Calculate the height & distance to the sun on the flat earth using angle only

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The sun never 'sets' on a flat earth. It goes out of view due to perspective. To calculate the height of the sun, you need to use the below math to take this fact into account.

Bust out excel.  Enter the following:

[C3] Type in Angle Starting with 89 going down to 1 [/C91]
[D3] =90-C3 [/D91]
[E3] =(40000/360)*(90-C3)[/E91]
[F3] =(40000/360)*(C3) [/F91]

[G3]***SPACER COLUMN***[/G91]

[H3] =(E3/40000)*C3 [/H91]
[I3] =(E3/40000)*D3 [/I91]
[J3] =H3+I3 [/J91]

[K3]***SPACER COLUMN***[/K91]

[L3] =(F3/40000)*C3 [/L91]
[M3] =(F3/40000)*D3 [/M91]
[N3] =L3+M3 [/N91]

[O3]***SPACER COLUMN***[/O91]

[P3] =J3+N3  [/P91]
[Q3] =D3*0.5 [/Q91]
[R3] =P3/Q3  [/R91]
[S3] =(DEGREES(ATAN(R3))) [/S91]
[T3] =TAN(RADIANS(S3))*E3 [/T91]


Notes: 

1) After setup, you can just insert a row and enter just the angle into column [C] and it will calculate the following:

Column E & Column F shows the distance to the 90 Degree Sun
Column P shows the unit of your perspective
Column S shows the actual angle taking into account your perspective
Coulmn T shows the flat earth sun height

" Remembered by God "


*

NAZA

  • 594
Quote
The sun never 'sets' on a flat earth. It goes out of view due to perspective


How does perspective explain this image?



Please explain the path the sunlight takes from the sun, that is supposed to be above the horizon, to the camera.  Where does perspective  bend it? Obviously it must bend it behind the sailboat...



Why does perspective lower the position of the sun, but not the sailboat?

I have tried several methods to calculate the angle of the sun but keep running into this problem.

The lowest I've been able to determine is 18° above the horizon.
What am I doing wrong?
Thanks
« Last Edit: October 24, 2017, 03:18:43 PM by NAZA »

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rabinoz

  • 24923
  • Real Earth Believer
1) After setup, you can just insert a row and enter just the angle into column [C] and it will calculate the following:

Column E & Column F shows the distance to the 90 Degree Sun
Sure.
Quote from: Silicon
Column P shows the unit of your perspective
What on earth does "the unit of your perspective" mean? It is so convenient that it is always 22.5°.
Quote from: Silicon
Column S shows the actual angle taking into account your perspective
Coulmn T shows the flat earth sun height

Could I be forgiven for suggesting that your whole magic fudge ends up with:
Sun's height = (equator to pole dist) x (1 - Lat/90)/(2 x (1 - Lat/90))
or Sun's height = (equator to pole dist)/2
Of course the answer is going to always end up as 5000 km.
Why not be up front and simply say the the sun's height = 5000 km but we haven't a clue about why the measured elevation don't fit?
Talk about knowing the answer before you start and forcing your "equations" to give your predetermined answer.
Stop wasting everybody's time with this garbage.

If you want to retain a shred of credibility, please justifiy why the sun should really appear to be at an elevation of 1.13° at 89° N 0°E on Sep 23rd, 2017 at solar noon, as in:
From: NOAA ESRL, Solar Position Calculator  and they know a lot more about the sun's angles than you do!

If you disagree, take a running jump off the edge of your fictional fantasy of a flat frittata that you try to pass off as the real earth,

1) After setup, you can just insert a row and enter just the angle into column [C] and it will calculate the following:

Column E & Column F shows the distance to the 90 Degree Sun
Sure.
Quote from: Silicon
Column P shows the unit of your perspective
What on earth does "the unit of your perspective" mean? It is so convenient that it is always 22.5°.
do you really want me to air the 'dirty laundry' in regards to what I've found about this?

Quote from: Silicon
Column S shows the actual angle taking into account your perspective
Coulmn T shows the flat earth sun height

Could I be forgiven for suggesting that your whole magic fudge ends up with:
Sun's height = (equator to pole dist) x (1 - Lat/90)/(2 x (1 - Lat/90))
or Sun's height = (equator to pole dist)/2
Of course the answer is going to always end up as 5000 km.
Why not be up front and simply say the the sun's height = 5000 km but we haven't a clue about why the measured elevation don't fit?
Talk about knowing the answer before you start and forcing your "equations" to give your predetermined answer.
Stop wasting everybody's time with this garbage.
Yes the formula is simple.  That's what lets you know its real. Your oversimplification obviously can not work.
   
If you want to retain a shred of credibility, please justifiy why the sun should really appear to be at an elevation of 1.13° at 89° N 0°E on Sep 23rd, 2017 at solar noon, as in:
From: NOAA ESRL, Solar Position Calculator  and they know a lot more about the sun's angles than you do!
Is that the best you can come up with?  I know you just about fell out of your chair running the numbers ;)  THIS CANNOT BE DENIED. 

If you disagree, take a running jump off the edge of your fictional fantasy of a flat frittata that you try to pass off as the real earth,

I'm hungry now... 

Where's mom. Why'd you throw this back out here with these nuts? They can't disprove this...
« Last Edit: October 24, 2017, 10:36:02 PM by Silicon »

Are you using the circumference of a globe (40000 km) to calculate the sun in relation to a flat earth?

Mike
Since it costs 1.82¢ to produce a penny, putting in your 2¢ if really worth 3.64¢.

If you want to retain a shred of credibility, please justifiy why the sun should really appear to be at an elevation of 1.13° at 89° N 0°E on Sep 23rd, 2017 at solar noon, as in:
From: NOAA ESRL, Solar Position Calculator  and they know a lot more about the sun's angles than you do!
Is that the best you can come up with?  I know you just about fell out of your chair running the numbers ;)  THIS CANNOT BE DENIED. 

 
So, you’re just going to dismiss the question rather than providing an answer.  I would love to see an answer because you spreadsheet seems to fall apart in this, and other, situations.

Mike
Since it costs 1.82¢ to produce a penny, putting in your 2¢ if really worth 3.64¢.

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ausGeoff

  • 6091
The sun never 'sets' on a flat earth. It goes out of view due to perspective...

