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Author Topic: Flat Earth  (Read 1074823 times)
Astargath
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June 25, 2019, 03:38:57 PM
 #15441

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  Cheesy

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?



Gravity has been tested for decades and it works, every single time, all the predictions done on gravity are accurate. The black hole that was recently "photographed" was also predicted to look that way.

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June 25, 2019, 04:01:56 PM
 #15442

^^^ You think it's fucking reasonable to believe that we're stuck to the surface of a spinning ball in a vacuum orbiting a million mile wide thermonuclear bomb in an endless virtually empty universe created by nothing exploding for no reason?

The only way anybody could possibly believe any of that shit is if they were indoctrinated and brainwashed from birth.

Oh, come on. Tell us what you are really doing with this thread.

There probably isn't anyone in the world that believes those things you say... spinning ball, thermonuclear bomb.

Yet you try to make us believe, that you believe, that there are people who believe that way.

You aren't that dumb. You are way too smart to believe that there are people who believe that way. We know it. You know it. Your FE buddies know it. So, why don't you tell us your real purpose in this thread? Or would telling us destroy your purpose?

Cool

Notice how notbatman ignores what I say above^^? All he's doing is playing a game with this entire thread.

Cool

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June 25, 2019, 04:05:19 PM
Last edit: June 25, 2019, 04:18:14 PM by lightlord
 #15443

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  Cheesy

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.



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June 25, 2019, 04:46:36 PM
 #15444

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?"

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June 25, 2019, 08:33:42 PM
 #15445

This theory is very confusing for me however I think about this theory I had the doubt so much how people can live in the flat Earth also if it possible then there is some end is also available in this world that's why it is so much confusion

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June 26, 2019, 02:45:04 AM
Last edit: June 26, 2019, 02:45:35 PM by notbatman
 #15446

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?"

NO!

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  Cheesy

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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June 26, 2019, 03:12:39 AM
 #15447

^^^ This is so humorous. You get someone who essentially agrees with notbatman, and he argues with them about it. How funny.

Cool

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June 26, 2019, 03:53:16 AM
Last edit: June 26, 2019, 04:10:27 AM by odolvlobo
 #15448

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.png
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! ...

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.



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June 26, 2019, 05:52:22 AM
 #15449

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.png
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! ...

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.




At this point im reconsidering the possibility of notbatman trolling us. He is literally contradicting himself and in the other post he said something like he passed out from calculating too much, sounded pretty trollish.

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June 26, 2019, 06:46:48 AM
 #15450

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.png
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! ...

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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June 26, 2019, 07:42:29 AM
 #15451

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.

I don't know these things and I can't measure the distance to the sun or its height or its width. You didn't measure these things either, so you must have figured them out somehow, or somebody must have showed you the reasoning or the evidence or the measurements. All I'm asking is that you share this information so that I can understand how you have come to your conclusions. Saying that I can't use simple trigonometry because you might be lying isn't helpful at all.

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June 26, 2019, 09:13:10 AM
Last edit: June 26, 2019, 02:39:28 PM by notbatman
 #15452

^^^ 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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June 26, 2019, 12:27:23 PM
 #15453

^^^ The problem with your calc is that it only works this way in one circumstance. If you drop the height of the guy to 5 feet rather than 6, the whole calculation changes, as well as the height and diameter of the sun.

If you give the guy a telescope so that he can see great distance beyond where his simple eye horizon exists, everything changes even more.

If you situate the sun directly overhead, The capability of doing your measurements disappears altogether.

You are missing a whole lot in the way you calculate.

Oh, but that's right. You are simply playing with people. So, I shouldn't be wasting my time with you. But I'll leave this here, for a while, anyway.

Cool

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June 26, 2019, 12:58:52 PM
Last edit: June 26, 2019, 02:39:51 PM by notbatman
 #15454

^^^ 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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June 26, 2019, 05:52:03 PM
 #15455

^^^ 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.

In a way, these posts are already an admission that you simply believed other people until now, you should have done these calculations long ago, no? You have been claiming the sun is 3068 miles away several times but it turns out that you have never done the calc?

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June 26, 2019, 07:04:48 PM
 #15456

^^^ 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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June 26, 2019, 08:31:36 PM
 #15457

^^^ 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.



Again admitting that all along you simply believed what other youtubers told you and yet you have been claiming to know it's close to us. You are a liar, exposing yourself at this point basically. Yes you believe those values are correct but why? What made you believe that? The youtubers?

''You just go with whatever your told by the men in small hats and never question any of it.'' And you just go with what youtubers tell you and never calculate or experiment for yourself, LOL.

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June 26, 2019, 08:52:08 PM
 #15458

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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June 27, 2019, 01:39:59 AM
 #15459

....., I just need to figure it out.

Yes my first attempt was horrific, very horrific, .....

It's simply a matter of time before I derive the correct equations and they give constant results with variable inputs. ......




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June 27, 2019, 10:26:47 AM
 #15460

^^^ smells like fear to me
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