Confirmed 1.3 mile Taliban kill w/ Barrett 50 Cal

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O.K. I ran some number using Pointblank as a ballistics calculator. Time of flight for a 750 grain Hornady A-Max (the only bullet I have fast access to info on) over 2500 yards is about 4.25 seconds.
Fast Ed



Sorry...

Your first comment is wrong.

On the time of flight issue - the the 2,500 yd shot was not made with a 750 A-Max - it was made with US Ball ammunition, at 2,500 fps.

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"... Now try to follow. If I am firing directly across the equator from south to north, 1250 yards south of the line to a target 1250 yards north of the line, my bullet travels 4.25 seconds before hitting anything. In 4.25 seconds (actual flight time with this example is 4.9 secs), the earth, rotating at 1037 mph at the equator, (24,901 miles/24 hours) or about 1521 fps, will be about 6,467 feet farther around than when the shot was fired, or 1.22 miles. The bullet would be that far off WITHOUT WIND OR OTHER EXTERNAL FACTORS as the bullet is no longer affected by the earth's rotation while in the air. Not true - the bullet fired north travels straight to the target - both the target and the rifle are traveling at the same speed - in a "spacial" view, the bullet is traveling north east at the same rate that the target is traveling and it chases the target, in a universal view, both the rifle and the target are traveling at about 12,000,000,000, miles an hour - but none of this makes any difference, because the motion of the bullet relative to the target is not effected by any of it - the bullet travels straight to the target. at HOWEVER, the "wind" is about 1521 fps in the opposite direction, give or take the relative wind speed. The question stated no wind - there is NO wind) I don't have a program to calculate the wind drift of a .50 bmg at 2500 yards in a 1037 mph wind or the effects of the east-to-west velocity of the gun/bullet mechanism, but I would have to guess that it is somewhere close to the 6,467 feet that the earth rotates while the bullet is in flight. There is NO wind - a wind of 1037 mph would be a catagory eleventy-seven hurricane! There IS NO WIND If not, we would have to do a heck of a lot more work shooting at shorter distances, too.

The curvature of the earth would have no effect because you are dealing with line-of-sight through the scope which doesn't care how the earth curves. Whether the earth curves up (mountain) or down (valley) the shooter doesn't care as long as he knows the relative elevation difference between him and the target. Since I am making this up, I have decided that the shooter and the target are at the same height above sea level, and since we can see the target, the curve doesn't matter. Range is determined straight line, not curve-of-the-earth. So as long as the scope doesn't bend the light around the earth's curve, scopes can't bend light it doesn't matter to the hold-over either.

Someone else can do the calculations for shooting east to west. Don't forget doppler shift and relativity.

Fast Ed



You have walked into one of the great classic "Red Herrings" of ballistics - none of the "data" has any effect on the bullet - it travels straight to the target, and is only effect by spin drift, which is not calculable, and can only be determined by empirical testing (and will be tiny).

The other two problems (45 degrees and the North pole) ARE practical problems with real answers, but the magnitude of the solutions are so small that they are lost in the "noise".

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Catshooter,
Its got to be the new math, because logic has left the building.

This stuff is so good.......I mean........ you cant make this S%!T up!!
 
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a wind of 1037 mph would be a catagory eleventy-seven hurricane!


/ubbthreads/images/graemlins/shocked.gif

Depending on the outcome of this thread, I may need to board some windows up!

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I got a real nice PM from a guy that was trying to solve the three questions...

... his wife won't speak to him anymore and his kids say, "Hey mommy, who's that guy tearing his hair out" /ubbthreads/images/graemlins/laugh.gif /ubbthreads/images/graemlins/laugh.gif

So after a few trying moments, I wrote this to him and thought it might be good to post it here.

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"Mr. Cat Shooter,
I really enjoyed your comments and information on this thread and it is obvious you know what you are saying.
I have been trying to answer the questions you presented, about determining hold over, at the equator etc.
If it is not too much trouble would you please send me the correct answers. I don't mind saying I am getting my butt kicked trying to answer these.
Also I did the time of flight calc. on the Hathcock shot and I think it is a bit over 7 seconds not the 4.25 that Fast Ed posted.
Thank you for the interesting post and your service to our country.
xxxxxx"

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Hello xxxx.

The answer to the one at the equator is "nothing"... see what I wrote.

The 45 degree question is complicated to calc.

But they are based on this.

Anything on the surface of the earth is moving east at a rate of 15 degrees of longitude per hour (degrees like in a circle of 360 degrees). Since there are 360 degrees of longitude and 24 hours, you travel (with the surface of the earth) 15 degrees in each hour of earth rotation.

