"Equator" problem

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cikljamas

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Re: "Equator" problem
« Reply #330 on: November 11, 2014, 06:04:22 AM »
ON GREAT FLOOD:

Did Noah's flood cover the whole earth or just areas where man lived?

Often when this question comes up it is because many, perhaps most, scientists these days are trying to say that if there was a flood of cataclysmic proportions, it was only a localized, or regional flood. That is, it occurred only where there was life, in particular, human life.

The scientists and others can't deny there was a great flood, because: 1) There is sea fossil life on even the summits of some of the world's highest mountains; and 2) Cave drawings, and other recorded art and written works by primitive civilizations around the globe record in their legends a great flood of some sort.

God's Word says:

"And the flood was forty days upon the earth; and the waters increased, and bare up the ark, and it was lift up above the earth. And the waters prevailed, and were increased greatly upon the earth; and the ark went upon the face of the waters. And the waters prevailed exceedingly upon the earth; and all the high hills, that were under the whole heaven, were covered. Fifteen cubits upward did the waters prevail; and the mountains were covered" (Gen. 7:17-20).

The Bible says further:

"And all flesh died that moved upon the earth, both of fowl, and of cattle, and of beast, and of every creeping thing that creepeth upon the earth, and every man: All in whose nostrils was the breath of life, of all that was in the dry land, died. And every living substance was destroyed which was upon the face of the ground, both man, and cattle, and the creeping things, and the fowl of the heaven; and they were destroyed from the earth: and Noah only remained alive, and they that were with him in the ark"
(Gen. 7:21-23).

Peter the apostle forewarns that the skeptics would be talking about such things in the last days.

"Knowing this first, that there shall come in the last days scoffers, walking after their own lusts, And saying, Where is the promise of his coming? for since the fathers fell asleep, all things continue as they were from the beginning of the creation. For this they willingly are ignorant of, that by the word of God the heavens were of old, and the earth standing out of the water and in the water: Whereby the world that then was, being overflowed with water, perished"[/b][/u] (2 Pet. 3: 3-6).

Believe the scientists and the other scoffers, or believe God, who cannot lie. We weren't there, so we must believe one or the other. It's a matter of faith, but the geological record is on God's side. The whole earth was covered at least 15 cubits above the highest mountain peak.
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Rama Set

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Re: "Equator" problem
« Reply #331 on: November 11, 2014, 06:25:40 AM »
On GOD:

"The cosmos is all that is, or ever was, or ever will be." Carl Sagan

According to above logic we should assume that one of next two options must be true:

1. The cosmos is eternal
2. The cosmos is not eternal, but still is "all that is, or ever was, or ever will be"

I choose to insert another option:

3. The cosmos is finite and before that nothing existed.

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If 1 then the cosmos is just another word for God! - What is wrong with such an idiotic identification?

No it isn't.  God is purported to be a sentient being whereas the cosmos has no such requirement.

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This is what is wrong: - The true meaning of a word "God" is: A being which cannot not to be!!!

Who declared this the true definition?  It sounds like you are asserting it is the truth for the expedience of your argument.

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Such Being can't be created since God has no need to have been created, He exists outside time, but the cosmos HAS NEED TO HAVE BEEN CREATED and IS submitted to the second law of thermodynamics!

No one knows if this is the truth or not.  This is merely conjecture.  Please be honest and treat it as such.

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If 2 then the cosmos popped into existence out of nothing, and one of the most basic philosophical principles is EX NIHILO NIHIL FIT which means that there is no such thing as "creation caused by nothing", or as "created entity that comes out of nothing."

That may not be true.  Please refer to the work of Lawrence Krauss.  At this time, Ex Nihilo creation cannot be definitively ruled out for creation of the universe.

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Of course, there is one necessary "exception": God himself!

Only necessary because you have asserted as such.  You literally have no evidence to show this.

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God is Absolute, Uncontingent Being, Essence of Existence, Actus Purus, First - Necessary Cause, Unmoved Mover, God exists forever and ever, He is a Giver of all contingent existences, but He Himself is Unlimited! God had introduced Himself to Moses using perfectly consistent philosophical definition: "I AM THAT I AM"! (IHWH).

