# "Equator" problem

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#### Alpha2Omega

• 3981
##### Re: "Equator" problem
« Reply #270 on: November 07, 2014, 03:27:43 PM »
Oh, for Dog's sake!

<bunch of stuff already addressed, in detail, in this thread>

Since you've changed the subject again, I take it that this means you're satisfied with the analysis of when and where Crux and stars in Cassiopeia are visible. It's entirely consistent with mainstream cosmology, which includes a spinning spherical earth. Glad you finally agree!

"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

#### 29silhouette

• 3304
##### Re: "Equator" problem
« Reply #271 on: November 07, 2014, 04:20:48 PM »
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:
http://i.imgur.com/Y3ro4i7.jpg?1
Now that picture helps quite a bit.  Looking on GE and at a few random pictures people have posted, I found a shot of that same spot from a different angle.  The edge of the concrete looks to be about 1 meter.  If you were standing when you took the picture, then the camera is more than 2 meters high.  The raised part with the railing and the utility door looks to be an additional meter high, and since your camera lines up with the top of the railing instead of the bottom (the section running almost parallel to your line of sight) then you're probably more than 2 meters (6'6") but not quite 3 (9'10"), perhaps 8-9 feet.

Anyway, using this info  http://www.sacred-texts.com/earth/za/za05.htm#page_9

The first hilltop above the water just to the right of the line of sight to the peak is about 440 feet and is 23 miles away. Minus the 3 miles (at least, based of only 6 feet camera elevation) you're looking at a drop for 20 miles. According to ENaG, that is 266 feet.  Probably not even 200 feet, as your picture really looks to have been taken from higher than 6 feet.  About the structure, is it a building, or maybe a boat in the distance?

Again with the mountain, it's elevation is 18510 ft.  With a distance of 120 miles, the drop is based off of approx. 117 miles, but that's based off a height of only 6 feet, so probably even less than 117 as it looks like you're higher than 6.  Maybe 9000 feet of drop, if that.  An extra couple feet makes a big difference over long distances.  Add in refraction.

Still curious how you got 15 degrees.  Did it look 15 degrees above the horizon when there in person with the naked eye?

#### 29silhouette

• 3304
##### Re: "Equator" problem
« Reply #272 on: November 07, 2014, 04:29:07 PM »
this error (equator line drawn a little bit more toward north) makes my point even stronger and more plausible!
No... not really.  With your red line perpendicular to where you should have drawn the equator, it would have been parallel to the axis, and thus lined up with both Polaris and Sigma Octantis.

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#### Saros

• 403
##### Re: "Equator" problem
« Reply #273 on: November 07, 2014, 05:22:11 PM »

Now that picture helps quite a bit.  Looking on GE and at a few random pictures people have posted, I found a shot of that same spot from a different angle.  The edge of the concrete looks to be about 1 meter.  If you were standing when you took the picture, then the camera is more than 2 meters high.  The raised part with the railing and the utility door looks to be an additional meter high, and since your camera lines up with the top of the railing instead of the bottom (the section running almost parallel to your line of sight) then you're probably more than 2 meters (6'6") but not quite 3 (9'10"), perhaps 8-9 feet.

Anyway, using this info  http://www.sacred-texts.com/earth/za/za05.htm#page_9

The first hilltop above the water just to the right of the line of sight to the peak is about 440 feet and is 23 miles away. Minus the 3 miles (at least, based of only 6 feet camera elevation) you're looking at a drop for 20 miles. According to ENaG, that is 266 feet.  Probably not even 200 feet, as your picture really looks to have been taken from higher than 6 feet.  About the structure, is it a building, or maybe a boat in the distance?

Again with the mountain, it's elevation is 18510 ft.  With a distance of 120 miles, the drop is based off of approx. 117 miles, but that's based off a height of only 6 feet, so probably even less than 117 as it looks like you're higher than 6.  Maybe 9000 feet of drop, if that.  An extra couple feet makes a big difference over long distances.  Add in refraction.

Still curious how you got 15 degrees.  Did it look 15 degrees above the horizon when there in person with the naked eye?

Yes, it looked much higher than the horizon. I am talking about Caucasus and especially Elbrus. I don't believe it should be visible that high from 200 km away. Seeing something at the horizon means exactly that. It means the very top of the thing is at the horizon. Clearly Mt.Elbrus is not at the horizon but much higher, and you definitely don't see its top 5 meters only. AusGeoff, by the way, mentioned that when he lived in Victoria he could see a mountain 364 m high from 90 km away across the bay from the shore. That example also contradicts the calculated horizon distance...
By the way, distant mountain observations or anything observed from a great distance are governed by the law of perspective. I don't think the reason why something is not visible is because of the curvature at all. Most likely it is simply too far away for its size to be visible and the angle of view is too small. My point is that perhaps it is not the curvature that hides stuff but the perspective or the changing water surface level due to waves and so on. If something is not very big it is unlikely to be visible through so many layers of air. After all you couldn't possibly see anything 1000 km away even if the surface is totally flat.

« Last Edit: November 07, 2014, 06:48:34 PM by Saros »

#### 29silhouette

• 3304
##### Re: "Equator" problem
« Reply #274 on: November 07, 2014, 06:00:12 PM »
Yes, it looked much higher than the horizon. I am talking about Caucasus and especially Elbrus. I don't believe it should be visible that high from 200 km away. Seeing something at the horizon means exactly that. It means the very top of the thing is at the horizon. Clearly Mt.Elbrus is not at the horizon but much higher, and you definitely don't see its top 5 meters only. AusGeoff, by the way, mentioned that when he lived in Victoria he could see a mountain 364 m high from 90 km away across the bay from the shore. That example also contradicts the calculated horizon distance...
By the way, distant mountain observations or anything observed from a great distance are governed by the law of perspective. I don't think the reason why something is not visible is because of the curvature at all. Most likely it is simply too far away for its size to be visible and the angle of view is too small. My point is that perhaps it is not the curvature that hides stuff but the perspective or the changing water surface level due to waves and so on. If something is not very big it is unlikely to be visible through so many layers of air. After all you couldn't possibly see anything 1000 km away even if the surface is totally flat.
Here's a calculator I just found for figuring the drop over a specific distance.
http://www.cohp.org/local_curvature.html
I estimated 9000 feet over 117 miles, but it's 9,130feet.  115 miles gives us 8,820.  The mountain is 18,510 feet, so that's probably why you see a lot more than just the top 5 meters.

What was Ausgeoff's 'eye level' elevation?

I've got pictures of a bridge and hillside 'sinking' below the horizon as my viewing elevation dropped, yet houses along the shoreline were still visible (although compressed) due to refraction.

