Rothbard

You should seriously consider becoming a writer. Excellent points!

The Russian cosmo-not was the best!

Those two statements are contradictory. If the bulge shrinks to a lesser degree based on distance, then it would undoubtedly still have an effect on a much further mountain as many cases may be. How much is the question. A bulge at 2.7 miles distance (5-foot observation) will certainly have a major impact on a 20,000-foot mountain that is 50 miles away which is now the size of less than 5 degrees (3 fingers) from the observer's camera's view. To say that it would have no effect on the mountain is just wrong. Also, I can affect the angular resolution of the bulge by zooming in with the camera. To what extent and to what effect, I don't know. For example, I can take a picture of the man's head (representing the bulge) (at 2:28 in the video) that is blocking the shrinking wall behind it but regardless of zooming in on the person's head, I cannot affect how much is hidden behind the head. It seems possible that the angular size of the bulge is meaningless to the ultimate angular size of the mountain in the distance. All of this needs experimentation and I guess that's the next step.

Another video of mine comparing children to adults:

Hey Jason! It's a valid question. I'm no expert, but here's a video I made on the subject. Check out this sunset footage my friend shot in Florida (compare to the warehouse expirement) Gudtims4all has been busy working on angular resolution experiments of light Here are a bunch more arguments and videos I wrote down if you're interested - http://tabooconspiracy.com/blog/flat-earth/the-sun-is-not-93-million-miles-away/

My latest

Here's an end-of-the world scenario for you. Releasing 20 million bacteria-infected mosquitoes in California, what could go wrong? https://www.bloomberg.com/news/articles/2017-07-14/20-million-mosquitoes-to-hit-fresno-that-s-a-good-thing-really Even more disturbing: Organizations like the Bill and Melinda Gates Foundation have been working on the bugs for more than a decade, running pilot projects in countries including Indonesia and Brazil. Verily’s contribution has been to create machines that automatically rear, count, and sort the mosquitoes by sex, making it possible to create vast quantities for large-scale projects. The Fresno project will be the biggest U.S. release of sterile mosquitoes to date, Verily says.

Just a little Space Shuttle compilation video I threw together for laughs ... enjoy ... or whatever else you do when confronted with NASA fakery

VonLud said, "The angular size (in height) of the 6000 Ft. mountain would be reduced to 2.6 Arcminutes at 1500 miles. (0.0434°)" For the calculator you click on the "Angle?" Box. Distance = 7,920,000 (1,500 x 5,280); size = 6,000. Click on "Minutes" Answer = 2.6 minutes The calculator matches VonLud's statement and so I'm guessing he's accurate. The formula (Angular size in degrees = (size * 57.29) / distance) is the simplified version of this one: tan (angle) = opposite/adjacent = Line CD/ Line AD = size/distance (see the website for the triangle drawing - http://www.1728.org/angsize.htm)

Good video in light of the latest discussions

This is absolutely not true and I've now lost all respect for you.

It took me forever but I finally found a simple angular size calculator for anyone who wants to make other calculations like this one above: http://www.1728.org/angsize.htm The formula is: Angular size in degrees = (size * 57.29) / distance I did have a question for @VonLud or anyone else who may know. An observer is looking at a mountain at 10,000 feet elevation at a distance of 230 miles from the observer who is also at 10,000 feet elevation. Under the official model, the horizon should be 122.5 miles away (http://www.ringbell.co.uk/info/hdist.htm) - meaning that's the point where the ground allegedly begins obscuring the distant mountain. However, the mountain continues to shrink from the observer's viewpoint beyond the 122.5 miles horizon line. According to the angular size calculator, the mountain would have been reduced to 0.88582 degrees from the horizon line (from 122.5 miles). How do I then calculate how many feet of the mountain should be obscured by the ground? The bulge has increased in height due to distance (hidden by the bulge = 7708.47 feet at 230 miles - https://www.metabunk.org/curve/) but the mountain is also decreasing in apparent height. How much more of the mountain would then be obscured when you add in the decreasing angular size of the mountain? Thanks.

After all of the blustering about having too many flat earth threads, you go ahead and post this? I just have to say that we've come a long way over the last 2 years (can you believe it's been 2 years?). Before, we had to argue about whether this Red Bull jump was the curvature of the earth: I still have family members who send me pictures like that to prove the curvature of the earth. Now, we're arguing whether you can see a degree of arc in this one photo of 20: Things have really changed haven't they?