically lowering the effectiveness of umbrellas and hoods, the problem is that as rain falls diagonally, a much greater surface of our body becomes exposed to it. For a rough idea, for my case it would be something like

$$1.8m \times 0.4m=0.72m^2$$

which is around 10 times the figure in the earlier example ($800cm^2=0.08m^2$)!

To calculate the body surface that’s exposed to the rain, which we call “effective body surface”, we will need to know the inclination with which the rain falls. This will depend on the wind among other things and it can get a little bit more complicated, probably out of the range of back of the envelope calculations, so I think it’s best to address it in another post. I’ll leave you with a question in the meantime: for our purposes, is it equivalent for the wind to be $5km/h$ and for us to be moving at $5km/h$ while the air is still?

Footnotes

1. Step by step it would be: $\\ 1m^2 · 0.5mm = 1 · 10^2dm^2 · 0.5 · 10^{-2}dm = 0.5dm^3.$ ↩