God told you this himself, or you read it in a thousands of years old book of questionable authorship?

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...

tl;dr
Aether is the  characteristic of action or inaction of charged  & noncharged particals.

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ausGeoff

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Re: "Equator" problem
« Reply #332 on: November 11, 2014, 08:20:17 AM »
It's actually quite astounding that in a scientifically enlightened 21st century, we still find the occasional diehard, fundamentalist god-botherer such as cikljamas attempting—in vain—to discredit every one of the natural sciences whose confirmatory data set has been built up over centuries by guys certainly smarter than any of us here.

The typically smug, condescending attitude of the bible-bashers is certainly repugnant to any rational person in 2014 that's for sure.  Bear in mind that these religious apologists use as a reference book a document cobbled together by a disparate group of ill-educated desert nomads more than 2,000 years ago over a 400-year period.  And those same guys believed in supernatural entities (such as their "god") and paranormal phenomena (such as "miracles").

And what's even more disconcerting is that someone like cikljamas—who should know better, and be better informed—still believes in the existence of gods and angels, and devils and miracles.

That being the case, it's virtually impossible to engage in any sort of meaningful, contemporary scientific debate with people like cikljamas who either choose to willfully ignore the scientific evidence, or who lack the necessary technical and/or logic skills to evaluate it.


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cikljamas

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Re: "Equator" problem
« Reply #333 on: November 11, 2014, 09:53:46 AM »
@ Rama Set, watch this :
Can God's Existence be Demonstrated? (William Lane Craig) : " class="bbc_link" target="_blank" rel="noopener noreferrer">

It's actually quite astounding that in a scientifically enlightened 21st century, we still find the occasional diehard, fundamentalist god-botherer such as cikljamas attempting—in vain—to discredit every one of the natural sciences whose confirmatory data set has been built up over centuries by guys certainly smarter than any of us here.

The typically smug, condescending attitude of the bible-bashers is certainly repugnant to any rational person in 2014 that's for sure.  Bear in mind that these religious apologists use as a reference book a document cobbled together by a disparate group of ill-educated desert nomads more than 2,000 years ago over a 400-year period.  And those same guys believed in supernatural entities (such as their "god") and paranormal phenomena (such as "miracles").

And what's even more disconcerting is that someone like cikljamas—who should know better, and be better informed—still believes in the existence of gods and angels, and devils and miracles.

That being the case, it's virtually impossible to engage in any sort of meaningful, contemporary scientific debate with people like cikljamas who either choose to willfully ignore the scientific evidence, or who lack the necessary technical and/or logic skills to evaluate it.


Evaluate this : Proof The Earth Is Young and Noah's Flood : " class="bbc_link" target="_blank" rel="noopener noreferrer">
« Last Edit: November 11, 2014, 10:00:07 AM by cikljamas »
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29silhouette

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Re: "Equator" problem
« Reply #334 on: November 11, 2014, 10:02:30 AM »
You guys are very confused.
I am confused now, and I don't know what we're discussing here anymore.

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You can see they are leaning away,
Can you point the indicators of this in your pictures?

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because your knowledge is not based on real, first-hand observations, experiments and analysis, but on mathematical equations, which assume unproven facts.
I conducted an experiment by taking pictures from different elevations, and then analyzed the results.  It shows me Earth isn't flat.

Enjoy your math world, which has nothing to do with reality. Now I am finally convinced you know nothing about the shape of the planet,
Would it help if the formula and/or calculation was in the form of a calculator?  You seem to be ok with those.

What do you believe the shape of the planet to be?  Flat or globe?

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ausGeoff

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Re: "Equator" problem
« Reply #335 on: November 11, 2014, 10:05:34 AM »
Evaluate this : Proof The Earth Is Young and Noah's Flood : " class="bbc_link" target="_blank" rel="noopener noreferrer">

I'm afraid I don't have the time or the inclination to sit through a fairy story proposed by William Lane Craig.  I've already seen more than enough of his religious drivel during an encounter with the late Christopher Hitchens, and during which Hitchens shot the guy down in flames.  Same goes for nearly an hour of the alleged "young age" of the earth video.

All these young-earth creationists must've come off the one assembly line LOL.  They're all certifiable whack-jobs.   

