[Curious Squirrel]

All starting numbers for this calculation will be pulled directly from the FES wiki. If there is an issue with any of them, take it up with the editor of the wiki.

On the equinox, the sun is directly 90° above a point on the equator. At 45° latitude N and at 45° latitude S along the same line of longitude, the sun is seen as being at 45° up from the horizontal. The distance between the equator and these two points is about 3000 miles. Thus, the sun is 3000 miles in the air. Right? Well, let's see what happens when we shift around some, shall we? The sun should show up at that distance no matter where we look from, right?

Let's cut our distance in half, going from 3000 miles away and a 45° angle, to 1500 miles away, and a 67.5° angle. This gives us a sun height of... 3621 miles. Oh. Well that's quite a bit different isn't it? Well how about if we go the other direction. Halfway from 45° to 0°, giving us an angle of 22.5° and a distance of 4500 miles. This gives us a sun height of.... 1864 miles. Oh dear, that's even more out of whack than our first one. Well what about if we just go out to the poles. That's about 6000 miles, but to make sure the angle is above 0 like it is there, we'll go to just 5900 miles and see what the angle is there. This should give us an angle well under 10° at most. So with a 3000 mile high sun we get an angle of... 27°

Welp, I see three options here. 1) There's something making the light from the sun bend to get the correct angles all across the globe. It's consistent in all types of weather, all across the Earth, every day of the year. The only thing I can think of for this, is magic. Nothing else makes sense. 2) The Earth isn't flat. You can't have a sun that is simultaneously 3621 miles up AND 1864 miles up AND 3000 miles up. That's not possible. Therefore the idea of Earth being flat with a close sun, cannot work. 3) We have incorrect measurements somewhere. I started with numbers in the wiki though, and every other number simply follows what holds to be true mathematically. So either the wiki is wrong, or we're back to one of the other two options. Actually even if the wiki is wrong, that would leave us at one of the other options being true still.

So FEers. What's it gonna be? You can't have light traveling in a straight line AND a flat Earth.

[Tom Bishop]

Let's cut our distance in half, going from 3000 miles away and a 45° angle, to 1500 miles away, and a 67.5° angle. This gives us a sun height of... 3621 miles. Oh. Well that's quite a bit different isn't it? Well how about if we go the other direction. Halfway from 45° to 0°, giving us an angle of 22.5° and a distance of 4500 miles. This gives us a sun height of.... 1864 miles. Oh dear, that's even more out of whack than our first one.

Who went out and made those observations of where the sun is at those latitudes on that day?

[Curious Squirrel]

Let's cut our distance in half, going from 3000 miles away and a 45° angle, to 1500 miles away, and a 67.5° angle. This gives us a sun height of... 3621 miles. Oh. Well that's quite a bit different isn't it? Well how about if we go the other direction. Halfway from 45° to 0°, giving us an angle of 22.5° and a distance of 4500 miles. This gives us a sun height of.... 1864 miles. Oh dear, that's even more out of whack than our first one.

Who went out and made those observations of where the sun is at those latitudes on that day?

That's how latitudes work, and latitudes are evenly spaced across the Earth. Your original author made no mention of measuring his 45° angle or the distance to said location. In fact, he uses 1° of declination being equal to 60 miles it would seem. Perhaps I should redo this using his more exact number of 2700 miles instead of the wiki's listed 3000 miles? Although I doubt that will change the results.

I'm simply using the exact same assumptions used in the reference material on the wiki.

Actually since he gives us some more precise measurements, let's test those. At 67.5° we get a distance of 1350 between it and the equator, and a sun height of 3259 compared to his 2700. At 22.5 we get a distance of 4050 miles to the equator, and a sun height of 1677 to his 2700. These don't appear to be working any better Tom.

[Tom Bishop]

Let's cut our distance in half, going from 3000 miles away and a 45° angle, to 1500 miles away, and a 67.5° angle. This gives us a sun height of... 3621 miles. Oh. Well that's quite a bit different isn't it? Well how about if we go the other direction. Halfway from 45° to 0°, giving us an angle of 22.5° and a distance of 4500 miles. This gives us a sun height of.... 1864 miles. Oh dear, that's even more out of whack than our first one.

Who went out and made those observations of where the sun is at those latitudes on that day?

That's how latitudes work, and latitudes are evenly spaced across the Earth. Your original author made no mention of measuring his 45° angle or the distance to said location. In fact, he uses 1° of declination being equal to 60 miles it would seem. Perhaps I should redo this using his more exact number of 2700 miles instead of the wiki's listed 3000 miles? Although I doubt that will change the results.

I'm simply using the exact same assumptions used in the reference material on the wiki.