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#### Saros

• 403
##### Re: "Equator" problem
« Reply #275 on: November 07, 2014, 06:36:53 PM »
Yes, it looked much higher than the horizon. I am talking about Caucasus and especially Elbrus. I don't believe it should be visible that high from 200 km away. Seeing something at the horizon means exactly that. It means the very top of the thing is at the horizon. Clearly Mt.Elbrus is not at the horizon but much higher, and you definitely don't see its top 5 meters only. AusGeoff, by the way, mentioned that when he lived in Victoria he could see a mountain 364 m high from 90 km away across the bay from the shore. That example also contradicts the calculated horizon distance...
By the way, distant mountain observations or anything observed from a great distance are governed by the law of perspective. I don't think the reason why something is not visible is because of the curvature at all. Most likely it is simply too far away for its size to be visible and the angle of view is too small. My point is that perhaps it is not the curvature that hides stuff but the perspective or the changing water surface level due to waves and so on. If something is not very big it is unlikely to be visible through so many layers of air. After all you couldn't possibly see anything 1000 km away even if the surface is totally flat.
Here's a calculator I just found for figuring the drop over a specific distance.
http://www.cohp.org/local_curvature.html
I estimated 9000 feet over 117 miles, but it's 9,130feet.  115 miles gives us 8,820.  The mountain is 18,510 feet, so that's probably why you see a lot more than just the top 5 meters.

What was Ausgeoff's 'eye level' elevation?

I've got pictures of a bridge and hillside 'sinking' below the horizon as my viewing elevation dropped, yet houses along the shoreline were still visible (although compressed) due to refraction.

This is a better calculator -> http://members.home.nl/7seas/radcalc.htm

Please be honest and admit you shouldn't see so much. Calculating the horizon distance doesn't mean you should see something from bottom to top, does it? It is 5642 meters, but even in my photos you can tell that the 2000 mark is above the horizon.

Ausgeoff's 'eye level' was 3 meters. You can read his comment. I hope he will provide more details about his observation though. Take a look at the new Elbrus picture I posted. I found it on the Internet. It was taken from Batumi beach. By the way, some of my photos can be found on the Internet on a website that deals with long distance observations.

« Last Edit: November 07, 2014, 06:59:21 PM by Saros »

#### cikljamas

• 2174
• Ex nihilo nihil fit
##### Re: "Equator" problem
« Reply #276 on: November 08, 2014, 02:02:57 AM »
@Alpha2Omega, i know that you know that you have no real arguments (just a mumbo jumbo type of arguments) to fight against obvious truth which says : The Earth is Flatly Flat! So, the final argument which is only capable to decide meritoriously about the real shape of the Earth is flatness of water's surface!

But before we get to the "TIDES ISSUE" & "NOAH'S ARK" A.K.A. "IMPOSSIBILITY OF A GREAT DELUGE TO HAPPEN ON THE GLOBULAR EARTH" ISSUE let's sum up what we have talked about so far:

1. In order to evade inevitable negative implications of lack of a real stellar parallax, heliocentric liars absolutely arbitrarily increased astronomical distances all the way up to idiotic numbers. Copernicus started with 3 000 000 miles, and now we should chew up 92 000 000 miles (alleged distance between the Earth and the Sun)!!! Chew it up if you like, but don't expect of any reasonable person to get over it as if it were a mere detail!

2. Has anybody ever tried to dispute this claim: "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."

3. When Heliocentrists failed to disprove the geocentric nature that we live in, they resorted to inventing assumptions, many of which are so absurd that the inventors themselves admit that they are unfalsifiable (by implication unscientific) thought-experiments. Some of these assumptions include:

-    the alleged tilt of the earth's axis,

-    the so called Copernican principle,

-    positive stellar parallax,

-    uniformitiy of the speed of light,

-    lengh contraction

-    time dilation

-    denial of inertia (only accepting an imaginary and isolated "chosen" inertial frame of reference)

-    the earth supposedly moving at a various speeds (in order to account for the observed eclipses)

These and many other assumptions are presented as evidence to each other. In other words one assumption is used in order to prove another assumption. In fact these assumptions are so fundamentally dependent on each other that one becomes meaningless without the other, which shows that heliocentrists don't refrain from applying deceit (circular reasoning in this case) in order to make their assertions believable.

4. There is not one single scientific proof that could corroborate alleged "scientific" fact of the alleged Earth's revolution around the Sun!
5. There is not one single scientific proof that could corroborate alleged "scientific" fact of the alleged Earth's rotation on it's axis! http://www.energeticforum.com/256388-post62.html
6. There is no axial tilt of the Earth, as well as no proof for such a nonsensical product (of deluded heliocentric mind) to be real or to be in accordance with anything in reality!
7. Whole universe is centred to the Earth! (Celestial equator is aligned with the Earth's equator). THERE IS NO SANE HELIOCENTRIC EXPLANATION FOR THIS ABSOLUTELY PROVEN SCIENTIFIC FACT!!!
8. There is no gravitation as such. Newton invented it with one and only purpose (to prove that the Earth revolves around the Sun)
9. http://theflatearthsociety.org/forum/index.php?topic=62199.msg1636031#msg1636031
10. CONCLUSIVE INFERENCE ABOUT THE EQUATION OF TIME ISSUE: http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637435#msg1637435

Now, to refresh your memory:

A) http://theflatearthsociety.org/forum/index.php?topic=62199.msg1636172#msg1636172

B) http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637695#msg1637695

C) http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637759#msg1637759

Are you ready for hitting the last nails ( TIDES & NOAH'S ARK issues) in heliocentric coffin?

edit:

Oh, i owe you conclusion on Crux - Casiopeia issue:

1. It is more than obvious that Casiopeia is a circumpolar constellation!
2. It is more than obvious that Crux is not a circumpolar constellation!
3. It is more than obvious that there is a lot of making up - inventing - adjusting data regarding "exact" numbers of different latitudes or "exact" positions of different stars, so it speaks for itself!!!
« Last Edit: November 08, 2014, 02:19:14 AM by cikljamas »
"I can't breathe" George Floyd RIP

#### cikljamas

• 2174
• Ex nihilo nihil fit
##### Re: "Equator" problem
« Reply #277 on: November 08, 2014, 06:02:13 AM »
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...?

Yes, why wouldn't we see Polaris if north-south tangent line at the equator would be exactly parallel to the Earth's axis of rotation?

Hm, strange question i must admit, pay attention to a green line in above picture and meditate some more on this issue...

Well, i just have finished my meditation on Polaris issue, and here is my conclusion:

Since Polaris declination is 89 degrees 19 ' even if we presumed that the distance between the Earth and Polaris is so idiotically great, we have to notice one problem associated with visibility of Polaris at the Equator:

Let's say that at midnight 1th January from the same point at the Equator we can see Polaris due to 0,8 degree (less) difference between 90 degree and 89 degree 19 ', this very same difference will be at midnight 1th June the reason with counter effect, am i right?

So, how come that there is no difference in visibility of Polaris from the same point at the Equator with respect to the constant half-annualy shifts of angles?

So, when someone says that we can see Polaris 1 or 2 degrees south of the Equator due to refraction, then that someone should take into account this 0,8 degree also!

Let alone seeing Polaris 12 degrees south of the Equator!
« Last Edit: November 08, 2014, 10:22:08 AM by cikljamas »
"I can't breathe" George Floyd RIP

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#### neimoka

• 738
##### Re: "Equator" problem
« Reply #278 on: November 08, 2014, 07:04:16 AM »
2. Has anybody ever tried to dispute this claim: "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."

