A question about sniping

[Hev]

Not quite related to the game but i was wondering how the coriolis effect actualy influences long range sniper fire. I understand the definition of it, just not how a sniper compensates for it.

If anyone can help here i'd love to learn something about this.

[the Fighting Seabee]

I don't think that snipers compensate for the coriolis effect. Let me do the math... near the equator the earth is 24000mi around. A bullet travels Xdistance in Yseconds. At this latitude, the earth is travelling roughly 1000mph.

1000mph/60/60=0.27mi/second, but the shooter is also moving at the same speed (roughly). I guess the variation in latitudinal speed between the shooter and the target (which is 1 mile north) is extremely small. Therefore, I think that adjustments and calculations made would be minimal, if not overcorrected by the smallest incremental adjustments. Something tells me shooting through a swarm of gnats would cause more deviation than the coriolis effect.

Dude, come on, I don't know the answer. But that is the best I could think of.

[Hev]

Sounds reasonable to me, i did a little research myself and like i said couldnt figure out what the deal was.

Thought this was a good place to ask the question. You know at least one person here likes cold food and has a fondness for hiding in bushes

[c3k]

Hmmm,

I did some very rough approximations.

Over a hypothetical 4 mile shot, using an average velocity of 1,000 ft/sec (so about 21.12 seconds of bullet flight time), the coriolis effect would induce the bullet to hit .01703 ft (.20 inches) to the spinward (East) of a target directly North/South of a firer on the equator.

That was more than I expected, even given exagerated distances.

Using those approximations, and reducing the flight time/distance to 1/4 (1 mile), the velocity difference at latitude would also be very roughly 1/4 the original, so would give 1/16 the original difference. 20" x 1/16 = .0125".

All very rough with a lot of initial conditions and only first order approximations, etc., etc.

I would recommend a series of actual shots. Let us know what you find out.

Regards,

Ken

edited to fix my decimal error

[ January 29, 2008, 09:52 AM: Message edited by: c3k ]

[YankeeDog]

Coriolis effect on something like a small arms round traveling a couple of kilometers over a few seconds is way to slight to have any significant effect; other issues like variations in air density and wind speed along the trajectory are far more of a concern.

If Coriolis effect could actually effect a sniper shot, then we would see all sorts of evidence of it in our everyday lives. For example, football quarterbacks would have to factor it in when throwing long passes -- time in the air is actually the more critical factor, not distance traveled.

Coriolis effect can come in to play in long range artillery calculations, especially if a high trajectory is used. This was first discovered by the Germans in WWI, with their big railway guns firing on Paris.

Cheers,

YD

[Thomm]

Without checking, I would assume that the Coriolis effect plays a role only for radial movements. Compared to the radius of the earth these radial movements (bullet drop?) are so miniscule that they will hardly induce any coriolis force.

Best regards,

Thomm

[Peter Cairns]

Well after we had our third kid my wife and I had......

Oh sorry I miss read that....

Peter.

[YankeeDog]

You guys are failing to factor in that, at the moment the bullet leaves the barrel, it and the air mass it is fired into are all traveling *with* the surface of the Earth in that 1,0000 mph rotation. Once the bullet leaves the barrel, things begin to change, but only gradually.

So the calculations are actually much more complicated than a simple measurement of the radial distance the rotating Earth travels while the bullet is in flight. I used to know how to do the calculations in detail, but it's been a long time and I'm not particularly inclined to figure it out again. Maybe Wikipedia can help you. I'm confident you'll find the effect is minimal.

Cheers,

YD

[Thomm]

Originally posted by YankeeDog:

... of the radial distance the rotating Earth travels while the bullet is in flight.

I guess you mean tangential? :confused:

Best regards,

Thomm

[YankeeDog]

Originally posted by Rollstoy:

</font><blockquote>quote:</font><hr />Originally posted by YankeeDog:

... of the radial distance the rotating Earth travels while the bullet is in flight.

I guess you mean tangential? :confused:

Best regards,

Thomm </font>

[c3k]

Well, in the maths I did, above, I assumed the shot was fired from the equator directly to north. Another assumption was that there was no effect from any type of crosswind, any deceleration due to drag, bullet rotation, tumbling, etc.

Assume the earth is a perfect sphere.

The rotational period at the equator is precisely 24 hours and the velocity is precisely 1,000 mph.

After the bullet leaves the barrel it maintains the rotational velocity imparted upon it, the gun, the firer, and the ground beneath the firer due to the equatorial rotation rate of 1,000mph.

That velocity has a direction pointing directly to the east.

Assume the target was directly in sight, and the gun was aimed directly towards the target. The target, being north of the equator, will rotate a lesser distance in the 24 hours per rotation.

The rotational velocities I found were:

Equator = 1,466.6666666666666667 ft/sec

4 miles north of Equator = 1,466.66586 ft/sec

(rotational period at equator is 24 hours; velocity of 1,000 mph, therefore equatorial circumference is 24,000 miles. C=2pi*r. Pi=22/7, solve for r. r(equator)= 3,818.1818181818 miles.

Next, the unknown r(4 miles north of equator) must be solved. Given r(equator), and a known distance 4 miles to the north. Approximate the angle between these two lines: from earth center to target point 4 miles north of equator; from target point 4 miles north of equator to firing point on equator: assume this is a 90 degree angle. I told you this is rough! That gives, using Pythagoras's theorum, r(target)= 3.818.1797 miles. The angle at earth center between firer point and target point is solved to be .028648 degrees.

