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Posted: 6/18/2009 9:56:18 AM EST
[Last Edit: 6/18/2009 10:07:18 AM EST by SniperCSA]
It's embarassing that I can't wrap my head around this at the moment.
The NRA hunter magazine had a quiz in it. It essentially asked if your rifle was zeroed for 300 yards and the target was 300 yards out at a 30 degree uphill, do you aim dead on, low, high or none of the above. I know the answer is low. Just like bow hunting from a stand, aim using the distance from the tree to the animal. Can anyone explain this in terms of physics, equations, diagrams? Intuitively, it seems as though the vertical component of the velocity vector is larger for the uphill shot. I'm stopping... just somebody explain it pretty plz. 


You use trigonometry to find the downhill distance.
You are fighting gravity shooting uphill also. ETA Check this site out. http://www.loadammo.com/Topics/April04.htm 

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Originally Posted By SniperCSA:
It's embarassing that I can't wrap my head around this at the moment. The NRA hunter magazine had a quiz in it. It essentially asked if your rifle was zeroed for 300 yards and the target was 300 yards out at a 30 degree uphill, do you aim dead on, low, high or none of the above. I know the answer is low. Just like bow hunting from a stand, aim using the distants from the tree to the animal. Can anyone explain this in terms of physics, equations, diagrams? Intuitively, it seems as though the vertical component of the velocity vector is larger for the uphill shot. I'm stopping... just somebody explain it pretty plz. The effective distance is 300 cos(30) yards. Aim slightly low, or better, use a zero with a Maximum Point Blank Range that will cover the critter you're after and hold dead on. And don't worry about "fighting gravity" in either direction, amazingly, it does not change in magnitude or direction whether traveling up hill or downhill. It's the same. 

It's true, Obama is the Leader of Fools deluded to believe, "Everything is going to change now".
As for me, I will embrace what is Right more tightly than ever. 1 lbf = 32.174 lbmft/sec^2 
Your range is 300, the cosine of 30d is .87.
.87 x 300 yds = 261 yards actual distance. Try this article for the explanation. Long Range Hunting 

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Originally Posted By SniperCSA:
It's embarassing that I can't wrap my head around this at the moment. The NRA hunter magazine had a quiz in it. It essentially asked if your rifle was zeroed for 300 yards and the target was 300 yards out at a 30 degree uphill, do you aim dead on, low, high or none of the above. I know the answer is low. Just like bow hunting from a stand, aim using the distants from the tree to the animal. Can anyone explain this in terms of physics, equations, diagrams? Intuitively, it seems as though the vertical component of the velocity vector is larger for the uphill shot. I'm stopping... just somebody explain it pretty plz. Only the horizontal distance matters, not the vertical component. If the target is 200 yards away (horizontal) and 300 feet high (vertical) only the 200 yards matters, not the apparent distance of ~223 yards straight line to the target. 


Originally Posted By AeroE:
Originally Posted By SniperCSA:
It's embarassing that I can't wrap my head around this at the moment. The NRA hunter magazine had a quiz in it. It essentially asked if your rifle was zeroed for 300 yards and the target was 300 yards out at a 30 degree uphill, do you aim dead on, low, high or none of the above. I know the answer is low. Just like bow hunting from a stand, aim using the distants from the tree to the animal. Can anyone explain this in terms of physics, equations, diagrams? Intuitively, it seems as though the vertical component of the velocity vector is larger for the uphill shot. I'm stopping... just somebody explain it pretty plz. The effective distance is 300 cos(30) yards. Aim slightly low, or better, use a zero with a Maximum Point Blank Range that will cover the critter you're after and hold dead on. And don't worry about "fighting gravity" in either direction, amazingly, it does not change in magnitude or direction whether traveling up hill or downhill. It's the same. I thought that the rule was to aim high whenever you are shooting uphill or downhill. 

