An excerpt from the Wiki:"your math is wrong and you'e lying" -Every flat earther who sees this thread. They don't have a way of calculating position. Just another hole in the flat earth theory. It's sad how something supposedly based on "common sense" over science is so misled in so many areas. For example, the Bishop experiment is one of their biggest proofs and it uses incorrect math and positioning.
"Latitude
To locate your latitude on the flat earth, it's important to know the following fact: The degrees of the earth's latitude are based upon the angle of the sun in the sky at noon equinox.
That's why 0° N/S sits on the equator where the sun is directly overhead, and why 90° N/S sits at the poles where the sun is at a right angle to the observer. At 45 North or South from the equator, the sun will sit at an angle 45° in the sky. The angle of the sun past zenith is our latitude.
Knowing that as you recede North or South from the equator at equinox, the sun will descend at a pace of one degree per 69.5 miles, we can even derive our distance from the equator based upon the position of the sun in the sky."
Have any of those who subscribe to the FE theory tested this? I did yesterday. I'm at North 44* 05'. The sun at its high point was 28 degrees elevation (62* from zenith) using the inclinometer on my Brunton compass. Using the spherical earth theory and considering the earth's axis tilt (solar declination) I calculated 44* 12'. This was using the formula that has been used for centuries. That's pretty close to my GPS and map position. The flat earth formula described above put me at 62* North which is clearly incorrect.
What is the explanation?
"your math is wrong and you'e lying" -Every flat earther who sees this thread. They don't have a way of calculating position. Just another hole in the flat earth theory. It's sad how something supposedly based on "common sense" over science is so misled in so many areas. For example, the Bishop experiment is one of their biggest proofs and it uses incorrect math and positioning.
An excerpt from the Wiki:I tried in The Sun's height from the method and distances in "the Wiki". « on: April 15, 2016, 01:58:07 AM » (https://forum.tfes.org/index.php?topic=4887.msg94442#msg94442) to use these distances and angles to calculate the sun's height.
"Latitude
To locate your latitude on the flat earth, it's important to know the following fact: The degrees of the earth's latitude are based upon the angle of the sun in the sky at noon equinox.
That's why 0° N/S sits on the equator where the sun is directly overhead, and why 90° N/S sits at the poles where the sun is at a right angle to the observer. At 45 North or South from the equator, the sun will sit at an angle 45° in the sky. The angle of the sun past zenith is our latitude.
Knowing that as you recede North or South from the equator at equinox, the sun will descend at a pace of one degree per 69.5 miles, we can even derive our distance from the equator based upon the position of the sun in the sky."
Have any of those who subscribe to the FE theory tested this? I did yesterday. I'm at North 44* 05'. The sun at its high point was 28 degrees elevation (62* from zenith) using the inclinometer on my Brunton compass. Using the spherical earth theory and considering the earth's axis tilt (solar declination) I calculated 44* 12'. This was using the formula that has been used for centuries. That's pretty close to my GPS and map position. The flat earth formula described above put me at 62* North which is clearly incorrect.
What is the explanation?
Why are you trying to use unverified ancient geometry/trigonometry as a proof of anything?
The Ancient Greeks did not verify that circles actually exist, and they did not verify that perspective lines actually stretch into infinity as they theorized.
You see, you're assuming that greek trigonometry is valid. You think that it works. But on a flat earth, it doesn't. Geometry is false on the flat earth.Mind you,if you look at the other posts from brainsandgravy you might guess that he was being very tongue-in-cheek in that post.
So, no one can explain the discrepancies? The Wiki article is simply incorrect. The statements do not fit the observations and are measurably inaccurate. Shouldn't it be modified?I have seen nothing better in "the Wiki" and no other explanations seem to be given except for "curved light".
The Flat EarthSee, easy. The whole article is worth reading, though not exactly complimentary to FE reasoning.
Early Ideas About the Shape of the Earth.The Flat Earth, by Donald E. Simanek (https://www.lhup.edu/~dsimanek/flat/flateart.htm)
But suppose you abandon Eratosthenes' two assumptions, and adopt instead the assumption that the earth is flat. Then, triangulation from the same data gives the distance to the sun: 3000 miles[1]! See how a simple change of assumptions can drastically alter the entire cosmos? However, the round earth was more than an arbitrary assumption for Eratosthenes, for him and his contemporaries had other very good reasons for knowing the earth was round. [Textbooks sometimes mislead by suggesting that his experiment was designed to prove the earth was a sphere. It was not, it was only intended to measure the size of the sphere.]