Star positions do change measurably and this is well documented. Proper motion (that's something you might want to look up) of some stars changes as much as 10 arc seconds per year.

1. It is more than obvious that Casiopeia is a circumpolar constellation!
2. It is more than obvious that Crux is not a circumpolar constellation!
3. It is more than obvious that there is a lot of making up - inventing - adjusting data regarding "exact" numbers of different latitudes or "exact" positions of different stars, so it speaks for itself!!!
What is circumpolar and what is not depends on observer's location, nothing else. Both are circumpolar in some areas but not in some other areas.

#### 29silhouette

• 3304
##### Re: "Equator" problem
« Reply #279 on: November 08, 2014, 09:52:31 AM »
This is a better calculator -> http://members.home.nl/7seas/radcalc.htm

Please be honest and admit you shouldn't see so much. Calculating the horizon distance doesn't mean you should see something from bottom to top, does it? It is 5642 meters, but even in my photos you can tell that the 2000 mark is above the horizon.

Ausgeoff's 'eye level' was 3 meters. You can read his comment. I hope he will provide more details about his observation though. Take a look at the new Elbrus picture I posted. I found it on the Internet. It was taken from Batumi beach. By the way, some of my photos can be found on the Internet on a website that deals with long distance observations.

http://www.barents.pl/sites/default/files/550988_356863567725394_578677839_n.jpg
The calculator I linked shows the distance to the horizon from whatever height you enter, and just below that it shows how much Earth curves over whatever distance you enter.  I'm not sure how to get that info from the calculator you linked.

Here's a few pictures I took a while ago.  It's 9 miles to the bridge, and 12 to the buildings along the shore.  The left half was from an elevation of around 20 feet (I could be a couple feet off), and the right half was taken from about 7 inches.  Even from a height of 20 feet however, the horizon is 5.5 miles out, so there is still a drop over the remaining distance, and the buildings are still only visible due to refraction.  I would have had to be about 80 feet high to put the horizon at the shoreline and have a completely 'unsunken' view.

The bridge deck, tree-line, and hillside in general have visibly 'sunk' beyond the horizon/waterline.  Yet the buildings are still visible, although with a compressed appearance, including the landslide, due to refraction.  The top of the hillside is about 300 feet.

How many degrees above the horizon would you say the tops of the trees are?

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#### Alpha2Omega

• 3981
##### Re: "Equator" problem
« Reply #280 on: November 08, 2014, 12:54:14 PM »
@Alpha2Omega, i know that you know that you have no real arguments (just a mumbo jumbo type of arguments) to fight against obvious truth which says : The Earth is Flatly Flat! So, the final argument which is only capable to decide meritoriously about the real shape of the Earth is flatness of water's surface!
"[ I] know that you know ..." No, you don't "know" that. It's what you want to believe, which is different. First thing in your post is a bogus claim. Not a good start.

Quote
But before we get to the "TIDES ISSUE" & "NOAH'S ARK" A.K.A. "IMPOSSIBILITY OF A GREAT DELUGE TO HAPPEN ON THE GLOBULAR EARTH" ISSUE let's sum up what we have talked about so far:
Apparently there are still a lot of issues, though.

Quote
1. In order to evade inevitable negative implications of lack of a real stellar parallax, heliocentric liars absolutely arbitrarily increased astronomical distances all the way up to idiotic numbers. Copernicus started with 3 000 000 miles, and now we should chew up 92 000 000 miles (alleged distance between the Earth and the Sun)!!! Chew it up if you like, but don't expect of any reasonable person to get over it as if it were a mere detail!
Note that increasing the radius of earth's orbit increases - not decreases - parallax. Clearly you have a misunderstanding what you're arguing about. Why would astronomers "arbitrarily" increase the length of the AU to "explain away" small values of stellar parallax? If that were their motive, they'd do the opposite. The first reasonably close measurement of the true value of the AU didn't occur until 1761 (as already discussed), well after Copernicus' (and Kepler's) time.

Oh, yes, you keep claiming that stellar parallax hasn't been measured. You're wrong.

"Liars". Again. In other words, you have no suitable explanation for this. Gotcha.

Quote
2. Has anybody ever tried to dispute this claim: "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."
"Has anybody ever tried to dispute this claim"? "Dispute"? All that's necessary is to point out that it's patently wrong.

Six months later earth will be 180,000,000 miles away, not 90,000 miles (using your approximate number). That 90,000,000 miles is the radius of the Earth's orbit, not its diameter, which is twice the radius; two times 90,000,000 miles is 180,000,000 miles. After half a year, the Earth has progressed halfway around it's orbit, so it's separated from that original point by the diameter of the orbit, not the radius. I've tried to avoid "mumbo-jumbo" here and explain it as best I can; if you don't understand this, I may be out of ideas - perhaps someone else can explain it better.

Anyway, if you had used the correct figure, it would have made your next point stronger (but still wrong).

What does "direction of the stars will have greatly changed, however small the angle of parallax maybe" mean? Isn't "direction ... will have greatly changed" the opposite of "however small the angle"?

"THAT THIS GREAT [or small] CHANGE IS NOWHERE APPARENT, AND HAS NEVER BEEN OBSERVED". Can you prove the assertion that stellar parallax "HAS NEVER BEEN OBSERVED"? You can't, of course. We'll talk about "proving" things in a bit. There are tons of published studies of stellar parallax, though, by many independent researchers, so plenty of evidence (not proof) that it can and has been measured. Whether you like it or not means exactly nothing.

Quote
3. When Heliocentrists failed to disprove the geocentric nature that we live in, they resorted to inventing assumptions, many of which are so absurd that the inventors themselves admit that they are unfalsifiable (by implication unscientific) thought-experiments. [Citation needed] Some of these assumptions include:

-    the alleged tilt of the earth's axis,
It explains apparent motion of the Sun and the seasons nicely. Also Precession of the equinoxes. This should be easily falsiable. Where's the evidence?

-    the so called Copernican principle,
Presuming otherwise has little value (other than, perhaps, spiritual, which is outside the realm of science) and makes many observations more difficult to explain.

-    positive stellar parallax,
Examined and answered in an earlier post.

-    uniformitiy of the speed of light,

-    lengh contraction

-    time dilation
These three are part of of the Theory of Relatively. It solved a 200-year mystery about the precession of the orbit of Mercury and many other observed and later-discovered phenomena. Although many have been tried, there have been no experiments that conclusively refute general relativity, but many that conclusively verify its predictions. It works so it is used.

-    denial of inertia (only accepting an imaginary and isolated "chosen" inertial frame of reference)
The universe is thought to be inertial. This means that it is not, as a whole, accelerating, which, as a corollary, means that it isn't spinning. This does not mean that nothing in the universe accelerates (or spins), but the whole of the universe itself is not accelerating. We use reference frames that make calculations easiest. Sometimes we use earth-fixed reference frames, others, like when calculating orbits (which operate in the inertial frame) we use the inertial frame. There is no denial that inertia exists - you're again confusing terms.