No.  This frequently-repeated FE claim about "perspective" indicates a total lack of knowledge about optical laws, nor the actual meaning of the word itself.  Perspective is defined as the art of representing three-dimensional objects on a two-dimensional surface so as to give the corrected concept of their height, width, depth, and location in relation to each other.  It has nothing to do with singular things purportedly disappearing in the far distance.

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rabinoz

  • 24923
  • Real Earth Believer
Is that the best you can come up with?  I know you just about fell out of your chair running the numbers ;)  THIS CANNOT BE DENIED. 
Of course it can be denied!

You managed, either accidently or on purpose, ended up with an expression for sun height that is completely independent of the latitude.

Look!
Sun's height = (equator to pole dist) x (1 - Lat/90)/(2 x (1 - Lat/90))
or Sun's height = (equator to pole dist)/2
Of course the answer is going to always end up as 5000 km.

And you have never shown anything is wrong with my original methods, so try again!

The following table gives the data for distance of each location from Vaupes and the elevations of the sun at solar noon on 20/Mar/2016.
All sun elevations were obtained from Sun Earth Tools as close as possible to the local midday on the last equinox. The time was UTC 20/Mar/2016 16:48.


Location   

Latitude   

Longitude   

Sun Elev   
Dist from   
Vaupes   

Flat Sun Ht   
Lat Diff from   
Vaupes   
Calc
Circum
Kimmirut, Canada   
62.847°   
-69.869°   
27.36°   
7,034 km   
3,609 km   
63.58°   
39,828 km
Santo Domingo   
18.486°   
-69.931°   
71.72°   
2,107 km   
6,077 km   
19.22°   
39,465 km
Vaupes, Colombia   
-0.565°   
-69.634°   
89.06°   
0 km   
------   
   
   
Chupa District, Peru   
-15.109°   
-69.998°   
74.69°   
1,610 km   
6,256 km   
14.37°   
40,334 km
Punta Arenas, Chile   
-53.164°   
-70.917°   
36.63°   
5,830 km   
4,388 km   
52.43°   
40,031 km

These locations and the directions to the sun on a flat earth are shown in the left hand  diagram below:
Once we have the angles from two sites the height of the sun can be calculated from: h = d/(cot(A1) + cot(A2)).


Sun Height on Flat Earth along 70°W Long
   

Sun Height on Globe Earth along 70°W Long

Using this method to find the height of the sun on the Flat earth gives measurements from 3609 km (for Kimmirut and Vaupes) to 6256 km (for Chupa District to Vaupes) depending on the spacing of the measurement sites.

In other words, claiming that the Flat Earth sun is at about 5,000 km altitude has no foundation whatever.

It is very telling when we note that when we plot these angles on a spherical earth the directions to the sun are all parallel.
Explain that!

Now, if instead of using these measurements to determine the Flat Earth sun height, we use them as Eratosthenes did, assuming a distant sun and use this data to calculate the circumference of the earth.

The circumference can be calculated from (distance from Vaupes) * 360°/(angle difference of sun from Vaupes)

This time, we get far better consistency.
              The estimated figures for the circumference of the earth range from 39,465 km to 40,334 km.
Certainly these figures would indicate that the earth is a globe with a distant sun.

And don't bring up the 89.06° again, that was used correctly in the calculations.

The application of perspective to the sun dipping below the horizon never made any sense to me.  A plane flying overhead just gets smaller until it “disappears”.  It does not stay the same size and then slip below the horizon little by little.  The perspective model has to apply to all things in the sky or it doesn't work.

What allows the sun to maintain its size at all times no matter how far away it travels?

Mike
Since it costs 1.82¢ to produce a penny, putting in your 2¢ if really worth 3.64¢.

Are you using the circumference of a globe (40000 km) to calculate the sun in relation to a flat earth?

Mike

I believe what they've done is just built the globe model around numbers based on the sun.  111.111km is not actually the distance per degree of lat, on a globe earth but rather 111.111km per degree of elevation of the sun.  It's not 10,000km as the distance to the equator, but rather the radius of the spotlight sun.

In other words, if I divide 8 by 2 or 40 by 10, do we not arrive at the same answer?

I suppose at the end of the day its all a matter of what you believe.

Is that the best you can come up with?  I know you just about fell out of your chair running the numbers ;)  THIS CANNOT BE DENIED. 
Of course it can be denied!

You managed, either accidently or on purpose, ended up with an expression for sun height that is completely independent of the latitude.

Look!
Sun's height = (equator to pole dist) x (1 - Lat/90)/(2 x (1 - Lat/90))
or Sun's height = (equator to pole dist)/2
Of course the answer is going to always end up as 5000 km.


Well I suppose you can do that, but in my opinion the spreadsheet I presented doubles up, using different math (see above post) which leaves zero doubt to the fact that the sun is 5,000km above the earth. 

Also if you plug in your Vaupes, Colombia sun elevation angles into my spreadsheet it all works just fine.

Is that the best you can come up with?  I know you just about fell out of your chair running the numbers ;)  THIS CANNOT BE DENIED. 
Of course it can be denied!

You managed, either accidently or on purpose, ended up with an expression for sun height that is completely independent of the latitude.

Look!
Sun's height = (equator to pole dist) x (1 - Lat/90)/(2 x (1 - Lat/90))
or Sun's height = (equator to pole dist)/2
Of course the answer is going to always end up as 5000 km.


Well I suppose you can do that, but in my opinion the spreadsheet I presented doubles up, using different math (see above post) which leaves zero doubt to the fact that the sun is 5,000km above the earth. 

Also if you plug in your Vaupes, Colombia sun elevation angles into my spreadsheet it all works just fine.
I don't think you can say it leaves "zero doubt".  The flat earth position of the sun and moon don't fit any real world observation of them.

For instance, you still haven't answered my question about perspective, sun set/rise and how it relates to other objects in the sky (e.g. planes).  Not to mention much of our perspective of distance objects if heavily influenced by atmospheric conditions.   Yet, our view of the transit of the sun isn't influenced but those same atmospheric conditions in the same ways that other objects in the sky and at the horizon are affected.

Add to this the fact that, in a flat earth, eclipses (solar & lunar) are impossible without some invisible object causing the eclipse...ignoring the oxymoronic idea that an invisible object blocks light.

