Hello gentlemen,
I dropped by to see the discussion, isn't this stuff exciting?
So, how did I create that scale? (I thought its obvious but maybe not)
Thanks, but the issue isn't with the how. The how WAS obvious. It's why you would consider such an approach valid that's the question.
I took the difference in height between the two peaks of Humphrey's Mtn, and created the scale based on that difference. (I got the elevations from Google Maps with Terrain data turned on. It's fairly accurate, but do consider that there is a slight downward looking angle of about 0.9 deg to Humphrey's Peak, per flat surface assumption)
I understand that, and with that you built a scale of vertical distance, but it's scaled for a distance of 200 miles away. Your scale would show where the summit of a 13,900' mountain that was at around 200 miles, not one that's 480 miles away. What I can't figure out is how you determine from that scale to say a 14,000' peak 480 miles away is aligned to the right height. You need to work out that angular dimensions to scale the linear measurements at 200 miles with those at 480 miles away.
Shockingly, the snow covered mountain peaks in the distance seem to come up to the correct elevation for a shallow angle of observation.
What's the rationale for reaching this conclusion? It's not found from your reasoning above.
The actual angle to the distant Colorado Mountains (flat earth surface assumption) is equal to atan( (31000 - 14000)/(5280*500mi) ) = 0.37 deg
This I can agree with. I call this "dip" angle and, using 480 miles vice 500, I got 0.38°. So we're in the same ballpark there.
I assume the mountains are about 14000 ft high (judging by the cluster of Mountain peaks seen and checking with Google Maps terrain data), and the observation distance is about 480 mi. Humphrey's Peak is at about 200 miles away, and the size of the snow capped Colorado Mountains, relative to Humphrey's peak seem to be about the right size per the laws of perspective. (i.e if twice the distance away, it should be half as small)
I follow. 480 miles is 2.4x 200 miles, so a 480-mile distant object of the same height as a 200-mile distant object should appear 0.42x as small. But has your scale shown that?'
By chance, or can we say providence, this turned out to be an excellent experiment, because two tall mountain peaks, that rise high above the more dense part off the atmosphere with its associated distortion, have lined up at that particular instant and we can judge their heights relative to each other. Shocking isn't it! I'm still reeling from this realization.
-JT
I hope I've conveyed to you why I don't follow your amazement.
For illustration, here's a rough diagram, not to scale but just showing the geometrical relationships. The linear values (mileage, heights) that I derived from your video vary a little bit from what you came up with, but just for the sake of this point, I'll use your numbers.
The angle circled in red we basically agree upon.
The angle circled in blue we're apart by only a tenth of a degree (you said -0.9°).
Without knowing where "eye level" is to measure declination angle from level, and not having a ground reference to work up from, we could take your approach and use the delta between the two Humphrey's peaks and, with a known distance, work out the angular difference and build a scale based on that, just like you did with the analysis of San Jacinto imagery using Le Meridien Delfina hotel. And then using that angular measure, see if that marries with what the geometry predicts for a 480-mile distant elevation.
I don't think it does. I agree that the distant peak shouldn't appear above the nearer peak based on spherical earth calculations with no refraction adjustments or even with standard atmospheric refraction. For whatever reason, the mere fact that we can see that distant peak means the earth appears to be less spherical than one with a radius of 3959 miles.
However, that doesn't mean "flat?" Comparing angular ratios, that distant peak is still lower than would be expected if flat earth geometry is assumed. Neither the curved earth calculation nor the flat earth calculation predict what you captured in your video. It's somewhere in between a flat earth and a globe earth of the size claimed. The challenge is to figure out why that is. Why would a flat earth appear convex? Why would a globe earth appear less convex?
The point of this particular question I had with your video was that I didn't agree with your declaration that the distant mountain aligned at the expected height for a flat earth. I didn't see how you could draw that conclusion from the way you appeared to be calculating. Hoping you can see what I'm saying and either confirm or tell me what I'm getting wrong and why you believe your method was correct.
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Even if we can't come to a meeting of the minds on this point, I'm hoping you'll stick around to address a few other inputs I have.
Oh, and while I'm at it, on another topic we got to talking about your Salton Sea cell tower sighting mentioned in your Malibu/San Jacinto video. Though there was some doubt from certain quarters, we think we id'ed the cell tower (T-Mobile tower in Salton City), but we couldn't figure out where your vantage had been to produce that picture. If you could resolve that for us, it would be much appreciated. And, of course, if we got the wrong cell tower, let us know that too. Thanks.