I believe the Earth is flat because it is flat,
I believe the Earth does not move because it does not move,
I believe the Sun is in motion because it is in motion and,
I believe the lights in sky are lights in the sky because they are lights in the sky.
The man in the small hat says he can weigh the red light in sky with heavy balls in a garden shed.
Well at least the man in the small hat DID an experiment, you on the other hand admitted to just believe what others tell you and we are still waiting for your final calculations. Where are they? Did you realize they were totally wrong hahaha what a fucking liar. What have I lied about? You're taking the fact I lack a complete understanding of some technical detail as evidence that I'm being dishonest. You're a fucking asshole. I'm analyzing how the atmospheric plane works as I'm gauging the Sun's distance at the point of maximum refractive magnification. I'm doing this because the Sun needs an object to be measured against; the horizon vanishing line. The horizon is at 90 degrees to the observers eye level thus it is the standard used to measure (with a sextant) the Sun is the measured (with a tape measure) height of the observer and the resolution limit (1 minute) of the eye. Now without being able to explain and prove how the refractive magnification caused by the atmospheric plane works I can't account for it (i.e. remove it mathematically) while calculating its size from the measured value. This is where I'm currently at in my "calculation".  |
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What is this? And why does it have so many pages...
It's a discussion about where we live and those bright objects we see in the sky. I suspect there are so many pages because these are important topics relevant to everybody. |
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Do you honestly believe that we're moving and spinning in multiple directions at thousands of miles per hour?
Have you lost your senses?
Are you stupid?
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You're hallucinating and psychotically delusional if you think the Earth is moving.
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I believe the Earth is flat because it is flat,
I believe the Earth does not move because it does not move,
I believe the Sun is in motion because it is in motion and,
I believe the lights in sky are lights in the sky because they are lights in the sky.
The man in the small hat says he can weigh the red light in sky with heavy balls in a garden shed.
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^^^ I'm not combining two models, I'm attempting to make the assumption that Earth is either a globe or flat not a requirement for calculating the distance to the Sun; the entire calculation is based on angular resolution limits thus the globe is completely BTFO.
The only way refraction is irreverent is if you're only taking a single measurement at 90 degrees, the entire sky plane is the result of refraction and showing/proving this is more important than the results of the distance calculation; we're already on a plain at this point.
It's time to stop staring at the ceiling and pretending we're on merry-go-round ride in a 10^-17 torr vacuum, we're not orbiting a million mile wide thermonuclear bomb. The entire Copernican model is so ridiculous and un-scientific it can be dismissed out-of-hand on face value alone.
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^^^ Wow, nine paragraphs including the statement that #1, the Earth is flat, and you failed to include refraction much less diffraction: light is changing angles due to an interaction with translucent atmospheric material.
You've addressed that it can and is eliminated at 90, and we're in agreement that on this.
We need a second measurement and I believe we're also in agreement on this.
You want to use two either two observers at the same time, or less desirably one observer at different times and locations. As is said, no dice; prove the earth either curved or flat; are those proofs beyond a reasonable doubt, do they express themselves numerically?
Do you see the problem here?
edit:
Keep in mind there's a concave mirror above the refractive planes.
edit2:
Do you see the angle I'm trying to calculate? It's zero; bring the Sun down so it touches the ground then remove the refraction.
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^^^ I know the Sun's exact size from the 90 degree measurement, it's simply a matter of proving/showing that the 1 minute per mile ratio used on the ground also applies to celestial objects.
The refractive layer is key here, I now think odolvlobo could have been on to something when he mentioned it. I just need to take step back and approach the problem from another (no pun intended) angle.
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...
Show me the fucking money!
