Do you have any critique of my proposed reticle/index, Tom?
Or do you have a defense for JTolen's?
I've given it a few days, so in the absence of any negative assessment, I'll proceed under the assumption my scale is correct and JTolen's is wrong.
That obscuring ridge is 450-480' above mean sea level, 19.9 miles away from the 150' vantage point on Malibu Bluffs and about 8 miles inland from the Santa Monica beach. I can provide supporting arguments for that if you like, but only if you want to contest it. (The closer ridge you can see is a mile closer and about 120' lower, but it's not an obscuring factor. It's just an aid in identification.
The 10,800' peak of San Jacinto lies 97 more miles beyond that ridge.
Without considering atmospheric refraction, I calculated that 1455' of San Jacinto should be visible from 150' Malibu Bluffs on a globe earth if that ridge is 450' high. 1293' if the ridge is 480' high. With standard refraction those figures would increase some, but a no-refraction calculation is the worst case, and even then JTolan would be wrong about San Jacinto not being visible if the earth is curved.
Working it out for a flat earth, 8925' of San Jacinto should be visible over that ridge if it's 450' and 8708' if the ridge is 480'. Again, I can show my work if you want.
How much of San Jacinto is, in fact, visible? By my index, 4.3 milliradians, or 0.25°. At 117 miles, that works out to be 2652' of San Jacinto is visible in that image.
That's around 1200-1350' more than predicted by the no-refraction globe earth calculation.
But it's 6000-6250 less than was a flat earth predicts.
JTolan's scale measures 7.5 milliradians (0.43°) which equates to 4636' at 117 miles, so that's closer but still only a little more than halfway (not to mention it doesn't jive with angular measurements in the foreground.)
I could be guilty of confirmation bias and forced pattern-matching, but if I'm right about these faint features in the IR image aligning with this view of the mountain:
Then that 117-mile perspective captured the area of San Jacinto's elevation seen in the dotted box here:
Which corresponds to elevations from below 8300' (Fullers Ridge visible) but above 7600' (Black Mountain not visible). That would means we're seeing somewhere around the upper 3000' of the mountain range there. My mrad scale figured 2652'. 348' at 117 miles is 1/2 a millirand. Perhaps my scale is still slightly off.
I had no success getting any images of the upper 4500' (and certainly not the 8700+ feet) of mountain range to match the IR image that might support a flat earth claim. Maybe someone more inclined toward flat earth can.
This still doesn't serve as a "win" for a globe earth since there's still a significant delta between the no-refraction prediction of San Jacinto visibility and what my analysis of the imagery is saying is visible. Even figuring for refraction (which I haven't done yet, but just estimating), I think globe earth only gains an addition 500-600' of predicted visibility, which is still not quite there.
There's a lot of estimating here, but it's not just shots in the dark. Whatever the margin for error, it's much, much closer to validating a globe earth than a flat earth. JTolan's scale is off. He positions it incorrectly. And he misinterprets just how much of San Jacinto is visible. He also incorrectly depicts that the mountain should be entirely obscured if the earth is curved. I argue that his imagery more closely shows what a globe earth predicts than a flat earth.
I don't think his errors are intentional. I think he's just mistaken, caught up in the excitement of making the distant mountains visible through the haze with his IR set up as if he's uncovered something that shouldn't be visible on a curved earth.
Maybe my calculations are wrong. Maybe I'm caught up a round earth premise and making my analysis fit what I already believe.
Maybe. But if so, show me where I've goofed up. I think I'm on the right track, but I welcome reasoned review.