Physics Asked on October 20, 2020
I’ve heard you must aim lower relative to the target (possibly below it??) when shooting from an elevated position versus shooting from the same level as the target. How can this be?
My initial thoughts were that shooting uphill or downhill would result in the projectile hitting below the target or no significant deviation. The general consensus on the internet at shooting related sites claim that one must aim below the target shooting both uphill and downhill.
Turning to the equations of projectile motion my calculations initially supported my conclusion.
X = Vxot = (VoCosθ)t
Y = Vyot-1/2gt^2=(VoSinθ)t-1/2gt^2
When sighting down the bore the projectile always has a greater deviation below the point of aim when shooting at a positive or negative angle relative to the horizontal. This results due to the reduction in the horizontal component of velocity (Vox=VoCosθ) with the change in angle θ. The reduction in Vox increases time (t) which results in greater acceleration downward (-1/2gt^2).
However, this didn’t take into account the fact that firearms are typically “zeroed in” at some range. This results in a difference between the sighting angle and the bore angle.
Taking this into account produced some effects which I hadn’t anticipated.
The angle α between the sighting angle and bore angle changes the results.
When shooting uphill the angle between the bore and horizontal is increased by α. When shooting downhill the angle between the bore and horizontal is reduced by α.
This results in lower values of Vox when shooting uphill vs. shooting downhill relative to the sighting angle. The net effect is that taking into account this differential between the sighting angle and bore angle shooting uphill will result in the bullet striking lower than the target and shooting downhill the bullet striking higher than the target when the horizontal shot is zeroed on the target.
With all this said the effect is minimal at normal angles of shot and zeroing ranges. The effect only become significant at extreme angles of shot and long zeroing ranges. Following are some calculations sighting at the horizontal, +60 degrees, and -60 degrees with the target at 100 M and the rifle zeroed at 100M. Vo is 800 M/S.
X 100 M
θ 0.043867086 Degrees
Vo 800 M/S
Yo (Rifle Height) 1.5 M
Vox 799.9997655 M/S
Voy 0.6125 M/S
t 0.125000037 S
Y -2.24398E-08 M
Yo plus Y 1.499999978 M
Target Height 1.5 M
Impact relative to target -2.24398E-08 M -8.83456E-07 Inch
R (Range) 99.99997069 M
α (Zero Angle Correction) 0.043867086 Degrees
Sighting Angle 0 Degrees
X 100 M
θ 60.04386709 Degrees
Vo 800 M/S
Yo (Rifle Height) 1.5 M
Vox 399.4694546 M/S
Voy 693.1263628 M/S
t 0.250332031 S
Y 173.2046663 M
Yo plus Y 174.7046663 M
Target Height 174.7050736 M
Impact relative to target -0.000407288 M -0.016034961 Inch
R (Range) 56506.69594 M
α (Zero Angle Correction) 0.043867086 Degrees
Sighting Angle 60 Degrees
X 100 M
θ -59.95613291 Degrees
Vo 800 M/S
Yo (Rifle Height) 1.5 M
Vox 400.5303357 M/S
Voy -692.5138628 M/S
t 0.249668979 S
Y -173.2046687 M
Yo plus Y -171.7046687 M
Target Height -171.7050736 M
Impact relative to target 0.000404954 M 0.015943081
R (Range) -56606.69591 M
α (Zero Angle Correction) 0.043867086 Degrees
Sighting Angle -60 Degrees
This is my analysis of the situation. I would welcome any comments or certainly any corrections to my methodology.
Answered by summe1001 on October 20, 2020
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