Thu, 12/21/2017 - 20:25 — sle

So on that day in Austin with the flat Earth model the sun should never have gotten closer than 20.6° to the horizon, but it dd get that close. Each sunrise and sunset the sun is 0°.

As a result of debating the nature of the sun with flat Earthers I decided to measure the sun in order see if the flat Earth model or the globe Earth model better describes the sun. Although I didn't expect the result to surprise me I wanted to be able to say I tried it myself and not just that I trust astronomy since flat Earthers are skeptical of astronomy. I'm particularly concerned with the angular size of the sun throughout the day and it's altitude (angle with the horizon). Feel free to jump to the conclusion at the end.

Although flat Earthers don't all agree it is common for them to believe that the earth is laid out like a North polar azimuthal projection with the North poll at the center around which the celestial bodies, including the moon and the sun, circle clockwise roughly 3,000 miles up. See the animation on this flat Earth website. My analysis assumes this flat Earth model.

If the flat Earth model is true then the sun should be closer at noon than it is at sunrise or sunset. In fact, if the sun is a sphere, as my photos suggest, then we'd expect the angular size to change almost exactly inversely with distance. Fortunately it's easy to calculate the distance to the sun from the observer on the flat Earth model with a bit of trig:

` sin(altitude) = 3000 / distance`

solving for distance:

` distance = 3000 / sin(altitude)`

At roughly 0.5° in size the relationship between angular size and width is almost exactly linear. The width will be measured in pixels since the zoom is constant. Therefore we can do a bit more math to find the expected linear size as function of the altitude relative to a reference distance, which will be noon:

` size = constant / distance`

` size = constant * sin(altitude) / 3000`

` size = (constant / 3000) * sin(altitude)`

So the size is proportional to the sine of the altitude, which means:

` size / noon_size = sin(altitude) / sin(noon_altitude)`

Multiplying both sides by noon_size gives produces the formula that the rest of this page refers to as the "size formula":

` size = noon_size * sin(altitude) / sin(noon_altitude)`

I bought a Meade EclipseView 82 for the 2017 August eclipse. I took pictures with my Nexus 6 phone by holding my phone up to the eyepiece. I took pictures of the sun directly using a free theodolite program named Dioptra, which I recommend. A you'll see in the pictures it recorded information about the location and orientation of the photo, but the altitude (angle to the horizon) on the right side is what matters for this discussion. All photos were taken December 21st 2017, which happens to be the Winter solstice:

*Morning*

Here's a photo where I removed the solar filter from the telescope and then held it up to my phone to take a picture. Note that you shouldn't take a picture of the sun without a filter for same the reason you shouldn't look at the sun without a filter - it can damage the camera. Also note that the sun is quite small with no zoom since it's only about 0.5° in size. Like all of the photos you can click on photo to see the original resolution (but still cropped). I rotated this image 90° clockwise since I took it in portrait mode, but that shouldn't matter. All of the Dioptra photos were taken with the same sun filter held up to my camera:

The sun at 2017:12:21 09:13:11 Austin time. We see from timeanddate.com that the sun should have an altitude (angle above the horizon) of about 19° at that time. Since the width of the sun at noon is 173 pixels and the angle at noon is 36 degrees with can use the the size formula from the Flat Earth Model section and find that the expected flat Earth sun width relative to noon is 173 * sin(19) / sin(36) = 95 pixels, which is shown with the green circle. The large outer circle that barely fits in the image is an artifact of the telescope eye piece. I find it reassuring that cropping to that large circle produces images that are approximately the same size. Click on the image to see the original but truncated image without the green circle:

The sun using Dioptra showing a difficult to see of altitude of 19.4° on the right side, so timeanddate.com is correct:

*Noon*

The sun at approximately noon at 2017:12:21 12:32:57. Notice that there's no green circle since this is the reference time - the flat Earth sun is assumed to be this size at this time:

The sun using Dioptra showing an altitude of 36.1° on the right side, so timeanddate.com is correct:

*Evening*

The sun in the evening at 2017:12:21 16:20:10. In this case the green circle representing the flat Earth sun is 173 * sin(13°) / sin(36°) = 66 pixels:

The sun using Dioptra showing an altitude of 12.2° on the right side, so timeanddate.com is correct:

There are flat Earth attempts to explain the sun maintaining a constant angular size in spite of varying distance as shown on this page. Many of the photos on the link given show lights spreading out due to lens flare. The shape of the light source can not be clearly seen which is in contrast to the sun with proper filtering. In the case of the sun with proper filtering we can see it's a circle or sphere with a distinct boundary. The article suggests that the magnification becomes more pronounce with weather which would be in contrast to the constant 0.5°. It also seems they should have photos of the sun instead of lights.

