Meh, I wouldn't have thought so. You have constant rainfall all round so it's going to be the same on top whatever speed you're doing. All you need is the frontal area of the car, the rate ("density") of rainfall), and the speed (in fact any speed will do, simply moving forward at all will show that the encounter rate increases). I work/ think better graphically than numerically, give me a short while and I'll see if I can sketch it out.
Just for you Please Register or Log in to view the hidden image! Please Register or Log in to view the hidden image!
Awww, thank you, Oli!! Please Register or Log in to view the hidden image! So it's the front area of the car that add up to the contact area of the car! Btw, what makes the rainfalls direction becomes diagonal when it is in contact with moving object? I notice that, too! Is it the same thing like when I am in a train station and then a train passes me, and then my hair somewhat blows approaching the train? (I hope you understand my English Please Register or Log in to view the hidden image!). Thanks again Please Register or Log in to view the hidden image!
Yep, it's the frontal area. Because it's relative - if the car is moving into vertical rain (easier if you think of it vertically first of all) then that movement, relative to the rain means that it's the same as the car being stationary and the rain "moving", i.e. the angle of descent alters. If the rain is already at an angle then that angle will increase as the car's speed increases. PS thanks for the kitten Please Register or Log in to view the hidden image!
Umm... Please Register or Log in to view the hidden image! This time I'm lost Please Register or Log in to view the hidden image! Also, I have to go shopping first, I'll think about it on the way, merci beaucoup!
With a stationary car and vertical rain; at any given time the distance (horizontally) between the raindrop and car is always the same. For example consider a rain drop 1 metre in front of the car, it's always 1 metre in front of the car. If the car is doing 1 metre per sec (2 mph) then at t=0 it's 1 metre away, at t=1 it's in contact (given sufficient height of car, of course). At 60 mph the car travels nearly 27 metres per second (or the rain travels toward the car - it's relative movement that counts - at 27 ms[sup]-1[/sup]). But rain tends to come in sequences of drops, i.e. consider each drop as a "beam" of water (it isn't, but it might as well be for the purposes of this exercise Please Register or Log in to view the hidden image!).
The same number of oranges are falling on the horizontal surfaces of the car, regardless of whether it's moving or not. To calculate the amount of oranges that hit the vertical surfaces of the car while moving you need the orange-rain density.
i say the same amount will hit the car. i can't see how a bucket pulled along the ground will catch more rain than a stationary bucket. this assumes a consistent rainfall.
The silhouette perpendicular to the direction of the rain is reduced when the car is moving forward. If you look at Oli's diagram above, the top of the rectangle is presenting a smaller target to the slanted rain. Think sine of the impact angle as measured from the car's roof. If the rain were more and more slanted, less and less would hit the top. The top of a car is larger than the front, and changes to the effective top area make a bigger difference. Also, what is gained in the front is lost in the back depending on the actual shape of the two ends. One way might be to calculate the volume of the space which contains the raindrops that will hit the car in a given time - when the car is stopped, a rectangular box extending straight up toward (or past) the clouds above. As the car begins to move, the box tilts, becoming both narrower and longer (to reach the same height). The degree of tilt depends on the relative speeds of the rain and the car.
I have noticed that if it's raining lightly and I'm driving in my convertable with the top down, little or no rain hits me in the front seat. I assume the air flow over the windshield deflects it. Now if it really starts to rain, the effect is lost. But there have been plenty of times I've been driving with the top down and found myself turning on the windshield wipers to clear the windshield despite the fact that I wasn't getting wet.
How is the problem so trivial? The diagram doesn't really help since you can't just model the car as a rectangle. I could easily model it as a triangle that gets hit by less rain. After a quick google, rain falls about 15mph. Obviously, when stationary, only the area of my car that can be seen from above is being hit by rain. Say I drive at 20mph. The rain will effectively hit me at 25mph at an angle of tan(3/4), ~53 degrees from the horizontal. So, when moving, I get hit by 25/15 = 5/3 times the amount of rain per unit area of my car that is being hit by the rain, which in this case is the area of my car when viewed from ~53 degrees. So if the area of my car when viewed from 53 degrees is <3/5 the area of the car when viewed from above then I will be hit by less rain when moving at 20mph. I don't think any cars fullfil that criteria but some must be fairly close.
the answer must be the same amount. instead of the car moving we can use a stationary raincell and a moving raincell instead. if both raincells produce the same amount of rain an hour then how could they be different?
The rectangle illustrates the answer: if it makes it easier turn it onto its side. Check again (or check your wording) if viewed at an angle the area will be LARGER. Moving = more rain. Please Register or Log in to view the hidden image!
Not necessarily - at least it looks like, for some angle range, less rain. It's the area of the rhomboid for the moving car compared with the area of the rectangle that represents stationary that we compare - the vertical height represents time, and the area represents a volume of raindrop filled air - the area in which the raindrops are, that will eventually hit the car. What is happening is that the aspect of the upper surface of the car is getting smaller as the car speeds up. The aspect of the front is getting larger, but it's a smaller area - in some range of speeds you are going to be losing more off the top than you gain in front, from the looks of things. If the car is imagined going very fast, only a thin sliver of rain volume would be set up to hit the car on top.
Okay. Let us change the question: Car always remains stationary. Not moving at all. It will stay under two different rainfall scenarios for, let's say 10 minutes. And it will be subject to same amount of rainfall per square mile. You know they calculate and we read on papers: Yearly rainfall per square mile there are country statistics, city, monthly, summer, winter blah blah statistics. We do not really know how do they really calculate it, but we believe in them, because we guess that they can do that. Scienceman, better than a Superman you know... We are looking for the effect of the speed. We need to keep other variables or inputs fixed (such as size of the car, time and amount of rainfall for per square mile - I like that expression-) so we can differentiate and see the effect of speed. How much time or how much rain per square mile are not important as long as same amounts are applied to both of these situations: One is without any help of wind. So drops are falling directly or near directly for 10 minutes. Second one is a thunder situation: Rain is falling with the help of 60 miles per hour rain. Angles, speed, caos and horror; everything. But again for 10 min. So the question is, under which situation (windy or not windy) our experimental subject car gets more rain?
Under the conditions you describe, you have specified that the car will receive more rain in the wind. That's the "rate per square mile" calculated to the fraction of the square mile that the car shadows the ground, and the wind makes the shadow bigger on the lee side than the car is. I'm beginning to feel persuaded, intuitively.