-    the earth supposedly moving at a various speeds (in order to account for the observed eclipses)
And yet the eclipses happen exactly as predicted decades in advance. If the model those predictions are based is wrong, how likely would that be? It appears that it has merit and the proof of the pudding is in the eating (that adage is often misquoted).

These and many other assumptions are presented as evidence to each other. In other words one assumption is used in order to prove another assumption. In fact these assumptions are so fundamentally dependent on each other that one becomes meaningless without the other, which shows that heliocentrists don't refrain from applying deceit (circular reasoning in this case) in order to make their assertions believable.
These models (you call them "assumptions") are verified by actual observations and measurements as we go, though. Sometimes better observations result in better measurements that require earlier models to be revised or occasionally thrown out altogether. This is how science works. If it were a simple matter of "it says this, therefore this is so" then things like the revision of the distance to Polaris by a significant factor wouldn't happen even after the more-reliable parallax data was obtained. This isn't an "embarrassment" for science - it's a good example how it works: better measurements replace less-good or less-reliable measurements.

Quote
4. There is not one single scientific proof that could corroborate alleged "scientific" fact of the alleged Earth's revolution around the Sun!
5. There is not one single scientific proof that could corroborate alleged "scientific" fact of the alleged Earth's rotation on it's axis! http://www.energeticforum.com/256388-post62.html
This is a good place to discuss the idea of "scientific proof". This is a term often tossed out by both sides, but, as a fact, nothing can be formally 'proven' scientifically, at least not in the same sense as a mathematical theorem is formally proven.

Quote from: Satoshi Kanazawa
Proofs have two features that do not exist in science:  They are final, and they are binary.  Once a theorem is proven, it will forever be true and there will be nothing in the future that will threaten its status as a proven theorem (unless a flaw is discovered in the proof).  Apart from a discovery of an error, a proven theorem will forever and always be a proven theorem.

In contrast, all scientific knowledge is tentative and provisional, and nothing is final.  There is no such thing as final proven knowledge in science.  The currently accepted theory of a phenomenon is simply the best explanation for it among all available alternatives.  Its status as the accepted theory is contingent on what other theories are available and might suddenly change tomorrow if there appears a better theory or new evidence that might challenge the accepted theory.  No knowledge or theory (which embodies scientific knowledge) is final.  That, by the way, is why science is so much fun.
From here

So you're technically right that neither of these things is 'proven', but, for the same reason, you'll never be able to 'prove' your model, either. While we can't 'prove' things, we can judge our models (or theories) based on how well they fit observations, and how well they can predict observations that haven't been made yet. A heliocentric solar system with elliptical Keplerian orbits (with a dash or general relativity when needed) is damn good at explaining the observed motion of solar system bodies and predicting future motion of these bodies. How's that explanation of retrograde motion of planets with a fixed earth and everything going around it coming along, by the way?

Many, many things are so well understood, characterized, and measured to within a micron of their lives that the likelihood that our understanding of them will change fundamentally is vanishingly small. Still a remote possibility, but exceedingly unlikely.
Quote
6. There is no axial tilt of the Earth, as well as no proof for such a nonsensical product (of deluded heliocentric mind) to be real or to be in accordance with anything in reality!
Here we go with the "nonsensical" and "deluded" schtick again, and see the above about 'proof'. If you really think this is necessary to buck yourself up in your belief like this, then you need to examine what it is you believe in. How do you propose to explain the annual motion of the Sun if the equator isn't tilted with respect to the ecliptic?

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7. Whole universe is centred to the Earth! (Celestial equator is aligned with the Earth's equator). THERE IS NO SANE HELIOCENTRIC EXPLANATION FOR THIS ABSOLUTELY PROVEN SCIENTIFIC FACT!!!
From our point of view it's convenient to treat it as such because the universe appears to rotate about us; it's an illusion, though. From the point of view from elsewhere, it appears different. As explained earlier in this post, since the universe is inertial, we can use any point we want as a center of reference, so treating the Earth (or SS Barycenter) as the origin of a coordinate system makes our analysis easier.
Quote
8. There is no gravitation as such. Newton invented it with one and only purpose (to prove that the Earth revolves around the Sun)
Newton deduced gravity while trying to understand why things in the heavens and on earth behaved as they do. He found a single force that explained both. Kepler applied Newton's discovery to explain, with great (but not perfect) accuracy, the motions of the planets.
Quote
9. http://theflatearthsociety.org/forum/index.php?topic=62199.msg1636031#msg1636031
10. CONCLUSIVE INFERENCE ABOUT THE EQUATION OF TIME ISSUE: http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637435#msg1637435

Now, to refresh your memory:

A) http://theflatearthsociety.org/forum/index.php?topic=62199.msg1636172#msg1636172

B) http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637695#msg1637695

C) http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637759#msg1637759

Are you ready for hitting the last nails ( TIDES & NOAH'S ARK issues) in heliocentric coffin?

B) and C) Let's finish this first before moving on. If you'll agree that you haven't disproved the Heliocentric model, and that people who believe that model is the best explanation for our observations aren't necessarily liars, cheats, idiots, Satan-worshippers, etc. because they of this, that would be a good start on wrapping this up. Note that doing so doesn't mean you agree with that model, only that you haven't disproved it and are willing to show some respect for those who disagree with you on this topic.

Quote
edit:

Oh, i owe you conclusion on Crux - Casiopeia issue:

1. It is more than obvious that Casiopeia is a circumpolar constellation!
2. It is more than obvious that Crux is not a circumpolar constellation!
3. It is more than obvious that there is a lot of making up - inventing - adjusting data regarding "exact" numbers of different latitudes or "exact" positions of different stars, so it speaks for itself!!!
1. Is true if you're far enough north.

2. Is most assuredly not true. Even you yourself said it was circumpolar south of 34° S in point 4. here.

3. About the only "adjustments" to a simple spherical geometric solution for what stars are visible where and when are the observers height above the horizon, which is simply a little more geometry and often isn't necessary, and atmospheric refraction, which can be estimated accurately enough for most situations using a standard model of the atmosphere (1/2° at the horizon from near sea level is a good first-order approximation). For really precise measurements we need a more detailed model of the atmosphere based on current local conditions and should might also need to account for oblateness of earth, but details like these are really "down in the noise" in discussions like this; these weren't mentioned before now, because the math is much more difficult and it would look like "mumbo-jumbo" to most reasonably-well educated people, and simply aren't significant at this level of detail.

Insofar as your whining about needing to use "exact" latitudes and declinations, remember that, based on an error of about 1 1/2° in the declination of a star (pi Cas), you "found" a discrepancy of about 1 1/2° in the expected location of this star from Auckland. This discrepancy vanished when the correct declination was used. The need to use accurate positions if you want accurate answers should be self evident. It's also self evident that for arm-waving they can sometimes be a total nuisance, so I see why you want to dismiss their importance.

You keep going over the same ground again and again. The answers aren't going to change.