Actually since he gives us some more precise measurements, let's test those. At 67.5° we get a distance of 1350 between it and the equator, and a sun height of 3259 compared to his 2700. At 22.5 we get a distance of 4050 miles to the equator, and a sun height of 1677 to his 2700. These don't appear to be working any better Tom.

If your argument is not based on any real observations, why should we conclude that you are correct?

[inquisitive]

Let's cut our distance in half, going from 3000 miles away and a 45° angle, to 1500 miles away, and a 67.5° angle. This gives us a sun height of... 3621 miles. Oh. Well that's quite a bit different isn't it? Well how about if we go the other direction. Halfway from 45° to 0°, giving us an angle of 22.5° and a distance of 4500 miles. This gives us a sun height of.... 1864 miles. Oh dear, that's even more out of whack than our first one.

Who went out and made those observations of where the sun is at those latitudes on that day?

That's how latitudes work, and latitudes are evenly spaced across the Earth. Your original author made no mention of measuring his 45° angle or the distance to said location. In fact, he uses 1° of declination being equal to 60 miles it would seem. Perhaps I should redo this using his more exact number of 2700 miles instead of the wiki's listed 3000 miles? Although I doubt that will change the results.

I'm simply using the exact same assumptions used in the reference material on the wiki.

Actually since he gives us some more precise measurements, let's test those. At 67.5° we get a distance of 1350 between it and the equator, and a sun height of 3259 compared to his 2700. At 22.5 we get a distance of 4050 miles to the equator, and a sun height of 1677 to his 2700. These don't appear to be working any better Tom.

If your argument is not based on any real observations, why should we conclude that you are correct?

Says he who will not make his own observations.

[Curious Squirrel]

Let's cut our distance in half, going from 3000 miles away and a 45° angle, to 1500 miles away, and a 67.5° angle. This gives us a sun height of... 3621 miles. Oh. Well that's quite a bit different isn't it? Well how about if we go the other direction. Halfway from 45° to 0°, giving us an angle of 22.5° and a distance of 4500 miles. This gives us a sun height of.... 1864 miles. Oh dear, that's even more out of whack than our first one.

Who went out and made those observations of where the sun is at those latitudes on that day?

That's how latitudes work, and latitudes are evenly spaced across the Earth. Your original author made no mention of measuring his 45° angle or the distance to said location. In fact, he uses 1° of declination being equal to 60 miles it would seem. Perhaps I should redo this using his more exact number of 2700 miles instead of the wiki's listed 3000 miles? Although I doubt that will change the results.

I'm simply using the exact same assumptions used in the reference material on the wiki.

Actually since he gives us some more precise measurements, let's test those. At 67.5° we get a distance of 1350 between it and the equator, and a sun height of 3259 compared to his 2700. At 22.5 we get a distance of 4050 miles to the equator, and a sun height of 1677 to his 2700. These don't appear to be working any better Tom.

If your argument is not based on any real observations, why should we conclude that you are correct?

The argument on your own wiki must therefore be incorrect as well. I used nothing but sources listed on the wiki of this website for my information. As I stated, in bold at the top of the first post so no one could miss it.

[StinkyOne]

Let's cut our distance in half, going from 3000 miles away and a 45° angle, to 1500 miles away, and a 67.5° angle. This gives us a sun height of... 3621 miles. Oh. Well that's quite a bit different isn't it? Well how about if we go the other direction. Halfway from 45° to 0°, giving us an angle of 22.5° and a distance of 4500 miles. This gives us a sun height of.... 1864 miles. Oh dear, that's even more out of whack than our first one.

Who went out and made those observations of where the sun is at those latitudes on that day?

That's how latitudes work, and latitudes are evenly spaced across the Earth. Your original author made no mention of measuring his 45° angle or the distance to said location. In fact, he uses 1° of declination being equal to 60 miles it would seem. Perhaps I should redo this using his more exact number of 2700 miles instead of the wiki's listed 3000 miles? Although I doubt that will change the results.

I'm simply using the exact same assumptions used in the reference material on the wiki.

Actually since he gives us some more precise measurements, let's test those. At 67.5° we get a distance of 1350 between it and the equator, and a sun height of 3259 compared to his 2700. At 22.5 we get a distance of 4050 miles to the equator, and a sun height of 1677 to his 2700. These don't appear to be working any better Tom.

If your argument is not based on any real observations, why should we conclude that you are correct?

And you base your worldview on what, exactly? I've yet to see any hard evidence that the world is flat. Beyond an unscientific experiment and photos that claim to show things over the horizon (both of which are explained by refraction), I fail to see any legitimate reason to even begin to think the Earth is flat.

I actually have an honest question, why? You seem intelligent enough and not too prone to conspiracy-think. (except for the military stuff) How did you come to the conclusion that the whole world was wrong and that the Earth is flat?

