Hi everyone, I had to register to chime in about an important point.
I understand some flat-earthers don’t accept claimed distances over water. So I want to demonstrate the proof introduced in this thread with smaller distances which can be covered by land.
Take these airports at the four corners of the continental US: Boston (BOS), Seattle (SEA), Miami (MIA) and Los Angeles (LAX). The same method as presented earlier will work even on such a small patch of land - no need to select cities as far apart as Johannesburg - Sydney, etc. - but of course the effect will be smaller.
http://www.webflyer.com/travel/mileage_calculator/ gives the following distances which can be more easily corroborated than (say) the “unknown” distance from New York to Paris since they do not cross any ocean:
SEA - MIA 2720 miles
SEA - BOS 2490 miles
BOS - MIA 1260 miles
SEA - LAX 954 miles
BOS - LAX 2600 miles
MIA - LAX 2340 miles
From
http://www.calculator.net/triangle-calculator.html then, we have:
BOS-SEA-LAX angle (85.773°) = BOS-SEA-MIA angle (27.538°) + MIA-SEA-LAX angle (56.916°)
The sum of the two angles on the right-hand side of this "equality" is 84.454°, over 1.3 degree less than the left-hand side.
From the Boston angle:
SEA-BOS-MIA (86.443°) = SEA-BOS-LAX (21.465°) + LAX-BOS-MIA (64.002°)
The sum on the right adds up to 85.467° and we're missing almost one degree.
With Los Angeles as the pivot the discrepancy is over two degrees (left as an exercise).
So my question is, do flat-earthers dispute even these within-landmass distances? If not... what's the rebuttal? Curved landmass surrounded by a flat ocean, maybe?