There's not nearly as much water in the air during a rain storm as some folks seem to believe.
Let's just use a simple thought process as an example. It may not be perfectly scientifically, mathematically correct but it's close enough to give you an idea.
Let's say it's raining at a very heavy 5 inches per hour. This is at least 20% higher than, and more likely 25x higher than, an average rain fall.
To get a really good number and put everybody to sleep, we'd have to use some fun calculus (Δt → 0) and other fun stuff.
I'm going to guesstiproximate instead.
Let's go with a 180gr .308 bullet at 2,600 fps MV. That bullet (according to JBM Ballistics) would take about 1.9 seconds to get to 1,000 yards. Let's call it 2 seconds for easy math.
Let's do some simple extrapolation,
A 5"/hour rain fall literally means that it takes one hour to fill a space with 5" water. There are 3,600 seconds in an hour. The bullet is in flight to 1,000 yards for 2 seconds. During that 2 seconds, only 2/3600ths of the total 1 hour rain will fall. That means that during the entire bullet flight time, the total amount of water in the air is 5 inches x 2 divided by 3,600 seconds. That's 1/360th of an inch (~0.0027777) of rain in that time. That amount is broken up evenly throughout the entire space.
Let's consider the space taken up by the bullet. Let's make it a 1,000 yard cylinder, 0.308 inches diameter. We have a cylinder 3,000 feet (36,000 inches) long with a .308" diameter. The total volume of that cylinder is 2,682 cubic inches. An average raindrop might be about 0.078 inches in diameter. That's a volume of ~0.0019877 in3.
In that 2 second flight time, we're only filling that entire cylinder with 1/360th of an inch of rain. That's a total volume of rain within that cylinder during time of flight of 0.000206955 in3.
Look at those two numbers, total volume and rain drop volume.... there's more zeros in the total volume than there are in a single rain drop!
That means that in the entire flight of the bullet, there will be less than one rain drop in it's entire path... ahead AND behind it!
Not just "less than one", a single drop of rain is almost 10 times the total volume of water in that column during the entire bullets flight!
You tell me. There's essentially 1 drop of water SOMEWHERE in the flight path of every 10th bullet. That's in front and behind. We could sort of almost consider that it would take 20 shots to get a single drop of rain to be somewhere in front of the bullet during it's time of flight.
You wanna take the bet that you're going to hit one?
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Still happily answering to the callsign Peetza.

The problem, as you so eloquently put it, is choice.
The Architect

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