I've noticed that, when I'm driving and it's raining, the faster I go the more rain I get on the windshield and the faster I have to run the wipers to compensate. When I'm stopped at a red light I can just about turn the wipers off, but once we get moving again I have to turn them up so I can see clearly.

Intuitively I would think that, assuming a statistically even distribution of rainfall throughout the local area, speed should not matter because for every raindrop I move into the path of, I'm moving out from under another one at the same time. But this is definitely not what I observe in actual driving conditions, so what's going on?

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    $\begingroup$ You are comparing total volume to volume flow. In going from A to B, the total volume of water remains unchanged no matter what your average speed is because you cover a given distance and the total volume of water in the air along your path should be fairly constant. But if you cover that distance faster, then more water must hit your windshield every second in order to allow the total amount of water to be the same. $\endgroup$ – Jim Jun 3 '15 at 15:26
  • $\begingroup$ @JimtheEnchanter that's basically my answer too $\endgroup$ – innisfree Jun 3 '15 at 15:28
  • $\begingroup$ Related: physics.stackexchange.com/q/19499/2451 $\endgroup$ – Qmechanic Jun 3 '15 at 15:28
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    $\begingroup$ This reminds me of the question: Is it better to walk or run in order to avoid rain? $\endgroup$ – Harshal Gajjar Aug 9 '15 at 14:44

Imagine a box where each side has an area $A$, and suppose that it contains a continuous flow of rain that falls vertically at a velocity $u$. The density of raindrops is $\rho$ (that's the number of drops per unit volume).

enter image description here

The rate at which the drops hit the bottom panel will be $\rho Au$ and the rate they hit the vertical (left) panel will be $\rho Av$, giving a total rate

$$ R=\rho A(u+v) $$

In the case of a car, you have a single windshield of total area $A$ at an angle $\theta$. Its horizontal cross-section is $A\cos\theta$ and its vertical cross-section is $A\sin\theta$. So the total rate will be

$$ R=\rho A(u\cos\theta + v\sin\theta) $$

where $v$ is the speed of your car.

Edit. The above comments that claim the same volume of rain hits your car for a given distance, irrespective of speed, are misleading. The longer you travel for, the more rain falls on your windshield. But the volume that you pick-up from moving is indeed only dependent on total distance travelled:

\begin{align} N&=\int_0^T R(t)\,dt \\ &=\rho A \left(uT\cos\theta + \sin\theta \int_0^T v(t)\,dt\right) \\ &=\rho A(uT\cos\theta + d\sin\theta) \end{align}

  • $\begingroup$ the comments are not wrong - they are simply discussing a simplified case in which the screen is vertical such that $\cos\theta=0$ and $\sin\theta=1$. It's perfectly clear that the slower you go, the more water will hit the car's roof etc. $\endgroup$ – innisfree Jun 3 '15 at 15:45

This is variation on a classic problem - if it is raining, will I get less wet if I run rather than walk home? In this case, however, we're discussing a car driving through rain (and we're not bothered about rain falling on the roof of the car). Let me make a simplification by supposing that your windscreen is vertical and that the rain is coming straight down (there is no wind).

The total amount of water water that hits your windscreen in a given journey is independent of your speed. Think of the rain water as a soft, solid object, like a big block of soft cheese. The amount of cheese you displace when you drive through it is independent of how fast you drive through it.

However, the rate at which the cheese hits your wind-screen is proportional to the speed at which you're driving through it. Going back to rain water, the faster you go, the faster water accumulates on your screen, and the faster your wipers have to work to remove it. For a given length of journey, though, the total amount of water hitting your windscreen is independent of your speed.

I encourage you to think of variations on this problem - what if your windscreen isn't totally vertical? What if the rain is coming down at an angle? Should you run home in the rain?

  • $\begingroup$ Well, seeing as how I drive a car and not a bus, my windshield isn't totally vertical; it slopes back at a significant angle. Does that have a significant effect on the math involved? $\endgroup$ – Mason Wheeler Jun 3 '15 at 15:19
  • $\begingroup$ Well I think you are only partly correct. The larger factor is the direction change in the rain than the speed. Look at my second point in the answer below. $\endgroup$ – gautam1168 Jun 3 '15 at 15:21
  • $\begingroup$ @gautam1168, sure, in that discussion I assume that the rain is coming straight down. I appreciate that a horizontal component of the rains velocity slightly complicates things. $\endgroup$ – innisfree Jun 3 '15 at 15:30
  • $\begingroup$ No, gautam has a point. A paper thin sheet parallel to the ground has virtually no volume of water along its path as it moves, but if it goes slowly, it'll get soaked. So clearly, the flux through the surface over the travel time plays an equal part in this as does the speed at which you cover the volume in your path $\endgroup$ – Jim Jun 3 '15 at 15:35
  • $\begingroup$ @Jim not sure I follow - is your paper getting soaked because rain's hitting the top of it? in this situation, that's rain hitting the car's roof, isn't it? (i said that the wind-screen was vertical) $\endgroup$ – innisfree Jun 3 '15 at 15:39

Well speed would matter. But really there are two things here.

