A new idea (I hope)...

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A new idea (I hope)...
« on: April 26, 2006, 10:38:13 PM »
Wait, I wonder if, say, I were to stand somewhere in China, on a perfectly clear day, with a powerful (non-manipulated :P ) telescope, could I see Mt. Everest?  In fact, could I theoretically see for thousands and thousands of miles in all directions?  

Yet this is not the case, therefore isnt some curvature on Earth's surface apparent?

A new idea (I hope)...
« Reply #1 on: April 26, 2006, 10:43:41 PM »
Quote from: "monkeyman"
Wait, I wonder if, say, I were to stand somewhere in China, on a perfectly clear day, with a powerful (non-manipulated  ) telescope, could I see Mt. Everest? In fact, could I theoretically see for thousands and thousands of miles in all directions?


Okay, you're defining an experiment that you can carry out on a clear day.  One argument against it is that there are no perfectly clear days (I won't argue this with you).  So, theoretically, you could see for thousands of miles on a flat earth.

Quote from: "monkeyman"
Yet this is not the case, therefore isnt some curvature on Earth's surface apparent?


Whoa!  Did you do the experiment already?!  That was fast.  You were like:  "Hey, what if I did this?  I'd be able to see thousands of miles in all directions right?"  And then two seconds later you were like "Yeah, turns out it didn't work.  So the earth must be curved."

Two flaws with your argument:
1.  You assumed the results of an experiment without having carried it out.
2.  You don't take into consideration that the flat earth is not a plane.  You said:  "therefore isnt some curvature on Earth's surface apparent?"  Well, of course.  I can look outside right now and tell you that.  There's a hill off in the distance, a "curvature" of the earth that is readily apparent.
ooyakasha!

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Erasmus

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Re: A new idea (I hope)...
« Reply #2 on: April 26, 2006, 10:46:08 PM »
Quote from: "monkeyman"
Wait, I wonder if, say, I were to stand somewhere in China, on a perfectly clear day, with a powerful (non-manipulated :P ) telescope, could I see Mt. Everest?  In fact, could I theoretically see for thousands and thousands of miles in all directions?


In Ideal Physics world -- where there is, for instance, no atmosphere -- then yes.  But the atmosphere isn't perfectly transparent.  If you've been hiking in the mountains, you know what I mean: when you get to the top and view the panorama, stuff in the distance looks fuzzy and, well, blue.  So even if the Earth were flat, your vision would be limited by air getting in the way.

Of course there's the caveat that Tibet isn't *that* big... maybe there's some place in China that isn't very far from Mt. Everest, and maybe you can see it from there; I dunno.

-Erasmus
Why did the chicken cross the Möbius strip?

A new idea (I hope)...
« Reply #3 on: April 26, 2006, 10:55:01 PM »
First off,Knight, when I meant curvature I was signifying on a huge scale (i.e. round earth) and I thought that maybe the people at this forum would be intelligent enough to deduce this fact.  Obviously I was proven wrong.

Anyways, couldn't someone carry out some mathematics equations to first figure out how far an observer can see on a "clear" day (on a flat earth), and then relate that to how far someone can see taking into account the curvature of the earth if spherical, and see if in fact you should be able to see, say, a mountain range at 1000 miles on a flat earth, yet according to the spherical earth theory, you can only see to 800 miles at sea level, etc, etc...does this make sense?

A new idea (I hope)...
« Reply #4 on: April 26, 2006, 10:59:08 PM »
Maybe if you tried the experiment on a boat and a really really powerful (laser maybe?) light source on land, a "light house". If you were to travel more than<sqrt(1.5h)> miles (with h being the height of the tower compared to sea level in feet) out to sea and still see the light on a clear day, you'd have a good case for flat earth.

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Erasmus

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A new idea (I hope)...
« Reply #5 on: April 26, 2006, 11:15:51 PM »
Quote from: "monkeyman"
First off,Knight, when I meant curvature I was signifying on a huge scale (i.e. round earth) and I thought that maybe the people at this forum would be intelligent enough to deduce this fact.  Obviously I was proven wrong.


Strong words!  What Knight meant is that things will always get in your way, on a round Earth or flat.  Things that are far away are usually not much bigger, on average, than things that are close.

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Anyways, couldn't someone carry out some mathematics equations to first figure out how far an observer can see on a "clear" day (on a flat earth),


First you insult us, then you ask us to do you a favour?

