Offline Nostra

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Some maths regarding sun displacement in the sky
« on: August 31, 2016, 02:33:28 PM »
As it has been already mentioned in many topics, there are many inconsistencies with the Sun and the Flat Earth theory, except if accepting some unicorn magical poop responsible for huge angular deviation.
For examples, during the fall/spring equinoxes:
        - if you stand on the equator, you will see the sunrise due East and the sunset due West. According to the flat Earth theory, you should see the sunset due North-East and the sunrise due North-West. For what I have read, there are yet no explanations for this 45° (!) angular deviation compared to observation in the flat earth theory.

        - if you stand on the equator, at the sunrise, the sun is at the horizon level (meaning height elevation of 0°). Basic trigonometry (provided on request) shows that in the FE theory, the sun has instead an elevation height of about 28°. Here FE explains that some huge light refraction (a.k.a unicorn magical poop) is responsible for this optical illusion.

I would like to provide some maths I've made regarding height elevation and the angular speed of the sun in the sky according to FE compared with RE.
My hypotheses are the followings:
1) (RE or FE) Equator is a circle with a radius of 3958 miles

2) (FE) As the sun height elevation during sunrise is obviously 0°, and 90° at noon, and 0° at sunset (during fall/spring equinoxes, on the equator), there should be some refraction law in FE making this optical illusion possible.
I presumed a linear refraction law with the absolute elevation height of the sun, i.e. when the sun if actually at a height of 28°, it appears to be at 0° and when it is actually at 90° it appears to be at 90°. (Vertical refraction)
(If some flat earther could indicate me what is the real refraction law, I would be sooooo happy to adapt my calculation to it!)

3) (FE) I presumed also that some unicorn magical poop enable to see the sun due west where it should appears North-West, and due East where it should appears North-East. (Horizontal refraction). This horizontal refraction is obviously linear with the elevation height of the sun otherwise the sun would not appear to make a perfect line in the sky on the equator during fall/spring equinoxes.

4) (RE) I didn’t take into account the refraction with RE, therefore the length of the day during fall/spring equinox during is slightly higher than 12 hours and also has slight impact on the elevation and angular speed of the sun during the first and last hours of the day. But this does not call into question the results presented here after, especially if you compare the results near noon.

Here below are the results of the height elevation of the Sun:
- In black is the actual height elevation of the sun in FE theory
- In red is the corrected height elevation with the linear refraction law to be compliant with sunset and sunrise at 0°
- In green is the sun elevation in RE "theory" (the elevation being only due to the constant earth rotational speed



In my infinite kindness I also provide some other refraction laws I tried with flat earth theory, as I have no idea how all of this works! The idea of the other laws is to be more compliant with RE, i.e. the refraction is only significant when the sun is low on the horizon:
-   In violet the observed elevation meets the real one 1 hour after sunrise (symmetrical for sunset)
-   In blue, the observed elevation meets the real on 2 hours after sunrise (symmetrical for sunset)


Knowing the elevation height of the sun through the day, one can easily calculate the angular speed of the sun. Keeping the same color scheme, here are the results:




Just by looking at the angular speed of the sun around noon, on round Earth, the angular speed is equal to the angular speed of the Earth, which is 0.25°/min (this is true whatever the sun height). On flat Earth, whatever the correction I use (but again, the "real" refraction law in FE theory would be well appreciated), the angular speed of the sun should be higher (between 0.33 and 0.48 °/min).

By looking on the results at the beginning or the end of the day, the differences are much more important, and absolutely not consistent with observations.

Therefore, the Earth is not flat.

NOTA: For all flat earthers, here is an interesting property of the air (or aether I don’t really understood who is responsible for the refraction during sunset/sunrise), when talking about refraction on flat earth model, don’t forget to mention that it is (hugely) anisotropic! (45° horizontally and 28° vertically at sunrise on the equator during sunset/sunrise!)
« Last Edit: August 31, 2016, 02:42:16 PM by Nostra »
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Re: Some maths regarding sun displacement in the sky
« Reply #1 on: September 01, 2016, 01:20:48 AM »
I have been meaning to do this exact series of charts for a while, nice work!
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Re: Some maths regarding sun displacement in the sky
« Reply #2 on: September 01, 2016, 03:24:56 AM »
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
My hypotheses are the followings:
1) (RE or FE) Equator is a circle with a radius of 3958 miles
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Great work, bur I must take issue on one point.