November 4th  FUCK OBAMA!
November 5th  FUCK! OBAMA! 
Originally Posted By AeroE:
Originally Posted By SniperCSA:
It's embarassing that I can't wrap my head around this at the moment. The NRA hunter magazine had a quiz in it. It essentially asked if your rifle was zeroed for 300 yards and the target was 300 yards out at a 30 degree uphill, do you aim dead on, low, high or none of the above. I know the answer is low. Just like bow hunting from a stand, aim using the distants from the tree to the animal. Can anyone explain this in terms of physics, equations, diagrams? Intuitively, it seems as though the vertical component of the velocity vector is larger for the uphill shot. I'm stopping... just somebody explain it pretty plz. The effective distance is 300 cos(30) yards. Aim slightly low, or better, use a zero with a Maximum Point Blank Range that will cover the critter you're after. And don't worry about "fighting gravity" in either direction, amazingly, it does not change in magnitude or direction whether traveling up hill or downhill. It's the same. The geometry is not the issue. What I'm after is a good physics explaination on "effective distance". The physics involved is what I'm wrestling with. 


Originally Posted By brickeyee:
Originally Posted By SniperCSA:
It's embarassing that I can't wrap my head around this at the moment. The NRA hunter magazine had a quiz in it. It essentially asked if your rifle was zeroed for 300 yards and the target was 300 yards out at a 30 degree uphill, do you aim dead on, low, high or none of the above. I know the answer is low. Just like bow hunting from a stand, aim using the distants from the tree to the animal. Can anyone explain this in terms of physics, equations, diagrams? Intuitively, it seems as though the vertical component of the velocity vector is larger for the uphill shot. I'm stopping... just somebody explain it pretty plz. Only the horizontal distance matters, not the vertical component. If the target is 200 yards away (horizontal) and 300 feet high (vertical) only the 200 yards matters, not the apparent distance of ~223 yards straight line to the target. That right there... why is the horizontal distance the only component affecting trajectory and hot the vertical component of initial velocity? 


Originally Posted By SniperCSA:
Originally Posted By AeroE:
Originally Posted By SniperCSA:
It's embarassing that I can't wrap my head around this at the moment. The NRA hunter magazine had a quiz in it. It essentially asked if your rifle was zeroed for 300 yards and the target was 300 yards out at a 30 degree uphill, do you aim dead on, low, high or none of the above. I know the answer is low. Just like bow hunting from a stand, aim using the distants from the tree to the animal. Can anyone explain this in terms of physics, equations, diagrams? Intuitively, it seems as though the vertical component of the velocity vector is larger for the uphill shot. I'm stopping... just somebody explain it pretty plz. The effective distance is 300 cos(30) yards. Aim slightly low, or better, use a zero with a Maximum Point Blank Range that will cover the critter you're after. And don't worry about "fighting gravity" in either direction, amazingly, it does not change in magnitude or direction whether traveling up hill or downhill. It's the same. The geometry is not the issue. What I'm after is a good physics explaination on "effective distance". The physics involved is what I'm wrestling with. Gravity pulls straight down. It's maximum affect on drop is observed when the muzzle is pointed perpendicular to gravity––that is when the bore is horizontal. As the bore moves away from horizontal, gravity's affect on drop (gravity times cos(angle)) becomes less. When the muzzle is vertical, angle = 90º, cos (90)=0 and drop goes to zero. Hold a fishing rod with a weight on the line horizonally. Note that it droops. Now, start to point it up, then down. Notice it droops less the more vertical it becomes––but regardless, it droops the same whether pointed up or down (always LESS than when it's horizontal). The affect of gravity in the direction of the bullet's flight is negligible, at least in the portion of the trajectory that we're interested in. We're talking about a fraction of 32ft/sec^2 acting on a bullet doing close to 3000fps for times well under a second. 


