Speed of a falling rain drop
- Thread starter wegs
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Breezes, wind, atmosphere, all are resistant to a tiny raindrop. So resistant that the drop may even be blown/forced upwards, gathering more moisture as the further up it is blown...then the lack of heat sees that drop starting to freeze, and depending on how much resistance it is feeling, can be continually blown upwards until so heavy that it falls as hail!
Seattle
Valued Senior Member
Air is dense, especially where there is rain, and a rain drop doesn't have a lot of mass.
Think of dust storms. The particles remain suspended for a long time.
Now come in from the rain, you're getting all wet.
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DaveC426913
Valued Senior Member
1.
A penny is much denser than water - about 8 times more dense. So for a given surface area (which is what creates drag), it will fall much faster.
2.
A rain drop is (mostly) spherical, so it always presents a maximum cross-section during its entire fall. A penny - even if tumbling - presents, on average, a much smaller surface area to the air, creating less drag.
In fact, it's worse than that. A raindrop gets distorted slightly, into a pseudo disc. But, because it's the air making it that shape, the droplet will always have its broadest cross section perpendicular to the air's movement. A rain drop cannot tumble to a more streamlined orientation like a penny can.
It would be as if you could drop a penny - but manage to keep it flat all the way down.
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Janus58
Valued Senior Member
Even a Penny dropped from a high altitude reaches terminal velocity between 30-50 mph.
To make things simple by assuming the same shape and size, something made of copper would have roughly 9 times the mass of something made of water. With all other things being equal, the terminal velocity goes up by the square-root of the mass.
Ergo, under such a scenario, a copper sphere would have a terminal velocity 3 times that of an equally sized water sphere.
A rough calculation gives ~ 75 mph for a 0.5 cm diameter copper sphere, which gives ~25 mph for the same size sphere of water. But as pointed out, the water drop would not keep a spherical shape, which in turn would increase its drag coefficient, decreasing its terminal velocity.
A smaller drop would have a lower terminal velocity. Terminal velocity is inversely proportional to the square-root of the cross section area. a .025 cm diameter sphere would have 1/4 the cross-section area, but only 1/8 the mass than the 0.5 cm sphere.
exchemist
Valued Senior Member
Added to which, 0.5cm is pretty huge for a raindrop. So the speed will generally be lower than this calculation suggests.
I have experienced being in a hailstorm with ~ 0.5cm hail and it is slightly painful.
I knew I could count on you guys to help me figure this out. Thanks!
So, from what I'm understanding now, air resistance strength/force (that happens during the fall of the rain drop or any object) balances out gravity, so this is why acceleration doesn't happen from the start to when the object (rain) hits the ground? (perpetually and continuously)
So, from what I'm understanding now, air resistance strength/force (that happens during the fall of the rain drop or any object) balances out gravity, so this is why acceleration doesn't happen from the start to when the object (rain) hits the ground? (perpetually and continuously)
Janus58
Valued Senior Member
Yes. Drag due to air resistance increases with speed. So at the onset, the object accelerates at g. But, as it picks up speed, the induced drag acts to counter-acts gravity, and the acceleration decreases. Eventually, the drag equals the force from gravity and the object no longer accelerates. The approach to terminal speed is asymptotic. For example, for a particular body it might take 3 seconds to reach 50% of terminal velocity, another 8 sec to reach 90%, and another 7 sec to reach 99%, etc. So terminal speed is teshnically more of a limit to how fast the object can fall than it is the actual speed at which it falls.
Out of curiosity what is the formula used to figure this out with accuracy? Is there a math equation?
DaveC426913
Valued Senior Member
There is, but it's not a simple relation. As we've pointed out, it's dependent on on shape and orientation, which is quite complex.
Depends on the droplet size. Small droplets fall slower. But above a certain size (about 4-5mm) the droplet usually breaks apart from chaotic drag forces on the drop. At their biggest they fall at about 20mph. Drizzle falls at about 5mph. Some droplets (mist, fog) never fall of course.
We (skydivers) used to joke that hitting rain in freefall hurts so much because we're hitting the pointy end of the drop. But of course it's just that we are hitting a 20mph drop with our 120mph bodies, so they have a closing speed of 100mph. And even a 3mm drop hurts at those speeds.
(Side note - I used to think that I was hitting frozen rain, or sleet, when I went through rain because it hurt so much. Then one day I actually went through sleet. Afterwards you could see where they hit me by the bruises - big ones under my chin where my skin was exposed, smaller ones under my jumpsuit, and even smaller ones under my T-shirt where there were two layers of protection.)
You could use Doppler radar to figure out their average speed.
DaveC426913
Valued Senior Member
Of course you can.
Any high-speed camera (and it doesn't have to be very high-speed) pointing at a flat background will show speed of passage.
Omg! Never would have considered pain (from rain lol) to be a part of a sky diver’s adventure!
The weight of an object is mg, where m is its mass and g is the acceleration due to gravity.
The drag force on an object is $\frac{1}{2}C\rho Av^2$, where $\rho$ is the density of the air (or other fluid) it is moving in, $A$ is its cross-sectional area (the effective area it presents to the fluid), $v$ is its speed and $C$ is the "drag coefficient" (a numerical factor that depends in part of the specific shape of the object).
The terminal velocity happens when the drag force on a falling object equals its weight, so the terminal velocity is approximately:
$v_{t} = \sqrt{\frac{2mg}{C\rho A}}$.
If you're interested, you can look up some typical values for raindrops, plug and play.