Test for Earth curvature:

Originally posted by Suzianne
Is this where you claim to be an atmospheric physicist or engineer?

I figure that's gotta be next.

No, babe: I just happen to know far more about the topic than you do now, or likely ever will.

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As stated, your game is pathetically weak.
14 year old me laughs derisively at your sorry attempts at shock.
That was 38 years ago, so you pretty much know where you stand, son.

The post that was quoted here has been removed

Sad, faggoty little donkey lover.
Keep swinging, hon.
You're bound to hit something eventually.

The post that was quoted here has been removed

Seriously?
You have one?
That's a mind blowing revelation.
That it is large enough to have it sucked?
Now you're just wishful thinking, micro-pepe.
Don't get too far ahead of yourself, hon.

Originally posted by FreakyKBH
Seriously?
You have one?
That's a mind blowing revelation.
That it is large enough to have it sucked?
Now you're just wishful thinking, micro-pepe.
Don't get too far ahead of yourself, hon.

So you figure if you see some distant object you would see it like tiny, if it was hundreds of miles away on a flat Earth how would you see it at all? You know about resolution? Since you claim to know more than anyone here, at 200 miles what would be the resolution of the human eye?

Here is a little bit of optics you undoubtedly know since you undoubtedly have a Phd in the subject, but for a minor refresher:

The human eye has a resolution of about 1 minute of arc, that is 60 arc seconds. Now as you also undoubtedly know, 1 second of arc splits a circle into 1,296,000 parts. The eye has only 1/60th of that or 21,600 parts of a circle.

So at 200 miles X 2 X Pi X 5280 is 6,635,043.684 feet circumference circle. Forgive me for showing you such simple stuff.

So 6.6 million divided by 21600 is a resolution of 307 feet. If an object is say 100 feet across, you would not even be able to make it out at all.

If it was 200 feet across you would not be able to make it out.

It has to be over 300 feet across to even see a dot. So at 600 feet wide you might see something.

So you would need a good telescope to see much at all even if it was above the horizon with no atmospheric distortions, refractions, diffractions or whatever AND no clouds.

Good luck seeing much at that distance un-aided.

So on a flat Earth you should be able to see 2000 miles which would require an object 3000 feet wide to even see as a dot by unaided human eyes.

With a 100X telescope, you would see an object 30 feet across. NOBODY sees ANYTHING like that ANYWHERE on Earth.

Flat Earth is wiped out. Period. You go on top of Mt. Everest with a 100 power telescope on a clear day you should be able to see stuff at LEAST 2000 miles way. You tell me ANYONE who has evidence of that. Good luck.

Originally posted by sonhouse
So you figure if you see some distant object you would see it like tiny, if it was hundreds of miles away on a flat Earth how would you see it at all? You know about resolution? Since you claim to know more than anyone here, at 200 miles what would be the resolution of the human eye?

Here is a little bit of optics you undoubtedly know since you undoubte ...[text shortened]... le to see stuff at LEAST 2000 miles way. You tell me ANYONE who has evidence of that. Good luck.

Ooøœ...
Numbers!
You bested me with numbers!
Too bad your math is wrong, but way to go on those numbers!

Originally posted by FreakyKBH
Ooøœ...
Numbers!
You bested me with numbers!
Too bad your math is wrong, but way to go on those numbers!

Prove me wrong with YOUR numbers. Good luck. Do you doubt for instance, 1 arc second cuts a circle into 1,296,000 parts? Or 60 arc seconds cuts a circle into 21,600 parts which is also one arc minute? Show me my mistake o optical maven. I just made big circles from a radius of 200 miles and 2000 miles read out in feet. If I am not mistaken there are 5280 feet per mile so for a 200 mile radius that would be 1,056,000 feet. Feel free to correct me here.

So a 1 million mile radius would be a two million mile diameter and multiply that times pi, or 3.14159 etc., you get a circle of 6,635,043 and change feet.

Divide that by 1 arc minute or 1 part in 21,600 and you get the smallest piece of the circle your eye can make out and that is 307 feet.

So show me my mistake if you dare. I mean, I bow to your superior knowledge in every way, since you clearly have a Phd in several disciplines including optics so show me o master.

Originally posted by sonhouse
Prove me wrong with YOUR numbers. Good luck. Do you doubt for instance, 1 arc second cuts a circle into 1,296,000 parts? Or 60 arc seconds cuts a circle into 21,600 parts which is also one arc minute? Show me my mistake o optical maven. I just made big circles from a radius of 200 miles and 2000 miles read out in feet. If I am not mistaken there are 5280 fe ...[text shortened]... y way, since you clearly have a Phd in several disciplines including optics so show me o master.

Do you doubt, for instance, that water is nearly 800 times more dense than air?
How about the fact that LIGHT refracts at measurable rates when there are temperature changes in the medium through which it passes?
Do you doubt that?
Now, account for the visibility of distant objects REGARDLESS of the temperature gradient of the air--- through which distant OBJECTS are clearly seen.
Further, account for the visibility of these same objects during nearly ALL atmospheric conditions.

