Coriolis effect in artillery debunks Flat Earth?

[Doctor Franger]

Being neither an expert on artillery nor planetary motion, I can give a mere layperson's explanation of an aspect of artillery laying which would seem to negate claims of a flat-disk earth. I don't support that claim but would be interested to hear feedback in case I have the totally wrong end of the stick.

If you are firing long-range artillery at a target to your north, you can't aim straight at it; you have to aim slightly to target's right because the earth is rotating under you as your shell flies through the air. The is I believe an aspect of the coriolis effect. The deflection factor for coriolis is to be increased, from zero if your target is at the north pole, and increased as you approach the equator where the factor is at its maximum, then gradually decreasing again to zero as you head to your additional enemies at the south pole.

If the earth is a flat disk, then zero deflection would used be at disk central, then increase as one reaches the peripheral administrative zones. Then end as one runs out of flat earth. Or do I misunderstand?

(edit added) Unless the flat earth does not spin, in which case there is the same coriolis deflection everywhere, zero.

(apologies if this has already been discussed on Metabunk. I probably won't contribute much because I have already exhausted my expert insights))

 

[Landru]

Being neither an expert on artillery nor planetary motion, I can give a mere layperson's explanation of an aspect of artillery laying which would seem to negate claims of a flat-disk earth. I don't support that claim but would be interested to hear feedback in case I have the totally wrong end of the stick.

If you are firing long-range artillery at a target to your north, you can't aim straight at it; you have to aim slightly to target's right because the earth is rotating under you as your shell flies through the air. The is I believe an aspect of the coriolis effect. The deflection factor for coriolis is to be increased, from zero if your target is at the north pole, and increased as you approach the equator where the factor is at its maximum, then gradually decreasing again to zero as you head to your additional enemies at the south pole.

If the earth is a flat disk, then zero deflection would used be at disk central, then increase as one reaches the peripheral administrative zones. Then end as one runs out of flat earth. Or do I misunderstand?

(apologies if this has already been discussed on Metabunk. I probably won't contribute much because I have already exhausted my expert insights))

It would be helpful (and required by the Posting Guidelines) to provide some evidence to back up your assertions.

 

[JMartJr]

If the earth is a flat disk, then zero deflection would used be at disk central, then increase as one reaches the peripheral administrative zones. Then end as one runs out of flat earth. Or do I misunderstand?

Flat Earth folks I have discussed this with say that the Earth is not only flat, but stationary (possibly disregarding eternal acceleration in an upwards direction to provide pseudo-gravity, as that is a feature in some flat Earth... well, not models, but attempts at a model.) Instead, there is some version of a rotating sky or at least one that appears to rotate. There is thus no Coriolis effect.

When they are asked why, then, hurricanes rotate in the fashin that they do, as well as wind or ocean currents, recourse is had to vague assertions about magnetic or "aether" currents, possibly associate with whatever they think the Sun is this week. I suppose magnetic effects might also have an impact on artillery shells, I am not qualified to even speculate on what effects currents in the aether might have, if it happened to exist.

 

[Doctor Franger]

Flat Earth folks I have discussed this with say that the Earth is not only flat, but stationary . . . There is thus no Coriolis effect.

I added a late edit about a non-spinning flat disk, in which there would be zero deflection factor anywhere. It sounds like the flat earthers are the ones to declare war on, they will never hit anything.

 

[flarkey]

As an ex Artilleryman I can confirm that the rotation of the earth is a contributing factor in the aiming of the guns. However, it is not something that is in the minds of the Gun Group or the Forward observers. Nowadays it's all done by computer, in the past the aiming calculations were all done using tables into which any coriolis effect was already considered.

Just as an example of other features taken into consideration - the droop of the barrel of the gun at different elevations is also a factor that goes into the projectile calculations

 

[Mendel]

The deflection factor for coriolis is to be increased, from zero if your target is at the north pole, and increased as you approach the equator where the factor is at its maximum, then gradually decreasing again to zero as you head to your additional enemies at the south pole.

It's the other way around, with the coriolis effect being zero at the equator, and different orientations north and south of it. It doesn't matter which direction you're aiming at.

