Kill target is a standing torso: range distance at which a sniper must factor in Earth curvature?

See subject. Clearly height above or below target is part of the question.

Thanks
Leo

No, not a need answer fast.

Never. If he can see the target in his scope, he already has a straight line between his gun and the target irrespective of the geometry of the earth. Earth’s gravity and rotation, he does have to take into account, though.

For earths rotation effects

Any long range gunnery like artillery and naval guns factor this into their aim as flight times can be 30+ seconds. With rifles, the rotation of the earth is measured in thousands of an inch over even long rifle ranges. Wind is a bigger deal.

What a great topic. From the cite:

Bullets at high velocity through the atmosphere
Because of the rotation of the earth in relationship to ballistics, the bullet does not fly straight although it may seem like it from the shooter’s perspective. The Coriolis effect changes the trajectory of the bullet slightly to give the path of the projectile a more arched shape. This situation only occurs at extremely long distances and therefore, is used to calculate a perfect shot by today’s trained snipers.

Does the OP mean having to take into account that a straight line of sight is cutting a chord across the arc that would mark a constant height?

Given the OP mentioned the height of the shooter is a factor, I think they’re asking about a target so far away that they dip “below the horizon”.

If so the relevant formulae can be found here but in short I don’t think it’s a factor for any rifle shooting (a rule of thumb from the article is that a standing person of average height can see 3 miles to the horizon, the very longest confirmed sniper kills are at half that distance or less).