Resolve this.

  • 1 Replies


  • 136
Resolve this.
« on: August 05, 2008, 10:52:33 PM »
In FE theory, the equator is a circle 12,450 miles in diameter. Now, the sun is also 3000 miles up (and don't try to change this distance. If it gets larger, it makes the sun more luminous. If the distance gets smaller, it becomes extremely easy to triangulate, even something FE'ers could do on a picnic) Now, on the other side of the earth on the equator, the sun should be 12,806 miles away, from our good friend Pythagoras (a^2+b^2=c^2.) Try this calculation out. Accept it or reject simple geometry you can simulate from that chair you fall out of so often. Now, what angle would we see it at? Use some simple trig functions (which you can also simulate with a protractor) and you have inversetan(3000/12450)=13.54 degrees. Hmmm... could it just be that trees are obstructing our view? No. If the largest tree is, let's say, 300 feet tall, just to make sure that you can see it, all you have to do is stand 1,245 feet back to see this (and I've been in clear fields larger than 1,245 feet pointing north.) So, we should be able to see it, right? Now, I love your calculations on it getting dimmer. The equation on luminosity is L=1/d^2. So the luminosity is 1/3000^2 when our good old sun is directly above you. So, when the sun is 12,806 miles away (as we've established), the new luminosity will be 1/12,806^2. Now, we divide that out and get the ratio as 3000^2/12806^2. Punch this into your Microsoft calculator, and you get the relative luminosity of .05, or 5%. Now tell me, when I'm in that clear field, looking 13.5 degrees into the sky, why don't I see a sun that's 5% the luminosity? This would outshine all the stars, and create a new moon-looking object.
Note: The following is about refraction that Tom put up in defense to my calculations.
Ok... I didn't want to do all the math that is required for a general proof, but I do think that this will kill everything.

I decided to do calculations with 3 indexes this time, all layered on one another. n_1=1, n_2=1.00015 (half of that of sealevel air), n_3=1.0003. I applied the equation 2 times to see the final angle. It gave me, exactly, .2359278779 radians, the exact value I got for the two layered calculation. you could argue that this could form a curved path, and it does, but the overall change is so small that it will be curved extremely minimally.

In short, if you guys are right, I should be able to look out and see the sun in the northern direction. There is virtually no denying it. I have just provided an objective proof that you guys can all demonstrate on your tabletop.

Somehow I don't think this will work with you guys, though.

EDIT: It wouldn't be nearly his model. For it to be his model, you have to find a quadratic equation that satisfies the observed degree (derivative at that point), starts at the sun and ends at the observer. These 3 criteria limit it to one, nearly straight parabola. Nice try, Tom.

I made this in paint, red is the real path between the sun and 7 o'clock, green is the lowest the sun can be seen with such a small angle change at the observer. The quadratic must be between those two lines...

EDIT: I've thought it over and, if the blocks of density change aren't linear, it isn't quadratic (e.g. there is more atmosphere that is less dense than sealevel than there is atmosphere at sealevel.) This is the case, and it can't be a quadratic. But this is actually worse for FE... It has a shorter path! The only criteria for a path through an atmosphere like that is that it is between those lines, and the longest path that is possible is the sum of the distance from the apparent sun to the real sun and the green line.

EDIT: I just calculated the maximum path, which is from the fact that we see this sun at an angle of .2349278779 radians, as I have proved. If the maximum path is calculated, it turns out to be 12,811.71792 miles, a mere 5.71792 miles longer than my original calculation that didn't include refraction. Apply the same luminosity formula, and you get 5% the luminosity still.

This is undeniable mathematics, not government conspiracy.

EDIT: I just calculated what Tom's illustration's density would have, and its index is 10.881, or the speed of light is 10.881x slower. To give some scale, the index of lead is ~2.6. (that's the highest the list I saw on the google search had... Although I've heard of technologies slowing it down to tens of meters a second experimentally)

I thought I just did, mathematically. Unless you can come up with a FE model that isn't susceptible to a perpetually shown sun, and give me a good mathematical reason (you can't use "conspiracy" in math) that what I said was wrong, I accept my calculations as a fatal blow to FET.

No you did not. In FET it's unknown how high the atmosphere stretches, or whether space is a perfect vacuum. Many FE (and RE) models suppose that empty space is filled with an aether medium. This accounts for the refraction of the sun's light.

Username has a few good calculations for aethethic refraction.

Aether is a slower medium than lead? (by 4x?) That is downright ridiculous. Firstly, there is an experiment that you can use to disprove aether:

You could probably buy the equipment necessary to do that with not too much money these days... But if you aren't satisfied by things you can see, just like a flat earth, let's find out how ridiculous your claim is.

Proof aether is immaterial: Let's do a fun experiment real fast that only involves a pane of glass, a vacuum pump, and a laser. For aether to compensate for the refraction we saw, it would have to have a refractive index of 10.881, right? Now, a vacuum must have a refractive index of 1. If the refraction results in this experiment do not compensate, then the only other option is for the aether to be immaterial, or present also in a vacuum.

Proof aether can not now produce the results you expect: If aether is omnipresent (which I have just proved has to be mutually accepted with the notion of aether), then there is aether in the vacuum where the sun emitted it. If the refractive index does not change throughout an experiment, there is no refraction that takes place. Therefore, your theory is invalid, via experiments doable with even a low budget.

I can say with quite a bit of certainty that I just destroyed FET for anyone who is sensible in mathematics at all.

I derived what the refractive index for this magical aether or whatever would have to be to make the assumption that the sun is visible half of the day, with "visible" meaning it can only be seen until its position is about 1 degree (this isn't always true, but it's a good range, and I could refine the calculations but these pretty much sum up everything.) Well, the refractive index would have to be 18.48. That is huge (the last one I took was just finding stuff off of Tom's illustration, not really calculating what it would really require.) Now, I applied this refractive index to see the expected angle in the sky for the sun at 11 or 1 o'clock (the sun shifts 2pi/24 radians in the circle.) The results were stunning. I should expect, with this aether, to see the sun at an angle of .0475 radians, or 2.727 degrees. That is at 1 o'clock... And you could barely see it given the assumption that 1 degree makes it invisible. Good bye, FES.

EDIT: If anyone is skeptical, I could show the mathematics behind it, but I decided to keep it out because it's a little lengthy to type... It isn't that hard to find with the refractive law and simple trigonometry.

That completely pwns all short-scale trig that Flat-Earthers pull... Needless to say small triangles are inconclusive therefore moot.
« Last Edit: August 07, 2008, 09:27:38 PM by mxmm »



  • 136
Re: Resolve this.
« Reply #1 on: August 06, 2008, 01:02:01 PM »
So, in not responding to this argument, I assume that you admit that this is a true, undebatable, irreconcilable flaw and FET. Am I correct?