 Clarification in last long paragraph. Fix minor typos.
« Last Edit: November 08, 2014, 01:11:06 PM by Alpha2Omega »
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

#### ausGeoff

• 6091
##### Re: "Equator" problem
« Reply #281 on: November 08, 2014, 08:45:15 PM »

Are you seriously tendering an evidential document that was written 150 years ago—about astrophysics?

Scientific knowledge has advanced logarithmically since then my friend.  You really need to catch up methinks LOL.

#### cikljamas

• 2174
• Ex nihilo nihil fit
##### Re: "Equator" problem
« Reply #282 on: November 09, 2014, 01:59:35 AM »
Alpha2Omega, one day you will be one of the best FET apologist, this day is not so far away...

Quote
"So you're technically right that neither of these things is 'proven', but, for the same reason, you'll never be able to 'prove' your model, either."

Sorry, but i am not just technically right, i am absolutely right!

Had there been any way to prove that the Earth is submitted to any kind of motion, scientists would have supplied us with these proofs up until now, and by doing this they would have provided immortal fame for themselves.

How about the proofs for the immobility of the Earth?

Allow me to present you some of them:

1. Observing the sun directly from the north pole the apparent motion of the sun would be straight line for days, and a camera should have to be slightly adjusted every few hours to cancel out scarcelly perceptible effect due to Earth's alleged rotation which speed is practically zero at North Pole.

Why haven't we seen until this day any such video which could have easily proved Earth's daily rotation? Lack of such video speaks for itself, and presents indirect but strong proof that there is no reason for making such video because such video would prove something else (that the Earth is at rest), and scientists are aware of that truth very well!

2. If the Earth was rotating about its axis, someone in Quito, Ecuador would be traveling twice as fast from west to east as someone in Oslo, Norway – at any moment, and at every moment. Meanwhile, someone looking at the proverbial North Pole, would hardly be moving at all! But is that reality?

Of course it is not reality, but this supposed fact of Earth's rotation now becomes deadliest error of all, concerning supposed differences of Earth rotational speeds at different latitudes.

If these differences were really the true fact then the speed of apparent motion of all celestial bodies would be twice greater for any observer on the equator than it would be for any observer on the latitude of Oslo.

How hard would be to make an experiment (measurement) of such kind???

3a) If the atmosphere were independent (non rotating but static) from Earth's daily rotation then we would have on the surface of the Earth permanent winds that blow 600 to 1600 km/h. Do you notice permanent winds which blow at such a speed?

3b) If the atmosphere were rotating along with the Earth the air flow at the surface of the Earth would have variable velocity (not the thermal), variable pressure (not the static), and variable density (not the normal). Such air flow and such air pressure regimes do not exist: http://www.energeticforum.com/256388-post62.html

4. If the Earth were suffering a daily rotation it would generate an incredible deflection on all flying matters in air atmosphere. 5) The inertial motion is terminated in air atmosphere, and thus all dropped objects should land behind their starting positions on a rotating Earth. http://www.energeticforum.com/255938-post21.html

5. To be pression or to be gravity? The choice of Earth’s rotation (the cause of pression), should repel the gravity from Earth. Consequently, the heliocentric model looses the most precious element. The choice of gravity should remove the concept of Earth’s rotation from the cosmos motion, consequently; the journey of the Earth around the sun becomes useless since half of the Earth should be always in darkness and the second half should be always in lightness.

6. No experiment has ever been performed with such excruciating persistence and meticulous precision, and in every conceivable manner, than that of trying to detect and measure the motion of the Earth. Yet they have all consistently and continually yielded a velocity for the Earth of exactly ZERO mph.

The toil of thousands of exasperated researchers, in the extremely varied experiments of Arago, De Coudre's induction, Fizeau, Fresnell drag, Hoek, Jaseja's lasers, Jenkins, Klinkerfuess, Michelson-Morley interferometry, Lord Rayleigh's polarimetry, Troughton-Noble torque, and the famous 'Airy's Failure' experiment, all conclusively failed to show any rotational or translational movement for the earth, whatsoever."

http://www.sacred-texts.com/earth/za/za21.htm

7. NO CAUSE OF EARTH'S ROTATION WHATSOEVER: retains the state of illusion. The most important element in heliocentric model is the Earth’s rotation about its polar axis. What is the cause of Earth’s rotation? No one has attributed the cause of Earth’s rotation to any type of action or force even though they have attributed the cause of orbital motion (revolution) to Newton’s law of gravity.

8. NO CAUSE OF THE ROTATION OF THE AIR-LAYER: 2) The rotation of the air-layer next to the rigid Earth is without cause, and lacks a technique and tool. Perhaps, one may envision the whole rigid sphere undergoes a rotation about its polar axis. But, how one can envision the air atmosphere (the surface layer) rotates with the rigid sphere without an engineering method (e.g.air foil). In addition, what maintains the air’s rotation for tens of thousands of years (we are practical people) without stop. The rotation of the background air is the greatest hoax ever invented by mankind.

9. CONCLUSIVE INFERENCE ABOUT THE EQUATION OF TIME ISSUE: http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637435#msg1637435

Arthur Eddington dared to contemplate that:

"There was just one alternative; the earth's true velocity through space might happen to have been nil."

Lincoln Barnett agrees:

"No physical experiment ever proved that the Earth actually is in motion."

So, when all attempts to prove any kind of motion of the Earth FAIL, what does it mean?

It means that the contrary is the fact: The Earth is at rest!

Every failure of all these attempts presents the proof to the contrary : The Earth is at rest!

If you don't want to make me laugh, then you cannot just say: O.K., we have failed to prove Earth's motion but there is still some chance that we could succeed to prove it in some distant future?

In how distant future? When we inhabit another galaxy? Come on, cut the crap, please!

Quote
Anyway, if you had used the correct figure, it would have made your next point stronger (but still wrong).

EXACTLY! Thanks for helping me make it TWICE stronger!

Quote
What does "direction of the stars will have greatly changed, however small the angle of parallax maybe" mean? Isn't "direction ... will have greatly changed" the opposite of "however small the angle"?

Here is the answer:

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.
« Last Edit: November 09, 2014, 02:06:50 AM by cikljamas »
"I can't breathe" George Floyd RIP

#### cikljamas

• 2174
• Ex nihilo nihil fit
##### Re: "Equator" problem
« Reply #283 on: November 09, 2014, 02:05:07 AM »
Quote
These models (you call them "assumptions") are verified by actual observations and measurements as we go, though. Sometimes better observations result in better measurements that require earlier models to be revised or occasionally thrown out altogether. This is how science works.

Bull s h i t! Don't call this crap "science"? What actual observations are you talking about?

Gravitation is a clever illustration of the art of hocus-pocus—heads I win, tails you lose ; Newton won his fame, and the people lost their senses.

Lord Beaconsfield wisely said—" A subject or system that will not bear discussion is doomed." Both Copernicus himself, who revived the theory of the heathen philosopher Pythagoras, and his great exponent Sir Isaac Newton, confessed that their system of a revolving Earth was only a possibility, and could not be proved by facts. It is only their followers who have decorated it with the name of an " exact science," yea, according to them, " the most exact of all the sciences."