And another complication, the setting/rising sun behind objects on the horizon such a ship, or the reflection of light from the setting/rising sun on the underside of clouds.  The fact that the calculations rely on some of the dimensions of the earth as a globe casts a lot of doubt.

We can usually make the math fit a single view of nearly anything we want.  But, the fact is, that math also has to fit in a complete analytical model to explain all the observations of whatever you’re trying to describe.  IOW, other than possibly viewing angle, the calculations don’t explain any observations of the sun. Therefore, the calculations are flawed, casting so much doubt as to invalidate your spreadsheet. 

Mike
Since it costs 1.82¢ to produce a penny, putting in your 2¢ if really worth 3.64¢.

Is that the best you can come up with?  I know you just about fell out of your chair running the numbers ;)  THIS CANNOT BE DENIED. 
Of course it can be denied!

You managed, either accidently or on purpose, ended up with an expression for sun height that is completely independent of the latitude.

Look!
Sun's height = (equator to pole dist) x (1 - Lat/90)/(2 x (1 - Lat/90))
or Sun's height = (equator to pole dist)/2
Of course the answer is going to always end up as 5000 km.


Well I suppose you can do that, but in my opinion the spreadsheet I presented doubles up, using different math (see above post) which leaves zero doubt to the fact that the sun is 5,000km above the earth. 

Also if you plug in your Vaupes, Colombia sun elevation angles into my spreadsheet it all works just fine.
I don't think you can say it leaves "zero doubt".  The flat earth position of the sun and moon don't fit any real world observation of them.

For instance, you still haven't answered my question about perspective, sun set/rise and how it relates to other objects in the sky (e.g. planes).  Not to mention much of our perspective of distance objects if heavily influenced by atmospheric conditions.   Yet, our view of the transit of the sun isn't influenced but those same atmospheric conditions in the same ways that other objects in the sky and at the horizon are affected.

Unfortunately (or fortunately) there is nothing like the sun that we know of to use for an apples to apples comparison.

Add to this the fact that, in a flat earth, eclipses (solar & lunar) are impossible without some invisible object causing the eclipse...ignoring the oxymoronic idea that an invisible object blocks light.

Why can't the moon just be passing in front of the sun for a solar eclipse on the FE?  As far as lunar, in my opinion it's equal on both sides for things that don't add up.  Too bad we'll just get the globe earth narrative, and not explore what is really happening.

And another complication, the setting/rising sun behind objects on the horizon such a ship, or the reflection of light from the setting/rising sun on the underside of clouds.  The fact that the calculations rely on some of the dimensions of the earth as a globe casts a lot of doubt.

I call these light tricks.  Haven't you seen clouds lit from below and above at the same time in the same view?

We can usually make the math fit a single view of nearly anything we want.  But, the fact is, that math also has to fit in a complete analytical model to explain all the observations of whatever you’re trying to describe.  IOW, other than possibly viewing angle, the calculations don’t explain any observations of the sun. Therefore, the calculations are flawed, casting so much doubt as to invalidate your spreadsheet. 

Well don't you think its ironic then, that we can calculate the height of the flat earth sun but cannot do the same thing on the globe model?

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Sentinel

  • 570
  • Open your eyes...
Why bother when it's all done and set anyway?

"No snowflake in an avalanche ever feels responsible."

Stanislaw Jerzy Lec

We can usually make the math fit a single view of nearly anything we want.  But, the fact is, that math also has to fit in a complete analytical model to explain all the observations of whatever you’re trying to describe.  IOW, other than possibly viewing angle, the calculations don’t explain any observations of the sun. Therefore, the calculations are flawed, casting so much doubt as to invalidate your spreadsheet. 

Well don't you think its ironic then, that we can calculate the height of the flat earth sun but cannot do the same thing on the globe model?
Well, actually you can calculate the distance in a globe model.  Sentinel posted an excellent video of one method that any layman can do.  Further, those calculations fit into an analytical model that fits all the observations I discussed in my previous post.

I call these light tricks.  Haven't you seen clouds lit from below and above at the same time in the same view?
No.  I haven’t.  Saying it’s a trick of light is one thing but the geometry doesn’t fit.  If the sun is 5000 above the earth physically impossible for there to be a situation where the sun lights only the underside of the clouds.  Yet that’s what we see so I have to disagree with you on that one.

Why can't the moon just be passing in front of the sun for a solar eclipse on the FE?  As far as lunar, in my opinion it's equal on both sides for things that don't add up.  Too bad we'll just get the globe earth narrative, and not explore what is really happening.
Because, it’s physically impossible in a flat earth model.  Especially considering all the paths.  The maps in the following link show eclipse paths that would only be possible in a heliocentric model.

https://eclipse.gsfc.nasa.gov/solar.html

Further, the following link shows all the math necessary to not only predict the occurrence of an eclipse but the path.  Here is everything used to calculate eclipses.
 
https://eclipse.gsfc.nasa.gov/5MCSE/5MCSE-Text11.pdf

BTW, what specifically is it that you think doesn’t add up on “both sides”.   
 
Mike
Since it costs 1.82¢ to produce a penny, putting in your 2¢ if really worth 3.64¢.

Why bother when it's all done and set anyway?



At some point "Sly Sparkane" will wind up here and have his day ruined because his video has been decimated, destroyed, and obliterated by this very thread.  I just wonder if he'll post another video showing the angle generated in column S of the spreadsheet perfectly lining up with the sun as it were.  Nah, probably not.
« Last Edit: October 25, 2017, 02:29:08 PM by Silicon »

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rabinoz

  • 24923
  • Real Earth Believer
Is that the best you can come up with?  I know you just about fell out of your chair running the numbers ;)  THIS CANNOT BE DENIED. 
Of course it can be denied!

You managed, either accidently or on purpose, ended up with an expression for sun height that is completely independent of the latitude.

Look!
Sun's height = (equator to pole dist) x (1 - Lat/90)/(2 x (1 - Lat/90))
or Sun's height = (equator to pole dist)/2
Of course the answer is going to always end up as 5000 km.


Well I suppose you can do that, but in my opinion the spreadsheet I presented doubles up, using different math (see above post) which leaves zero doubt to the fact that the sun is 5,000km above the earth. 

