# Running in the Rain

Do you get wetter walking or running in the rain? I remember people arguing this question long before the internet. It seems to be one of those silly arguments that never goes away, like whether it’s better to add the cream to the coffee early or late, or whether you should add tea to milk or milk to tea (both of which are in my blog queue).

I googled it and got plenty of hits to recent articles discussing it, so the conversation has obviously not died down, even though this article says that mathematician David Bell worked out the mathematics in 1976.

For me the answer is obvious: move as fast as possible to spend as little time in the rain as possible. And I arrive at that conclusion by reductio ad adsurdum, taking the argument to the extreme: The slowest you can possibly move is to stand still. And if you stand still, there is no limit to the amount of rain that is going to hit your poor head. It’s infinite.

I think the passionate argument about the pro-walking proponents is something like “if you walk you only get the rain on your head but if you run, you intersect all of those and also you run into drops. So you get more.”

Neither of those is a calculation, so it’s pointless to carry that particular hypothetical barroom conversation any farther. Let’s leave our hypothetical debaters to argue amongst themselves at the bar, while we do some physics.

Because this is a fun opportunity to practice developing a mathematical model. I don’t have Prof. Bell’s article handy (that’s my excuse) but really I just want to take a crack at it because modeling is my idea of fun.

#### How wet is “wet”? Defining the variables

First thing is to establish exactly what number we’re going to calculate. What do we mean by “how wet you get?” I’m going to say it’s the total volume of water that strikes your surface, whether it’s on your head, your face, your back, etc. Se we define our first variable $W$, the volume of water that hits you.

What else goes into our model? The rate of rainfall obviously. Total rainfall is measured in terms of a height (usually inches in the US). If I say 1 inch fell, that means whether I use a large container or a small one, the water is 1 inch deep. The rate of rainfall, how fast it is falling, will be a height per time, for instance mm per hour.

We need a measure of how heavily it’s raining. Total rainfall is measured in terms of the height, for instance in inches. Presumably any container wither narrow or wide, would fill up to that many inches. We’ll define a rainfall rate as some height of rain per time (for instance mm/hour). Let’s call that $\rho$. We do love our Greek letters.

We need the speed of motion at which our person is moving, let’s call that $v$. They want to travel a total distance $D$ in the rain, which will take time $t = D/v$

The rain might be coming down at an angle. So let’s define $\theta$ as that angle relative to the vertical. It probably matters whether it’s slanted toward you or away from you so we’ll say $\theta$ is positive when the rain is slanted toward you, and negative when slanted away.

It’s also going to turn out to be useful to consider the horizontal and vertical components of the rain’s velocity separately. We’ll call those $u_x$ and $u_y$ respectively, with the same convention of positive if pointing toward the person and negative if away.

From the person’s point of view, the horizontal velocity is different, so the rain angle in their frame of reference is a different angle, which we’ll call $\phi$.

Two more things I can think of: Rather than get into the complexities of shape, we’ll imagine a human who’s kind of blocky in the sense that there are only two surfaces to consider: The top of the head, cross section $A_{head}$ (covering the horizontal surfaces of head and shoulders), and the front of the body, cross section $A_{body}$ (covering all vertical surfaces)

I think that will do it.

The rain on the head is pretty straightforward. If you imagine walking with a container balanced on the head, then no matter where you’re standing, the container will fill up at a rate of $\rho$ per unit time. The volume that hits a horizontal area $A_{head}$ in time $t$ is $\rho A_{head} t$ no matter where you’re standing. So it’s also true as you move from place to place.

So that means the volume is proportional to how much time you spend in the rain. This part, as you might expect, will be less the less time you can spend in the rain. Walking is good, running is better.

But of course that’s not the whole story.

#### Rain in the face

The angle matters. If the rain is falling straight down (and you’re standing still), none will hit your front. The volume is 0. If you’re walking in a hurricane and the rain is practically horizontal, obviously quite a lot of that rate $\rho$ is going to be on your face.

