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Angle Shooting

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Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: May/01/2008 at 09:33
supertool73 View Drop Down
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What are the physics behind angle shooting that makes it work the way it does.  At a rifle class I took last weekend we discussed the math of how to do it and those things, but I did not feel like the physics behind how it works was explained very well.  I don't think anyone really new, they just new it changes things and you have to do "this and this" to make hits.

Say you are shooting 600 yards at a 40 degree angle so you set the dope on your gun for 460 yards.  Is it just the gravity is pulling the bullet different because of the slope? 
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: May/01/2008 at 12:32
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The slope has less effect on the bullets density when shooting down or up hill than when shooting flat line, of course shooting up hill does increase the drag on the bullet as well so only up to a certain point will the math be the same as for shooting down slope.
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: May/01/2008 at 14:39
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Think of it this way, the greater the angle away from horizontal, the less the bullet drop will be. Uphill / downhill has nothing to do with it. If you shot your rifle straight up in the air 500 yards or straight down 500 yards, the bullet drop would be zero.
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: May/01/2008 at 20:41
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Is it just the gravity is pulling the bullet different because of the slope?

at - (downhill) inclinations the bullet seems to get there sooner, shorter distance less effect from gravity

at + (uphill) inclinations it seems to take longer, thus the distance seems to be longer.

it also depends on bc and velocity, lack of time of flight is everything
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: May/02/2008 at 07:15
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Another thing you could try is checking your bullet drop with one of the new range finders which has the angle compensator feature built in. You might be surprised by the results.
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: May/02/2008 at 07:51
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Uphill and down hill corrections are the same.  YOu are dealing with two different ranges in angle shooting;
ONE is the visual range to target, what you see thru the scope.
TWO is the range to target in which gravity acts on the bullet. 
 In your 600 yard example the actual range to target is something  approximating 415 yards on a flat surface.  Gravity acts on the bullet for 415 yards when you take that shot.
In 10-15 degrees and less, there's just about no sight correction.
15-25 degrees is a 10% correction as an approximation
30-45 degrees is about a 20% sight correction.
 
The numbers are approximate and if we could do graphs on here it would make more sense.
Mike
 
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This is a rather complex answer. In the end is has to do with the time that gravity effects the bullet at a 90 deg angle to earth.
Should you fire an air rifle and a 300 Win Mag with both barrels level(90 deg to the earth) and both barrels are exactly the same distance from the earth, and should both projectiles leave the barrel at the same moment, both projectiles will strike the ground at the same time. Now let us say it took the projectiles took 1 sec to reach the ground, the air rifle travelling at 500ft per second will be 500 ft from the starting point, and the 300WinMag at 2900ft per second will be at 2900ft from the starting point. Both will reach the ground and their destination at the exact same time. Remember the old experiment of dropping a 10lb and a 1lb weight from a building. They both hit the ground at the same time.
Now, when you shoot at an angle, the gravity is not 90 deg to the level. So the effect of gravity is less. Should you say shoot at a 45 deg incline or decline, then the bullet only spends half its normal time with gravity at 90 deg to level. So the bullet drops slower and appears to go further. So the coorect way is to calculate your horizontal distance, and that is the amount of time your bullet will spend at a 90 deg pull. Should the angular distance be 600yds, but the horizontal distance only 300yds, then bullet drop should be calculated for 300yds.
I have simplified the matter. The mathematical formulae is best left to Einstein.
Mostly on ranges of up to 250yds and a 30 Deg slope, actual bullet rise (or lack of drop) is minimal and can be ignored. Try it on a field target and see for yourself.
The upshoot of this is that heavier bullets do not drop faster then lighter bullets. Slower bullets drop more then faster bullets, due to spending more time at 90 deg to the pull of gravity.
 
Eureka!!!!!Newton cried when the apple fell on his head!!!
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wallis was newtons tutor that actually did the formulas-anyway the formulas 8 shot is referring to use the cosine of the inclination, projectile velocity and trig. -- another way is to use the differential equations for projectile flight and take the derivatives at any time period for the drop along the x axis. the later is easier when done numerically as a difference equation.
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: May/02/2008 at 10:04
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Thanks guys, this is exactly what I was looking for.  I just could not get my head around it with the explanations I got from other students ( to many of them were way off and I got lost) since the instructors did not spend much time explaining it to us.  They just taught us the math on how to figure it, but not the reasoning behind it.
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: May/02/2008 at 10:23
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Originally posted by supertool73 supertool73 wrote:

Thanks guys, this is exactly what I was looking for.  I just could not get my head around it with the explanations I got from other students ( to many of them were way off and I got lost) since the instructors did not spend much time explaining it to us.  They just taught us the math on how to figure it, but not the reasoning behind it.
 