Also if you plug in your Vaupes, Colombia sun elevation angles into my spreadsheet it all works just fine.
No, it does not leave zero doubt at all!
Your spreadsheet has no physical or theoretical basis and it is just a set of calculations that forces to result that you want.

Of course if " :P works fine :P" and if you replace all occurrences of Cxx by ABS(Cxx) if " :P works perfectly :P" in the Southern Hemisphere as well.

It works simply because:
  • On the real earth, what ever its shape might be, at solar noon on an equinox the sun's elevation = 90° - abs(Latitude)[1].

  • You managed to drag up a set of calculations that forced the sun height to be (Pole to Equator distance)/2 when the sun's elevation = 90° - abs(Latitude).
    Real smart trick that!
My calculations for both the flat earth and the Globe are based on nothing more than light travelling in straight lines and simple, easily justified trigonometry.

Now, as I said before, if want to retain any credibility you will explain exactly the basis of your calculations.

[1] Funny that,
      maybe it's because the earth is really a Globe and the elevation angle of the sun is simply due to the curve of the earth's surface, as in:

Sun Height on Globe Earth along 70°W Long
      It's so simple, a child could see it!

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rabinoz

  • 24923
  • Real Earth Believer
Why bother when it's all done and set anyway?



At some point "Sly Sparkane" will wind up here and have his day ruined because his video has been decimated, destroyed, and obliterated by this very thread.  I just wonder if he'll post another video showing the angle generated in column S of the spreadsheet perfectly lining up with the sun as it were.  Nah, probably not.
Well, even though Sly Sparkane probably hasn't the slightest that you or this society even exist, you go ahead and
ruin my day by decimating, destroying, and obliterating that video by this very thread.
i think Sly Sparkane drags it out a bit, but it is based on simple measurements, made by real people and demonstrated by simple geometry,

And don't try your silly totally fabricated spreadsheet on it, unless you can justify your calculations.

And justify your unsubstantiated statements like:
"The sun never 'sets' on a flat earth. It goes out of view due to perspective. To calculate the height of the sun, you need to use the below math to take this fact into account."

"Column P shows the unit of your perspective, Column S shows the actual angle taking into account your perspective"

Perspective does not come into it! That's something you dreamed up you explain the unexplainable.


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NAZA

  • 594

Quote
The sun never 'sets' on a flat earth. It goes out of view due to perspective

I'll ask again but will probably be ignored again.
How does perspective explain this image?



Please explain the path the sunlight takes from the sun, that is supposed to be above the horizon, to the camera. 

For instance on a globe it is explained  like this:

Photons from the distant sun travel in a straight to the camera.  Some are blocked by the sailboat, hence the silhouette of the sailboat.  Some are blocked by the earth itself, hence the missing bottom of the sun.

What path does the sunlight take in your model?

Where and how does perspective  bend it? Obviously it must bend it behind the sailboat...



Why does perspective lower the position of the sun, but not the sailboat?

Thanks

Why bother when it's all done and set anyway?



At some point "Sly Sparkane" will wind up here and have his day ruined because his video has been decimated, destroyed, and obliterated by this very thread.  I just wonder if he'll post another video showing the angle generated in column S of the spreadsheet perfectly lining up with the sun as it were.  Nah, probably not.
   
His day will not be ruined.  You can use any of a dozen websites that give you the azimuth of the sun on any given day/time to verify the numbers in the video.  I used one to determine the dimensions of the awning on my deck in the backyard and they were exact.

I also verified the numbers from the video and they were all within ≈0.5°.  Since the data is verified, yours is the day ruined.  You can verify the elevation angles and see for yourself that your spreadsheet doesn’t match reality.

BTW, you seem to be ignoring my question about perspective.
 
Mike
Since it costs 1.82¢ to produce a penny, putting in your 2¢ if really worth 3.64¢.

Is that the best you can come up with?  I know you just about fell out of your chair running the numbers ;)  THIS CANNOT BE DENIED. 
Of course it can be denied!

You managed, either accidently or on purpose, ended up with an expression for sun height that is completely independent of the latitude.

Look!
Sun's height = (equator to pole dist) x (1 - Lat/90)/(2 x (1 - Lat/90))
or Sun's height = (equator to pole dist)/2
Of course the answer is going to always end up as 5000 km.

And you have never shown anything is wrong with my original methods, so try again!

The following table gives the data for distance of each location from Vaupes and the elevations of the sun at solar noon on 20/Mar/2016.
All sun elevations were obtained from Sun Earth Tools as close as possible to the local midday on the last equinox. The time was UTC 20/Mar/2016 16:48.


Location   

Latitude   

Longitude   

Sun Elev   
Dist from   
Vaupes   

Flat Sun Ht   
Lat Diff from   
Vaupes   
Calc
Circum
Kimmirut, Canada   
62.847°   
-69.869°   
27.36°   
7,034 km   
3,609 km   
63.58°   
39,828 km
Santo Domingo   
18.486°   
-69.931°   
71.72°   
2,107 km   
6,077 km   
19.22°   
39,465 km
Vaupes, Colombia   
-0.565°   
-69.634°   
89.06°   
0 km   
------   
   
   
Chupa District, Peru   
-15.109°   
-69.998°   
74.69°   
1,610 km   
6,256 km   
14.37°   
40,334 km
Punta Arenas, Chile   
-53.164°   
-70.917°   
36.63°   
5,830 km   
4,388 km   
52.43°   
40,031 km

These locations and the directions to the sun on a flat earth are shown in the left hand  diagram below:
Once we have the angles from two sites the height of the sun can be calculated from: h = d/(cot(A1) + cot(A2)).


Sun Height on Flat Earth along 70°W Long
   

Sun Height on Globe Earth along 70°W Long

Using this method to find the height of the sun on the Flat earth gives measurements from 3609 km (for Kimmirut and Vaupes) to 6256 km (for Chupa District to Vaupes) depending on the spacing of the measurement sites.

In other words, claiming that the Flat Earth sun is at about 5,000 km altitude has no foundation whatever.

It is very telling when we note that when we plot these angles on a spherical earth the directions to the sun are all parallel.
Explain that!

Now, if instead of using these measurements to determine the Flat Earth sun height, we use them as Eratosthenes did, assuming a distant sun and use this data to calculate the circumference of the earth.

