1. How sigma Octantis, Polaris and the rotating stars are explained on a flat earth.
2. Here are some observations of the sun’s angular size reducing as it rotates away at ‘sunset’
3. This is the clip of the sun setting at 6 degrees west of North and rising 6 degrees East degrees of North an hour later . This was observed from Boda, Norway. This is an impossibility on the sphere earth . Watch how the sun is rotating above the stationary earth and how the sun becomes larger at the end of the clip as it starts to head due south
4. Here is another time lapse clip of the rotating sun
5. Here is the clip that proves the earth cannot be a rotating sphere. The sun is returned to the sky through magnification. Magnification increases the visual range / distance to the vanishing point
6. Evidence that earth ‘photos’ are actually graphic designs. These images have been sent to Michael to share.
7. More fake images of the earth
8. The Famous Earthrise is a fake
9. A clip every sphere earther must see
10. Boats don’t disappear down a curve beyond the horizon. The principle of perspective and vanishing point explains what actually occurs
11. Polaris is directly overhead at the ‘North Pole’ (90° of latitude); in other words, the angle between Polaris and the horizon at the North Pole is 90°. This angle is called “the altitude of Polaris”. The altitude of Polaris equals the latitude of the observer eg Paris is 48 degrees North, so Polaris will be 48 degrees above the Northern horizon. This is also the observers angle of elevation. At the Equator (0° of latitude), the Polaris is on the horizon, at an angle of 0°. So obviously it’s not going to be seen by observers south of the equator , just as observers north of the equator don’t see sigma octantis. It’s not that hard. The fact is , it works on a sphere earth if you believe that Polaris is 434 light-years away and on a flat earth with Polaris just over 6000 miles away. The confusion is dispelled by understanding the difference between ‘actual height’ and ‘apparent height’ which varies with the observers distance from the object. The actual height of the lights in this tunnel does not change. The apparent height does depending on the observers distance from them.
12.The following question was asked . “At what distance will a 10 foot lamppost disappear below a perfectly flat road?”I believe this question was asked In the context of point 11, actual Versus apparent height. However, the answer doesn’t come from that principle. This article explains that (if the earth was flat) we could see a candle flickering at a distance of 30 miles.
However, we know this is an impossibility because the horizon line is the vanishing point of objects (how much actually vanishes is dependant on the size of the object eg: a lamppost will vanish but only a portion of a mountain will vanish at a distance of 3 miles).
So the answer is, the light will be visible until your angle of elevation reaches zero. This occurs at the horizon line, which from ground level is 3 miles away. So the light will no longer be visible at 3 miles
13.This question was asked “Hourly, the sun is measured to deliver about 430 quintillion Joules of energy to earth. What is the mass of the flat earth sun, what is the method of energy production and fuel?” Show the math. There is no disagreement about the process by which the sun produces it’s energy. Solar energy is created by nuclear fusion that occurs when hydrogen atoms collide in the sun’s core and fuse to create a helium atom. The question is partly correct, The sun’s luminosity is about 3.8 x 10^26 Joules a second which equates to, in terms of mass, about 4,000,000 tons every second. HOWEVER, the research shows that we only receive about 4.5 pounds per second of that energy.
The mass of the sun in the heliocentric model is 1.989 × 10^30 kg and it’s volume is 1,409,272,569,059,860,000 cubic kilometers. The diameter of the sun rotating above the flat earth is 57km. It’s volume is 96966.83 cubic kilometres. See below formula So the mass of the flat earth sun is calculated by; (Volume of FE sun/ Volume of heliocentric sun) x mass of the heliocentric sun This equates to (96,967/ 1.40927256905990 × 10^18) x 1.989 × 10^30 = 1.37241 × 10^17kg