[ElTrancy]

This equation is not correct, and I just want to say this to everybody who doesn't know this already.

http://earthcurvature.com/

[Max_Almond]

You're saying "the drop below the horizontal is not 8 inches per mile squared"?

Here's a much better curve calculator, by the way:

www.metabunk.org/curve/

[ElTrancy]

You're saying "the drop below the horizontal is not 8 inches per mile squared"?

Here's a much better curve calculator, by the way:

www.metabunk.org/curve/

Yes, I am saying that.

[Max_Almond]

Okay. Well I say it is correct - though I will also say that it isn't very useful ("hidden amount" is much more useful, for flat earth purposes).

What is the correct figure and equation then?

[Tom Bishop]

You guys just posted links to two calculators, created by individual authors, which give different results to each other.

Yet that somehow demonstrates Rowbotham and the entire scientific community of the 1800's to be wrong?

[ElTrancy]

You guys just posted links to two calculators, created by individual authors, which give different results to each other.

Yet that somehow demonstrates Rowbotham and the entire scientific community of the 1800's to be wrong?

Both calculators give the exact same result. Use any number in both calculators. Each giving different results? Did you even use them?

[ElTrancy]

Okay. Well I say it is correct - though I will also say that it isn't very useful ("hidden amount" is much more useful, for flat earth purposes).

What is the correct figure and equation then?

Are you a Flat Earther or Round Earther? Either way, you contradict both of them.

[Max_Almond]

You guys just posted links to two calculators, created by individual authors, which give different results to each other.

Yet that somehow demonstrates Rowbotham and the entire scientific community of the 1800's to be wrong?

Both calculators give the exact same result. Use any number in both calculators.

Actually, both calculators give similar results: the one I linked to uses slightly more sophisticated and therefore accurate trigonometry.

Both calculators also agree, more or less, with "8 inches per mile squared" - thereby validating Rowbotham's simplified, 'ball park' equation.

"8 inches per mile squared" is only useful up to about 100 miles, though - after which something more precise is required.

For most practical purposes, though, it suffices for measuring "the drop".

Though, as stated above, "the drop" isn't that useful: "hidden amount" more so.

Are you a Flat Earther or Round Earther? Either way, you contradict both of them.

How can I contradict them with a question? (I.e., "what is the correct equation?")

I'm a 'Round Earther'.

[ElTrancy]

You guys just posted links to two calculators, created by individual authors, which give different results to each other.

Yet that somehow demonstrates Rowbotham and the entire scientific community of the 1800's to be wrong?

Actually, both calculators give similar results: the one I linked to uses a slightly more sophisticated and accurate equation.

Both calculators also agree, more or less, with "8 inches per mile squared" - thereby validating Rowbotham's simplified equation.

"8 inches per mile squared" is only useful up to about 100 miles, though - after which something more precise is required.

Generally, though, it suffices for measuring "the drop".

Though, as stated above, "the drop" isn't often that useful: "hidden amount" more so.

Are you a Flat Earther or Round Earther? Either way, you contradict both of them.

How can I contradict them with a question? (I.e., "what is the correct equation?")

I'm a 'Round Earther'.

Nevermind.

[Max_Almond]

Nevermind.

Please tell me you're not going to leave here without posting "the correct equation".

[AATW]

Nevermind.

Please tell me you're not going to leave here without posting "the correct equation".

If my maths is right (highly questionable!) then the drop in inches over a distance of 'm' miles should be:

(1 - cos(360m/24938)) x 250,842,240

24938 = 2pi x radius of earth in miles
250,842,240 = radius of earth in inches

This gives 7.96 over 1 mile, 31.85 over 2. So I think it's pretty much right.

I do have some scribble where I derived this, can tidy it up and post it if anyone is interested.

[Max_Almond]

Exactly. It's actually a shade under "8 inches" - but most of us find it easier to multiply by 8 than by 7.96.

It's 'correct enough' - though disappointing to see how many people use it for calculating the 'hidden amount' of a distant object, which is something different.

[ElTrancy]

Nevermind.

Please tell me you're not going to leave here without posting "the correct equation".

I had to leave that day, and I realized that maybe I was wrong, I apologize for my short response. 

[ElTrancy]

To sum up what I was saying. Slight inaccuracies can lead to huge mistakes. (Do I really need to make a reference here?) It's just better not to take everything and round it up or down, because although it isn't much at first, it can make a difference.

[Max_Almond]

Very true. After about 100 miles, "8 inches per mile squared" isn't accurate enough to reflect the true figure.