[3DGeek]

Let's cut our distance in half, going from 3000 miles away and a 45° angle, to 1500 miles away, and a 67.5° angle. This gives us a sun height of... 3621 miles. Oh. Well that's quite a bit different isn't it? Well how about if we go the other direction. Halfway from 45° to 0°, giving us an angle of 22.5° and a distance of 4500 miles. This gives us a sun height of.... 1864 miles. Oh dear, that's even more out of whack than our first one.

Who went out and made those observations of where the sun is at those latitudes on that day?

Just about every seaman who ever did celestial navigation - from about 1600 to the 1960's.

If there was a discrepancy of more than a small fraction of a degree - you could be VERY sure we'd know about it.

[LuggerSailor]

Let's cut our distance in half, going from 3000 miles away and a 45° angle, to 1500 miles away, and a 67.5° angle. This gives us a sun height of... 3621 miles. Oh. Well that's quite a bit different isn't it? Well how about if we go the other direction. Halfway from 45° to 0°, giving us an angle of 22.5° and a distance of 4500 miles. This gives us a sun height of.... 1864 miles. Oh dear, that's even more out of whack than our first one.

Who went out and made those observations of where the sun is at those latitudes on that day?

We can collectively make observations;

No one else observed the sun's elevation at the solstice?

The geometry is quite interesting. I'm about 77.25° North of the Tropic of Capricorn. Each degree is 60 Nautical miles which works out that I'm 5334 miles North of the Tropic of Capricorn. Using trigonometry to calculate the vertical height of the sun above the Tropic on a flat plane, 5334*tan(13) = 1234. That doesn't fit with a 3000 mile height of the sun above a flat plane.
The 13° elevation does fit with the 77° angle between the Tropic and my location on a globe with a far distant sun (ok, I might have been 1/4 of a degree off in my measurement of the sun's elevation)

Autumn Equinox in a couple of weeks, who's up for measuring the elevation angle of the sun at their local noon (adjusting for daylight saving time) and posting the angle and their latitude?

[StinkyOne]

Let's cut our distance in half, going from 3000 miles away and a 45° angle, to 1500 miles away, and a 67.5° angle. This gives us a sun height of... 3621 miles. Oh. Well that's quite a bit different isn't it? Well how about if we go the other direction. Halfway from 45° to 0°, giving us an angle of 22.5° and a distance of 4500 miles. This gives us a sun height of.... 1864 miles. Oh dear, that's even more out of whack than our first one.

Who went out and made those observations of where the sun is at those latitudes on that day?

We can collectively make observations;

No one else observed the sun's elevation at the solstice?

The geometry is quite interesting. I'm about 77.25° North of the Tropic of Capricorn. Each degree is 60 Nautical miles which works out that I'm 5334 miles North of the Tropic of Capricorn. Using trigonometry to calculate the vertical height of the sun above the Tropic on a flat plane, 5334*tan(13) = 1234. That doesn't fit with a 3000 mile height of the sun above a flat plane.
The 13° elevation does fit with the 77° angle between the Tropic and my location on a globe with a far distant sun (ok, I might have been 1/4 of a degree off in my measurement of the sun's elevation)

Autumn Equinox in a couple of weeks, who's up for measuring the elevation angle of the sun at their local noon (adjusting for daylight saving time) and posting the angle and their latitude?

I like this idea. Easy way to do a little group science project and confirm or deny a little RET.

http://designcoalition.org/kids/energyhouse/sunangles.htm

[3DGeek]

I like this idea. Easy way to do a little group science project and confirm or deny a little RET.

http://designcoalition.org/kids/energyhouse/sunangles.htm

The trouble with this idea is that Tom denies that the "laws of perspective" are correct...which means that any data you *DO* come up with will immediately be ignored on grounds that perspective did...well....um....er...something...it's not quite clear what.

Since he's simultaneously saying that light travels in straight lines *AND* that there can be sunsets (which are STRONGLY contradictory claims) - it's not beyond the realms of his confusion to roll in a distortion of the angles you measure.

What's needed at this point is to FULLY understand what he's saying about perspective.

I **KNOW** he's wrong because the pinhole camera thought-experiment proves it conclusively - but it's hard to nail down exactly what he's claiming because all we get is wooly thinking.

[Curious Squirrel]

I like this idea. Easy way to do a little group science project and confirm or deny a little RET.

http://designcoalition.org/kids/energyhouse/sunangles.htm

The trouble with this idea is that Tom denies that the "laws of perspective" are correct...which means that any data you *DO* come up with will immediately be ignored on grounds that perspective did...well....um....er...something...it's not quite clear what.

Since he's simultaneously saying that light travels in straight lines *AND* that there can be sunsets (which are STRONGLY contradictory claims) - it's not beyond the realms of his confusion to roll in a distortion of the angles you measure.