More background on the artillery:

Article:

Source: https://flatearth.ws/artillery-manual

If the earth is a flat disk, then zero deflection would used be at disk central, then increase as one reaches the peripheral administrative zones. Then end as one runs out of flat earth. Or do I misunderstand?

Flat Earthers generally hold Earth to be motionless, with no coriolis forces.
If it wasn't, the deflection should be the same everywhere.

Previous Metabunk threads:
https://www.metabunk.org/threads/the-missing-coriolis-effect.8625/
https://www.metabunk.org/threads/de...-14-ways-the-flat-earth-theory-is-false.7148/ (number 14 on the list)

 

[Doctor Franger]

It would be helpful (and required by the Posting Guidelines) to provide some evidence to back up your assertions.

Apologies, I will make a better effort than this one in future once I figure out the correct method, my own fault and shortcomings as a computer neophyte (believe it!)

"In the final year of World War I, when the German military pointed its largest artillery at Paris from a distance of 75 miles, the troops adjusted the trajectory for many factors that could be ignored with less powerful guns. In particular, a subtle influence from the rotation of the earth—the Coriolis effect or force—would have shifted all their shots by about half a mile."
https://www.scientificamerican.com/article/coriolis-effect/

"Does the rotation of the earth have any effect upon artillery fire?​

For example, if a target is out to 20,000 meters, as the earth turns below while the shell is in the air, would this not affect where the shell would land?​

Yes - Trajectory calculation of long range projectiles must account for the Coriolis Effect.

Explanation:​

In the northern hemisphere a projectile fired northward (away from equator) would appear to drift eastward due to coriollis effect and those that are fired southward (towards the equator) would appear to drift westward."

https://socratic.org/questions/does-the-rotation-of-the-earth-have-any-effect-upon-artillery-fire​

Content from External Source

 

[Doctor Franger]

It's the other way around, with the coriolis effect being zero at the equator, and different orientations north and south of it. It doesn't matter which direction you're aiming at.

Then as I feared, I don't get it. Or maybe 'coriolis' doesn't apply and the factor has another name.

The example I gave was of firing artillery north or south. My understanding was that the adjustment factor for such fire was zero at the poles, maximum at the equator, sliding scale between. With the adjustment being to aim left or right of target.

For firing east or west, it's my understanding the adjustment is to aim short or long of your target (respectively), the factor at its maximum on the equator and reducing north or south.

Adjustment for bearings such as SSE are taken from complicated firing tables using combinations.

Whatever the deflection effect is called, I understand that it is about your target moving away from where it was when you fired at it, being on the surface of a rotating sphere.

 

[Mendel]

The example I gave was of firing artillery north or south. My understanding was that the adjustment factor for such fire was zero at the poles, maximum at the equator, sliding scale between.

The gun rotates the most when you place it at a pole.
Think of Focault's pendulum: it will do 360⁰/day if at a pole, but 0⁰/day at the equator.

For firing east or west, it's my understanding the adjustment is to aim short or long of your target (respectively), the factor at its maximum on the equator and reducing north or south.

That's the Eötvös effect.

On the Northern hemisphere, shots are deflected to the right no matter which direction you aim; the reason is that they're not parallel to the axis of rotation.

 

[Doctor Franger]

The gun rotates the most when you place it at a pole.
Think of Focault's pendulum: it will do 360⁰/day if at a pole, but 0⁰/day at the equator.



That's the Eötvös effect.

On the Northern hemisphere, shots are deflected to the right no matter which direction you aim; the reason is that they're not parallel to the axis of rotation.

Sorry, I wasn't communicating clearly. Re the poles, I meant the poles as target, gun position as anywhere other than a pole = zero deflection.

Regarding the second point, I think I really really don't get it.

 

[Mendel]

Sorry, I wasn't communicating clearly. Re the poles, I meant the poles as target, gun position as anywhere other than a pole = zero deflection.

Yes. It doesn't matter, the Coriolis effect is strongest at the poles.

 

[DavidB66]

The gun rotates the most when you place it at a pole.

The angular velocity (rotation around the earth's axis) may be greatest at the poles, but is that what matters for the Coriolis effect? The actual velocity of (say) an artillery shell in a W-E direction is least at the poles and greatest at the equator. Doesn't that mean the deflection is also greatest? And if we look at weather systems, the Coriolis effect is not obviously weaker at the equator, what with all those tropical cyclones and so on. (Of course, the energy in the systems is also greater.)

