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.

On September 23rd 2018 I added Additional Sun Photos that I took with a Nikon P900 camera. It includes an analysis of the angular size of the sun in those photos which is then compared to Stellarium.

Flat Earth Model

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 pole 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:

Sun Photos


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 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 is correct:


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 is correct:


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 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 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°.

Additional Sun Photos

This section contains two additional photos taken by a Nikon P900 camera with a Thousand Oaks Optical solar filter. Always use a proper solar filer when photographing the sun to protect your camera and your eyes.

For both photos maximum optical zoom was used which, as can be seen in the EXIF (properties that are stored in JPEG images) information, is the equivalent of 2000 mm for 35 mm film:

exif:FocalLength: 3570/10
exif:FocalLengthIn35mmFilm: 2000

This photo was taken two minutes before solar noon (when the sun is highest in the sky) on September 2nd 2018 at 1:27:15 pm:

It does not have EXIF GPS location information, but the next photo does.

This photo was taken in the evening of September 18th at 6:35:45 pm about 56 minutes before sunset:

I didn't measure the sun's altitude directly this time, but it was certainly much lower in the second photo.

It's possible to use the 2000 mm 35 mm equivalent focal length to calculate the relationship between pixels, and the angular size of the sun. The relationship between pixels and angle is not quite a simple as there being a fixed number of degrees per pixel for a given zoom level. It is the slope per pixel (I'll call it the "SPP") that is constant. Imagine placing the camera level at the base of road that slopes upward with a known slope going up a hill. When a photo was taken the top of the hill would be a certain number of pixel above the horizontal center line. If we divide that gives us the SPP. When the same experiment is performed with another hill with a different slope the top of the hill appears a different distance above the horizontal center line, but with the same ratio. The first step is calculate the SPP.

Since photo sensors are compared to 35 mm using their diagonal length we start by calculating the diagonal for 35 mm film. 35 mm film is 36x24 mm so

diagonal_35mm = sqrt(36^2 + 24^2) = 43.27 mm

The lens is effectively 2000 mm above the center of the sensor, so from the center to a corner is 43.27 / 2 = 21.6 mm. Considering the center of the sensor level the slope up to the corner is 21.6 mm / 2000 mm = 0.0108. Since images at maximum resolution for the P900 are 4608x3456 the number of pixels in the same half diagonal distance is sqrt(4608^2 + 3456^2) / 2 = 2880. The SPP is then:

SPP = 0.0108 / 2880 = 3.75e-06

The HFOV (horizontal field of view) is twice the angle from the vertical center line to one side, which is:

HFOV = 2 * atan(SPP * (4608 / 2)) = 0.990°

Similarly for the VFOV:

VFOV = 2 * atan(SPP * (3456 / 2)) = 0.743°

Similarly for each sun image the angular size, which is measured horizontally due to possible refraction, the angle is measured in two parts which are added together:

angular_size = angular_size_left_of_center + angular_size_right_of_center

Measuring in pixels with the vertical center line at x=4608/2 and applying the SPP concept derived above this becomes:

angular_size = atan(SPP * ((4608 / 2) - left_pixel)) + atan(SPP * (right_pixel - (4608 / 2)))

Stellanium is free open source planetarium software. We can view the sun for the same time and location and take screenshot, which will be included in the table below.

Image Link Left Pixel Right Pixel Image Sun Angular Size Image Error Stellarium Sun Angular Size Stellarium Sun Altitude Stellarium Screenshot Link
Solar Noon 1071 3515 0.5251° -0.62% 0.5284° 67.35° Stellarium Solar Noon
Evening 1055 3500 0.5253° -1.00% 0.5306° 11.50° Stellarium Evening

From the above its clear that the sun being much closer to the horizon an hour before sunset does not cause it to be larger. Actually, it's a tiny bit bigger due to Earth's elliptical orbit. From this link for the sun in Austin that September we can see that from September 2nd to September 16th the sun's distance (right most column) dropped from 93.783 million miles to 93.397 million miles. Using a small angle approximation this relates to the expected change in size:

expected_angular_size_change = 100 * (93.783 - 93.397) / 93.783 = 0.4116%

which can be compared to the angular size observed in the table above:

actual_angular_size_change = 100 * (0.5253° - 0.5251°) / 0.5251° = 0.0381%

I'd like to better understand the error, but it's 1% at most, a small amount compared to the discrepancy one would see in the flat Earth model. It probably has to do with a slight mechanical error with the focal length in the camera, but I'll investigate further.

It would be nice if the expected and actual change in the angular size was closer, but that wasn't the goal. I'll take another photo after a greater change in type and see if I get a better result.


A photo of sunset from the top of a small hill "Cross of the Martyrs" in Santa Fe New Mexico at the expected sunset time for that location (taken 2019-10-24 18:17:02 Santa Fe time). Since the center is slightly below the horizon it's possible to draw an ellipse to get an estimate of the full size. Note that the ellipse is longer than tall one would expect from standard atmospheric refraction for objects on the horizon. From the ellipse horizontally the sun goes from x=1223 to x=3738. Using the "angular_size" formula above with SPP=3.7558e-06 since it's the same magnification as the other photos (maximum optical zoom on a Nikon P900) we find that the angular size is 0.5667° which compares somewhat to the 0.5360° angular size from the Stellarium image from the same time and location. Some error is probably introduced by applying the ellipse, but it's just a 5.7% error.


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.