Celestial sphere

Come on admins, you have seen the OP now, but you haven't answered, this proves RE right 

Yes you would, read the link in the OP. And learn geometry.
At the north poll you could see all stars that are north of the planet, any that are south would be blocked by the planet.
Edit: Wow, for trying to get the admins to admit they are wrong they are threatening to ban me.
Morton's demon much!!!
I have conclusive prove that the earth is round, it's relevant to the whole site.

It seems that you are the one who needs to learn Geometry.  The OP said this. 

The stars are mapped onto the sphere. You can only see the stars in the half of the sphere centred around you. No more, no less (excluding light pollution).

Well, it looks like I need to draw a picture to show that your statement is wrong.  In the picture below, a person standing at point A will see 180 degrees from horizon to horizon and the person at point B will as well.  They can not see each other's stars; however, they would also not be able to see the stars directly above the person at point C.  Therefore, your statement was wrong to begin with.  They would not expect to see exactly half of the stars.

Come on admins, you have seen the OP now, but you haven't answered, this proves RE right 

I think that I just proved that you roundies have no spatial awareness. 

Consider two planes parallel to the ground at A and B, the only stars not visible to A or B are between the two planes. If you were to say all the stars were 4 light years away, the fraction if the stars between the two planes is negligible. Most stars are futher than 4 light years, so even less will be between the two planes.
So it is very close to half, more stars will be blocked by local geology than be between the two planes. Ill calculate the exact percentage later.

Negligible is a subjective word, but it does not mean the same thing as exact. 

0. 00000005 percent of the stars are not viable to A or B if they are all 4 light years away, thats what I meant by negligible.

If your math skills are anything close to your buddy EternalHoid's geometry skills, then I am sure those numbers are spot on.    lol

The formula is sin-1 (radius of earth/distance of stars)*2 to get the fraction of the sky both A and B  can't see. Simple trigonometry.

I think you might have some kind of delusional problems, like Luke 22:35-38 when he claimed in post 24 that the hours get shorter in Australia during the winter. 

I think you might have some kind of delusional problems, like Luke 22:35-38 when he claimed in post 24 that the hours get shorter in Australia during the winter. 

Fewer hours of daylight, as you know.

I think you might have some kind of delusional problems, like Luke 22:35-38 when he claimed in post 24 that the hours get shorter in Australia during the winter. 

What delusion do I have, that trig works?

My cut on this is - it is very simple.

The axis of the Earth is from the N.Pole to the S.Pole. The extensions of this on the Celestial Sphere are the N./S. Celestial Poles (NCP/SCP, the center of star trail pics). On a flat disk, the true S.Pole is directly below the N.Pole on the underside of the disk. Therefore, the SCP can not be viewed from above the disk. PERIOD.

Here is an impossible FE picture of the SCP taken from the Harker Glacier on the island of South Georgia in the S. Atlantic (lat -54.3667)(http://news.nationalgeographic.com/news/2014/03/140319-antarctica-big-bang-inflation-telescope-south-pole-astronomy/)

Sorry I got the equation wrong, it gives you the angle of the sky that can't be seen by A or B, not the fraction,
sin-1 (radius of earth/distance of stars)*2 gives you the angle
sin-1 (radius of earth/distance of stars)/180 gives you the fraction of the sky not visable to A or B
0.000007% of the stars aren't visable to A or B.
https://drive.google.com/open?id=0By3xuzrccYkqV0NsamZVVVo1UVU
R is the radius of the planet, S is the distance to the stars, the green bit is the angle arcsin(R/S)
Here I have set S to 4 lightyears, distance of the nearest star.
49.9999965% of the sky is close enough to half. It is not noticeable.
Still no answer on why the celestial sphere map works on a flat earth.
And there never will be, because FE=WRONG.

Because FEer are lazy I better make my point the last post again so they will notice it.

The celestial sphere is a map of the stars; you can take half of the sphere based on your position and time of night, and use it to get an exact map of what you can see in the sky, still no answer on how the hell this works on a flat earth. Come on guys, if you have no answer to this then the earth can’t be flat. Game over.