Finally, the angular size of the sun is 0.5°. Using this fact with a distance to the sun of 3000 miles, gives the sun's diameter: 32 miles. It, therefore, appears that the flat-earther's figures are based on sun elevation data at just two particular latitudes, perhaps even Eratosthenes' values. I speculate that flat earthers may have picked these out of some book, and when the calculation was finished, they looked no further. For if they had done the calculation with a variety of latitudes, including large latitude differences, conflicting results would have been obtained.
The diagram at the right shows how this works. The angle that the rays strike the earth's surface is correct, matching the left diagram.
To complete their path from the sun to the earth the rays must curve to strike the earth at the correct (observed) angle. The curvature of the rays for latitude differences of less than 50° hardly shows on the diagram. Of course this result can be obtained in various ways. The curvature could be confined to the region near the earth, even within the atmosphere. The diagram shows circular arcs, but other shapes might be used as well. (https://www.lhup.edu/~dsimanek/flat/refract.gif)
How light refracts near the earth.
Left: conventional physics. Right: flat-earther's physics.
What is the explanation?
The spherical earth formula and explanation does work. The observations and measurements I made resulted in my latitude according my gps, Google Earth, and USGS map location. The formula and explanation given in the TFES wiki yielded an incorrect latitude calculation. Honestly, I can't figure out how the wiki explanation can even be plausible with any flat earth theory I have seen or the one explained in the wiki. For the wiki latitude calculation to work the sun would have to be over the north pole and at a significant distance from the earth. Definitely not 3,000 miles....unless you subscribe to the curving light theory.What is the explanation?
If i understand your post correctly i think that you would be correct in your calculations , I mean it seems perfectly plausible that using basic math and then checking it against itself would result in knowing for sure that you have calculated your position on earth correctly . Are you wrong ? Could you have missed something ? . I can't seem to get any of my calculations to work out my position to be accurate on a flat earth map as compared to a globe . I too must be missing something .
An excerpt from the Wiki:
"Latitude
To locate your latitude on the flat earth, it's important to know the following fact: The degrees of the earth's latitude are based upon the angle of the sun in the sky at noon equinox.
That's why 0° N/S sits on the equator where the sun is directly overhead, and why 90° N/S sits at the poles where the sun is at a right angle to the observer. At 45 North or South from the equator, the sun will sit at an angle 45° in the sky. The angle of the sun past zenith is our latitude.
Knowing that as you recede North or South from the equator at equinox, the sun will descend at a pace of one degree per 69.5 miles, we can even derive our distance from the equator based upon the position of the sun in the sky."
Have any of those who subscribe to the FE theory tested this? I did yesterday. I'm at North 44* 05'. The sun at its high point was 28 degrees elevation (62* from zenith) using the inclinometer on my Brunton compass. Using the spherical earth theory and considering the earth's axis tilt (solar declination) I calculated 44* 12'. This was using the formula that has been used for centuries. That's pretty close to my GPS and map position. The flat earth formula described above put me at 62* North which is clearly incorrect.
What is the explanation?
Are you saying that days of an equinox are the only days that the FE version of the experiment ever work properly? Why is it that the RE version works just fine on any day of the year?An excerpt from the Wiki:
"Latitude
To locate your latitude on the flat earth, it's important to know the following fact: The degrees of the earth's latitude are based upon the angle of the sun in the sky at noon equinox.
That's why 0° N/S sits on the equator where the sun is directly overhead, and why 90° N/S sits at the poles where the sun is at a right angle to the observer. At 45 North or South from the equator, the sun will sit at an angle 45° in the sky. The angle of the sun past zenith is our latitude.
Knowing that as you recede North or South from the equator at equinox, the sun will descend at a pace of one degree per 69.5 miles, we can even derive our distance from the equator based upon the position of the sun in the sky."