The circumference can be calculated from (distance from Vaupes) * 360°/(angle difference of sun from Vaupes)

This time, we get far better consistency.
              The estimated figures for the circumference of the earth range from 39,465 km to 40,334 km.
Certainly these figures would indicate that the earth is a globe with a distant sun.

And don't bring up the 89.06° again, that was used correctly in the calculations.

Rab, can you use his excell sheet to calculate observations based on your figures for the above?

Or do you mean his equations simply remove latitude from calculations. If it does, kindly show where that happens.

*

rabinoz

  • 24923
  • Real Earth Believer
The following table gives the data for distance of each location from Vaupes and the elevations of the sun at solar noon on 20/Mar/2016.
All sun elevations were obtained from Sun Earth Tools as close as possible to the local midday on the last equinox. The time was UTC 20/Mar/2016 16:48.


Location   

Latitude   

Longitude   

Sun Elev   
Dist from   
Vaupes   

Flat Sun Ht   
Lat Diff from   
Vaupes   
Calc
Circum
Kimmirut, Canada   
62.847°   
-69.869°   
27.36°   
7,034 km   
3,609 km   
63.58°   
39,828 km
Santo Domingo   
18.486°   
-69.931°   
71.72°   
2,107 km   
6,077 km   
19.22°   
39,465 km
Vaupes, Colombia   
-0.565°   
-69.634°   
89.06°   
0 km   
------   
   
   
Chupa District, Peru   
-15.109°   
-69.998°   
74.69°   
1,610 km   
6,256 km   
14.37°   
40,334 km
Punta Arenas, Chile   
-53.164°   
-70.917°   
36.63°   
5,830 km   
4,388 km   
52.43°   
40,031 km
Rab, can you use his excel sheet to calculate observations based on your figures for the above?

Or do you mean his equations simply remove latitude from calculations. If it does, kindly show where that happens.
Silicon's equations do not quite remove latitude from calculations.

Ideally on the Globe at solar noon on either equinox the sun's elevation is 90° - abs(Lat).
If the elevation is exactly this ideal value, Silicon's calculations do make the sun's height always exactly 5,000 km.

In practice, the actual elevations deviate slightly from the ideal value because of refraction (a very minor effect) any the local solar noon not being exactly at the precise time of the equinox.

Silicon's equations are calculated from the actual elevations, which differ from the ideal values by only fractions of a degree.
The trouble is that, from what he has told us, there is no physical or theoretical basis for his calculations.
On the other hand mine (and JackBlack's) are simply the height calculated from the elevation from two points using simple trigonometry.

Here is the same table with Silicon's calculation in the right column.

Location   

Latitude   

Longitude   

Sun Elev   
Dist from   
Vaupes   

Flat Sun Ht   
Lat Diff from   
Vaupes   
Silicon's
Sun Height
Kimmirut, Canada   
62.847°   
-69.869°   
27.36°   
7,034 km   
3,609 km   
63.58°   
4,974 km
Santo Domingo   
18.486°   
-69.931°   
71.72°   
2,107 km   
6,077 km   
19.22°   
4,997 km
Vaupes, Colombia   
-0.565°   
-69.634°   
89.06°   
0 km   
------   
   
5,000 km
Chupa District, Peru   
-15.109°   
-69.998°   
74.69°   
1,610 km   
6,256 km   
14.37°   
5,002 km
Punta Arenas, Chile   
-53.164°   
-70.917°   
36.63°   
5,830 km   
4,388 km   
52.43°   
5,017 km

The sun never 'sets' on a flat earth. It goes out of view due to perspective. To calculate the height of the sun, you need to use the below math to take this fact into account.
No, Perspective wont make things appear to drop below the horizon.
It makes them get closer to the horizon, and appear smaller.

it is also completely impossible to calculate the height of something given only a single angle (or even 2).
It is effectively trying to solve a right angle triangle with just a single angle (other than the right angle).
This will give you the shape, but not the scale.
It will tell you the sun is somewhere along a line, but not how far along it.

The sun isn't special. If the formula works on the sun, then it should work for everything else with similar circumstances.
There is no condition here that the sun is so far away that paralax does not matter like there is for the RE case, so this should work for anything.
If I want to know the height of a building, I just note the angle to the top and then stick it in the formula and find that out.
But that shows a tiny building to be 5000 km high.
That can't be right.

Same for a fence, it says it is 5000 km high, but I can easily climb over it in a few seconds.

Something tells me your math is pure BS.

And as per usual, you just provide a bunch of math, with no justification at all, so there is absolutely no basis for this being based upon a flat Earth.

I'll work backwards from your number, and I'll ditch the row part of the cell reference, and I will ditch the conversion between degrees and radians (i.e. the trig functions will use degrees, not radians). I will also use Z=10000, 4Z=40000, Y=90, 4Y=360, to make some of the writing out easier:
So you have the height given by:
T=TAN(S)*E

But S is simply given by:
S=ATAN(R)

As you are limited between 0 and 90 degrees, then tan(atan(x))=x.

This means you effectively have:
T=R*E
with one layer of extraneous BS removed.

Then subbing in
R=P/Q
E=(40000/360)*(90-C)=(4Z/4Y)*(Y-C)=(Z/Y)*(Y-C)
Thus:
T=(P/Q)*((Z/Y)*(Y-C))

Subbing in
P=J+N
Q=D*0.5
T=((J+N)/(D*0.5))*((Z/Y)*(Y-C))
T=2*(J+N)*(Z/Y)*(Y-C)/D

Then subbing in:
D=90-C=Y-C
T=2*(J+N)*(Z/Y)*(Y-C)/(Y-C)
T=2*(J+N)*(Z/Y)

So there goes a lot more of your extraneous BS.

Now, I'll just try and quickly work with (J+N) separately.
J+N=H+I+L+M

H=(E/4Z)*C
I=(E/4Z)*D
L=(F/4Z)*C
M=(F/4Z)*D

Then remembering from before that D=Y-C, this simplifies to:
H=(E/4Z)*C
I=(E/4Z)*(Y-C)
L=(F/4Z)*C
M=(F/4Z)*(Y-C)

And with that I can already see another layer being removed:
H+I=(E/4Z)*C+(E/4Z)*(Y-C)=(E/4Z)*(C+Y-C)=(E/4Z)*Y
Similarly:
L+M=(F/4Z)*Y
So J+N=(E/4Z)*Y+(F/4Z)*Y
J+N=(Y/4Z)*(E+F)

So subbing that back into T:
T=2*(J+N)*(Z/Y)
T=(2/4)*(E+F)=(E+F)/2

So that is now getting much simpler.