Originally Posted By AeroE:
Originally Posted By SniperCSA:
Originally Posted By AeroE:
Originally Posted By SniperCSA:
It's embarassing that I can't wrap my head around this at the moment. The NRA hunter magazine had a quiz in it. It essentially asked if your rifle was zeroed for 300 yards and the target was 300 yards out at a 30 degree uphill, do you aim dead on, low, high or none of the above. I know the answer is low. Just like bow hunting from a stand, aim using the distants from the tree to the animal. Can anyone explain this in terms of physics, equations, diagrams? Intuitively, it seems as though the vertical component of the velocity vector is larger for the uphill shot. I'm stopping... just somebody explain it pretty plz. The effective distance is 300 cos(30) yards. Aim slightly low, or better, use a zero with a Maximum Point Blank Range that will cover the critter you're after. And don't worry about "fighting gravity" in either direction, amazingly, it does not change in magnitude or direction whether traveling up hill or downhill. It's the same. The geometry is not the issue. What I'm after is a good physics explaination on "effective distance". The physics involved is what I'm wrestling with. Think of it this way  the bullet is exposed to gravity only over the distance that is normal to gravity. The vertical component of travel is irrelevent. Cool, so if I'm holding a bullet and let go, it won't fall since it won't travel any distance! 


Level shooting is 90 degrees to the gravitational force and the bullet is subject to the maximum force (gravity).
Gravitational force effects (curving the bullet path) are lessened at angles other than 90 degrees (level shooting). Angle shots equal high strikes. 

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tag
eta: This has always been confusing to me. 

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Originally Posted By SniperCSA:
It's embarassing that I can't wrap my head around this at the moment. The NRA hunter magazine had a quiz in it. It essentially asked if your rifle was zeroed for 300 yards and the target was 300 yards out at a 30 degree uphill, do you aim dead on, low, high or none of the above. I know the answer is low. Just like bow hunting from a stand, aim using the distants from the tree to the animal. Can anyone explain this in terms of physics, equations, diagrams? Intuitively, it seems as though the vertical component of the velocity vector is larger for the uphill shot. I'm stopping... just somebody explain it pretty plz. Basically, the range to target (slant range) has to be corrected for a shorter horizontal distance to target. Gravity is only a function of the vertical vector. Since the effect of gravity is compounded by the vertical vector component in either case (whether the target is above gun [TAG] or below the gun [TBG] ), the effect will be in the same direction (higher PoI), but more (TBG) or less (TAG) degree depending on the time of flight. At shorter distances the differences in TAG and TBG correction are negligible. At indirect fire distances a la Field Artillery, they become very significant. In FA parlance, this is called computing "SITE." Like in artillery, firearms computations for shooting angled fire are dependent upon projectile ballistics (muzzle velocity, ballistic coeff, proj wt) and range to target. Two copies of the same article. http://www.snipertools.com/article5.htm http://longrangehunting.com/articles/angleshooting.php 


It's about vectors and breaking it into a vertical and horizontal component. If you change the angle, you are reducing the horizontal velocity no matter which direction you hold it. Gravity remains the same at all times and only affects the vertical velocity regardless of angle. If you're only affecting the horizontal component by changing the angle, then you only have to compensate based on the horizontal component (effective distance).



Originally Posted By 30Caliber:
Originally Posted By SniperCSA:
Originally Posted By AeroE:
Originally Posted By SniperCSA:
It's embarassing that I can't wrap my head around this at the moment. The NRA hunter magazine had a quiz in it. It essentially asked if your rifle was zeroed for 300 yards and the target was 300 yards out at a 30 degree uphill, do you aim dead on, low, high or none of the above. I know the answer is low. Just like bow hunting from a stand, aim using the distants from the tree to the animal. Can anyone explain this in terms of physics, equations, diagrams? Intuitively, it seems as though the vertical component of the velocity vector is larger for the uphill shot. I'm stopping... just somebody explain it pretty plz. The effective distance is 300 cos(30) yards. Aim slightly low, or better, use a zero with a Maximum Point Blank Range that will cover the critter you're after. And don't worry about "fighting gravity" in either direction, amazingly, it does not change in magnitude or direction whether traveling up hill or downhill. It's the same. The geometry is not the issue. What I'm after is a good physics explaination on "effective distance". The physics involved is what I'm wrestling with. Gravity pulls straight down. It's maximum affect on drop is observed when the muzzle is pointed perpendicular to gravity––that is when the bore is horizontal. As the bore moves away from horizontal, gravity's affect on drop (gravity times cos(angle)) becomes less. When the muzzle is vertical, angle = 90º, cos (90)=0 and drop goes to zero. ROTFLMFAO!! I know what you mean, but you are saying that a shot fired vertical will never fall back to Earth. 