  1. When you drive then in your reference frame the rain falls faster and at an angle. Since it falls faster you collect more of it.

  2. Since it falls at an angle that is directed at your windshield, the volume of rain it can collect increases. If you are initiated in maths : you just need to project the area of the windshield perpendicular to the rain to see why this happens. If you are not: then imagine yourself holding a pane of glass vertically in a rain. It will barely collect anything. Now if you hold it parallel to the ground... See? And when you drive the rain hits the shield more perpendicularly than when you don't.


I will give it a shot (actually it is my first seriously answer with math, ha! Sorry for my lousy English!).



$\sigma$ is the density of rain.


  1. your front windshield is vertical.
  2. the raining is uniform at the scale of the area of your windshield.
  3. the raining goes vertically.

such that when your car is at rest, the rain that your windshield catches is $(\vec{J}\cdot \vec{a})t=0\cdot t=0 \ldots (1) $

whereas $a$ is your windsheild's area; the direction is pointing out same as your car's velocity.

let your car's speed is $v_\text{car}$; To make things simpler: considering in the frame of you sitting in the car. additionally assuming:

  1. Your car is at constant speed.

then by relative motion, the rain drops your car's windshield will get is:

$(\vec{J'}\cdot \vec{a})t=\sigma(\vec{v}_\text{rain}-\vec{v}_\text{car})\cdot \vec{a}t$

by (1)

$=-\sigma \vec{v}_\text{car}\cdot \vec{a}t$

given the assumption (1.) and (3.) , then it becomes:

$=-(\sigma at) v_\text{car} $

So it is exactly proportional your car's speed!!! That's why you got that feeling. but wait a second.

$=-(\sigma at) \frac{L}{T} =-\sigma aL $

It is independent now! Why? Because higher speed means faster to get to your destination. As it turns out, these two factors cancel each other exactly like I have just showed.

But wait a second..... again.

The rain drops will NOT stay on your windshield till the cows come home. As it turns out, you are RIGHT!! You drive faster, the windshield will get "more rain drops", because the rain drops don't get enough time to slip down!

considering the extra reality, such as wind, and the degree of decline of windshield, will have some effect, but that doesn't change the physics I have just typed, but adding details.

Say, if there is wind, then its effect on the rain drop's velocity horizontal part survive and add up as:

$$(\vec{J'}\cdot \vec{a})t=\sigma(\vec{v'}_{rain}+\vec{v}_\text{car})\cdot \vec{a}t=\sigma(\vec{v}_\text{rain}+\vec{v}_\text{wh}-\vec{v}_\text{car})\cdot \vec{a}t =\sigma\vec{v}_{wh}\cdot \vec{a}t-\sigma aL $$

$\vec{v}_\text{wh}\cdot \vec{a}<0$ means your windshield gets more rain drops if there is harder wind against your car's motion. Meanwhile if $\vec{v}_\text{wh}\cdot \vec{a} >0$ the if your car goes with the same direction with the wind, your windshield will get less rain drops.


It is pretty simple.

The faster you move, the larger the total volume your windshield will pass through per unit time. You can combine your speed with the speed that the raindrops fall. The larger the volume, the more rain that you windshield passes through.

The combination of your speeds combine as vectors for a total effective velocity.

The perpendicular cross section of the area your windshield to this effective velocity give the cross sectional area your windshield cuts through. So the total volume per unit time is simply the perpendicular cross sectional area times the distance covered at your effective velocity.

Then using the density of the rain per unit volume give you how much rain will hit your windshield per unit time.

Clearly this volume increases with increased speed because your effective speed always increases monotonically with your speed. The cross section maximizes when the effective speed is perpendicular to the slant of the windshield and will lessen a little as your speed becomes much greater than the rain speed. But generally the total number of raindrops tend to increase even though the cross section lessens a bit as the effective speed tends toward horizontal at high speeds, but the increase in speed overcomes the relatively slight lessening.


Big raindrops fall at 10 m/s. Let's say you drive a big car, with a windscreen 2 m high. Your car is also very fast, and you like driving in circles. So much so in fact that you do it at 5 rounds per second (I know, just assume it's possible). You also like rain, so you removed your windscreen, capturing all the rain that would otherwise hit it. At 5 rounds per second, the rain can only fall the length of your absent windscreen, so you catch all the drops, keeping the ground under you perfectly dry. The faster you drive, the larger you can make the circle while still doing 5 rounds per second. Obviously, the larger the circle, the greater the surface area of the annulus you are keeping dry. So you must be catching more rain when driving faster, no?


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