"Equator" problem

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rottingroom

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Re: "Equator" problem
« Reply #420 on: November 16, 2014, 12:16:28 PM »

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Thanks for the supporting evidence.

Thanks for acknowledgment that the Earth is flat!

Pretty sure your evidence supported the opposing notion.

Re: "Equator" problem
« Reply #421 on: November 16, 2014, 12:25:32 PM »

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Thanks for the supporting evidence.

Thanks for acknowledgment that the Earth is flat!

Pretty sure your evidence supported the opposing notion.

Me, too. What is it that makes you think this, cikljamas? I used your approximation for the "dip" due to curvature; it checks as pretty close for small distances like this.
"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 #422 on: November 16, 2014, 12:31:12 PM »

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Thanks for the supporting evidence.

Thanks for acknowledgment that the Earth is flat!

Pretty sure your evidence supported the opposing notion.

You mean, obvious lack of any evidences for exponential drop of the ground from a plane's perspective proves that the Earth is round? How can you be so stupid?

As for Alpha2Omega reckoning, it was just one more typical example of heliocentric mumbo jumbo, hocus pocus bull s h i t ting...
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rottingroom

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Re: "Equator" problem
« Reply #423 on: November 16, 2014, 12:32:08 PM »

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Thanks for the supporting evidence.

Thanks for acknowledgment that the Earth is flat!

Pretty sure your evidence supported the opposing notion.

You mean, obvious lack of any evidences for exponential drop of the ground from a plane's perspective proves that the Earth is round? How can you be so stupid?

As for Alpha2Omega reckoning, it was just one more typical example of heliocentric mumbo jumbo, hocus pocus bull s h i t ting...

No, that isn't what i mean. You've lost, again.

Re: "Equator" problem
« Reply #424 on: November 16, 2014, 12:40:04 PM »

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Thanks for the supporting evidence.

Thanks for acknowledgment that the Earth is flat!

Pretty sure your evidence supported the opposing notion.

You mean, obvious lack of any evidences for exponential drop of the ground from a plane's perspective proves that the Earth is round? How can you be so stupid?

As for Alpha2Omega reckoning, it was just one more typical example of heliocentric mumbo jumbo, hocus pocus bull s h i t ting...

No, that isn't what i mean. You've lost, again.
You were using math trying to prove your point. If I use math to show where your math is incorrect, it's "mumbo jumbo"?

If you don't understand the math, just say so. Not everyone does so there's no shame in that.
"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 #425 on: November 16, 2014, 01:08:00 PM »
Alpha2Omega, what is your answer to this question:

Does obvious lack of any evidences for exponential drop of the ground from a plane's perspective prove that the Earth is round?

Regarding your math, we shall discuss this further after i receive concrete answer to above question...
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rottingroom

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Re: "Equator" problem
« Reply #426 on: November 16, 2014, 01:52:55 PM »
From the tip of everest the horizon is 288 miles away. This is 8/1000 of the circumference of the earth which is merely 2.88° of the entire sphere. I'm bit sure how much of a drop you expect to intuitively notice from that but I can tell you it'd be difficult to discern.

Re: "Equator" problem
« Reply #427 on: November 16, 2014, 05:41:15 PM »
Alpha2Omega, what is your answer to this question:

Does obvious lack of any evidences for exponential drop of the ground from a plane's perspective prove that the Earth is round?

Regarding your math, we shall discuss this further after i receive concrete answer to above question...

As I said before, it's not exponential, it's circular. For short distances (less than about 600 miles or so) the distance squared function that you suggested is an adequate approximation, but its answers degrade rapidly as distance increases from that, and utterly fails when the distance reaches 1/4 of the circumference.

I don't agree that there's a lack of evidence for this at all. Why do you think there is?

There's your answer. Bring on what you think the math is. If the prior try is an example, it's probably wrong, but let's give it a look.
"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 #428 on: November 17, 2014, 01:30:54 AM »
As I said before, it's not exponential, it's circular. For short distances (less than about 600 miles or so) the distance squared function that you suggested is an adequate approximation, but its answers degrade rapidly as distance increases from that, and utterly fails when the distance reaches 1/4 of the circumference.

Bull s h i t t!!!

I don't agree that there's a lack of evidence for this at all. Why do you think there is?

Putin is trying to tell you something : http://theflatearthsociety.org/forum/index.php?topic=62318.msg1639792#msg1639792



And that is not all, watch this video and tell me, do you notice any curvature OR EXPONENTIAL DROP OF THE GROUND from 27 km altitude : MIG-25 EARTHVIEW AMAZING : " class="bbc_link" target="_blank" rel="noopener noreferrer">
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ausGeoff

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Re: "Equator" problem
« Reply #429 on: November 17, 2014, 04:38:15 AM »
How can you be so stupid?