 

[Doctor Franger]

Yes. It doesn't matter, the Coriolis effect is strongest at the poles.

It seems counterintuitive, but I know that's a thing. Maybe it's the terminology I have gotten wrong, but I can't understand it other than as I have explained it. Here is my most clearly stated version of what I have understood.

If your target is at the N or S pole, you aim for it, you don't have to adjust at all, and you will hit it.

If your target is on the equator and you aim directly at it:
1. firing south at it from the north, your shell will hit something on target's right (gun's POV)
2. firing north at it from the south, your shell will hit something on target's left (gun's POV)
3. firing east at it from the west, your shell will fall beyond target.
4. firing west at it from the east, your shell will fall short of target.

Phew.

 

[Landru]

Evidence? Anybody?

 

[Doctor Franger]

Point being, this only works on a spherical earth. Different rules on a flat disk, either spinning or stationary.
Franger over and out.

 

[Mendel]

Evidence? Anybody?

See the video in my first post.

 

[Landru]

See the video in my first post.

How about this:


published by the Naval Surface Weapons Center, Dahlgren Laboratory, in 1975.
https://apps.dtic.mil/dtic/tr/fulltext/u2/a010816.pdf

 

[Mendel]

If your target is on the equator and you aim directly at it:
1. firing south at it from the north, your shell will hit something on target's right (gun's POV)
2. firing north at it from the south, your shell will hit something on target's left (gun's POV)
3. firing east at it from the west, your shell will fall beyond target.
4. firing west at it from the east, your shell will fall short of target.

Yes.

https://armypubs.army.mil/epubs/DR_pubs/DR_a/pdf/web/tc3_09x81.pdf

 

[Mendel]

If your target is on the equator and you aim directly at it:
1. firing south at it from the north, your shell will hit something on target's right (gun's POV)

This is true across the Northern hemisphere.

2. firing north at it from the south, your shell will hit something on target's left (gun's POV)

This is true across the Southern hemisphere.

3. firing east at it from the west, your shell will fall beyond target.
4. firing west at it from the east, your shell will fall short of target.

This is true everywhere on Earth, but less so nearer the poles.

 

[Doctor Franger]

Back on air unexpectedly; thank you very much, Landru and Mendel, very helpful.
Exactly the right info to confirm my outsider's lay understanding. I don't have the interweb chops yet to find this stuff quickly, it's not just my laziness. I can't boast anything except typing ability and outdated paper-based research skills, but like they say, "a little knowledge. . .".

JMartJr in post#3 suggests that are several <schools of thought> or models in the flat disk earth community, perhaps I should go away and find out what the main ones are, in case this coriolis/deflection objection doesn't apply to all of them.

(edit) I just didn't want to start out with an erroneous malunderstood model of my own.

 

[Doctor Franger]

This is true across the Northern hemisphere.

This is true across the Southern hemisphere.

This is true everywhere on Earth, but less so nearer the poles.

My post #13 reflected a clearer expression of my original understanding, so the problem was in my initial communication. Apologies.

 

[FatPhil]

My understanding was that the adjustment factor for such fire was zero at the poles, maximum at the equator, sliding scale between. With the adjustment being to aim left or right of target.

A small region around one of the poles is effectively a roundabout (adopting the flat earth approximation ). Go to your local children's playground and try to play catch on a spinning roundabout. If you fail horribly and hilariously, then that's because there's so much corealis force because of the rotation of the roundabout. If you can't find a local playground, persuade your pet mice to play basketball on a lazy susan. If even they won't play ball, technically I guess you could roll a ballbearing down a short ramp with a go-pro on it situated on a turntable platter (for the youngsters in the room who don't know what I'm talking about - wow! (and flutter)). The go-pro will record the ballbearing diving off to the right as soon as it's unconstrained by the ramp (and then hopefully darting back to the left again to hit the spindle, assuming you were aiming at the spindle). In summary: I'd just go to the park, it's easier.