Anyway, the geodesic distance you can see on a round Earth, discounting atmospheric occlusion, is D=arcsec(1+H), where H is your height above sea level.  All units are Earth radii here.  The line-of-sight distance is L=tan(D).

I don't know how to work atmospheric occlusion into the calculations, which means I also can't compute the visibility on a flat Earth.

-Erasmus
Why did the chicken cross the Möbius strip?

Re: A new idea (I hope)...
« Reply #6 on: April 27, 2006, 05:54:01 AM »
Quote from: "Erasmus"
Quote from: "monkeyman"
Wait, I wonder if, say, I were to stand somewhere in China, on a perfectly clear day, with a powerful (non-manipulated :P ) telescope, could I see Mt. Everest?  In fact, could I theoretically see for thousands and thousands of miles in all directions?


In Ideal Physics world -- where there is, for instance, no atmosphere -- then yes.  But the atmosphere isn't perfectly transparent.  If you've been hiking in the mountains, you know what I mean: when you get to the top and view the panorama, stuff in the distance looks fuzzy and, well, blue.  So even if the Earth were flat, your vision would be limited by air getting in the way.

Of course there's the caveat that Tibet isn't *that* big... maybe there's some place in China that isn't very far from Mt. Everest, and maybe you can see it from there; I dunno.

-Erasmus


how can an atmospher exist on a flat earth? wouldn't the clouds just float off the edge or whatever?
i]On this issue -- my default assumption is that all members of this forum are male.  I usually expect women to have more sense than to waste their time arguing trivialities over the internet.
[/i]
-Erasmus

A new idea (I hope)...
« Reply #7 on: April 27, 2006, 05:58:14 AM »
I'm guessing, based on observation, that the distance at which the atmosphere typically obscures vision is roughly D = arcsec(1+H), where H is your height above sea level. Units are in the constant "R", which is derived from atmospheric density and which RE scientists misinterpret to represent the radius of the earth, but that's pretty silly since the earth is flat.

Again, just based on observation.

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Chaltier

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Re: A new idea (I hope)...
« Reply #8 on: April 27, 2006, 09:52:43 AM »
Quote from: "monkeyman"
First off,Knight, when I meant curvature I was signifying on a huge scale (i.e. round earth) and I thought that maybe the people at this forum would be intelligent enough to deduce this fact.  Obviously I was proven wrong.


What you meant to say doesn't matter. What you say does. Be specific.

Quote from: "Marshy"
Quote from: "Erasmus"
how can an atmospher exist on a flat earth? wouldn't the clouds just float off the edge or whatever?


This has been brought up before. I suggest that there's a "dome" of sorts that keeps it in. This dome doesn't necessarily have to be tangible to us, but it certainly affects the atmosphere. Just a theory, but does anyone have a better idea?


--Chal

A new idea (I hope)...
« Reply #9 on: April 27, 2006, 11:22:47 AM »
Yeah, a round earth.

All FEers have to prove in their case are some scientific facts which *could* be correct, then whenever a REer mentions gravity, plane trips, etc, you decide to "just forget the theory of relativity, etc." and use the "good ol' government conspiracy" route.

Hmmm....

A new idea (I hope)...
« Reply #10 on: April 27, 2006, 11:27:58 AM »
What, pray tell, do plane trips have to do with relativity?

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Chaltier

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A new idea (I hope)...
« Reply #11 on: April 27, 2006, 11:39:49 AM »
Quote from: "monkeyman"
Yeah, a round earth.


I knew this comment was coming. Thank you for justifying my starting a similar topic on the FE board.

And, as Unimportant said, um, what do plane trips have to do with relativity? You've got me curious now.


--Chal

A new idea (I hope)...
« Reply #12 on: April 27, 2006, 11:45:58 AM »
Guys I know absolutely nothing about relativity... but doesn't everything have to do with relativity?
ooyakasha!

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Erasmus

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A new idea (I hope)...
« Reply #13 on: April 27, 2006, 12:09:11 PM »
Quote from: "Unimportant"
I'm guessing, based on observation, that the distance at which the atmosphere typically obscures vision is roughly D = arcsec(1+H), where H is your height above sea level. Units are in the constant "R", which is derived from atmospheric density and which RE scientists misinterpret to represent the radius of the earth, but that's pretty silly since the earth is flat.


The difference: you are making a hypothesis about how to interpret a formula that agrees with observations (actually, it probably doesn't agree with observations), whereas I performed a derivation from first principles.

