# The Flat Earth Society

## Flat Earth Discussion Boards => Flat Earth Theory => Topic started by: Regicide on August 26, 2020, 04:19:24 PM

Title: The elevation of Polaris. Yes, I read the Wiki
Post by: Regicide on August 26, 2020, 04:19:24 PM
I brought this up in a question about the visibility of Polaris a while ago, and didn’t really get an answer. This is really difficult to explain with a FE mindset and super simple with RE, so here it goes.

Mariners have, for centuries, used Polaris to ascertain latitude. The reason for this is that, for every 1 degree of latitude you go away from the North Pole, Polaris appears 1 degree lower in the sky. This is measurable: if you have a measuring wheel/ car with an odometer, a sextant, and a compass, you can do this experiment yourself. Simply find a location that is relatively flat, and observe the elevation of polaris. Travel 69 miles north or south as the crow flies, and it should be 1 degree higher or lower. This works anywhere on Earth that Polaris can be observed. It works over shorter distances too. 1/60th of a degree of elevation per nautical mile north or south. Now, let’s start with a flat earth. 1 degree of latitude is equal to 69 miles. If you travel 1 degree away from the north pole in any direction, and then observe the elevation of Polaris, it is at an elevation of 89 degrees. tan(angle) = opposite over adjacent. That’s basic trig. I’ll solve it here for everyone’s sake
tan(89)*69= x
57.290*69=x
3953.007=x
So, that’s that. Presuming a flat earth, a trigonometric evaluation at a point 1 degree away from the north pole puts Polaris at a height of 3953.007 miles. Now, let’s travel another degree south and repeat this. We are now 138 miles away from the North pole, and we are observing Polaris at an elevation of 88 degrees. Once again, I’ll do the math for you, but feel free to check my work.
tan(88)*138= x
28.636*138=x
3951.803=x
Huh. We now have a difference of 1.204 miles between the results.(Please note, I’ve rounded off all the figures here to 3 places, but I’m using a calculator with more places accounted for. All last decimal place discrepancies are due to this.)
This isn’t too huge, so let’s keep going. I’ll refrain from writing out the calculation every time, but once again feel free to check it yourself. Also, I don’t want to calculate for all 90 degrees, and you don’t want to read that, so I’ll skip some.
At 87 degrees north, the calculations put the height of Polaris at 3949.79
At 86 degrees north, the calculations put the height of Polaris at 3946.983
At 85 degrees north, the calculations put the height of Polaris at 3943.368
At 80 degrees north, the calculations put the height of Polaris at 3913.184
At 70 degrees north, the calculations put the height of Polaris at 3791.519
At 60 degrees north, the calculations put the height of Polaris at 3585.345
At 50 degrees north, the calculations put the height of Polaris at 3289.240
At 40 degrees north, the calculations put the height of Polaris at 2894.893
At 30 degrees north, the calculations put the height of Polaris at 2390.230
At 20 degrees north, the calculations put the height of Polaris at 1757.976
At 10 degrees north, the calculations put the height of Polaris at 973.325
Finally, at 0 degrees North, for Polaris to appear on the horizon, it would have to be on the ground at the north pole. This could not possibly be refraction.
If a flat earth is assumed, it is difficult to see how this is possible. However, with a spherical earth, all those problems go away. If Polaris is assumed, as in the RE model, to be 434 light years away, then it’s quite simple. There’s barely any observable difference with a star that far away. On a flat earth, it would always appear to be directly above you. On a round earth, though, the horizon, and therefore your idea of something being directly up, is changing as you travel the earth. At 90 degrees north, Polaris is indeed directly up. At 89 degrees north, what is directly above you has shifted by 1 degree, and so polaris appears to be at an elevation of 89 degrees. This continues across the earth, and so the observations of Polaris, the ones that can be directly observed with great ease, and have been proven by mariners, fit the Round Earth conjecture, and (as far as I can see) can not coexist with any FE model. I welcome your thoughts.

P.S. I wrote this in an external word processor, and I almost posted it before I thought to check the Flat Earth Wiki. Thank goodness I did, because there’s one more thing to clear up. I am aware that objects seem smaller as you travel further away. That’s why I showed that the observations of Polaris don’t fit with this. If it grew closer to the horizon at a normal rate, then all of my calculations would have put it at the same height, but they didn’t. They put it at varying heights, that grew more and more different as you travel away from it. 90 degrees of latitude is 6210 miles. If Polaris’s changing elevation was due to the fact that things appear lower the further away you travel, Polaris would still be at 32.479 degrees of elevation at the equator, if the height of Polaris was taken from the first calculation. I’m surprised that Rowbotham overlooked this. Then again, I'm really not. He overlooked a lot.
Title: Re: The elevation of Polaris. Yes, I read the Wiki
Post by: existoid on August 28, 2020, 09:54:44 PM
Could the introduction of the EA theory explain this?  In other words, Polaris remains at the same height all the time (whatever that height is), but its light bending in the exact way needed for this phenomenon.

So now, the question for someone who can do math far better than me (like you!), would the observations of Polaris as you've described, and the observations of the moon match each other - are they mathematically coherent with each other?  If so, then EA could explain both, even if we don't know the actual formula for EA due to the Bishop Constant.

