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
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?
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.
FjLee
I wonder if any of the software packages that calculate trajectories will integrate and find the arc length?
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?
I believe that what he's asking.
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
FjLee
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.
NAVLAV8R...."THANK YOU" for any input.....yes, you have interpreted my query correctly.
FjLee
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.
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))
FjLee
Denver,colorado
It isn't parabolic either when there's drag.
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.
FjLee
Denver,CO
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
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.
Have you done such a calc?
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
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.
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.
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
Wow! And I thought bullets took some time to get to the targets when I used to shoot pistol silhouette! That must seem like you could shoot and go have lunch before the bullet hit the target.
mathman- even with the rainbow trajectory of the OP's parameters I wouldn't be surprised if the total flight distance is not a lot greater than the straight line distance. No calculations involved, just SWAG.
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
Wow! And I thought bullets took some time to get to the targets when I used to shoot pistol silhouette! That must seem like you could shoot and go have lunch before the bullet hit the target.
mathman- even with the rainbow trajectory of the OP's parameters I wouldn't be surprised if the total flight distance is not a lot greater than the straight line distance. No calculations involved, just SWAG.
And I am certain that if it really mattered one iota Dr. Mann, Hatcher, Whelen or some other super dude would have published a formula for such.
Mathman......
Heh, heh....I'm working with an "outside source"....."OS"...and after some starts and some stops.....got an answer.
The OS answer/solution.....has me..... I guess I am skeptical. Therefore, I also won't reveal what the OS says. At least not right now.
I don't want to taint or prejudice any future efforts by anyone.
And their method, which was presented in detail.....got lost when my computer blinked. I never got it printed out. Now I seem to be unable to make contact with my "outside source".....but I really really do want to print/record their effort.
We'll see.....and I do appreciate the interest and comments. It helps me think!!!!
FjLee in Denver, CO
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.
Mine too lately, surely it would be easier to calculate in metric. Hah!
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
Wow! And I thought bullets took some time to get to the targets when I used to shoot pistol silhouette! That must seem like you could shoot and go have lunch before the bullet hit the target.
mathman- even with the rainbow trajectory of the OP's parameters I wouldn't be surprised if the total flight distance is not a lot greater than the straight line distance. No calculations involved, just SWAG.
Would it be something like adding the midrange height, the drop after the midrange height, the correction distance needed for the wind from max windage at both ends calculated on arced tapers?
Unnamed OS, hah. Let’s just change field goals from 50 yards to 61.5 yards due to arc. How’s about we change the 22 ft 1 inch 3 point shot to 24.6 ft due to arc of the bball. Change the distance of the pitch from the mound due arc. Worthless.
Beginning to think Larry Root.
I think you could come close by imagining a straight line between the muzzle and the target. Then drop another line that goes from the highest point along the flight path to intersect the muzzle-target line at a right angle.
That would give two right triangles and by adding the hypotenuse of the two right triangles you could approximate the flight distance.
At the range you’re dealing with, those triangles are going to be very tiny slivers. So tiny as to be inconsequential.
Estimating with a circle with end points at 0, 600yds and the 3rd pt as (342,2.9) [the numbers from the above post] the arc length is 600.04 yds. or 600yds + 1.44"
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.
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.
Yep you be a grumpy azz hole. Throw out a I am better than you and yes I will call em like I see them. I called it a minuscule matter not worthy. And according to the last post it isn’t. Whether he is correct or not I am still not surprised as to how unimportant it really is.
Estimating with a circle with end points at 0, 600yds and the 3rd pt as (342,2.9) [the numbers from the above post] the arc length is 600.04 yds. or 600yds + 1.44"
I think you could come close by imagining a straight line between the muzzle and the target. Then drop another line that goes from the highest point along the flight path to intersect the muzzle-target line at a right angle.
That would give two right triangles and by adding the hypotenuse of the two right triangles you could approximate the flight distance.
At the range you’re dealing with, those triangles are going to be very tiny slivers. So tiny as to be inconsequential.
You've approximated the curve with two chords. What I did was use a ballistics calculator to give me the heights at 25 yard increments calculated 24 chord lengths to sum. The lengthening added to the 600 yard straight line distance does seem very small.
Yep you be a grumpy azz hole. Throw out a I am better than you (Where did I do that? I asked if you had done a similar calculation to see if you found out a similar small figure. That's all.) and yes I will call em like I see them. I called it a minuscule matter (which I also think it is if my number is correct) not worthy (Worthy of what? Thinking about? Are you the arbiter of what questions people can look into?). And according to the last post it isn’t. Whether he is correct or not I am still not surprised as to how unimportant it really is.
I'm the guy who posted the original question. I think I omitted one parameter.
I don't know if "standard atmospheric conditions" assumes zero wind velocity.
Therefore my possible omission is that my stated problem assumes zero wind velocity.
Therefore, no "come ups" or windage "hold offs" involved in my query.
Folks, "Thank you" for the interesting responses.
I hope there are more.
FjLee Denver,CO
I assumed zero wind drift for my approximation. That way the trajectory would lie in a plane and I could easily calculate the lengths of the approximating chords.
Yep you be a grumpy azz hole. Throw out a I am better than you (Where did I do that? I asked if you had done a similar calculation to see if you found out a similar small figure. That's all.) and yes I will call em like I see them. I called it a minuscule matter (which I also think it is if my number is correct) not worthy (Worthy of what? Thinking about? Are you the arbiter of what questions people can look into?). And according to the last post it isn’t. Whether he is correct or not I am still not surprised as to how unimportant it really is.
Not worthy of my time, don’t give a damn about arbitrating the question. I had to fire up my laptop anyways so I ran the number in ballistic explorer and Infinity, graphed it out by the yard, I could see it didn’t mean squat. You and the OP both refused to say what your numbers were and all I said was I wouldn’t be surprised, why because I could see it in the graph.
As soon as I get to the other computer I'll put up what I found. It is very small.
all I said was I wouldn’t be surprised, why because I could see it in the graph.
That would have been a nice earlier addition to the conversation.
I think you could come close by imagining a straight line between the muzzle and the target. Then drop another line that goes from the highest point along the flight path to intersect the muzzle-target line at a right angle.
That would give two right triangles and by adding the hypotenuse of the two right triangles you could approximate the flight distance.
At the range you’re dealing with, those triangles are going to be very tiny slivers. So tiny as to be inconsequential.
You've approximated the curve with two chords. What I did was use a ballistics calculator to give me the heights at 25 yard increments calculated 24 chord lengths to sum. The lengthening added to the 600 yard straight line distance does seem very small.
Yep. That was my intent and just trying to keep it simple for my brain. 24 data points would be more precise like shooting more shots in a group, huh? 😁
all I said was I wouldn’t be surprised, why because I could see it in the graph.
That would have been a nice earlier addition to the conversation.
How’s about we call a truce 😁
Sounds good to me!
My estimate turns out to be a whole 1.5" extra on the 600 yards.
Sounds good to me!
My estimate turns out to be a whole 1.5" extra on the 600 yards.
👍
It’s good to have the math wizards around when we really get down in the weeds. 😁
Estimating with a circle with end points at 0, 600yds and the 3rd pt as (342,2.9) [the numbers from the above post] the arc length is 600.04 yds. or 600yds + 1.44"
My estimate turns out to be a whole 1.5" extra on the 600 yards.
My circle estimate was 1.44"
Pretty close for the 2 methods.