"Equator" problem

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Re: "Equator" problem
« Reply #240 on: November 06, 2014, 07:36:41 AM »



I love when crazies start their posting in a seemingly intelligent, if misinformed, manner, and once instructed on their mistaken information, they devolve into full on pant-on-head crazy.

I've noticed this very well in my short time here so far and frankly I am disappointed. I thought that when confronted with a true challenge to their beliefs they would respond in a more rigorous fashion involving some form of data rather than the childish. "I'm right and you're wrong, that's all there is to it" (complete with tin foil hat and poutty face)

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cikljamas

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Re: "Equator" problem
« Reply #241 on: November 06, 2014, 10:10:24 AM »
Mr. Laing  tells us:

" The conclusions of science are irresistible, and old forms of faith, however venerable and however endeared by a thousand associations, have no more chance in a collision with science than George Stephenson's cow had, if it stood on the rails and tried to stop the progress of a locomotive."

From purely practical data we have already seen that "the conclusions of science" are as unreasonable and fallacious as it is possible for the human mind to conceive. A mixture of infidel superstitions and gross absurdities constitute the most of present-day science respecting the world we live on. Its relation to truth is as darkness to light. Science has as much chance in a collision with TRUTH as a rotten ship would have in a collision with an ironclad.

Even professedly Christian people are hoodwinked and befogged by modern hypothetical science. A. Giberne in " Sun, Moon and Stars," says, when speaking of the Moon :

"All is dead, motionless, still. Is this verily a blasted world? Has it fallen under the breath of Almighty wrath, coming out scorched and seared ?"


The " lesser light " that God declares He made to " rule the night" is set down as a blasted world, and that by a professed Christian ! To this end the teaching of modern astronomy tends to " attract " all who receive its dicta, and cannot, therefore, be retained in the same mind with the Bible. A noteworthy feature of the present day is the fact that many so-called Christian ministers are joining hands with the enemies of the Bible to teach the people that the Old Book is so very unscientific that it can no longer be regarded in the light of a word from God at all.

In the Christian World Pulpit," of 14th June, 1893, the Rev. C .F. Aked is reported as saying, at Pembroke Chapel, Liverpool, that:

"No student of science is able to believe that any such flood as that recorded in the early chapters of Genesis ever took place in the history of the human race . , . . The Flood story IS A MYTH, 'not history'".


This gentleman has arrived at this conclusion by supposing that science is truth, and he is logically forced to believe that the Bible is a myth. Then what say the avowed
enemies of the Book of God ? Says the Freethinker, of 16th October, 1892 :

"There is something in Christianity calculated to make it hostile to science. Its sacred books are defaced by a puerile cosmogony, and a vast number of physical absurdities ; while its whole atmosphere,in the New as well as in the Old Testament, is in the highest degree unscientific. The Bible gives a false account of the origin of the world; a foolish account of the origin of man ; a ridiculous account of the origin of languages. It tells us of a universal flood which never happened. And all these falsities are bound up with essential doctrines, such as the fall of man and the atonement of Christ ; with important moral teachings and social regulations. It was therefore inevitable that the Church, deeming itself the divinely-appointed guardian of Revelation, should oppose such sciences as astronomy, geology, and biology, which could not add to the authority of the Scripture, but might very easily weaken it. Falsehood was in possession, and truth was in exile or a prisoner."

This is clinched by the Public Press which teaches people to think. Reynolds' Newspaper of 13th October, 1895, says:

" The most noteworthy feature of the British Association this year is that the assembled savants — representing religion, science, philosophy, politics — have surrendered hands down to views which, if accepted by anyone ten years ago, would be sneered at as a mark of disgrace. The Church has had to give in because geology and biology have been too strong for the Book of Genesis, which is no longer to be accepted as a real account of the Creation, but merely a symbolical one. The incontestable experiments and
experiences of the practical scientists have proved that Darwin was right, and that evolution is as certain a law as that of gravitation. What a number of the ' learned ' books of a few years ago opposing evolution must now be ignominiously withdrawn from circulation ? And how small must the controversial parson and the lay evangelist, who would prove to you in  two jiffies that science was all bosh,' feel at the thunders of competent scholars ! "


While the Press is filled with suchlike articles, the people who do not think for themselves take for granted that science is right, and as a consequence, reject the Bible.

In the "Earth Review" for January, 1893, the following is found :

"      HONEST AND NOBLE CONFESSIONS.
When we consider that the advocates of the earth's stationary and central position can account for, and explain the celestial phenomena as accurately, to their own thinking, as we can ours, in adition to which they have the evidence of their senses, and SCRIPTURE and FACTS in their favour. WHICH WE HAVE NOT : it is not without a show of reason that they maintain the superiority of their system .... However perfect our theory may appear in our estimation, and however simply and satisfactorily the Newtonian hypothesis may seem to us to account for all the celestial phenomena, yet we are here compelled to admit the astounding truth that, IF OUR PREMISES BE DISPUTED AND OUR FACTS CHALLENGED, THE WHOLE RANGE OF ASTRONOMY DOES NOT CONTAIN THE PROOFS OF ITS OWN ACCURACY.Dr. Woodhouse, a late Professor of Astronomy at Cambridge." 

Those who believe the plain and provable facts of the Bible are set down as lunatics, but the above shows where the lunacy really lies. John Wesley did not believe in the teachings of the men of the modern astronomical school, although most of his followers do. In his Journal he writes :

 "The more I consider them, the more I doubt of all systems of astronomy .... Even with regard to the distance of the sun from the earth, some affirm it lo be only three, and others ninety millions of miles."

Thus, this admittedly great and good man stands out in bold contrast with many of the present day " reverend " gentlemen. The Bishop of Peterborough is another notable  example. He says:

" I have no fear whatever, that the Bible will be found, In the long run, to contain more science than all the theories of philosophers put together."
"I can't breathe" George Floyd RIP

Re: "Equator" problem
« Reply #242 on: November 06, 2014, 10:15:17 AM »
Answers in italics.