" We declare that this motion is all mere ' bosh,' and that the arguments which uphold it are, when examined by an eye that seeks Truth, mere nonsense and childish absurdity."

Now what confidence can any man place in a science which gives promissory notes of such extravagance as these? They are simply bankrupt bills, not worth the paper on which they are written. And yet, strange to say, many foolish people endorse them as if they were good, the reason being that they are too lazy to think for themselves, and, to their own sad cost, accept the bogus notes as if they had been issued by a Rothschild."

"      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."

@ Ausgeoff, the testimonies of a noble men who had spotted Polaris at 12 degree South, and even all the way down to 23 degree South are still worthy of our attention, and always will be, no matter how far science of lying gets by developing various technics of deceiving and lying.

Quote
Scientific knowledge has advanced logarithmically since then my friend.

Well, my friend, all you have to do is to read these two very educational and sobering texts:

http://www.energeticforum.com/253864-post240.html

http://www.energeticforum.com/255678-post304.html

It (reading it) will do good to you but only if you are not 200 % troll, we already know that you are 100 % troll!
« Last Edit: November 09, 2014, 02:07:29 AM by cikljamas »
"I can't breathe" George Floyd RIP

?

#### Saros

• 403
##### Re: "Equator" problem
« Reply #284 on: November 09, 2014, 02:17:09 AM »
This is a better calculator -> http://members.home.nl/7seas/radcalc.htm

Please be honest and admit you shouldn't see so much. Calculating the horizon distance doesn't mean you should see something from bottom to top, does it? It is 5642 meters, but even in my photos you can tell that the 2000 mark is above the horizon.

Ausgeoff's 'eye level' was 3 meters. You can read his comment. I hope he will provide more details about his observation though. Take a look at the new Elbrus picture I posted. I found it on the Internet. It was taken from Batumi beach. By the way, some of my photos can be found on the Internet on a website that deals with long distance observations.

http://www.barents.pl/sites/default/files/550988_356863567725394_578677839_n.jpg
The calculator I linked shows the distance to the horizon from whatever height you enter, and just below that it shows how much Earth curves over whatever distance you enter.  I'm not sure how to get that info from the calculator you linked.

Here's a few pictures I took a while ago.  It's 9 miles to the bridge, and 12 to the buildings along the shore.  The left half was from an elevation of around 20 feet (I could be a couple feet off), and the right half was taken from about 7 inches.  Even from a height of 20 feet however, the horizon is 5.5 miles out, so there is still a drop over the remaining distance, and the buildings are still only visible due to refraction.  I would have had to be about 80 feet high to put the horizon at the shoreline and have a completely 'unsunken' view.

The bridge deck, tree-line, and hillside in general have visibly 'sunk' beyond the horizon/waterline.  Yet the buildings are still visible, although with a compressed appearance, including the landslide, due to refraction.  The top of the hillside is about 300 feet.

How many degrees above the horizon would you say the tops of the trees are?
The calculator I linked is much better. You have the option to enter your height(H1) and then the height of the observed object (H2), and see the horizon distance as it changes. If you enter just H1 you would get the general horizon distance without taking into account the observed object's height, it basically assumes it's 0. Please stop attributing everything anomalous to refraction as it is not true. If you see something above the calculated horizon and it is not distorted, the refraction is not anomalous but within the normal range. Anomalous refraction causes distortion like in fata morgana or mirages. By the way, if I apply your calculations to my photo, the horizon distance from 7 ft is 3.24 miles, so bearing in mind that Elbrus is 5642 m (3.5 miles) does it mean that it is 0.26 miles (418 meters) above the horizon? I don't think it is calculated like  that, but if it were then Elbrus shouldn't be so much protruded above the horizon at all. It would literally mean that you can only see its top 418 meters. Are you sure this is what you're claiming? I don't think this makes sense, that is why you should enter both heights. The horizon distance from a height of 2 meters is 5 km, but the horizon distance from 5642 meters is close to 270 km. That is why there is a formula, and that is why it is necessary to enter both heights. Of course, the formula might be wrong as well. I am just showing you how the calculator works.
« Last Edit: November 09, 2014, 02:50:16 AM by Saros »

#### ausGeoff

• 6091
##### Re: "Equator" problem
« Reply #285 on: November 09, 2014, 06:07:31 AM »
@ Ausgeoff, the testimonies of a noble men who had spotted Polaris at 12 degree South, and even all the way down to 23 degree South are still worthy of our attention, and always will be, no matter how far science of lying gets by developing various technics of deceiving and lying.
Quote
Scientific knowledge has advanced logarithmically since then my friend.
Well, my friend, all you have to do is to read these two very educational and sobering texts:
http://www.energeticforum.com/253864-post240.html

http://www.energeticforum.com/255678-post304.html

It (reading it) will do good to you but only if you are not 200 % troll, we already know that you are 100 % troll!
Can you please refrain from repeatedly posting links to your delusional opinions on the Energetic Forum site and referring to your own pseudo-scientific drivel as "educational".

And why is it that so many flat earthers—when confronted with actual contemporary science—start calling any opposition "trolls"—like petulant little schoolkids?

You can call me any names you like sonny, but it ain't gonna win you any friends here.

#### JimmyTheCrab

• 9445
##### Re: "Equator" problem
« Reply #286 on: November 09, 2014, 06:12:01 AM »

Yes, it looked much higher than the horizon. I am talking about Caucasus and especially Elbrus. I don't believe it should be visible that high from 200 km away. Seeing something at the horizon means exactly that. It means the very top of the thing is at the horizon. Clearly Mt.Elbrus is not at the horizon but much higher, and you definitely don't see its top 5 meters only.
@Saros, where do you get 5 metres from?  You should be able to see the top 2700m of the mountain - which is an awful lot of mountain still.

You can try the calculation here:

http://www.cactus2000.de/uk/unit/masshor.shtml

I entered the elevation as 3m, the distance as 200km.  This calculates the smallest object you can see as 2947m.  Which means if the mountain is 5642m we should still be able to see 2695m of mountain.  This is nearly 2.7km of mountain - ie loads.  I'm not sure what the problem is?
Quote from: mikeman7918
a single photon can pass through two sluts

Quote from: Chicken Fried Clucker
if Donald Trump stuck his penis in me after trying on clothes I would have that date and time burned in my head.

?

#### Saros

• 403
##### Re: "Equator" problem
« Reply #287 on: November 09, 2014, 07:53:43 AM »

Yes, it looked much higher than the horizon. I am talking about Caucasus and especially Elbrus. I don't believe it should be visible that high from 200 km away. Seeing something at the horizon means exactly that. It means the very top of the thing is at the horizon. Clearly Mt.Elbrus is not at the horizon but much higher, and you definitely don't see its top 5 meters only.
@Saros, where do you get 5 metres from?  You should be able to see the top 2700m of the mountain - which is an awful lot of mountain still.

