I think the thread starter should clarify certain points in the original question. Because topic is escalating to surface tension or maybe even interference of angels. What does it "get" means? Does it mean "absorb" or "collect" the water (like a spunch), or number of hits that car will receive? Up until this point, more, less, and same rain theories are all considered by different suggestions. However we still could not find an answer for thid speculation.
Well since the number of cars that actually absorb water are few and far between it seems sort of obvious to most people. An answer has been given. Just because you give credence to the wrong solutions also does not make them valid. It gets more water, and the reason for so doing has been explained in detail and at length.
i still say they get the same amount. if the stationary cell is positioned ahead of the stationary car with the wind blowing the rain to produce the slant you suggest be compared to a moving cell raining on a moving car with the same slant with both cell producing the same rain per hour it is impossible for one car to get more rain than the other.
Leopold, you would be right if cars were infinitesimally thin horizontal plates. Cars of course are not thin horizontal plates. They have some height to them, and because of this the effective surface area is greater when the rain falls at an angle on the car versus when it falls straight down. Whether this angled rain fall results because the car is moving, the rainstorm is moving, or both are moving is irrelevant. All that matters is that the relative velocity is not zero.
If you have a look from the beginning of this thread, you will see that I am trying to understand all possible suggestions. But I am not convinced. Plus, you have not started this thread, and how do you know what it meant in the first place. You are probably speculating over an assumption. Check out internet for "Do you get more rain if you run or walk" subject. You will see many different ideas and formulas on the issue. This topic is not something that is agreed by everyone. I am trying to clarify the arguments, I want examples, resources or something in that kind. No body explained anything in length and detail. Just handful of people mentioned the time factor. What I saw were couple of primary school drawings and super scientific statements: "I just explained, you did not understand" and so on, and so for... I am not going to come back to this thread, anybody can believe anything they like, this must be the proper way of finding the truth.
i believe my model is correct except it doesn't account for water splashed and slung by the tires. it seems impossible to include it in my model.
The OP "car" is specific. "rain" is specific. "on it" is specific. No. And again: an incorrect answer is an incorrect answer. It doesn't matter how many people believe 2+2=5, it doesn't make it right. Because the time factor is irrelevant, unless you want to claim that the question meant "stationary for an hour, moving at 60 mph for 5 minutes". And the fact that you dismiss the drawings as "primary school" says a good deal about your understanding of what is necessary to illustrate the particular point being made. Ignoring the correct answer is the "proper way of finding truth"? Way to go...Please Register or Log in to view the hidden image!
Granted I can see that at some point aerodynamics has an influence, but with a standard car surely this is not a substantial factor. Consider the different parts: With increasing speed. The amount on the front would increase greatly. The amount on the roof would be about the same The amount on the rear would be marginally less. The amount on the base would be zero. The amount on the sides would be about the same, perhaps marginally more.
Once rain splatters, it doesn't take a lot of relative wind to carry the resulting droplets around a shape, especially a streamlined one. Where airflow is not turbulent, the motion of the lighter droplets follows the contours of the shape. Inertial separators (the windshield of an open car is one) trip air around an abrupt corner, while precipitation is carried somewhat beyond. At any speed, aerodyamics (and Newtonian reaction) do come into play when we talk about how much rain strikes a particular opening or facet of a vehicle. It's already been fairly well explained what's going on, if we eliminate impact and slipstream effects- and that may be what the OP was vaguely asking. Let's visualize a formation of billions of average-size raindrops, that are all falling straight down at 100mph (we'll say). The velocity-vector onto a motionless vehicle with is 100mph, 90 degrees from the ground plane in front of the vehicle. Now let's get the vehicle moving. As it accelerates, a trigonometric solution expresses the velocity-vector of the rain, with the angle tilting forward and the speed (and frequency of droplet impacts) increasing. Things are getting generally wetter. With the vehicle moving perpendicular to the falling rain at 100mph (for easy visualization) the vector triangle would be an isosceles shape, with equal vertical and horizontal legs in a vector diagram of the relationship. By measuring or calculating the third leg, we have the rain vector- the relative angle and speed at which raindrops actually approach the car in motion. 100mph rainfall and 100mph forward speed results in a rain vector of 45º and 141.4 mph impact speed on an exposed driver's face (ouch!). Slowing our car to 50, the rain vector is 111.8 mph, coming at the car at a 63º angle. The velocity of the rain vector is what best answers the OP, if we wish to ignore aerodynamics and impact interactions. Higher rainspeed (a combined vector of rainspeed and vehicle speed) does mean more rain impacting the vehicle. Just consider a "vehicle" consisting of only a hoop moving along in a horizontal (flat) orientation. The rainspeed, or number of drops passing through the simple hoop in a given time period increases with the hoop's forward speed. Playing around with the trig visualizer linked below, its easier to understand how the vectors combine: Fill in the green boxes at left and bottom of the triangle with values representing rainfall and vehicle speeds, respectively. The resultant vector (sloping part of the triangle) is the angle and speed at which the rain approaches our car (or hoop) in motion. If you experiment with this for a couple of minutes, it will become easy to visualize how more rain passes through our hoop as we move it faster through the rain- even if the plane of the opening is parallel with the direction of motion. Again, this is eliminating the complex flow effects around a moving shape. VisualTrig.com - Trigonometry Angle Calculator Edit: Oh, Really? There is an aspect I have been neglecting to visualize here. It was obvious to others including Oli who contributed explanations here: The cross-section presented to the oncoming rain, if we're strictly interested in the amount of rain entering our hoop example, representing an opening in the top of a car. While trigging out I neglected this basic consideration. As we speed up (and all other dynamics aside) the rain vector slopes forward, the cross-section of our opening is gets smaller, changing from a circle into a narrowing ellipse (from the point of view of the oncoming rain, that is). Although the rain is passing through the hoop faster, the hoop presents a smaller cross-section, the shallower the rain's angle becomes with speed. Do the two cancel each other out? Intuition says yes, and that those here who assert that the vertical component of the rain is a constant are correct: The rain that would enter our flat wire hoop- whether it is stationary, or moving rapidly in a perpendicular direction to the rain, would not change. Muddling through this thought-experiment the challenge seems more in framing the question than in formulating an answer: We need first to isolate what it is that we're after. jmpet asked: "car driving in the rain at 60 miles an hour. Does it get less rain on it, more rain or the same amount of rain if it were standing still?" The answer depends on the vertical and head-on cross-sections of our car. If the car were somehow a two-dimensional shape moving along edgewise, the answer to the question is that the raindrop impacts are unchanged in frequency and volume at any speed- it's the same amount of rain at any vehicle speed. But when we add a vertical dimension to the vehicle, then the answer changes, in proportion to that dimension: Now, the car as a whole does get wetter with speed. How much wetter is not a matter of simple math, especially if the questioner wishes to include other considerations such as how raindrops spatter, how the airflow carries the spattered raindrops around a moving vehicle, how tires interact with a wet road, etc. I often find situations like this an interesting opportunity for clearing up my own thoughts. Vague questions can be great opportunities for thinking through and clarifying things.
