FE Photographer believes he's captured no-curve imagery. But has he?
I say no, he hasn't. And here's why.
This is representative of the channel marker FE Photographer makes a point of showing us is along the line of sight, 6.45 miles down range. And it is fully visible in the IR image, along a sight line that intersects Clark Island at a distance of 19.59 miles.
I judge that buoy to sit about 12' above the waterline, based on personal knowledge of the typical size and the gull sitting on the base in the photo. Maybe it's 10'. I can do that math for that too, but for now I'm going with 12'.
A height of 12' at a distance of 6.45 miles results in an vertical angle of 0.02°.
At a distance of 19.5 miles, 0.02° represents 35 feet.
FE Photographer cites the highest elevation of Clark Island as 95' which would be at the knoll in-line with the channel marker sight line. Tree vegetation on the island would add height to the land, which he figures at a 'generous' +100'.
Why, then, if the earth is truly flat, is the top of the treeline cresting at only 70'. If the earth is flat, and the island is 95' and trees adding, at best, 100' to the island profile, why isn't the IR image seeing the tops of the trees at 195'? Or even 145' if the trees are even half the height he generously allows?
At the 3:06 mark of the video, he claims to point out rocky shore line features:
Yet, the rocky eastward-facing bluffs on that side of the island rise steeply 20-40' as depicted in the topographical chart and seen in these images:
He correctly assumes that the dark sections of the IR image would be rocks, but his image doesn't depict the vertical displacement of rocky bluffs that should be 1x or 2x that of the buoy in the foreground.
And here's a picture of the tree canopy along that hump. 100' is not merely "generous."
95' elevation + 100' vegetation height? There's 100-120' of island missing in this "flat earth" image. Where is it?
FE Photography used
this Earth Curve Calculator to figure that 141' would be the hidden height on a globe earth, which would leave but the top of the trees visible "at best:"
Why does he draw the line there? The buoy has given us a vertical gauge to let us deduce that we're seeing 35-50' of almost all vegetation.
Not only that, but that earth curve calculator doesn't take refraction into account. If we do, using a standard index of refraction, we get 110-115' that would be hidden. Not 140'.
It sure seems to me his IR images captured exactly what one would expect on a globe of the claimed radius of earth, and you don't even need all that much refraction to account for what was captured. On the other hand, to call it a picture of a flat earth you have to imagine you're seeing a rocky shoreline and you have to come up with an explanation for why the angular dimension of the buoy is not in proportion to the island's elevation. For what we're seeing of the island, to be 150-195' on a flat earth, that buoy in the foreground would have to be 50-60' tall.
It's not, and that IR imagery doesn't depict a flat earth.