Offline Flatout

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Latitude
« on: February 01, 2017, 02:09:56 AM »
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?   
« Last Edit: February 01, 2017, 02:14:51 AM by Flatout »

Offline Rekt

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Re: Latitude
« Reply #1 on: February 01, 2017, 01:50:20 PM »
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?
"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.

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Offline junker

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Re: Latitude
« Reply #2 on: February 01, 2017, 03:11:09 PM »
"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.

You have been warned multiple times about low-content posting in the upper fora. Have a few days off to review the rules.
« Last Edit: February 01, 2017, 03:12:46 PM by junker »
Please make sure to check out these resources to ensure that your time at tfes.org is enjoyable and productive.

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Offline Flatout

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Re: Latitude
« Reply #3 on: February 01, 2017, 04:03:10 PM »
Junker, what is your take on the discrepancy between actual observations and the information stated in TFES Wiki about latitude?

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Offline rabinoz

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Re: Latitude
« Reply #4 on: February 02, 2017, 08:51:34 AM »
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?
I tried  in The Sun's height from the method and distances in "the Wiki". « on: April 15, 2016, 01:58:07 AM » to use these distances and angles to calculate the sun's height.

The only answer I got on this forum on a similar question was:
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.

And to that post
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.

Offline Flatout

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Re: Latitude
« Reply #5 on: February 03, 2017, 03:30:32 AM »
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?

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Offline rabinoz

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Re: Latitude
« Reply #6 on: February 03, 2017, 06:59:03 AM »
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".

You just need to use "suitable physics", in this case this "curved light". The following article explains it, though it is not sympathetic to the notion of a "Flat Earth".
Quote from: Donald E. Simanek
The Flat Earth
Early Ideas About the Shape of the Earth.


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.
   

How light refracts near the earth.
Left: conventional physics. Right: flat-earther's physics.
The Flat Earth, by Donald E. Simanek
See, easy. The whole article is worth reading, though not exactly complimentary to FE reasoning.

[1] Actually, it doesn't! The accepted distance from Alexandria to Syene as estimated by Erostothanes is 500 miles, giving a sun height = 500/tan(7.2°) = 3,960 miles.

Re: Latitude
« Reply #7 on: February 11, 2017, 10:58:03 AM »
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 .   

Offline Flatout

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Re: Latitude
« Reply #8 on: February 12, 2017, 12:02:05 AM »
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 .
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.   

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Offline Tom Bishop

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Re: Latitude
« Reply #9 on: February 12, 2017, 01:47:27 AM »
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.

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Offline markjo

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Re: Latitude
« Reply #10 on: February 12, 2017, 02:38:30 AM »
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?
Abandon hope all ye who press enter here.

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Offline Tom Bishop

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Re: Latitude
« Reply #11 on: February 12, 2017, 02:50:45 AM »
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.

Offline Flatout

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Re: Latitude
« Reply #12 on: February 12, 2017, 07:06:36 AM »

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.

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Offline rabinoz

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Re: Latitude
« Reply #13 on: February 12, 2017, 08:46:09 AM »

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.
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".

But, as you continue on to ask, how does the Flat Earth explain the angles at the poles and
to complicate matters Tom Bishop believes that the correct FE continental layout is the bipolar one as in:

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.
But getting information on the sun's movement means wading through
But, be warned, you might it a rough ride with all these "Monstrous Hypothetical Motions" and some "interesting" ideas about astronomy.

The diagram on p 30 of Sea-Earth Globe and the "explanation" on pp 32,33 explain some of the sun's movement:
Quote
TWO POLES.

Fig, 25

The question has been asked, If the sun crosses from the
northern circuit to the southern, how is it so little difference
is observable in its positions? The above diagram (Fig. 25)
will help the student to understand this more intricate part
of the subject; but we must remember that there is a great
difference between the motions of the solar orb, and the
motions of light which proceed in every direction away from
it. The motions of the celestial bodies we have already
explained in connection w ith Fig. 22; and we have also
shewn that the equator is a broad belt of vertical rays, and
not a mere “imaginary line.”

We will refer to Fig. 25. A t the vernal equinox the sun is
at E in the morning at 6 a.m. Its height travelling round with
the etherial currents, is seen at the same moment by an
observer at A. Now an observer always sees an object in
the direction of the rays entering the eye; and the curve
of about 6,000 miles from E to A is so great, that for the
last few miles the rays seem to come to A in a straight line
in the direction from H. Hence he sees the sun’s image
rise “due east,” not north-east, proving that light travels
in great curves.

In the same way observers at a, and at M, see their dffferent
sun images at I and at T ; but it is self-evident that the orb
of the sun itself cannot be in these various positions at one
and the same time. Six hours later the sun itself arrives
from E to A, and it may happen that then its swirl outwards
from N drives it into the southern current, and it goes round
with that current in the direction of the arrow until it arrives
at p, when its light, preceding it in a great curve, the sun’s
image is again seen at H from A.
It then goes round with the southern currents, daily,
contracting its circle in a fine spiral until it arrives at 23 1/2° S.
when, having lost its further southern tendency or swirl,
electrical and magnetic forces, doubtless under intelligent
supervision, drive it again northwards. Similar explanations
apply to the moon, and to the planets, but with different
periods, owing to their different altitudes, as already explained
in a former article.


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Offline markjo

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Re: Latitude
« Reply #14 on: February 13, 2017, 01:27:32 AM »
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 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?
Abandon hope all ye who press enter here.

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Offline Tom Bishop

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Re: Latitude
« Reply #15 on: February 13, 2017, 10:14:18 PM »

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.

The sun is at or near the horizon at the poles. This is due to other effects which limits its duration, otherwise the earth would be in perpetual daylight.

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Offline Tom Bishop

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Re: Latitude
« Reply #16 on: February 13, 2017, 10:14:57 PM »
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 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?

The method being discussed doesn't work for RET on any day of the year.

Re: Latitude
« Reply #17 on: February 13, 2017, 11:21:10 PM »
Just an FYI: If you continue down this road of reasoning with Tom, you will eventually reach a point where Tom claims that basic geometry doesn't work when it comes to predicting the angle of the sun. His excuse for why the sun isn't where it should be in the flat earth model based on basic trigonometry boils down to: perspective stops working at long distances. Fair warning.

Sources for the brave:

http://forum.tfes.org/index.php?topic=4953.0
http://forum.tfes.org/index.php?topic=4994.msg96829#msg96829
http://forum.tfes.org/index.php?topic=5346.0

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Offline markjo

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Re: Latitude
« Reply #18 on: February 14, 2017, 02:23:09 PM »
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 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?

The method being discussed doesn't work for RET on any day of the year.
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.
« Last Edit: February 15, 2017, 02:06:48 PM by markjo »
Abandon hope all ye who press enter here.

Offline Flatout

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Re: Latitude
« Reply #19 on: February 14, 2017, 07:36:08 PM »
Yea, it's called a Solar Analemma. It's on nearly every globe.   It's plotable based on the orbital mechanics of the earth around the sun.  It's seems as if TFES is unable to calculate or predict beyond the equinox.  Is the science of the TFES generating something usable to navigate?