[Opeo]

Does that picture have an elevation and a location that it was shot at?
I am not computer smart enough to access it if it had. maybe you are. I am old.

If we can find those then maybe we can figure out how long that horizon is.
At 8 in drop per mile then 10 miles of horizon should have a 6.66ft drop. (number of the beast)

I would think that the horizon in the photo is at least 100 miles long.
If it is then that is a 66.6ft difference.

We can estimate it pretty easily. The horizon is 950 pixels long, with 6 pixels of drop on either side. That picture appears to be from low Earth orbit, which means the distance to the horizon is about 1,400 miles.

Breaking out the simple geometry, you can find the radius of an arc based on height and width via:

r = H / 2 + W^2 / 8H

Plugging in my pixel-counting values:

r = 6p / 2 + (950p)^2 / 8(6p) = 18,805 p

This lets us turn pixels into distance since we have the radius in both values:

1,400 mi / 18,805 p = length / 950 p

Length = 70 miles

We should then see about 815 ft of curvature on either side of the picture away from the center given that both are 35 miles away from the middle. 815ft / (70mi*5280ft/mi) = 0.2% of the total distance.

That's why it's barely visible.

[retlaw]

I think you are right and the arc will not be detected at that high.

Lets go to the last picture from NASA. I measured roughly maybe you can measure more accurate with pixels.

Measure the atmosphere then you get the 60 mile perspective.
With that perspective the minimum horizon length shows to be 1320 miles.
I simply divide the atmosphere longest distance as 60 miles into the horizon length to get 1320 miles.
Not 100% but close.

Then measure the arc distance the earth has on the photo by placing a flat edge from horizon to horizon.
The distance you get of the arc is roughly 4 times the distance of the atmosphere.
60 miles divided by 4 gives you 15 miles.
So the arc in a distance of 1320 mile long horizon is 15 miles as the map shows.
But at 8 in per mile at 1320 miles give you an arc of 880ft. That is .1666 of a mile.
That's to big a difference.

This NASA picture has made the arc to big.

[Opeo]

I think you are right and the arc will not be detected at that high.

Lets go to the last picture from NASA. I measured roughly maybe you can measure more accurate with pixels.

Measure the atmosphere then you get the 60 mile perspective.
With that perspective the minimum horizon length shows to be 1320 miles.
I simply divide the atmosphere longest distance as 60 miles into the horizon length to get 1320 miles.
Not 100% but close.

Then measure the arc distance the earth has on the photo by placing a flat edge from horizon to horizon.
The distance you get of the arc is roughly 4 times the distance of the atmosphere.
60 miles divided by 4 gives you 15 miles.
So the arc in a distance of 1320 mile long horizon is 15 miles as the map shows.
But at 8 in per mile at 1320 miles give you an arc of 880ft. That is .1666 of a mile.
That's to big a difference.

This NASA picture has made the arc to big.

Last one I'll do because it's time consuming and I can't imagine it's convincing anyone:


This one isn't high enough resolution for fine pixel counting, but that 2.9% looks about right from what we can see.

Anyway, moving away from that, why do you think the flat Earth movement hasn't been able to provide a single universally agreed upon flat Earth map? All round Earthers have agreed on one single globe for hundreds and hundreds of years. Yet even the most popular FE model has huge discrepancies (see my post here: https://forum.tfes.org/index.php?topic=8995.0)

[TomInAustin]

So when you say orbit do you mean its going round and around the world like the rings of Saturn?

Is that addressed to the room, or to someone in particular?

To who's calling the station orbiting. Its hard to answer when one doesn't know what they refer to as an orbit.

Orbit is a well used and well-known term.  What part don't you get?

https://www.google.com/search?q=orbit&oq=orbit&aqs=chrome..0j69i57j69i65j69i61j0l2.3023j0j7&sourceid=chrome&ie=UTF-8

[retlaw]

"460 miles leads to 26.7 miles of curve away from the center."

I am confused at that number.

460 miles @ 8 inch drop per mile works out to be 306.66 feet drop over 460 miles.

[Opeo]

"460 miles leads to 26.7 miles of curve away from the center."

I am confused at that number.

460 miles @ 8 inch drop per mile works out to be 306.66 feet drop over 460 miles.

A flat 8" drop per mile would mean the horizon is a triangle. The high point at the center of your vision and a slight constant negative drop away. Instead you need to pull out a pen and do some simple geometry:



Here the circumference of the Earth is 24,875 mi. Thus:

a = (360° / 24,875) * d
h = 3,959 * (1 - cos a)

Then to find h, just plug and chug.

[douglips]

"460 miles leads to 26.7 miles of curve away from the center."

I am confused at that number.

460 miles @ 8 inch drop per mile works out to be 306.66 feet drop over 460 miles.

First, the rule of thumb is 8 inches per miles SQUARED, not just 8 inches per mile. Second, it's just a rule of thumb approximation because the real formula would involve trigonometry.

[wRadion]

I just watched the ISS do a orbit around earth on http://www.n2yo.com/

I printed out a FE map and then marked the ISS orbit on a FE.

If you look at the orbit of the ISS on an ball earth you will see it does waves. Why and how it does it I don't know.

EDIT : Oh, I just saw that this video also explain why Greenland and Africa looks similar in size in some maps (if this wasn't explained enough already).