In short, is EA operating "the same" with regards to both Polaris and the Moon?  If not, then the FE community may have to come up with a new theory to explain the observations of Polaris.
Title: Re: The elevation of Polaris. Yes, I read the Wiki
Post by: Regicide on August 29, 2020, 07:46:24 PM
I don't know, could you bring me up to speed on EA?

I considered it, and I don’t really see any possible explanation for this.  It could be that there is some force pulling all light away from the center of the earth, but that would have to affect the light from the sun, too. That would mean that, if the sun is traveling in a circle over the equator, then the people in the Northern Hemisphere would see it much lower in the sky than people in the southern hemisphere, which is not observed. No flat earth proponents have responded yet (suspicious), but I can’t really see any way out of this.
Title: Re: The elevation of Polaris. Yes, I read the Wiki
Post by: existoid on August 31, 2020, 06:37:06 PM
I don't know, could you bring me up to speed on EA?

I considered it, and I don’t really see any possible explanation for this.  It could be that there is some force pulling all light away from the center of the earth, but that would have to affect the light from the sun, too. That would mean that, if the sun is traveling in a circle over the equator, then the people in the Northern Hemisphere would see it much lower in the sky than people in the southern hemisphere, which is not observed. No flat earth proponents have responded yet (suspicious), but I can’t really see any way out of this.

When I first discovered this forum (in March or April or May of this year, it's fuzzy), I joined a thread regarding the equation of EA - the one that is supposed to describe HOW the light of the moon bends. It ultimately went nowhere due to semantics, basically, and then it was locked by Pete.

You can read on the Wiki here: https://wiki.tfes.org/Electromagnetic_Acceleration

In any case, the basic idea of EA is that sunlight and moonlight bend in exactly the right way, to specifically account for time zones and moon phases as observed across the earth.

My guess is that it could explain away Polaris as well.

Title: Re: The elevation of Polaris. Yes, I read the Wiki
Post by: Bikini Polaris on September 02, 2020, 02:19:12 PM
My guess is that it could explain away Polaris as well.

Not sure about this. The observation that any model need to be account for is that sailors travelling in a straight line (presumably with some physical constraint, not only with a compass) see Polaris always at the same height. Same for planes.This cannot be explained by some FE maps, even adding a light bending effect.
Title: Re: The elevation of Polaris. Yes, I read the Wiki
Post by: Tom Bishop on September 06, 2020, 09:20:20 AM
My guess is that it could explain away Polaris as well.

Not sure about this. The observation that any model need to be account for is that sailors travelling in a straight line (presumably with some physical constraint, not only with a compass) see Polaris always at the same height. Same for planes.This cannot be explained by some FE maps, even adding a light bending effect.

Traveling East and West on an RE would only be a straight heading on the equator. Consider what occurs when traveling on a constant Eastward or Westward heading 100 feet from the North Pole.
Title: Re: The elevation of Polaris. Yes, I read the Wiki
Post by: GreatATuin on September 06, 2020, 12:16:31 PM
My guess is that it could explain away Polaris as well.

Not sure about this. The observation that any model need to be account for is that sailors travelling in a straight line (presumably with some physical constraint, not only with a compass) see Polaris always at the same height. Same for planes.This cannot be explained by some FE maps, even adding a light bending effect.

Traveling East and West on an RE would only be a straight heading on the equator. Consider what occurs when traveling on a constant Eastward or Westward heading 100 feet from the North Pole.

There isn't really such as a thing as a "straight" or "not straight" heading. Heading is not a line, it's a compass direction that can be measured as an angle. If you travel with a constant East or West heading, you're not traveling along a great circle except in the special case of the Equator. If you travel due North or due South, you do travel on a great circle (a meridian). In other cases, you travel along a rhumb line.

Ships usually prefer to travel along rhumb lines - lines of constant bearing - because it's easier to keep a constant bearing than to travel along a great circle. The Mercator projection was designed to have rhumb lines appear as straight lines.

So, more accurately sailors have to travel along a parallel (a line of constant latitude) to see Polaris always at the same height. That's a special case of a rhumb line. But it's only "straight" on a Mercator map.
Title: Re: The elevation of Polaris. Yes, I read the Wiki
Post by: Tom Bishop on September 06, 2020, 04:42:56 PM
There isn't really such as a thing as a "straight" or "not straight" heading. Heading is not a line, it's a compass direction that can be measured as an angle.

Incorrect. It doesn't always mean in reference to a compass.

Quote

2. the direction or path along which something moves or along which it lies

Synonyms: aim, bearing

Types: tack

the heading or position of a vessel relative to the trim of its sails

Type of: direction, way

a line leading to a place or point
Title: Re: The elevation of Polaris. Yes, I read the Wiki
Post by: GreatATuin on September 06, 2020, 05:10:05 PM
There isn't really such as a thing as a "straight" or "not straight" heading. Heading is not a line, it's a compass direction that can be measured as an angle.

Incorrect. It doesn't always mean in reference to a compass.

Quote

2. the direction or path along which something moves or along which it lies

Synonyms: aim, bearing

Types: tack

the heading or position of a vessel relative to the trim of its sails

Type of: direction, way

a line leading to a place or point

It does always mean a direction. And heading West or East is definitely a compass direction.

But anyway, that misses the point. Travel due East or due West, and Polaris will stay at the same elevation in the sky.