Sniper tells of long-shot success(1/2 mile shot in the chest)

http://www.timesonline.co.uk ^ | 4-06-03 | OXENinFLA

[spunkets]

" it would be nice to be able to enter your own problems."

I saw that program, it was nice. I got it from one of your posts. I'm writing a program to do just that, the ability to enter your own loads and other data. I also want to be able to change the scene, pick out stuff and shoot it and add moving targets. That will take a bit longer though.

[Shooter 2.5]

Nice.

Let me know when it's finished. I would love to take a look at it.

[ctdonath2]

Noting that the shot was made in a "fierce gale", assuming a .308 175grain bullet (typical sniper), and doing some quick numbers using my training from Storm Mountain, I figure if the guy had to aim 56 feet to the left he must have made the shot across a 100 MPH wind. The barrel was also pointing about 30 feet over the target.

Yes, skeptics, that's FEET both ways.

[spunkets]

I will. I'll let you know when I've got a beta.

[the_doc]

Thanks for your post! It clears up everything for me....sort of.

I guess I'm going to have to holler "uncle." Based on your tables, the shock wave (or perhaps the spiraling effect for the slug?), must be causing a bizarre susceptibility to crosswind effects after about 300 yards. (This is not easy to see, perhaps, but I graphed the points.)

At some point between 300 yards and 400 yards for the case of a 40-mph crosswind, the slug apparently goes bonkers with the wind effects. By the 600-yard mark, it is moving offcourse a lot faster than the wind is blowing it sideways. It covers the last 300 yards in .064 seconds, but it moves 19.6 feet sideways over the same very short interval. That means it has an average sideways velocity over that distance of over 200 mph. This is a whole lot faster than the wind is blowing, by a factor of five, of course!

So, if the tables are correct, then the law of the conservation of energy is telling us that the wind is not directly accelerating the bullet to the side. The bullet is NOT behaving like a golf ball would--not at all. Somehow the crosswind is stealing the forward energy of the bullet and converting that kinetic energy directly into a rather incredible rate of velocity at right angles to the forward direction!

I frankly don't understand this phenomenon. As I said in one of my earlier posts, maybe the features of the shock wave--or the gyroscopic nature of the bullet--involves something which my model can't address. I will now freely admit that this seems to be the case.

(Hmmm....energy transfers at right angles does sound like a gyroscopic phenomenon of some kind to me!)

[the_doc]

See my post, above.

[spunkets]

A 40mph wind is 58.3 ft/sec. This tbl. shows the avg v for the wind deflection in 100 yd increments. Sure there's not some conversion factor error? The spin has no effect as long as it stablizes the bullet. Spin keeps the bullets long axis tangent to the direction of flight. If the spin is insufficient, coning and/or yaw will occur. That increases the projected frontal area, increasing the drag and decreasing the B.C. The higher the B.C., the less the projectile is effected by the wind. The effect of the wind is proportional to the B.C., that's all.

What is happening here is the bullet's v, normal to the sighting axis, is approaching, but will never excede the wind v. At ~750yds, the bullet has attained ~1/2 the wind speed.

Range	wind deflection ft	bullet sideways v ft/sec
0	0	0
100	0.2	1.67
200	0.84	4.90
300	1.96	7.98
400	3.66	11.0
500	6.02	14.0
600	9.12	16.7
700	13.1	19.2
800	18.0	21.6
900	23.8	23.6

[Squantos]

The egg crate is called kill flash, stops a reflective flash from the shiny part of the lens that can give away the shooter. Looks a lot like a mesh of beehive. Inserts right into the front over the lens and under a protective cover.

The rifle is made by Accuracy International. The designer is the "late" Malcomb Cooper who was an olympic grade marksman who didn't like what was on the shelf so he built him a shooting iron hisself.

I have a friend that works in their Oak Ridge Shop and have shot the .338 lapua and .50 caliber BMG versions. I've laid hands on the 7.62 AWP but never got any trigger time with it...........guns and girls......my piss poor luck.

