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# Sun’s angular elevation

It’s easy to calculate the expected angular elevation of the sun for the flat earth. Flat Earthers often claim the sun is 3,000 miles above the flat earth, sometimes more, sometimes, less. We will use 3,000 miles. It doesn’t change the outcome much, use your own numbers if you prefer.

To calculate the angular elevation we use a simple right triangle. We place the sun at point “B”, and the observer at point “A”. We use the elevation of the sun, 3,000 miles for side “a” and measure the distance the observer is from the point under the sun for side “b”.

Use “Side Angle Side” to calculate all the sides and angles of the triangle manually. It’s easier to use an online calculator like this one. For this problem, use calculator #2 on that web site.

The AE map is used for this example, but any flat earth map imaginable could be used with similar results. I have used Walter Bislin’s Creating Flight Plans for Flat Earth web page to calculate distances on the flat earth AE map. If you have a different method to calculate distances on the flat earth, send feedback to mctoon@mctoon.net and I’ll include alternative measurements in all my calculations.

Let’s examine a couple sample locations and also pick the most extreme possible observer location in favor of a flat earth.

### Example 1: Quito, Ecuador, Equinox

Quito is on the equator, at 0°N 78.5W. We will calculate the minimum angular elevation of the sun on the day of the equinox. At local solar noon the sun is directly overhead so the angle is 90°. At local solar midnight the sun will be directly over Indonesia on the opposite side of the world, still on the equator. The distance is 12,436 miles. Plugging in to the triangle calculator we get an angular elevation of 13.56°.

### Example 2: Minneapolis, Northern Solstice

Minneapolis, MN, USA is at 45°N 93W. On the Northern Solstice the sun travels over the Tropic of Cancer at 23.5° North latitude. At local solar noon the sun is the closest to Minneapolis it ever will be all year, directly over the Gulf of Mexico, 1,486 miles away. At local solar midnight, the sun is directly over Myanmar, 7,676 miles away. Plugging into the triangle calculator we get an angular elevation of 21.35°.

This means, according to flat earth theory, in the middle of the night in Minneapolis, the sun’s elevation is 21.35° above the horizon.

### Example 3: Puerto Williams, Southern Solstice

Puerto Williams is on the far southern tip of Chile at 54°56′S 67°37′W. It’s one of the southernmost cities on the earth. On the day of the southern solstice, at local solar midnight, the sun is over the western tip of Australia, 17,861 miles away from Puerto Williams. Plugging that distance into the triangle calculator we get an angular elevation of 9.53°

This is the absolute lowest angular elevation possible for the sun on the AE map from land other than Antarctica. Everywhere else on Earth will always have a higher angular elevation at local solar midnight.

### Example 4: Forcing the outcome, Total imagination

What conditions are required for the sun to get really close to the horizon if the earth were flat? How far away and how low would the sun need to be to make it work?

- Lower the sun to 1000 miles and use the same distance as Example 3, the angular elevation is 3.20°, still too high to even consider to be setting.
- How about 1,000 miles high and 30,000 miles away? That is 1.91°, still too much. For reference, the angular size of the moon is 0.5°. That’s more than 3 moon widths above the horizon.
- How about 1000 miles high and 100,000 miles away? The elevation is 0.57°, a little more that the width of the moon. Still a clear noticeable gap between the horizon and the sun.
- The visible limit of the human eye is about 1 arcminute or 0.02°, this is considered the vanishing point. In order for the sun and the horizon to be within 0.02° of each other, the sun must be 1,000,000 miles away and only 350 miles high.
- What if we left the sun at 3,000 miles elevation and forced the sun to be in the vanishing point of the flat earth horizon? How far away would it need to be? The sun would need to be 8,500,000 miles away to appear on the horizon.

### Physical Model

This is pretty easy to model physically. Find a hallway, measure it’s length and height so you know the limits of what you can model. Select the angular elevation you want to model and plug the numbers into a triangle to find the length and height.

I have a 120-foot long hallway that is 8 feet high. To model the example that is the most in favor of a flat earth, Puerto Williams, we need an angular elevation of 9.53°. I have a light with a magnet that will stick to the ceiling, it’s 6 inches long. Calculating a triangle with a 7.5-foot elevation and 9.53° the observer needs to be 45 feet away from the point directly under the stand-in sun.

When I did this, the fluorescent light on the ceiling completely out-glared the light I stuck to the ceiling. As you can see in the picture the stick light is not visible. For this model, this is highly in favor of the flat earth theory. The light source is significantly larger than the sun would be, yet it is still a long way from the floor. There is still a large visible gap between the light and the floor.

### Special Case: Dome

A lot of people see this impossibility and reach for the dome claim to resolve it. I figure the thought is “the sun is literally high up, the dome is curved, that must make it appear lower, duh, MCToon, you’re stupid”.

Of course, there is no evidence of a dome at all, but people really love their dome. So, I’ll make an exception for the dome: model the dome. Do the math, calculate the angles, plot it out. Use 3D apps, whatever. Do it for the three examples above. Then, use suncalc.org to get the observed sunrise, local solar noon and sunset locations for the three examples above and include these observations into the same dome model. Then use the same dome model to model the angular size of the sun to match the observed size of the sun, about 30 arcminutes, all day every day.

If you can model the dome to account for just these three examples, for just these three observations, I’m in. I’ll post a video showing the model. I’ll contact all my globe head friends and beg them to mirror the details on this revolutionary information.

If you can’t model the dome, provide evidence of the dome. How far is it? How was the distance measured? What is it made of? What are the refractive properties? Are the sun, moon, and stars projected on it from inside or outside? Provide real, substantial evidence.

If you can’t model the dome or provide evidence of the dome, don’t bring it up. I’m not gullible, you shouldn’t have been.

### Conclusion

At the location most beneficial to flat earth, the lowest angular elevation of the sun in the middle of the night is 9.53°. This is too high for any known phenomenon to account for a setting sun. There is absolutely no way this could be considered a sunset.

When someone claims the sun sets due to:

- Perspective
- Circular perspective
- Atmospheric lensing
- Refraction
- Diffraction
- Reflecting off the dome
- Bendy light
- Electromagnetic acceleration
- Angle of attack
- Rayleigh criterion
- Projections
- Holograms
- Luminaries
- Or anything else

Just ask them to explain the phenomenon responsible for dropping the apparent location of the sun by at least 9.53°. Since this is likely an as-yet unknown phenomenon, be sure to request formulas, derivations, and calculations to account for this angular declination.

Are you are a flat earther and think I’m wrong? I’m open to your ideas. Please send your response to mctoon@mctoon.net. Any polite emails will be added directly below.

## Feedback

Twitter user @T__E__S__L__A: “What should I prove you wrong about? That webpage speaks about linear perspective and it does not apply to Sun.”

If I understand this odd claim, he is suggesting that the sun is inexplicably exempt from the regular behavior of literally everything else humans have ever seen or experienced. Instead, the sun follows “circular perspective”. He has an image as well posted her for reference. Unfortunately, there is zero supporting evidence for this claim. When asked for this supporting evidence, he was unable to provide any. No formulas, no derivations, nothing to support this. Since there’s no evidence, it gets dismissed.

Back to Sunset and Moonset on Flat Earth

Great discussion, afraid it will shoot way over the heads of the flerfers, or else in one ear and out the other, with no stops in between. Nothing in between of course!

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If you really want to make their heads hurt, ask them to explain how an equatorial telescope mount works on their FE “model”. Of course it can’t possibly work, but then again I doubt they can even understand how they work on a correct spherical earth, in a heliocentric solar system.

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