On the only FE maps we have the Equator is a circle with the radius being the distance from the Equator to the North Pole.

This is well documented as very close to 10,000 km or 6214 miles.
The original definition of the metre was (Equator to the North Pole distance on the longitude through Paris)/10,000,000,

This probably won't affect the general shape of what you have done.

The TFES Wiki essential agrees with this, as in
Quote from: the Wiki
Finding your Latitude and Longitude
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.

This "one degree per 69.5 miles" makes the Equator to North Pole distance 6,255 miles - pretty close.

Of course, this exposes a paradox for the Flat Earth Theory!
This figure for the Equator radius makes the circumference of the Equator Circle 39044 miles, but the actual circumference of the earth is about 24,900 miles!
But I can't get any Flat Earther to actually accept it. The usual answer is "Have you personally measured the circumference of the Equator?" Well, no!

Offline Nostra

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Re: Some maths regarding sun displacement in the sky
« Reply #3 on: September 01, 2016, 10:51:51 AM »
Thanks for this information.

I haven't realized that the equator was not the same size on FE and RE... This is getting weirder and weirder...

Anyway, here are the updated graphs, it gets even more wrong with this equator lenght!

Elevation height (with the same color scheme):







Angular speed (with the same color scheme):





The updated conclusion are worst (for FE) :

On flat Earth, whatever the correction I use (but again, the "real" refraction law in FE theory would be well appreciated), the angular speed of the sun is between 0.52 and 0.66 °/min, between 2 and 3 times the actual angular speed of the Sun (0.25°/min)!

Also, the biggest deviation between RE and FE sun angular speed are close to noon with this length of equator, meaning that any "refraction" argument is not acceptable...

Therefore, the Earth, really is not flat!

EDIT : This also change the "real" height elevation of the sun during sunrise or sunset to 18° instead of 28°. Therefore you need a little less of unicorn magical poop to be responsible for making the sun appears at 0°. But this quite increase the anisotropy of the refraction in the air/aether (18° vertically, 45° horizontally)
« Last Edit: September 01, 2016, 11:50:29 AM by Nostra »
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Offline rabinoz

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Re: Some maths regarding sun displacement in the sky
« Reply #4 on: September 01, 2016, 12:09:39 PM »
Thanks for this information.

I haven't realized that the equator was not the same size on FE and RE... This is getting weirder and weirder...

<< I removed the graphs for compactness >>

The updated conclusion are worst (for FE) :

On flat Earth, whatever the correction I use (but again, the "real" refraction law in FE theory would be well appreciated), the angular speed of the sun is between 0.52 and 0.69 °/min, between 2 and 3 times the actual angular speed of the Sun (0.25°/min)!

Also, the biggest deviation between RE and FE sun angular speed are close to noon with this length of equator, meaning that any "refraction" argument is not acceptable...

Therefore, the Earth, really is not flat!
Yes, it gets weirder and weirder. You begin to appreciate how Alice felt "Through the Looking Glass".

Another bit about size:
Quote from: the Wiki
The Ice Wall
The figure of 24,900 miles is the diameter of the known world; the area which the light from the sun affects.

Which more or less agrees with the 10,000 km equator to north pole.

But when you get onto "the 'real' refraction law in FE theory would be well appreciated"! Here be dragons! Refraction "in FE theory" is whatever it takes and in whatever direction is needed to make the sun "appear to set behind the horizon". Now NO amount of refraction can do that, but no matter!

I imagine the distance to and the apparent size of the sun would be easy to work out from your work, because I had a thread with photos showing that the angular size of the sun does not change during the day, right down to about 3° from the horizon.

Here is the thread on The Flat Earth Society site The Constancy of the Angular size of the Sun « on: August 24, 2016, 08:12:47 PM ».

And here is a post on the site with a bit less explanation: Re: Astronomy debunk: The sun is clearly moving away « Reply #13 on: July 12, 2016, 01:45:31 AM ».

