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.