You can try the calculation here:

http://www.cactus2000.de/uk/unit/masshor.shtml

I entered the elevation as 3m, the distance as 200km.  This calculates the smallest object you can see as 2947m.  Which means if the mountain is 5642m we should still be able to see 2695m of mountain.  This is nearly 2.7km of mountain - ie loads.  I'm not sure what the problem is?
Yeah, this is correct. I am not arguing against that. I am saying that if it were to be visible barely at the horizon you would only see its very top. In the photo it is only 200 km away, so it is not at the horizon but higher, which is normal indeed even though such long distance observations are not very common.

You can always argue how high the mountain should appear to be given the distance. I cannot provide any strong argument about this though.

What I am curious about is why the mountains observed from big distances are not inclined? I mean, assuming the Earth is round and they go straight up, if you look at them from say 200 km, they shouldn't be pointing upward 90 degrees from your perspective, but there should be an incline, no? Just making a point. At least the horizon distance diagrams appear to show just that. Do you know anything about this? Same should be true for tall buildings. If you look at one from a distance they shouldn't appear straight but like the leaning tower of Pisa, right?
« Last Edit: November 09, 2014, 08:05:11 AM by Saros »

#### JimmyTheCrab

• 9445
##### Re: "Equator" problem
« Reply #288 on: November 09, 2014, 08:33:18 AM »
@Saros, I understand what you mean about them "leaning away", but how much do you think this would be by?

More importantly, how would you ever perceive it?  How would a large object 200km away leaning slightly away from you look any different from one that wasn't?
Quote from: mikeman7918
a single photon can pass through two sluts

Quote from: Chicken Fried Clucker
if Donald Trump stuck his penis in me after trying on clothes I would have that date and time burned in my head.

#### cikljamas

• 2174
• Ex nihilo nihil fit
##### Re: "Equator" problem
« Reply #289 on: November 09, 2014, 09:17:54 AM »
@Saros, I understand what you mean about them "leaning away", but how much do you think this would be by?

More importantly, how would you ever perceive it?  How would a large object 200km away leaning slightly away from you look any different from one that wasn't?

Well, you should live on a round planet in order to see how would it be if it were to be...But since you are clever boy you can try (and only try) to imagine such a nonsensical  world (in terms of functionality of basic physical laws as we (don't) know them)

In first link that i suggested for reading to AusGeoff you can find this description of hypothetical world of ours and how such hypothetical world would look like if a fairy tales made up by heliocentrists were true description of our reality:

According to the theory, there should be a distance beyond every edge of every galaxy and every star where the light behind is bent just the right amount to reach us here on Earth. All objects that we can see have other objects behind them. Every star we see has stars and/or galaxies behind it, and many objects we see are eclipsing objects of considerable brightness. If bending and lensing were true, we would expect every single object in the sky to be fully haloed. No, more than that: we should expect the entire sky to be filled with bent light.

Every object we see has an object behind it or near it, and every object has a distance of bending beyond every edge where the angle would be right to bend the light to us. Therefore the night sky should be filled from corner to corner with multiple images. According to the theory of light bending, there shouldn’t be a dark dot in the sky.

« Last Edit: November 09, 2014, 09:20:55 AM by cikljamas »
"I can't breathe" George Floyd RIP

#### 29silhouette

• 3304
##### Re: "Equator" problem
« Reply #290 on: November 09, 2014, 09:40:36 AM »
The calculator I linked is much better. You have the option to enter your height(H1) and then the height of the observed object (H2), and see the horizon distance as it changes. If you enter just H1 you would get the general horizon distance without taking into account the observed object's height, it basically assumes it's 0.
Using my picture of the bridge and hillside, I entered;
20ft for h1
300ft for h2
It gives me;
radar horizon of 30.82 miles
visual horizon of 26.69 miles.

How does that figure into my picture of a hillside 12 miles away?

Quote
Please stop attributing everything anomalous to refraction as it is not true. If you see something above the calculated horizon and it is not distorted, the refraction is not anomalous but within the normal range. Anomalous refraction causes distortion like in fata morgana or mirages.
The refraction is obvious in my picture since I took it from two different elevations.  The shoreline buildings that dropped below the horizon are still visible, but compressed, gradually diminishing the higher the object is.  It's quite common.  I've got other pictures taken from the same spot looking through a mirage.  The results are quite different.

Quote
By the way, if I apply your calculations to my photo, the horizon distance from 7 ft is 3.24 miles, so bearing in mind that Elbrus is 5642 m (3.5 miles) does it mean that it is 0.26 miles (418 meters) above the horizon? I don't think it is calculated like  that, but if it were then Elbrus shouldn't be so much protruded above the horizon at all. It would literally mean that you can only see its top 418 meters. Are you sure this is what you're claiming?
I thought you were the one claiming it should only be barely visible.  My calculations show a couple thousand meters should still be visible.

Quote
I don't think this makes sense, that is why you should enter both heights. The horizon distance from a height of 2 meters is 5 km, but the horizon distance from 5642 meters is close to 270 km. That is why there is a formula, and that is why it is necessary to enter both heights. Of course, the formula might be wrong as well. I am just showing you how the calculator works.
So an observer at the peak would see the horizon about 70km beyond Batumi. Observer's height, distance to horizon, curvature drop along the visual line of sight between the horizon and observed object, and common refraction all explain our pictures pretty well.

#### 29silhouette

• 3304
##### Re: "Equator" problem
« Reply #291 on: November 09, 2014, 10:40:19 AM »
How about the proofs for the immobility of the Earth?

Allow me to present you some of them:
1. At the equinox from the pole it would skim the horizon all day.  The camera would have to be spun around in a slow circle for 24 hours to track it.  There are videos such as this.  One in particular wasn't exactly at the pole though, so the sun moves up and down as it circles the horizon.

2.  Notice the 'star trail' pictures in which the stars closest to the celestial poles have the shortest trails versus those closer to the celestial equator?

3a.  The air pretty much rotates with the surface.

3b.  Linking to your own flawed post with your own self quote doesn't mean much if you can't figure out that if a person or measuring device is moving along at the exact same speed as the wind to be measured, it's not going to measure much wind.

4.  Not sure exactly what you mean, but if you're trying to say things would fly off, no, one rotation every 24 hours isn't fast enough.

5.  The object would still have enertia as it falls.

other 5.  Earth shouldn't have gravity because it rotates?

6.  No.  There are experiments to show it is rotating.

7.  Formation of the planet.

8.  Surface was rotating as atmosphere formed.  What is there outside atmosphere to slow it?

Anyway, speaking of all these examples of evidence, did you ever find any visual proof yet southern stars moving overhead indicative of circling a big disk as opposed to circling a celestial pole of a globe as is commonly seen?

?

#### Saros

• 403
##### Re: "Equator" problem
« Reply #292 on: November 09, 2014, 10:47:16 AM »
The calculator I linked is much better. You have the option to enter your height(H1) and then the height of the observed object (H2), and see the horizon distance as it changes. If you enter just H1 you would get the general horizon distance without taking into account the observed object's height, it basically assumes it's 0.
Using my picture of the bridge and hillside, I entered;
20ft for h1
300ft for h2
It gives me;
radar horizon of 30.82 miles
visual horizon of 26.69 miles.

