[StinkyOne]

Scientific validity requires repeatability.  Despite repeated requests, Tom has never identified the specific beach from which he conducted the observations, so we cannot repeat it.

I managed to get "500x" out of him regarding the specs of the telescope.

This tells me he was using a department store telescope. To get true 500x magnification, he would have to be using a scope with ~10" aperture to get a clear picture.

[3DGeek]

i'll save you both some time.  it doesn't make any difference at the lengths you're talking about.  no one is doing any bedford-style experiments over 100+ mile distances.  visibility affects your measurements long before the errors accumulate.

How about just use the right equation?

[devils advocate]

Here...let me save you some time...



The green circle is the correct result...your "eight inches per mile squared" is in red - both axes are in miles.

Like I said, it's the WRONG equation.  Kinda-sorta-maybe-vaguely-right for VERY short distances...but not definitive.

The error doesn't look much on this graph - but bear in mind - the vertical scale is 500 *miles* to each grid square - and we're arguing about 10's of feet.

It's typical of an FE'er to CAREFULLY trim the graph so that his horrible error is just off the edges - and to hope that the rest of us are too stupid not to know the correct answer.   Honestly...you must have peeked another 1000 miles off the the side...seen your problem...hoped none of us would call you on it.

Well, guess what Boots?   We're not as dumb as you think we are!

Thank you 3D, I thought you'd be the one to make it make sense :-) so basically Tom has fudged his figures (again) and his experiment proves only his lack of knowledge......Come on Tom Bishop, this one's got your name on it!.....

[3DGeek]

Here...let me save you some time...



The green circle is the correct result...your "eight inches per mile squared" is in red - both axes are in miles.

Like I said, it's the WRONG equation.  Kinda-sorta-maybe-vaguely-right for VERY short distances...but not definitive.

The error doesn't look much on this graph - but bear in mind - the vertical scale is 500 *miles* to each grid square - and we're arguing about 10's of feet.

It's typical of an FE'er to CAREFULLY trim the graph so that his horrible error is just off the edges - and to hope that the rest of us are too stupid not to know the correct answer.   Honestly...you must have peeked another 1000 miles off the the side...seen your problem...hoped none of us would call you on it.

Well, guess what Boots?   We're not as dumb as you think we are!

Thank you 3D, I thought you'd be the one to make it make sense :-) so basically Tom has fudged his figures (again) and his experiment proves only his lack of knowledge......Come on Tom Bishop, this one's got your name on it!.....

I've seen this 8 inches per mile-squared number before - not even in relation to Flat Earthism - it was mentioned in a research paper on flight simulation a decade or so ago.  I was doing a peer review on it and had to insist that the equation be fixed for publication.

I knew it had to be inaccurate over longer distances - so I researched where it comes from.

It was actually published in a Napoleonic Wars era manual for British naval artillery battery commanders.  Since their weapons only ranged out to 3,200 meters (about 2 miles) - the rule implies an impact point 32" *above* what you'd expect due to Earth curvature - but its hard to see why this rule-of-thumb was even important compared to the 'fall of shot' due to gravity.  But that's the first reference to this approximation I could find.

[mtnman]

I haven't dug into the math behind it, but there is a calculator online at https://dizzib.github.io/earth/curve-calc

It's open source, so anyone can review it.

Just doing a couple of test calculations on it, it does give the results that match the 8 inches/mile².

But I see a very basic thing that people quoting the 8 inch do wrong. They don't understand that there are two inputs to the equation. The distance of course, but also the the viewing height. They imagine standing on the beach and looking at something ten miles away. 10 miles squared * 8 inches = 66.6867 feet obscured. But by doing the calculation that way, they have implicitly set the viewing height to zero.

Say you're standing on a parking lot at the beach, maybe your eye level is 15 feet above Earth/sea level. The same calculation then results in 18.4317 feet obscured. A significant difference.

[garygreen]

i'll save you both some time.  it doesn't make any difference at the lengths you're talking about.  no one is doing any bedford-style experiments over 100+ mile distances.  visibility affects your measurements long before the errors accumulate.