It was just one more typical example of heliocentric mumbo jumbo, hocus pocus bull shitting.

Whenever people resort to responding as childishly as this, you can be pretty sure that at last, they're starting to have the tiniest doubts that their flat earth stance may not be the correct one.  It's invariably a giveaway, and the nail in the coffin of their failure is the inevitable ad hominem.

For this reason alone, it's a waste of everybody's time attempting to debate anything at a mature, reasoned level with this guy; he's simply being intransigent for the sake of it—like a petulant little kid.


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cikljamas

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Re: "Equator" problem
« Reply #430 on: November 17, 2014, 05:39:05 AM »
Australian troll, of course, the only thing that we could have expected from you is such a crap pseudo-psychological analysis. However, If anyone else were to ask himself/herself why my answer to Alpha2Omega's (bull s h i t) assertion, wasn't very polite ("bull s h i t"), this is how he/she can get satisfaction:



So, if the Earth were round, after 100 miles the ground would drop 2000 meters, and in order to be able to see certain low profile figure/point at the horizon you should climb up 2000 meters also. Now, if somebody claims something like this...
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"As I said before, it's not exponential, it's circular. For short distances (less than about 600 miles or so) the distance squared function that you suggested is an adequate approximation, but its answers degrade rapidly as distance increases from that, and utterly fails when the distance reaches 1/4 of the circumference."

...i just can't refrain myself to answer simply and illustratively (like this):

Bull s h i t t!!!

Isn't this question quite legitimate : Should i refrain myself (in the first place) from giving such kind of answers, given that i have to deal with constant lying and mumbo jumbo mathematical calculations which have no value and no common ground with reality whatsoever?
« Last Edit: November 17, 2014, 05:42:41 AM by cikljamas »
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ausGeoff

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Re: "Equator" problem
« Reply #431 on: November 17, 2014, 06:05:55 AM »
Australian troll, of course, the only thing that we could have expected from you is such a crap pseudo-psychological analysis.

LOL..... I've obviously touched a raw nerve.

Re: "Equator" problem
« Reply #432 on: November 17, 2014, 08:44:24 AM »
<rant>


So? What is the significance of Point B? It's a point in space that is level with Point C above a spot on Point C's horizon. There's nothing there but thin air.

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So, if the Earth were round, after 100 miles the ground would drop 2000 meters, and in order to be able to see certain low profile figure/point at the horizon you should climb up 2000 meters also.
2000 m looks like a reasonable approximation, and that interpretation is reasonable. Remember, though, that 100 miles is not very far.
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Now, if somebody claims something like this...
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"As I said before, it's not exponential, it's circular. For short distances (less than about 600 miles or so) the distance squared function that you suggested is an adequate approximation, but its answers degrade rapidly as distance increases from that, and utterly fails when the distance reaches 1/4 of the circumference."

...i just can't refrain myself to answer simply and illustratively (like this):

Bull s h i t t!!!
Maybe you should refrain from glib retorts and actually examine your ideas when a problem is pointed out. If, after examination, you think there is an error with the reply, show specifically what the flaw is. If you can't find an error (or don't understand it), admit it. Simply calling "BS" sounds like you're giving up.

Using the approximation formula on p. 9 of this exhibit (thanks, Aus!) that you referenced here - and used, let's look at a more distant case. Since you deny there's a problem with it, let's calculate its value for the dip one-quarter of the circumference away and compare with more accurate calculations. Fig. 14 from the above is replicated here for convenience.



"In calculating the amount of curvature, or dip below the eye-line of the observer, we have a simple rule, ignoring some small decimal points, namely :— Square the number of miles given as the distance, and multiply the product by eight inches, and divide by twelve, which will give in feet the depth of the dip from the observer’s line of sight. This is true for a globe of 25,000 miles circumference ; thus in six miles there would be a dip of 24 feet, and in twelve miles a dip of 96 feet."

The formula is:

Dip = D2 * 8 / 12
Where Dip is in feet if D is the distance in miles.

Dip = (6250)2 * 8 / 12
 = 39,062,500 * 8 / 12
 = 26,041,666.67 feet
 = 4932.1 miles

The author gives two different meanings for "Dip" on the following page, referring to Fig. 14.

"And this is true whether we reckon the dip towards the centre of the globe in the direction of G L, or at right-angles from the line of sight G M."
 
Examining the second meaning of "Dip" - right angles to the sightline - first, since it's easier:

For a location one-quarter of the way around the circle, the "Dip" would clearly be equal to the radius of the circle.

R = 25,000 miles / (2 pi) = 3978.9 miles

This is an error of more than 953 miles.

The author's first meaning of Dip is even worse - far worse.