Originally Posted By AmericanPatriot:
Level shooting is 90 degrees to the gravitational force and the bullet is subject to the maximum force (gravity). Gravitational force effects (curving the bullet path) are lessened at angles other than 90 degrees (level shooting). Angle shots equal high strikes. So one aims low 


Originally Posted By TaxPhd:
Originally Posted By 30Caliber:
Originally Posted By SniperCSA:
Originally Posted By AeroE:
Originally Posted By SniperCSA:
It's embarassing that I can't wrap my head around this at the moment. The NRA hunter magazine had a quiz in it. It essentially asked if your rifle was zeroed for 300 yards and the target was 300 yards out at a 30 degree uphill, do you aim dead on, low, high or none of the above. I know the answer is low. Just like bow hunting from a stand, aim using the distants from the tree to the animal. Can anyone explain this in terms of physics, equations, diagrams? Intuitively, it seems as though the vertical component of the velocity vector is larger for the uphill shot. I'm stopping... just somebody explain it pretty plz. The effective distance is 300 cos(30) yards. Aim slightly low, or better, use a zero with a Maximum Point Blank Range that will cover the critter you're after. And don't worry about "fighting gravity" in either direction, amazingly, it does not change in magnitude or direction whether traveling up hill or downhill. It's the same. The geometry is not the issue. What I'm after is a good physics explaination on "effective distance". The physics involved is what I'm wrestling with. Gravity pulls straight down. It's maximum affect on drop is observed when the muzzle is pointed perpendicular to gravity––that is when the bore is horizontal. As the bore moves away from horizontal, gravity's affect on drop (gravity times cos(angle)) becomes less. When the muzzle is vertical, angle = 90º, cos (90)=0 and drop goes to zero. ROTFLMFAO!! I know what you mean, but you are saying that a shot fired vertical will never fall back to Earth. Negatron; we're not talking about the same "drop". The bullet will fall back to Earth, but Bullet Drop (the vertical distance between the bullet and the boresight line) will be teh zero. 


Originally Posted By AeroE:
Originally Posted By SniperCSA:
Originally Posted By AeroE:
Originally Posted By SniperCSA:
It's embarassing that I can't wrap my head around this at the moment. The NRA hunter magazine had a quiz in it. It essentially asked if your rifle was zeroed for 300 yards and the target was 300 yards out at a 30 degree uphill, do you aim dead on, low, high or none of the above. I know the answer is low. Just like bow hunting from a stand, aim using the distants from the tree to the animal. Can anyone explain this in terms of physics, equations, diagrams? Intuitively, it seems as though the vertical component of the velocity vector is larger for the uphill shot. I'm stopping... just somebody explain it pretty plz. The effective distance is 300 cos(30) yards. Aim slightly low, or better, use a zero with a Maximum Point Blank Range that will cover the critter you're after. And don't worry about "fighting gravity" in either direction, amazingly, it does not change in magnitude or direction whether traveling up hill or downhill. It's the same. The geometry is not the issue. What I'm after is a good physics explaination on "effective distance". The physics involved is what I'm wrestling with. Think of it this way  the bullet is exposed to gravity only over the distance that is normal to gravity. The vertical component of travel in the gravity field has no effect on the trajectory due to gravity. [Clarified for the funny guy below that doesn't recognize the difference between the horizontal component of the bullet's speed and acceleration of a bullet with no velocity component normal to the gravity vector.] http://media.ar15.com/media/viewFile.html?i=10663 I just talked to my boss about this for a half hour. He kept insisting that there's no change in trajectory whether the 300 yards is horizontal or the hypotenus(sp?) of teh 30 degree right triangel. What's funny is that our jobs.............. this topic fits better than OJ's glove. That's why it's so embarassing. No, it's not firearms related but ballistics nonetheless. 


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