But what about E and F:
E=(40000/360)*(90-C)=(4Z/4Y)*(Y-C)=(Z/Y)*(Y-C)
F=(40000/360)*(C)=(4Z/4Y)*(C)=(Z/Y)*(C)

So:
E+F=(Z/Y)*(Y-C)+(Z/Y)*(C)
E+F=(Z/Y)*(Y-C+C)=(Z/Y)*Y
E+F=Z

Now subbing that back into T:
T=(E+F)/2
T=Z/2

Much simpler.
So why not do this as your formula to calculate the height of the sun based upon the angle:
Height=5000 km.
No need for any extraneous BS.

Or if you would like to pretend it is there, how about this one:
5000 km * (1+angle-angle).

It is effectively the same thing, your way is filled with more convoluted BS.
You know the height you want, and are simply wiping out the angle from the equation.


Why do you insist on making it so convoluted?
Is it so you can pretend that it might actually be complex math based upon reality, rather than BSing numbers?
Do you work in a sideshow somewhere where you "read people's minds" by asking them to pick a number then continually manipulating it (without them easily seeing the pattern) to then reveal what the final number is, even though there could only be one possibility for the final number?


So where is the FE based math?
All I see there is height=5000 km, a baseless assertion.

We can also experiment with the value of Z and get whatever height we want.

Column E & Column F shows the distance to the 90 Degree Sun
Column P shows the unit of your perspective
Column S shows the actual angle taking into account your perspective
Coulmn T shows the flat earth sun height
Column E shows the distance to the sub-solar point, based upon RE math.
Column F shows the distance to the closest point 90 degrees away from the sub-solar point, again based upon RE math.
Column P is 22.5, it doesn't show any magic "unit of perspective".
Column S shows the angle that the sun should be at if the sun was actually 5000 km above Earth, rather than the observed angle in column C. This is the angle you would expect due to the effect of perspective, i.e. this is the angle that you should be observing the sun at. The sun not being at this angle shows the FE model to be BS.
Column T shows your asserted, and unjustified height of 5000 km.
« Last Edit: December 29, 2017, 11:15:24 PM by JackBlack »

do you really want me to air the 'dirty laundry' in regards to what I've found about this?
You mean with you repeatedly spouting pure bullshit and repeatedly getting your ass handed to you?

Yes the formula is simple.  That's what lets you know its real. Your oversimplification obviously can not work.
No, the formula you provided is not simple.
The formula you provided is overly convoluted to pretend it might match reality and actually do something with the angle that is input.
In reality, that extraneous BS can be removed to provide what Rab has provided.
It is not an oversimplification. But it does show your dishonesty. You have simply fixed the sun's height to be 5000 km, or more technically 40000 km/8.

In reality, assuming a flat Earth, the formula to determine the sun's height from 2 angles of elevation, and the distance between 2 locations (where the sub-solar point and the 2 locations are in a straight line with both locations on the same side of the sub-solar point) is quite simple:
h=tan(a1)*tan(a2)*d/(tan(a1)-tan(a2)), noting that a1 is the angle of elevation for the location closer to the sun.

The problem for you is that it shows the FE model is wrong because it produces widely different heights for the sun.

Is that the best you can come up with?  I know you just about fell out of your chair running the numbers ;)  THIS CANNOT BE DENIED.
You have baselessly asserted that the height of the sun is 5000 km with no justification at all. That can quite easily be rejected.

I believe what they've done is just built the globe model around numbers based on the sun.
Yes, by first working out Earth is round and the sun is very far away then using the angle difference between 2 locations by using the sun as a measuring tool to determine the circumference of Earth.

It's not 10,000km as the distance to the equator, but rather the radius of the spotlight sun.
Except the radius of the spotlight sun should have no bearing on its height.
You can have a spotlight mounted 50 m away from a stage and still only have a 1 m wide circle on the stage.

I suppose at the end of the day its all a matter of what you believe.
No, it is a matter of which has justification.
The RE model doesn, your crap does not.

but in my opinion the spreadsheet I presented doubles up, using different math (see above post) which leaves zero doubt to the fact that the sun is 5,000km above the earth.
And like so often, your opinion is pure bullshit.
You haven't doubled up and used different math.
You have simply made an overly convoluted formula which discards the input value and outputs 5000 km.
That leaves just as much doubt as simply asserting the sun is 5000 km above Earth.
There is no justification at all and thus no reason to remove any doubt.

Unfortunately (or fortunately) there is nothing like the sun that we know of to use for an apples to apples comparison.
Sure there is.
For reality, i.e. the RE model, it will depend upon what you want to do for a comparison.
For this discussion, you can use a small ball and the sun itself. Or taking the place of the sun you can simply use a distant light source.
The reason it is so hard to find a simple model for an apples to apples comparison is the vast distance to the sun compared to Earth.

But for a FE it is easy.
Get a table that is 2 m in diameter and put a small light 50 cm above it. That is a scaled down FE model.
Then try moving the sun around to various positions, or move an observation point (using shadows cast by a stick can be good) to see what should happen if the FE model was reality.

This is because scaling the entire system doesn't affect what perspective does at all.
Something located 5000 km above (measured from the height of your eye/observation device) a point 10000 km away will appear in the same position as an object located 50 cm above a point 1 m away.

So apples to apples comparisons for the FE is extremely simple. The problem for you is it just doesn't match reality.

Why can't the moon just be passing in front of the sun for a solar eclipse on the FE?
This has been explained to you repeatedly.
We get both annular solar eclipses and total solar eclipses that are larger than the moon in the FE model.
In order to get annular solar eclipses the moon needs to be smaller than the sun. In order to get total solar eclipses that are larger than the moon, the moon needs to be larger than the sun.

As far as lunar, in my opinion it's equal on both sides for things that don't add up.
And once again, your opinion is pure BS.
What are some of these things you think "don't add up" for the lunar eclipse in a RE model?
Are you going to bring up your lies about it allegedly being impossible to see both the eclipsed moon and the sun, where you ignore refraction and observer height yet again?