How far you travel (in miles) depends on where you are - on the equator, you are traveling over 1,000 miles an hour - at the north and south poles, you are traveling "0" miles an hour, but turning around at 360 degrees per day.

OK so far???

At the equator, both you and the target are on the same diameter, so there is no rotational correction - you shoot at the target and the bullet goes straight to it.

On the north pole, you are standing still, and the world is rotating under you. So you take the range as the radius of a circle - lets say 1,000 yds. You calc the circumference of the circle (D x Pi), or 6283 yds - convert that into feet and then inches and you get 226,188 inches in circumference.

Divide the circle into minutes and then seconds of arc - there are 360 degrees, and each degree has 60 minutes, and each minute has 60 seconds)... so a circle has 360x60x60 seconds of arc = 1,296,000 - OK so far??

The 750 A-Max will get to the edge of the circle in 1.4 seconds.

The earth will rotate the edge of the circle, with you in the center... so it will move 0.174528" in one second, so the bullet will be 0.174528 x 1.4 (travel time) = 0.24432" off target to the right.

Since the equator is the least case of error, and the poles are the most case error, you can see how silly the people that make these arguments are being - they are trying to impress the "kids"... it is all BS.

As to the 45 degree problem - logic says that if the equator is "0", and the pole is "0.24432", then half way between would ~ 0.12216" (give or take a few thou) and that is true.

The proper way to calc the 45 degree problem is to compute the diameter of the earth at 45 degrees, and then calc the eastward speed of the shooter in feet per second - then calc the diameter earth at the target that is 1,500 yds away... and it will be a number that is about:

0.9999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999...

... less than the first number, then you calc the eastward speed of the surface of the earth there, and the difference in speed, in inches per second, times the time of flight is the answer. The error will be to the right. If you turn around and shoot south from 45 degrees, and calc's are kinda the same except that the target will be moving faster than the shooter, and the error will be to the right (are you seeing a pattern here )

You can see now how ridiculous these arguments are, but the people that read these things on the web think they "are on to something"... and will never shut up.

If this doesn't make sense, please write me and ask what you don't understand - I LOVE to teach shooting and ballistics - it is a passion with me.

Meow

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So, for those that think that they can argue Coriolis in small arms fire, you are kidding yourself while trying to impress the children and newbees.

If you are talking 16" guns with 25 mile ranges and 15 mile apexes, that's a little different (and includes different effects - like the radius effect), and when fired east and west, the launch angle effect times the sine of the angle off the north/south line.

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Quote:
wiehet wounjdh chousen hw withenks withnked.

CatShooter do you know what I just said in the above sentence?



I think it's something to do with "nekid" women.

You gotta ask "GJ2nd", cuz he can read that stuff, I can't

/ubbthreads/images/graemlins/laugh.gif /ubbthreads/images/graemlins/laugh.gif /ubbthreads/images/graemlins/laugh.gif

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I learned things, and understand things that had never occurred to me, better than before. Thanks Catshooter!
Now, with the changing axis of the earth and alot of wonderful ballistic information, I am far better equipped for that 50 yard shot at a responding coyote!!!!!!!!
LOL
This is a great thread, an american hero, knowledge, debate, how much more "american" does it get!!!
Carl
 
Man I should win the Predator Masters Soap Opera award for starting this post. Happy New Year to all and God Bless the Troops. Oh by the way I just love these new Miller High Life camo cans.
 
So a hunter walks one mile south......spots a bear and follows one mile west, fires a shot and misses and realizes he didn't bring any more ammo, so he walks one mile north to the hunting camp......what color is the bear?

Pop
 
Hey Cat,
Since our brains are smoking already................


The only seemingly legit recorded instance I have been able to find regarding the "Coriolis Effect" relates to heavy artillery fired at approximately 75 miles. I have to think that may be true, given an extremely long time of flight. Is that for real?

I would think that ICBM's and the like (like plains)(sorry, couldn't help it) would be exempt from this due to the fact that they are self propelled and point of impact is an issue of timing in relation to the earth's rotation. Not an issue of the rotational force of the earth having an effect on the trajectory of a projectile. At what point does it really matter?

Why do Exbal, Nightforce, and FFS ballistic programs even consider this?

I already have a friggin' headache, lay it on me..........
/ubbthreads/images/graemlins/smile.gif
 
Quote:
If you are talking 16" guns with 25 mile ranges and 15 mile apexes, that's a little different (and includes different effects - like the radius effect), and when fired east and west, the launch angle effect times the sine of the angle off the north/south line.



Got my answer. Sorry, the slightly smooth grey matter took a beating in this thread.....