I am tired of heliocentric lies!

Says the guy who can't refute the hard data he doesn't want to hear.

1. On the celestial sphere, Crux is exactly opposite the constellation Cassiopeia. (source : http://www.constellation-guide.com/constellation-list/crux-constellation/)

This is pretty close.

2.  In the Northern hemisphere Cassiopeia never sets below the horizon, as a result it is visible all year in the night sky. (source : http://www.solarsystemquick.com/universe/cassiopeia-constellation.htm)

Yep, they say that, but not everything published is meaningful. The official constellation boundary extends as far north as about +78 declination, but the commonly-recognized asterism - the 'W' - is much further south, in the range of about +57° to +64° declination. There are no stars brighter than fourth magnitude in the constellation north of Segin (Epsilon Cas), northernmost star of the 'W' asterism. Technically, the northernmost part of the official constellation is circumpolar down to about 12° N latitude, but that is basically empty sky with very dim stars and dim Milky Way. Part of the 'W' sets if you're south of about 33° N and the whole thing will completely set at least briefly each day south of about 26° N.

3. Some stars within the far northern constellations, such as Cassiopeia, Cepheus, Ursa Major, and Ursa Minor, roughly north of the Tropic of Cancer (+23½°), will be circumpolar stars that never rise or set. (source : http://en.wikipedia.org/wiki/Circumpolar_star)

Note that they say "some stars" within these constellations. See the above. It's interesting that they omit Draco and Camelopardalis, both of which officially extend further north than "Cass" and Ursa Major. Could it be that these dim constellations aren't well known because they are hard to see, so don't count? See the part above about no bright stars in "Cass" north of the 'W'.

4. The constellation is circumpolar south of 34 degrees S latitude and visible every night of the year, though it’s best viewed high overhead in the early evening from April through June.  During these months, south of the Tropic of Cancer (23.5 degrees N latitude), northern-hemisphere stargazers can glimpse the Southern Cross– just barely, above the southern horizon. - See more at: http://oneminuteastronomer.com/1976/legends-southern-cross/#sthash.ZmOW3e4S.dpuf

The Southern Cross is visible evenings low in the south along the Tropic of Cancer April - June. It gets higher as you go further south, even while still in the northern hemisphere. It can be seen early in the morning much earlier in the year.

5. Cassiopeia is visible on any clear night of the year from most locations in the northern hemisphere. For amateur stargazers, September through March are the optimal months for spotting Cassiopeia. In the southern hemisphere, the constellation is visible from May to August at locations north of the Tropic of Capricorn (23.5 degrees south latitude). (source : http://www.sciences360.com/index.php/how-to-identify-the-constellation-of-cassiopeia-in-the-night-sky-2648/)

The 'W' of "Cass" isn't visible in May or most of June from the Tropic of Capricorn; it doesn't fully rise before dawn drowns it out until late June, and isn't visible in the evening from there until November.

A) So, we can see Casiopeia all year long above the tropic of cancer (at least) which is 23,5 degree N, but north of 34 degree S Crux is not circumpolar constellation, how come?

Crux is the smallest constellation. Its official boundaries barely extend beyond the Cross asterism itself. Cassiopeia's official boundaries extend much closer to the pole, so you can technically say part of the constellation Cassiopeia is still circumpolar much further south than the asterism that's associated with the constellation is. If you're talking about what people recognize in the sky as Cassiopeia, it is not circumpolar south of about 33 N. See the answer to 2., above (again).

B) In the southern "hemisphere" Casiopeia is visible from may to august at location north of 23,5 degree S, but in northern hemisphere Crux is visible from april to june (tropical regions), but from tropic of cancer only from may to june! Where this huge difference comes from?

Where did you get those dates? Each of these constellations is visible from the opposite tropic at some time of night for much longer than that. Your dates for "Cass" from the Tropic of Capricorn are simply wrong. Crux rises just before dawn in December, and sets just after sunset in late June, from the Tropic of Cancer.

C) "Tonight (14/5/04), the Southern Cross will be tilted to the left in the early evening, straight up and down at around 9:45 pm and tilted over to the right by midnight.". This is not half a circle you say? Of course it's not, since Crux is not circumpolar constellation at all!

Didn't we already both agree that Crux isn't circumpolar from the latitude of Honolulu? "Tilted to the left" doesn't mean laying on its side, it means it's tilted, as in, not vertical. Ditto for "tilted to the right". So, no, it's not half its circle. Crux is circumpolar south of about 34° S latitude. Didn't we also agree on this?

Now, stop lying and come back to the Earth!

Would you please cut the "you're lying" crap and discuss facts. Just ignoring mine and saying I'm lying makes it look like you have no facts to argue, which, I think actually is the case. Are there any other possibilities?

Explain us how the great deluge could have happened on the round Earth?

Let's not start on yet another topic until we've finished this one, then get finished with tides, OK?

[I see you just posted something else...]
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

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Socratic Amusement

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Re: "Equator" problem
« Reply #243 on: November 06, 2014, 10:15:28 AM »
See what I mean? Like clockwork...
"As for me, all I know is that I know nothing."

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cikljamas

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Re: "Equator" problem
« Reply #244 on: November 06, 2014, 10:27:25 AM »
By the little help from my friend:



Accompanying text:

This is a photo which I took last year from Batumi, Georgia. You can actually see in it Mt.Elbrus which is 202 km away !!! The photo was taken from the port, so it was taken from 1-2 meters above the sea level. Notice that you can see some buildings across the sea. They are more than 50 km away. The very next day I walked out and looked in the same direction, but neither Elbrus nor the buildings were visible. It appeared as if the horizon was much closer. There was no fog or anything, but nothing of what I saw the day before was visible. I wish I could spend more time there. I am sure one day I might have been able to actually take a picture of the coast on the other side...
« Last Edit: November 07, 2014, 09:03:13 AM by cikljamas »
"I can't breathe" George Floyd RIP

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cikljamas

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Re: "Equator" problem
« Reply #245 on: November 06, 2014, 10:58:15 AM »
@Alpha2Omega, i will study your post tomorrow, o.k.?... now i have to go...
"I can't breathe" George Floyd RIP

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ausGeoff

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Re: "Equator" problem
« Reply #246 on: November 06, 2014, 11:08:29 AM »
The above image(s) may well be genuine, but the person who wrote the accompanying text was either confused, drug or alcohol affected, or lying.

