Telescope Question

  • 52 Replies
  • 13761 Views
?

Tоm Bishоp

  • 72
  • Petitio Principii Pirate
Telescope Question
« on: November 25, 2008, 02:21:09 PM »
According to Tom Bishop (:P), a powerful telescope can restore a ship that appears to sink over the horizon. Without a telescope, I can see a distance of about 3 miles to the horizon. So by that (Tom's) logic, if I picked up one of these for Christmas, surely the 750x magnification should allow me to see Cuba from Florida (about 90 miles according to Tom Bishop) on a clear day. But according to friends in Florida that I've asked (including one who told me she owns a 600x reflecting telescope), you can see neither Cuba nor Andros Island from Florida on clear day, with or without a telescope. I also asked around the internet, and I've consistently gotten this same answer (yet to see a single person who disagrees). What's the excuse this time, muddy air that 'bends the light'? The atmosphere is over 22 miles thick, yet most telescopes have no problem seeing through it without objects 'disappearing behind the focal point' (as Tom Bishop says). So what's the explanation for this?
« Last Edit: November 25, 2008, 02:35:28 PM by TomFool »
the earth is flat, hence we can see distant lighthouses, hence the earth is flat

*

Tom Bishop

  • Flat Earth Believer
  • 17538
Re: Telescope Question
« Reply #1 on: November 25, 2008, 02:25:38 PM »
Quote
According to Tom Bishop (:P), a powerful telescope can restore a ship that appears to sink over the horizon. Without a telescope, I can see a distance of about 3 miles to the horizon. So by that (Tom's) logic, if I picked up one of these for Christmas, surely the 750x magnification should allow me to see Cuba from Florida (only about 25 miles) on a clear day.

Wrong. From Key West to Cuba is about 90 miles.

Quote
But according to friends in Florida that I've asked (including one who told me she owns a 600x reflecting telescope), you can see neither Cuba nor Andros Island from Florida on clear day, with or without a telescope. I also asked around the internet, and I've consistently gotten this same answer (yet to see a single person who disagrees). What's the excuse this time, muddy air that 'bends the light'? The atmosphere is over 22 miles thick, yet most telescopes have no problem seeing through it without objects 'disappearing behind the focal point' (as Tom Bishop says). So what's the explanation for this?

Firstly, what makes you think it's possible to even see a low laying island from a distance of 90 miles? It's hard to see a large object from even a mile away.

Secondly, the atmosphere is not perfectly transparent, so you would have to factor whether the amount of atmosphere between you and your target is optimal for viewing.
« Last Edit: November 25, 2008, 02:28:27 PM by Tom Bishop »

?

Tоm Bishоp

  • 72
  • Petitio Principii Pirate
Re: Telescope Question
« Reply #2 on: November 25, 2008, 02:35:04 PM »
Quote
According to Tom Bishop (:P), a powerful telescope can restore a ship that appears to sink over the horizon. Without a telescope, I can see a distance of about 3 miles to the horizon. So by that (Tom's) logic, if I picked up one of these for Christmas, surely the 750x magnification should allow me to see Cuba from Florida (only about 25 miles) on a clear day.

Wrong. From Key West to Cuba is about 90 miles.
The telescope I linked to can easily see the surface of other planets. Why should 90 miles be a problem?

Quote
But according to friends in Florida that I've asked (including one who told me she owns a 600x reflecting telescope), you can see neither Cuba nor Andros Island from Florida on clear day, with or without a telescope. I also asked around the internet, and I've consistently gotten this same answer (yet to see a single person who disagrees). What's the excuse this time, muddy air that 'bends the light'? The atmosphere is over 22 miles thick, yet most telescopes have no problem seeing through it without objects 'disappearing behind the focal point' (as Tom Bishop says). So what's the explanation for this?

Firstly, what makes you think it's possible to even see a low laying island from a distance of 90 miles? It's hard to see a large object from even a mile away.
No it's not, I can see plenty of low-lying hotels 2 miles down the Yucatan Peninsula from a beach in Cancun, Mexico without a telescope.

Secondly, the atmosphere is not perfectly transparent, so you would have to factor whether the amount of atmosphere between you and your target is optimal for viewing.
I've already addressed this point.
the earth is flat, hence we can see distant lighthouses, hence the earth is flat

*

Tom Bishop

  • Flat Earth Believer
  • 17538
Re: Telescope Question
« Reply #3 on: November 25, 2008, 03:57:31 PM »
Quote
The telescope I linked to can easily see the surface of other planets. Why should 90 miles be a problem?

When you look up at celestial bodies you are looking through less atmosphere than when you are looking horizontally across the surface of the earth.