How does that figure into my picture of a hillside 12 miles away?

Quote
Please stop attributing everything anomalous to refraction as it is not true. If you see something above the calculated horizon and it is not distorted, the refraction is not anomalous but within the normal range. Anomalous refraction causes distortion like in fata morgana or mirages.
The refraction is obvious in my picture since I took it from two different elevations.  The shoreline buildings that dropped below the horizon are still visible, but compressed, gradually diminishing the higher the object is.  It's quite common.  I've got other pictures taken from the same spot looking through a mirage.  The results are quite different.

Quote
By the way, if I apply your calculations to my photo, the horizon distance from 7 ft is 3.24 miles, so bearing in mind that Elbrus is 5642 m (3.5 miles) does it mean that it is 0.26 miles (418 meters) above the horizon? I don't think it is calculated like  that, but if it were then Elbrus shouldn't be so much protruded above the horizon at all. It would literally mean that you can only see its top 418 meters. Are you sure this is what you're claiming?
I thought you were the one claiming it should only be barely visible.  My calculations show a couple thousand meters should still be visible.

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I don't think this makes sense, that is why you should enter both heights. The horizon distance from a height of 2 meters is 5 km, but the horizon distance from 5642 meters is close to 270 km. That is why there is a formula, and that is why it is necessary to enter both heights. Of course, the formula might be wrong as well. I am just showing you how the calculator works.
So an observer at the peak would see the horizon about 70km beyond Batumi. Observer's height, distance to horizon, curvature drop along the visual line of sight between the horizon and observed object, and common refraction all explain our pictures pretty well.

I am sorry, but I don't understand your point. You gave me the picture. I don't know anything about it. I just told you that the picture is fine. I didn't say you're looking at a mirage. There is no anomalous refraction in it at first glance. This is what I said. The refraction is normal. You're simply changing the angle of view. The refraction is completely normal as there is no distortion in any of the observed objects. You're the one who gave me another calculator claiming it does a better job when in fact it produces the same results. What is the argument about?  I am confused now, and I don't know what we're discussing here anymore. Please stop addressing every word I say, but focus on the big picture.

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#### Saros

• 403
##### Re: "Equator" problem
« Reply #293 on: November 09, 2014, 10:51:29 AM »
@Saros, I understand what you mean about them "leaning away", but how much do you think this would be by?

More importantly, how would you ever perceive it?  How would a large object 200km away leaning slightly away from you look any different from one that wasn't?

Are you sure it should be leaning only slightly away? How about buildings which are smaller objects than peaks ? Shouldn't the leaning away be visible? From 15-20 km, there should be enough leaning away to be observed, right? I am just saying. Can't give you the exact math.

#### cikljamas

• 2174
• Ex nihilo nihil fit
##### Re: "Equator" problem
« Reply #294 on: November 09, 2014, 11:54:26 AM »
Anyway, speaking of all these examples of evidence, did you ever find any visual proof yet southern stars moving overhead indicative of circling a big disk as opposed to circling a celestial pole of a globe as is commonly seen?

How come that you always turn discussion in wrong (celestial) direction whenever we point to the right direction which is Earth itself?

If we discuss shape of the Earth, why don't you stick with the Earth and her properties instead of hiding behind the celestial phenomena and thus diverging discussion in wrong direction?

Btw, have you ever asked yourself how did we get uniform water LEVEL on something like this:

"I can't breathe" George Floyd RIP

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#### neimoka

• 738
##### Re: "Equator" problem
« Reply #295 on: November 09, 2014, 12:14:09 PM »
topography not at all exaggerated

#### 29silhouette

• 3304
##### Re: "Equator" problem
« Reply #296 on: November 09, 2014, 12:19:43 PM »
I am sorry, but I don't understand your point. You gave me the picture. I don't know anything about it.
I gave details.  If you need to know more, just ask.

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I just told you that the picture is fine. I didn't say you're looking at a mirage. There is no anomalous refraction in it at first glance. This is what I said. The refraction is normal. You're simply changing the angle of view.
Moving 20 feet will not change the angle of something 12 miles away enough to result in that much difference.

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The refraction is completely normal as there is no distortion in any of the observed objects.
Except for the compression of the lower objects.

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You're the one who gave me another calculator claiming it does a better job when in fact it produces the same results.
No, I provided a link to a site with calculators that shows the distance to the horizon based on elevation and the amount of curvature drop over a distance.  You're the one who provided another calculator claiming it does a better job when in fact it produces the same results for finding the distance to the horizon, but not the drop over the distance.

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What is the argument about?  I am confused now, and I don't know what we're discussing here anymore. Please stop addressing every word I say,
Hmmm... ok. I'm not addressing every word, just the pertinent sentences.

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but focus on the big picture.
Ok.  Your pictures are possible with a curved surface, and mine indicate a curved surface.

#### 29silhouette

• 3304
##### Re: "Equator" problem
« Reply #297 on: November 09, 2014, 12:30:10 PM »
Anyway, speaking of all these examples of evidence, did you ever find any visual proof yet southern stars moving overhead indicative of circling a big disk as opposed to circling a celestial pole of a globe as is commonly seen?

How come that you always turn discussion in wrong (celestial) direction whenever we point to the right direction which is Earth itself?

If we discuss shape of the Earth, why don't you stick with the Earth and her properties instead of hiding behind the celestial phenomena and thus diverging discussion in wrong direction?
Lol...Because you derailed your own thread onto the discussion of celestial objects with reply#10 back on the first page.

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Btw, have you ever asked yourself how did we get uniform water LEVEL on something like this:

http://i.imgur.com/D4ZoqyH.jpg
Go top off an ice-cube tray and get back to us.

#### ausGeoff

• 6091
##### Re: "Equator" problem
« Reply #298 on: November 09, 2014, 12:46:44 PM »
Btw, have you ever asked yourself how did we get uniform water LEVEL on something like this:

I'm pleased that you said something "like" this.  Your image is nothing more than a computer-generated rendition of the planet, with a scale perpendicular to its surface that's totally distorted.  And you still seem to think that the water in the oceans and lakes is "level" (or flat?).  It's not.  Its mean surface is curved.

—Please don't bother posting any more George Lucas-style cartoons.

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#### Saros

• 403
##### Re: "Equator" problem
« Reply #299 on: November 09, 2014, 01:22:48 PM »
Btw, have you ever asked yourself how did we get uniform water LEVEL on something like this:

I'm pleased that you said something "like" this.  Your image is nothing more than a computer-generated rendition of the planet, with a scale perpendicular to its surface that's totally distorted.  And you still seem to think that the water in the oceans and lakes is "level" (or flat?).  It's not.  Its mean surface is curved.

—Please don't bother posting any more George Lucas-style cartoons.

OK, you are entitled to think so, but do you have any proof for that besides mere words? Has any experiment been done to confirm that the ocean surface is curved? For instance, a relatively easy experiment could be extending a straight metal rod across a bay. I have never heard of anything like that. All the evidence that it is curved is mathematical and observational(mainly astronomical). How about measuring the Earth itself?