There are some cases when the atmosphere magnifies objects horizontally due to a change in the density gradient (second order gradient), but it would be odd for there to change in the gradient horizontally.

If we're considering a systematic distortion that causes the sun to appear to be same size in spite of it's distance changing drastically we may as well consider a complete point to point mapping that maps light from where it's supposed to be in the flat Earth model to where it is actually seen in the heliocentric model. Such a mapping would be very complicated. For example, the distortion wouldn't simply be a function of the object's altitude, it would also be a function of the viewer's location. I doubt it's possible to augment the refraction equation so that the mapping would be a local property (not just that the system magically does it as a whole) of the air (or any ether like medium flat Earthers might want to imagine). As it is at any point the amount of refraction is proportional to and in the direction of the air density gradient at a right angle to the light, which is relatively simple and understandable. How could we have such a simple understandable local mechanism for the required mapping?

There is also the fact that the brightness, or apparent magnitude, of the sun does not appear to change as a function of altitude. For example, the solar panel industry assumes that, weather permitting, only the angle the sunlight makes with the solar panel matters, not the distance to the sun, as can be seen on this page. Magnification does not by itself solve this problem as it dose not increase the light energy. As can be seen with telescopes there is a trade off between brightness and magnification.

One way of solving this brightness variation is in addition to the point to point mapping mentioned in the magnification section would be for the brightness of objects to change as function of the viewing angle, but it's not clear why this would be, and in particular it's not clear why the variation just happens to produce a constant brightness after the assumed point to point mapping.

The pictures were taken on December 21st 2017 which was the Winter solstice which means the sun was over the tropic of Capricorn. We can use this information to calculate the minimum altitude the sun for the sun in the flat Earth model given that timeanddate.com says that December 21st was 10:11:40 in length which is 10.194 hours. The sun should be due South at maximum altitude (roughly noon) with the sunrise and sunset 10.194 / 2 = 5.097 hours before and after noon. 5.097 hours corresponds to 360 * 5.097 / 24 = 76.46° on either side of due South. Imagine rotating the flat Earth so that Austin is directly below the North pole and then overlaying an X-Y coordinate system with the origin at the North pole, The Y coordinate pointing down toward Austin and the X coordinate pointing to the right. We can now calculate the coordinates of Austin at the flat Earth sun at sunrise.

Xaustin = 0 miles Yaustin = 69.2 * (90 - 30.27) = 4133 miles Xsun = 69.2 * (90 + 23.4) * sin(76.46) = 7629 miles Ysun = 69.2 * (90 + 23.4) * cos(76.46) = 1837 miles distance = sqrt((Xaustin - Xsun)^2 + (Yaustin - Ysun)^2) = 7967 miles altitude = atan(3000 / 7967) = 20.6°

So on that day in Austin with the flat Earth model the sun should never have gotten closer than 20.6° to the horizon, but it dd get that close. Each sunrise and sunset the sun is 0°.

From the original larger telescope photos of the sun we can see that the smallest width is in the morning at 378 pixels and the largest width is at noon at 384 pixels, so the noon sun is 1.0158 times as wide, or about 1.6% bigger. Although admittedly the flat Earth model expects the noon sun to be larger than the morning or evening sun we can use the size formula in the flat earth model section to see that the size is in proportion to the sine of the altitude, so for the evening sun, which had the lowest altitude of 13 degrees, the expected difference in size should be sin(36) / sin(13) = 2.6129 times as big at noon as in the evening, or 161.3% bigger. The observed 1.6% variation seem well within the margin of error given that I was holding my cell phone up to the telescope eye piece. However, 161.3% bigger is not within the margin of error. No such variation was seen. Given that and the lack of a plausible magnification mechanism to keep the size constant, or plausible brightness mechanism to keep the brightness constant the flat Earth model seems impossible.