But you will soon learn the most Flat Earthers seem to work in an EFZ[1], so this stuff won't impress them one little bit!


[1] EFZ = Equation Free Zone.

Offline Nostra

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Re: Some maths regarding sun displacement in the sky
« Reply #5 on: September 01, 2016, 12:58:02 PM »
Indeed, the sun distance is even an input for my calculations, here it is (no comparison with RE because this make no sense!) :



And yes, you are right, it is quite easy to provide the angular diameter of the sun through the day (still fall/spring equinoxes, on the equator), here it is, with comparison with average RE angular diameter of the sun (I am actually too busy to provide the actual angular diameter of the sun in RE, but it varies between 0.5244° and 0.5422° through the year, so I took 0.535°, which is close to the actual angular diameter of the sun during equinoxes and it is only for comparison purpose)



You can clearly see that the sun should appears more than 3 times bigger at noon compared with sunset or sunrise. But I read some crap explaining a magnification effect...
What I actually find kind of sad here, is that with the hypothesis of the flat earth, it would be quite easy to provide some mathematical laws (for refraction or magnification for example as it is the subject here, but also many other topics) to explain the observation. Of course these would be just made-up laws to correct the way things should be (in FE theory) and made them appears to be like in the observations (just a reverse work compared with what I provided here). But of course, asking people with not even a little of scientific background to provide such a thing is probably overestimating them...

And yes, I understand that any proof that the Earth is not flat will no be taken as it is by all flat earthers here (I have read quite a lot before starting posting). However, for some weird reasons this site provide me some fun and also if some "undecided" people come here to find the truth, maybe this post as well as all others proving the earth is not flat, will help them not believing the Earth is flat, and this would be a little win for human kind!
« Last Edit: September 01, 2016, 01:08:53 PM by Nostra »
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Offline rabinoz

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Re: Some maths regarding sun displacement in the sky
« Reply #6 on: September 01, 2016, 11:18:06 PM »
Indeed, the sun distance is even an input for my calculations, here it is (no comparison with RE because this make no sense!) :
And yes, you are right, it is quite easy to provide the angular diameter of the sun through the day (still fall/spring equinoxes, on the equator), here it is, with comparison with average RE angular diameter of the sun (I am actually too busy to provide the actual angular diameter of the sun in RE, but it varies between 0.5244° and 0.5422° through the year, so I took 0.535°, which is close to the actual angular diameter of the sun during equinoxes and it is only for comparison purpose)
You can clearly see that the sun should appears more than 3 times bigger at noon compared with sunset or sunrise. But I read some crap explaining a magnification effect...
What I actually find kind of sad here, is that with the hypothesis of the flat earth, it would be quite easy to provide some mathematical laws (for refraction or magnification for example as it is the subject here, but also many other topics) to explain the observation. Of course these would be just made-up laws to correct the way things should be (in FE theory) and made them appears to be like in the observations (just a reverse work compared with what I provided here). But of course, asking people with not even a little of scientific background to provide such a thing is probably overestimating them...

And yes, I understand that any proof that the Earth is not flat will no be taken as it is by all flat earthers here (I have read quite a lot before starting posting). However, for some weird reasons this site provide me some fun and also if some "undecided" people come here to find the truth, maybe this post as well as all others proving the earth is not flat, will help them not believing the Earth is flat, and this would be a little win for human kind!
Yes, "some crap explaining a magnification effect..."  describes pretty well this bit:
Quote from: the Wiki
Magnification and Shrinking
Q: If the sun is disappearing to perspective, shouldn't it get smaller as it recedes?
Wasn't  The sun remains the same size as it recedes into the distance due to a known magnification effect caused by the intense rays of light passing through the strata of the atmolayer.
I have posted some of my own photos on the moon's angular size staying almost constant, as in:
Re: Size of the Sun and the "Known Magnification Effect" « Reply #40 on: June 08, 2016, 05:27:52 AM »
Re: Size of the Sun and the "Known Magnification Effect" « Reply #58 on: June 15, 2016, 11:50:39 AM »

One reason I post is to give new-comers another explanation. Some new-comers, like yourself, clearly don't need it.
What riles me is they will not tackle these "simple" issues that we can so easily observe.

