Snipers and the Coriolis effect

I read part of the way through the thread, and then zonked out. Too much math.

As someone who’s been taught to shoot at moderately long ranges (taught, never said I was any good at it…), I’ve never heard anyone mention anything about worrying about the Earths rotation.

I’ve heard people worry about air temperature, wind speed, even gyroscopic drift (which I never quiet understood… but it’s there, apparently). But I’ve never heard of anyone taking into account the rotation of the Earth itself. Just from a purely nonscientific, practical point of view.

This, however, may be an area I’m no really all that experienced in, because I don’t shoot long range very often – or ever, anymore.

Thanks for clearing that up. Want to go shooting some time?

If the barrel contracts wouldn’t this create more friction and a slower bullet?

Also, regarding spin drift (Magnus effect) it’s a different issue but not to be ignored. It’s what causes golf balls to hook and slice for example.

On the same topic but writ large (to illustrate the effect), naval gunnery has been plagued by the Coriolis effect since the early 20th Century. The rapid increase in gun range led to a new need for centralized gunnery control, and led the development of pre-plotted tables. The British navy famously suffered when engaging German vessels in the South Atlantic; their gunnery tables had been written for the North Sea, and found that the Coriolis corrections were backward south of the Equator.

Answer: The difference between a wounding shot and a kill shot, quite likely.

I also suspect that top military snipers have the cheat-tables memorized for the general lat/lon at which they intend to be operating on a given mission.

Column says:

Bolding added. Ed, this is still funked up. If you are shooting directly east, then the ground is rising to meet you. You have to aim lower, because the ground at the target will be higher by the time the bullet arrives.

mathius_dragoon said:

That would be true if all motion was linear motion. The problem is that the Earth’s surface is curved and rotating. The motion is not straight line. The bullet is no longer in contact with the Earth’s surface.

The bullet will have the inertia of the starting point, the last point it was effected by the barrel. It will have a component of motion that continues in a straight line from that point. The problem is that the surface of the Earth will move in a curved path with respect to that straight line. The shooter will move “down” from the line of that inertia. The target will move “up” with respect to that line. This is because the Earth’s surface between the point of firing and the point of impact is an arc, not a plane.

Barking Dog said:

Methinks you are confusing latitude with altitude.

Bookkeeper said:

The wind is not necessarily going the same speed and direction as the surface, especially for higher altitude/longer range shots like artillery.

Cecil will be cross with me if I messed this up twice, but I think you’re making the same mistake I originally did. If you’re shooting east, the ground is falling away from you. By the time the bullet arrives, the target will be farther away and lower relative to your line of sight at the instant the gun was fired.

I don’t see how it’s messed up. The Earth rotates such that east moves ahead of west. As one fires to the east, items to the east are rotating ahead, which means they are moving on the surface arc in the same direction as the bullet. I think Cecil is still correct on this.

Crapsicles. You are correct, Ed. I keep modeling it and then tying my target to a fixed distant point instead of the moving surface. Yes, the target runs away from the bullet, and moves down.

Correcting my comment to mathius_dragoon

Ok, I don’t see the East/West postulation, because if you calculate the rotation of the earth into the movement of the target you also have to figure that into the movement of the shooter, thereby increasing or decreasing the velocity of the shot fired. A bullet fired from a jet EXCEEDS the speed of the jet by its own muzzle velocity, thus you add the speed of the jet and the muzzle velocity to get the true bullet speed. (otherwise bullets couldn’t catch jets at the ranges fired) So, you add the speed of the rotation of the earth, approximately 1,900,800,000 ft/sec, to a shot fired due east and you negate the same movement of the target. And on a bullet fired due West, you would subtract the speed of the rotation of the earth from the bullet’s velocity and you see that the bullet speed is actually a minus number in true space… but we don’t shoot in true space, we shoot in relative space. Everything we do is relative to the earth’s position. So, while we are constantly moving with the rotation of the earth, that rotation has no impact on the RELATIVE relationships between two different men, or a shooter and his target thus would not need to be calculated.

Nor does the impact of the curvature of the earth on a bullet’s flight require calculation as the line of sight is straight (no curvature) and the bullet drop compensation for the range already compensates for curvature. Changes in elevation are a different factor.

You are not allowing for the curvature of the earth’s movement.

[Ed McMahon]You are correct, sir![/EMM]

voyagernok, you would be correct if the Earth’s surface were a flat plane. But it is not, it is a globe.

Assume the target is in line of sight. Let’s even ignore gravity and pretend that a bullet’s trajectory is line of sight (like a slow laser - bullets have travel time). Therefore, there is no need to compensate altitude for distance. You see the target, you aim directly at the target, you shoot.

During the time that the bullet is traveling to the target, the target is moving away. Aha, you say, but you, the shooter, were moving forward at the same velocity, so they should cancel.

Yes, the forward velocity difference cancels. But here is the trick - the curved surface of Earth rotating means that the target you were shooting at moves down (or up) as the surface rotates away.

Try this model - sit in a rotating chair. Anchor a target (e.g. your finger) to the arm of the chair, and align that target with some distant point (e.g. another chair leg, a door handle, etc) that is not connected to the chair. Now that is the aiming point, the time of firing. Now keep your eyes on the distant point to simulate the bullet’s flight path, and slowly rotate the chair as the “bullet” flies from your eyes to the target. Notice how the target (finger) moves away from the line of bullet flight (line to distant background). The rotation of the chair moves the target away from the flight path of the bullet.

This is exactly the same effect occurring with the bullet and the rotating Earth’s surface. The curved surface is rotating, making the actual location of the target different with respect to the flight path than the initial target location.

Just chanced upon this site and wow, what an interesting proposition this one. Lots of math thrown around but sometimes a little randomly, I think, without really getting to the bones of the argument.