Too bad we'll just get the globe earth narrative, and not explore what is really happening.
No, no narrative, explanations of how things work which you dismiss, and don't want to discuss.


I call these light tricks.  Haven't you seen clouds lit from below and above at the same time in the same view?
Yes, because the sun is larger than the clouds, and thus is capable of illuminating both sides at once.
However this only works if the sun is at the same "height" as the clouds.
But there are plenty of times when the sun is just lit from below, not above. That requires the sun to be BELOW the clouds.


Well don't you think its ironic then, that we can calculate the height of the flat earth sun but cannot do the same thing on the globe model?
No you can't.
We have shown this repeatedly.
You can baselessly assert the height is 5000 km, or you can make measurements and do the math and get wildly varying heights.
So no, you cannot calculate the height of the sun above a flat Earth, at least not in any meaningful consistent way.

And no, your math in the OP is not calculating the height of the sun, it is baselessly asserting it as 5000.

It has also been explained repeatedly why you can't do the same on a RE.
Due to just how far away the sun is you cannot measure the distance to the required precision.
However there are plenty of simple observations which show the sun needs to be very far away (relative to the sun of Earth anyway).

*

Macarios

  • 1896
No, Perspective wont make things appear to drop below the horizon.
It makes them get closer to the horizon, and appear smaller.

it is also completely impossible to calculate the height of something given only a single angle (or even 2).
It is effectively trying to solve a right angle triangle with just a single angle (other than the right angle).
This will give you the shape, but not the scale.
It will tell you the sun is somewhere along a line, but not how far along it.

If Sun is above Equator, and you know how far Equator is, then you not only know the angle, but also the side of the right triangle.

The problem is, measured from diferent distances from Equator, in FE model you can't get consistent results. In GE model everything fits.

Durham, NC, equinox, solar noon.
Sun angle 54.001 degrees.
Distance to Equator 4029km.
Easy to calculate height of the Sun above Equator: 4029 * tan(36) = 5546 km

But at the same time from Toronto the Sun angle is 46.33 degrees.
Distance from Toronto toe Equator is 4876 km.
It means Sun above ground is 4876 * tan(46.33) = 5107 km

Doesn't differnce of 439 km look a bit too much of a discrepancy?
Can Sun be at both heights at once?

There is third distance:
Hall Beach, Canada.
Sun angle 21.23 degrees.
Distance to Equator 7667 km.
Sun height above Equator: 7667 * tan(21.23) = 2978 km.

How big is discrepancy now?

EDIT: From Miami, FL the Sun is 64.17 degrees up, and from Miami to Equator is 2874 km.
2874 * tan(64.17) = 5937 km.
« Last Edit: December 30, 2017, 03:05:56 AM by Macarios »
I don't have to fight about anything.
These things are not about me.
When one points facts out, they speak for themselves.
The main goal in all that is simplicity.

*

EvolvedMantisShrimp

  • 914
  • Physical Comedian
Are you using the circumference of a globe (40000 km) to calculate the sun in relation to a flat earth?

Mike

I believe what they've done is just built the globe model around numbers based on the sun.  111.111km is not actually the distance per degree of lat, on a globe earth but rather 111.111km per degree of elevation of the sun.  It's not 10,000km as the distance to the equator, but rather the radius of the spotlight sun.

In other words, if I divide 8 by 2 or 40 by 10, do we not arrive at the same answer?

I suppose at the end of the day its all a matter of what you believe.

No, it isn't.

How could a round earth model base it's measurements on the shape of the spotlight when the spotlight has no defined shape? On the Winter Solstice(Dec 21st), the Sun can rise from that Southeast in Tierra Del Fuego and still be a couple hours from setting in the Southwest in Perth. Notice I said SOUTHeast and SOUTHwest. Meanwhile, the Sun can be seen from anywhere in Antarctica at the same time! What is the shape of this spotlight?

When you can create a mathematical model that can account for the position and height of the Sun on the Winter Solstice from Tierra Del Fuego at sunrise, let me know.
« Last Edit: December 30, 2017, 04:13:57 AM by EvolvedMantisShrimp »
Nullius in Verba

*

rabinoz

  • 24923
  • Real Earth Believer
Are you using the circumference of a globe (40000 km) to calculate the sun in relation to a flat earth?

Mike

I believe what they've done is just built the globe model around numbers based on the sun.  111.111km is not actually the distance per degree of lat, on a globe earth but rather 111.111km per degree of elevation of the sun.  It's not 10,000km as the distance to the equator, but rather the radius of the spotlight sun.

Rubbish! Even "the Wiki" has essentially the same information,
Quote from: The Flat Earth Society 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.
From: Finding your Latitude and Longitude


Look at, "The angle of the sun past zenith is our latitude." and
"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 69.5 miles per degree is 111.8 km/° - the 111.1 km/° is a bit more accurate.
Then your claim of "It's not 10,000km as the distance to the equator" is also rubbish, again the Wiki says that
Quote from: The Flat Earth Society 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. Along the edge of our local area exists a massive 150 foot Ice Wall.
And this 24,900 miles is 40073 km, close enough to 40,000 km - making the distance from the North Pole to the equator 10,000 km.

*

EvolvedMantisShrimp

  • 914
  • Physical Comedian
Are you using the circumference of a globe (40000 km) to calculate the sun in relation to a flat earth?

Mike

I believe what they've done is just built the globe model around numbers based on the sun.  111.111km is not actually the distance per degree of lat, on a globe earth but rather 111.111km per degree of elevation of the sun.  It's not 10,000km as the distance to the equator, but rather the radius of the spotlight sun.

Rubbish! Even "the Wiki" has essentially the same information,
Quote from: The Flat Earth Society 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.
From: Finding your Latitude and Longitude


Look at, "The angle of the sun past zenith is our latitude." and
"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 69.5 miles per degree is 111.8 km/° - the 111.1 km/° is a bit more accurate.
Then your claim of "It's not 10,000km as the distance to the equator" is also rubbish, again the Wiki says that
Quote from: The Flat Earth Society 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. Along the edge of our local area exists a massive 150 foot Ice Wall.
And this 24,900 miles is 40073 km, close enough to 40,000 km - making the distance from the North Pole to the equator 10,000 km.