I was going to stay keep on topic, but the FEer are never going to answer this, its impossible.
But who is this Jack that edited the third post, why didn't he reply to the topic.

A war metaphor doesn't really work though. This thread is like a nuke to their theory, but they can just ignore it through the power of stupidity, in a war ignoring something doesn't work.

Quote from: EternalHoid on December 28, 2015, 03:13:18 AM

Yes you would, read the link in the OP. And learn geometry.
At the north poll you could see all stars that are north of the planet, any that are south would be blocked by the planet.
Edit: Wow, for trying to get the admins to admit they are wrong they are threatening to ban me.
Morton's demon much!!!
I have conclusive prove that the earth is round, it's relevant to the whole site.

It seems that you are the one who needs to learn Geometry.  The OP said this. 

Quote from: EternalHoid on December 26, 2015, 03:31:05 PM

The stars are mapped onto the sphere. You can only see the stars in the half of the sphere centred around you. No more, no less (excluding light pollution).

Well, it looks like I need to draw a picture to show that your statement is wrong.  In the picture below, a person standing at point A will see 180 degrees from horizon to horizon and the person at point B will as well.  They can not see each other's stars; however, they would also not be able to see the stars directly above the person at point C.  Therefore, your statement was wrong to begin with.  They would not expect to see exactly half of the stars.

This is the correct diagram



stop thinking in 2D

Green Observation from North pole
Red Observable from Equator
Blue Observable from south pole

Obviosly seasons and tilt of the axis come into play

Your drawing is just lines placed where ever you want them to go, even if they go through the Earth.  That is mighty deceitful of you.  Mine is an actual 2D depiction.  I can create a real 3D rendering of what the feilds of view would be, if you want for me to do so. 

Your drawing is just lines placed where ever you want them to go, even if they go through the Earth.  That is mighty deceitful of you.  Mine is an actual 2D depiction.  I can create a real 3D rendering of what the feilds of view would be, if you want for me to do so.

I'm here to admit i'm wrong...... at best you can see 184 degrees due to refraction at sea.


Hey flatties , I can admit when i'm wrong, can you. Keep it real

That is mighty big of you to say.  We should all be as honest as you. 

Well, it looks like I need to draw a picture to show that your statement is wrong.  In the picture below, a person standing at point A will see 180 degrees from horizon to horizon and the person at point B will as well.  They can not see each other's stars; however, they would also not be able to see the stars directly above the person at point C.  Therefore, your statement was wrong to begin with.  They would not expect to see exactly half of the stars.

I suppose if the people at A and B were lying down with their eyes as close to the ground as possible, and the stars directly above the person at C are actually smaller than Earth. 

According to RET however, a person who's eyes are at 5-6 feet will have a line of sight at a slight downward angle to the horizon.  That gap between A and B would eventually shrink to nothing.  Also, stars (in RET, and excluding neutron stars, etc) are bigger than Earth, so any star directly above C, will still be within the line of sight of A or B.

According to RET however, a person who's eyes are at 5-6 feet will have a line of sight at a slight downward angle to the horizon.  That gap between A and B would eventually shrink to nothing.  Also, stars (in RET, and excluding neutron stars, etc) are bigger than Earth, so any star directly above C, will still be within the line of sight of A or B.

The gap shrinks to nothing

6371002*5100/2=16,246,055,100m

away from both A and B. this is just 1/9 of the distance to the nearest star - the Sun.

So yes, both A and B see 50% of stars at night (ignoring ground obstacles).

Another failed argument defeated.

Your drawing is just lines placed where ever you want them to go, even if they go through the Earth.  That is mighty deceitful of you.  Mine is an actual 2D depiction.  I can create a real 3D rendering of what the feilds of view would be, if you want for me to do so.