How about just use the right equation?

lol how about learn2backoftheenvelope.  the answer is the same at the distances we're talking about, and i can't do sqrt(3959^2 + 6^2) - 3959 in my head.

also sigfigs are going to cut off your errors anyway.  first order approximations are 🔥🔥🔥💯😂👍👍👌👌👍💯💯🔥💯 bro. 

[model 29]

Scientific validity requires repeatability.  Despite repeated requests, Tom has never identified the specific beach from which he conducted the observations, so we cannot repeat it.

I managed to get "500x" out of him regarding the specs of the telescope.

This tells me he was using a department store telescope. To get true 500x magnification, he would have to be using a scope with ~10" aperture to get a clear picture.

Yeah, that's a pretty big telescope.  I did the calculation for my reflector, and if I get a 2xbarlow, I can get 480x I think.  It's a Meade 4501, 4 or 4.5" mirror.  I'll have to double check all that. 

[Boots]

Here...let me save you some time...



The green circle is the correct result...your "eight inches per mile squared" is in red - both axes are in miles.

Like I said, it's the WRONG equation.  Kinda-sorta-maybe-vaguely-right for VERY short distances...but not definitive.

The error doesn't look much on this graph - but bear in mind - the vertical scale is 500 *miles* to each grid square - and we're arguing about 10's of feet.

It's typical of an FE'er to CAREFULLY trim the graph so that his horrible error is just off the edges - and to hope that the rest of us are too stupid not to know the correct answer.   Honestly...you must have peeked another 1000 miles off the the side...seen your problem...hoped none of us would call you on it.

Well, guess what Boots?   We're not as dumb as you think we are!

I'm not sure what you're on about. Of course it's a quadratic equation and will graph as a parabola. And of course it gets less and less accurate as the distance increases. I have never thought otherwise or tried to convince anyone otherwise. But it's a fine approximation over short distances. What are you all in a wad about? I bet you think I'm an FEer too. Maybe that's what's got you all riled up. I suggest you calm down. Life's not as serious as you think it is!

[3DGeek]

I haven't dug into the math behind it, but there is a calculator online at https://dizzib.github.io/earth/curve-calc

It's open source, so anyone can review it.

Just doing a couple of test calculations on it, it does give the results that match the 8 inches/mile².

But I see a very basic thing that people quoting the 8 inch do wrong. They don't understand that there are two inputs to the equation. The distance of course, but also the the viewing height. They imagine standing on the beach and looking at something ten miles away. 10 miles squared * 8 inches = 66.6867 feet obscured. But by doing the calculation that way, they have implicitly set the viewing height to zero.

Say you're standing on a parking lot at the beach, maybe your eye level is 15 feet above Earth/sea level. The same calculation then results in 18.4317 feet obscured. A significant difference.

Yes - we were having a  hard time convincing Tom of this in a couple of other threads too.

The answer (using the CORRECT equation) is extremely sensitive to eye height over the first 100 or so feet.

So when people on either side of the fence say "I just took this photo of the far shore of a lake/bay and you can see XYZ" - then unless the eye height above the water is known very exactly, the result tells you almost nothing - and the whole "eight inches per mile-squared" thing is junk...it's not even approximately right unless eye height is zero.  We'd know if your eye height was exactly zero because half of the camera lens would be underwater...and that close to the ground/ocean, even an inch of error makes a massive difference.

Worse still, people use even the "correct" equation incorrectly by discussing distant objects that are above ground level - and that adds a THIRD input - and yet a different equation again.

What convinces Tom that the Earth is flat is that he's taking a photo from the lake-shore 10' to 15' above sea level - and looking at some target that's probably 50' above sea level.

Add in the effects of grazing angle refraction with temperature and humidity "inversion" layers, tides and waves (all of which are critical because even a foot of difference matters) and you have results that are going to be junk 99% of the time.

This is why I don't use view-over-water experiments to debunk (or prove) FET.   The math is *SO* sensitive to the smallest error that it's impossible to verify anyone's results.

Just look at the mess that was the Bedford Level Experiment and the efforts to reproduce its' results...two said "Flat", three or four said "Spherical" and one person said "Concave".

To some extent, the fault here lies with naive Round Earthers who repeatedly say: "Look how ship hulls disappear over the horizon!" - thereby enabling Flat Earthers to convince themselves that they are right all over again.

It's much better to call this one an inconclusive result and talk about sunsets instead.