Similarly, a small region around the equator's effectively just a sideways sliding platform (using the same FE approximation). Go to your local airport, and juggle on the travellator - no friends necessary for this demo[*]. Everything works fine, does it not? That's because as you're launching your ball you're already giving it the additional forward velocity it needs in order to reach your other hand, which will also have moved under that same additional velocity for the same length of time, so the same distance - everything's where it should be. If you can't find an airport with a travellator, let your pet butterflies out in a train carriage, and notice how they don't have to set off at unusual angles to get from one side of the carriage to the other; whilst there, prove special relativity too!

How can I, a certified[**] mathematician, get away with these flat earth approximations for explanations/demonstrations intended to disprove the flat earth? The special sauce is the restriction to small regions. The deviation from flatness in a small region is utterly negligible, so you don't need to bother about it. However, a small region near a pole is very different from a small region around around a point on the equator (but with a lovely smooth sine wave telling you by how much each small region in between would see the effect).

[*] Technically no friends are necessary on the roundabout either - you can throw towards the centre of the roundabout, such that it arrives at your future self at the appropriate time. To you it looks like you've thrown in a loop to yourself.
[**] or at least certifiable.

 

[Ann K]

effect).

[*] Technically no friends are necessary on the roundabout either - you can throw towards the centre of the roundabout, such that it arrives at your future self at the appropriate time. To you it looks like you've thrown in a loop to yourself.

Have you been banned from all the local playgrounds yet?

 

[Mendel]

Source: https://m.youtube.com/watch?v=o5m7bNziVLg

 

[Ann K]

@Mendel
Another YouTube video that comes up after this is particularly good, as it explains the motion of the object on a turntable in some detail. Steve Mould has a lot of good nerd-stuff.


Source: https://www.youtube.com/watch?v=3oM7hX3UUEU

 

[FatPhil]

Have you been banned from all the local playgrounds yet?

Well there was the infamous flamethrower experiment, but I don't like to talk about it.

 

[FatPhil]

Source: https://m.youtube.com/watch?v=o5m7bNziVLg

*None* of that was the PoV of someone in the rotating frame, and it was almost entirely friction driven - it's pretty but it's entirely irrelevant. That's why I *explicitly* mentioned putting a go-pro on the ramp feeding the ballbearing onto the turntable, it's the rotating frame of reference that sees the pseudoforce.

The best I can find right now is:

Source: https://www.youtube.com/watch?v=ugDmEHWhz6A


There used to be a vid on youtube of some actual jugglers actually juggling on an actual roundabout as it was actually spinning - it was mindblowing. I can't find it now, you're just going to have to imagine how amazing it was.

 

[Doctor Franger]

I love those vids.
I think my original understanding of the effect I described is unchanged, except for my poor choice of technical jargon which may not have applied (but as I said in the OP ". . . it is I believe an aspect of the coriolis effect. "). An unwise decision to appear fancy.

In its least scientific expression, and for whatever reason, long range artillery has to treat every surface target like a moving target. Target is moving left or right or closer or farther from Gun depending on the direction Gun is firing on Target. Because Target is on a moving surface (earth).

This effect is almost zero at one pole, smoothly increases to its maximum at the equator, reduces again the closer to the opposite pole where it returns to zero. Because of Sphere.

This effect (zero to maximum and back to zero) is absent in a FE. For a spinning FE, the effect increases from zero at the centre to its maximum at the rim and then ends; for a stationary FE the effect is totally absent. The only way this effect is present on a FE is if it is a double-sided disk with seas and land mass on both sides like an LP, and I don't know anyone is arguing for that (hey have I accidently modified and improved the ultimate model of a flat disk earth? Oops.)

 

[Mendel]

This effect is almost zero at one pole, smoothly increases to its maximum at the equator, reduces again the closer to the opposite pole where it returns to zero. Because of Sphere.

Still wrong.

The Eötvös effect, affecting the range, does this, but the Coriolis effect does not. Consider: if you shoot across the pole, your target is moving in the opposite direction from the gun! Consider: your shot deviates right in the North, but deviates left in the South: can this sign change really occur at maximum, or should it occur at zero?

For a spinning FE, the effect increases from zero at the centre to its maximum at the rim and then ends;

No, for a spinning FE the Coriolis effect is the same everywhere because the angular speed of the surface is constant, and there would be no Eötvös effect because there is no vertical rotation.

