Flat horizons everywhere I look at every altitude. How many flat horizons does it take to make a sphere?~desMEGA

How many "Flat Horizons" does it take to make your horizon go 360° around you? Anyone standing on a peninsula can observe this:

AirPano Caucasus Mountains |

How does a "Flat Horizon" manage to curve around you at the same time? You'll laugh that off but you know you cannot explain or address it. If the horizon were TRULY flat it could NOT go all the way around you. It LOOKS FLAT because you are looking at a curve nearly edge on and it happens to fill most of your limited human vision.

Can any Flat Earther honestly address this?

Moving on...

Get yourself a Canon EOS 6D (Full Frame DSLR, 5,472 × 3,648 pixels, ~$1400) and a Canon 11-24mm F/4L lens (~$2700, you can rent these things also) which, at 11mm, has a 126° 5′ field of view and is NOT fisheye (is one of the widest-angle rectilinear lenses on the market). Go up to 40,000 feet and take an image at 11mm with the horizon passing through DEAD CENTER of the lens with NO obstructions of the horizon view (including clouds, meaning you need a very clear/not hazy day). You aren't allowed to PRETEND the horizon extends beyond where you can see it because YOU are the one ASSERTING that you see SEE IT AS FLAT.

You should see approximately 103 pixels of horizon sagitta (the "droop" of the horizon circle as it curves around you).

This version uses a balloon with a curvilinear fisheye lens but this is the METHOD I'm talking about, here the balloon is up higher but we have a much smaller field of view and lower resolution so we only get about 38 pixels, from about 33124 meters.

w × d/R × tan(fov/4)/2

Where 'h' is observer height, 'd' is the distance to the horizon (d=sqrt(h(h+2×R))), 'R' is Earth radius (in same units as 'd'!), 'w' is the width in pixels (assumes square pixels), 'fov' is the field of view in appropriate units for your tan() function (usually radians unless you selected degrees).

At the same time take a water level up and take an image where we can see where the water in the water level matches against the horizon so we can see where LEVEL truly is.

Water level, like so:

You can see my own measurements here with an iPhone app Theodolite where I show that these measurements agree with the Globe model.

You must supply RAW images at full resolution exactly as detailed above (horizon dead center or close to it) and the EXIF data must be 100% intact and unaltered in any way.

So -- is any Flat Earther willing to do the work to demonstrate the horizon curve to themselves? If not, then you are being DISHONEST.

Moving on...

Get yourself a Canon EOS 6D (Full Frame DSLR, 5,472 × 3,648 pixels, ~$1400) and a Canon 11-24mm F/4L lens (~$2700, you can rent these things also) which, at 11mm, has a 126° 5′ field of view and is NOT fisheye (is one of the widest-angle rectilinear lenses on the market). Go up to 40,000 feet and take an image at 11mm with the horizon passing through DEAD CENTER of the lens with NO obstructions of the horizon view (including clouds, meaning you need a very clear/not hazy day). You aren't allowed to PRETEND the horizon extends beyond where you can see it because YOU are the one ASSERTING that you see SEE IT AS FLAT.

You should see approximately 103 pixels of horizon sagitta (the "droop" of the horizon circle as it curves around you).

This version uses a balloon with a curvilinear fisheye lens but this is the METHOD I'm talking about, here the balloon is up higher but we have a much smaller field of view and lower resolution so we only get about 38 pixels, from about 33124 meters.

Flat Earth Insanity: High Altitude Balloon analysis of the Rotaflight Balloon footage See also: further analysis of location |

This distinction between METHOD and CLAIM is something that Flat Earthers seem to have a hard time with. I'm not saying this PROVES anything -- I'm showing you that you are LIKELY wrong and

However, here is a very simple way to show that the curvature above isn't due to the fisheye lens (NEVERMIND that below lens center the distortion curves straight lines the OPPOSITE WAY than this curvature)... we can compare the same horizon at different altitudes all going through lens center:

The keys here are ALTITUDE, HIGH RESOLUTION, and WIDE ANGLE. I show you EXACTLY how to calculate this on a spheroid in my post. In simple terms this "sagitta" is well approximated by the rather astonishingly simple equation:**HOW**YOU CAN TEST IT FOR YOURSELF.However, here is a very simple way to show that the curvature above isn't due to the fisheye lens (NEVERMIND that below lens center the distortion curves straight lines the OPPOSITE WAY than this curvature)... we can compare the same horizon at different altitudes all going through lens center:

Rotaflight at lower-altitude shows nearly flat horizon |

Rotaflight at approx 13520 meters,. shows increasing curvature |

w × d/R × tan(fov/4)/2

Where 'h' is observer height, 'd' is the distance to the horizon (d=sqrt(h(h+2×R))), 'R' is Earth radius (in same units as 'd'!), 'w' is the width in pixels (assumes square pixels), 'fov' is the field of view in appropriate units for your tan() function (usually radians unless you selected degrees).

At the same time take a water level up and take an image where we can see where the water in the water level matches against the horizon so we can see where LEVEL truly is.

Water level, like so:

You can see my own measurements here with an iPhone app Theodolite where I show that these measurements agree with the Globe model.

You must supply RAW images at full resolution exactly as detailed above (horizon dead center or close to it) and the EXIF data must be 100% intact and unaltered in any way.

So -- is any Flat Earther willing to do the work to demonstrate the horizon curve to themselves? If not, then you are being DISHONEST.

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