Your 2D model is fine, but you have come to the wrong conclusion. For a star to not be visable, it would have to be between the two planes parrallel to the ground at A and B, stars are much larger than the earth so you couldn't  fit a star between the two planes. Combined with the fact that A or B can each see close to 50% of the sky at 4 lightyears away, means we see 50% of the sky.
Seeing 50% of the sky makes no sense on a FE model.

And a lot of stars are somehow in the south of the sky in both South America and Australia and Africa, that makes no sense if the earth was flat.

Here is a step by step prove that the earth isn't flat, just say which bit is wrong.
1. The working map of the stars is spherical.
2. The map can be easily verified, so you could easily show it if it was fake.
4. A spherical map of the stars wouldn't work in the southern hemisphere if there was no SCP (south celestial poll)
5. On a flat earth, there is no SCP, because there is no single place that is most south.
6.  Since the sphere map works, there is no SCP, so the earth isn't flat.
What point is wrong?
Any FEer that reads this and doesn't answer, is a coward and an idiot.

This question has been answered before:

http://www.theflatearthsociety.org/forum/index.php?topic=64009.msg1701825#msg1701825


As for the content of the last message...

http://www.theflatearthsociety.org/forum/index.php?topic=65330.msg1743837#msg1743837

This question has been answered before:

http://www.theflatearthsociety.org/forum/index.php?topic=64009.msg1701825#msg1701825

This does not answer anything, it only proves lack of understaing of basic geometry or your ignorance over the fact that that argument fails to disprove anything. And yet you are linking it here as an answer/proof.

brouwer, let us remember that you had no idea how to calculate the annular size of a planet.

http://www.theflatearthsociety.org/forum/index.php?topic=64777.msg1727677#msg1727677

You certainly are in no position to dare to criticize anybody: each time we met in the past, it was a total defeat for you, do not kid yourself.


My answer stands correct.

“The astronomers' theory of a globular Earth necessitates the conclusion that, if we travel south of the equator, to see the North Star is an impossibility. Yet it is well known this star has been seen by navigators when they have been more than 20 degrees south of the equator. This fact, like hundreds of other facts, puts the theory to shame, and gives us a proof that the Earth is not a globe.” -William Carpenter, “100 Proofs the Earth is Not a Globe”



"For instance, Ursa Major, very close to Polaris, can be seen from 90 degrees North latitude (the North Pole) all the way down to 30 degrees South latitude. The constellation Vulpecula can be seen from 90 degrees North latitude, all the way to 55 degrees South latitude. Taurus, Pisces and Leo can be seen from 90 degrees North all the way to 65 degrees South. Aquarius and Libra can be seen from 65 degrees North to 90 degrees South! The constellation Virgo is visible from 80 degrees North down to 80 degrees South, and Orion can be seen from 85 degrees North all the way to 75 degrees South latitude! An observer on a ball-Earth, regardless of any tilt or inclination, should not logically be able to see this far."



Please refer to my work done on the Tunguska explosion, seen all the way from London, Antwerp, Berlin, Stockholm, instantaneously...



The issue raised in the opening message of this thread USED THE WRONG MAP, the unipolar map which is wrong.

With the correct map in mind, the bipolar map, the one I have brought here to the FES (but nodoby wants to listen), the question is answered easily.

brouwer, let us remember that you had no idea how to calculate the annular size of a planet.

http://www.theflatearthsociety.org/forum/index.php?topic=64777.msg1727677#msg1727677

Let me remind you that I have already explained how to find a angular size of a planet without knowing its distance and its size. You seem to have no idea how to do it. All it takes is to understand triangle geometry and properties of similar triangles.

You certainly are in no position to dare to criticize anybody: each time we met in the past, it was a total defeat for you, do not kid yourself.

You seem to refer to non-existing things.

My answer stands correct.

