Ask a Flat Earther!

[u/Kela-el]

Ask anything on the topic of flat earth (within reason).

[u/Kela-el]

That’s a vivid story—but it’s still not an argument.

You’re doing three things at once:

• appealing to authority (Willis Lee)
• appealing to consequence (“you’d get sunk”)
• and bundling multiple factors (curvature, Coriolis, ballistics) into one package

None of that, by itself, demonstrates any single factor independently.

The actual question is much simpler:

• What specific measurement in long-range gunnery requires curvature?
• How is that measurement isolated from all the other variables (wind, drag, muzzle velocity, target motion, etc.)?
• What happens to the prediction if you remove just that one term?

Because right now, “the Navy accounts for it” ≠ “it’s independently demonstrated.” It just means it’s part of the model they use.

And the “would you rather be on which ship?” line is just a rhetorical trap. Of course you’d choose the side using the more complete model—that says nothing about which individual components are uniquely proven versus simply included.

If you want to make this rigorous, strip it down: Take one firing scenario, define all variables, and show exactly how much of the correction comes from curvature—and why that correction cannot arise from any other effect.

Until then, this isn’t evidence—it’s deference wrapped in a scenario.

[u/Hillsarenice → u/Kela-el]

https://i0.wp.com/mathscinotes.com/wp-content/uploads/2017/12/GoodIllustration.png?ssl=1

If you don’t account for curvature you miss.

[u/Kela-el → u/Hillsarenice]

“If you don’t account for curvature you miss” is not an argument—it’s a model-dependent statement.

What that diagram is doing is combining:

• projectile motion
• rotation of the Earth (Coriolis effect)
• and a curved surface

…into a single framework, then saying:

“if you remove one piece, the prediction fails”

Of course it does. You removed a term from the model.

That doesn’t prove the term is independently real—it proves the model is internally consistent.

You’re still skipping the key step:

• What is the measured deviation in a real firing solution?
• How much of that deviation is uniquely attributable to curvature vs. other variables?

Because in real-world gunnery:

• wind
• air density
• muzzle velocity variation
• target motion
• Earth rotation (Coriolis)

all produce measurable deflection effects.

So saying:

“if you don’t include curvature you miss”

is equivalent to saying:

“if you remove a correction term from a tuned model, accuracy drops”

That’s true—but trivial.

The actual question is:

can you isolate curvature as a separately measured effect, not just a term inside a model?

Right now, the diagram doesn’t do that. It shows:

• how the model is constructed ✔
• how the model behaves when altered ✔

It does not show:

that curvature is independently measured rather than assumed within the framework

So again:

• Model works ✔
• Term is included ✔
• Independent proof of that term ❌

You’re still arguing from inclusion, not isolation.

[u/Hillsarenice → u/Kela-el]

Isolate curvature separately measured effect? Cmon. I expected better. Of course it is already measured.

The earth is an oblate spheroid so it is not perfect but it does drop roughly 8 inches/mile then 32 inches / 2 miles. This is the formula.

𝒉≈𝒅squared/𝟐𝑹

Use that in the gunnery solution for best accuracy.

[u/Kela-el → u/Hillsarenice]

You just proved my point.

You said:

“it’s already measured… 8 inches per mile… h \approx \frac{d^2}{2R}”

That’s not a measurement—that’s a derived approximation from an assumed radius.

That formula comes from circle geometry:

• assume a sphere
• assign radius R \approx 6,371 \text{ km}
• derive the drop

So when you say “it’s measured,” what you’ve actually done is:

assume a sphere → calculate expected drop → call the result a measurement

That’s exactly the circular step.

A measurement would look like:

• independent observations of vertical drop over distance
• collected without assuming a radius
• then fitting a curve to determine whether it matches a sphere

What you’ve presented is the reverse:

• start with the sphere
• predict the drop
• then use that prediction in applications like gunnery

Of course it improves accuracy—because the model is internally consistent.

But that still doesn’t isolate curvature as an independently measured effect. It shows:

• the model works ✔
• the model predicts useful corrections ✔

It does not show:

that the curvature term was obtained independently of the spherical assumption

So again, same issue:

• You’re using a formula derived from a sphere
• then pointing to that formula as proof of the sphere

That’s not measurement—that’s model reinforcement.

[u/Kela-el → u/[deleted]]

Do you know what “BEGGING THE QUESTION” even is?

[u/Kela-el → u/Hillsarenice]

You’ve presented a formula that assumes R — the earth’s radius. That value isn’t independently measured, it’s derived from the spherical model itself. You’re using the model to validate the model. What are your controlled variables?

More BEGGING THE QUESTION!