# Velocity required for Horizontal Rain

Related to my previous question - Change in appearance of liquid drop due to gravity...

Ok, I think we all have noticed this practical phenomena (a kind of illusion though)... During rain (while we're at rest), rain drops would definitely fall in the form of somewhat perfect vertical lines (assuming there's no wind). And When we're in motion, those "lines of rain" would be inclined to some $\theta$. That inclination depends on our velocity (what our eyes perceive at our speed). A friend of mine drove his car to some 120 km/h. Unfortunately, I didn't observe any horizontal line.

So, Is there any way we could observe "rain lines" horizontal..? What would be the velocity required for it? I think we don't have to break any barriers for that perception :-)

I googled it. But, I can't find anything regarding lines. Any explanation or perhaps a link would be good.

(As the climate would be rainy for atleast 2 months, I could test it in some grounds)

• Am I missing something? Isn't the tangent of the angle of deflection equal to the ratio of your horizontal velocity to the drop's vertical velocity? Then they'll never be perfectly horizontal. And whether or not you see streaks depends on lighting. – user10851 Oct 31 '12 at 17:10
• If your objective is to observe horizontal rain lines, just put a heavy weight in the back of the car, to make it tilt up. Then drive the right speed. – Mike Dunlavey Nov 2 '12 at 0:08

Let's say your eye integrates everything it sees over a time $\Delta t$. In that time, a raindrop will have fallen vertically a distance $v_\text{rain} \Delta t$. It will also appear, to you, to have moved horizontally a distance $v_\text{you} \Delta t$. You can draw a right triangle with these two distances as the legs. The hypotenuse is the perceived raindrop streak.
Let $\theta$ be the angle opposite the horizontal side. This measures deflection away from vertical. We have $$\theta = \tan^{-1}\left(\frac{v_\text{you}}{v_\text{rain}}\right)$$ from basic trigonometry. This is independent of $\Delta t$. $\theta$ asymptotically approaches $90^\circ$ as your speed approaches infinity, while for low speeds, $\theta$ is approximately proportional to your velocity. The length $l$ of the streak follows from Pythagoras: $$l = \Delta t \sqrt{v_\text{you}^2 + v_\text{rain}^2}.$$
• Certainly you could never quite reach the speed of light. Even as you did so, the equations would alter. Basically, the raindrop would appear to fall slower (time dilation), making the streak more horizontal. If you do the SR correctly, you'll find that as your speed approaches $c$, the streak approaches horizontal. – user10851 Nov 1 '12 at 19:39