Quote
No it's not, I can see plenty of low-lying hotels 2 miles down the Yucatan Peninsula from a beach in Cancun, Mexico without a telescope.

Right, now how tiny would that hotel be if it were 90 miles away?

?

Tоm Bishоp

  • 72
  • Petitio Principii Pirate
Re: Telescope Question
« Reply #4 on: November 25, 2008, 04:06:43 PM »
When you look up at celestial bodies you are looking through less atmosphere than when you are looking horizontally across the surface of the earth.
So explain why this doesn't work even at closer distances, say 20 miles of atmosphere (vs. 22.7-mile-thick vertically).

Right, now how tiny would that hotel be if it were 90 miles away?
Big enough to be seen at 750x magnification.
the earth is flat, hence we can see distant lighthouses, hence the earth is flat

?

Tоm Bishоp

  • 72
  • Petitio Principii Pirate
Re: Telescope Question
« Reply #5 on: November 25, 2008, 09:21:17 PM »
The vertical vs. horizontal atmospheric density argument doesn't work either. 50% of atmospheric mass is contained in the lower 5.6 km (3.48 miles) of the troposphere (what we see through horizontally with a telescope in this argument). That means that since a telescope can see fine through the 22.7 miles of vertical atmosphere, it should at LEAST be able to see through 6.96 miles of the troposphere. Yet, you cannot see any point on the ocean past (1.17√a)+(1.17√b) nautical miles [where a is your height b is the height of the object you are looking at, in feet] with or without a telescope. If the earth was flat, height should have no effect on your ability to see something within those 6.96 miles. A hypothetical experiment: assuming you were 5 feet 6 inches tall looking at an object 5 feet high (a Rowbotham experiment number): (1.17√5.5)+(1.17√5)=5.36 nautical miles=6.17 miles. That means that on a clear day, at LEAST 0.79 miles beyond the object should be clearly visible with a telescope, which you will find does not hold true in the real world. Before you attack my 'hypothetical experiment' with 'no proof', here's a real example: http://theflatearthsociety.org/forum/index.php?topic=24914.msg552253#msg552253 Rowbotham also fails to explain why the bottom of objects disappear first as shown in the link (no 'similar' colors there, it's pretty obvious that the lower half of the platform is obscured by the water).

Edit: And he's gone!
« Last Edit: November 25, 2008, 09:37:20 PM by Tоm Bishоp »
the earth is flat, hence we can see distant lighthouses, hence the earth is flat

*

Tom Bishop

  • Flat Earth Believer
  • 17538
Re: Telescope Question
« Reply #6 on: November 26, 2008, 11:33:48 AM »
Quote
So explain why this doesn't work even at closer distances, say 20 miles of atmosphere (vs. 22.7-mile-thick vertically).

It does work. If you've ever seen a body more than five or so miles away you would know that it's suddenly becomes a shade of blue, shaded by the thickness of the atmosphere.

Just look at a few distant mountains sometime:

http://www.freenaturepictures.com/assets/images/lores/distantmountain1.jpg
http://www.kauaivacationrentals.com/propimages/hbv19_lan2.jpg
http://lh4.ggpht.com/_veCoV_WjXBM/SEt4vBRkXpI/AAAAAAAABZg/tf7fHPU0WgI/Processed-IMG_0183.JPG

Notice how all of the mountains in the background are shaded blue. The farther the body is the more obscured by the atmosphere it becomes.

Quote
Big enough to be seen at 750x magnification.

Proof?
« Last Edit: November 26, 2008, 11:56:37 AM by Tom Bishop »

*

Tom Bishop

  • Flat Earth Believer
  • 17538
Re: Telescope Question
« Reply #7 on: November 26, 2008, 11:47:36 AM »
The vertical vs. horizontal atmospheric density argument doesn't work either. 50% of atmospheric mass is contained in the lower 5.6 km (3.48 miles) of the troposphere (what we see through horizontally with a telescope in this argument). That means that since a telescope can see fine through the 22.7 miles of vertical atmosphere, it should at LEAST be able to see through 6.96 miles of the troposphere. Yet, you cannot see any point on the ocean past (1.17√a)+(1.17√b) nautical miles [where a is your height b is the height of the object you are looking at, in feet] with or without a telescope. If the earth was flat, height should have no effect on your ability to see something within those 6.96 miles. A hypothetical experiment: assuming you were 5 feet 6 inches tall looking at an object 5 feet high (a Rowbotham experiment number): (1.17√5.5)+(1.17√5)=5.36 nautical miles=6.17 miles. That means that on a clear day, at LEAST 0.79 miles beyond the object should be clearly visible with a telescope, which you will find does not hold true in the real world. Before you attack my 'hypothetical experiment' with 'no proof', here's a real example: http://theflatearthsociety.org/forum/index.php?topic=24914.msg552253#msg552253 Rowbotham also fails to explain why the bottom of objects disappear first as shown in the link (no 'similar' colors there, it's pretty obvious that the lower half of the platform is obscured by the water).