Keep trying.

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Offline cel

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Re: Some maths regarding sun displacement in the sky
« Reply #7 on: September 02, 2016, 07:27:37 AM »
DON'T BE TOO HARD ON YOURSELVES, PEOPLE! DON'T MAKE SIMPLE THINGS DIFFICULT. YOU DON'T HAVE TO DRAW OR ANALYZE IT YOURSELF IN KNOWING IF THE SUN IS 93M AWAY FROM EARTH OR IF IT IS THAT BIG (AND ALSO NOT MOVING? :))  ANYWAY, JUST WATCH THIS VIDEO, AND DON'T LET YOURSELF BE CONFUSED AND DECEIVED BY SOME PEOPLE HERE TRYING TO USE GRAPHS AND LINES, BUT WITH INSUFFICIENT EVIDENCE, EMPIRICAL DATA AND LOGICAL REASONING/EXPLANATION.

ENJOY WATCHING, LEARN, KNOW AND ASK QUESTIONS, OR BETTER BE ENTICED TO SEEK THE TRUTH. BE A TRUTH SEEKER THEN... .. :)





YOU'RE LUCKY TO KNOW THESE IN YOUR LIFE TIME... AT LEAST YOU CAN HAVE THIS RARE CHANCE TO KNOW SOMETHING WORTH KNOWING! :)

 
You may wish to decipher how many squares are there in the 4x4 matrix of my profile image. If you do, tell me! That way I can tell if you really have an imaginative/creative mind that knows how to think out of the box. If you got it right, you've got great potential of becoming a genuine Truth Seeker! Welcome then to the Truth Seeker's group!

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Re: Some maths regarding sun displacement in the sky
« Reply #8 on: September 02, 2016, 08:18:14 AM »
[...]
 AND DON'T LET YOURSELF BE CONFUSED AND DECEIVED BY SOME PEOPLE HERE TRYING TO USE GRAPHS AND LINES, BUT WITH INSUFFICIENT EVIDENCE, EMPIRICAL DATA AND LOGICAL REASONING/EXPLANATION.
[...]

If using math and science to prove things is not authorized in FE theory, then... Well...

Could you just inform me what kind of "insufficient evidence"; "enprirical data" or "logical reasoning" you need to fully understand my post, I'll be glad to provide it to you!

Another think I find very funny is how the sentence I quoted should actually refere to flat earthers...

[...]
YOU DON'T HAVE TO DRAW OR ANALYZE IT YOURSELF IN KNOWING IF THE SUN IS 93M AWAY FROM EARTH OR IF IT IS THAT BIG (AND ALSO NOT MOVING? :))  ANYWAY, JUST WATCH THIS VIDEO
[...]

What a great advice! This litterally means : "Don't think by yourself, don't try to analyze the world yourself and take my word for the truth, but please, don't try to check it!"

I'll give a better advice to all people reading this. Draw and analyze yourself, search and explore for information, and try to understand the world instead of taking the words of other for the truth. Form your own opinion!
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Offline rabinoz

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Re: Some maths regarding sun displacement in the sky
« Reply #9 on: September 02, 2016, 08:42:15 AM »
DON'T BE TOO HARD ON YOURSELVES, PEOPLE! DON'T MAKE SIMPLE THINGS DIFFICULT. YOU DON'T HAVE TO DRAW OR ANALYZE IT YOURSELF IN KNOWING IF THE SUN IS 93M AWAY FROM EARTH OR IF IT IS THAT BIG (AND ALSO NOT MOVING? :))  ANYWAY, JUST WATCH THIS VIDEO, AND DON'T LET YOURSELF BE CONFUSED AND DECEIVED BY SOME PEOPLE HERE TRYING TO USE GRAPHS AND LINES, BUT WITH INSUFFICIENT EVIDENCE, EMPIRICAL DATA AND LOGICAL REASONING/EXPLANATION.

ENJOY WATCHING, LEARN, KNOW AND ASK QUESTIONS, OR BETTER BE ENTICED TO SEEK THE TRUTH. BE A TRUTH SEEKER THEN... .. :)

 
It is very hard getting all the details of what angles are being given when he just races through it in Autocad. I think he's simply trying to skite at how wonderful he is at using Autocad.

