I thought I'd do these calculations, seeing that my Physics exam just got over.

From Wikipedia, I found that the maximum radius of a pure rainwater drop is 5mm(5x10^-3m)

and its terminal velocity while falling in air is 16.5m/s.

So, mass of the droplet= Volume x Density

=4/3 x pi x r cubed x 1000kg/m^3

=0.000523kg.

The momentum of the droplet= Mass x Velocity = 0.000523 x 16.5 = 0.00863 kg m/s

The area this force will be spread over = area of the circle with radius of the droplet

= pi x r^2

=0.0000785 meter squared

So, the force applied = dp/dt (rate of change in momentum)

(I'm taking that the velocity of the water becomes 0 after impact)

I'm not too sure about the time it takes for a droplet to stop, but I took it as 0.5 s.

F= 0.00863/.5 =0.01726N

Pressure = Force/Area = 0.01726/0.0000785 = 220 Pascals.

This is equivalent to the force exerted by 'y' m of water.

220=y x density x acc. due to gravity

y= 0.02m = 2cm

After adding atm. pressure, the WR needed to be rain proof is 10.02m.

I think this value is too low. Probably the time it takes for the drop to stop moving is less than 0.5 s.

Does anyone have any idea what the value for time is?

Thanks for reading!

PS: When I take time as 0.1s, then the pressure comes out to be 11cm of water. Which would mean a WR of 10.11m is needed.

So does this mean that 30m WR resistance is enough for rain?