You actually do the maths???!!!Whacko Sharp%20Shooter  Shocked
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: May/02/2008 at 10:29
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No, they taught us the math.  I use my mildot master Big%20Smile  Much easier, I love that thing.
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: May/02/2008 at 10:33
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It was pretty cool to see it work.  After they taught us the math we went out to a place that had about a 20degree incline and they had these little tiny bottles full of explosives so when you shot them they would explode.  They were so small and so far away that if you did not figure it right and get the right dope you would miss.  They were 340 yards line of sight, most people was not able to hit them at first.  I shot 6 or the 10 before everyone else got on, it was pretty fun.
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: May/02/2008 at 11:00
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Sounds great. What training are you busy with?
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: May/02/2008 at 11:08
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I took two different Precision rifle class in March and April.  Pretty fun stuff.  The head instructor was a Marine sniper for 9 years and served 14 tours in the middle east.  Man he was knowledgeable and was a fantastic teacher.  One of the other guys was a master tracker so the two of them together was able to teach us a ton about movement and field craft.  Most of the field craft was just examples and talking since Front Sight is a defensive school and all. Wink  But it was still great stuff to learn about.
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: May/02/2008 at 11:22
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tannerite??
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: May/02/2008 at 11:30
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Yup,t hats the stuff, I could not remember the name

Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: May/02/2008 at 17:32
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Mike's description of what's happening is correct.  Imagine a right triangle laying on it's side, and the angle you're shooting is the hypotenuse.  It doesn't matter whether you're shooting up or downhill.  The only thing that matters is the linear distance to the target parallel to the horizon, because gravity has the same effect on bullet drop as if you were indeed shooting horizontally.  If you know the angle you're shooting relative to the horizon and your distance to the target along the angle, your actual corrected distance to the target is:
 
b= c X cos A
 
where b = actual distance to target;
c = distance to target along shooting angle (hypotenuse of the triangle);
A = angle of the shot relative to the horizon, either up or down.
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: May/03/2008 at 11:51
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It is a little more complicated then calculating the horizontal distance. It has to do with the amount of time the bullet spends on the 90 deg pull of gravity. Remember, a bullet travels in an arc, not a straight line.
The formula actually looks like this:

where y0 is the initial height, in this case 4 feet; vy0 is the initial velocity in the y (vertical) direction, in this case 0 since all the initial velocity was horizontal; and ay is the acceleration in the y direction, in this case the only acceleration is gravity at -32 f/s2; and t is the point in time where we want to know the position, in this case 0.5 sec.

Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: May/03/2008 at 12:09
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note to self:
 Never ask these people what time it is.
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: May/03/2008 at 12:21
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Guess it depends Mike if you're looking for Metric Vs Imperial time
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: May/03/2008 at 20:08
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using 8 shots formula and finding the atan of the of vx/xy will give you the angle of motion over time which when graphed will show when the traj falls thru the 0 axis and contact is made with the target, and or course is the derivative of how fast it does it.
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: May/04/2008 at 08:09
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Originally posted by Mike McDonald Mike McDonald wrote:

note to self:
 Never ask these people what time it is.
 
Oh, for crying out loud, Mike! The big hand is on 2 and the little hand is on 8...Bucky
 
 Of course I'm just cracking on ya! It's all gibberish to me too. I hold for the horizontal distance, and squeeze the trigger. Seems to get pretty good results.
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: May/04/2008 at 08:51
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oh no-- you don't get off that easy--- yell uncle
make it stop!!! make it stop!!!!
 
the trig. method can be incorporated into the differential equation and solved using identity matrices and solving blocks, time can then be decimated and the difference equation ititerated to result in a smooth curve.
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: May/04/2008 at 12:33
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Gents, you are making this more complicated than it has to be.  The cosine calculation is sufficient.  It doesn't matter that the bullet's traveling in an arc, the cosine formula merely gives you the corrected horizontal distance to the target.  Unless you're shooting at extreme angles, it truly doesn't have to be any more complicated than that.
 
See this link:
 
P.S.; it's now 12:34 p.m. Central Standard time (U.S.)


Edited by RifleDude - May/04/2008 at 12:34
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: May/04/2008 at 12:40
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