Question

Fluid mechanics.
Hydrodynamics.
There are people who say that to get wet the least when it's raining and one does not have an umbrella is better to run, and others say it's better to go slowly. Suppose it rains vertically with a flow of 10^-5 m^3/s per square meter, and that you have to walk 100 m in the rain. The drops have an average size of 1 mm^3. Modeling an adult person as a box 2m high, 1m wide and 0.5m deep, determine if in this situation it is better to walk at 1m/s or run at 4m/s. Do we have to take into account other circumstances or physical phenomena?
Explain in detail your procedure and reasoning to solve this problem.

Answer 1

So, the given discharge of rainfall = 10^-5 m³/s per square meter area.
Now, it's given that the volume of one raindrop = 1 mm³
Therefore, Number of droplets falling every second on every square meter area
= 10^-5 m³ x 10⁹ mm³/m³/1 mm³ = 10000 drops.
Also, volume of 1 droplet = 1 mm³
Therefore, Radius = 0.62 mm, which makes it a drizzle.
Average fall velocity of drizzle = 4 m/s
Therefore, in any given span of one second, there are 10000 droplets in a 4 m high, 1 m wide and 1 m deep box of space (Volume = 4 m³). i.e. Number of droplets in a 1 m³ volume for one second span = 10000/4 = 2500 droplets/m³/s
Our assumed human model is a box of top area = 1 x 0.5 = 0.5 m² and frontal area = 1 m x 2 m = 2 m²

Now, Since, the rain is falling vertically, no matter how fast we move, the same amount of raindrops will always fall on the top of the human model per unit time. This is because, while moving, if we dodge a droplet as we moved forward, we catch another droplet which was in front of us assuming uniform dispersion of raindrops across 3D space.

Case 1: Moving at a speed of 1 m/s
In this case, in one second, we move 1 m forward. Volume covered by the model's front area = 2 m² x 1 m = 2 m³
Total time taken to reach our destination = 100/1 = 100 s

No. of raindrops fallen on the top of model = 10000 drops/m² /s x 0.5 m² x 100s = 500000 drops.

Now, since we know, there are 2500 droplets in a span of 1 s in every 1 m³ of 3D space.
Total droplets fallen on the frontal area of the model = 2500 drops/m³/s x 2m³ x 100 s = 500000 drops.
Therefore, total number of drops fallen on the model = 1000000 drops.

Case 2: Moving at a speed of 4 m/s
Distance covered in 1 s = 4m Volume covered by frontal area = 2 m² x 4 m = 8 m³
Total time taken to reach our destination = 100/4 = 25 s.

No. of raindrops fallen on the top of model = 10000 drops/m² /s x 0.5 m² x 25s = 125000 drops.
Total droplets fallen on the frontal area of the model = 2500 drops/m³/s x 8m³ x 25 s = 500000 drops.
Therefore, total number of drops fallen on the model = 625000 drops.

The above calculations show that it's better to run than walk in case of a rainfall if you have a destination near you.

However, these calculations are based on the assumptions that our model walks/runs being completely straight and rigid, which is usually not the case with actual humans. If one could incline their body in the direction of net resultant of the falling raindrops which can be calculated by the vector addition of rainfall velocity and our walking/running speed, they can significantly reduce the frontal area wetting. There are some other factors in consideration as well such as the wind speed and direction, splashing due to flooding, etc.

Thank you! Upvotes are appreciated.