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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Flat Earth – The Horizon, Curvature and Angles

Over the past few months I’ve been working on the curvature equation for a circle.  It doesn’t seem like very exciting stuff but it has enormous implications.  During this process I thought I had found this equation but it turned out to be incorrect.  I had fixed what I thought was the error but that turned out to be incorrect as well.

The nagging problem stems from the way one would experience curvature if they truly lived on a ball.  All the current methods of finding curvature don’t really have a good explanation and upon further study are shown to be calculating something other than curvature.  So what do we mean by curvature?  I will answer this question and provide a new and rational equation for curvature and show why the other equations do not work.

Ball Earth Math

This famous document that has made the rounds in the flat earth community is not so much incorrect as it calculates an irrelevant number.  Though it does properly calculate the value for X (or the “drop” along the axis), X is not the value we are looking for.

As you can see, the line X is tilted so it is parallel to the axis.  This is where the calculation becomes misleading.  What we really need to do is extend the line in order to intersect the line of site projected from point 0 (in the Ball Earth Map document).  Please see my curvature image to see an example of this.

BallEarthMath

 

Distance² x 8 / 12

The equation “distance² x 8 inches / 12” does not solve for curvature but only for the hypotenuse of the distance: “Height = R-R*(cos(θ))” and “Length = sin(θ)*R”.  Since these two values only represent the length and height of the distance travelled along the hypotenuse; and, since you cannot have a circle without height and length of equal value, you cannot have an arc without height and length of unequal value; therefore, curvature cannot simply be height (or the amount of “drop” along the axis) nor the value of the hypotenuse.

As an example, 1° of circumference is equal to 24901/360°=69.1 miles.  If we plug in that value to the equation we get “√((3959-3959*(cos(1°))² + (sin(1°)*3959))² = 69.09 miles.  This is not the value for curvature.  However, this does resolve the hypotenuse.  The equation was being used to measure curvature by plugging in the arc distance (the distance travelled along the ball) not the hypotenuse; therefore, the resultant value will only solve the value of the hypotenuse.  All we are calculating with this equation is the point at which two different lines of site intersect on a ball not the height required for an object to be visible from point A.

Two points would intersect if we “forced the line” and built along the ground from each point (A and G).  The two lines would have to be 3959 miles long.  This is equivalent to a building at point D being 1639 miles in height.  Since we build either along the surface or perpendicular to the Earth’s surface (i.e. buildings), we need to calculate something else.

Line of Site, the Hypotenuse and θ

So what are we actually trying to calculate?  If the hypotenuse is not the distance nor the length along the axis, then what is?  The confounding problem is related to the line of site of the observer.  Once we establish the position of the observer, all the other pieces fall into place.  If you examine the image below, you will notice that a line of site moves away from point A towards infinity.  So we need to take the observer as being at point A and does not move.

Next, several dashed lines at various degrees have been drawn until they intersect with the line of site originating at point A.  The equation ((1/(COS(θ)/R))-R) calculates the hypotenuse from θ to the point at which it intersects with the line of site from point A minus the radius.  What we are calculating is the height necessary for an object to be above the surface, at a particular arc distance away or at angle θ, in order for it to be visible to the observer at point A.  This is the key.

As an example (if the ball theory is to work), as a ship goes over the horizon, the mast (and the rest of the boat) will begin to tilt as per angle θ.  Though this is a tiny angle at first, it nonetheless must follow that angle.  The mast does not start tilting back to stay parallel with the axis of the ball, it stays fixed to the boat.   As the boat continues along the circumference, the mast would need to extend in height to remain visible to the observer at point A.  This is an effect of curvature.

In the diagram below, I put the original curvature calculation beside the equation I have proposed.  You can see that at smaller distances the two values are very similar.  However, as you move past 2° the values begin to diverge more rapidly.

