"Horizontal distance" is irrelevant
As was posted in another thread:
Just to be painfully technical, the horizontal distance is irrelevant to the discussion. If you are reading for entertainment, you may want to stop here.
Please bear with me as I explain without the benefit of the necessary diagrams and way too much time elapsed since high school physics and engineering school.
The major error commonly made is stating that the horizontal distance to the target affects the bullet's drop. In fact, the horizontal distance is irrelevant. For our purposes, keeping miscellaneous variables such as atmospheric conditions, etc. equal, there is only one force causing bullets to drop - gravity. The only other variable in amount of drop is the time that gravity has to act upon the bullet in flight, i.e., time of flight (TOF). That is because gravity is a vector, acting vertically downward with an acceleration of about 32.2 feet per second per second.
Here is an example. Suppose a pair of shots, both equal distances from the muzzle, say 100 yards. The first is a perfectly horizontal shot. The second is at the mathematically convenient angle of 45 degrees upward (could be downward as well) from horizontal. The horizontal distance of the second shot is less than the first, measuring only 70.7 yards, but this is irrelevant.
Now here is where one must grasp the physics. Both shots are 100 yard shots; the first is a horizontal 100 yards, and the second is 100 yards at an angle of 45 degrees, but still 100 yards from muzzle to target. Since both bullets travel 100 yards at the same average velocity, the times of flight of both shots are equal. Since the TOF are equal for both shots, gravity has an equal time to act upon both bullets, meaning the drop must be the same, right?
Wrong. Bear with me.
While it is true that gravity has the same time to act upon both bullets, the acceleration of gravity does not act equally on them relevant to the direction of drop. Remember, gravity is a vector. In this example, the bullet is a vector also. Vectors have both a magnitude (in the case of gravity, an acceleration of 32 fps/s) and a direction (vertically downward). Vector mechanics involves both the angles and the magnitudes of the involved vectors.
Now back to the horizontal bullet fired in shot one. By definition, horizontal is perpendicular to gravity. Therefore, gravity acts in a downward direction that happens to be perpendicular to the original path of this particular bullet. Therefore, the drop of the bullet is related directly to the TOF and the effect of the full magnitude of gravity (32 fps/s) because gravity in this case acts in exactly the same direction as drop. Simple concept; everybody understands this.
For the bullet fired in shot two it gets more complicated. You see that bullet has a vector direction of 45 degrees to horizontal, and to vertical. While gravity still acts vertically upon the bullet, the bullet is not traveling horizontally or perpendicular to gravity, so gravity no longer acts perpendicularly to the original path of the bullet, but at the 45 degree angle. (This is easier with a diagram.) In other words, gravity pulls the bullet vertically downward, but vertically downward in this case is not the same as drop. Drop, as I understand the definition, is the amount the bullet falls perpendicularly (but not necessarily vertically) away from the line of the original path (barrel). Since the barrel in this case is not horizontal, drop is not vertical.
Now let's do the math. Picture shot two, upward at an angle of 45 degrees from horizontal, with a muzzle-to-target distance of 100 yards.
Gravity is acting at an angle of 45 degrees from the path of the bullet, rather than perpendicular to it. The component of gravity that is acting in the direction of drop, or perpendicular to the path of the bullet, is described by the equation: (Gravity) x (sine 45 degrees). You will notice that due to the angle this amount is only 0.707 times the horizontal shot's gravity effect, or about 23 fps/s. The other component of gravity is acting perpendicularly to the first. It acts exactly parallel, but opposite (for an uphill shot) in direction to the path of the bullet and the magnitude is described by the equation: (Gravity) x (cosine 45 degrees). This component can be legitimately ignored as insignificant relative to the velocity and TOF of the bullet.
Just for fun, let's consider a third shot. Vertical. Up or down, I don't care. There is zero drop; the bullet stays on its original path with no drop from gravity, because all of gravity is acting exactly parallel to the path of the bullet, and not at any angle to pull the bullet off that path.
In summary, the drop depends upon TOF and angle of bullet path from vertical (gravity). For two equidistant shots with the same load, the TOF are equal, leaving only the different angles to determine the difference in drop.
I hope this has clarified rather than confused.
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