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Socratic Amusement

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Re: "Equator" problem
« Reply #336 on: November 11, 2014, 10:07:39 AM »
HA! William Lane Craig...

Wow, those are some pretty old, thoroughly destroyed arguments you are linking to, cikljamas.

Jeez. I haven't seen someone use WLC since 2008. I thought everyone knew how intellectually bankrupt he is, including Christians?
"As for me, all I know is that I know nothing."

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JimmyTheCrab

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Re: "Equator" problem
« Reply #337 on: November 11, 2014, 10:35:50 AM »
Would it help if the formula and/or calculation was in the form of a calculator?  You seem to be ok with those.

There is a calculator here:

http://www.1728.org/radians.htm
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Alpha2Omega

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Re: "Equator" problem
« Reply #338 on: November 11, 2014, 11:25:11 AM »
You guys are very confused. Using a formula you get some results and of course you believe them immediately. It is mathematics after all. This formula is wrong, and it is common sense that if there is a curvature the objects will lean more especially if they are tall enough when observed from a big distance. You go tell this 1.9% to someone else, I am not buying that math, as it is wrong to anyone with half a brain, but you apparently. That might be useful in high school to get an A but not in real life.
You seem to be the one who's confused.

Since cikljamas has gone off to a completely religious diatribe which I wouldn't touch with a 10-foot pole, there's time for this sub-discussion. The question at hand seems to have originated with this:

@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 ;D? 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.
Here Saros surmises that buildings 15-20 km distant should lean away from the observer enough to be perceived and suggests that math can give the answer, but says he doesn't know how to do the math. That's fair enough; most people don't know how to approach this problem. But now, after being shown how to do the math - two different ways - he has a temper tantrum and complains that it isn't a math problem. Who's confused here?

Revisiting Saros' specific question about whether buildings 15-20 km away should be visibly tilting or not, and his assertion that they should, let's see what we get:

a = 20 km / 6371 km = 0.00314 radians
0.00314 radians * 180° / pi radians = 0.18°

A plumb building 20 km from you will be leaning directly away from you by 0.18°. If this building is 100m tall, its top will be 31 cm further from you than if it were square with your sightline. (hint: you can multiply that radian measure of the angle by the height and arrive directly at the answer here). Do you think you can detect that change in distance - directly away from you - in a 30-story building from 20 km away? Considering the building itself would subtend only a bit more than 1/4° (if you could see the whole thing), this seems exceedingly unlikely.

How is that formula wrong? You said you didn't know the exact math, didn't you. If you've learned how to do this since your original post, what do you think the correct formula is?

Not to mention, that I was misleading you by claiming that you can't see they are leaning away. I just wanted to see how confused you are. You can see they are leaning away, but you guys are so big fans of math that you just proved you know nothing about reality, and supposedly you just proved that any leaning away shouldn't be visible, when in fact it is. Good job! Great math. Great logic. Congratulations!!! Enjoy your math world, which has nothing to do with reality. Now I am finally convinced you know nothing about the shape of the planet, because your knowledge is not based on real, first-hand observations, experiments and analysis, but on mathematical equations, which assume unproven facts.
If they are leaning away, doesn't that indicate the Earth is curved? Or did I miss something while trying to catch up with this branch of the thread? So, how much do your first-hand observations, experiments and analysis show they are leaning away from you? Can you describe the analysis? Does it involve math?

As an aside, I remember reading about building the Verrazano Narrows Bridge in NYC when I was a kid. It had the longest suspension-bridge span in the world when it was built and the fact that they had to account for the curvature of the Earth was mentioned at the time. It's not much, but, apparently, was enough to be considered in the engineering.

Quote from: Wikipedia
Because of the height of the towers (693 ft or 211 m) and their distance apart (4,260 ft or 1,298 m), the curvature of the Earth's surface had to be taken into account when designing the bridge—the towers are 1 5⁄8 inches (41.275 mm) farther apart at their tops than at their bases.
I get a bit less than 43 mm using the technique above, but they may be using a slightly different value for the radius (6371 km is the volumetric mean), and the difference is only about 1/16", so may mostly be round-off error.
"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 #339 on: November 11, 2014, 12:06:55 PM »
1. How about being able to see a whole figure of 200 km distant mountains from bottom to top? How come that more than a half of these distant mountains is not out of range of our sight due to alleged curvature of the Earth?