A simple way to calculate the “distance to horizon” is to divide the distance travelled by 69.17 miles which equals θ and plug that into ((1/(COS(θ)/R))-R). 

flatplanetrig_newv9
Click me

Curvature

So what do we mean by curvature?  For example, if I travelled from point A to G, I would have experienced 6225 miles of the total circumference but there is not 6225 miles of curvature between point A and G.  If you look at line D1 – D0, you will see that there is only 1159 miles of arc height (or the maximum height of the arc between two points).  It also happens to be the value of X at 45° or 90°/2.  This works for any arc length.

If you know the distance travelled, you can calculate the equivalent θ travelled.  Using a variation of the Ball Earth Math equation, R-R(COS(θ/2)), we can solve for X which gives us the maximum height of the arc.

At a distance of 6225 miles (or ¼ of the globe) the maximum height of the arc has been shown to be 1159 miles.  In a similar fashion, the ship and the observer would have to lift off the surface at 45° and travel for 1159 miles to see each other.  However, we can see that all this is doing is altering the line of site for both the boat and the observer.

Conclusion

The definition of curvature is the degree to which a curve deviates from a straight line, or a curved surface deviates from a plane.  The curvature of a circle is defined mathematically as the reciprocal of the radius (ie. κ = 1/r).  All this tells us is that as the circle becomes larger κ becomes smaller.  This does not help us figure out what the effect of curvature is to a person living on a ball.

However, by showing rational examples I have demonstrated the effect of curvature is the height necessary for an object to be above the surface, at a particular arc distance away or at angle θ, in order for it to be visible to the observer at point X.  Without altering the line of site of the observer and keeping the object (in this case a boat) on the surface of the earth, we can measure the effect of curvature.

I would propose then, that what we are looking for is effective curvature and it is defined as “The height necessary for an object to be above the surface, at a particular arc distance away or at angle θ, in order for it to be visible to the observer at point X and X being a stationary position.”

 

Polaris Paradox – Flat Earth Trigonometry

The problem of Polaris has been a stubborn one for those investigating the Flat Earth. The crux of this problem has to do with viewing angles, distances and elevation.

Using trigonometry, one should be able to measure the height of any object from any particular distance. Unfortunately, the math just doesn’t seem to add up. Using the supposed radius of the Earth – which is 3959.16 miles – we should be able to figure out the height of Polaris based upon the viewing angle and distance from the North Pole. The two assumptions – the radius of Earth and the distance from the North Pole – are generally agreed to values from both FE (Flat Earth) and GE (Globe Earth) people.

The viewing angle (“VA”) is where the problem starts – and ultimately will be resolved.   GE theory states that the viewing angle of Polaris is equal to the particular latitude the observer views Polaris (ie. 49th parallel has a viewing angle of 49°). The distance from the 49th parallel to the North Pole is 2,597.55 miles or the radius of the Earth at that parallel.   In the GE theory, the viewing angle is dependant upon the curvature of the Earth.

In the traditional FE view, Polaris is approximately 3600 miles above the North Pole.   However, the viewing angle from the Equator is supposed to be 1° but according to traditional trigonometry, the viewing angle should be around 42° – Hence the paradox (or in GE theory, proof of a globe).

In examining this problem, I began by using a classic trigonometry set and drew, in 10° increments, the viewing angles from an object at 3600 miles above the North Pole. Several interesting anomalies appeared that, in the end, helped me resolve this problem.

20160320_105030

You can see from this image that the distances between the viewing angles are not equal. In GE theory, the distances between viewing angles are equal since the curvature is doing the work. I summed up these observations as follows:

  • distance between degrees on a sphere/circle are equal (degrees of parallel)
  • distance between degrees on a flat plane are not equal (this is important)
  • As the height of an object from a flat plane decreases, the angle of view decreases and tends towards infinity (law of perspective using geometry). As the observer increases distance from the object, the angle of view decreases.
  • The viewing angle is inversely proportional to the distance from the object. As the viewing angle doubles, the distance to the object is reduced by half.
  • An object of 3959 miles from a flat plane would have a viewing angle of 10° and would be at a distance of 22,962.2 miles.