(After all, math and numbers are totally abstract systems - I’ll flesh that out a little later).

And sometimes we are all answering a different question - so what should the question be?

Irishman, I like your thinking here (last post), but it’s a teeny bit flawed. If we leave in the effect of the air (which is dragged around by the earth), and gravity, and fire in a random direction, we are truly lost in space. This complicates matters enormously from a mathematical point of view.

So a more sensible question might be: ‘What happens if a rifle is fired in a vacuum, at the equator on a massless rotating sphere the size of the earth, both with and against the direction of rotation (E-W then W-E)?’

And the answer to this question is - NOTHING. It wouldn’t make the slightest bit of difference if we fired E-W or W-E. Why? Because the Earth’s motion is added to or subtracted from the bullet’s. Exactly as it is with the target.

Think about the classic ‘fly in a railroad carriage’ (or a person) the fly can’t tell that the carriage is moving (at constant velocity). And nor would YOU if the visual and auditory etc. cues are taken away. If you were on the top of the carriage in a vacuum (ok so you’re wearing a spacesuit) and you fired end to end at a target in either direction there would be no difference RELATIVE TO YOU at all. This is what got Einstein so excited when he thought about light.

We can even throw in gravity. No difference. The projectile necessarily describes a parabolic curve as it travels but that doesn’t change the argument.

If the rifle is fired in a random direction we then have 2 vectors to consider so it gets interesting. Someone may want to work out an example.
Anyway thats enough of that for now. What about the expanding / contracting barrel?

When heated, all points in a material move away from each other and the centre point, even if it doesn’t exist because it’s been hollowed out, as in the gun barrel. I just don’t see the i.d. contracting on a thermal differential because it’s ‘held’ by the outer part.

Interestingly, going back to the math part, any calculation here, like so many, could never be absolutely correct. Why? Because considering a section of the barrel, we would have to imagine an infinite number of infinitely thin concentric rings. And every time we use math with infinities (like integration and differentiation) we can never have an absolute answer.

I can’t remember which greek philoshopher said that we can have 1 of something but never 2 of anything because 1 apple (or whatever) can never be the same as the next. He had a very good point.

Anyway I’ve digressed a little here but I hope it’s got people thinking and ready to challenge!

In the ‘fly in a railroad carriage’ example, both ends of the carriage are travelling at the same velocity. If the ‘bullet on the Earth’ example, both points on the Earth are at slightly different velocities–same speed (if directly east-west), but different directions and so different velocities. This is more analagous to sitting atop a train that’s travelling around a bend in the track. If you were to shoot toward an object at the other end of the train, you’d have to “lead” the target. Same thing on the Earth.

This is absolutely untrue.

Different instantaneous directions doesn’t matter for the purposes of this argument. At the instant of firing West both the rifle and the target have the same angular velocity. At the instant of firing East both the rifle and the target have the same angular velocity. However, when firing East the bullet has increased velocity from the motion of the earth and when firing West the bullet has reduced velocity, relative to the Earth, cancelling out the effect of Earth’s rotation, yes?

About the mathematics:

Like to add to that?

Not really. Quoting Irishman:

Again, this is analagous to sitting atop a train that’s travelling around a bend in the track and shooting toward an object at the other end of the train. Do you think you can score a hit without leading the target?

The implication that infinite sums or calculus operations is not exact is incorrect. The most obvious example I can think of is the well-known 0.999 repeating (a decimal point followed by an infinite number of nines). This is by definition the sum of 9/10 + 9/100 + 9/1000 + 9/10000 and so on, and exactly equals one.

Ok, ok, ok. We’re getting into some very deep water here…
Phew. What are we really saying?


  1. If we shoot W-E then the bullet has further to travel than E-W, agreed?

  2. The bullet has higher total velocity from the slingshot effect, we’ve established, but only relative to a point in space, not relative to the target, so we can discount this effect.

  3. The target is moving down relatively during the bullet’s flight time.

  4. Air molecules close to the surface are on average moving along with the surface, adding to the bullet’s velocity. This must necessarily make the bullet hit high, as increasing the velocity must straighten it’s parabolic arc.

So, whilst 3 and 4 conspire to make the bullet hit too high then to some extent 1 will compensate. How exactly this would work we don’t know.

Do we agree on this?

Zut, sorry old chap but it doesn’t matter how many iterations are performed, it can never exactly equal 1. Thats the whole point and thats why mathematicians have gone mad over infinity!

Yes, in fact, it does. But you don’t have to take my word for it; here’s the Straight Dope column on 0.999 repeating.

However, if you have any interest in discussing that, it would probably be better to open a new thread, as it’s been discussed ad nauseum already.

It doesn’t require a new thread, and presenting a hypothesis as a fact is very dangerous indeed.

Mathematical assertions require proof, and if anyone can prove that 0.999~ = 1 I’ll give them my house. On the other hand, it is self evident that:

0.9 does not equal 1
0.99 does not equal 1
0.999 does not equal 1… And so on.

School books show that 0.333~ = 1/3 for the obvious reason - it is aimed at school kids.

1/3 is a formula which invites us to convert to decimal at some level of accuracy.

Mathematics is indeed abstract, not ‘real’ in any sense. Going back to calculus, integration, for example, requires that we calculate the area under a curve. Newton, I think, realized that the best way to do this would be to divide up this area into rectangular strips and add together each strip area to achieve the total.
If we use, say, 10 strips, we have a certain level of accuracy. If 100 then we are more accurate still, so calculus attempts to use an infinite number of strips. It cannot give an absolute result, albeit a very very close approximation. Decimal and binary systems further hammer a nail into the ‘absolute’ coffin.