The Sun never sets on the Flat Earth

The title of this blog might seem, at first, to be debunking the flat earth; it is however, a vital hypothesis towards a proper theory of the earth’s shape.  My previous posts have presented various concepts that try to show many inconsistencies with the heliocentric model.  The problem with such an effort is that a proper model for the flat earth that fits the empirical data is never properly developed.  My goal is to present a model that does, indeed, do this.

What do I mean by the ‘Sun never sets’?  Well, in the heliocentric model the sun is literally “setting” or disappearing below the horizon.  Due to the hypothesized ball shape of the earth and its associated spin, the sun will eventually become obscured by the curved surface of the ball (i.e. you can’t see through solid ground).

In the flat earth, the sun is always at the same height (with some minor changes as will be explained later), regardless of the date or time, at all locations on the plane; hence, it never sets.  So how does the sun function on a flat plane to create seasons, timezones, night, day, etc?  Many current models have been presented and many of these questions have been answered to varying degrees of satisfaction.  The most confounding problem that faces the flat earth model is the trigonometric disconnect between the latitudes, angle of incidence and height of the sun – the sun is just not where it’s supposed to be!

This problem provides heliocentric proponents with the most devastating argument against the flat earth because even with a obviously lack of curvature, the trigonometry matches the ball model better than the flat model.  So we are in a kind of stalemate – neither model can claim victory.

However, I have developed a hypothesis for the sun that not only fits the flat earth model but also a methodology for mapping the earth we live on.

I wanted to add a video by Jeranism which perfectly visualizes what I’m trying to present.  Please watch this video before reading the rest of this post since it will help you understand how such a phenomenon is possible.

Hypothesis of the Sun

  • The sun is electromagnetic
  • The maximum radius of the sun’s radiation is approximately 6225 miles in all directions (gases in atmosphere are excited by the presence of the sun and then go into quiescence as the sun moves away) hence you get a glow in the distance even after sun is no longer in view.  Please watch this video by Dan Dimension:
  • At a distance of 6225 miles from the observer, due to laws of perspective and refraction, the sun will appear to be at the horizon (~ 1°).  At a distance greater than 6225 miles the sun will begin to “set” or “rise” as it approaches.
  • Since the sun is at the same height (with specific increases and decreases with respect to seasons), regardless of the position of the observer, we must apply the laws of perspective to obtain an accurate altitude of the sun.
  • The sun is between 4200 – 4600 miles above the plane depending on the date.  It is this fluctuation that produces the analemma.solar-analemma-070000-UTC
  • The curvature calculation (Ball Earth Math) is really a perspective calculation (The Effect of Perspective).  It calculates the reduction in size of an object as it moves away from the observer.  This value must be subtracted from the trigonometric calculation.  This is due to the fact that standard trigonometry defines 2D objects while we exist in a 3D environment; hence, perspective and angular resolution must be taken into account.  Please watch this video by Curious Life:
  • For example, at 49.2827° N, 123.1207° W (Vancouver, Canada), the sun appears to be approximately 40.6° above the horizon.  Since there are 69.15 miles per degree of latitude, this places Vancouver ~ 3388.35 miles from the equator.  On the spring equinox at solar noon, standard trigonometry would calculate the altitude as 53.4°.  However, we must also subtract the effect of perspective:

[tan-1 (Height of Sun / distance to equator)] = Altitude (in degrees)

tan-1 (4400 / 3388) = 53.4°


(Height of Sun – (Height of Sun(cos(θ))) = Effect of Perspective

(4400 – (4400(cos(49.2°))) = 1,525 miles


Height of Sun – Effect of Perspective = Perceived height

4400 – 1525 = 2875 miles


tan-1 (2875 / 3388) = 40.3°

Scaled Perspective

I’m certain that many will object to this model by arguing that objects at a shorter distance do not present this kind of affect.  In other words, an object that is 100 feet away at 10 feet in height will not display the effect of perspective.  That the effect is so infinitesimal does not mean it’s not present.  For example, an observer 67.15 miles from an object 2 miles in height would only measure an effect of perspective of 1.6 feet – That’s a 1.6 foot difference over 10,560 feet.  In other words, if it’s almost imperceptible at 67.15 miles, it’s definitely imperceptible at 100 feet.

So what am I basing this scale on?  Well…the sun, moon and stars.  They are the only objects that are at a sufficient distance and height to show the effect.  Ultimately, the position, height and motion of the celestial objects in the sky are either due to perspective or curvature.  Since no curvature has been show to exist (at least empirically), then we are left with perspective.

This video by Wide Wake perfectly shows how the sun becomes distorted as it moves away from the observer (it becomes egg shaped).  The video by Jeranism shows the mechanism by which this affect occurs and why it appears to drop.

I’ve uploaded the Excel workbook for anyone to use.  I was able to use both Date & Time and Stellarium values for the Sun’s height at various location on the earth.  The values match almost perfectly with the hypothesis.  There are 4 tabs in the workbook as I had to calculate the equinox and both solstices.  I added a short-distance tab to calculate objects closer than 1°.  The result is that the effect of perspective is accumulative with both distance AND height.  In other words, as the object increases in height and distance, the effect is more pronounced.  In the example above, a mountain that is 67.15 miles away and at a height of 2 miles will have an insignificant amount of this affect since the accumulated affect is limited to relatively small values.

The workbook is a little too complex to put directly into the post so I’ve provided it for people to download.  Just click on image below to download:

Screen Shot 2018-04-29 at 10.29.24 AM

With an accurate way to measure the distance and heights of the celestial objects, we should be able to map the surface of the earth using the formula above.

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