What one sees from a specific vantage point on one day doesn't—and cannot—change the following day (assuming swell and tide levels are constant).  Mt Erebus is 5,600m in height, and had obviously "disappeared" in the mist on the following day.  No mystery at all.  (Was that a pun?)

Re: "Equator" problem
« Reply #247 on: November 06, 2014, 11:29:12 AM »
A) So, we can see Casiopeia all year long above the tropic of cancer (at least) which is 23,5 degree N, but north of 34 degree S Crux is not circumpolar constellation, how come?
The observer is too far north.  Did you know that for an observer at either pole, pretty much all stars down to the celestial equator become circumpolar?

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B) In the southern "hemisphere" Casiopeia is visible from may to august at location north of 23,5 degree S, but in northern hemisphere Crux is visible from april to june (tropical regions), but from tropic of cancer only from may to june! Where this huge difference comes from?
Maybe because one is much bigger than the other, not exactly opposite of each other, and those are probably 'optimal' viewing dates from two different sites.

Quote
C) "Tonight (14/5/04), the Southern Cross will be tilted to the left in the early evening, straight up and down at around 9:45 pm and tilted over to the right by midnight.". This is not half a circle you say? Of course it's not, since Crux is not circumpolar constellation at all!
Tilted how much?

Also, did you find any visual evidence yet, whether photographic or video, of the southern stars moving overhead in a manner indicative of being 3,000 miles high and traveling around the outer part of a giant disk?

Re: "Equator" problem
« Reply #248 on: November 06, 2014, 11:38:04 AM »
See what I mean? Like clockwork...
And just 12 minutes later... on to the next.  This must be a record for someone derailing their own derailings of their own thread.

Re: "Equator" problem
« Reply #249 on: November 06, 2014, 11:43:23 AM »
The above image(s) may well be genuine, but the person who wrote the accompanying text was either confused, drug or alcohol affected, or lying.

What one sees from a specific vantage point on one day doesn't—and cannot—change the following day (assuming swell and tide levels are constant).  Mt Erebus is 5,600m in height, and had obviously "disappeared" in the mist on the following day.  No mystery at all.  (Was that a pun?)

You have no idea what you're talking about. Normally even without any fog and clear skies you cannot see anything into the sea which is 50 km away. The mountains indeed can be seen sometimes due to their height, but in this particular picture you're actually seeing Mt.Elbrus from 202 km while at the same time the photo was taken from 1-2 m, do you see the difference? Even with its height of 5642 m, Mt.Elbrus is physically impossible to be seen from more than 270 km away under most perfect conditions and observation height of 2 meters, and you should see it then barely at the horizon. Does it look it's getting close to the horizon? It is at least 15 degrees above it. Even if it were 2000 meters high though it would still be seen at the horizon under those atmospheric conditions. You actually see parts of it which are much lower than 5642 m, you don't just see the very top 5 meters. Anyway, there are many more examples of long distance observations which are challenging the maximum possible visibility. By the way, forget about the peak, look at the building tops, so you think it is okay to see buildings 50 km into the sea? What about Earth's curvature, shouldn't it be much bigger for that distance? Yes, it should. In fact, you shouldn't see a 20-30 m tall building from 1-2 m height from farther than 20 km away, and we're talking seeing it at the horizon here, while in the photo they are above the horizon and the distance is about 50 km.

And in case you're wondering where Elbrus is follow the arrow in the photo below. The ship is actually the one on the left in the photo which Cikljamas submitted.
« Last Edit: November 06, 2014, 12:16:21 PM by Saros »

Re: "Equator" problem
« Reply #250 on: November 06, 2014, 03:03:19 PM »
B) In the southern "hemisphere" Casiopeia is visible from may to august at location north of 23,5 degree S, but in northern hemisphere Crux is visible from april to june (tropical regions), but from tropic of cancer only from may to june! Where this huge difference comes from?
Maybe because one is much bigger than the other, not exactly opposite of each other, and those are probably 'optimal' viewing dates from two different sites.

It's easier than that. His dates for Cassiopeia are simply wrong. The dates for Crux are approximately when it's visible in the evening from that far north, so you have a point about convenience. Crux is visible from the south coast of Kauai (about 22° N) in the morning before sunrise in January.
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

Re: "Equator" problem
« Reply #251 on: November 06, 2014, 05:17:56 PM »
You have no idea what you're talking about. Normally even without any fog and clear skies you cannot see anything into the sea which is 50 km away. The mountains indeed can be seen sometimes due to their height, but in this particular picture you're actually seeing Mt.Elbrus from 202 km while at the same time the photo was taken from 1-2 m, do you see the difference? Even with its height of 5642 m, Mt.Elbrus is physically impossible to be seen from more than 270 km away under most perfect conditions and observation height of 2 meters, and you should see it then barely at the horizon. Does it look it's getting close to the horizon? It is at least 15 degrees above it. Even if it were 2000 meters high though it would still be seen at the horizon under those atmospheric conditions. You actually see parts of it which are much lower than 5642 m, you don't just see the very top 5 meters. Anyway, there are many more examples of long distance observations which are challenging the maximum possible visibility. By the way, forget about the peak, look at the building tops, so you think it is okay to see buildings 50 km into the sea? What about Earth's curvature, shouldn't it be much bigger for that distance? Yes, it should. In fact, you shouldn't see a 20-30 m tall building from 1-2 m height from farther than 20 km away, and we're talking seeing it at the horizon here, while in the photo they are above the horizon and the distance is about 50 km.