Edit: And he's gone!

The vertical atmosphere is a gradient which is thinnest at the top and thickest at the bottom. It's not a constant medium as you are supposing. Even the lower 5.6 km of the atmosphere is a gradient of thickness. There's quite a difference when looking up at the sky and across the surface of the earth.

Furthermore, it's possible for a sufficiently bright light to shine through the density of the atmosphere, which is why on a foggy day it's possible to see the light of a streetlamp in the distance, but not its pole, the buildings, trees, or scenery around it.

The celestial bodies above our heads are sufficiently bright enough to shine through the vertical atmosphere. Otherwise they would not be seen.
« Last Edit: November 26, 2008, 11:55:41 AM by Tom Bishop »

Re: Telescope Question
« Reply #8 on: November 26, 2008, 11:48:55 AM »
Quote
So explain why this doesn't work even at closer distances, say 20 miles of atmosphere (vs. 22.7-mile-thick vertically).

It does work. If you've ever seen a body more than five miles miles away you would know that it's suddenly becomes a shade of blue, shaded by the thickness of the atmosphere.

Just look at a few distant mountains sometime:

http://www.freenaturepictures.com/assets/images/lores/distantmountain1.jpg
http://www.kauaivacationrentals.com/propimages/hbv19_lan2.jpg
http://image56.webshots.com/156/8/7/26/2694807260058778447elkKlO_fs.jpg

Notice how all of the mountains in the background are shaded blue. The farther the body the more obscured by the atmosphere it becomes.

Quote
Big enough to be seen at 750x magnification.

Proof?

How about some proof those photos weren't edited in Photoshop.  I call conspiracy unless you can prove you took those pictures yourself or provide examples of photos you yourself have taken.
The Earth is Round.

?

Tоm Bishоp

  • 72
  • Petitio Principii Pirate
Re: Telescope Question
« Reply #9 on: November 26, 2008, 12:05:10 PM »
Quote
So explain why this doesn't work even at closer distances, say 20 miles of atmosphere (vs. 22.7-mile-thick vertically).

It does work. If you've ever seen a body more than ten or so miles away you would know that it's suddenly becomes a shade of blue, shaded by the thickness of the atmosphere.

Just look at a few distant mountains sometime:

http://www.freenaturepictures.com/assets/images/lores/distantmountain1.jpg
http://www.kauaivacationrentals.com/propimages/hbv19_lan2.jpg
http://image56.webshots.com/156/8/7/26/2694807260058778447elkKlO_fs.jpg

Notice how all of the mountains in the background are shaded blue. The farther the body the more obscured by the atmosphere it becomes.
So why can the surface of other planets be seen clearly through a backyard reflecting telescope with a light filter?

Quote
Big enough to be seen at 750x magnification.

Proof?
You can't prove it, because the curvature of the earth does not allow you to see an object 90 miles away, unless the total elevation of you and the object equal almost 10,000 feet (elevation of the object being at the highest point on the object). If you are going to ask for proof again, I really can't help you unless I can convince my friend to send me pics from Florida. I'm sure that a test will show that my statements hold true. If I knew enough optics, I could technically figure out the size of something at a magnification and distance, I'm sure.


The vertical vs. horizontal atmospheric density argument doesn't work either. 50% of atmospheric mass is contained in the lower 5.6 km (3.48 miles) of the troposphere (what we see through horizontally with a telescope in this argument). That means that since a telescope can see fine through the 22.7 miles of vertical atmosphere, it should at LEAST be able to see through 6.96 miles of the troposphere. Yet, you cannot see any point on the ocean past (1.17√a)+(1.17√b) nautical miles [where a is your height b is the height of the object you are looking at, in feet] with or without a telescope. If the earth was flat, height should have no effect on your ability to see something within those 6.96 miles. A hypothetical experiment: assuming you were 5 feet 6 inches tall looking at an object 5 feet high (a Rowbotham experiment number): (1.17√5.5)+(1.17√5)=5.36 nautical miles=6.17 miles. That means that on a clear day, at LEAST 0.79 miles beyond the object should be clearly visible with a telescope, which you will find does not hold true in the real world. Before you attack my 'hypothetical experiment' with 'no proof', here's a real example: http://theflatearthsociety.org/forum/index.php?topic=24914.msg552253#msg552253 Rowbotham also fails to explain why the bottom of objects disappear first as shown in the link (no 'similar' colors there, it's pretty obvious that the lower half of the platform is obscured by the water).

