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Flat Earth Theory / Re: New idea on observing Sigma Octantis from multiple locations
« on: April 15, 2024, 11:16:03 PM »
Where is Sigma Octantus?
If the earth is a spheroid. light travels straight in a vacuum, and Sigma Octantus is a star 294 light years away and 1 degree off a line extended from the south pole, then all over the southern hemisphere it will be visible at any point in the southern hemisphere almost directly south and at an angle above the horizon equal to the latitude of the observer's location. This is explained in textbooks, web sites, videos, etc, consistently and unambiguously. Navigators have used this and observers have confirmed this. In this respect, the earth appears round and the geometry is consistent. Any RE will tell the exact same story. In this respect, the earth appears to be round.
If the earth is flat, we know the light is bending but do not know why or how. The bending can only be determined by what bending needs to occur for the round appearance to actually be flat. We don't know whether there is one pole or two. We don't know why at the same time people in the north see entirely different stars than people in the south. Since we don't know how the light bends, we don't know where Sigma Octantus actually is.
It is not just the azimuth of Sigma Octantus but also elevation (angle above the horizon). It is stated above (hypothesized? speculated?) that SIgma Octantus is directly above the south pole in the bi-polar model. Since Sigma Octantus is on the horizon when viewed from the equator, that makes it appear to be directly on the south pole while far to the south it appears to be far above it. So the light bends vertically as well as horizontally.
In the monopole disc model Sigma Octantus is in every direction, always directly opposite to the north pole. Seems like it would be visible from the northern hemisphere. It has the same elevation problem as bi-polar model.
So that leaves us with: RE has an explanation that is known, consistent with observations, and identical in all RE info sources. FE does not know which model and has no equations, explanations, or verification experiments to explain observations. Yet some believe the earth is flat. Sure would like to hear the details of EA, but so far the definition is "whatever it has to be to make the appearance of RE be actually FE".
Interestingly, if we know the light bends in various directions but do not know exactly how, Sigma Octantus could be anywhere. I claim that to know where Sigma Octantus is, we have to know the forces and equations of how the light bends. If it does not bend, the earth is round.
I hope we can all agree with everything I said above. Please advise if I said anything that isn't true.
If the earth is a spheroid. light travels straight in a vacuum, and Sigma Octantus is a star 294 light years away and 1 degree off a line extended from the south pole, then all over the southern hemisphere it will be visible at any point in the southern hemisphere almost directly south and at an angle above the horizon equal to the latitude of the observer's location. This is explained in textbooks, web sites, videos, etc, consistently and unambiguously. Navigators have used this and observers have confirmed this. In this respect, the earth appears round and the geometry is consistent. Any RE will tell the exact same story. In this respect, the earth appears to be round.
If the earth is flat, we know the light is bending but do not know why or how. The bending can only be determined by what bending needs to occur for the round appearance to actually be flat. We don't know whether there is one pole or two. We don't know why at the same time people in the north see entirely different stars than people in the south. Since we don't know how the light bends, we don't know where Sigma Octantus actually is.
It is not just the azimuth of Sigma Octantus but also elevation (angle above the horizon). It is stated above (hypothesized? speculated?) that SIgma Octantus is directly above the south pole in the bi-polar model. Since Sigma Octantus is on the horizon when viewed from the equator, that makes it appear to be directly on the south pole while far to the south it appears to be far above it. So the light bends vertically as well as horizontally.
In the monopole disc model Sigma Octantus is in every direction, always directly opposite to the north pole. Seems like it would be visible from the northern hemisphere. It has the same elevation problem as bi-polar model.
So that leaves us with: RE has an explanation that is known, consistent with observations, and identical in all RE info sources. FE does not know which model and has no equations, explanations, or verification experiments to explain observations. Yet some believe the earth is flat. Sure would like to hear the details of EA, but so far the definition is "whatever it has to be to make the appearance of RE be actually FE".
Interestingly, if we know the light bends in various directions but do not know exactly how, Sigma Octantus could be anywhere. I claim that to know where Sigma Octantus is, we have to know the forces and equations of how the light bends. If it does not bend, the earth is round.
I hope we can all agree with everything I said above. Please advise if I said anything that isn't true.