Alpha2Omega in post #126 writes:

See that sharp line "cutting off" bottom of the mountains in the distance, commonly called the horizon? That's the "brow" of your "hill of water". It's fairly close (and so not as "high") to the photographer since he's at a relatively low height above the water. It's further from the mountains in the distance and higher than their bases by your 2000' since they're farther away. You still see their tops because they're higher than the "hill of water". This diagram may explain it better.



Alpha2Omega offers us totally wrong explanation (as always), but his diagram is much closer to the supposed reality of allegedly globular Earth!
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Alpha2Omega

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Re: "Equator" problem
« Reply #340 on: November 11, 2014, 01:36:32 PM »
1. How about being able to see a whole figure of 200 km distant mountains from bottom to top? How come that more than a half of these distant mountains is not out of range of our sight due to alleged curvature of the Earth?

Alpha2Omega in post #126 writes:

See that sharp line "cutting off" bottom of the mountains in the distance, commonly called the horizon? That's the "brow" of your "hill of water". It's fairly close (and so not as "high") to the photographer since he's at a relatively low height above the water. It's further from the mountains in the distance and higher than their bases by your 2000' since they're farther away. You still see their tops because they're higher than the "hill of water". This diagram may explain it better.



Alpha2Omega offers us totally wrong explanation (as always), but his diagram is much closer to the supposed reality of allegedly globular Earth!
What's wrong with the explanation? Where does it not match the diagram?
"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 #341 on: November 11, 2014, 02:39:12 PM »
1. How about being able to see a whole figure of 200 km distant mountains from bottom to top? How come that more than a half of these distant mountains is not out of range of our sight due to alleged curvature of the Earth?

Alpha2Omega in post #126 writes:

See that sharp line "cutting off" bottom of the mountains in the distance, commonly called the horizon? That's the "brow" of your "hill of water". It's fairly close (and so not as "high") to the photographer since he's at a relatively low height above the water. It's further from the mountains in the distance and higher than their bases by your 2000' since they're farther away. You still see their tops because they're higher than the "hill of water". This diagram may explain it better.



Alpha2Omega offers us totally wrong explanation (as always), but his diagram is much closer to the supposed reality of allegedly globular Earth!
What's wrong with the explanation? Where does it not match the diagram?

Here:



Zoomed shot of the same mountains taken from the same spot as in the first picture above:



Accompanying words of Saros:

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.

Now, try to conciliate your diagram with what is clearly shown in above picture and with what implies my question which i repeat once more:

1. How about being able to see a whole figure of 200 km distant mountains from bottom to top? How come that more than a half of these distant mountains is not out of range of our sight due to alleged curvature of the Earth?

Show me the "brow" of my "hill of water" in above pictures!

ONE PICTURE IS WORTH MORE THAN A THOUSAND WORDS!!!
« Last Edit: November 11, 2014, 02:41:42 PM by cikljamas »
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Alpha2Omega

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Re: "Equator" problem
« Reply #342 on: November 11, 2014, 08:35:31 PM »
1. How about being able to see a whole figure of 200 km distant mountains from bottom to top? How come that more than a half of these distant mountains is not out of range of our sight due to alleged curvature of the Earth?

Alpha2Omega in post #126 writes:

See that sharp line "cutting off" bottom of the mountains in the distance, commonly called the horizon? That's the "brow" of your "hill of water". It's fairly close (and so not as "high") to the photographer since he's at a relatively low height above the water. It's further from the mountains in the distance and higher than their bases by your 2000' since they're farther away. You still see their tops because they're higher than the "hill of water". This diagram may explain it better.



Alpha2Omega offers us totally wrong explanation (as always), but his diagram is much closer to the supposed reality of allegedly globular Earth!
What's wrong with the explanation? Where does it not match the diagram?

Here:



Zoomed shot of the same mountains taken from the same spot as in the first picture above:


These aren't the picture the diagram was drawn to represent.