Have any of those who subscribe to the FE theory tested this? I did yesterday. I'm at North 44* 05'. The sun at its high point was 28 degrees elevation (62* from zenith) using the inclinometer on my Brunton compass. Using the spherical earth theory and considering the earth's axis tilt (solar declination) I calculated 44* 12'. This was using the formula that has been used for centuries. That's pretty close to my GPS and map position. The flat earth formula described above put me at 62* North which is clearly incorrect.
What is the explanation?
The explanation is that you have a very hard time following instructions. It clearly says to do it at noon equinox.
Are you saying that days of an equinox are the only days that the FE version of the experiment ever work properly? Why is it that the RE version works just fine on any day of the year?
Are you saying that days of an equinox are the only days that the FE version of the experiment ever work properly? Why is it that the RE version works just fine on any day of the year?
Actually, the method described in the article works for RET as well. There are several ways to get your latitude, that is only one of them.
Yes, I didn't read you OP carefully enough. I believe that the posts I made were correct, but I missed the "Knowing that as you recede North or South from the equator at equinox".
Are you saying that days of an equinox are the only days that the FE version of the experiment ever work properly? Why is it that the RE version works just fine on any day of the year?
Actually, the method described in the article works for RET as well. There are several ways to get your latitude, that is only one of them.
You are right, Tom, it does say equinox. I mis-applied the wiki latitude formula to a time that wasn't at equinox. My bad.
Can you please explain how it works on a flat earth? The north pole for example is 6215 statute miles from the equator. If the sun is 3,000 miles above the flat earth right over the equator on equinox then its elevation angle would be 25.8 degrees at the poles. Real observations show the sun to be right on the horizon during the equinox at both poles simultaneously.
How does the flat earth model explain it? How has TFES derived its latitude formula?
To me, the observations are explained by the spherical earth heliocentric model.
(http://wiki.tfes.org/images/c/c2/Altmap.png) Another alternative model descripting Antarctica as a distinct continent. There is still an "ice wall" in this model, but it not Antarctica. Beyond the rays of the sun the waters will naturally freeze. |
Quote TWO POLES. |
Of course it works for RET. It works for RET on pretty much any day of the year and from pretty much any latitude. The problem is that it only works for FET on certain days of the year and from certain latitudes. Why is that?Are you saying that days of an equinox are the only days that the FE version of the experiment ever work properly? Why is it that the RE version works just fine on any day of the year?
Actually, the method described in the article works for RET as well. There are several ways to get your latitude, that is only one of them.
Are you saying that days of an equinox are the only days that the FE version of the experiment ever work properly? Why is it that the RE version works just fine on any day of the year?
Actually, the method described in the article works for RET as well. There are several ways to get your latitude, that is only one of them.
You are right, Tom, it does say equinox. I mis-applied the wiki latitude formula to a time that wasn't at equinox. My bad.
Can you please explain how it works on a flat earth? The north pole for example is 6215 statute miles from the equator. If the sun is 3,000 miles above the flat earth right over the equator on equinox then its elevation angle would be 25.8 degrees at the poles. Real observations show the sun to be right on the horizon during the equinox at both poles simultaneously.
How does the flat earth model explain it? How has TFES derived its latitude formula?
To me, the observations are explained by the spherical earht heliocentric model.
Of course it works for RET. It works for RET on pretty much any day of the year and from pretty much any latitude. The problem is that it only works for FET on certain days of the year and from certain latitudes. Why is that?Are you saying that days of an equinox are the only days that the FE version of the experiment ever work properly? Why is it that the RE version works just fine on any day of the year?
Actually, the method described in the article works for RET as well. There are several ways to get your latitude, that is only one of them.
Of course it does. Sailors at sea did it pretty much every day. It's just a little more work on days other than the equinox.Of course it works for RET. It works for RET on pretty much any day of the year and from pretty much any latitude. The problem is that it only works for FET on certain days of the year and from certain latitudes. Why is that?Are you saying that days of an equinox are the only days that the FE version of the experiment ever work properly? Why is it that the RE version works just fine on any day of the year?
Actually, the method described in the article works for RET as well. There are several ways to get your latitude, that is only one of them.
The method being discussed doesn't work for RET on any day of the year.