What's needed at this point is to FULLY understand what he's saying about perspective.

I **KNOW** he's wrong because the pinhole camera thought-experiment proves it conclusively - but it's hard to nail down exactly what he's claiming because all we get is wooly thinking.

I think this is what we need at this point. Precise definitions of what perspective does, and how it works. Because as noted in another thread, the version of perspective we have at present also won't allow someone to see further the higher they go. That doesn't work with how perspective is supposed to function in order for the sun to set. I think? Once again, a reason why we need to get this bit of jello frozen so we can properly nail it down.

But again, all of my measurements are based on claims in the wiki. This is one of those science things that zeteticism seems to have a problem with. Each person doesn't have to reinvent the wheel every time someone new comes along and wants to roll something. I can base my claims on the information provided by others that are proven accurate. Since the source is in the wiki, I have to assume it's an accurate source, at least to begin with.

[inquisitive]

Perspective is about drawing and how things look in the distance, nor relevant here.

[3DGeek]

Perspective is about drawing and how things look in the distance, nor relevant here.

You'd think so - but we have:

* https://wiki.tfes.org/Basic_Perspective -- "A fact of basic perspective is that the line of the horizon is always at eye level with the observer."
* https://wiki.tfes.org/Experimental_Evidence -- "It is proven that the ship does not sink behind a hill of water, but that it is actually perspective which hides it."
* https://wiki.tfes.org/Constant_Speed_of_the_Sun -- "The sun moves constant speed into the horizon at sunset because it is at such a height that already beyond the apex of perspective lines. It has maximized the possible broadness of the lines of perspective in relation to the earth. It is intersecting the earth at a very broad angle."
* https://wiki.tfes.org/Viewing_Distance -- "since man cannot perceive infinity due to human limitations, the perspective lines are modified and placed a finite distance away from the observer"
* https://wiki.tfes.org/Magnification_of_the_Sun_at_Sunset -- "The sun remains the same size as it recedes into the distance due to a known magnification effect caused by the intense rays of light passing through the strata of the atmolayer."
* https://wiki.tfes.org/Optics -- Talks about "the electromagnetic accelerator" - which Tom said he does not believe.  But it also says that the limits of the resolution of the human eye are an explanation...which is weird because 4kx4k TV cameras, telescopes and old-fashioned pinhole cameras all produce images that match what the "limited" human eye can see?!
* https://wiki.tfes.org/Shifting_Constellations -- has a long "Rowbotham" quote about how the pole star vanishes as you approach the equator because of perspective.
* https://wiki.tfes.org/Sinking_Ship_Effect -- more of the same basic nonsense.

...I count more than a dozen references to this peculiar effect - but none of them pass the "Pinhole camera test".

So it this claimed to be a human failing?   If so, how come cameras suffer from the same effect?

[Rounder]

Who went out and made those observations of where the sun is at those latitudes on that day?

Well, in your favorite book Rowbotham reported observations from which "it is perfectly safe to affirm that the under edge of the sun is considerably less than 700 statute miles above the earth."  His words, not mine.

[3DGeek]

Who went out and made those observations of where the sun is at those latitudes on that day?

Well, in your favorite book Rowbotham reported observations from which "it is perfectly safe to affirm that the under edge of the sun is considerably less than 700 statute miles above the earth."  His words, not mine.

If it were only 700 miles up, at least sunsets would be a bit more credible than if it's 3000 miles up.   Of course many other things would go horribly wrong.

The figure of 3000 miles (I think they should use 4000) comes about because that's the height it needs to be to explain a few other phenomenon.  The fact that the number is roughly the radius of the (round) earth is no coincidence because it comes about from the same "length of shadow" experiment that Eratosthenes used back in 250BC:

    https://wiki.tfes.org/Erathostenes_on_Diameter

(You REALLY would hope that someone would fix the spelling of "Eratosthenes" on that page!)

If you assume that the Earth is flat, then Eratosthenes was actually measuring the height of the sun, not the radius of the Earth...and the answer would be the same.

The height of the FE sun MUST equal the radius of the RE Earth in order for "stick and shadow" experiments to come out right.

Eratosthenes came up with a number that's between 10% and 15% larger than the modern value.  But a careful recreation of his experiment done in 2012 came up with the 4,000 mile figure with an error of only 0.16% compared to the "accepted" number.

If you take the Rowbotham 700 mile figure - you can trivially disprove it with sticks and shadows,

The 3,000 mile number found in the Wiki and used in all discussions here can be disproved in the same manner - but it takes quite a bit of effort to do the experiment accurately enough without using GPS and other "forbidden" RET techniques.

So if I were an FE'er - I'd be claiming that the sun was 4,000 miles above the ground...but 3,000 miles is every bit as hard to defend.