The earth is a flat, non spinning realm, not a planet in an infinite universe

Is the Earth Flat or a Globe


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Brushy Bill

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I did, ****.
 

Brushy Bill

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Somebody didn't learn to improvise when they miss their cue. [roll]

 
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Posted by Pilot Lx
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Pilot Lx
· April 12 at 4:05am ·
Today 37,000ft, full moon high above the horizon after sunrising.


For my own curiosity, I wanted to check how much curvature one should see at 37,000 feet above the Earth. So I opened up a drafting program and drew a circle 20.9 units in radius (Where one unit here represents one million feet, as the Earth has a 20.9 million foot radius). See on the left, 4th box from the bottom, that this is a circle of radius 20.9 units.



Deselect the circle and you can see under "View" that this is a view height of 0.065 units - 65,000 feet. At this height, the amount of curvature results in a drop of one pixel from the center to the edge on a viewing area 738 pixels high.



Finally, zoom in to a view height of 0.0416 units - 41,600 feet. The amount of curvature is less than one pixel. The circle displays as a straight line. Which is to be expected. At this zoom, each pixel represents about 56 feet. The viewing width is 0.0750 units, so the distance from the center to edge is 0.0375 units - 37,500 feet, or 7.1 miles. The drop over 7.1 miles at 8" times 7.1 miles squared is 403.5 inches, or 33.6 feet. An amount too small to be seen at 720p resolution, and approximately equal to a single pixel at 1080p.



At a height of 100,000 feet you'd see a drop from center to edge of about 200 feet. Roughly 4 pixels on a 720p video.The Earth is really, really, really damn big. Even the "extreme" height of 100,000 feet, you're the equivalent of 1.5 millimeters above the surface of a 1-foot globe.

Fun exercise.
 

Brushy Bill

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For my own curiosity, I wanted to check how much curvature one should see at 37,000 feet above the Earth. So I opened up a drafting program and drew a circle 20.9 units in radius (Where one unit here represents one million feet, as the Earth has a 20.9 million foot radius). See on the left, 4th box from the bottom, that this is a circle of radius 20.9 units.



Deselect the circle and you can see under "View" that this is a view height of 0.065 units - 65,000 feet. At this height, the amount of curvature results in a drop of one pixel from the center to the edge on a viewing area 738 pixels high.



Finally, zoom in to a view height of 0.0416 units - 41,600 feet. The amount of curvature is less than one pixel. The circle displays as a straight line. Which is to be expected. At this zoom, each pixel represents about 56 feet. The viewing width is 0.0750 units, so the distance from the center to edge is 0.0375 units - 37,500 feet, or 7.1 miles. The drop over 7.1 miles at 8" times 7.1 miles squared is 403.5 inches, or 33.6 feet. An amount too small to be seen at 720p resolution, and approximately equal to a single pixel at 1080p.



At a height of 100,000 feet you'd see a drop from center to edge of about 200 feet. Roughly 4 pixels on a 720p video.The Earth is really, really, really damn big. Even the "extreme" height of 100,000 feet, you're the equivalent of 1.5 millimeters above the surface of a 1-foot globe.

Fun exercise.

OK. Now how do you see Chicago all the way across the lake from Michigan?
 
Mar 13, 2004
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Not getting into games with a troll (or moron) again, just wanted to go through the exercise and show the demonstration to any curious onlookers.
 

Brushy Bill

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Not getting into games with a troll (or moron) again, just wanted to go through the exercise and show the demonstration to any curious onlookers.

Thanks for the actual input. Still doesn't prove your point but at least you tried.
 

Tinker Dan

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Thanks for the actual input. Still doesn't prove your point but at least you tried.
Willy used a point awhile back in the political thread. Paraphrasing here.... "they do not want to like Trump...."

I think that thought applies here. You do not want to believe the earth is round. So to your original question. No. No one can prove to you the earth is round. You do not want to believe it.
 
Feb 24, 2009
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For my own curiosity, I wanted to check how much curvature one should see at 37,000 feet above the Earth. So I opened up a drafting program and drew a circle 20.9 units in radius (Where one unit here represents one million feet, as the Earth has a 20.9 million foot radius). See on the left, 4th box from the bottom, that this is a circle of radius 20.9 units.



Deselect the circle and you can see under "View" that this is a view height of 0.065 units - 65,000 feet. At this height, the amount of curvature results in a drop of one pixel from the center to the edge on a viewing area 738 pixels high.