Define the point 'O' (capital letter O, not zero) in Fig. 14 as the center of the circle "shewn". It's the intersection of the lines through EK and JL. This defines the right triangle EHO with the right angle at point E and line HO the hypotenuse. Line EO is the radius of our circle. Line HJ is the "Dip" under the second meaning. HJ is the length of the hypotenuse of our right triangle, HO, minus the length of line JO, which is the radius of the circle.

First calculate the length of hypotenuse, HO. This is the adjacent side, EO, divided by the cosine of the angle opposite side EH. Let's call this angle 'a'.
HO = EO / cos(a)

Since EO is the radius R, then

HO = R / cos(a)

Since our point J is one-quarter of the way around the circumference of circle O, a is exactly 90 degrees, and recall that R = 3978.9 miles, so

HO = 3978.9 miles / cos(90)

Uh, oh! cos(90) is zero! Mr. Math may quibble with me here since division by zero is undefined, but this effectively means that line HO is infinitely long. Subtracting our radius R, 3978.9 miles, from infinity is still infinity.  Your estimated dip using the approximation formula, 4932.1 miles, is infinitely less than the more accurate calculated value (infinity). This is what's meant by "blowing up".

No bullshit, just facts and a little elementary trig. Instead of calling this "mumbo jumbo", instead ask yourself why you didn't try harder in high-school math. A bit more effort there would have saved you from falling for the quite lame arguments the likes of Rowbotham and "Zetetes" foist off on the ignorant and gullible with some apparent success, even today.

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Isn't this question quite legitimate : Should i refrain myself (in the first place) from giving such kind of answers, given that i have to deal with constant lying and mumbo jumbo mathematical calculations which have no value and no common ground with reality whatsoever?
No, that's not a legitimate question since the premise it's based on is wrong. It would seem to be a good idea to refrain yourself for other reasons, though.

[Edit] Adjust width of top figure.
« Last Edit: November 17, 2014, 08:46:41 AM by Alpha2Omega »
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

Re: "Equator" problem
« Reply #433 on: November 17, 2014, 09:03:26 AM »
As I said before, it's not exponential, it's circular. For short distances (less than about 600 miles or so) the distance squared function that you suggested is an adequate approximation, but its answers degrade rapidly as distance increases from that, and utterly fails when the distance reaches 1/4 of the circumference.

Bull s h i t t!!!
See above.

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I don't agree that there's a lack of evidence for this at all. Why do you think there is?

Putin is trying to tell you something : http://theflatearthsociety.org/forum/index.php?topic=62318.msg1639792#msg1639792

<MiG pix>

And that is not all, watch this video and tell me, do you notice any curvature OR EXPONENTIAL DROP OF THE GROUND from 27 km altitude : MIG-25 EARTHVIEW AMAZING : " class="bbc_link" target="_blank" rel="noopener noreferrer">

Too many unknowns. Do we know what Putin's actual flights were, or are you just using great-circle or other distances measured from maps? How did these compare to the return flight times?

The shape of the horizon thru the MiG windscreen in those pictures is either inconclusive due to the clouds and haze, or clearly convex. I see nothing that suggests other than a spherical earth here.


"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 #434 on: November 17, 2014, 10:03:28 AM »
Distance 50 miles --- dip 500 meters
Distance 100 miles --- dip 2000 meters
Distance 200 miles --- dip 8000 meters
Distance 400 miles --- dip 32000 meters

What kind of mathematical progression is this? Arithmetical or GEOMETRICAL?

So, if you were a pilot in MIG 25 flying at 18 miles altitude, you could see (if atmospheric conditions were favorable) the ground dropping 28880 meters at distance of 380 miles (ahead and below you). 

Would the horizon line be (in that case) at eye level?

What do you see in this picture, how the horizon line looks like to you? ROTUND?



Do you notice any sign of EXPONENTIAL dropping of the ground in this picture?

Why objects get smaller at a FIXED RATE instead of EXPONENTIALLY???

How about the inclination of a distant objects?

How about synergy of all above factors combined?

Well, it seems that you have already answered:
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"The shape of the horizon thru the MiG windscreen in those pictures is either inconclusive due to the clouds and haze, or clearly convex. I see nothing that suggests other than a spherical earth here."

So, you see a spherical earth here?

Congratulations!!!
« Last Edit: November 17, 2014, 10:10:16 AM by cikljamas »
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Re: "Equator" problem
« Reply #435 on: November 17, 2014, 02:36:25 PM »
Distance 50 miles --- dip 500 meters
Distance 100 miles --- dip 2000 meters
Distance 200 miles --- dip 8000 meters
Distance 400 miles --- dip 32000 meters

What kind of mathematical progression is this? Arithmetical or GEOMETRICAL?

