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It's embarassing that I can't wrap my head around this at the moment.
The NRA hunter magazine had a quiz in it. It essentially asked if your rifle was zeroed for 300 yards and the target was 300 yards out at a 30 degree uphill, do you aim dead on, low, high or none of the above. I know the answer is low. Just like bow hunting from a stand, aim using the distants from the tree to the animal.
Can anyone explain this in terms of physics, equations, diagrams? Intuitively, it seems as though the vertical component of the velocity vector is larger for the uphill shot. I'm stopping... just somebody explain it pretty plz.
The effective distance is 300 cos(30) yards.
Aim slightly low, or better, use a zero with a Maximum Point Blank Range that will cover the critter you're after. And don't worry about "fighting gravity" in either direction, amazingly, it does not change in magnitude or direction whether traveling up hill
or downhill. It's the same.
The geometry is not the issue. What I'm after is a good physics explaination on "effective distance". The physics involved is what I'm wrestling with.
Gravity pulls straight down. It's maximum affect on drop is observed when the muzzle is pointed perpendicular to gravity––that is when the bore is horizontal. As the bore moves away from horizontal, gravity's affect on drop (gravity times cos(angle)) becomes less. When the muzzle is vertical, angle = 90º, cos (90)=0 and drop goes to zero.
Hold a fishing rod with a weight on the line horizonally. Note that it droops. Now, start to point it up, then down. Notice it droops less the more vertical it becomes––but regardless, it droops the same whether pointed up or down (always LESS than when it's horizontal).
The affect of gravity in the direction of the bullet's flight is negligible, at least in the portion of the trajectory that we're interested in. We're talking about a fraction of 32ft/sec^2 acting on a bullet doing close to 3000fps for times well under a second.