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In External Ballistics – Elevation Part 1, I introduced the concept of error derived from shooting at an angle, and gave you a few practical examples. In this article we will examine the real reason for point of impact shift.


In many texts around the web, you’ll find one common incorrect explanation of this phenomenon: a simplistic approach derived from the formula used to calculate compensation. You may have read that your shot will land high because, when shooting at an angle, gravity has less effect on the bullet trajectory, since the horizontal distance traveled by the bullet it’s shorter. That’s not exactly the case. The amount of drop due to gravity is not a function of distance, but a function of time of flight, which remains constant because the linear distance to the target is the same. Actually, when shooting at an angle, there is a small change in drop due to gravity, which slows down a climbing bullet and accelerates a descending bullet, but its amount is negligible.


The accurate explanation for why the bullet flies high derives from a perspective fact. To fully comprehend this, you need to understand the difference between bullet drop and bullet path. As you read in the terms explanation articles, bullet drop is the distance from the line of departure to the bullet trajectory at a given distance. Drop is measured vertically(as with a plumb-bob), irrespective of the line of departure angle.


Bullet path on the other hand, effectively is where you would see the bullet along its trajectory, through your aligned sights. The distance that we can measure between line of sight and bullet trajectory (so it can be positive or negative), but it’s always measured perpendicular to the line of sight. This difference is the key of the concept, and you can see why analyzing the image below:

As you can see, when you zero for a distance (R) on a level range, you set the line of sight at a certain angle relative to the line of departure (in reality it is generally less than 1°, here it is exaggerated for clarity) to compensate for the bullet drop at that exact distance. In this case, the bullet path at distance R is equal to 0 (zero) because the line of sight intersects the trajectory. In other words, the point of aim corresponds to the point of impact.


When raising the elevation angle by 45° to shoot a target above the firing position, you can see how the angle between line of sight and line of departure remains the same, because of sight regulation. Trajectory also remains the same. What actually changes is bullet drop. Since it is measured vertically, the distance measurement for R falls on the nearest point along the bullet trajectory. That point is the point of the trajectory that you must now aim to be able to hit the target. As you can see, at that specific point, the trajectory is above the line of sight (that is zeroed to a point further along the trajectory) to a degree, which means that when aiming dead center, the shot hits high. Hence, you must adjust your sights to the amount of bullet path to be able to hit the target.


The same concept applies for depressed angles in the same way. There is only a subtle difference in trajectory between uphill and downhill shooting, due to the effect of drag acting respectively downward and upward, causing the bullet to drop more or less quickly. The difference in trajectory is slight—roughly 5in at 1000yds with an elevation or depression angle of 60°.


In the future “how to” section I will show you how to to gauge elevation/depression angle, and how to calculate compensation for it. But in my next article, I’ll focus on the effects of light and mirage, and how they affect point of aim and, consequentially, point of impact.

More: Long Range Shooting: External Ballistics - Elevation Part 2 | The Arms Guide
 

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Thanks for this graphic. I didn't read the text but it was all I needed to understand how the drop is in relation to shooting angles.
 
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