Isn't weird that the 'spotlight' on the equinox is almost exactly a semicircle that illuminates almost exactly half the planet? What are the odds?
Nullius in Verba

The sun never 'sets' on a flat earth. It goes out of view due to perspective. To calculate the height of the sun, you need to use the below math to take this fact into account.
No, Perspective wont make things appear to drop below the horizon.
It makes them get closer to the horizon, and appear smaller.
How do you know what a 32 mile wide object circling 5000km above the earth will look like exactly?

it is also completely impossible to calculate the height of something given only a single angle (or even 2).
It is effectively trying to solve a right angle triangle with just a single angle (other than the right angle).
This will give you the shape, but not the scale.
It will tell you the sun is somewhere along a line, but not how far along it.
Just stop. There are other numbers involved.

The sun isn't special. If the formula works on the sun, then it should work for everything else with similar circumstances.
There is no condition here that the sun is so far away that paralax does not matter like there is for the RE case, so this should work for anything.
If I want to know the height of a building, I just note the angle to the top and then stick it in the formula and find that out.
But that shows a tiny building to be 5000 km high.
That can't be right.

Same for a fence, it says it is 5000 km high, but I can easily climb over it in a few seconds.

Something tells me your math is pure BS.


Something tells me you really didn't think this one through.  Just typing as fast as your fingers can go!  Or maybe someone in your shill team isn't quite up to the task.

And as per usual, you just provide a bunch of math, with no justification at all, so there is absolutely no basis for this being based upon a flat Earth.


As posted elsewhere, I could post further details in believers section.

I'll work backwards from your number, and I'll ditch the row part of the cell reference, and I will ditch the conversion between degrees and radians (i.e. the trig functions will use degrees, not radians). I will also use Z=10000, 4Z=40000, Y=90, 4Y=360, to make some of the writing out easier:
So you have the height given by:
T=TAN(S)*E

But S is simply given by:
S=ATAN(R)

As you are limited between 0 and 90 degrees, then tan(atan(x))=x.

This means you effectively have:
T=R*E
with one layer of extraneous BS removed.

Then subbing in
R=P/Q
E=(40000/360)*(90-C)=(4Z/4Y)*(Y-C)=(Z/Y)*(Y-C)
Thus:
T=(P/Q)*((Z/Y)*(Y-C))

Subbing in
P=J+N
Q=D*0.5
T=((J+N)/(D*0.5))*((Z/Y)*(Y-C))
T=2*(J+N)*(Z/Y)*(Y-C)/D

Then subbing in:
D=90-C=Y-C
T=2*(J+N)*(Z/Y)*(Y-C)/(Y-C)
T=2*(J+N)*(Z/Y)

So there goes a lot more of your extraneous BS.

Now, I'll just try and quickly work with (J+N) separately.
J+N=H+I+L+M

H=(E/4Z)*C
I=(E/4Z)*D
L=(F/4Z)*C
M=(F/4Z)*D

Then remembering from before that D=Y-C, this simplifies to:
H=(E/4Z)*C
I=(E/4Z)*(Y-C)
L=(F/4Z)*C
M=(F/4Z)*(Y-C)

And with that I can already see another layer being removed:
H+I=(E/4Z)*C+(E/4Z)*(Y-C)=(E/4Z)*(C+Y-C)=(E/4Z)*Y
Similarly:
L+M=(F/4Z)*Y
So J+N=(E/4Z)*Y+(F/4Z)*Y
J+N=(Y/4Z)*(E+F)

So subbing that back into T:
T=2*(J+N)*(Z/Y)
T=(2/4)*(E+F)=(E+F)/2

So that is now getting much simpler.

But what about E and F:
E=(40000/360)*(90-C)=(4Z/4Y)*(Y-C)=(Z/Y)*(Y-C)
F=(40000/360)*(C)=(4Z/4Y)*(C)=(Z/Y)*(C)

So:
E+F=(Z/Y)*(Y-C)+(Z/Y)*(C)
E+F=(Z/Y)*(Y-C+C)=(Z/Y)*Y
E+F=Z

Now subbing that back into T:
T=(E+F)/2
T=Z/2

Much simpler.
So why not do this as your formula to calculate the height of the sun based upon the angle:
Height=5000 km.
No need for any extraneous BS.

Or if you would like to pretend it is there, how about this one:
5000 km * (1+angle-angle).

It is effectively the same thing, your way is filled with more convoluted BS.
You know the height you want, and are simply wiping out the angle from the equation.


Why do you insist on making it so convoluted?
Is it so you can pretend that it might actually be complex math based upon reality, rather than BSing numbers?
Do you work in a sideshow somewhere where you "read people's minds" by asking them to pick a number then continually manipulating it (without them easily seeing the pattern) to then reveal what the final number is, even though there could only be one possibility for the final number?


So where is the FE based math?
All I see there is height=5000 km, a baseless assertion.

It's an identity right? 5000 = 5000.  Ok did you know the math for the spherical earth is deducted using exact same thing?  I'll use the supposed circumference since we can't do any calculation to see how far away the sun is on the globe earth model. So 40,075 = 40,075! 

Your globe earth number is simply an identity!

Are you using the circumference of a globe (40000 km) to calculate the sun in relation to a flat earth?

Mike

I believe what they've done is just built the globe model around numbers based on the sun.  111.111km is not actually the distance per degree of lat, on a globe earth but rather 111.111km per degree of elevation of the sun.  It's not 10,000km as the distance to the equator, but rather the radius of the spotlight sun.

In other words, if I divide 8 by 2 or 40 by 10, do we not arrive at the same answer?

I suppose at the end of the day its all a matter of what you believe.

No, it isn't.

How could a round earth model base it's measurements on the shape of the spotlight when the spotlight has no defined shape? On the Winter Solstice(Dec 21st), the Sun can rise from that Southeast in Tierra Del Fuego and still be a couple hours from setting in the Southwest in Perth. Notice I said SOUTHeast and SOUTHwest. Meanwhile, the Sun can be seen from anywhere in Antarctica at the same time! What is the shape of this spotlight?

When you can create a mathematical model that can account for the position and height of the Sun on the Winter Solstice from Tierra Del Fuego at sunrise, let me know.

I could probably do it