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Posted: 6/5/2010 11:54:29 AM EDT
At the rifle range that I frequent 50 yard targets are approximately 6 feet below the shooting stations. If I’m not clear imagine yourself at the stadium sitting in the 4th or 5th row and shooting at something in the center of the playing field. My question is : how the zero obtained under these conditions will differ from “regular” zero ? Please don't offer to visit different range or tell me that real men don't shoot bench rested or at something closer than 1 mile.
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| This is sort of a guess, but I was picturing the range as you posted. Being that on level ground the trajectory of the bullet includes full effect of gravity I would imagine that shooting downward the effect would be slightly less. I really don't think you will notice much difference. With a military zero of 25 meters your point of aim vs. impact at 50yds would only be like 1.5" high. The slightly downward angle of your range I would predict the result to be around 1.6" or so....hardly noticeable. Typically when shooting at less than 90deg (horizontal) the bullet has reduced effects of flight time, gravity, etc... so it will tend to be a tad high. You should experiement with a level shooting area and see how it plays out vs your range. I would bet the difference is nominal at only 50yds. and much more pronounced at extended range. |
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That's basically it. When shooting up or down at extreme angles, the rifle will shoot higher than expected. In essence, it is only the horizontal vector that matters even though the actual slant range is longer.
But it is really only an issue at fairly large angles and over fairly long distances. In this case with a 6 ft drop (the opposite side of the triangle) over 150 ft, you have an angle of only 2.3 degrees. Not much and not worth worrying about. The "shorter" adjacent side of the triangle representing the horizontal distance would be 149.88 feet rather than a full 150 feet of the "longer" hypotenuse side of the triangle representing the slant range to the target. The 1.4" difference is not going to matter. |
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You needn't worry about it.
I'm guessing a drop of 6 feet at 50 yards is around 10 degrees. A Govt 55gr M193 load zeroed at 100 yards drops 0.28 inches at 50 yards With a 10 degree angle of sight. There is no difference in drop with that load, at that distance, with a zero degree angle of sight. Even at longer ranges the difference with a 10 degree angle of sight is negligible. At 300 yards the bullet drop is 11.01" (10 degree angle) and 11.30" (0 degree angle). Only at steeper angles and longer distances do you usually have to worry about it. At 300 yards that same M193 load, zeroed at 100 yards, will drop 5.57" with 45 degree angle, compared with 11.30" with a zero degree angle. Even with that fairly extreme 45 degree angle, at 50 yards the difference is quite small - 0.16" versus 0.28" with a zero degree angle. You're good. Edit: I see Dakota did the math to determine the actual angle of sight. You've certainly got nothing to worry about. |
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Quoted: At the rifle range that I frequent 50 yard targets are approximately 6 feet below the shooting stations. If I’m not clear imagine yourself at the stadium sitting in the 4th or 5th row and shooting at something in the center of the playing field. My question is : how the zero obtained under these conditions will differ from “regular” zero ? Please don't offer to visit different range or tell me that real men don't shoot bench rested or at something closer than 1 mile. According to my calculations, for a 100 yard zero, your bullet will hit .0039 inches high at 50 yards. So I think you're ok. |
| Horizontal distance is the only thing that matters. Imagine you're on top of a 6 story building shooting at a target 50 yards from the base. You would end up aiming exactly like you would on a perfectly level range at 50 yards (true ballistic range, not line of sight range). |
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Quoted:
Horizontal distance is the only thing that matters. Imagine you're on top of a 6 story building shooting at a target 50 yards from the base. You would end up aiming exactly like you would on a perfectly level range at 50 yards (true ballistic range, not line of sight range). We need to clarify a few things here: 1. If you shoot a 55 grain FMJ at a horizontal range of 250m out of an AR-15 with 0, 45 and negative 45 degree angles, the difference in bullet drop will be less than 1/2" between all three trajectories. In other words, bullet drop depends only on the horizontal vector where ravity is acting perpendicular to it, not the actual slant range. If that were the only thing going on, the 45 and negative 45 degree angle rounds would NOT shoot high. But since that is not the case, you have to consider what else is happening. 2. In all rifles, and in particular in an AR15 where there is often substantial distance between the line of sight (the line you see along when looking through the sights or scope) and the line of departure (the line along which the bullet departs as it leaves the barrel. The distance between the line of sight and the line of departure also remains the same regardless of angle of elevation. (And, as we noted in 1. the bullet drop is also constant at any horizontal range regardless of elevation angle.) 3. Visualize a straight line rising at a slight angle from the muzzle of the weapon. This is the line of departure. Visualize a second line going straight out through the sights. This is the line of sight. In between them you will have a bullet path starting along the line of departure and arcing down to meet the line of sight at the range at which the weapon is zeroed. This is the bullet path. (simplified somewhat for our purposes as the path will normally rise through the line of sight then arc back down to it creating 2 ranges at which the rifle is zeroed.) 4. Now, the problem is that in the zero angle of elevation shot, the true bullet drop is perpendicular to the line of departure which is essentially horizontal. However in the high and low angle shots, the true bullet drop is still perpendicular to a horizontal line but that is now at a sharp angle to the line of departure. It's hard to describe without drawing a picture, but now, instead of bullet drop being on the short side of a right triangle between the line of departure and the actual bullet path, the same amount of bullet drop is now on the much longer hypotenuse of the triangle. This means the short side of the triangle can no longer reach all the way down to the line of sight, In other words, the bullet drop is no longer sufficient to bring the pullet's path all the way down to the line of sight so the weapon shoots high. 5. How high it shoots depedns on the angle and the total bullet drop. This is the total drop that occur blow the line of departure - the line you'd see if you looked trough the bore. For a 55 gr FMJ at around 3280 fps, this total bullet drop below the line of departure (not the "drop" below the line of sight with the rifle zeroed at a 250m) will be approximately 3.5" at 150m, 7" at 200m, 12.5" at 250m, 19" at 300m, 30" at 350m and about 43" at 400m. To find how high you'll shoot you multiply the total drop at a given range times a constant that depends on the angle of elevation. 15 degree = 0.034 30 degrees = 0.134 40 degrees = 0.234 50 degrees = 0.357 60 degrees = 0.500 So if the rifle is zeroed at 250m and I am shooting at an angle of: 15 degrees = 0.425 high (not worth bothering with) 30 degrees = 1.65" high 40 degrees = 2.93" high (starting to make a noticeable difference) 60 degrees = 6.25" high. At 400m, where you have 43 inches of total drop, it now makes a lot more difference. 15 degrees = 1.46 high (still not worth bothering with) 30 degrees = 5.76" high 40 degrees = 10.06" high (more than enough to miss) 60 degrees = 21.5" high. So in effect it is not an issue until the ranges get long and the angles get steep, but the effect is worth knowing. |
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