1. Web Hosting Master
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## Measuring Rain Density

Okay, this is totally weird and insane, I'll admit.

I started wondering the other day about how much rain there was. You can easily measure rainfall, but what I wanted to know was... When it's raining, if you took a square foot of air, how much would be raindrops, and how much would be water?

I'd assume this number would change a bit (drizzle vs. downpour), but I still don't have the foggiest clue as to what range the numbers would be in.

Does anyone know this? Has someone else actaully wasted their time before measuring this?

I'm thinking of constructing a big box. The top and bottom would be 'open,' but could be slammed shut. (Provided that the top and bottom are always in teh same position as each other, you don't have to close the box too quickly to get an accurate reading?) You could then empty the contents of the box and see how much water there was, and then compare that with the size of the box to get a number.

Am I nut? Has anyone else wondered this? Is this a commonly-known number that I don't need to waste my time measuring?

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## Re: Measuring Rain Density

Originally posted by fog

Am I nut?
In short, yes. In long, yes. In anything in between, yes.
But the again we are all a little crazy
I've wondered similiar things before and i'm pretty sure there isn't a scientific formulae for it!

Jord

3. rogue element
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## Re: Measuring Rain Density

Originally posted by fog
...if you took a square foot of air...
A square foot of air doesn't contain anything.

The density of rain is 1.

Perhaps you wish to measure the volume of rain in a cubic foot of air?

Sorry, I can't help being a smart***!

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Hehe, you're right. I knew that, too.

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A square foot of air doesn't contain anything.
Don't you just hate smart people - especially when they are right

Since it has already been established that the terminal velocity of raindrops varies between 20mph and 0.02mph ( http://www.grow.arizona.edu/Grow--Gr...ResourceId=146 ) all you need to do is take a box of a known size (preferably a perfect cube as this will make calculation easier) judge the size of the rain droplets and establish their terminal velocity then let the box collect rain for a measured period of time.

By dividing the box height into the terminal velocity you can calculate the time a raindrop would take to enter the box and land in its base so, measuring the amount of water collected and again dividing by the measured period of time and then mutiplying by the calculated period to travel the depth of the box you can arrive at the amount of water in any cubic area at any one moment.

.......I think lol

6. rogue element
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Originally posted by psihost
...judge the size of the rain droplets and establish their terminal velocity then let the box collect rain for a measured period of time.
Well, as long as I'm hated... Terminal velocity is 120mph, ignoring wind resistance. Mass doesn't factor into the equation -- heavier skydivers don't fall any faster than lighter ones.

Thus, two objects with the same wind resistance will fall at the same speed. Since a raindrop is, by definition, the most aerodynamic form possible, any two raindrops share the same value for their aerodynamic coefficient -- zero. Compare to the average passenger car, at around 0.4.

7. rogue element
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There are ≈7.5 gallons in a cubic foot of water.

Two inches of rain per hour is about as hard as it rains, except in extremely rare cases. Imagine a 12" per side, cube-shaped glass box with no top as a rain gauge.

That two inches of rain is 1.25 gallons of water, per 120 minutes. That's about a tenth of a gallon per minute falling in our box.

I don't think you can ask how much water is in a cubic foot of space during any particular instant, I think you have to specify a timeframe. My reasoning on this, is it rains at different intensities.

So my answer is, in a hard storm it's possible to have .016 gallons of water falling through a cubic-foot space per second.

I guess, divide that by the velocity of the raindrops to get the real answer to the OP's query.

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In Chicago it is "0", NOTHING, NADA, NILCH.

We have not had rain in so long I forgot what it looks like.

"Does anyone know this? Has someone else actaully wasted their time before measuring this? "

I would love to waste some time if we had a drop :-(

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Originally posted by BigBison
So my answer is, in a hard storm it's possible to have .016 gallons of water falling through a cubic-foot space per second.
As you said, though, that's not quite instantaneous. I'm still sold on my slammable box idea, which would be near-instantaneous.

Isn't that interesting, though? .016 gallons of water out of about 7.5 gallons. That's 0.2133% rain.

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Originally posted by BigBison
Well, as long as I'm hated... Terminal velocity is 120mph, ignoring wind resistance. Mass doesn't factor into the equation -- heavier skydivers don't fall any faster than lighter ones.

Thus, two objects with the same wind resistance will fall at the same speed. Since a raindrop is, by definition, the most aerodynamic form possible, any two raindrops share the same value for their aerodynamic coefficient -- zero. Compare to the average passenger car, at around 0.4.
Terminal velocity is when the resistance on an object becomes equal to the object's mass, so the mass of an object is indeed a factor. A Styrofoam ball and a lead ball both of equal size would have a very different terminal velocity.

11. rogue element
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Originally posted by The Islander
Terminal velocity is when the resistance on an object becomes equal to the object's mass, so the mass of an object is indeed a factor. A Styrofoam ball and a lead ball both of equal size would have a very different terminal velocity.
Not in a vacuum. You are right, though, weight is a factor when wind resistance is considered:

http://en.wikipedia.org/wiki/Terminal_velocity

So a bigger raindrop, falls at the same speed of a smaller raindrop. The increased mass is offset by the increased wind resistance.

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