[retlaw]

Put the satellites path on a FE map and they to go around in big circles.

[pablozablo]

Put the satellites path on a FE map and they to go around in big circles.

Cool. Can you show me the FE map? The one on the wiki seems in dispute, and I read something from the FE Tom guy along the lines of "we don't have the technology to produce accurate FE maps". Even better if you can show the satellite paths on it. Thanks.

[Frocious]

Put the satellites path on a FE map and they to go around in big circles.

First of all, I think it's awesome the amount of work and thought you are putting into this. It seems like you actually want to find the truth!

There is a problem with your methods, however. Flat Earthers do not have a map of the flat earth -- they admit this freely.

The "FE map" you are using is a projection of a globe onto a 2D surface.

[wRadion]

Put the satellites path on a FE map and they to go around in big circles.

They are circles in round earth too.
Did you even watch the video I posted?

[Opeo]

Put the satellites path on a FE map and they to go around in big circles.

But here's the problem. There's no aspect of the FE theory that says they should have to go in circles. For heliocentricism, ellipses are the only stable orbits, so without fail everything needs to be orbiting in an ellipse unless it actively has some external force working on it. It's the mathematical byproduct of an object with some large fixed momentum constantly accelerating towards the center of a massive body. That explains why all satellites move like that.

FE on the other hand has no such rules. I've seen multiple explanations for visible satellites: drones flown by NASA, holograms, weather balloons, et cetera, and none of them need a circular orbit. It would be way better for you if you discovered their orbits weren't elliptical, since there's no reason for them to be in your model.

[wRadion]

I just noticed that that doesn't even make sense, because I thought FE reject the idea of the existence of satellite (or grounded satellite whatever).
So why are you arguing with us about that?

[retlaw]

All satellites use Polaris as there center.
They do not use earth.

Everything obits around Polaris.
Everything. Every star, moon, planet, web site, turd, liberal, etc.

Its easy to make something obit when everything orbits. Just put it up there and it will obit.

Why do people comment that FE don't believe in satellites?
Some do and some don't.

Hard to doubt them when you can see them but hey, 2 pac still plays concerts right?

If there is a dome then what else is a satellite to do?
How many satellites drifted off course into outer space?
Not one ever.

Why when the RE moves at ridiculous speeds is there no drag on the dark side of earth?
We learn that the earth is moving about our sun in a very nearly circular orbit. It covers this route at a speed of nearly 30 kilometers per second, or 67,000 miles per hour.

A satellite is at its outer most limits with just enough energy to keep orbit for decades around earth but in that sensitive environment has no impact from being in front of an object that traveling at thousands of a miles per hours compared to being at the tail of the object traveling thousands of miles per hours.

That does not happen in real life. All you have to do to prove it is ride in a roller coaster in the front seat and then do the same ride in the back set. We have the same gravity with different g force. Why doesn't that happen on a RE with satellites?

I believe a true FE map can be draw using the satellites imagines they give.
And yes I don't think there is a true flat earth map or any for that matter. Well there is one or two but its in the bottom of the Vatican library. Thank you book burning days. Vatican also has the best telescope on the planet yet they give zero info out.

If one took the time and marked the orbital path of a satellite and when it crosses over a land mass on the RE model and then transferred the info to a FE map and rewrote the FE boundaries as a orbital path on a FE map then I think we would have an accurate map.

FE or RE I don't care, I just want the truth.

[Curious Squirrel]

Your very first point is wrong according to most FE ideas. Everything does not orbit Polaris. Planets orbit around the sun. To claim they don't is rather easily shown to have holes with retrograde motion. As well, stars in the southern hemisphere appear to circle/orbit the southern Cross, not Polaris.

[wRadion]

A satellite is at its outer most limits with just enough energy to keep orbit for decades around earth but in that sensitive environment has no impact from being in front of an object that traveling at thousands of a miles per hours compared to being at the tail of the object traveling thousands of miles per hours.

That does not happen in real life. All you have to do to prove it is ride in a roller coaster in the front seat and then do the same ride in the back set. We have the same gravity with different g force. Why doesn't that happen on a RE with satellites?

I don't know what you mean.
The satellites are orbiting a planet, so it going the same speed as the planet.

A very simple observation would be to take the train, stand up, and just notice that you are not ejected from the train.
Another one would be the hula hoop. If you start hooping and then move, the ring just keep spinning the same way, without being influenced by your movement.
Another one would be the "spill not". I'll just let the video talk for me on this one (move to around 3:00 for the actual demo):


It's called momentum and it has been proved to be a reliable physic concept.

The same happens for satellite in orbit.

[Dr David Thork]

Why would you market that thing to people at home?

Surely you'd manufacture larger ones and sell them to pubs, so people can carry 8 pints to their table without spilling them?

[wRadion]

It more like a science gadget than anything else. But this is absolutely not the point of my post.

[nickrulercreator]

All satellites use Polaris as there center.
They do not use earth.

Everything obits around Polaris.
Everything. Every star, moon, planet, web site, turd, liberal, etc.

But Polaris isn't even at true North. It's about .75 degrees off true North. To say everything orbits Polaris would be crazy because Polaris isn't motionless in the sky relative to observers, it moves slightly.