And in case you're wondering where Elbrus is follow the arrow in the photo below. The ship is actually the one on the left in the photo which Cikljamas submitted.
http://i.imgur.com/L3y1Hgl.jpg?1
Measuring between the port of Batumi and Mt. Elbrus on google earth, I get 200km .  The coast is angled in relation to the line of sight, but, depending where in the port the picture was taken from, measures about about 31km (19 miles) away at the point directly in line with the peak and the port.

According to the chart in ENaG, the drop after 20 miles is 266 feet and 10 miles is 66 feet.  With a height of 2m (6 feet), the drop along the line of sight doesn't start for 3 miles (again, according to the chart and table in ENaG), so we are left with a curvature drop based on 16 miles for the opposite shore, and 117 out of 120 miles (200km) for the mountain (100 miles = drop of 6,666 feet, 120 miles = 9600 feet.)  If someone knows the math to figure out the drop for 16 and 117 miles, go for it.

The land along the coast there slopes upward with several small buildings and a few larger looking structures scattered up to a couple hundred feet above the shoreline. 

I'll throw and estimate of 3000 meters of drop along the line of sight to the mountain, which has an elevation of 5642m.  Refraction is a common occurance too.  Had a picture also been taken from higher up, any refraction and/or recovery of objects hidden beyond the curvature would have been seen.

As for being 15 degrees above the horizon, how did you come up with that?  How tall would something need to be to be 15 degrees above the horizon 200km away?  The original picture looks like magnification was used versus 1x.  The mountains wouldn't have been very high looking in a 1x shot.

Re: "Equator" problem
« Reply #252 on: November 06, 2014, 11:30:23 PM »
You have no idea what you're talking about. Normally even without any fog and clear skies you cannot see anything into the sea which is 50 km away. The mountains indeed can be seen sometimes due to their height, but in this particular picture you're actually seeing Mt.Elbrus from 202 km while at the same time the photo was taken from 1-2 m, do you see the difference? Even with its height of 5642 m, Mt.Elbrus is physically impossible to be seen from more than 270 km away under most perfect conditions and observation height of 2 meters, and you should see it then barely at the horizon. Does it look it's getting close to the horizon? It is at least 15 degrees above it. Even if it were 2000 meters high though it would still be seen at the horizon under those atmospheric conditions. You actually see parts of it which are much lower than 5642 m, you don't just see the very top 5 meters. Anyway, there are many more examples of long distance observations which are challenging the maximum possible visibility. By the way, forget about the peak, look at the building tops, so you think it is okay to see buildings 50 km into the sea? What about Earth's curvature, shouldn't it be much bigger for that distance? Yes, it should. In fact, you shouldn't see a 20-30 m tall building from 1-2 m height from farther than 20 km away, and we're talking seeing it at the horizon here, while in the photo they are above the horizon and the distance is about 50 km.

And in case you're wondering where Elbrus is follow the arrow in the photo below. The ship is actually the one on the left in the photo which Cikljamas submitted.
http://i.imgur.com/L3y1Hgl.jpg?1
Measuring between the port of Batumi and Mt. Elbrus on google earth, I get 200km .  The coast is angled in relation to the line of sight, but, depending where in the port the picture was taken from, measures about about 31km (19 miles) away at the point directly in line with the peak and the port.

According to the chart in ENaG, the drop after 20 miles is 266 feet and 10 miles is 66 feet.  With a height of 2m (6 feet), the drop along the line of sight doesn't start for 3 miles (again, according to the chart and table in ENaG), so we are left with a curvature drop based on 16 miles for the opposite shore, and 117 out of 120 miles (200km) for the mountain (100 miles = drop of 6,666 feet, 120 miles = 9600 feet.)  If someone knows the math to figure out the drop for 16 and 117 miles, go for it.

The land along the coast there slopes upward with several small buildings and a few larger looking structures scattered up to a couple hundred feet above the shoreline. 

I'll throw and estimate of 3000 meters of drop along the line of sight to the mountain, which has an elevation of 5642m.  Refraction is a common occurance too.  Had a picture also been taken from higher up, any refraction and/or recovery of objects hidden beyond the curvature would have been seen.

As for being 15 degrees above the horizon, how did you come up with that?  How tall would something need to be to be 15 degrees above the horizon 200km away?  The original picture looks like magnification was used versus 1x.  The mountains wouldn't have been very high looking in a 1x shot.

Thanks for the analysis. Indeed there was 3x magnification, but anyway without magnification if you take a picture it looks much farther than looking at the object with the naked eye. You should remember that the eye has an average focal length of 40-50 mm while most cameras start at 20 mm. You actually need to go up to 2x even 3x approximately in order to see it as it really appears to you in real life.

Additionally, I am just pointing out that the maximum visible horizon is around 270 km for something 5642 m high if you're 2 m above the sea level. Then you would see it barely looming at the horizon. As you see in the photo it is not at the horizon at all. It is above it. It is not 270 km away either, but 202 km is far enough.

You're right about the distance of the buildings, the one in the cropped imaged I posted must be a little over 30 km away not 50 km away. But still it shouldn't be visible at all from 2 m height. The distance was wrongly computed due to the fact that those towns are indeed over 50 km away, but not when the distance is measured directly across the sea.

I would imagine though there are better days to take pictures of Elbrus when the visibility is much better. I was there for 3 days only, so I doubt I was so lucky to catch the best visibility ever. My guess is something is wrong with the horizon  distance formula or the Earth doesn't curve at the same rate everywhere if it does at all.

People often get it wrong when they use the horizon calculator and conclude it matches reality. They forget that when they see something they rarely see it only 1 degree above the horizon, but often much higher than that.

You might not believe me but the photo was really taken from about 1-2 meters above the sea level.