 

[Mauro]

This effect is almost zero at one pole, smoothly increases to its maximum at the equator, reduces again the closer to the opposite pole where it returns to zero. Because of Sphere.

It makes no sense to say the Coriolis 'effect' depends on a particular position on the globe, this because the Coriolis 'effect' arises from moving on a rotating body. At the equator the peripheral speed of the Earth is at a maximum, it then decreases to zero going towards a pole. So if you start, say, at a certain point on the northern emisphere, moving at a certain peripheral speed, and move North, you are bound to drift East, because the Earth under you moves slower and slower as you go northwards. Instead, if you start from the North pole and go towards the equator you will drift West (Of course you must not be in contact with the Earth, or this will automatically adapt your tangential speed to whatever speed the Earth is rotating under you. This is why Coriolis applies to weather systems, airplanes and artillery, but not to human beings or cars).

And, by the way, you can have the Coriolis force also on a flat (but spinning) disk. It's actually easier to understand it thinking of a disk rather than a sphere: the center of the disk correspinds to the pole (zero peripheral speed) while the circumference corresponds to the equator (maximum peripheral speed). Ie.: if you are at the center and move towards the circumference you need to 'gain' tangential speed, or you will drift the opposite direction the disk is rotating.

Notice the math is different in the two cases, so by measuring the Coriolis 'effect' a spinning disk can be distinguished form a spinning sphere (I will never regret enough having sold my copy of the Feynman book used in the physics course I took at university or I could be more informative. Bad mistake, but I really needed the money at the time. Sigh)

 

[Doctor Franger]

Once again I feel like I should adjust my psych med dosage.
It was my mistake to call it Coriolis, probably also to call it an Effect; I don't know what to call it or if it has a name. At its basis it is a need for the adjustment of aim.
When fired upon by long range artillery, all stationary surface-fixed targets must be aimed at as if they are moving targets, and a gun's aim adjusted accordingly. The extent and direction of the adjustment varies, depending on the direction of fire and the location of the target (aim not at where your target is, but where it will be when the shell arrives). During the time it takes for the shell to travel through the air, the target has moved on, as has the gun but after firing it has no further influence.
The only place where almost no adjustment is necessary is when the target is at a pole (i.e., on the axis of Earth rotation). The extent of the adjustment goes from zero and increases to its maximum then back to zero, as the gun is moved from firing on a polar target to an equatorial target and beyond to the opposite polar target.
My claim was that this is the behaviour of artillery on the surface of a sphere but not the surface of a flat plane.

I am not going to feign any more doubt, uncertainty or confusion (which was done out of politeness), because I am pretty bloody sure that what I am saying is correct or I would not have said it. This is my SOP which is why I usually don't go anywhere pre-weighed down with evidence, a habit I am ready and willing to change for Metabunk. When I find myself alternately agreed with or contradicted by the same person depending on how I paraphrase myself, then clearly the problem is in how I am expressing the details of my claim.
As Flarkey mentioned in post#5, I understand that it is not an active consideration for gun crews or spotters, it is accounted for along with many other factors in the detailed firing tables, provided to gunners to arrive at a firing solution.
I am most familiar with the firing tables for the A4/V2 rocket, which with its five-minute travel time and mostly eastwards firing direction, had to take extreme account of the earth's rotation when trying to hit a target.

But to reiterate, my claim was that this is the behaviour of artillery on the surface of a sphere but not a flat plane, thereby possibly negating the FE model.

 

[Mendel]

But to reiterate, my claim was that this is the behaviour of artillery on the surface of a sphere but not a flat plane, thereby possibly negating the FE model.

There's general agreement on this; I hope my first reply above makes this clear.

When fired upon by long range artillery, all stationary surface-fixed targets must be aimed at as if they are moving targets, and a gun's aim adjusted accordingly. The extent and direction of the adjustment varies, depending on the direction of fire and the location of the target (aim not at where your target is, but where it will be when the shell arrives). During the time it takes for the shell to travel through the air, the target has moved on, as has the gun but after firing it has no further influence.
The only place where almost no adjustment is necessary is when the target is at a pole (i.e., on the axis of Earth rotation). The extent of the adjustment goes from zero and increases to its maximum then back to zero, as the gun is moved from firing on a polar target to an equatorial target and beyond to the opposite polar target.