Edit: And he's gone!

It's possible to see bodies through the horizontal atmosphere which are at least 40 miles away. I'm not sure where you got your figure of 22.7 miles from.
Only if the total elevation of you and the object are over 1000 feet (elevation of the object being at the highest point on the object). Once again, I can't prove this without a telescope of course, unless I can get someone with a telescope to send me pics. I'm sure it will hold true if you try.

Also, the vertical atmosphere is a gradient which is thinnest at the top and thickest at the bottom. It's not a constant medium as you are supposing. Even the lower 5.6 km of the atmosphere is a gradient of thickness. There's quite a difference when looking up at the sky and across the surface of the earth.

Furthermore, it's possible for a sufficiently bright light to shine through the density of the atmosphere, which is why on a foggy day it's possible to see the light of a lamp in the distance, but not its pole, the buildings, trees, or scenery around the streetlamp. The celestial bodies above our heads are sufficiently bright to shine through the vertical atmosphere above our heads. Otherwise they would not be seen.
No. The 'thickness' excuse still does not hold, because looking through the entire atmosphere vertically does not explain why you cannot see past 3 miles to a point on the horizon if you are 5 feet 6 inches (notice I said point, a solid object will increase this distance at a decreasing rate based on its height). The gradient you are talking about is nowhere steep enough for this to occur:

http://en.wikipedia.org/wiki/Image:Atmosphere_model.png

The total vertical mass mathematically must be at least slightly greater than the horizontal distance you can see to the horizon. The mass percentage of the atmosphere at ground level must be 50% or greater for your theory to hold, which it isn't, because the 50% must be spread throughout the first 5.6km. Your light explanation also does not hold, because I'm talking about a clear day. Fog is water vapor, the atmosphere is not always foggy in all places. The surface of Mars and other planets can be seen clearly with a light filter through backyard reflecting telescopes, but the light filter will not remove fog, so this does not explain why objects disappear at less than 6.96 miles on the horizon, through a telescope, on a clear day. Here's pictures of Mars through a simple backyard telescope:

http://img152.imageshack.us/img152/9800/owens1xr6.jpg
http://img211.imageshack.us/img211/1111/grafton1im4.jpg

Why hasn't it vanished or become even partially obscured through the atmosphere?
« Last Edit: November 26, 2008, 01:03:27 PM by Tоm Bishоp »
the earth is flat, hence we can see distant lighthouses, hence the earth is flat

*

Moon squirter

  • 1405
  • Ding dong!
Re: Telescope Question
« Reply #10 on: November 27, 2008, 02:24:00 AM »
Tom,

The picture below of the CN Tower in Toronto clearly shows how the bottom half is not visible (I've superimposed a scaled picture of the real tower).   Can you honestly say the tower and buildings will be restored with a "good telescope" ?

I haven't performed it and I've never claimed to. I've have trouble being in two places at the same time.

*

Tom Bishop

  • Flat Earth Believer
  • 17538
Re: Telescope Question
« Reply #11 on: November 27, 2008, 02:29:00 AM »
Tom,

The picture below of the CN Tower in Toronto clearly shows how the bottom half is not visible (I've superimposed a scaled picture of the real tower).   Can you honestly say the tower and buildings will be restored with a "good telescope" ?



It's a perspective effect. Read Earth Not a Globe.

Re: Telescope Question
« Reply #12 on: November 27, 2008, 02:46:46 AM »
Tom,

The picture below of the CN Tower in Toronto clearly shows how the bottom half is not visible (I've superimposed a scaled picture of the real tower).   Can you honestly say the tower and buildings will be restored with a "good telescope" ?



It's a perspective effect. Read Earth Not a Globe.

How do you cope in everyday life?? Surely your willingness to live in this fantasy land is having serious affects on your health and personal relationships? The things you spout and denial of everything else are akin to religious fundamentalism.

*

Moon squirter

  • 1405
  • Ding dong!
Re: Telescope Question
« Reply #13 on: November 27, 2008, 02:59:08 AM »
Tom,

The picture below of the CN Tower in Toronto clearly shows how the bottom half is not visible (I've superimposed a scaled picture of the real tower).   Can you honestly say the tower and buildings will be restored with a "good telescope" ?



It's a perspective effect. Read Earth Not a Globe.

