Earth rotation formula?

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photog_shooter

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Tried doing a quick search but didn't find anything. I've seen a lot of discussions regarding bullet drop and wind compensation but nothing about the rotational effects of the earth. The earth spins at 1670 kph (according to NASA), so if you're taking these super long shots, then wouldn't the earth rotation be a factor as well? Is there some formula to help calculate this rotation? Like earth rotation x distance to target / bullet speed or something (half asleep right now so the x and / might be off)?
 
Doesn't the bullet already have momentum in the direction of the earth's rotation when it exits the barrel?
 
Its my understanding that the calculation for the earths rotation only matters at ranges artillery pieces shoot at. Its completely negligible for small arms IIRC.
 
Andy said:
Doesn't the bullet already have momentum in the direction of the earth's rotation when it exits the barrel?
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Kind of, if you shoot straight east or west. If you shoot north or south, then the bullet gets left behind. If you ever had the chance to ride one of those platform style kids playground merry go rounds, and tried to walk a straight line from the centre out to the edge you will have felt the coriolis effect.
 
Sniper_485 said:
Its my understanding that the calculation for the earths rotation only matters at ranges artillery pieces shoot at. Its completely negligible for small arms IIRC.
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I'm sure that it would have more of an effect on artillery, but wouldn't it still be of some importance to big toys like the 50bmg where the round could potentially travel in excess of 5-6 seconds?
 
photog_shooter said:
Tried doing a quick search but didn't find anything. I've seen a lot of discussions regarding bullet drop and wind compensation but nothing about the rotational effects of the earth. The earth spins at 1670 kph (according to NASA), so if you're taking these super long shots, then wouldn't the earth rotation be a factor as well? Is there some formula to help calculate this rotation? Like earth rotation x distance to target / bullet speed or something (half asleep right now so the x and / might be off)?
Click to expand...

I'm not aware of a formula as I let the ballistics program figure that out - I use a kestrel. You'll probably be able to notice it yourself if you were able to set up a couple targets north, east, south and west of your shooting spot from about 1000 yards and out - bullet flight time dependent.

If shooting west the earth is rotating towards you and since it's a sphere your target wil "rise" from your aiming point and if you had used dope from a north or south zero you'll hit low. Opposite for shooting east as your target is dropping away from you will hit high. Shooting north you'll hit left and south right. That's for coriolis effect and then there's spin drift too...

Sure I'll be corrected on something but I believe that's the theory anyways......
 
Here's a tidbit on the subject, with a link for anyone who wishes to read further:

"Despite being associated with Coriolis, the phenomenon that actually affects the vertical component of the trajectory is called Eötvös Effect. The rotation of the Earth generates a centrifugal force, the same that pushes you to the side when you make a sharp turn with your car. This force acts perpendicular to the Earth rotatory axis, adding or subtracting to the gravity force. When an object flies eastward, in the same direction of Earth’s rotation, centrifugal force acts opposite of gravity, pushing it away from the Earth’s surface. If the object flies westward, in the opposite direction of the Earth rotation, centrifugal force pushes the object toward the ground concurrently to gravity force. Thus, bullets fired to the east always fly a little higher, and, conversely, bullets fired to the west always travel somewhat low." h t t p s://thearmsguide.com/5329/external-ballistics-the-coriolis-effect-6-theory-section/
 
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