Look at it this way. Find a piece of flat land somewhere, say, Kansas. Now, you will have to prepare this ahead of time. Pick a spot where you know the sun will be directly overhead, say, at about noon. Since the sun travels in an east-west direction, get somebody to stand 16 nm east of the spot where the sun will be directly overhead, and get another person to stand 16 nm directly west of the spot. Stand them so that if you draw a straight line between the two people, the line would pass through the spot where the sun is directly overhead, and this spot would be right in the middle of the line, kinda like this >>> ·+· . (The two dots are the people. The vertical line shows the center of the horizontal distance the people are apart. The intersect is where the sun is directly overhead.) If the two people look straight up, they should each be looking directly at opposite edges of the sun. But they aren't. Even transits or telescopes will show that they are still looking at the center of the sun. This means that the sun must be far wider than you suggest, or that there is some contradiction in you calculations somewhere. Because of the size of the earth, any atmospheric distortion would be too small to notice looking straight up. Perspective isn't included, because the guys are looking directly at the sun, even with the telescopes, not at some width where they need to focus perspectively.  I've already considered this scenario, no dice. The Sun is ~32 miles wide and the observers need to be farther than that apart. At that distance+ the question of whether the earth is flat or curves at 8" per mile^2 becomes an issue, and the globalist will start injecting the refraction is bending the light around the globe argument. The globalist (with a small hat) will kick the can so far down the road posing one augment after another for why the Earth is actually bent, that he's able to muddy the water while wasting time and effort and completely shut it down. After much analysis I've come to realize that the plane where the Sun's light is refracted is well, a plain; a level layer of dense gas. This (the plane) and the linear refractive magnifying effect is why the one minute per nautical mile applies to celestial objects. The fact there's zero refraction at 90 degrees allows for the Sun's true diameter to be measured. ...
Show me the fucking money!
Exactly, you are trying to prove it but havent done so how do you know if you havent proved it yet? Did you just beljeve it or what? You might want to check the last 773 pages, I'm fairly sure you'll find I've posted some valid reasons on why the Copernican model is a crock of shit. |
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^^^ You're a massive piece of shit, you know that right? I know the Sun is small, close and in motion and you start screaming that I'm liar for trying to prove it? Just fuck off and die already. e
The Sun measures 32 minutes in diameter and there's one nautical mile per measured minute. If this is wrong then fucking prove it you worthless sack of shit. Ask any mariner who knows how to use a sextant or any book on the subject and they'll tell you 1 nm = 1 min.
Show me the fucking money!
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^^^ All I need to do is prove that the ratio of 1 minute to 1 nautical mile applies to the apparent measured size of the Sun. Can you prove that it doesn't?
If you want to call bullshit then show us the money dickweed.
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^^^ smells like fear to me
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I'm attempting to calculate this myself and you're accusing me of not calculating things myself, fuck off. You're a fucking coward!
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^^^ Fuck, like you've ever measured and calculated the distance to sun yourself and got 93,000,000 miles? Everybody just trusts the fucking liars at NASA to provide factual information.
There's enough information from the angular measurements and the known distance to the horizon to make the calculation, I just need to figure it out.
Yes my first attempt was horrific, very horrific, but I'm including the apparent size this time and I think I'm getting closer... The only value I'm assuming is correct is the angular size limit of the eye, and considering the nautical mile is based on this value, and that I have peer reviewed documentation it's not really much of an assumption.
It's simply a matter of time before I derive the correct equations and they give constant results with variable inputs. Yes I believe the 32 mile diameter and 3,100 mile distance values are correct but I'm actually trying to prove it. You just go with whatever your told by the men in small hats and never question any of it.
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^^^ Yeah I'm sure you'll say anything to make people believe that only NASA or a Jesuit trained astronomer can calculate the distance to the Sun.
Distance to the horizon and the Sun's elevation in degrees above the horizon is enough information (maybe) for anybody to make the calculation. If not the addition of the Sun's apparent size and possibly another object of known distance and size will fill the gap.
At this point I'm making an attempt, my calc for "A" may need work but the distance is 3 miles for a 6 foot person so I'll check that formula over later.
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^^^ The distance to the sun (my preliminary calculation) is 3068.4 nautical miles and I detail exactly how I measure and calculate this value:
[image]
As you can see I've calculated it based on the distance to the horizon and the sun's elevation angle.
edit:
I noticed the calculation for Z was missing, it's the standard right angle trig calculation: Z=D*tan(P). I updated the image.
This is obviously a work in progress, I'm still checking it over and I'm going to re-draw the whole graphic a bit differently and hopefully clarify how the calculation is made.
edit:
I've removed the image/calculations due to horrific errors, I must have mixed up my notes.
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Sure you can. It is simple trigonometry. Given the either the distance or the size, it is easy to compute the other. https://i.imgur.com/pUPdRLk.pngAs for calculating the effect of refraction, I meant this: "Given the angle above the horizon and the distance, what is the amount of magnification?" You don't know either the actual size or the distance! ... In your diagram you label the sun's diameter as 32 nautical miles (nm) how do you know this? How do you know that the 1 nautical mile per 1 minute applies to the 32 minuets you measured the sun at? You have said many times that the sun is 32 nm wide because it is 32 minutes wide and 1 minute is 1 nm. I'm using your numbers. Now you say it isn't 32 nm wide? ... Hold a dime at arms length and it's size is 32 minuets wide, does that mean it's 3,000 miles away? Fuck no! ...