Don't even bother posting, as anyone who has even taken a cursory view of the topic knows: you don't have a leg to stand on.
You simply cannot explain the visibility of distant objects by replying on atmospheric refraction because it simply isn't applicable.

Originally posted by FreakyKBH
Do you doubt, for instance, that water is nearly 800 times more dense than air?
How about the fact that LIGHT refracts at measurable rates when there are temperature changes in the medium through which it passes?
Do you doubt that?
Now, account for the visibility of distant objects REGARDLESS of the temperature gradient of the air--- through which dista ...[text shortened]... ity of distant objects by replying on atmospheric refraction because it simply isn't applicable.

So you are totally ignoring my post about resolution of eyeballs. Nice. You say my numbers are wrong and when I reiterate them you just move the goalpost again.

I just showed you at 200 miles you have to have an object 300 odd feet wide just to make out a dot. And 2000 miles on a clear day, you would need a good telescope which can give maybe 30 feet of res at that range.

If you had actually read my post about my suggested experiment you would see I found out the max a laser beam can go underwater is about 50 meters, not even close to a solution for my experiment so I changed it to an above water test which would require multiple tests at different times of day to peg down the atmospheric distortions of the laser beam path. It would not be a simple job. Like I said, those casual laser experiments were flawed from day one because of atmospheric refraction's and diffraction.

So prove my numbers wrong. You can't can you.

Originally posted by sonhouse
So you are totally ignoring my post about resolution of eyeballs. Nice. You say my numbers are wrong and when I reiterate them you just move the goalpost again.

I just showed you at 200 miles you have to have an object 300 odd feet wide just to make out a dot. And 2000 miles on a clear day, you would need a good telescope which can give maybe 30 feet of ...[text shortened]... e of atmospheric refraction's and diffraction.

So prove my numbers wrong. You can't can you.

As soon as you can respond to the FACTS put to you, I'll humor yours.

Originally posted by FreakyKBH
As soon as you can respond to the FACTS put to you, I'll humor yours.

Look at this link about horizons and especially the paragraph on atmospheric refraction.

https://en.wikipedia.org/wiki/Horizon#Effect_of_atmospheric_refraction

BTW, from ISS, about 240 miles up, the horizon is about 2500 miles. That is only 1/10th the way round the planet and if Earth were flat it would see the entire thing.

http://blogs.discovermagazine.com/badastronomy/2009/01/15/how-far-away-is-the-horizon/#.V48yivkrKJA

And this from PBS:

http://www.pbs.org/newshour/updates/7-diy-experiments-b-o-b-the-earth-is-round/

Look at the video showing Coriolis effect, it can be shown quite clearly on a merry go round with video shot from above.

Sorry you are so brainwashed you can't see the forest for the trees.

NOW tell me exactly where my numbers are wrong about what you can see with human eyes, the resolution that my numbers say you cannot see an object under 300 feet at 200 miles without a telescope.

From Chicago for instance, on a flat Earth, you should be able to see the Rocky Mountains while sitting flat on the ground on a clear day. I dare you to say you could ever see the Rockies from Chicago. Your cloud objection would not hold against the entire Rocky mountain range it goes north and south for a thousand miles so there would be some part of it visible almost all the time.

Prove THAT wrong.

Originally posted by sonhouse
Look at this link about horizons and especially the paragraph on atmospheric refraction.

https://en.wikipedia.org/wiki/Horizon#Effect_of_atmospheric_refraction

BTW, from ISS, about 240 miles up, the horizon is about 2500 miles. That is only 1/10th the way round the planet and if Earth were flat it would see the entire thing.

http://blogs.discoverm ...[text shortened]... usand miles so there would be some part of it visible almost all the time.

Prove THAT wrong.

I don't know why you insist on returning to something so patently absurd and then attempt to use it as though it bolsters your position.
The further one moves away from ANY object, the decrease in appearance of said object: it continues to diminish until it merges with the convergence of the receding lines.
The Rocky Mountains don't fall from view on account of the curvature of the earth.
If there were a completely flat plain on the eastern side of the range which extended for untold miles, eventually the RM would become merged with the horizon.
Remove even one of the mountains in that range (Mount Elbert) from its base and catapult the same into the sky above the earth: you'd be able to see that 14,440' peak hundreds and hundreds of miles off the surface of the earth.
But...
not so on the surface, which limits the vision on account of atmospheric convergence at the horizon.

Are you even really trying?

Originally posted by FreakyKBH
I don't know why you insist on returning to something so patently absurd and then attempt to use it as though it bolsters your position.
The further one moves away from ANY object, the decrease in appearance of said object: it continues to diminish until it merges with the convergence of the receding lines.
The Rocky Mountains don't fall from view on acc ...[text shortened]... the vision on account of atmospheric convergence at the horizon.

Are you even really trying?

Explain your 'atmospheric convergence at the horizon'. Why would you not be able to see the RM's with a telescope even if they are not in view of eyeballs?