Stay Safe !

[the_doc]

You're right. I was misreading your chart. (I was getting too tired, and I was using the silly calculator on my computer. Worst of all, it has been been about 12 years since I have done any tedious engineering calculations!)

What is still very intriguing to me is that your analysis forces me to the conclusion that the wind acting on the shockwave makes the bullet move sideways something like a chunk of balsa wood--i.e., not much like what I would expect with a slug of jacketed lead. So, I still don't understand the mechanics, I guess.

Let me try (?) to use your most recent chart to show you what I mean. At 750 yards, the wind deflection is about 15.5 feet. Based on the deflection at 400 yards, it has moved 11.8 feet sideways between 400 yards and 750 yards.

According to your #75, the travel times at 400 yards and 750 yards are about 0.55 and 1.22 seconds, respectively. This yields a time lapse of 0.67 seconds.

Now, if we have a constant sideways acceleration effect from the crosswind, then the distance the bullet moves sideways between the 400-yard mark and the 750-yard should be given by

distance = 11.8 = 0.5 x (acceleration due to crosswind) x (square of the time lapse) + (lateral velocity at 400 yards) x (time lapse)

Plugging in the times and the initial sideways velocity, we can solve for the lateral acceleration:

11.8 = 0.5 x (acceleration) x (0.67 x 0.67) + (11.0) x (0.67)

11.8 = 0.224 x (acceleration) + 7.37

acceleration = 19.8 ft/sec/sec

The interesting thing about this figure is that it represents a surprisingly large crosswind friction. (These charts are based on a mere 30-mph wind.) To illustrate, let's assume that you are standing in a 30-mph wind and you drop a bullet from a height of 6 feet. It should take 0.61 seconds to hit the ground at your feet. But if the crosswind produces a lateral acceleration of 19.8 feet/sec/sec, then it will land 44 inches away from where your dropped it.

This tells us that the crosswind is having a much larger effect on the supersonic bullet than I would have guessed. The shockwave is somehow allowing the bullet to be pushed around by the 30-mph wind very easily!

The picture gets even more interesting when we do the crosswind acceleration calculations for the bullet over the different parts of its flight. Using your 30-mph charts again, the lateral acceleration over the first 200 yards is 13.4 feet/sec/sec. The lateral acceleration from the 700-yard mark to the 900-yard mark is 15.2 feet/sec/sec. But the lateral acceleration between the 400-yard and 600-yard marks is 24.6 feet/sec/sec.

You are more than welcome to check my calculations (I need all the help I can get!), but I think these are correct analyses of your data for the 30-mph crosswind case. And the numbers are telling us that the wind effects become downright severe in about the middle of the bullet's path. Using our earlier example of dropping a bullet from 6 feet, a lateral acceleration of 24.6 feet would cause the bullet to land 55 inches from where it was dropped from a height of only 72 inches.

This is completely counterintuitive to me. I would not have expected a heavy metallic bullet to behave like a low-density object which is so easily tossed around by a mere 30-mph wind. And I would not have expected the worst effects to occur in the middle of the bullet's path.

Again, I welcome your double-checking of my calculations. I guess I am getting senile. I offer my sincere apologies for my earlier freaky calculations!

[the_doc]

P.S. I wonder if you have read David Mason's Shadow of Babylon. It's a very good fictional work on sniper technology.

One of the things which Mason mentions is the tendency of the bullet to veer at long range due to the spinning effect. Do you have tables on this. It sounds like a gyroscopic effect under the effects of gravity.

[spunkets]

I think you left out the v already attained. Calc. the acceleration over an interval with this:

v_initial - a*t = v_final

Then you'll end up with the decreasing acceleration you should end up with, that approaches zero when the bullet speed equals the wind speed. Note the greater distance moved sideways at the long ranges is due to the speed the bullet has already aquired, not any greater acceleration out there.