With the given radius(target), the circumference is given by C=2pi*r. C=23,999.9867 miles.

That, divided by a period of 24 hours per revolution gives the target point velocity.)

Therefore, if the bullet maintains its rotational velocity, it will have a greater lateral velocity than the target (given equal gravitational forces and no drag, etc...). That extra lateral velocity is approximately .0008067 ft/sec directly East.

Now, compute the flight time of the bullet. I used a constant 1,000 ft/sec. I know; that is a horrid approximation for a 4 mile shot.

That 4 mile shot (5,280 ft/mile * 4 miles = 21,120 ft) would take 21.12 seconds.

Now it's simply 21.12 seconds * .0008067 ft/sec = .0017037 ft. Or, .20 inches. That's for 4 miles.

Again, taking that approximation and multiplying by 1/16 (1/4th distance = 1/4 flight time; 1/4th distance = 1/4 lateral velocity difference; distance = velocity * times; 1 mile distance = 1/4 * 1/4 4 mile distance)gives the approximation for 1 mile; .0125".

Obviously, this is the lateral translation to the East.

Again, let us know what results are given by experimentation.

Thanks,

Ken

edited to fix my decimal placement errors!

[ January 29, 2008, 09:51 AM: Message edited by: c3k ]

[YankeeDog]

Ken,

I don't have time to run check all your calcs right now, but something's fishy. You initially show a difference in rotational velocity between the point on the equator, and the point 4mi north of the equator as less than 1/1000th of a ft/sec (1,466.6666666666666667 ft/sec vs. 1,466.66586 ft/sec).

Then, later you end up with a lateral velocity difference between the bullet and the target as nearly 1/10th of a ft/sec (.0867 ft/sec). . . unless I'm grossly misunderstanding what you're doing, that can't be right. . . perhaps you've got the decimal in the wrong place on the .0867 ft/sec figure?

Cheers,

YD

[hammelman]

If ever there is a shortfall in Artillery candidates, send in the Battlefront Brigade.

[c3k]

YankeeDog,

Thanks! You are correct. Silly me; I don't even have the excuse that I was drinking at the time. I shall rectify both the decimal placement AND the lack of drinking...

Carry on!

Regards,

Ken

[YankeeDog]

Ken,

No problem; that's what peer review is for...

Anyway, I see the decimal place is rectified (I trust you are hard at work on the drinking as well), and looking over the rest of your process, everything makes sense to me.

So .2 inch deviation, for an extremely long shot, with an extremely slow bullet. Since most sniper shots considerably shorter than 4 miles, and most sniper rounds much faster than 1,000fps, and furthermore deflection would be less for all shots that were not on a due N-S axis, I think it's safe to say that this is a worst case scenario for direct fire ballistics; most real world results would be far less and therefore quite insignificant.

As I mentioned before, long range indirect fire can be a different matter -- once you have projectiles going tens of miles, with a flight time of minutes, it becomes a much greater consideration.

Cheers,

YD

[Hev]

Thanks all, this has been a great thread and very informative, and my hat goes of to all the match people with bigger noggins than me

[Apexicus]

Well the calculations made above seem to be good approximations, but the problem is that they assume you are on the equator. Unfortunately there the Coriolis effect is at it's weakest and it is strongest near the poles.

I'll take the same distance to the target 1.6 kilometers (~ 1 mile) and time of flight (5.3 seconds) as above. Now assuming that the shot is fired near one of the poles, I get a deviation of approximately 0.6 meters, which is roughly 24 inches. This is definitely of relevance when taking the shot.

If you want to know is the effect is still relevant in Syria, an approximate formula for the deviation when firing on a N-S-axis would be

x = 2*pi*d*t*sin(y)/(24*3600),

where d is the distance to target, t is the time of flight (in seconds) and y is the latitude. The formula is simple as it neglects the (local) curvature of the earth (and some other stuff) and thus won't work near the equator, but in Syria with a latitude of 33 degrees it should work fine.

So in Syria the deviation would be roughly half of the deviation I got for the poles. So it is still of a relevant degree of magnitude and I would argue that taking the Coriolis effect into account is necessary when making long sniper shots. Of course you have to take into account a lot more stuff and the Coriolis effect won't be the biggest cause for error.

Well, that my two cents to the subject.

[ January 30, 2008, 05:59 AM: Message edited by: Apexicus ]

[Bertram]

Apexicus,

How do you get the .6 meters?

I get a larger (!) effect: firing form the pole (defined as the spot where the rotational axis crosses the earth surface) a target a 4 miles (6.4 km) out makes a daily journey of 2 * pi * 6.4 km = 40.192 km. This translates in a speed of 0.465 m/sec, or with 5.3 seconds flight time 2.46 *meters* !!!

I am very surprised at this high number!!

Of course this isnt the effect you see on the bullet, as the shooter will be rotating as well, and so the bullet will start out with a tangential vector.

Bertram

[Apexicus]

Well I used 1 mile as the distance, not four. I guess I expressed it poorly.

[wunwinglow]

Perhaps snipers should sight their weapons in all 4 cardinal directions, and at maybe 4 different latitudes, then write out a correction card, like they have on magnetic compasses.

It was weird seeing metric inches, by the way! Isn't it wonderful that the country that threw off the shackles of British Oppression in 1776 is now the only country in the world still using Imperialist feet and inches... I must show my kids, they will be highly amused!

Tim