Observations 1-4 are all perfectly logical and fit well with the FE model. However, the 5 observation does not fit with known distances whether FE or GE. There is the possibility the FE model is incorrect but direct observations have shown that there is no curvature. We are right back in the middle of the paradox.

In an effort to resolve this confounding riddle, I began to model distances, heights and viewing angles in Excel and look for patterns or answers of some kind. After a few weeks of tinkering I developed this model:

https://docs.google.com/spreadsheets/d/11rVPrOIpXaej5OZxiUo13NdKwXUuHrI3Ul8IwBmOqaE/edit?usp=sharing

There are 2 assumptions in the model:

  • The radius of the Earth (3.959.16 miles). All other numbers are generated using standard trigonometry and are without opinion or conjecture.
  • There are 90° between the North Pole and the Equator

The model is defined by 1° increments (1-89) and uses TAN, COS and ATAN functions to obtain either an angle or a distance. There are two main sections separated by a blue line. The left hand section takes each viewing angle (starting at 1°) use a TAN function (H/TAN(VA)) to derive the distance. For example, and object that is 69.101 miles above the observer, would have a VA of 1° and a distance of 3958.79 miles. This equation is applied to each VA up to 89°.

I noticed that the VA and the distance are related to each other (see observations 3 & 4) up to and including the 32°. After that, the relationship doubles and the distance an observer is required to travel to double the VA is 4x the distance. I added a column that calculates the distance whenever the distance doubles starting at 1°. The distances correlate well but not perfectly. Plus, any differences increase as the distance decreases up to the 32° and then returns to normal after that. The distances are variable and change as the height of the object changes. By doing this, the actual VA is maintained and the distance alters the equation mentioned above. Another column (Apparent VA) was finally added but I will return to that one later as it is directly related to the solving of the paradox.

The right hand column uses the radius of the Earth as the fix value (rather then the VA in the left side). To obtain the actual viewing angle based upon distance from object, I used an ATAN function – ATAN(H/R).   The common value between both sides is the object height. I derived the radius of the Earth at each degree by the following equation [Radius of Earth*COS((Degree)/180*3.14159)]. This essentially flattens out the Earth into a series of concentric circles.

Once all these relationships were in place, all I had to do was change one single value – the height – to see how the entire model behaved. The biggest pattern that I observed was the “compression” of VA as the distance and height increased. For example, at a relatively low height of 2 miles, all of the distances and VA matched the degree relative to the equator. However, as the height increased the VA began to “compress” at the lower VA values. As I increased the height, the VA differential (difference between the degrees from the equator and the actual VA) increased. You can observe the graph “Angle Differential” begin to form a SIN wave as the height increases. I haven’t taken the VA into decimal increments at the top and the bottom but my guess is that the pattern repeats.

So how important is this VA “compression”? As it turns out, it makes all the difference in the world. The model suggests that VA on a flat plane do not operate the same as VA on curved surfaces. As the distance from the object increases, the change in the actual viewing angle per degree increases at a slower rate.   Take the above example of 69.101 miles – the rate of change from 1° to 16° is only 1° of VA (33° to 34°). As we can see, the VA does not increase at the same rate as the degrees from the equator. Therefore, on a flat plane we would expect a variable VA per degree from the equator whereas on a curved surface it would not be variable.

At 69.101 the VA from the equator is equal to 1°. However, as the height increases from this point, the VA becomes “compressed”. What do I mean by “compresses”? If you examine the “actual viewing angle” column on the right side of the model, you will notice that for the first 70° there is only a 3° change in VA. Within those first 3° the observer will notice very little change in the height of the object even over a great distance. It is only in the last few miles (from 76° to 89°) that any real movement in the object would be noticeable.

The phenomenon becomes even more exaggerated as the object increases in height. As I continually increased the height, I found that the height of Polaris would be 2,513.5 miles above the North Pole. At this height, the VA from the equator would actually be between 32° and 33°; all the remaining degrees are hidden from view since they are “compressed” into a small area below that degree. Of course no actual compression is happening but it is a phenomenon of perspective on a flat plane.