But the angle is affected by your motion. If the rain is falling straight down when you’re standing still, then as you move, it will have an angle relative to you. We defined the components of the rain’s velocity as $u_x$ and $u_y$, so $\tan \theta = u_x/u_y$.

Relative to you if you are moving at speed $v$, the horizontal component is $u_x + v$, so the angle relative to you is given by $\tan \phi = (u_x + v)/u_y$. The relation between the angles is:

$\displaystyle \tan(\phi) = \frac{u_x + v}{u_y} \\ = \frac{u_x}{u_y} + \frac{v}{u_y} \\ = \tan(\theta) + \frac{v}{u_y}$

##### Well, this is a drag (force)

Hold that thought. We need to know how fast the rain is moving in order to work with this. So we need a short aside about drag terminal velocity.

Things falling in air are not in free fall. They experience a downward force $mg$ which acts to accelerate them, but they also experience an upward drag force (air resistance) which counters the gravitational force. Drag increases with speed, so a slow object will keep increasing in speed until the drag increases to the point that it balances gravity. That speed where the forces balance is called the terminal velocity, and the mathematics of calculating that aren’t relevant here. Only that there is one.

If it was originally moving faster than the terminal velocity, then the drag force is more than the gravitational force and it will act to slow the object down, until it slows down to the terminal velocity.

Rain, like everything else, has a terminal velocity. The most common figure I see online is 10 m/s. Lighter drizzle, consisting of much smaller drops has a lower velocity but let’s work with 10 m/s.

But here’s the thing about terminal velocity: The drag force is the same in every direction, not just vertically. Drops at an angle are being moved by gravity but also by the wind. Should I say that their vertical velocity is 10 m/s, or their total velocity?

After carefully pondering this question I arrived at this conclusion: I don’t know. I want to keep the model simple but reasonably accurate. So I’m going to arbitrarily say that the vertical velocity of the rain is 10 m/s, and the horizontal velocity is the wind velocity. The assumption I’m making is that the wind is gusty and the drops have a chance to fall vertically for much of their fall. So I’m hypothesizing that the wind speed is added to the falling speed at the last minute.

I think this is probably a pretty good assumption in hurricanes, where the wind no doubt dominates gravity. Perhaps less so for a lighter rain, but you don’t care about light rain, right? You’re not going to get that wet anyway.

By the way, what I’m talking about here is the very important concept of model fidelity. When you’re building models, you have to make deliberate choices about how much accuracy you’re going for. High fidelity models with a minimum of simplifying assumptions are more accurate, but only if you have a correct model of the high-fidelity physics and good data to put into your model. For instance, the earth is often modeled as sphere of uniform density, which we know darned well it is not. Taking account of the density profile requires more complicated equations and also requires knowing what the profile is.

Is my assumption good? It’s not off by an order of magnitude, it’s good enough. Probably. Let’s see where it goes. We can always come back later with a higher-fidelity model.

So that will be our model: That $u_y$ is 10 m/s, and $u_x$ is the wind speed. Note: 10 m/s is 22 mph or 36 kph, a fairly modest wind speed. So the wind can easily equal or exceed the downward terminal velocity.

##### Back to our regularly scheduled calculation

So we have the equations to figure out the angle. Now we need to figure out how the angle affects the water hitting you.

Consider the figure below. Two drops indicated by the arrows pass by the horizontal surface (dashed line) a distance $d$ apart.

They strike the vertical surface a distance $a$ apart, where $d/a = \tan \phi$, or $d = a \tan \phi$. The rain is spread out on the vertical surface.

This does not happen in the perpendicular direction. Two drops which are $d$ apart in the direction parallel to the wall (perpendicular to the page) will remain $d$ apart. The stretching happens in one dimension only.

As a result, if we consider drops striking a horizontal area $A_h$, those drops will be spread over an area $A_v = A_h /\tan \phi$. Equivalently, $A_h = A_v \tan \phi$.