Please place a link to posts you want to reference rather than a reply number. Navigating around long threads like this one is a pain (there is a shortcut instead of using the page numbers, but direct links are easier). Here's how to do that:

The Very top line of each post shows the title of the post, probably Re: <thread name>. That title is a link to the post itself. This one is
Re: "Equator" problem

In some browsers on Windows systems, if you right click that title, it will give you the option to "Copy link address" or something like that (Chrome does). If it's available, just select it. If you don't have the option to do that, just click the link and copy the URL from your browser's address line. Either way, paste it into your reply and make it a URL by highlighting it and clicking the little SPHERICAL EARTH and document icon    above the reply area. Ha! They really should do something about that! :)
like this:
From http://theflatearthsociety.org/forum/index.php?topic=62199.msg1636303#msg1636303

instead of
Alpha2Omega in post #126 writes:

Anyway...
That picture showed a mountain peak dropping directly below the sea-level horizon. This shows a distant peak behind a nearer ridge which is behind an even nearer low ridge for part of it, both of which fall below the sea-level horizon. The same principle applies in general, though.

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Accompanying words of Saros:

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.
How did he determine that 15°? I recall seeing this question before, but don't think he answered it. If the Earth were flat and something 200 km appeared 15° up, it would have to be more than 50 km high. Does he think this mountain is that high? Angles like this are notoriously hard to estimate; I think Saros is simply wrong when he says "at least 15 degrees".
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Even if it were 2000 meters high though it would still be seen at the horizon under those atmospheric conditions.
What?

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Now, try to conciliate your diagram with what is clearly shown in above picture and with what implies my question which i repeat once more:

1. How about being able to see a whole figure of 200 km distant mountains from bottom to top? How come that more than a half of these distant mountains is not out of range of our sight due to alleged curvature of the Earth?
It looks like there is a ridgeline between the photographer and the distant peak. That keeps you from seeing anything but the highest elevations of the more distant peak. It's hard to tell from these pictures what's going on with the lower reaches of the closer ridge, or how far away it is.  At any rate, that ridge appears to drop below the horizon.

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Show me the "brow" of my "hill of water" in above pictures!

ONE PICTURE IS WORTH MORE THAN A THOUSAND WORDS!!!
Are you kidding? The "brow" of the "hill of water" is the sharp line between dark blue (water) and lighter blue (distant hillsides) where the bottom of the ships disappear. What else would it be?

The words conveyed by the picture don't mean anything if you don't understand what the picture is saying.
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

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29silhouette

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Re: "Equator" problem
« Reply #343 on: November 11, 2014, 10:17:54 PM »
Accompanying words of Saros:
You two sure were arguing though on the energeticforum.  And now you're quoting him to support your argument? 

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1. How about being able to see a whole figure of 200 km distant mountains from bottom to top? How come that more than a half of these distant mountains is not out of range of our sight due to alleged curvature of the Earth?
You're not seeing the whole mountain.

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Show me the "brow" of my "hill of water" in above pictures!
Do you see the waterline where the hills appear?

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ONE PICTURE IS WORTH MORE THAN A THOUSAND WORDS!!!
My pictures shows the hills sinking below the curvature.


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29silhouette

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Re: "Equator" problem
« Reply #344 on: November 11, 2014, 10:45:52 PM »
How did he determine that 15°? I recall seeing this question before, but don't think he answered it. If the Earth were flat and something 200 km appeared 15° up, it would have to be more than 50 km high. Does he think this mountain is that high? Angles like this are notoriously hard to estimate; I think Saros is simply wrong when he says "at least 15 degrees".
I never did see an answer after I asked.  I even asked how high he figured the tree line to be in my picture.

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It looks like there is a ridgeline between the photographer and the distant peak. That keeps you from seeing anything but the highest elevations of the more distant peak. It's hard to tell from these pictures what's going on with the lower reaches of the closer ridge, or how far away it is.  At any rate, that ridge appears to drop below the horizon.
  There's a small dark spot just below the peak (the arrow is pointing directly toward it).  It's at the 10,000 foot level (10,300ft total just to the right) of some peaks 37 miles closer.  Mt Elbrus just pokes above it.  Just below this ridge is another ridge visible.  There's a lighter area (devoid of forest), sort of oblong shaped with the right end in line with the arrow.  The bottom is 5400 feet and the top is 6600 feet.  It's about another 5 miles closer.