“The astronomers' theory of a globular Earth necessitates the conclusion that, if we travel south of the equator, to see the North Star is an impossibility. Yet it is well known this star has been seen by navigators when they have been more than 20 degrees south of the equator. This fact, like hundreds of other facts, puts the theory to shame, and gives us a proof that the Earth is not a globe.” -William Carpenter, “100 Proofs the Earth is Not a Globe”



"For instance, Ursa Major, very close to Polaris, can be seen from 90 degrees North latitude (the North Pole) all the way down to 30 degrees South latitude. The constellation Vulpecula can be seen from 90 degrees North latitude, all the way to 55 degrees South latitude. Taurus, Pisces and Leo can be seen from 90 degrees North all the way to 65 degrees South. Aquarius and Libra can be seen from 65 degrees North to 90 degrees South! The constellation Virgo is visible from 80 degrees North down to 80 degrees South, and Orion can be seen from 85 degrees North all the way to 75 degrees South latitude! An observer on a ball-Earth, regardless of any tilt or inclination, should not logically be able to see this far."

If you believe these arguments are valid, then you do not understand the geometry. All those examples are easily explained under "ball-Earth".

Please refer to my work done on the Tunguska explosion, seen all the way from London, Antwerp, Berlin, Stockholm, instantaneously...

Why should I? This is not the topic of this discussion, neither was of any I took a part in.

The issue raised in the opening message of this thread USED THE WRONG MAP, the unipolar map which is wrong.

With the correct map in mind, the bipolar map, the one I have brought here to the FES (but nodoby wants to listen), the question is answered easily.

Your bipolar map is wrong too, but you will not want to listen. You see, you fall into the same hole.

brouwer, your scientific reach is way beyond your scientific grasp.

Let me remind you that I have already explained how to find a angular size of a planet without knowing its distance and its size.

But you haven't explained anything in fact.

Let us go back to your very argument.

http://www.theflatearthsociety.org/forum/index.php?topic=64777.0#.VoPks7Z961s


Using simple trigonometry we can find that, during the noon, where the Sun is above us, hence it is the closest and has the largest angular size, we have the relation d=104.131456R.
Same calculations lead us to d=107.7909868R for the case where the Sun is the furthest away, setting (or rising).

Then, you have a single equation and two unknowns: the very reason I reminded you to go back to the textbook on angular size/diameter calculations:

The angular sizes of objects show how much of the sky an object appears to cover. Angular size does not, however, say anything about the actual size of an object.

How large an object appears depends not only on its size, but also on its distance.


Then, all of a sudden, you dropped this kind of delusional argument on all of us:

In 6h, the Sun will move 9003.1632km away horizontally (1/4th of the circumference, adjusted to straight line - a diagonal of a square).

Using Pythagoras theorem we find that R=323.2912km, and so d=33653.7747km.

The path of the Sun is a curve, not a straight line.

You needed, at the very least, to draw some kind of a map, to try to explain your mathematical setting.


This is why one of the RE had the honesty to state:

You are correct. You can not determine size or distance from just the angular size of the sun. You need its actual diameter or its actual distance to calculate the other.

That is why your armchair calculations are wrong: you have an equation with two basic unknowns, for which you provided no explanation at all on how to obtain a proper value for at least one of them.


Moreover, your basic assumption that the Sun is a sphere (d+R) is wrong: I told you that you do not know basic RE science.


Let me remind you that indeed the sun has the shape of a disk.

Impossibility of a round Sun shape:

The atmospheric pressure of the sun, instead of being 27.47 times greater than the atmospheric pressure of the earth (as expected because of the gravitational pull of the large solar mass), is much smaller: the pressure there varies according to the layers of the atmosphere from one-tenth to one-thousandth of the barometric pressure on the earth; at the base of the reversing layer the pressure is 0.005 of the atmospheric pressure at sea level on the earth; in the sunspots, the pressure drops to one ten-thousandth of the pressure on the earth.

The pressure of light is sometimes referred to as to explain the low atmospheric pressure on the sun. At the surface of the sun, the pressure of light must be 2.75 milligrams per square centimeter; a cubic centimeter of one gram weight at the surface of the earth would weigh 27.47 grams at the surface of the sun. Thus the attraction by the solar mass is 10,000 times greater than the repulsion of the solar light. Recourse is taken to the supposition that if the pull and the pressure are calculated for very small masses, the pressure exceeds the pull, one acting in proportion to the surface, the other in proportion to the volume. But if this is so, why is the lowest pressure of the solar atmosphere observed over the sunspots where the light pressure is least?