Range yds	acceleration ft/sec2
100	8.33
200	5.07
300	2.74
400	1.79
500	1.24
600	0.88
700	0.65
800	0.47
900	0.34

As long as the bullet has sufficient angular momentum to stabilize it, there are no effects from the rotation. If there are any effects due to the rotation , or lack of, then the bullet is not worth shooting because it won't fly straight, or be easily predictable. Also the shock wave is due to the bullet passing through the air. It is always a bullet/air interaction, never an air/shockwave interaction. The drag of the bullet is proportional to v², but using a std bullet and trajectory for reference allows a constant coeff, the BC to be used.

When a bullet's v passes through the speed of sound, in whatever conditions are present, there's an asymtotic jump in the value of the BC. it's concave up on the high side and concave down on the low side. At that point depending on it's spin, geometry and weight it can make a relatively significant and unpredictable move. That is the major reason the range for anyload is limited to where the bullet attains the speed of sound. After that range the group size starts to blow up.

As far as the wind creating aerodynamic lift effects on a spinning bullet, that's only important at long ranges with light bullets that don't work out that far anyway, or the spin is insufficient to begin with.

"David Mason's Shadow of Babylon."

Never read that.

[spunkets]

"using a std bullet and trajectory for reference allows a constant coeff, the BC to be used."

That's because a ratio is formed: drag of std/ drag of actual, then v² drops out.

[the_doc]

Thanks for the additional info/

I didn't leave out the starting velocity. I just didn't use your elegant nomenclature (partly because I am html compromised).

The formula I used in calculating the crosswind's acceleration over the interval from 400 yards to 750 yards was:

distance = 11.8 = 0.5 x (acceleration due to crosswind) x (square of the time lapse) + (lateral velocity at 400 yards) x (time lapse)

Plugging in the times and the initial sideways velocity, we can solve for the lateral acceleration... This is how I came up with 19.8 as the effective acceleration over the interval. Please correct me if I have made a mistake in the math.

[KC_Conspirator]

I still hope Aresenal eins the cup though.

[spunkets]

Now, if we have a constant sideways acceleration effect from the crosswind, then the distance the bullet moves sideways between the 400-yard mark and the 750-yard should be given by...

There is no constant acceleration. That table in #111 shows the average acceleration(a) in 100 yd increments. The correct formula is given there also. This one:

x = 1/2*a*t,² +v*t

can't be used, because a is not constant over the interval. a depends on the force of the wind. ie. the drag of the projectile:

C_{drag> = 1/2* drag coeff * area(A)* air density * v²}

So you can see the force, and the corresponding a, will depend on v². Notice in the table in #111, that a is highest at the muzzle and decreases to zero as the bullet speed approaches the wind v. Side note: smaller increments of range and an intercept at v=0, would give a better curve and be exact as delta x-> 0. The a curve is parabolic, because a = a(v). ie. a depends on v. Sorry, I didn't notice right away why that was the wrong eq.

Normally BCs are used, then the v² drops out and all that's left is a constant and a table of wind deflection values based on the constant, BC. The constant can be further corrected for environmental conditions for a better shot.

v² = v²

[spunkets]

also v_initial = v_initial

[spunkets]

C_drag = the force the air exerts on the bullet.

[the_doc]

I realize that the acceleration is not constant. I was figuring out what the magnitude of a putative constant acceleration would have to be in order to accomplish the lateral displacements which you presented in your data.

The reason why I was interested in what I have called the putative constant acceleration is because it serves as a surrogate for something which I do have a feel for--and that's the friction force exerted by the wind.

As I suggested in my earlier post, I conclude that the lateral force being exerted by the crosswind is a lot more than a non-fired bullet.

[hoyaloya]

Welcome to FreeRepublic.

Wow. I see you're really on top of breaking news. Stick around and you may actually be raised out of your naivete. In the real world, things aren't always tidy and pretty like your tafetta skirt.

Eventually, you'll come to thank those courageous enough to address the Middle East's problems. In the meantime, go back to watching soaps.