Equation for distance traveled

[Roslyn]

Would appreciate someone making a stab at this question.

Assume a shot is executed at 1000 yards (or any distance, but using 1K as the example), with (example again) a 7.0 mils elevation and zero wind hold; and that the shot hits the center X. The bullet will have traveled through its arc a specific distance, call that X inches. Further assume the round exits the muzzle at 3100 fps -- and if you need additional data, let's use a 6 mm 105 hybrid as the bullet.

Assume another shot at 1000 yards, but the required wind hold is, say, 2 mils in order for the shot to hit the center X. That bullet will have traveled further than the original shot, since the lateral wind hold introduces additional distance of travel. That should mean that the elevation hold has to be higher than 7.0 mils, to account for the additional distance and time of flight.

At least, that is what my intuition tells me. That said, my brain locks up just about here, so I am looking for a way to guess how much additional elevation is necessary to account for the wind hold, or even better an equation that mere mortals could use to make such a guess.

Any help -- either mathematical or simply experiential -- much appreciated.

 

[JPeelen]

The difference between the straight line and the curved trajectory is ubelievably small.

Provided that your 3100 ft/s and a G7 form factor of 0.909 (according to Litz) are realistic, I arrive at an elevation of 8.19 mils to get to 1000 yd (in ICAO atmosphere).
Computing at 0.001 s intervals (1/1000 s) and taking the bullet path between intervals as straight lines, the path of the bullet in this case is only 0.63 inch longer than the straight line to 1000 yd. Maximum trajectory height is 2.5 yd, or 0.0025 of distance.
Considering that a 1000 yd range could well be 1010 or 990 yards in reality, the different path length is drowned in much larger surveying errors.

Edit:
To get a rough estimate of the lengths involved, you could look up the formulas for a "circular segment". Assume the trajectory curve approximated by a circular path, with the range being the "chord length" and the maximum trajectory height being the "height" (or sagitta) of the segment.

 

[Geronimo Jim]

Are you guys trying to give me a headache?

 

[Roslyn]

Thanks for the replies, much appreciated. I believe the curve describing the trajectory of a shot is not a circular function. The high point of the trajectory usually is around 65-70% of the distance down range, because velocity of the round slows over time, therefore the curve is not symmetrical. No idea whether that makes it para- or hyperbolic.

In the example we started with, the elevation in my Applied Ballistics app at the density altitude as I write this is 7.5 mils, let's call that 22.5 feet at the top of the curve, since the elevation dialed is the tangent to that point. Presumably the windage adjustment of 2 mils, call that 6 feet, has to be added to distance traveled.

My brain still locks up at this point and I accept the thought that the delta between a zero wind call and a 2 mil wind call may be minimal... but would be interesting to know the difference. Presumably at longer ranges, say 2000 yards, the wind deflection would have a material impact on time of flight and therefore the required elevation.

Will go to school on the link provided, thanks for that as well.

 

[Bryan Litz Ballistics]

A properly written ballistic solver such as a point mass simulation will model the actual path length of the bullet. So the answer you get is valid for the conditions of the shot. As stated above, the difference is incredibly small and wouldn't impact the actual solution if it were ignored but the fact is a point mass solver cannot 'ignore' the actual path.

The Applied Ballistics point mass solvers that run on the apps, kestrel, etc even model the change in air density for the height of the bullet. This is a minor effect on a level shot (10 or 20 feet higher altitude at apogee) but on an angled fire shot (10-20 degree look-up or look-down), the bullet can traverse a great deal of air density and DA along its path. The AB solver models this.

I understand the curiosity and it's a good question. But in all practical reality, there's bigger fish to fry if you're concerned about hitting targets.

Take care,
-Bryan

 

[JPeelen]

To avoid any misunderstanding: I did not want to give the impression that a trajectory is circular.

I just wanted to point to the circular segment as a geometrically similar problem that is easy to compute. It will give you, as I wrote, "a rough estimate" of the relation of trajecory path length versus length of a straight line to the target.

When I first computed path length myself, I could not believe how small the difference was, and was convinced that my program was in error. Then I used the circular segment as an orientation about the size of the numbers to expect.

 

[Roslyn]

This has been helpful, I really appreciate all the contributions. And as usual, it is fundamental to do the arithmetic. My intuition would have led me to conclude the delta at 1K was more substantial.

In the bliss that I can now ignore this artifact of shooting, will instead concentrate on my wind call....

 

[Mozella]

To avoid any misunderstanding: I did not want to give the impression that a trajectory is circular.

I just wanted to point to the circular segment as a geometrically similar problem that is easy to compute. .......... snip.........

Well, I for one, understood the point you were trying to make. But this is the Internet, so all kinds of misunderstanding, misinterpretation, and general confusion is not only possible, but to be expected.

 

[XTR]

This has been helpful, I really appreciate all the contributions. And as usual, it is fundamental to do the arithmetic. My intuition would have led me to conclude the delta at 1K was more substantial.

In the bliss that I can now ignore this artifact of shooting, will instead concentrate on my wind call....

Real life shooting at 1000 yards will tell you that you can't see the variable that you are trying to calculate. I've shot quite a number of Xs on a ½MOA F class target holding off 2MOA on the water line; on the other hand, BR shooters will tell you that they have to account for the up right/down left effect of spin and a cross wind to shoot the iddy biddy groups needed to win in that game. (meaning you hold up left and down right).

It can be fun but I think you are trying to weigh pixie wings.