This, however, is not true.

Unless your firing distance is zero, either the gun or the target are moving, even at the poles, and that will throw the aim off.
Thd only exception is when the gun and the target are moving in parallel.

Article:

Eine nennenswerte Rolle spielt dabei nur die zur Erdoberfläche parallele Komponente der Corioliskraft, weshalb diese in den Geowissenschaften vereinfachend oft als „die Corioliskraft“ bezeichnet wird. Ihre Stärke hängt von der geographischen Breite ab. Sie verschwindet am Äquator und ist am stärksten an den Polen.

Source: https://de.m.wikipedia.org/wiki/Corioliskraft

Article:

The horizontal deflection effect is greater near the poles, since the effective rotation rate about a local vertical axis is largest there, and decreases to zero at the equator.

Source: https://en.m.wikipedia.org/wiki/Coriolis_force

 

[Doctor Franger]

According to my grasp, this is what happens when you shell your enemy's igloo at the north pole, from several miles distant:

You aim precisely at your target, in both range and bearing. You fire, and the shell (if visible like a tracer) veers to your left but swings gradually back to hit the igloo (gunner's POV).

The igloo POV would be of a shell travelling in a differently graded curve. If the igloo enemy survived the impact, s/he might see and hear gunner's celebration of a direct hit, from somewhere other than where the shell seemed to originate from. (edit: but kinda not. The puff of smoke from the initial firing if rising but otherwise stationary, would appear to drift in a sidewind from igloo POV and be reasonably if erroneously attributed thus.)

 

[Mendel]

According to my grasp, this is what happens when you shell your enemy's igloo at the north pole, from several miles distant:

You aim precisely at your target, in both range and bearing. You fire, and the shell (if visible like a tracer) veers to your left

Please explain why the shell veers left.

but swings gradually back to hit the igloo (gunner's POV).

Please explain why the shell swings back.

What happens when you double the distance?

Spoiler

Article:

This force causes moving objects on the surface of the Earth to be deflected to the right (with respect to the direction of travel) in the Northern Hemisphere

Source: https://en.m.wikipedia.org/wiki/Coriolis_force

 

[Doctor Franger]

I'm on the verge of slapping my forehead and saying Madonn'

I'm glad if there is general agreement about my central claim, but it leaves me to wonder why I am being nitpicked over details of a subject on which I never claimed, expressly denied, being an expert.

If I inadvertently typed left for right or vice versa, it's because it's my 15th attempt to explain what I mean in the face of contradictory criticism. But yes, it's a triumphal gotcha moment.

 

[Mendel]

I'm glad if there is general agreement about my central claim, but it leaves me to wonder why I am being nitpicked over details of a subject on which I never claimed, expressly denied, being an expert.

This isn't nitpicking. You're saying there is no net deflection at the pole, which is wrong. The horizontal deflection is strongest there, while it is zero at the equator. If you argue this with a FEer who knows physics (there are some), they'll crucify you over this, probably using the toilet water spin myth, which is wrong in the particular but correct in principle, and requires that there be no spin/horizontal reflection at the equator. You might then be tempted to say, "Metabunk said I'm right", and then where would we be?

You can consult wikipedia and many other sources on this if you like; or you could think about the questions in my previous post, which I'd be happy to discuss with you; or you can decide you don't really want to know. It's your choice entirely.

You really need to consider the motion of both the gun and the target.

 

[Mauro]

According to my grasp, this is what happens when you shell your enemy's igloo at the north pole, from several miles distant:

You aim precisely at your target, in both range and bearing. You fire, and the shell (if visible like a tracer) veers to your left but swings gradually back to hit the igloo (gunner's POV).

The igloo POV would be of a shell travelling in a differently graded curve. If the igloo enemy survived the impact, s/he might see and hear gunner's celebration of a direct hit, from somewhere other than where the shell seemed to originate from. (edit: but kinda not. The puff of smoke from the initial firing if rising but otherwise stationary, would appear to drift in a sidewind from igloo POV and be reasonably if erroneously attributed thus.)