So how much rain strikes the vertical surface $A_{body}$ of our running man in time $t$? The same rain that strikes a horizontal area $A_{body} \tan \phi$, which is $\rho\,A_{body}\,t \tan \phi$.

We need to handle the case when $\phi < 0$ (the rain is hitting your back) as well. $\tan \phi < 0$ but of course the physical area, both vertical and horizontal, are positive. So in general the horizontal area is $A_{body} \left | \tan \phi \right|$ and the amount of rain striking $A_{body}$ in time $t$ is $\rho\,A_{body}\,t \left| \tan \phi \right |$.

#### Putting it all together

In time $t$ the rain on the the head is $\rho A_{head} t$. The rain on the face and front of the body is

$\displaystyle \rho A_{body}\,t\,\left | \tan \phi \right | = \rho A_{body}\,t \left | \tan \theta + \frac{v}{u_y} \right |$

Recall that $t = D/v$ and we have fixed $u_y$ at 10 m/s. So summing the head and body terms and pulling out the common factors, we find the total water volume W is:

$\displaystyle W = \frac {\rho D}{v} \left ( A_{head} + A_{body} \left |\tan \theta + \frac{v}{10} \right | \right ) = \rho\, D \left ( \frac{A_{head}}{v} + A_{body} \left |\frac{\tan \theta}{v} + \frac{1}{10} \right | \right )$

The first term clearly decreases with $v$. If $\theta$ is positive (you’re running into the rain), then the second term also decreases with $v$. The only possibility for different behavior is if $\theta < 0$, i.e. the rain is hitting your back and slanting away from you in front.

The details will depend on the relative sizes of the two cross sections $A_{head}$ and $A_{body}$. For purposes of analysis, I will arbitrarily estimate that the vertical body area is about 4 times the horizontal head/shoulders area. (Obviously that’s one of several free parameters you could play with.) Let’s see what that gives us.

What’s a reasonable value of $v$? Usain Bolt ran the 100 m in 9.58 s, an average speed of 10.4 m/s. But world-class sprinters are accelerating for most of a 100 m race. He was clocked at well over 12 m/s in that world-record race. So we’ll let $v$ vary from 0 to 12. But for an ordinary human, a comfortable walking pace is probably closer to 1 m/s (2.2 mph), with 3 m/s being a jogging pace,

I’m just interested in the part that changes with $v$, so I’ll plot

$\displaystyle \frac{W}{\rho D A_{head}} = \frac{1}{v}\left [ 1 + 4 \left (\tan \theta + \frac{v}{10} \right ) \right ]$

Here are the results.

Here is the graph for various positive (in your face) angles. As you can see, if the rain is at a pretty severe angle, running at 3 or 4 m/s is significantly better than walking at 1 m/s. The graphs are a lot flatter for angles of 30 degrees or less. Let’s look at those graphs more closely.

Even for no wind, rain at 0 degrees, it makes some difference to run (compare 3 m/s to 1 m/s) simply because you’re spending less time in the rain. But it makes much less difference to do a full-out sprint at Usain Bolt speeds compared to an ordinary jog.

Now let’s look at the negative angles, the rain hitting your back. This picture is a little more interesting.

Here you can see that there’s an optimum speed. That’s going to be about where you’re keeping up with the horizontal speed of the rain, so from your point of view it’s falling straight down with almost none hitting your front or back. Run much faster than that, and you’re starting to run into the rain and actually making things worse. But up to the jogging pace of 3-4 m/s, you’re probably all right in most conditions.

#### Conclusion

Is it better to run in the rain? Science says yes, big difference between a walk and a jog. But between a light jog and a full-out sprint, not so much. Also, a full out sprint is going to increase your chance of hurting yourself.

(Some people might point out that there’s an invention called an umbrella which removes the entire question of walking vs. running. To which I say, where’s the fun in that? Also, I’ve always hated carrying umbrellas, but that’s just me.)

Anyway yes, go ahead and run. Carefully.