I think that it has been asked before, but I can't find it, so I'll ask again.
Say it's raining, in order to get less wet should I run or walk?
From one hand, if I run, my surface area expands but I reach my destination in less time. On the other hand, if I walk I have less surface area but i'll reach my destination in more time.

I know that it is very complicated and has many variables.
So lets say that the rain is falling in a constant velocity V, and I'm running in a constant velocity v₁ and walking in v₂ ,
and in one second on a surface area of 1 m^2, n raindrops fall (not always on the same point on the surface, so if one wishes we can also say that we have the probabily P that a raindrop will fall on a certain point on the surface in a certain time)
I'm a 2 dimensional body, (square) having an area of A [m^2] and I make an angle of a with the ground.

It makes the question more specific.
In this paticular case, what rather I do, Run or Walk.

And does anyone have any idea how to expand it to more complicate bodies...

Yatir

i don't have a mathematical proof for this but intuitionally you can observe that...

the amount by which the body gets wet(A) depends upon two things......
1> the time for which the body is in contact with the rain(t)
2> the amount of drops falling on the body...(n)

i.e
$A\propto t\times n$

if $n$ is assumed to be constant, then $A\propto t$
so i would certainly prefer running..
love arun

But when we run more raindrops come in contact with us (it isn't the same n)

Yatir

Surely that depends on the angle at which the rain falls?

I think this is the way to approach the problem for a general body. Suppose that the rain is moving at a constant velocity vector r , the body can walk at velocity v ₁ or run at velocity v ₂ . Also assume that in both cases we travel the same distance in a straight line, and that s_i =|v _i |. Now let w _i =r -v _i , this is the relative velocity of the rain from the frame of the moving body. Imagine looking at the body from the point of view of the rain, i.e. an orthogonal projection along w _i . If the 2D area of this shape, as seen by the rain, is A_i then the rate of getting wet will be A_i . So, we want to minimise A_i /s_i in order to minimise the amount of rain collected, taking into account different speeds. So, you need to work out how A_i depends on s_i for a human body to answer the question. More later...

In particular, if the rain is coming at an angle of just over 45 degrees to the horizontal then walking is probably your best bet unless the rain speed is really massive. You'd also think (if you looked at a graph of $s(r,\theta )$ for $0<\theta <90$ degrees that for $\theta$ just under 90 degrees you'd also want to walk slowly. However, this is probably only true if you are very, very thin (a rectangle in fact). To make the calculations more realistic for a person when the angle of the rain to the horizontal is large, we could perhaps assume the person is a cuboid. Who wants to do that calculation?

By the way, is everyone happy with the explanation in my first post? I wrote it in a bit of a hurry (I was being called to a social engagement of sorts). If not, I can probably elaborate with a diagram.

Please do.

Yatir

Yatir,
i forgot to make a correction...
n-> the amount of drops falling on the body per second....

this n sure maybe assumed to be constant..

love arun

Dan,
those are some interesting deductions you have made.....would certainly like to see your idea through a picture....

love arun

Here's the picture (I've just done the 2D case because it's easier to draw):



The black line is the ground, the red line is the person and the parallel blue lines are the rain.

Suppose the person is standing still, how much rain is hitting them per second? It will be proportional to A (is this clear?), the length of the green line which is perpendicular to the blue lines. In 3D, A would be an area rather than a length - the area of the projection of the person's body along the rain vector.

So, we have that for someone standing still with rain coming down along a vector r the rate at which they get wet is proportional to A, the area of the 2D projection of the person's body along r .

Now how about if the person is moving? Well, in that case we change frame (to some physics people this will be enough explanation). For people who haven't done frames of reference in physics, imagine that you have a video camera recording the person in the rain. Now imagine that the video camera is moving so that the person seems to be staying still, i.e. the camera moves at the same velocity as the person. If the rain was originally moving at velocity r and the camera is now moving at velocity v (the velocity of the person) then it will seem, from the camera's point of view, as though the rain is moving with velocity r -v . For example, if the camera was moving at the same speed as the rain, v =r so r -v =0, the rain will seem as though it is not moving to the camera.

Right, so from the camera's point of view the person is not moving and the rain is moving with velocity r -v . But we have already solved this problem above, the rate of getting wet is proportional to A, the area of the projection of the person's body along r -v (the apparent velocity of the rain).

So, if we define v =v ₀ s where v ₀ is a unit vector and s is a scalar, the speed of the person, and we define A_s to be the area of the person's body projected along r -v ₀ s, then the rate of getting wet if the person runs at speed s is A_s . If the distance to be run is d then the time taken will be d/s. The total amount of wetness will be proportional to A_s d/s, but d is a constant so it will be proportional to A_s /s. In other words, we want to minimise A_s /s.

What would we change if our body is not perpendicular to the ground?
Yatir