# Will you get less wet running in the rain?

This is something I did one morning while waiting to take an exam that
afternoon on mathematical methods and fluid mechanics. I couldn't bring
myself to do any more revision but needed to keep my mind on maths. As
it was raining outside I started to wonder whether it was possible to
build a simple mathematical model for an object moving in the rain. What
follows is a write up of my thinking that morning.

Common sense would say that one will get less wet running in the rain
than walking, but this might not necessarily be true. The slower the
speed travelled the more water will land on one's your head, but the
faster the speed the greater the water hitting one's front. Given that,
for most people, the surface area pointing vertically upwards (top of
head, shoulders and chest) is much smaller than that of their surface
area pointing forwards (face, front of torso and limbs), it is possible
that travelling at greater speeds could in fact make one wetter. To
investigate this I build a simple mathematical model for an object
moving through rain and start by looking at the case when the rain is
falling directly downwards (there is no wind). Later I develop the model
to include the effects of wind.

## Assumptions

1. Rain can be modelled by a constant density vector field.
2. Rain travels at a constant velocity downwards.
3. A person can be modelled by a regular cuboid.
4. The person moves at constant velocity through the rain.
5. Rain that comes in to contact with the person is assumed to be
perfectly absorbed.
6. Other than the person there are no other surface effects that need
to be considered.

The first assumption states that rather than modelling the rain as
individual drops it is possible to model it spread out evenly throughout
space. This appears reasonable as long as the rain droplets are
significantly smaller than the object moving through them and there is
no clumpiness; i.e. they are evenly spaced out. This the key assumption
of the model, without this the following maths is invalid.

The second assumption asserts that rain moves at a constant speed
towards the floor and doesn't change direction. While in the real world
the wind will affect the direction and speed of rain for the sake of
simplicity this model will ignore these effects. This assumption will
make the model far simpler than including wind direction, but is an
unrealistic assumption, any further models should include wind effect.

The third assumption appears to be fine for out basic initial model,
further investigation is needed to confirm this for more advanced
models.

The fourth assumption states that we won't be taking the acceleration
of the object into account. As the velocities involved are quite small I
think this is fair. Again, a more developed model may want to take this
into account.

The fifth and sixth assumptions are similar; none of the water hitting
the object bounces off and there is no spray onto the object from other
surfaces. Both of these are appear to be okay for our simplistic model,
but they don't realistically represent a person running in the rain
where there will be some spray effects.

## The Model

Let the rain be described by the vector field

$\frac{d\mathbf{R}}{dt}=-\alpha \mathbf{k}$ $\left(\alpha >0\right)$

where:

• $\mathbf{R}$ is the rain vector field
• and $\alpha$ is the downwards speed of the rain

The body, $O$, moving through the rain is a regular cuboid with corners located at $\left(0,0,0\right)$, $\left(x,0,0\right)$, $\left(0,y,0\right)$, $\left(x,y,0\right)$, $\left(0,0,z\right)$, $\left(x,0,z\right)$, $\left(0,y,z\right)$ and $\left(x,y,z\right)$.

The equation for the velocity of $O$ through the rain is

$\frac{d\mathbf{s}}{dt}=\beta \mathbf{i}$ $\left(\beta >0\right)$

where:

• $\mathbf{s}$ is the displacement vector
• and $\beta$ is the speed

Given the above mathematical definitions it is possible to calculate the flux of the vector field $\mathbf{R}$ for each face of $O$ as it moves at velocity $\frac{d\mathbf{s}}{dt}$
by calculating the surface integrals, giving the volume flow rate
through each surface. In other words how wet each surface gets per unit
of time. The model has been created so only two surfaces need to be
considered; the top and the front faces. That is surface ${S}_{\mathrm{top}}$ with corners located at $\left(0,0,z\right)$, $\left(x,0,z\right)$, $\left(0,y,z\right)$ and $\left(x,y,z\right)$, and the surface ${S}_{\mathrm{front}}$ with corners located at $\left(x,0,0\right)$, $\left(x,y,0\right)$, $\left(x,0,z\right)$ and $\left(x,y,z\right)$.

The surface integral for a face $S$ is defined as

${\int }_{S}\mathbf{F}·\mathbf{n}\phantom{\rule{0.3em}{0ex}}dA$

where:

• $\mathbf{F}$ is the vector field
• $\mathbf{n}$ is the unit normal vector for the surface
• and $A$ is the area of the surface

For our model it is necessary to calculate how wet each face gets through time. The equation for this is

$\frac{d{w}_{S}}{dt}={\int }_{S}\mathbf{F}·\mathbf{n}\phantom{\rule{0.3em}{0ex}}dA$

where:

• ${w}_{S}$ is the wetness of surface $S$.

As the body is moving through a moving field the two vector equations describing velocity need to be combined, hence

$\mathbf{F}=\frac{d\mathbf{s}}{dt}+\frac{d\mathbf{R}}{dt}$
$\phantom{\rule{1em}{0ex}}=\beta \mathbf{i}-\alpha \mathbf{k}$