I've read ENAG.  It's wrong about perspective:

1.  Rowbotham gets geometric "vanishing points" confused with "limits of resolution". 
2.  He tries to draw perspective lines from the viewpoint of a 3rd person
     (perspective is a first-person only phenomenon).
3.  Objects don't just "vanish" when behond resolution.  For example brighter objects are still detectable
     at large distances, even if the object's shape cannot be determined.

Are you suggesting the bottom 1/3 of the tower is just "not resolved", when it quite obviously intersects the horizon?   I really don't think you can defend the indefensible.
« Last Edit: November 27, 2008, 06:43:10 AM by Moon squirter »
I haven't performed it and I've never claimed to. I've have trouble being in two places at the same time.

*

Tom Bishop

  • Flat Earth Believer
  • 17538
Re: Telescope Question
« Reply #14 on: November 27, 2008, 10:55:10 AM »
Quote
I've read ENAG.  It's wrong about perspective:

1.  Rowbotham gets geometric "vanishing points" confused with "limits of resolution". 
2.  He tries to draw perspective lines from the viewpoint of a 3rd person
     (perspective is a first-person only phenomenon).
3.  Objects don't just "vanish" when behond resolution.  For example brighter objects are still detectable
     at large distances, even if the object's shape cannot be determined.

That's not at all what it says. Read Earth Not a Globe a few more times and let us know when you're ready for a real debate.
« Last Edit: November 27, 2008, 12:27:47 PM by Tom Bishop »

*

Perfect Circle

  • 734
  • You are a pirate!
Re: Telescope Question
« Reply #15 on: November 27, 2008, 12:29:49 PM »
Quote
I've read ENAG.  It's wrong about perspective:

1.  Rowbotham gets geometric "vanishing points" confused with "limits of resolution". 
2.  He tries to draw perspective lines from the viewpoint of a 3rd person
     (perspective is a first-person only phenomenon).
3.  Objects don't just "vanish" when behond resolution.  For example brighter objects are still detectable
     at large distances, even if the object's shape cannot be determined.

That's not at all what it says. Read Earth Not a Globe a few more times and let us know when you're ready for a real debate.
I've read the whole thing. Your explanation is trash, because it conflicts with the results of some of his experiments.
Like the sun, the stars are also expanding and contracting their diameter as they spin around the hub every six months.

*

Moon squirter

  • 1405
  • Ding dong!
Re: Telescope Question
« Reply #16 on: November 27, 2008, 01:07:26 PM »
Quote
I've read ENAG.  It's wrong about perspective:

1.  Rowbotham gets geometric "vanishing points" confused with "limits of resolution". 
2.  He tries to draw perspective lines from the viewpoint of a 3rd person
     (perspective is a first-person only phenomenon).
3.  Objects don't just "vanish" when behond resolution.  For example brighter objects are still detectable
     at large distances, even if the object's shape cannot be determined.

That's not at all what it says. Read Earth Not a Globe a few more times and let us know when you're ready for a real debate.

Can you please address each of my points.  Otherwise we will assume you have not understood Robothem's shortcomings.
I haven't performed it and I've never claimed to. I've have trouble being in two places at the same time.

*

Perfect Circle

  • 734
  • You are a pirate!
Re: Telescope Question
« Reply #17 on: November 27, 2008, 01:28:42 PM »
Rowbotham says the round Earth is 25,000 English statute miles in circumference, which I assume (looking at his chart) is what he based his calculations on. The mean circumference of the round Earth is actually closer to 24880.62 English statute miles (significantly)--that's over 100 English statute miles off. Being that his circumference is only two significant digits, his results would have an error margin of +-10000000 feet. Even if he knew the correct 24880.62 English statute miles figure (which I highly doubt, since this was the 19th century), he would have an error margin of +-100 feet. All this assumes that circumference of the round Earth is consistent at all angles and degrees of longitude and latitude, which is not true according to RET. So how did Rowbotham calculate his experimental round Earth numbers to the nearest inch? Even if the results he got were accurate (his experiments have actually been tried and disproved several times by several different people), this would not be enough evidence to throw out the RET with that margin of error. What does the real Tom Bishop have to say about this?
Like the sun, the stars are also expanding and contracting their diameter as they spin around the hub every six months.

?

dim

  • 404
  • More overpowered than Aristotel.
Re: Telescope Question
« Reply #18 on: November 27, 2008, 02:22:46 PM »
He was not sure about 25.000 miles, he said it like this cuz he knew he would not have anymeans to know the exact number, cuz noone is able to reach the beyond point of Ross Ice Shelf, and Ross Ice Shelf could easy be some 100 grand thousands more of ice, or some would say, that it is infinite plate.