Take a look:
« Last Edit: November 07, 2014, 02:17:43 AM by Saros »

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cikljamas

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Re: "Equator" problem
« Reply #253 on: November 07, 2014, 04:34:04 AM »
@Alpha2Omega, remember one thing:

HELIOCENTRISM = SATANISM

I am very sorry, but as long as you defend this evil-perverted "science" of lying you cannot (by default) be anything but liar. I don't say that you deliberately lie, but since i am very aware how much intelligent and educated you are, you just don't leave me much choice...

Anyway, i will do my best to refrain myself of calling you "a liar"...

Answers in italics.

1. On the celestial sphere, Crux is exactly opposite the constellation Cassiopeia. (source : http://www.constellation-guide.com/constellation-list/crux-constellation/)

This is pretty close.

Doesn't this say it all?

Well, according to heliocentrists, Cassiopeia is circumpolar constellation, and Crux is ALMOST circumpolar constellation ! Why is this so, since they are exactly opposite positioned constellations? Now, who do you think you are fooling?

Once more, (quotes from more sources):

- Cassiopeia is circumpolar, and is visible all year round in the Northern Hemisphere. (http://www.eaas.co.uk/cms/index.php?option=com_content&view=article&id=24:andromeda-cassiopeia-and-perseus&catid=5:learning-zone&Itemid=8)

- Cassiopeia, the Queen, is visible in the Northern Hemisphere all year long. (http://www.botproductions.com/stellar/cassiopeia.html)

- Cassiopeia is visible all year round, though the best time to see her is in the Autumn when she is high overhead. (http://www.skyguide.org.uk/constellations/cassiopeia/cassiopeia.htm)

Now, this is going to be quite interesting for our readers:

The most southerly parts of Cassiopeia are visible from the North Island of New Zealand. They are highest in mid November at about 10.30 pm (NZDT). α Cas remains about 3° below the horizon as seen from Auckland and is immediately below (north of) M31, the Andromeda Galaxy. The most southerly bright star of the constellation, π Cas (mag 5.0), rises to 6° at Auckland, but is only just above the horizon at Wellington. Read more: http://www.rasnz.org.nz/Stars/Cassiopeia.htm

New Zealand latitude:



Now something very interesting about the picture provided by my friend Saros:



18,7 * 18,7 (miles) = 351,56 * 8 (inches) * 2,5 cm = 7031 cm / 100 = 70 meters (high hill of water)

If we assumed that this ship is right in the middle of this bay which is 18,7 miles long, and if we assumed that the deck of this tanker is 17,5 meters above the water (which is generous enough assumption) then we could with all right in the world ask this question:

Doesn't "horizon calculator formula" somehow mislead in favor of RET?

Maybe next illustration can help you to see what i mean:










"I can't breathe" George Floyd RIP

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Socratic Amusement

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Re: "Equator" problem
« Reply #254 on: November 07, 2014, 05:58:19 AM »
HELIOCENTRISM = SATANISM

You have failed to disprove the heliocentric model, so to claim it is merely a facet of the dogma of a sub-sect of Christians without providing any evidence for this claim is disingenuous at best.
"As for me, all I know is that I know nothing."

Re: "Equator" problem
« Reply #255 on: November 07, 2014, 07:45:56 AM »
@Alpha2Omega, remember one thing:

HELIOCENTRISM = SATANISM

I am very sorry, but as long as you defend this evil-perverted "science" of lying you cannot (by default) be anything but liar. I don't say that you deliberately lie, but since i am very aware how much intelligent and educated you are, you just don't leave me much choice...

Anyway, i will do my best to refrain myself of calling you "a liar"...
I'm not going to argue about religion with you. I will stick with facts.

Quote

Answers in italics.

1. On the celestial sphere, Crux is exactly opposite the constellation Cassiopeia. (source : http://www.constellation-guide.com/constellation-list/crux-constellation/)

This is pretty close.

Doesn't this say it all?
It says when you can see all of one you can't see all (probably not any) of the other. What else do you think this says?

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Well, according to heliocentrists, Cassiopeia is circumpolar constellation, and Crux is ALMOST circumpolar constellation ! Why is this so, since they are exactly opposite positioned constellations? Now, who do you think you are fooling?
In what context do astronomers say that Crux is "almost circumpolar"? Whether it is or not depends on the latitude, and all agree that it is circumpolar if you're far enough south. In fact, right here

4. The constellation is circumpolar south of 34 degrees S latitude and visible every night of the year, though it’s best viewed high overhead in the early evening from April through June.  During these months, south of the Tropic of Cancer (23.5 degrees N latitude), northern-hemisphere stargazers can glimpse the Southern Cross– just barely, above the southern horizon. - See more at: http://oneminuteastronomer.com/1976/legends-southern-cross/#sthash.ZmOW3e4S.dpuf
you say it is circumpolar. I don't think anyone disagrees with that statement.

For that matter, Whether "Cass" is circumpolar or not also depends on where you are.

It would be helpful if, for Cassiopeia, you distinguished between the official constellation boundary, which is an area of sky defined in, I think, the 1870s, or the asterism (group of bright stars forming the familiar 'W' shape). This is less essential for Crux because its boundaries are barely larger than the familiar Southern Cross asterism.

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Once more, (quotes from more sources):

- Cassiopeia is circumpolar, and is visible all year round in the Northern Hemisphere. (http://www.eaas.co.uk/cms/index.php?option=com_content&view=article&id=24:andromeda-cassiopeia-and-perseus&catid=5:learning-zone&Itemid=8)

- Cassiopeia, the Queen, is visible in the Northern Hemisphere all year long. (http://www.botproductions.com/stellar/cassiopeia.html)

- Cassiopeia is visible all year round, though the best time to see her is in the Autumn when she is high overhead. (http://www.skyguide.org.uk/constellations/cassiopeia/cassiopeia.htm)
Yes, yes. But these descriptions aren't very rigorous, so using them to try to prove a point in a technical discussion is weak. The second refers to "the Queen" suggesting they're talking about the group of stars including and immediately around the 'W' (or "throne"), so at least there's that, and it probably safe to assume the others are, as well. Note that none of these say that Cassiopeia is visible all night all year which would be necessary to meet the strict definition of circumpolar. For informal use, this might be good enough, but don't try to draw too many technical conclusions from the informal descriptive text.