Finally, zoom in to a view height of 0.0416 units - 41,600 feet. The amount of curvature is less than one pixel. The circle displays as a straight line. Which is to be expected. At this zoom, each pixel represents about 56 feet. The viewing width is 0.0750 units, so the distance from the center to edge is 0.0375 units - 37,500 feet, or 7.1 miles. The drop over 7.1 miles at 8" times 7.1 miles squared is 403.5 inches, or 33.6 feet. An amount too small to be seen at 720p resolution, and approximately equal to a single pixel at 1080p.


At a height of 100,000 feet you'd see a drop from center to edge of about 200 feet. Roughly 4 pixels on a 720p video.The Earth is really, really, really damn big. Even the "extreme" height of 100,000 feet, you're the equivalent of 1.5 millimeters above the surface of a 1-foot globe.

Fun exercise.
I'll take this one, Billy.



 

LadyCaytIL

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There is far more evidence this world is a computer simulation. Check into it..... There are a lot of scientists who give this theory legs. Even the ones who say its a 50/50 chance of it being true.......... that still speaks volumes.

the flat earth theory by itself though is absolutely stupid...... if its aliens or actually god behind it..... why not put us on a sphere planet like we see in space...... and then the lie wouldnt be needed??
 

Brushy Bill

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The viewing width is 0.0750 units, so the distance from the center to edge is 0.0375 units - 37,500 feet, or 7.1 miles.

Changes is elevation will also create changes in the visible width of your view (the higher you go the wider your view) but it doesn't seem that you accounted for this variable. At elevations of 37,000 ft up to 100,000 ft your viewing width from edge to edge or center to edge would be much greater than 14.2 e to e, or 7.1 c to e and therefor the visible "curvature" would be greater than 200 ft as it would have to be calculated over the entire visible distance, edge to edge or center to edge. Even at 37,000 feet your view width limitation of 14.2 miles doesn't hold up.
 
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Changes is elevation will also create changes in the visible width of your view (the higher you go the wider your view) but it doesn't seem that you accounted for this variable. At elevations of 37,000 ft up to 100,000 ft your viewing width from edge to edge or center to edge would be much greater than 14.2 e to e, or 7.1 c to e and therefor the visible "curvature" would be greater than 200 ft as it would have to be calculated over the entire visible distance, edge to edge or center to edge. Even at 37,000 feet your view width limitation of 14.2 miles doesn't hold up.

My viewing area was roughly the same shape as the frame of that video, so it was accounted for.
 

Brushy Bill

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OK. So there's bushy bill and bRushy bill. What are your other 4 CatPaws usernames?
I voted once and haven't posted under the other account since Rivals fixed the issue. Quit trying to play off like I've been duplicitous in any way shape or form.
 

Brushy Bill

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I voted once and haven't posted under the other account since Rivals fixed the issue. Quit trying to play off like I've been duplicitous in any way shape or form.

No it wasn't, first off he pans the camera the camera in at least a 180 degree arc. Also the initial reference view is unknown but I can assure you it is more than 14.2 miles in width.
 

BigBlueFanGA

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It's only obvious to those not looking for the truth.
So, I'm not going to waste my time trying to go through 43 pages of this. What are your beliefs exactly?

Do you believe the earth is flat?
Do you believe that the edge is a wall of ice and earth is basically the shape of a cup coaster?
Do you believe in "dark energy"?
What do you believe is on the backside of earth?
Do you believe the moon and sun are both small objects in an elliptical orbit above earth?

Thats a decent start I think. Once you answer I'll understand your position better.
 

Brushy Bill

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Do you believe the earth is flat? Yep
Do you believe that the edge is a wall of ice and earth is basically the shape of a cup coaster? flat realm circular ice wall. Antarctica as a continent doesn't exist.
Do you believe in "dark energy"? No outer space, no need for dark energy
What do you believe is on the backside of earth? who TF knows, no one's ever been there.
Do you believe the moon and sun are both small objects in an elliptical orbit above earth? Yep

Now, make it interesting by asking a unique question or a question that I'm interested in otherwise I'll simply refer you to my already posted body of work.
 

BigBlueFanGA

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Do you believe the earth is flat? Yep
Do you believe that the edge is a wall of ice and earth is basically the shape of a cup coaster? flat realm circular ice wall. Antarctica as a continent doesn't exist.
Do you believe in "dark energy"? No outer space, no need for dark energy
What do you believe is on the backside of earth? who TF knows, no one's ever been there.
Do you believe the moon and sun are both small objects in an elliptical orbit above earth? Yep

Now, make it interesting by asking a unique question or a question that I'm interested in otherwise I'll simply refer you to my already posted body of work.
Interesting. So how high is this ice wall?
 
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