And Robowothom wasnt that math fanatic, like astronemers count distances to stars, using all those meaningless error margins +-1000000(dont know how to say better, all those "smart" math things that are used only in the paper, and nobody use them for something really useful). So, he said 25.000 approximatlelly.

Really, do you really believe there is edge somewhere beyond ice wall and not too far away? I think, that ice after ice wall could be just strenghten for 25.000 more English statue miles, and no body wasn't flying that far in ices.



pic: Ross Ice Shelf
« Last Edit: November 27, 2008, 02:24:39 PM by dim »

*

Perfect Circle

  • 734
  • You are a pirate!
Re: Telescope Question
« Reply #19 on: November 27, 2008, 02:27:06 PM »
He was not sure about 25.000 miles, he said it like this cuz he knew he would not have anymeans to know the exact number, cuz noone is able to reach the beyond point of Ross Ice Shelf, and Ross Ice Shelf could easy be some 100 grand thousands more of ice, or some would say, that it is infinite plate.

And Robowothom wasnt that math fanatic, like astronemers count distances to stars, using all those meaningless error margins +-1000000(dont know how to say better, all those "smart" math things that are used only in the paper, and nobody use them for something really useful). So, he said 25.000 approximatlelly.

Really, do you really believe there is edge somewhere beyond ice wall and not too far away? I think, that ice after ice wall could be just strenghten for 25.000 more English statue miles, and no body wasn't flying that far in ices.


His experiments assumed the earth was round, so what does the ice wall have to do with this? All I want to know is how did he know that something should have been '11 feet 8 inches' below the horizon?
Like the sun, the stars are also expanding and contracting their diameter as they spin around the hub every six months.

?

dim

  • 404
  • More overpowered than Aristotel.
Re: Telescope Question
« Reply #20 on: November 27, 2008, 02:40:31 PM »
His experiments assumed the earth was flat.
And i saying just about he couldn't know the exact earth diameter(or what it can be if it is infinite plate). He never told us that.
And about 11 feet 8 inch below the horizon - it must be a very easy thing how to use this, but i am not familiar with this counting method.

*

Perfect Circle

  • 734
  • You are a pirate!
Re: Telescope Question
« Reply #21 on: November 27, 2008, 02:57:55 PM »
His experiments assumed the earth was flat.
And i saying just about he couldn't know the exact earth diameter(or what it can be if it is infinite plate). He never told us that.
And about 11 feet 8 inch below the horizon - it must be a very easy thing how to use this, but i am not familiar with this counting method.
His experiments assumed the earth was round, or he wouldn't be calculating curvature. I'm asking for how he calculated curvature with such vague numbers.
Like the sun, the stars are also expanding and contracting their diameter as they spin around the hub every six months.

*

markjo

  • Content Nazi
  • The Elder Ones
  • 39416
Re: Telescope Question
« Reply #22 on: November 27, 2008, 08:38:08 PM »
Tom,

The picture below of the CN Tower in Toronto clearly shows how the bottom half is not visible (I've superimposed a scaled picture of the real tower).   Can you honestly say the tower and buildings will be restored with a "good telescope" ?



It's a perspective effect. Read Earth Not a Globe.

Tom (or any other FE'er), can you cite a credible reference (i.e. not FE literature) that supports your claim that not being able to see the bottom of the CN Tower in that photo is a perspective effect?  Or is it possible that Rowbotham has a his own unique perspective on the topic of perspective?
Science is what happens when preconception meets verification.
Quote from: Robosteve
Besides, perhaps FET is a conspiracy too.
Quote from: bullhorn
It is just the way it is, you understanding it doesn't concern me.

?

dim

  • 404
  • More overpowered than Aristotel.
Re: Telescope Question
« Reply #23 on: November 28, 2008, 09:53:35 AM »
Guys, if you telling that half of the CN tower is not seen because of the curvature, then we should agree - there is curvature in this perspective. But if there is curvature, then we should be able to see the curvature of the horizon too! Not in this picture, but at least from the plane flying at 10.000m. But how come, I dont see the curvature from the plane, but see it in this CN tower example. Then it just means, that it is not the curvature proof you telling us about. Some other effect happens here, probably the one about Rowbotham wrote how about dissapering ship.
« Last Edit: November 28, 2008, 09:56:02 AM by dim »