So, maybe you could go through the video are get the figures he is using. We need to know:
  • The Latitude and Longitude of each location.
  • The exact time at which the sun elevations are given. 
  • The solar noon in each location.
I ask for these because he claims that Cape Town and Budapest are at the longitude (they are, close enough), yet he has the solar noon is an hour different. That is not possible. If you look carefully you will find that he is quite wrong.
I have worked through the figures, but haven't had time finish the diagram yet. No, not in Autocad, just lowly Corel draw.
But the bottom line is the video is wrong!

Quote from: cel
 

YOU'RE LUCKY TO KNOW THESE IN YOUR LIFE TIME... AT LEAST YOU CAN HAVE THIS RARE CHANCE TO KNOW SOMETHING WORTH KNOWING! :)
How many times does this have to be answered? It's bunkum.

Again, if you summarise the main points I will tackle it them, but rest assured, it's wrong!

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Offline cel

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Re: Some maths regarding sun displacement in the sky
« Reply #10 on: September 02, 2016, 09:50:45 AM »
BEFORE YOU SAY THAT THE CAD PRESENTATION IS "DEAD WRONG", BETTER PRESENT SOMETHING CONCRETE AND VERIFIABLE DATA. YOUR JUDGEMENT IS HASTILY DONE, IMPAIRED! ARE GLOBE EARTHERS OR SUPPORTERS LIKE THAT? WHEN RUNNING OUT OF EVIDENCE AND ARGUMENT, THEY WOULD JUST RESORT INTO SWEEPING AND HASTE CONCLUSION THAT SOMETHING, THOUGH HOW CLEARLY PRESENTED TO PROVE SOMETHING, IS "DEAD WRONG", JUST LIKE THAT? O C'MMON, WHO WILL BELIEVE AND LIKE YOU, MAN? NO ONE I GUESS.. :)

OK, SHOW YOUR COREL, AND LET THIS TECHSIE GUY OR ENGINEER REBUT YOUR ACCUSATIONS AND ALLEGATIONS. WELL, LET THE BALL GO ROLLING. LET'S SEE WHAT'S THE OUTCOME. I JUST HOPE EVERYTHING WOULD TURN INTO BENEFITS FOR TRUTH SEEKERS.

MEANWHILE, REFRAIN FROM HASTY CONCLUSION, YOU HAVEN'T PRESENTED OR PROVED ANYTHING YET. IF I WERE YOU, BETTER GET MOVING YOUR ASS AND PRESENT SOMETHING IN YOUTUBE FOR OTHERS TO SEE YOUR STUFF.

GOOD LUCK, MAN! :)
« Last Edit: September 02, 2016, 09:56:34 AM by cel »
You may wish to decipher how many squares are there in the 4x4 matrix of my profile image. If you do, tell me! That way I can tell if you really have an imaginative/creative mind that knows how to think out of the box. If you got it right, you've got great potential of becoming a genuine Truth Seeker! Welcome then to the Truth Seeker's group!

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Offline rabinoz

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Re: Some maths regarding sun displacement in the sky
« Reply #11 on: September 02, 2016, 10:13:54 AM »
BEFORE YOU SAY THAT THE CAD PRESENTATION IS "DEAD WRONG", BETTER PRESENT SOMETHING CONCRETE AND VERIFIABLE DATA. YOUR JUDGEMENT IS HASTILY DONE, IMPAIRED! ARE GLOBE EARTHERS OR SUPPORTERS LIKE THAT? WHEN RUNNING OUT OF EVIDENCE AND ARGUMENT, THEY WOULD JUST RESORT INTO SWEEPING AND HASTE CONCLUSION THAT SOMETHING, THOUGH HOW CLEARLY PRESENTED TO PROVE SOMETHING, IS "DEAD WRONG", JUST LIKE THAT? O C'MMON, WHO WILL BELIEVE AND LIKE YOU, MAN? NO ONE I GUESS.. :)

OK, SHOW YOUR COREL, AND LET THIS TECHSIE GUY OR ENGINEER REBUT YOUR ACCUSATIONS AND ALLEGATIONS. WELL, LET THE BALL GO ROLLING. LET'S SEE WHAT'S THE OUTCOME. I JUST HOPE EVERYTHING WOULD TURN INTO BENEFITS FOR TRUTH SEEKERS.