Because of its swift rotation, the gaseous sun should have the latitudinal axis greater than the longitudinal, but it does not have it. The sun is one million times larger than the earth, and its day is but twenty-six times longer than the terrestrial day; the swiftness of its rotation at its equator is over 125 km. per minute; at the poles, the velocity approaches zero. Yet the solar disk is not oval but round: the majority of observers even find a small excess in the longitudinal axis of the sun. The planets act in the same manner as the rotation of the sun, imposing a latitudinal pull on the luminary.

Gravitation that acts in all directions equally leaves unexplained the spherical shape of the sun. As we saw in the preceding section, the gases of the solar atmosphere are not under a strong pressure, but under a very weak one. Therefore, the computation, according to which the ellipsoidity of the sun, that is lacking, should be slight, is not correct either. Since the gases are under a very low gravitational pressure, the centrifugal force of rotation must have formed quite a flat sun.

Near the polar regions of the sun, streamers of the corona are observed, which prolong still more the axial length of the sun.

If planets and satellites were once molten masses, as cosmological theories assume, they would not have been able to obtain a spherical form, especially those which do not rotate, as Mercury or the moon (with respect to its primary).



Solar Atmosph. Pressure as a Function of Depth (official science information)

Depth (km) % Light from this Depth Temperature (K) Pressure (bars)

0 99.5 4465 6.8 x 10-3
100 97 4780 1.7 x 10-2
200 89 5180 3.9 x 10-2
250 80 5455 5.8 x 10-2
300 64 5840 8.3 x 10-2
350 37 6420 1.2 x 10-1
375 18 6910 1.4 x 10-1
400 4 7610 1.6 x 10-1

This table indicates that the solar atmosphere changes from being almost completely transparent to being almost opaque over a distance of about 400 km. Notice also that in this region the temperature drops rapidly as we near the surface, and that the pressure (measured in bars, where one bar is the average atmospheric pressure at the surface of the Earth) is very low - generally 1% or less of Earth surface atmospheric pressure.

Let me make it easier for you.

If you claim that you have discovered A NEW WAY to calculate the distance to the Sun and its diameter, based strictly on its angular size and similar triangles, then you must provide much more information on the subject.

Using simple trigonometry we can find that, during the noon, where the Sun is above us, hence it is the closest and has the largest angular size, we have the relation d=104.131456R.
Same calculations lead us to d=107.7909868R for the case where the Sun is the furthest away, setting (or rising).


According to which simple trigonometry? Where are the actual calculations? Your bibliographical references?

Are those actual upper bounds? Is it possible to obtain even larger values?


Let me give an example.

The angular size of the Moon has an average value of 31'27". Lower bound: 29'23". Upper bound: 33'32". However, if the Moon is overhead (observer at equator at noon), the value can increase to 34'9".


So, you need to provide us with these calculations/bibliographical references.


This triangle is formed by the observator point, the Sun at noon and at set/rise. From Pythagoras we get R, and then d.

Really? Are your measurements so precise as to actually assume you have a right triangle? What if you are off by a single degree? Then you will no longer have a right triangle, so you will need to apply the law of cosines.


So, you have to prove that you have A RIGHT TRIANGLE to start with.


In 6h, the Sun will move 9003.1632km away horizontally (1/4th of the circumference, adjusted to straight line - a diagonal of a square).

Where is your diagram? You think that if it was this simple, somebody else would not have thought of it by now (to dispense with the requirement that you need either the distance to the Sun or its diameter to get anything done)?

The path of the Sun is a curve, not a straight line. To "adjust" to straight line would mean to have the actual value of the curve itself, and then calculate the value of the straight segment that joins the two points on the curve itself. Again, you are assuming a square, when I have just reminded you that your triangle cannot be a right triangle, and you will have to use the law of cosines to proceed further.