A dime is 18 mm wide, so if it is 32 minutes wide, it is 0.018 / 2 / tan(0.533 / 2) = 1.9 meters away. I'm a random idiot on the internet, I could be wrong or lying. How do you know that 1 minute = 1 nautical mile applies to the 32 minute apparent diameter of the sun? I know that in the case of the sun it does but how do you know? Just because I said so is not a valid reason. My point is you need measure the distance to the sun first, then you have a distance to calculate its actual diameter with. |
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Keep in mind that I'm combining angles from both physical space and optical space and the diagram is less than clear in this regard. R & T also need to be moved over to the left side.
First point, the distance to the sun can't be calculated from the sun's apparent size alone. As you can see I didn't use "I" or "U" at all in my calculation at all, only "Q" and "A".
You can if you know the effect of refraction. Do you know how to compute that? ...
Finally the sun is always 32 minutes wide due to refraction and maintains the same diameter as if it was at 90° because of it. As you can see the calculations for the angular size (X) have not been completed yet. In the case of the sun and moon here X will equal U and refraction plays a role such that a refracted optical space is created in addition to the physical and optical spaces.
I see. The sun appears to have the same size because of refraction. That raises a question for me. Why are the sun and moon affected by refraction in this way, but other objects are not? No, no, no and no! You can not calculate the distance to the sun or moon from just apparent size you have to have another object, in this case I used the horizon. Atmospheric refraction is not another object. Refraction can be determined by calculating the sun or moon's apparent size and position then comparing them with measured values. The difference will be the effect of refraction. The sun and moon are affected by atmospheric refraction because there are layers of different density gases below them and they causes the light to change direction. Sure you can. It is simple trigonometry. Given the either the distance or the size, it is easy to compute the other.  As for calculating the effect of refraction, I meant this: "Given the angle above the horizon and the distance, what is the amount of magnification?" You don't know either the actual size or the distance! Hold a dime at arms length and it's size is 32 minuets wide, does that mean it's 3,000 miles away? Fuck no! In your diagram you label the sun's diameter as 32 nautical miles (nm) how do you know this? How do you know that the 1 nautical mile per 1 minute applies to the 32 minuets you measured the sun at?
If you want to prove a pressurized atmosphere can be maintained within a hard vacuum environment without a container be my guest.
Rather simple, Gravity but then again you don't believe in gravity  So if I have a 1L beaker that contains 1L of air at standard temperature and pressure, and I place it in a vacuum chamber with a hard vacuum, the force of gravity will keep the air from rising up and escaping the beaker?  Quite the misconception, a vacuum is the ABSENT of anything, it doesn't exert any force, and doesn't pull. Two Principles, Inversion and non-Inversion 1#) A vacuum chamber on earth with compromised Integrity, can implode in on itself, due to air pressure equalization. 2#) A air'd in the container in space with 1 atmospheric pressure/surface with compromised Integrity, can Explode, as positive pressure, and escape outwards. If the pressure is at 0.2 atmospheric pressure, less outward force. Like the atmosphere around Earth, the higher up you go from the sea, the lower the atmospheric pressure. On Mountains it's lower, go up even higher it's even lower. Thus less outward force. The Force of gravity is stronger then the outward force of the much lower pressure, hence it doesn't escape into space. Edit: *Actually some does escape but much slooower due to the less pressure, and replacement is on-par with the loss, thus our planet isn't like mars, bad example: But think of it like this: An hose that inserts air into the beaker, and the vacuum chamber maintains it's vacuum by sucking air out. While air rushes out of the beaker, it still gets replaced at par with the loss, so it stays in the beaker. However, the scale is much longer for the earth. Where does the hose leading into the earth come from? |
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If you want to prove a pressurized atmosphere can be maintained within a hard vacuum environment without a container be my guest.
Rather simple, Gravity but then again you don't believe in gravity So if I have a 1L beaker that contains 1L of air at standard temperature and pressure, and I place it in a vacuum chamber with a hard vacuum, the force of gravity will keep the air from rising up and escaping the beaker? |
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^^^ You're wandering off-topic and I've been advised not to cast pearls before swine, so I'll just let you wallow in your own crapulence. If you want to prove a pressurized atmosphere can be maintained within a hard vacuum environment without a container be my guest. Here's Stephen Hawking pictured on Jeffrey Epsteins Island of Sin:  |
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