MEANWHILE, REFRAIN FROM HASTY JUDGEMENTAL AND CONCLUSION, YOU HAVEN'T PRESENTED OR PROVED ANYTHING YET. IF I WERE YOU, BETTER GET MOVING YOUR ASS AND PRESENT SOMETHING IN YOU TUBE FOR OTHERS TO SEE YOUR STUFF.

GOOD LUCK, MAN! :)
As I said before, when I get around to it. You're not the only one around here.  And it would help no end if you put in writing the information I asked for. At least then I can't be accused of not listening to the video properly.

By the way, how is your explanation going of,
the sun appearing to set behind the horizon and
the sun rising almost due east everywhere on earth (except very close to either pole) at each equinox.

When you start explaining simple things we might be more inclined to listen.

By the way, I don't need luck, my beliefs don't depend on debunking a few videos. I believe the way I do because it fits with all I see.

It would make things a lot easier if you posted a high resolution version of your map so I can check distances that I know first hand.

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Re: Some maths regarding sun displacement in the sky
« Reply #12 on: September 02, 2016, 01:06:21 PM »
BEFORE YOU SAY THAT THE CAD PRESENTATION IS "DEAD WRONG", BETTER PRESENT SOMETHING CONCRETE AND VERIFIABLE DATA. YOUR JUDGEMENT IS HASTILY DONE, IMPAIRED! ARE GLOBE EARTHERS OR SUPPORTERS LIKE THAT? WHEN RUNNING OUT OF EVIDENCE AND ARGUMENT, THEY WOULD JUST RESORT INTO SWEEPING AND HASTE CONCLUSION THAT SOMETHING, THOUGH HOW CLEARLY PRESENTED TO PROVE SOMETHING, IS "DEAD WRONG", JUST LIKE THAT? O C'MMON, WHO WILL BELIEVE AND LIKE YOU, MAN? NO ONE I GUESS.. :)

OK, SHOW YOUR COREL, AND LET THIS TECHSIE GUY OR ENGINEER REBUT YOUR ACCUSATIONS AND ALLEGATIONS. WELL, LET THE BALL GO ROLLING. LET'S SEE WHAT'S THE OUTCOME. I JUST HOPE EVERYTHING WOULD TURN INTO BENEFITS FOR TRUTH SEEKERS.

You should apply to yourself what you are advising to the others, because what you wrote here sums up quite well the flat earther, and particularly your behavior. I provided, "corel", and "data", and yet you don't seems to have even look at them or produce any constructive criticism on it...

MEANWHILE, REFRAIN FROM HASTY CONCLUSION, YOU HAVEN'T PRESENTED OR PROVED ANYTHING YET. IF I WERE YOU, BETTER GET MOVING YOUR ASS AND PRESENT SOMETHING IN YOUTUBE FOR OTHERS TO SEE YOUR STUFF.

GOOD LUCK, MAN! :)

Once again, are you sure you are not talking about yourself here?
And seriously, is youtube the only acceptable media to proove something nowadays?

And please, stop writing in capital letter, this is really unpleasant to read.
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Re: Some maths regarding sun displacement in the sky
« Reply #13 on: September 02, 2016, 08:11:18 PM »
DON'T LET YOURSELF BE CONFUSED AND DECEIVED BY SOME PEOPLE HERE TRYING TO USE GRAPHS AND LINES, BUT WITH INSUFFICIENT EVIDENCE, EMPIRICAL DATA AND LOGICAL REASONING/EXPLANATION.
I addressed CEL's pair of videos in another thread, I won't repeat it all here.  If you're interested, click the link.  The TL;DR version: Video 1 narrator is an idiot, Video 2 narrator is a liar.