Also note that the last link is clearly aimed at a British audience. All of Great Britain is far enough north that "Cass" is visible all night all year, but it makes no claim that this is true everywhere in the northern hemisphere.

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Now, this is going to be quite interesting for our readers:

The most southerly parts of Cassiopeia are visible from the North Island of New Zealand. They are highest in mid November at about 10.30 pm (NZDT). α Cas remains about 3° below the horizon as seen from Auckland and is immediately below (north of) M31, the Andromeda Galaxy. The most southerly bright star of the constellation, π Cas (mag 5.0), rises to 6° at Auckland, but is only just above the horizon at Wellington. Read more: http://www.rasnz.org.nz/Stars/Cassiopeia.htm

New Zealand latitude:

<image showing New Zealand's North Island spanning about from about 34° S to 42° S latitude>

The first sentence in that link says:

Quote from: Royal Astronomical Society of New Zealand
Only the most southerly parts of Cassiopeia rise above the horizon in northern New Zealand. None of the brighter stars are visible.
We're clearly talking about the official constellation and not the familiar asterism ("None of the brighter stars..."). This is confirmed if you look at the included chart. Just as the official boundaries of Cassiopeia extend much farther north than the 'W' asterism, the southern boundary extends almost another 10° in the other direction, down to about +47°.

π Cas is just a smidgen north of declination +47°, Auckland and Wellington are at latitude 37.8° 36.8° S and 41.25° S, respectively. The reported maximum elevations of 6° and 1° for this star sound right for those latitudes.

All of this exactly matches what we would expect for a spherical earth. What is your point?

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Now something very interesting about the picture provided by my friend Saros:

<off topic pictures and discussion containing some bad assumptions>
Later, if it's still active when we finish what's already on our plate.

[Edit] Correct Auckland latitude.
« Last Edit: November 07, 2014, 10:49:37 AM by Alpha2Omega »
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

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ausGeoff

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Re: "Equator" problem
« Reply #256 on: November 07, 2014, 08:09:12 AM »
You have no idea what you're talking about.

I thank you for this erudite answer LOL.  It proves absolutely my ignorance, and indisputably proves your vastly superior knowledge. [sic]

Where I lived in Victoria (Australia) I could more than easily see a mountain (LOL) with an elevation of 364 metres from a distance of over 90km.  This was across 52km of land and a 50km wide bay from my elevation of 3 metres at the shoreline of the bay.

Your guesstimations for the alleged visibility of Mt Erebus are laughably incorrect.  Please do better next time.


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cikljamas

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Re: "Equator" problem
« Reply #257 on: November 07, 2014, 08:44:33 AM »
Cassiopeia Declination : 77,6923 - 48,6632
Auckland latitude : 36,84 S

36,84 + 6 degree = 42,84
48,6632 + 42,84 = 91,5 degree

What is my point?

This is my point:



Now, if you carefully observe above tangent lines and their geometrical implications for the observers on the alleged globular Earth you could easily figure that from the equator (if the Earth were really a globe) you wouldn't be able to see even Polaris, let alone being able to see any star of the Cassiopeia constellation from New Zealand...

This is another example of fooling people who haven't even begun to think for themselves!
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markjo

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Re: "Equator" problem
« Reply #258 on: November 07, 2014, 09:10:17 AM »
What is my point?

This is my point:


Just as an FYI: the equator in your drawing is about 15 degrees too far north.
Science is what happens when preconception meets verification.
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Besides, perhaps FET is a conspiracy too.
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ausGeoff

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Re: "Equator" problem
« Reply #259 on: November 07, 2014, 09:15:39 AM »
Just as an FYI: the equator in your drawing is about 15 degrees too far north.

You should know by now that flat earthers never let the facts get in the way of a good story LOL.


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cikljamas

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Re: "Equator" problem
« Reply #260 on: November 07, 2014, 09:28:17 AM »
What is my point?

This is my point:



Now, if you carefully observe above tangent lines and their geometrical implications for the observers on the alleged globular Earth you could easily figure that from the equator (if the Earth were really a globe) you wouldn't be able to see even Polaris, let alone being able to see any star of the Cassiopeia constellation from New Zealand...

This is another example of fooling people who haven't even begun to think for themselves!
Just as an FYI: the equator in your drawing is about 15 degrees too far north.

Thanks for your observation, this error (equator line drawn a little bit more toward north) makes my point even stronger and more plausible!
« Last Edit: November 07, 2014, 09:30:13 AM by cikljamas »
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ausGeoff

  • 6091
Re: "Equator" problem
« Reply #261 on: November 07, 2014, 10:02:10 AM »
Thanks for your observation, this error (equator line drawn a little bit more toward north) makes my point even stronger and more plausible!

LOL... typical flat earther.  Just shift the earth's equator to wherever you want it to be in order to improve your story.

Sad really.    ::)


Re: "Equator" problem
« Reply #262 on: November 07, 2014, 10:37:36 AM »
You have no idea what you're talking about.

I thank you for this erudite answer LOL.  It proves absolutely my ignorance, and indisputably proves your vastly superior knowledge. [sic]

Where I lived in Victoria (Australia) I could more than easily see a mountain (LOL) with an elevation of 364 metres from a distance of over 90km.  This was across 52km of land and a 50km wide bay from my elevation of 3 metres at the shoreline of the bay.

Your guesstimations for the alleged visibility of Mt Erebus are laughably incorrect.  Please do better next time.

Mount Elbrus(not Erebus, are you Japanese?) can be seen maximum from 270 km if you're 2 m above the sea level.
Then it will be seen just barely at the very horizon. In the picture I showed it is not at the horizon, but above it, and it is not 270 km away, but only 202 km.

