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Flat Earth Theory / Re: Predictions
« on: June 14, 2018, 01:13:23 PM »
I predict that latitude lines (i.e. lines determinied by the angle of observation of celestial poles) will be equidistant.
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Flat Earth Theory / Meridians on a flat earth
« on: December 01, 2017, 03:55:15 PM »
How do you explain meridians on a flat earth? Assuming nothing about the shape of the earth, let's define a meridian as the set of locations that experience the same time of day during the equinox. We should all agree that on or around March 21 and September 22, people on a given meridian will all observe sunset and sunrise at the same time.
On a globe earth, this is pretty straightforward: a meridian is a half-great-circle joining the two poles.
On a flat earth, what is the shape of meridians? Are they straight lines, or do they have some curve to them? If they are curved, how can the sun shed daylight on a convex area on one side of a meridian, but then light the opposite side 12 hours later (at equinox) which is concave?
On a globe earth, this is pretty straightforward: a meridian is a half-great-circle joining the two poles.
On a flat earth, what is the shape of meridians? Are they straight lines, or do they have some curve to them? If they are curved, how can the sun shed daylight on a convex area on one side of a meridian, but then light the opposite side 12 hours later (at equinox) which is concave?
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Flat Earth Theory / Re: What happens when the ice wall melts?
« on: November 22, 2017, 07:05:03 PM »
Expected FE response: If they can fake the space program, how much easier is it to fake this "global warming" hoax?
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Flat Earth Theory / Re: Using airline flight data.
« on: November 17, 2017, 07:05:47 PM »
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?
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?
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