ENJOY WATCHING
No, I really didn't.
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Re: Some maths regarding sun displacement in the sky
« Reply #14 on: September 05, 2016, 09:44:06 PM »
I finally got around to doing some follow up on this.  As we all know, the FE sun moves around the geographic north pole in an essentially flat path, never actually going below the horizon.  We are told that perspective and / or atmospheric effects cause it to APPEAR to go below the horizon at a distance.  We are given this information in a manner suggesting that it is enough to explain the apparent movement of the sun, while they seem to utterly ignore the fact that the apparent elevation angle of the sun is not the only problem with this model: there is also an apparent left-to-right shift happening.

I calculated the actual position of the nearby sun above the flat earth from the viewpoint of an observer located at 45° North latitude on the day of the September Solstice.  I did this using simple trigonometry, as illustrated below, calculating the distance and direction to the sub-solar point on the earth's surface and then the elevation angle up to the sun knowing the distance and height.  I did this ignoring any atmospheric or perspective effects, which matches how Rowbotham and other FE calculate the sun's height.  (Notice in the wiki, for example, that during the section on calculating the height of the sun, no allowance is made for the sun appearing to be at 45° elevation angle while actually being somewhere else.  No, it is assumed to actually be where it appears to be, at 45° elevation)  I did ten minute increments all night and day, midnight to midnight.

I then pulled information from the US Naval Observatory web site listing the projected direction and elevation angles for the same geographic location and also in ten minute increments.  I then calculated the difference between observation and calculation, and graphed the result.  This number is the magnitude of the FE proposed perspective and atmospheric effects.  The X axis is the difference left or right between the direction where the sun should be to where the sun is observed to be at that time of day.  The Y axis is the difference between how high above the horizon the sun should be and where it is observed to be, absent the proposed perspective and atmospheric effects.  This curve and the one that follows are both slightly asymmetrical due to my local apparent solar noon not lining up exactly with clock noon, so the USNO numbers are coming in slightly ahead of my calculated numbers.


One more simple equation allows us to combine the left/right offset effect with up/down offset effect to calculate a total offset.  When the sun rises/sets the absolute effect is at its maximum, and the result is a vector effectively moving the sun's apparent position a total of 45° from its calculated position.  That is 1/4 of the way across the sky!  I suspect the typical FE who has not done the math would expect the atmospheric effects to be at a minimum at noon and maximum at sunrise/sunset, and he or she would be correct.  I also suspect the typical FE would imagine these effects changing in a pattern that makes intuitive sense, maybe a straight line or a sine wave; in this he or she would be incorrect.  Observe the total effect curve:


The point of all this: those of you proposing optical effects to explain the difference between observed sun and moon locations and where those bodies should be above your flat earth?  THIS is what your optical effect must achieve.
« Last Edit: September 06, 2016, 02:12:58 AM by Rounder »
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Re: Some maths regarding sun displacement in the sky
« Reply #15 on: September 06, 2016, 03:38:03 PM »
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The point of all this: those of you proposing optical effects to explain the difference between observed sun and moon locations and where those bodies should be above your flat earth?  THIS is what your optical effect must achieve.

And this is true only from the place you calculated it. If you do this for other latitude, you will find out that this effect also depends of the latitude. So there is not even a universal law of optical deviation (which should be the minimum... I expect the FE deviation to be the same if the total length of the air/aether got through by the sun light whatever the latitude...
Said differently, in the FE theory, the deviation of the sun at the same "real" elevation angle differ if you are at New york or Sidney, whereas the light is passing through the same amount of air/aether, with the same angle. Therefore the law of deviation of the light is not universal ans you will have a really bad time trying to explain that.
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Re: Some maths regarding sun displacement in the sky
« Reply #16 on: September 06, 2016, 04:53:44 PM »
And this is true only from the place you calculated it. If you do this for other latitude, you will find out that this effect also depends of the latitude. So there is not even a universal law of optical deviation (which should be the minimum...Said differently, in the FE theory, the deviation of the sun at the same "real" elevation angle differ if you are at New york or Sidney, whereas the light is passing through the same amount of air/aether, with the same angle. Therefore the law of deviation of the light is not universal ans you will have a really bad time trying to explain that.
Yes, I plan to illustrate this next.  Should be an interesting set of "seagull plots" at varying latitude.  The only place on earth that will ever get a zero error ever, and then only at exactly local noon, will be the spot under the sub-solar point.
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