Somebody with more knowledge than me could derive a formula from first principles of atmospheric optics as well, and until they do so, you may not commandeer my formula -- especially since you have not actually made the observations that you claim support your formula (I know you haven't because my formula doesn't take atmospheric effects into account; the real formula would be different.)

Just a hint that my formula will not work on a flat Earth: it's basically a geometric definition of the secant through the centre of a circle.  So really, it's more of a first principle itself than it is a derivation.

Also: if the atmosphere limits visibility, then being higher up in the air means there's more air between you and the ground, which means your visibility of the ground should decrease, not increase.  Basically, there would be some sphere of visibility around you; the ground that you could see would be the intersection of the flat Earth with that sphere.  As the distance between the centre of that sphere and the surface of the Earth increases, the intersection gets smaller.  My formula predicts exactly the opposite, so.... yeah, there's an easily verifiable test for sphericity.

-Erasmus
Why did the chicken cross the Möbius strip?

A new idea (I hope)...
« Reply #14 on: April 27, 2006, 12:45:13 PM »
Quote from: "Erasmus"
The difference: you are making a hypothesis about how to interpret a formula that agrees with observations (actually, it probably doesn't agree with observations), whereas I performed a derivation from first principles.

You said your formula was based on observation, although I'm sure the formulas you provided are those used to determine the distance between a tangent line's point of intersection with the surface of a sphere and a point on the line height H above the sphere;  that's what would make the most sense on a spherical earth anyways.

If they were based on observation, though, and observable on the real earth, I would assume what I said before. I can't argue that you see what you see - which is the case of a formula derived from observation - just why you're seeing it. So I guess in that sense my point was hypothetical, because the formulas you provided clearly aren't observational.
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Somebody with more knowledge than me could derive a formula from first principles of atmospheric optics as well, and until they do so, you may not commandeer my formula -- especially since you have not actually made the observations that you claim support your formula

I guess I've already mentioned this, but I find it hard to believe you've made the observations to verify your formula either, especially since it applies to a perfect sphere. Like you said, you applied fundamental geometric principles to the concept of a spherical earth. Nothing to do with observation here.
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Also: if the atmosphere limits visibility, then being higher up in the air means there's more air between you and the ground, which means your visibility of the ground should decrease, not increase.

It seems logical that, as the atmosphere gets thinner at higher elevations, it's obfuscatory properties might diminish as well. If your formulas were observational (which we have established they are not), then I would assume the H value is inversly proportional to the rate at which the atmosphere limits visibility.

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Erasmus

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« Reply #15 on: April 27, 2006, 02:08:27 PM »
Quote from: "Unimportant"
You said your formula was based on observation,


I said no such thing.

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although I'm sure the formulas you provided are those used to determine the distance between a tangent line's point of intersection with the surface of a sphere and a point on the line height H above the sphere;


Yup; right on.

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I guess I've already mentioned this, but I find it hard to believe you've made the observations to verify your formula either, especially since it applies to a perfect sphere. Like you said, you applied fundamental geometric principles to the concept of a spherical earth. Nothing to do with observation here.


Really, I mean it: I have neither made the observations, nor claimed to have done so, nor claimed to have based my formula on observations (as far as I can remember).

My guess however is that observations will agree with the formula within experimental error: I expect (without justification) that neither atmospheric limitations on visibility nor the oblateness of the Earth will have a noticeable affect on measurements.

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It seems logical that, as the atmosphere gets thinner at higher elevations, it's obfuscatory properties might diminish as well.


Certainly!  But just because you're adding thinner atmosphere, doesn't mean you have removed any of the thicker atmosphere.  You still have to look through the atmosphere near the ground, in order to see stuff on the ground, no matter how high you are.

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If your formulas were observational (which we have established they are not), then I would assume the H value is inversly proportional to the rate at which the atmosphere limits visibility.


Um... I'm not sure what you mean by this.  H is a variable that I plug numbers into; like, I climb to such-and-such a height, and then set H to be that height.

Anyway, point is, whereas neither of us has observational evidence to support our formulae, only one of us has a formula.  I promise that I will seek opportunities to test my formula, but it may have to wait until I'm in a less mountainy area.  If somebody would care to derive a formula for maximum geodesic visibility as a function of height and the rate of dropoff of atmospheric density (this is a constant, since the dropoff is linear), we can go about testing that as well, and checking to see if (a) it works at all, and (b) there's any "R" term which coincidentally makes radius-based formulas work out as well.

