Should you run in the rain?May 29, 2013
An interactive recreation of the BBC magazine piece, "Do you get less wet if you run in the rain?". I made this primarily as an experiment in using web technology to make math and science problems more open-ended and exploreable. To some extent it is also an experiment in the authoring of these types of documents, to gauge the level of expertise needed to build them, and to identify the main obstacles in creating a large body of free, remixable learning material online.
Theory
Let's call the total amount of water that has struck a person walking in the rain \(W\), for "wetness." The rate at which the person is getting wet (wetness per unit time) is \(dW/dt\) and can be expressed as the surface integral of the rain's density times its velocity over the person's body:
$$\frac{{dW}}{{dt}} = - \int \rho \vec{v} \cdot d \vec{A}$$ |
\(dW/dt\) \(\rho\) \(v\) \(A\) |
rate person is getting wet density of raindrops in atmosphere velocity of raindrops (vector) surface area of person |
kg/s kg/m^3 m/s m^2 |
\(\vec{V}\) is the velocity of the rain relative to the person, \((\vec{v_r}-\vec{v_p})\) or \(-(\vec{v_p}-\vec{v_r})\) if we pull out a minus sign. Plugging that into the integral we get \(\frac{{dW}}{{dt}} = \int \rho (\vec{v_p}-\vec{v_r}) \cdot d \vec{A}\).
Simplification
Now, we're not going to be able to take the integral over a human body—it's not a shape that we can really represent as a function. Instead, we are going to assume that all the horizontal parts (shoulders, etc.) have area \(a\), and the vertical parts (torso, face, etc.) have area \(A\). We also assume that the rain is falling vertically exclusively and the person is travelling horizontally exclusively (e.g., \(v_r = 0 \vec{i} + a \vec{j}\) and \(v_p = b \vec{i} + 0 \vec{j}\)). Thus, \(v_r\) only multiplies \(a\) in the integral, and \(v_p\) only multiplies \(A\). Integrating with respect to time:
$$W = \rho (a v_r + A v_p) t$$ |
\(W\) \(\rho\) \(v_r\) \(v_p\) \(a\) \(A\) \(t\) |
total mass of water that strikes person density of raindrops in atmosphere velocity of raindrops (scalar) velocity of person (scalar) surface area, top surface area, front time elapsed |
kg kg/m^3 m/s m/s m^2 m^2 s |
Clearly we can't control the velocity of the raindrops, so it seems that no matter what wetness is proportional to the velocity of the person, \(W \sim v_p\). The faster you run, the wetter you get.
Known Distance to Destination
Now consider the possibility that the person's destination is a distance \(d\) away. Assuming an average pace \(v_p\), the walker can cover that distance in a time \(t=d/v_p\). Sub that into the wetness equation above:
$$W = \rho (a \frac{v_r}{v_p} + A) d$$ |
\(W\) \(\rho\) \(v_r\) \(v_p\) \(a\) \(A\) \(d\) |
total mass of water that strikes person density of raindrops in atmosphere velocity of raindrops (scalar) velocity of person (scalar) surface area, top surface area, front distance to shelter |
kg kg/m^3 m/s m/s m^2 m^2 m |
In this case, wetness is proportional to distance and can be reduced by increasing \(v_p\). The absolute minimum wetness \(W_{min} = \rho A d\) occurs at \(v_p = \infty\), a scenario in which, as soon as the first bit of rain reaches their head, the walker is whisked off to shelter at infinite speed, with no further rain reaching their head but only their front. This is an extreme case, and in general, the insight is that the faster you walk, the the sooner you will get to shelter and the less wet you will end up.
Complications
However, there a lot of other factors that might come into play here:
Suppose you have a maximum speed \(v_{max}\). And suppose that there's a certain "saturation wetness" \(W_{sat}\), after which, you no longer care how wet you get. At some distance \(d_{max}\), you won't be able to reach shelter in time to avoid reaching your saturation wetness, so it doesn't matter if you ran there or took a leisurely stroll. The following simulation lets you explore such possibilites.
Simulation
First, let's try and get a handle on these values by solving for them in terms of some recognizable units. In the United States, rainfall intensity is typically measured in inches per hour (in/hr). Average intensity in most places is about 0.25 in/hr, and 1 or 2 in/hr would be a heavy thunderstorm. If we know the raindrop velocity, we can use the rainfall intensity to back-calculate the rain density as \(\rho = I/v_r \) (example). According to Figure 2 in this paper, terminal velocity of a raindrop under normal conditions maxes out at 9 m/s. This page shows typical values for various intensities and drop sizes. Combining these two sources, it looks like the range we'd expect to encounter is about 2–9 m/s.
Manipulate the values below to test out different scenarios.
$$W = \rho (a \frac{v_r}{v_p} + A) d$$ |
\(I\) \(v_r\) \(\rho\) \(v_p\) \(a\) \(A\) \(d\) \(W\) |
in/hr m/s mg/mL m/s m^2 m^2 m g |
intensity of rainfall velocity of raindrops (scalar) density of raindrops in atmosphere velocity of person (scalar) surface area, top surface area, front distance to shelter total mass of water that strikes person |
$$V = W \rho_{water}$$ |
\(\rho_{water}\) \(V\) |
1.0 g/mL L |
density of water total volume of water that strikes person |
Libraries Used
Tangle.js for the slider
things,
MathJax for equation
rendering.
Some Comments
- This demo builds directly on the BBC article at the top of the piece, and hasn't been checked experimentally in any way. Mythbusters did an "investigation" of this, reaching opposite conclusions.
- If you find an error in this page, feel free to fork the page on GitHub and send me a 'Pull' request with your fix.