How can this be possible?

I was contemplating the problem of “compressed” VA but I could find anything that worked with the trigonometry – until I saw this awesome video by p-brane:

This video provides the mechanism with which the VA becomes “compressed” for objects near the horizon.

The human eye and perspective

An important piece to this puzzle is within the nature of the human eye. I have included two major references that the reader can take the time to read. The first is from Ian P. Howard (Perceiving in Depth, Volume 3: Other Mechanisms of Depth Perception, Volume 3 Chapter 26.4 .1(Effect of Height in the field of view)) and the second is Zetetic Astronomy, by ‘Parallax’ (pseud. Samuel Birley Rowbotham), [1881] chapter 14.

When looking at the horizon with the naked eye, (as opposed to using a telescope or binoculars) there are various laws of perspective that need to be considered.

Let’s take another example with an object at a height of 2.571 miles above the observer. To achieve a VA of 1° the observer would need to be 147.29 miles from the object. The observer would then have to move half that distance – 73.65 miles – to achieve a VA of 2°. However, if the observer traveled half the distance again – 36.82 miles – the VA would become 4°. This continues at the same rate until the VA is 32° at which the observer is merely 4.60 miles from the object. To achieve a 64° VA the observer would have to travel 4 times the distance – 1.15 miles from the object.

As we can see, the non-linear changes would have a direct impact on the VA based upon distance. We can put this model into practice through the observation and measurement of distant objects in relation to their height. For example, from Vancouver, BC, the distance to Mount Baker is approximately 68.39 miles. This would mean the VA would be approximately 1.71°. As an aside, if we assume a curvature of the planet, 1/3rd of Mt. Baker should be below the horizon when observed from Vancouver. In fact, 3,184 feet of 10,781 feet of the mountain would be below the horizon. As anyone from Vancouver has seen with their own eyes, the entire height of Mt. Baker can been seen (from base to peak) from 68.39 miles.

Take into consideration that 97% of the distance between Vancouver and Mt. Baker is traveled in the first 3rd of the VA (up to the 32°). The remaining 68° of VA occur during the last 3% of the journey. This will have an impact on how the human eye perceives objects at a distance. The VA is not constant. Because of this objects in the sky will appear higher or lower than they really are.

For example, an airplane flying overhead at 500 miles/hr will approach the observer slowly at first and then begin to accelerate as the distance decreases.   The airplane will reach maximum velocity (from the point of view of the observer) when it is directly overhead. The plane will then begin to decrease velocity as it moves away. We all know that the speed of the plane has not changed but the VA is changing. The more distance the plane gets the slower it appears to go. If we take this example and apply it to static objects (like a mountain) the same rules apply. However, the change in VA is due to the observer. As the observer approaches the mountain, the VA changes but at an inconsistent rate. At a great distance the mountain will appear to “rise” up from the horizon at a very slow rate until the first 32° of VA are completed. After that the mountain will begin to “rise” at an accelerated rate.

If we now apply these observations to Polaris, we can see that 97% of the VA is far behind the Equator. According to the model, the star will rise at a much faster and consistent rate after the 32° (which the actual degree of parallel of the Equator). You have to imagine an observer that is 147,298 miles from the North Pole. At that distance the actual VA is 1°. It would take 97% of the journey before the star will begin to “rise” up from the horizon. After that point, the star “rises” and a relatively consistent rate (albeit not at 1° per degree of parallel – but close).

Another important observation that needs to be taken into consideration is the human eye itself. It is documented that the total VA that the eye is able to perceive if the observer is looking directly at the horizon is 60°. In other words, the total field of view is only 60°. Also, the field of view also contains the ground beneath our feet. So within that 60° we have 100% of the field of view.

It is important to note that the model being presented is scalable for any object at any height (assuming the radius of earth being valid). This means we can use this model to accurately map the surface of the Earth using an object at a constant height (ie. Polaris).