Measuring the few distinct features on these ridges that match up in Ge, their full height extends below the waterline/horizon/peak of the hill of water quite a bit.  It's hard to get an exact measurement, as the lower elevations are closer than the higher elevations, which means the higher elevations are dropped that much more due to the long distance.  Beside which, just how close were these actually to the horizon and so how affected by refraction is everything?  And, is the waterline really exactly where it appears, or is it a tad higher in reality?  Judging by the way the hill slopes down to meet the water, and the small white dot, I think there's a tiny bit of mirage. 

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rottingroom

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Re: "Equator" problem
« Reply #345 on: November 12, 2014, 06:19:26 AM »
You guys are very confused. Using a formula you get some results and of course you believe them immediately. It is mathematics after all. This formula is wrong, and it is common sense that if there is a curvature the objects will lean more especially if they are tall enough when observed from a big distance. You go tell this 1.9% to someone else, I am not buying that math, as it is wrong to anyone with half a brain, but you apparently. That might be useful in high school to get an A but not in real life. Not to mention, that I was misleading you by claiming that you can't see they are leaning away. I just wanted to see how confused you are. You can see they are leaning away, but you guys are so big fans of math that you just proved you know nothing about reality, and supposedly you just proved that any leaning away shouldn't be visible, when in fact it is. Good job! Great math. Great logic. Congratulations!!! Enjoy your math world, which has nothing to do with reality. Now I am finally convinced you know nothing about the shape of the planet, because your knowledge is not based on real, first-hand observations, experiments and analysis, but on mathematical equations, which assume unproven facts.

We can do this so many ways and we keep coming up with the same number. Here is another way.

A circle is 360°.

The circumference of the round earth is 40,000km.

200km/40,000km = x/360°

Solve for x.

0.005 = x/360°

0.005*360° = x

1.8° = x

---------------------------------------

If you have a problem with the math then let's check it for something intuitive. Suppose there is a skyscraper at both the north pole and at the equator. If the earth is round then the skyscraper at the equator should be tilted 90° away from the skyscraper at the north pole.

According to this google search result (https://www.google.com/search?q=distance+from+north+pole+to+the+equator), the distance from the north pole to the equator is roughly 10,000 km, which incidentally, is what originally defined the metric system.

Again, the circumference of earth is 40,000km and there are 360° in a circle. If these numbers are plugged in it should all check out, right?

10,000km/40,000km = 90°/360°

1/4 = 1/4

Looks good!





« Last Edit: November 12, 2014, 06:21:37 AM by rottingroom »

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cikljamas

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Re: "Equator" problem
« Reply #346 on: November 12, 2014, 08:57:53 AM »



"I can't breathe" George Floyd RIP

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rottingroom

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Re: "Equator" problem
« Reply #347 on: November 12, 2014, 09:04:05 AM »
what part of your intuition is irrelevant is hard for you to grasp?

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rottingroom

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Re: "Equator" problem
« Reply #348 on: November 12, 2014, 09:38:13 AM »
clickijamas,

A while back a forum member by the name of Scintific Method made a thread called measure a mountain. Here is the thread.

http://theflatearthsociety.org/forum/index.php?topic=59240.0#.VGOZ8sljWmw

I invite you to do his experiment and see what you come up with. Here is the relevant parts of the original post.

Equipment: 1 spirit level, 1 set square, and 1 ruler. You might also want something to set the spirit level up on, and some packing to level it.

Method: pick a mountain that is easily identifiable from a distance (I have a few to pick from where I am). Find a point on the map 30km away, 40km away, and 50km away (or any other largish distances, as long as you can measure them accurately. This is part of what negates any doctoring of the maps: picking arbitrary points to measure from). Set up your spirit level so that it points toward the mountain you picked, ensuring it is level. Set your ruler at a specific distance along the top of the level, standing up at right angles to it (that's what the set square is for). Sight from the edge of the level furthest from the mountain to the mountain top, and note where your line of sight crosses the ruler. The ratio of the measurement on the ruler to it's distance along the top of the level would be equal to the ratio of the mountain's height above your observation point to your distance to it (simple triangular geometry).