*

Perfect Circle

  • 734
  • You are a pirate!
Re: Telescope Question
« Reply #24 on: November 28, 2008, 02:47:34 PM »
Guys, if you telling that half of the CN tower is not seen because of the curvature, then we should agree - there is curvature in this perspective. But if there is curvature, then we should be able to see the curvature of the horizon too! Not in this picture, but at least from the plane flying at 10.000m. But how come, I dont see the curvature from the plane, but see it in this CN tower example. Then it just means, that it is not the curvature proof you telling us about. Some other effect happens here, probably the one about Rowbotham wrote how about dissapering ship.
Rowbotham's disappearing ship experiment is exactly what this disproves--that something obscured by the horizon can not be magically restored with a telescope, as Rowbotham suggests. That, coupled with the fact that light can only appear to bend with a gradient index-of-refraction (which would be measurable and not produce this effect) or with a curved geodesic as a result of gravity bending the space path (and FET does not include gravity), means that 'bendy light' cannot explain this. You clearly underestimate the size of the earth. At 10000 meters, you'd be able to see a distance along the surface of the earth of about 218 miles. In that distance, the surface would only drop 6 miles from a flat horizontal. 6/218=0.0275 mile vertical drop per horizontal mile, which would be an unnoticeable curve on the horizon from 10000 meters, not even counting terrain differences and the fact that at 10000 meters, your view would be obscured be clouds.
« Last Edit: November 28, 2008, 03:41:46 PM by Perfect Circle »
Like the sun, the stars are also expanding and contracting their diameter as they spin around the hub every six months.

?

dim

  • 404
  • More overpowered than Aristotel.
Re: Telescope Question
« Reply #25 on: November 29, 2008, 03:35:33 AM »
Well, if I can't see the curvature from 10.000 meters high, then how com CN tower is obscured by the horizon? What is the distance from picture taker to the CN tower in this picture - more then 218 miles? That's why we see the curvature(=horizon obscuring part of the tower) ?

Rowbotham wrote:
"IF the earth is a globe, and is 25,000 English statute miles in circumference, the surface of all standing water must have a certain degree of convexity--every part must be an arc of a circle. From the summit of any such arc there will exist a curvature or declination of 8 inches in the first statute mile. In the second mile the fall will be 32 inches; in the third mile, 72 inches, or 6 feet" (c) ENAG

So, 218 miles would be 31682 feet and every part of it is an arc of a circle. It should be noticable curavture even for unaided eye. But there is none.

And actually, FE is understandable w/o numbers at all. It is easy. One more time - if we see the curavture in the CN tower example picture(and according to you we see it, cuz part of the tower is "obscured" by the horizon, so called "vertical" curavature", then we should be ABLE to see the curvature of the horizon at least somewhere on the Earth under normal conditions. But how come - we see dissapearing ship("obscured" by the horizon as RET says) and never see the curvature of the horizon?

No numbers - just observations.

*

Perfect Circle

  • 734
  • You are a pirate!
Re: Telescope Question
« Reply #26 on: November 29, 2008, 10:41:10 AM »
Well, if I can't see the curvature from 10.000 meters high, then how com CN tower is obscured by the horizon? What is the distance from picture taker to the CN tower in this picture - more then 218 miles? That's why we see the curvature(=horizon obscuring part of the tower) ?

Rowbotham wrote:
"IF the earth is a globe, and is 25,000 English statute miles in circumference, the surface of all standing water must have a certain degree of convexity--every part must be an arc of a circle. From the summit of any such arc there will exist a curvature or declination of 8 inches in the first statute mile. In the second mile the fall will be 32 inches; in the third mile, 72 inches, or 6 feet" (c) ENAG

So, 218 miles would be 31682 feet and every part of it is an arc of a circle. It should be noticable curavture even for unaided eye. But there is none.

And actually, FE is understandable w/o numbers at all. It is easy. One more time - if we see the curavture in the CN tower example picture(and according to you we see it, cuz part of the tower is "obscured" by the horizon, so called "vertical" curavature", then we should be ABLE to see the curvature of the horizon at least somewhere on the Earth under normal conditions. But how come - we see dissapearing ship("obscured" by the horizon as RET says) and never see the curvature of the horizon?

No numbers - just observations.
You really aren't too bright. 6 miles is not a noticeable dip when your view is 218 miles. However, not only is your view in this picture much less, but the CN Tower is not 12 miles tall. Being that the CN Tower is 1815.39 feet tall, and it is about half obscured in the pictures, the curvature at that distance would mean a dip of only about 900 feet. And like I said, if you know someone who is a fighter pilot, or you have ridden the Concorde, the curvature of the horizon is clearly visible.
Like the sun, the stars are also expanding and contracting their diameter as they spin around the hub every six months.