Ah maybe I understood what you mean: if you shoot at the pole then you don't need any correction, because the pole only rotates, it does not translate. But if you shoot at a point on the equator then the needed correction is maximal because the tangential speed at the equator is the highest. This looks correct, as I think is correct that a missile launched straight north from the equator to hit the pole will be seen (in the Earth reference frame, say from the launch point) to loop around before hitting (and this is the Coriolis force), while in the missile reference frame it just follows a straight trajectory due north.

 

[FatPhil]

According to my grasp, this is what happens when you shell your enemy's igloo at the north pole, from several miles distant:

You aim precisely at your target, in both range and bearing. You fire, and the shell (if visible like a tracer) veers to your left but swings gradually back to hit the igloo (gunner's POV).

The igloo POV would be of a shell travelling in a differently graded curve. If the igloo enemy survived the impact, s/he might see and hear gunner's celebration of a direct hit, from somewhere other than where the shell seemed to originate from. (edit: but kinda not. The puff of smoke from the initial firing if rising but otherwise stationary, would appear to drift in a sidewind from igloo POV and be reasonably if erroneously attributed thus.)

To understand the coriolis effect it's best to understand it in the simplest context where it appears, namely the rotating plane (where there's no 3rd dimension in which gravity will act.
The poles would be a decent approximation to this were it not for gravity, as sin(x)~=tan(x)~=x and cos(x)~=1 for small x - for the ranges your talking about, you've got 5 digits of accuracy.

A projectile aimed right at the axis of rotation (north pole) doesn't have just the radial vector as its initial velocity, it also has the instantanious sideways motion caused by the rotation. Therefore you will miss the igloo, to the right as you view it at the point of firing. What the miss looks like to you at the time it misses will depend on the muzzle velocity - slow enough and it might look like the bullet passed right in front of the front of the igloo (and is now starting to come back towards you).

If you correct for that east-moving feature of your rotation, and do start the projectile on a path directly towards the axis of rotation, by aiming west of it, then you will indeed see the bullet swing left, then right, and then hit the target, nothing can change that, the only question that remains is which side of the igloo you might hit, as it will have rotated by the time the projectile arrives.

 

[Scaramanga]

because the earth is rotating under you as your shell flies through the air.

Um, no....that cannot be 'the' reason because one would then be negating the answer to one of the other favourite flat earther misperceptions. After all, a common claim by flat earthers is that if the Earth is rotating then why doesn't the ground zoom under us at 1000mph if we jump up in the air. We respond that our momentum is what prevents that scenario happening.

Coriolis is actually caused by the difference in angular momentum across latitude. Conservation of mass/energy and all that. It is a vector thing, rather than the ground moving 'under' the object. If Coriolis was caused by the ground moving under an object then the effect would be largest at the equator....when in fact the reverse is the case.

 

[Scaramanga]

Being neither an expert on artillery nor planetary motion, I can give a mere layperson's explanation of an aspect of artillery laying which would seem to negate claims of a flat-disk earth. I don't support that claim but would be interested to hear feedback in case I have the totally wrong end of the stick.

If you are firing long-range artillery at a target to your north, you can't aim straight at it; you have to aim slightly to target's right because the earth is rotating under you as your shell flies through the air. The is I believe an aspect of the coriolis effect. The deflection factor for coriolis is to be increased, from zero if your target is at the north pole, and increased as you approach the equator where the factor is at its maximum, then gradually decreasing again to zero as you head to your additional enemies at the south pole.

If the earth is a flat disk, then zero deflection would used be at disk central, then increase as one reaches the peripheral administrative zones. Then end as one runs out of flat earth. Or do I misunderstand?

(edit added) Unless the flat earth does not spin, in which case there is the same coriolis deflection everywhere, zero.

(apologies if this has already been discussed on Metabunk. I probably won't contribute much because I have already exhausted my expert insights))

This is a good opportunity to throw in something tangentially ( excuse the pun ) related that defies what most people regard as common sense. It's another of those related issues that goes totally against what one would expect.

The old 'object flies off in a straight line' ( 90 degree tangent to center ) issue...which so many believe in and is even taught in science lessons.....but is not true....


Source: https://www.youtube.com/watch?v=AL2Chc6p_Kk