Older thread, but just found it while killing time awake tonight...

The reason the baseline distance is used, vice the straight line of sight distance is as follows...

If a bullet is fired horizontal to the ground (no slope) gravity acts on the bullet in the straight downward direction at roughly 32ft/sec^2, so for this example, essentially, all the force of gravity is applied to change the bullet's trajectory, and none to changing its velocity. Consider this firing a bullet along the x-axis of a graph.

Now if you fired a bullet vertically, straight up (they Y-axis of a graph) the bullet is experiencing the force of gravity as a change in its velocity (again, at roughly 32ft/sec^2) but having no effect on its trajectory (the bullet is flying straight up, will continue to do so until it uses up all of its velocity, and fall straight back down). There is no "drop" from straight line distance in this plane.

Now think of a bullet fired on a trajectory somewhere between the Y-axis (straight up) and the x-axis (straight out). This will present a combination of the 2 effects, deceleration and change of trajectory. Obviously, the change in trajectory is 0 at one angle (vertical) and maximum at the other angle (horizontal) for a given distance/initial speed. And to make these two forces unify, its some in between value, for a bullet fired at an angle.

Now finally, assume you took a segment along the x-axis = 300 yards, and measured the bullet change in trajectory along that path. Then take the same 300 yard segment along the y-axis, and note there is no change in trajectory from the straight-line path. Now take the same 300 yard segment, at an angle in between the x and y-axis, and the change in trajectory should be between the full value along the x-axis and the zero value along the y-axis.

This is where you grade school geometry comes in... Remember, for the 3 measurements you just made, the straight line distance was constant between them. So if the x-axis is full value of deflection, the angled shot in between has some lesser angle of deflection, and this is where you draw your triangle, from the origin of the graph (0,0). Assume its a 45 degree up angle shot, so draw your line with a slope of 1 (rise over run = 1/1 =1). when you get to your 300 yard segment, draw a straight line down to intersect with your x-axis, and this shorter than 300 yard point on the x-axis is the actual drop distance that gravity affected it, per the above discussion, at a lesser than full drop value.

Hope this helps... its after 5am here and i've been at work all night, 12 hour shift... but it makes perfect sense to me right now as i'm writing it!