The height of the location the measurement is taken from is important, as you will get different results from different elevations at a given distance. Another reason why this cannot be faked or doctored.

In the below diagram, the ratio of h/d should be equal to r/s if the earth is flat, and should always be that way, no matter how far away from the mountain you are.



When I did this experiment though, there was a difference between h/d and r/s, and it got bigger and bigger the further away I got from the mountain. I did some extra maths to work out where the mountain top would be for a round earth:

apparent height on a round earth = square root of ((earth's radius + h)2 - d2) - (earth's radius + altitude of your position)

When I tested the results against those calculations, they matched perfectly. Remember: maths and geometry don't care what shape the earth is; they will always tell you the truth.

Results:

Mountain (Castle Top, in the Nandewar Ranges): 1075m AMSL
Point 1 (my back yard): 30.44km from peak, 215m AMSL
Point 2 (Kamilaroi Hwy): 40.46km from peak, 205m ASMSL
Point 3 (Wee Waa levee): 55.57km from peak, 192m AMSL

Using a 1100mm baseline on the spirit level, the following apparent heights were recorded:
Point 1: 28mm
Point 2: 18mm
Point 3: 11mm
Let's be pessimistic and say they're only within 1mm either way.

Here's what they should have been for a flat earth:
Point 1: 31mm
Point 2: 21mm
Point 3: 16mm
Note the increasing difference. This is due to the mountain 'going over the edge'.

Conclusion: The data collected disagrees with FE predictions*, and almost exactly matches RE predictions. The experiment is easy to conduct, and not prone to significant error, so the results can be considered reliable.

* Rowbotham's 'perspective' has been cited as a possible cause of the results. However, a passing familiarity with geometry should be sufficient to show that this 'effect' is not responsible for the rate at which the actual measurements diverge from what they should be for a flat earth.

Interesting Note: if you use the following equation, you can work out the difference between the apparent height for a round earth and a flat earth (with respect to the diagram):

rf - rr = s * tan((d / ce) * 360 / 2)

where rf is the apparent height on a flat earth, rr is the apparent height on a round earth, and ce is the circumference of the round earth. d and ce need to be in the same units as each other, and rf - rr will be in the same units as s.


Sorry for the long post! I thought it would be best to give plenty of detail, so that others can reproduce this experiment and obtain accurate results.


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Alpha2Omega

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Re: "Equator" problem
« Reply #349 on: November 12, 2014, 11:42:53 AM »




It's Mt. Elbrus, not Mt. Erebus, isn't it?

I think those lines are at about the right places on your annotated picture. They are saying that a point a bit more than halfway between sea level and the top of Elbrus is about horizon level from your vantage point 200 km away (2989 is 53%  of 5642, approximately half). Extrapolating that downward gives an idea where sea level directly below the peak of Elbrus is below your local horizon.

What's funny about it, and where else would you expect it to be on a spherical planet?
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

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Saros

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Re: "Equator" problem
« Reply #350 on: November 12, 2014, 01:02:46 PM »

It's Mt. Elbrus, not Mt. Erebus, isn't it?

I think those lines are at about the right places on your annotated picture. They are saying that a point a bit more than halfway between sea level and the top of Elbrus is about horizon level from your vantage point 200 km away (2989 is 53%  of 5642, approximately half). Extrapolating that downward gives an idea where sea level directly below the peak of Elbrus is below your local horizon.

What's funny about it, and where else would you expect it to be on a spherical planet?

Here is what it should look like from the same coordinates and 20 meters elevation.


Source:http://www.peakfinder.org/?lat=41.6487&lng=41.645&off=20&ele=4

And this is what it looked like with absolutely no optical zoom(note that with the naked eye it appeared about twice bigger than what you see in the photo), and  the camera had only 3x optical zoom.
« Last Edit: November 12, 2014, 02:26:15 PM by Saros »

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JimmyTheCrab

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Re: "Equator" problem
« Reply #351 on: November 12, 2014, 01:51:38 PM »
@Saros, I've no idea what your point is??

"And this is what it looked like with absolutely no zoom"

So what was the focal length?
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.

