That probably makes as much sense as any of your other nonsense.
That probably makes as much sense as any of your other nonsense.
Posted by Pilot Lx
23,926 Views
23,926 Views
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.
Asked and answered
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.
Willy used a point awhile back in the political thread. Paraphrasing here.... "they do not want to like Trump...."Thanks for the actual input. Still doesn't prove your point but at least you tried.
I'll take this one, Billy.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 hope that isn't for me, William.
OK. Now how do you see Chicago all the way across the lake from Michigan?
Actually, it hasn't been answered, and I've been paying VERY close attention.
Thanks for the actual input. Still doesn't prove your point but at least you tried.
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.
6 votes for
66 votes against
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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.
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.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.
Solid contributions from the new guy folks. Maybe he'll put down his crayons and pull the pacifier out of his mouth and join the conversation soon.
Is this guy serious? 43 pages of insisting people prove to him the obvious?
So, I'm not going to waste my time trying to go through 43 pages of this. What are your beliefs exactly?It's only obvious to those not looking for the truth.
Interesting. So how high is this ice wall?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?