For ${S}_{\mathrm{top}}$ the face normal points directly up, so

$\mathbf{n}=\mathbf{k}$

therefore

$\frac{d{w}_{\mathrm{top}}}{dt}={\int }_{{S}_{\mathrm{top}}}\left(\beta \mathbf{i}-\alpha \mathbf{k}\right)·\mathbf{k}\phantom{\rule{0.3em}{0ex}}dA$
$\phantom{\rule{2.5em}{0ex}}={\int }_{x}^{0}{\int }_{0}^{y}\left(\beta \mathbf{i}-\alpha \mathbf{k}\right)·\mathbf{k}\phantom{\rule{0.3em}{0ex}}dy\phantom{\rule{0.3em}{0ex}}dx$
$\phantom{\rule{2.5em}{0ex}}={\int }_{x}^{0}{\int }_{0}^{y}-\alpha \phantom{\rule{0.3em}{0ex}}dy\phantom{\rule{0.3em}{0ex}}dx$
$\phantom{\rule{2.5em}{0ex}}={\int }_{x}^{0}-\alpha y\phantom{\rule{0.3em}{0ex}}dx$
$\phantom{\rule{2.5em}{0ex}}=\alpha xy$

and

${w}_{\mathrm{top}}=\int \alpha xy\phantom{\rule{0.3em}{0ex}}dt$
$\phantom{\rule{2em}{0ex}}=\alpha xyt+C$

as ${w}_{\mathrm{top}}=0$ when $t=0$ hence $C=0$, the surface is dry at the start,

therefore

${w}_{\mathrm{top}}=\alpha xyt$

For ${S}_{\mathrm{front}}$ the face normal points directly outwards, so

$\mathbf{n}=\mathbf{i}$

therefore

$\frac{d{w}_{\mathrm{front}}}{dt}={\int }_{{S}_{\mathrm{front}}}\left(\beta \mathbf{i}-\alpha \mathbf{k}\right)·\mathbf{i}\phantom{\rule{0.3em}{0ex}}dA$
$\phantom{\rule{3em}{0ex}}={\int }_{0}^{y}{\int }_{0}^{z}\left(\beta \mathbf{i}-\alpha \mathbf{k}\right)·\mathbf{i}\phantom{\rule{0.3em}{0ex}}dz\phantom{\rule{0.3em}{0ex}}dy$
$\phantom{\rule{3em}{0ex}}={\int }_{0}^{y}{\int }_{0}^{z}\beta \phantom{\rule{0.3em}{0ex}}dz\phantom{\rule{0.3em}{0ex}}dy$
$\phantom{\rule{3em}{0ex}}={\int }_{0}^{y}\beta z\phantom{\rule{0.3em}{0ex}}dy$
$\phantom{\rule{3em}{0ex}}=\beta yz$

and

${w}_{\mathrm{front}}=\int \beta yz\phantom{\rule{0.3em}{0ex}}dt$
$\phantom{\rule{2.5em}{0ex}}=\beta yzt+D$

as ${w}_{\mathrm{front}}=0$ when $t=0$ hence $D=0$, again, the surface is dry at the start,

therefore

${w}_{\mathrm{front}}=\beta yzt$

The total wetness of the body if found by summing the wetness of all the faces.

So

$w=\sum {w}_{i}$
$\phantom{\rule{0.9em}{0ex}}={w}_{\mathrm{top}}+{w}_{\mathrm{front}}$
$\phantom{\rule{0.9em}{0ex}}=\alpha xyt+\beta yzt$
$\phantom{\rule{0.9em}{0ex}}=\left(\alpha x+\beta z\right)yt$(1)

The ratio of rain landing on the top surface to rain hitting the front surface is

${W}_{\mathrm{ratio}}=\frac{{w}_{\mathrm{top}}}{{w}_{\mathrm{front}}}$
$\phantom{\rule{2.5em}{0ex}}=\frac{\alpha xyt}{\beta yzt}$
$\phantom{\rule{2.5em}{0ex}}=\frac{\alpha x}{\beta z}$(2)

## Initial Analysis

The model has produced two equations that will help to shed light on the
initial question of how the speed travelled in the rain effects how wet
a body gets. These are equation (1), giving the total wetness of the
body at a specified time, and equation (2), giving a ratio for the
proportion of rain landing on the top compared to the front of the body.

### Ratio of Surface Wetness, $W_{ratio}$

Equation (2) gives a ratio for the proportion of rain landing on the top
compared to that hitting the front, and depends on 4 independent
variables: the speed of the rain, $\alpha$; the speed of the body,
$\beta$; the thickness of the body, $x$; and the height of the body,
$z$. This ratio shows that, according to the model, the two factors
affecting how wet the top of the body gets, compared to the front, are
the speed of the rain and the thickness of the body. Likewise, the two
factors affecting the front compared to the top are the height and the
speed of the body. By trying to find standard values for the rain speed,
and body width and height it is possible to see how this ratio changes
according to the speed the body travels at.

• According to

the speed of falling rain depends on the size of the raindrop.
Drizzle falls at 2 m/s, a 2 mm raindrop falls at 6.5 m/s, while a 5
mm raindrop falls at 9 m/s. To compare the effect of different rain
speed I'll take a low speed of 3 m/s, a mid speed of 6 m/s, and a
high speed of 9 m/s.
• The mean height of an adult in England in 2008 was 1.68 m
.
To compare the effects of height I'll take a short person to be 0.3
m lower than this and a tall person 0.3 m higher. Given that people
aren't cuboid taking 80% of these values should give a fair value
for the height components. This gives the heights 1.10 m, 1.34 m,
and 1.58 m for a short, average and tall body, respectively.
• To estimate people's thickness, finding statistics has been hard.
There is data for waist and chest circumferences, but turning these
into a reliable figure for thickness is nigh on impossible. The
figures I've chosen are arbitrary but hopefully represent a
reasonable estimate for the thickness of an individual. I will take
the thickness for a slim, average and large person as 0.2 m, 0.35 m
and 0.5 m, respectively.
• The average walking speed is 3 miles per hour
,
or about 1.3 m/s. An Olympic sprinter can run 100m in under 10 s
,
an average speed of under 10 m/s.

## References

21st February 2013