If the car is imagined as standing still, and the rain being driven at various angles, then the extra rain hitting the car can be calculated by the increased area of the car's rain shadow - the area of ground not being hit by rain, equal to the horizontal outline of the car plus the shielded ground on the lee end or side.
All the arguments seem to be based on considering the amount of rain hitting the car per unit of time, ignoring the effect of speed on the time spent being rained on. If I am driving 60MPH to a specific destination, my car will be rained on for 25% of the time compared to a car at 15MPH. The problem of being in the rain for 10 minutes @ 60 MPH or @ 15MPH or standing still is a different problem. Obviously for vertical rain, the faster you go the more rain per minute hits the front of the car. I have no intuition for the effect of air flow with a slanted window. BTW: I think that if I had to go from my front door to my neighbor's house across the street, I would not get as wet running as walking slowly in light rain. In a real downpour, I would feel drenched either walking or running & would not notice a difference.
All the arguments seem to be based on considering the amount of rain hitting the car per unit of time, ignoring the effect of speed on the time spent being rained on. Yes, of course, that's correct. Bloody obvious really. Please Register or Log in to view the hidden image! It's amazing how with these problems, what you believe to be the common sense answer often turns out to be incorrect.
Somewhere in internet answers, someone mentioned Vertical velocity and horizontal velocity gets to Hypotenuse of the right triangle etc., is what amount it contributes to the increased rain observed. But I rather agree with Oli's first post. We are viewing the rain in front windshield, that is where most of the air is compacted and more rain contained becomes compacted as well, hence increased rain is observed. Imagine a flat 1/2 inch plywood sheet of about size of a car, move in rain say @60 mph, would collect about the same amount of rain water, with that had it been stationary. because it doesn't have any windshield, and no air to compact in front of it. So it is to do with the shape of the vehicle, how much air does it pushes upwards to compact in front is what matters. Higher the car speed, more air is pushed in front. Similar to the flat plywood, riding in a flat relatively non-aerodynamic City Bus will appear much less increased rain on its windshield, compared to a Sedan or SUV with a snout in its front, moving at same speed right next to that City Bus. Imagine if high winds are consistently blowing in opposit direction, hitting rare of the car, sort of helping sail the car forward, and not resisting, you might experience reduction in rain while driving, compared to had the car been standing stationary. I would appreciate if anybody can come up with a formula for a given rate of horizontal speed of car and vertical speed of ran (just 90 deg, no angels, no weather wind speed) and some kind of unit of air-compression (due to aero dynamicity) what percent would increase rain would be experienced on wind-shield?
I think he just said about what I was. The projected area of a typical car to rain falling straight down goes down with increased velocity so the car impact FEWER drops/sec at speed.
More. When the relative speed of the rain increases, more rain hits the car. (More raindrops per second.) If rain has a normal downward speed of 10mph, then the resulting rain speed will be 61mph. Thus 6x as much rain will hit the car based on speed alone. When the relative angle of the rain increases, more or less rain hits the car depending on the new surface area. This will not increase rain by 6x. Thus more rain, overall, will hit the car.
I doubt very many people would argue that less rain would hit the car if it were standing still in a wind-driven rain. Still less would they tend to argue, I think, that a given car shaped outline on the ground would be hit by less rain if the rain were being blown sideways by a wind as it fell. All of the rainfall per square inch hitting a car sized outline on the ground in a 60 mph wind will hit the car sized horizontal outline of the car itself in a 60mph wind, plus the rain that would have hit the lee ground beside the car in the wind - its rain shadow, where you could sit and stay dry in the windblown rainstorm. The total amount of that rain hitting the car is given by the rate of fall per unit area multiplied by the total time spent being rained on and the total area of intercepted rain - car outline plus rain shadow. The advantage of going fast (creating a relative wind speed) is that it reduces the total time factor in that multiplication.The disadvantage is that it increases the total area (by the rain shadow).