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neimoka

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Re: "Equator" problem
« Reply #352 on: November 12, 2014, 01:57:50 PM »
And this is what it looked like with absolutely no zoom(note that with the naked eye it appeared about twice bigger), and  the camera had only 3x zoom.
"3x zoom" means absolutely nothing. It's like saying that something is three times bigger. Also, cameras do not have zoom. Ever. It's a feature of the lens. With the naked eye it appeared about twice bigger than what? Are you looking at the picture on a cell phone or a 50" screen or or what?

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cikljamas

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Re: "Equator" problem
« Reply #353 on: November 12, 2014, 01:59:16 PM »




It's Mt. Elbrus, not Mt. Erebus, isn't it?

Whatever, it doesn't make any difference however! It is still VERY INTUITIVE!!!

I am absolutely amazed with your ability to perceive and discern between large and small scale features.

Do you notice any difference between these two dogs:





Your attitude, not your aptitude, will determine your altitude.


This is a bad attitude:





"I can't breathe" George Floyd RIP

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Rama Set

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Re: "Equator" problem
« Reply #354 on: November 12, 2014, 02:01:55 PM »
Well that was unexpected.
Aether is the  characteristic of action or inaction of charged  & noncharged particals.

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cikljamas

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Re: "Equator" problem
« Reply #355 on: November 12, 2014, 02:06:24 PM »
Well that was unexpected.

You mean Einstein?

Well...


"I can't breathe" George Floyd RIP

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Saros

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Re: "Equator" problem
« Reply #356 on: November 12, 2014, 02:16:48 PM »
@Saros, I've no idea what your point is??

"And this is what it looked like with absolutely no zoom"

So what was the focal length?

This is the camera used http://www.manualslib.com/manual/26037/Casio-Exilim-Ex-Z19.html?page=135

Lens/Focal Distance
F2.8 (W) to 5.2 (T) f= 6.2 to 18.6 mm(equivalent to 37.5 to 112.5 mm in 35 mm format)Six lenses in five groups, including aspherical lens.

By the way for your information the peak just in front of Elbrus is 3103 m - Gora Digdali-Dudi. Actually, I have no idea why you're still looking at these pictures?! I thought you believe the Earth is 100% round? Why bother? Why are you wasting your time here?


« Last Edit: November 12, 2014, 02:22:16 PM by Saros »

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neimoka

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Re: "Equator" problem
« Reply #357 on: November 12, 2014, 02:33:52 PM »
@Saros, I've no idea what your point is??

"And this is what it looked like with absolutely no zoom"

So what was the focal length?

This is the camera used http://www.manualslib.com/manual/26037/Casio-Exilim-Ex-Z19.html?page=135

Lens/Focal Distance
F2.8 (W) to 5.2 (T) f= 6.2 to 18.6 mm(equivalent to 37.5 to 112.5 mm in 35 mm format)Six lenses in five groups, including aspherical lens.

By the way for your information the peak just in front of Elbrus is 3103 m - Gora Digdali-Dudi. Actually, I have no idea why you're still looking at these pictures?! I thought you believe the Earth is 100% round? Why bother? Why are you wasting your time here?
Focal length in itself doesn't say much either; size of the imaging device is important. It's easiest to state either the field of view in degrees or use the 35mm format equivalents - those are stated in that spec sheet and most people understand them fairly well.

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sokarul

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Re: "Equator" problem
« Reply #358 on: November 12, 2014, 02:42:23 PM »

By the way for your information the peak just in front of Elbrus is 3103 m - Gora Digdali-Dudi. Actually, I have no idea why you're still looking at these pictures?! I thought you believe the Earth is 100% round? Why bother? Why are you wasting your time here?
The peak in the picture that should be way under 2989m?
ANNIHILATOR OF  SHIFTER

It's no slur if it's fact.

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Saros

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Re: "Equator" problem
« Reply #359 on: November 12, 2014, 02:58:44 PM »

By the way for your information the peak just in front of Elbrus is 3103 m - Gora Digdali-Dudi. Actually, I have no idea why you're still looking at these pictures?! I thought you believe the Earth is 100% round? Why bother? Why are you wasting your time here?
The peak in the picture that should be way under 2989m?



Enhanced image:





 
« Last Edit: November 12, 2014, 03:51:15 PM by Saros »