Walk or run in the rain
Better to walk or run in the rain?
Model
- A person has to go from point \(A\) to point \(B\) covering a distance \(d\ [m]\).
- The person is modeled as a cuboid with dimensions \(w\), \(l\), \(h\) \([m]\).
- The rain is falling with speed \(v_r\ [m/s]\) at a rate of \(\Phi\ [mm/h]\propto[l/m^2s]\).
- Find the amount of water that falls on the person walking with constant speed \(v\ [m/s]\).
Solution
Assuming that the person moves horizontally from \(A\) to \(B\), only the top of the head and the body front of the person will get wet. The total amount of water will be the sum of the water collected from the top and the front.
The amount of rain \(V_h\) collected from the top of the head with area \(w \cdot l\) for a period of time \(t=d/v\) is $$V_h = \Phi\cdot \ w \cdot l\cdot \frac{d}{v}\ \ [l].$$ This equation tells us that the faster the person walks, the less wet the person will get.
The amount of rain \(V_b\) collected from the body front with area \(w \cdot h\) traveling for a distance \(d\) through a region with rain density \(\Phi/v_r\ [l/m^3]\) is $$V_b = \frac{\Phi}{v_r}\cdot \ w \cdot h\ \cdot d\ \ [l].$$ This equation does not depend on the speed of the person, but only on the distance.
The total amount of rain \(V\) is $$V = V_h + V_b = \left(\Phi\cdot \ w \cdot l\cdot \frac{d}{v}\right) + \left(\frac{\Phi}{v_r}\cdot \ w \cdot h\ \cdot d\right)\ \ [l].$$ We can consider trhe normalized amount of water \(\bar V\) per unit of distance $$\bar V = \frac{V}{d} = \left(\frac{\Phi}{v}\cdot \ w \cdot l\right) + \left(\frac{\Phi}{v_r}\cdot \ w \cdot h\right)\ \ \left[\frac{l}{m}\right].$$
For a given rain speed \(v_r\) and rate \(\Phi\), \(\bar V\) is minimized for \(v\to\infty\) and it is equal to \(\bar V_b\).