-Erasmus
Why did the chicken cross the Möbius strip?

A new idea (I hope)...
« Reply #16 on: April 27, 2006, 02:58:47 PM »
Quote from: "Erasmus"

I said no such thing

No you didn't. I think I maye have looked at my quote nested within your quote and confused myself.

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Certainly!  But just because you're adding thinner atmosphere, doesn't mean you have removed any of the thicker atmosphere.  You still have to look through the atmosphere near the ground, in order to see stuff on the ground, no matter how high you are.

Yes, but you'll have to look through less of it. If there are only two kinds of air, "thin air" and "thick air", and they meet at a certain height 'H', then if you are at a height 2H you will only be looking through "thick air" for half the distance. Obviously atmospheric density is a gradient so there is no such boundry, but the same principle holds true.

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Um... I'm not sure what you mean by this.  H is a variable that I plug numbers into; like, I climb to such-and-such a height, and then set H to be that height.

And as height H increases, the air density decreases. Inversely proportional (assuming linearity of density dropoff, which may not be true).

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Anyway, point is, whereas neither of us has observational evidence to support our formulae, only one of us has a formula.

Nothing I can say about that. I'm sure such formulas exist, it they would likely just involve plugging in an atmospheric constant. I'll keep my eyes open.

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Erasmus

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A new idea (I hope)...
« Reply #17 on: April 27, 2006, 04:05:51 PM »
Quote from: "Unimportant"
Yes, but you'll have to look through less of it. If there are only two kinds of air, "thin air" and "thick air", and they meet at a certain height 'H', then if you are at a height 2H you will only be looking through "thick air" for half the distance. Obviously atmospheric density is a gradient so there is no such boundry, but the same principle holds true.


I see what you mean.  You're not exactly looking through half the thick air (if you're just in Thick Air, you're looking through sqrt(H^2+D^2) of it; if you're in Thin Air, you're looking through sqrt(4*H^2+D^2) of Thick Air; the latter quantity is not twice the former), but I agree that you are looking through less thick air.

Let's try and come up with a formula, shall we?  I suggest we use your thick/thin model, but expand it to N layers, each of whose densities decreases by a constant amount K as you go higher up.  Then we can just take the limit as N goes to infinity and K goes to zero, where  NK=(surface density).

-Erasmus
Why did the chicken cross the Möbius strip?

A new idea (I hope)...
« Reply #18 on: April 27, 2006, 04:20:21 PM »
Quote from: "Erasmus"
I see what you mean.  You're not exactly looking through half the thick air (if you're just in Thick Air, you're looking through sqrt(H^2+D^2) of it; if you're in Thin Air, you're looking through sqrt(4*H^2+D^2) of Thick Air; the latter quantity is not twice the former), but I agree that you are looking through less thick air.

Do you mean sqrt[(.5H)^2 + (.5D)^2] for the distance through thick air? Becuase otherwise I'm missing something.

I'll consider this on the way home, it's a long drive.

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Erasmus

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A new idea (I hope)...
« Reply #19 on: April 27, 2006, 04:51:16 PM »
Quote from: "Unimportant"
Do you mean sqrt[(.5H)^2 + (.5D)^2] for the distance through thick air? Becuase otherwise I'm missing something.


I'm pretty sure I don't mean that:

erk.  Photobucket is down.  Will have to wait for later.  In the meantime:

Code: [Select]

             . 2H
            /|
           / |
          /  |
         /   |
        /    |
     A /     + H
      /    _-|
     /  B_-  |
    /  _-    |
   / _-      |
  /_-        |
 +-----------+ 0
       D


A = sqrt((2H)^2  + D^2)
B = sqrt(H^2 + D^2)

-Erasmus.
Why did the chicken cross the Möbius strip?

A new idea (I hope)...
« Reply #20 on: April 27, 2006, 09:04:22 PM »
Ok, for what you're saying that's right. I was describing the distance the light has to travel through the "thick air" from a height of H with a seperation between thick and thin at 1/2 H, and apparently you weren't.

A new idea (I hope)...
« Reply #21 on: April 28, 2006, 03:33:08 AM »
ok, now you are claiming that the density of the air as you are closer to the earth is causing the effect? if that were true, then one would see the object fading into the ground. the object does not fade into the ground, it's cut off sharply. this does not support your idea that the denser air is causing some kind of bending of the light.[/code]