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To: dangerdoc
I don’t have the math chops anymore but I would like to see the actual calculation for 1000 yards and less than 2 second flight time.

Well, let's see...

Earth rotates 360° in 24 hours, which is 86,400 seconds. Old-fashioned division gives us 0.0042 degrees/second, or (interestingly) fifteen seconds per second, although the first "second" is arc-seconds, and the second "second" is actual seconds.

If you take my meaning.

Anyway, during a two-second flight time, Earth moves under us 0.0084 degrees. At a distance of 1000 yards, and if you're at the equator, that's a distance of

1000yd*sin(0.0084°) = 0.147 yd

which is a bit more than five inches! 5.27788 inches, to be more precise.

Of course, as pointed out in the article, that "movement" will be "up" if you're aiming due West, and "down" if you're aiming due East. If you're aiming North or South, there's no effect at the equator.

If you're at the North Pole, and you aim in any direction, you're aiming South. In that case, the 5+ inch error accumulates left-to-right, so you will hit a spot 5.27788 inches to the right of your aim point. The direction would be reversed at the South pole, of course.

At any point in between, you've got to bring more sines and cosines into the situation.

42 posted on 11/29/2016 12:23:30 PM PST by Steely Tom ([VOTE FRAUD] == [CIVIL WAR])
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To: Steely Tom

I’d look at it a different way - the earth is about 24,000 miles in circumference, and rotates in 24 hours. So a point at the equator moves at about a thousand miles an hour.

A high velocity bullet will travel between about 2500 and 3200 feet per second, say roughly a mile in two seconds, or roughly a thousand yards in about a second.

1000 mph is 16 mpm or .28 mps, or approx. 1250 feet per second.

The flight of a high speed bullet to a target 1000 yards away is about a second.

In that second the target moves about 1250 feet.

Test: 1250 FPS x 60 sec x 60 min x 24hrs / 5250 feet per mile - about 20,571 miles, or within 20% of the assumed diameter of the earth.


49 posted on 11/29/2016 1:03:58 PM PST by Fido969
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To: Steely Tom

My math skills are corroded by time, but just thinking about it, shooting east/west and vice versa, the rifle, bullet and target are all traveling in the same frame, Applying Newton’s theorums I’m not seeing how the trajectory would be affected.

Shooting north or south, coriolis effect would apply but would be greater over the equator compared to standing on one of the poles. The speed of rotation at the equator is faster than it would be either 1000 meters north or south but the variation would be small and it would take me hours to remember the math to calculate the difference.


50 posted on 11/29/2016 1:06:12 PM PST by dangerdoc ((this space for rent))
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To: Steely Tom
which is a bit more than five inches! 5.27788 inches, to be more precise. Of course, as pointed out in the article, that "movement" will be "up" if you're aiming due West, and "down" if you're aiming due East.

But that movement of 5.7788 inches would be lateral, almost a right angle to an imaginary stationary line radiating at a right angle from the axis of the earth. At the equator that would be a plumb line since it would radiate from the center of the earth. At any other latitude it would not be plumb since it would radiate from the same latitude on the axis.

The apparent "up" or "down" movement would be a lot less than 5.7788 inches as it would be a factor of the curvature of the earth and how much "up" or "down" deviation occurred at 1,000 yards with plus or minus 5.7788 inches of lateral movement along that arc.

The specific arc of the curvature would be different at any other latitude other than the equator as well. But then so would the lateral movement.

54 posted on 11/29/2016 1:28:25 PM PST by TigersEye (Congratulations, President Donald J. Trump! - Let's MAGA!!!)
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