?

dim

  • 404
  • More overpowered than Aristotel.
Re: Telescope Question
« Reply #27 on: November 29, 2008, 11:04:31 AM »
Well, if I can't see the curvature from 10.000 meters high, then how com CN tower is obscured by the horizon? What is the distance from picture taker to the CN tower in this picture - more then 218 miles? That's why we see the curvature(=horizon obscuring part of the tower) ?

Rowbotham wrote:
"IF the earth is a globe, and is 25,000 English statute miles in circumference, the surface of all standing water must have a certain degree of convexity--every part must be an arc of a circle. From the summit of any such arc there will exist a curvature or declination of 8 inches in the first statute mile. In the second mile the fall will be 32 inches; in the third mile, 72 inches, or 6 feet" (c) ENAG

So, 218 miles would be 31682 feet and every part of it is an arc of a circle. It should be noticable curavture even for unaided eye. But there is none.

And actually, FE is understandable w/o numbers at all. It is easy. One more time - if we see the curavture in the CN tower example picture(and according to you we see it, cuz part of the tower is "obscured" by the horizon, so called "vertical" curavature", then we should be ABLE to see the curvature of the horizon at least somewhere on the Earth under normal conditions. But how come - we see dissapearing ship("obscured" by the horizon as RET says) and never see the curvature of the horizon?

No numbers - just observations.
You really aren't too bright. 6 miles is not a noticeable dip when your view is 218 miles. However, not only is your view in this picture much less, but the CN Tower is not 12 miles tall. Being that the CN Tower is 1815.39 feet tall, and it is about half obscured in the pictures, the curvature at that distance would mean a dip of only about 900 feet. And like I said, if you know someone who is a fighter pilot, or you have ridden the Concorde, the curvature of the horizon is clearly visible.

Well, I dont know someone who is a fighter pilot. I was only at the altitude of 10.000 m and never seen the curvature. I even made pictures(which are much more qualitative than the pictures you are presenting) and there is no any any deviation in the straight line of the horizon. Not a little bit! Does it perspective effect too?

*

Perfect Circle

  • 734
  • You are a pirate!
Re: Telescope Question
« Reply #28 on: November 29, 2008, 11:08:37 AM »
Well, if I can't see the curvature from 10.000 meters high, then how com CN tower is obscured by the horizon? What is the distance from picture taker to the CN tower in this picture - more then 218 miles? That's why we see the curvature(=horizon obscuring part of the tower) ?

Rowbotham wrote:
"IF the earth is a globe, and is 25,000 English statute miles in circumference, the surface of all standing water must have a certain degree of convexity--every part must be an arc of a circle. From the summit of any such arc there will exist a curvature or declination of 8 inches in the first statute mile. In the second mile the fall will be 32 inches; in the third mile, 72 inches, or 6 feet" (c) ENAG

So, 218 miles would be 31682 feet and every part of it is an arc of a circle. It should be noticable curavture even for unaided eye. But there is none.

And actually, FE is understandable w/o numbers at all. It is easy. One more time - if we see the curavture in the CN tower example picture(and according to you we see it, cuz part of the tower is "obscured" by the horizon, so called "vertical" curavature", then we should be ABLE to see the curvature of the horizon at least somewhere on the Earth under normal conditions. But how come - we see dissapearing ship("obscured" by the horizon as RET says) and never see the curvature of the horizon?

No numbers - just observations.
You really aren't too bright. 6 miles is not a noticeable dip when your view is 218 miles. However, not only is your view in this picture much less, but the CN Tower is not 12 miles tall. Being that the CN Tower is 1815.39 feet tall, and it is about half obscured in the pictures, the curvature at that distance would mean a dip of only about 900 feet. And like I said, if you know someone who is a fighter pilot, or you have ridden the Concorde, the curvature of the horizon is clearly visible.

Well, I dont know someone who is a fighter pilot. I was only at the altitude of 10.000 m and never seen the curvature. I even made pictures(which are much more qualitative than the pictures you are presenting) and there is no any any deviation in the straight line of the horizon. Not a little bit! Does it perspective effect too?

In your narrow viewport of a plane window, and at an altitude of only 10000m, you wouldn't see any deviation with a curvature that slight. Unless you were outside the plane as to have a full angle view, or were flying at 18000m, you wouldn't be able to tell. Your pictures are all photoshop anyway.
Like the sun, the stars are also expanding and contracting their diameter as they spin around the hub every six months.

?

dim

  • 404
  • More overpowered than Aristotel.
Re: Telescope Question
« Reply #29 on: November 29, 2008, 11:13:39 AM »
Ahaha! My pictures are all photoshopped! :)) That's a great joke, you know, I am laughing here.

And pilots in the cockpit have wider viewangle, and they never happened to see the curvature too. But unfortunately I dont know them in private :)