George Rebane

Time for little respite from matters political and geo-strategic. Years ago I came up with a TechTest problem that asked if there were an optimal speed that minimized the amount of rain coming through the area of a car’s missing windshield. A few days ago I ran into another form of this problem in one of my techie subscriptions. The question this time was whether you would get more wet or less wet if you ran or walked some distance to get to shelter. These are essentially the same problem, and I learned that the problem has a long history and debated solutions among techies versed in quantitative STEM.

Well, if you frame the problem correctly, the solution readily reveals itself along with an interesting twist. So, I decided to re-solve it using the missing windshield version because that makes all the variables and resulting math unambiguous and intuitively clear. We start with an area *A*, at angle *φ* to the vertical, moving horizontally at speed *v _{A}* through a steady rainfall in which the drops fall at speed

*v*at an angle

_{R}*θ*off vertical as shown in the figure below. When you add these two velocities (speeds with angles), you get

*v*at angle

_{RA}*α*, the windshield-relative velocity of rain coming down. We need one more parameter to characterize the problem, and that is the water density

*ρ*that characterizes the intensity of the falling rain. Important to note here is that no matter the resulting

*v*speed and direction, the water density (e.g. gals/ft^3) in the volume that

_{RA}*A*, the hole of the missing windshield, sweeps out remains unchanged.

The problem then becomes twofold – as a function of the vehicle speed *v _{A}*, what is the rate that water comes in through the hole, and if you had a distance

*D*to travel to shelter, how much water will have come through the hole.

Solving this problem requires a bit of trigonometry and some straightforward squiggly pushing that I include in the box below. Out of this comes the formula for the rate *r _{R}* (through the area that is perpendicular to

*v*), and the formula for

_{RA}*V*(

_{R}*D*,

*v*), the volume of water through the hole until distance

_{A}*D*is covered at speed

*v*.

_{A}To illustrate the answer to the stated problem, I’ve put together a little scenario with actual numerical values as indicated in the figure, and programmed it up in MS Excel™. The quick answers are shown in the graphs. The blue line shows how *r _{R}* (gals/sec) increases linearly with vehicle speed

*v*as expected. The faster you go, the more water comes pouring through the whole. As

_{A}*v*goes to infinity, so does

_{A}*r*as long as the hole’s orientation

_{R}*φ*presents some frontal area to

*v*which approaches

_{RA}*v*as the vehicle’s speed gets very large. Note that if we make the hole face parallel to

_{A}*v*(i.e. -

_{RA}*φ*=

*α*) then

*r*goes to zero, again as expected.

_{R}But if you only have a certain distance *D* to go to reach shelter, the orange curve shows that the faster you go (i.e. the higher the *v _{A}*) the less water

*V*you take through the hole. The limiting value is just the volume of rain at density

_{R}*ρ*swept out when your speed reaches infinity. In the illustrated scenario this volume of rain

*V*(

_{R}*D*,∞) = 3,733.5 gallons. So the answer for 'should you run to get out of the rain?' is a definite YES; however, if you want to minimize the rate of water coming through the hole, then don’t move (i.e.

*v*= 0). As the

_{A}*r*equation shows, setting

_{R}*v*= 0 reduces the gals/sec to just the volume to the area perpendicular to and increasing at the rate of

_{A}*v*. And its easy to also see that

_{R}*V*will go to infinity since you will be standing still and making no progress to the shelter at distance

_{R}*D*. (Isn’t math wonderful?) Bottom line – run, don’t walk to get out of the rain.

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