Anyway, that is not the point. I am not making up anything. I am just saying that the horizon distance calculators produce lower horizon distance results than what is observed in practice. The building tops also shouldn't be visible from over 30 km across the bay. You can check in the calculator that from a height of 2 meters you can see something which is 30 meters high maximum from 24 km away. You would only see the very top though. Do you understand? If you don't, study the horizon distance calculators first.

By the way, for your reference, your own example also contradicts the horizon distance formula.

Look it up http://members.home.nl/7seas/radcalc.htm

From 3 meters the visual horizon, if you're looking at something 364 meters tall, is maximum 74 km!!! If you can see a 364 meters mountain from 90 km away just fine and you're only 3 meters above the sea level that means A. the horizon distance calculators are wrong B. the Earth is flat C. You are proving exactly what you wanted to refute D. You made it up

So, thank you very much for that great example, which if true, should point to the possibility that the Earth is flat. You're not even aware that what you think is normal is not. That is why I suggest use the calculator first. Then come back and tell me how exactly you can see a 364 m mountain from 90 km away, if you're at 3 m. You should have asked yourself this question a long time ago. Good you mentioned it here.
« Last Edit: November 07, 2014, 10:40:34 AM by Saros »

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rottingroom

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Re: "Equator" problem
« Reply #263 on: November 07, 2014, 10:52:50 AM »
Those tangential lines in clicky's diagram are pointing to a spot just above the north pole / south pole. That makes no sense considering that polaris is 2.55009322 × 10^15 miles away.

As long as you can see below the horizon ever so slightly more than exactly 90° then you will be about to see farther and farther past the horizon for things that are farther away.

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ausGeoff

  • 6091
Re: "Equator" problem
« Reply #264 on: November 07, 2014, 11:04:09 AM »
So, thank you very much for that great example, which if true, should point to the possibility that the Earth is flat.

Of course there's a "possibility" that the earth is flat.  There's also a possibility that leprechauns exist too.  But science indicates that this possibility is so infinitesimally tiny that it's totally insignificant in any real-world scenario.

And it always amuses me how literally the flat earthers treat these distance to horizon calculations (when it suits them LOL).  So a 16km discrepancy (74km v. 90km) really proves the earth is flat.  Gee, then it must be true.

But... is that really the best you can do?    ::)

Re: "Equator" problem
« Reply #265 on: November 07, 2014, 11:08:38 AM »
So, thank you very much for that great example, which if true, should point to the possibility that the Earth is flat.

Of course there's a "possibility" that the earth is flat.  There's also a possibility that leprechauns exist too.  But science indicates that this possibility is so infinitesimally tiny that it's totally insignificant in any real-world scenario.

And it always amuses me how literally the flat earthers treat these distance to horizon calculations (when it suits them LOL).  So a 16km discrepancy (74km v. 90km) really proves the earth is flat.  Gee, then it must be true.

But... is that really the best you can do?    ::)

I am not doing anything. Simply pointing out there is a discrepancy. It doesn't prove the Earth is flat, but makes you think why the formula doesn't really match reality. You choose how to interpret it. Explain why there is a discrepancy please. I would like to learn something new.

Re: "Equator" problem
« Reply #266 on: November 07, 2014, 11:28:34 AM »
Cassiopeia Declination : 77,6923 - 48,6632
Citation needed.

I get the southernmost boundary of Cassiopeia at declination about +46.75°. You're off by almost 2°.   
Sky & Telescope's Pocket Sky Atlas, ISBN 1-931559-31-7; Stellarium v.0.11.0.

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Auckland latitude : 36,84 S

36,84 + 6 degree = 42,84
48,6632 + 42,84 = 91,5 degree

Oops... I had Auckland at 37.8°, not 36.8° S. Corrected.

The claim was that pi Cas was 6° above the horizon, not the dead southernmost boundary of Cassiopeia. pi Cas has declination +47° 06' (now; it was +47° 01' at epoch J2000) according to Stellarium and consistent with Pocket Sky Atlas.

36.84° + 6° = 42.84°
47.1° + 42.84° = 89.94°.

Pretty darn close.

The southern boundary of the constellation would be a bit higher than this.

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What is my point?

This is my point:

<diagram based on incorrect data>

Now, if you carefully observe above tangent lines and their geometrical implications for the observers on the alleged globular Earth you could easily figure that from the equator (if the Earth were really a globe) you wouldn't be able to see even Polaris,
A north-south tangent line at the equator would be exactly parallel to the Earth's axis of rotation. Why wouldn't you see Polaris, or, more to the point, the North (and South) Celestial Pole(s) when viewing any distance (like, say, 5 feet) above the surface? This is neglecting refraction, which would help make it appear even higher above the horizon.

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let alone being able to see any star of the Cassiopeia constellation from New Zealand...
Not true. See above.

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This is another example of fooling people who haven't even begun to think for themselves!

I could come back with some smartass retort, but you are at least trying to think for yourself, which is good. What is not so good is that you're building your ideas on very shaky foundations (and I'm not referring to your religious beliefs, but instead to some of your "factual" ones), ignoring reliable data, and only accepting (blindly, apparently) whatever you can find that appears to support the idea you're already convinced of.
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

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cikljamas

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Re: "Equator" problem
« Reply #267 on: November 07, 2014, 02:46:45 PM »
Those tangential lines in clicky's diagram are pointing to a spot just above the north pole / south pole. That makes no sense considering that polaris is 2.55009322 × 10^15 miles away.

As long as you can see below the horizon ever so slightly more than exactly 90° then you will be about to see farther and farther past the horizon for things that are farther away.

Prove that Polaris is so far away!

Before you make any effort in this direction maybe you would like to refresh your memory with just a few facts:

1.