/ubbthreads/images/graemlins/laugh.gif /ubbthreads/images/graemlins/laugh.gif
 
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Hey Cat,
Since our brains are smoking already................


The only seemingly legit recorded instance I have been able to find regarding the "Coriolis Effect" relates to heavy artillery fired at approximately 75 miles. I have to think that may be true, given an extremely long time of flight. Is that for real?



Yes and no...

When you get into heavy artillery, there are more things that enter the solution.

One is the radius effect (which no one ever talks about, except big gun guys) - if you shoot a "big" gun straight up (not at the north or south pole), the bullet will not come down on your head - it will fall west of you.

The reason is this - at the surface, the gun, the bullet, and you are traveling east at the same speed. Lets say you are in New Yawk and traveling at 500 mph east. you shoot a bigassed gun straight up and it goes up 25 miles. At 25 miles, the distance "around" the earth is more than at the surface - think of it as the distance from the center to the surface being 4,000 miles, and the distance to the height of the shell as 4,025 miles. In order to stay on top of you, the shell has to travel faster than at the surface - like satellites have to travel much faster to stay in orbit over our head.

But the shell isn't going east faster - it is still traveling at 500 mph, so it falls behind and winds up coming down to the west, because the earth under it went east.

Now, it gets REAL complicated when you are shooting north with big guns - you shoot 35 miles north - the place you are shooting from is traveling at 500 mph, but the target might be traveling at 499.9, cuz now we are talking big distances, not 1,000 yds. So the shell wants to go to the right... BUT since it will have an apogee (peak of arc) that might be 20 miles high, it will be further from the center of the earth, and so while it is up high, it can't keep up with the surface, and falls behind, which is west.

So Coriolis errors are always to the right in the northern hemisphere and to the left in the southern hemisphere (but too small with rifles to makes difference).

ALL radius errors are ALWAYS to the west, no matter where you are, and are more pronounced as you are closer to the equator, and less pronounced as you approach the poles, and are ~0 at the poles.

I don't know if this is understandable - it is easier with a blackboard.

Quote:
I would think that ICBM's and the like (like plains)(sorry, couldn't help it) would be exempt from this due to the fact that they are self propelled and point of impact is an issue of timing in relation to the earth's rotation.



It IS an issue with propelled missiles, which is why guidance is necessary for long range.

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Not an issue of the rotational force of the earth having an effect on the trajectory of a projectile. At what point does it really matter?



It matters when the errors are large enough to effect hitting the target. With rifles, the errors are so small that they are lost in the other errors (noise), but with large artillery, the errors become larger than the targets. With 16" guns, with a range of 25 to 32 miles, the errors are larger than other ships or forts at long range, and it MUST be figured into the shooting solution.

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Why do Exbal, Nightforce, and FFS ballistic programs even consider this?



Because it's sooooooooo sexy!

I mean, you are a new guy and looking at BC programs and have heard of these mysterious things and Voila... here is a program that figures out these mysteries for you - heck, it's worth the extra money.

If you need a program to figure out an error that is less than 1/2" at 1,000 yds, then do I have some goodies to sell to you /ubbthreads/images/graemlins/laugh.gif /ubbthreads/images/graemlins/laugh.gif /ubbthreads/images/graemlins/laugh.gif

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I already have a friggin' headache, lay it on me..........
/ubbthreads/images/graemlins/smile.gif



Me too - I need a napie /ubbthreads/images/graemlins/laugh.gif

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You 'da man Cat!

As much as you sometimes rub people the wrong way, (cats hate that!), you sure make a guy think about if he really knows what he's talking about or not. It's good to always have a forcefully objective sounding board.........

Thanks for the education man!!
 
I fell stupid for forgetting the spotter.

And yes I know about the creep, stalk, shooting, walks, more shooting, looking, and so on.

Cat I respect you because your older and have more experience than I do.

I back down.

Feel ashamed for arguing with an elder.

Happy New Years every one.

GJ
 
Ok mister Cat, I have one more for ya.

A plane is traveling at 3000 fps and a projectile is fired from the rear of the plane traveling 2800 FPS, What happens to the projectile as soon as it clears the barrel and is no longer being propelled from the gasses created by the powder. Does it fall straight down, or begin to chase the plane as it falls?
 
A plane flying 2,045.46 miles per hour. (unless I screwed up the math)


The initial influence on the projectile would coincide with the motion of the plane plus the projected trajectory vector, minus wind drag and gravity depending on the departure angle.

Okay, now I have hurt myself and can't continue.... /ubbthreads/images/graemlins/laugh.gif /ubbthreads/images/graemlins/laugh.gif

Guess you had better handle this one Paul......
 
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