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fjlee Offline OP
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Too much time on my hands, yes. However:

The lasered horizontal distance to a target is 600 yards. A rifle sighted in correctly, to have a "second zero" of 600 yards, fires a .200 G1 BC bullet at 2,200 fps. Standard atmospheric conditions.

How far would that bullet actually travel before it hits the target. Obviously, it hasta travel more than 600 yards, as it arcs towards the target.

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Originally Posted by fjlee
Too much time on my hands, yes. However:

The lasered horizontal distance to a target is 600 yards. A rifle sighted in correctly, to have a "second zero" of 600 yards, fires a .200 G1 BC bullet at 2,200 fps. Standard atmospheric conditions.

How far would that bullet actually travel before it hits the target. Obviously, it hasta travel more than 600 yards, as it arcs towards the target.

FjLee

Aren’t you actually pointing the bore of the rifle X amount above the target?



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fjlee Offline OP
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Sure nuff.....

But iffen yer second zero is at 600 yards....your line of site to the target would be the same as the laser beam path. 600 yds. And the upward angled bore would have the bullet arcing upards for a part of the path towards the target. Then the bullet starts arcing downwards. I think it's called MRT....mid range trajectory....which is not a good descriptive term, in this case.

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I wonder if any of the software packages that calculate trajectories will integrate and find the arc length?

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Are you talking about the length of the flight path of the bullet? IOW you’re asking how long is the arc that the bullet travels as it travels through the line of sight, to its MRT and then drops into the target?


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I believe that what he's asking.

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fjlee Offline OP
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Mathman.......I don't know. That's a parameter that doesn't seem to be tossed around very much..Possibly the "Long range crowd" would be more heavily involved. I have QL, and it's fairly "techno-weenie"...but I did not check that out with QL

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Sierra infinity will break it down to 1 yard intervals but not interested in firing up the laptop. His other consideration is the windage arc for total distance traveled in the arc.



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fjlee Offline OP
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NAVLAV8R...."THANK YOU" for any input.....yes, you have interpreted my query correctly.

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I'm not set up to do a precise analysis, but I'm thinking of a rough approximation I can figure up. It will underestimate the distance, but it will be a first cut at it.

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fjlee Offline OP
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The arc that the bullet makes is not a "linear "circular arc....it's "parabolic", mostly cuz of the rapid velocity decay, especially after MRT.

((I think...... I am 80 years old, and don't have a definitive grasp of exterior ballistics))

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It isn't parabolic either when there's drag.

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fjlee Offline OP
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Mathman.....I certainly welcome your input! ...and what you said about "drag" makes sense...and I'm Now feeling there are some other things to be factored in if one wants a precise answer.

My gut tells me the true correct solution is NOT NOT trivial.

It'd be nice if Bryan....or is it Brian?.... Litz checked in about now, he knows a lot, based upon empirical stuff that he's done.

Lottsa intelligent people on "the FIRE".....I'm kinda puzzled that a related or similar question hasn't cropped up before.

I guess it's not too pertinent in everyday "loony" stuff...but it's nevertheless sorta interesting.

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As per the OP’s parameters

A .200 G1 bullet leaving at 2200 will reach maximum elevation of 103.98 inches at 342 yards with a 600 yard zero. Total time of flight to 600 1.45 seconds.

Addendum time of flight to 342 yards = .651 seconds

Last edited by Swifty52; 05/10/23.


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I did a calculation to approximate the arc length of that curve. It was done using chords connecting points along the trajectory. I'm not sure I believe the figure I arrived at. I know the approximation I used will under estimate, but I'm still uneasy about how little the curve added to the straight line distance.

I'll not put up the number until I'm able to fiddle with this a bit more.

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I’m not.



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Have you done such a calc?

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Interesting for sure. Went to a bpcr one mile shoot, the distance line of sight was a surveyed exact 1 mile. It took 354 moa elevation and varying amounts of windage of which 2.5 moa were just for precession to connect with my 45-110. 540 gr paper patched bullet 108 grs Fg Goex with an mv of about 1350. How much that windage and elevation added to the length I don't know but it would take right at 7 seconds to get there. Yeah it was fun .mb


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Originally Posted by mathman
Have you done such a calc?

Nope, I let fools like you delve into minuscule matters. It’s actually not a valid worry. 600 yard zero, I worry more about windage than elevation. With a 5 mph wind it added 58.6 inches windage in other words over 50% of the elevation.



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There was no worry or concern, it was just an interesting question I thought I'd take a whack at answering.

It seems like coming across as a grumpy asshole has become part of your modus operandi.

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