2. Modern astronomers have lengthened the sun's distance by nearly a hundred millions of miles, which has necessarily increased the earth's supposed orbit more than 300 000 000 of MILES!!! But this extreme alteration is neither acknowledged nor permitted to detract from the great name of Kepler, lest it might also reflect upon the "science" of astronomy; for in this exact "science" the alteration of MILLIONS of MILES is "a mere detail!"

Copernicus wrote: " It is not necessary that hypotheses be true or even probable ; it is sufficient that they lead to results of calculation which agree with calculation. . . . Neither let any one, as far as hypotheses are concerned, expect anything certain from Astronomy, since that science can afford nothing of the kind, lest in case he should adopt for truth things feigned for another purpose, he should leave the science more foolish than when he came.. . . The hypothesis of the terrestrial motion was nothing but an hypothesis, valuable only so far as it explained phenomena not considered with reference to absolute truth or falsehood."

If such was the conviction of Copernicus, the reviver of the old Pagan system of Pythagoras, and of Newton, its chief expounder, what right have Modem Astronomers to assert that a theory, which was given only as a possibility, is a fact, especially when they differ so much among themselves even as regards the very first elements of the problem—the distance of the Sun from the Earth ? Copernicus computed it as being only three millions, while Meyer enlarged it to one hundred and four millions of miles, and there are many estimates between these two extremes.

John Wesley did not believe in the teachings of the men of the modern astronomical school, although most of his followers do. In his Journal he writes :

"The more I consider them, the more I doubt of all systems of astronomy .... Even with regard to the distance of the sun from the earth, some affirm it lo be only three, and others ninety millions of miles."

3.  The distance to the north star was recently downgraded from 424 light years to 324 years!

All this shows that scientists still hold no key to the universe, they have no clue about the real distances between celestial objects, it's all just guessing at best...

4. Again and again have their theories been combated and exposed, but as often have the majority, who do not think for themselves, accepted the popular thing. No less an authority in his time than the celebrated Danish astronomer, Tycho Brahe, argued that if the earth revolves in an orbit round the sun, the change in the relative position of the stars thus necessarily occasioned, could not fail to be noticed. In the " History of the Conflict between Religion and Science," by Dr. Draper, pages 175 and 176, the matter is referred to m the following words :

" Among the arguments brought forward against the Copernican system at the time of its promulgation, was one by the great Danish astronomer, Tycho Brahe, originally urged by Aristarchus against the Pythagorean system, to the effect that if, as was alleged, the earth moves round the sun, there ought to he a change in the relative position of the stars ; they should seem to separate as we approach them, or to close together as we recede from them... At that time the sun's distance was greatly under-estimated. Had it been known, as it is now, that the distance exceeds 90 million miles, or that the diameter of the orbit is more than 180 million, that argument would doubtless have had very great weight. In reply to Tycho, it was said that, since the parallax of a body diminishes as its distance increases, a star may be so far off that its parallax may be imperceptible. THIS ANSWER PROVED TO BE CORRECT."

To the uninitiated, the words " this answer proved to be correct," might seem to settle the matter, and while it must be admitted that parallax is diminished or increased according as the star is distant or near, parallax and direction are very different terms and convey quite different meanings. Tycho stated that the direction of the stars would be altered ; his critics replied that the distance gave no sensible difference of parallax. This maybe set down as ingenious, but it is no answer to the proposition, which has remained unanswered to this hour, and is unanswerable.

If the earth is at a given point in space on say January 1st, and according to present-day science, at a distance of 90,000,000 miles from that point six months afterwards, it follows that the relative position and direction of the stars will have greatly changed, however small the angle of parallax maybe. THAT THIS GREAT CHANGE IS NOWHERE APPARENT, AND HAS NEVER BEEN OBSERVED, incontestably proves that the earth is at rest — that it does not "move in an orbit round the sun."

5. The other way around, when holding on to their galactic conjectures, they are at a loss how to account for a steady 20”.5 stellar aberration. For in that scheme our earth, dragged along by the sun, joins in this minor star's 250 km/sec revolution around the center of the Milky Way. If, for instance, in March we indeed would be moving parallel to the sun's motion, our velocity would become 250+30 = 280 km/sec, and in September 250-30 = 220 km/sec. The “aberration of starlight,” according to post- Copernican doctrine, depends on the ratio of the velocity of the earth to the speed of light. As that velocity changes the ratio changes. Hence Bradley's 20”.496 should change, too. But it does not. Therefore, there is truly a fly in this astronomical ointment, paraded and promoted as a truth.

So, in order to be able to explain how we could see celestial objects which are sometimes declined from the point of observation more than 90 degrees your last resort is a miraculous refraction, but before you could hide behind a miraculous refraction resort, you have to prove that these fabulous distances make any sense at all...


@Alpha2Omega, it goes for you too...

Fabulous distances doesn't add up when we look for something that supposed to be (but it is not since doesn't exist in the first place) curvature of water's surface in the pictures which show entire figures of 200 km distant mountains!!!
"I can't breathe" George Floyd RIP

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rottingroom

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Re: "Equator" problem
« Reply #268 on: November 07, 2014, 03:00:00 PM »
Is this an astrophysics class? I have no obligation to prove polaris' distance to you. You are making a claim with a faulty diagram showing tangential lines that incorrectly show a problem because those lines are pointing to the wrong spot above the celestial north pole. The claim you are trying to refute hinges on what scientists claim to be the distance, which is 2.55009322 × 10^15 miles or 433.8 light years away. So you need to show why the horizon is a problem GIVEN those massive distances. The distance to these objects is paramount to your claims.

By the way, again, I'm not going to type out a lengthy explanation about the distance to polaris or the sun but I can at least point you in the right direction.

The sun's distance is determined with the help of venus and polaris' distance can be determined using a technique called spectroscopic parallax.

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cikljamas

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Re: "Equator" problem
« Reply #269 on: November 07, 2014, 03:05:44 PM »
You mean, spectroscopic my ass?
"I can't breathe" George Floyd RIP