They're both geometric progressions because you chose a geometric progression of distances; each number is a constant times the number before it (the constant is 2 for the distances, and 4, which is 2 squared, for the dips).

If you had chosen 50, 100, 150, 200, 250, ... for the distances, they would have been an arithmetic progression; each number is a constant (50) plus the number before. The dip sequence, however:
500, 2000, 4500, 8000, 12500... is neither arithmetic nor geometric; each member is simply proportional to the square of its counterpart in the arithmetic progression.

In both cases, the dips increase as the square of the distances.

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So, if you were a pilot in MIG 25 flying at 18 miles altitude, you could see (if atmospheric conditions were favorable) the ground dropping 28880 meters at distance of 380 miles (ahead and below you). 

Would the horizon line be (in that case) at eye level?

No.

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What do you see in this picture, how the horizon line looks like to you? ROTUND?

<MiG pix>

Clearly convex in some of the pictures, inconclusive in others. These are the same pictures as before, aren't they? Were you expecting a different answer?

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Do you notice any sign of EXPONENTIAL dropping of the ground in this picture?

No, because:   
1) The "drop" isn't exponential, it's squared.[nb]Exponential would be of the form ad; what we have is of the form d2, where a is a constant and d is the distance.[/nb] [nb]Cool! I just learned how to footnote![/nb].
2) We don't have a good horizontal reference.

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Why objects get smaller at a FIXED RATE instead of EXPONENTIALLY???

Because the sightlines to different points of an object are straight lines, so the size (apparent linear distance between them) drops off linearly with distance. The area drops off as the square of distance, however, if that makes you feel any better.

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How about the inclination of a distant objects?

If the horizon is 380 miles away, that's about 5.5 degrees. Any object on the horizon that was plumb would appear to tilt away from you at an angle of 5.5 degrees. But it would also be 380 miles away, so that would be very hard to see, even if it was really, really, really tall!

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How about synergy of all above factors combined?

No problems with a spherical earth with a radius just under 4,000 miles.

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Well, it seems that you have already answered:

"The shape of the horizon thru the MiG windscreen in those pictures is either inconclusive due to the clouds and haze, or clearly convex. I see nothing that suggests other than a spherical earth here."
[nb]Nesting quotes causes all the footnotes above to be in the wrong place, so it's not as cool as it might be  :(.[/nb]

Yes, so why did you ask again earlier in this post?

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So, you see a spherical earth here?

Yes. Clearly in some of the pictures, inconclusive in others.

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Congratulations!!!

Thanks!!

For what? Finally answering your questions to your satisfaction? Shucks... glad to help  :).

[Edit] Corrected first reply section as noted with strikethrough underscore.

« Last Edit: November 17, 2014, 04:10:06 PM by Alpha2Omega »
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cikljamas

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Re: "Equator" problem
« Reply #436 on: November 18, 2014, 04:47:29 AM »
No.

So?

The view from the height of 8650 m from the ridge of Mt Everest.



On the right MAKALU (8463m - distance 12 miles)
In the back KANCHANJUNGA (8456m - distance 78 miles)

Dip for MAKALU : 29 meters
Dip for KANCHANJUNGA : 1 217 meters

HOW COME THAT KANCHANJUNGA IS AT EYE LEVEL AND ABOVE THE LEVEL OF MAKALU???

WHERE IS THE GROUND DROPPING FOR 1 217 METERS AT A DISTANCE OF KANCHANJUNGA?

Now, you eagerly want to see this video, don't you: FLAT EARTH COMPASS CONFUSION : " class="bbc_link" target="_blank" rel="noopener noreferrer">

In addition:

« Last Edit: November 18, 2014, 08:20:42 AM by cikljamas »
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rottingroom

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Re: "Equator" problem
« Reply #437 on: November 18, 2014, 06:00:48 AM »
The view from the height of 8650 m from the ridge of Mt Everest.





While the curvature of earth may be detectable, with a large FOV, as low 20,000 ft, (but more reliably at as much as 35,000 ft) the picture you've brought forth of Everest isn't searching for a curvature of Earth and is instead looking at the curvature of the top of clouds. We might as well be looking at a picture from a dinky hilltop.
« Last Edit: November 18, 2014, 06:02:47 AM by rottingroom »

Re: "Equator" problem
« Reply #438 on: November 18, 2014, 09:31:40 AM »
HOW COME THAT KANCHANJUNGA IS AT EYE LEVEL AND ABOVE THE LEVEL OF MAKALU???

WHERE IS THE GROUND DROPPING FOR 1 217 METERS AT A DISTANCE OF KANCHANJUNGA?
Why are you yelling?  Can you provide your source for the height of Kangchenjunga?  I found a height of 8586m.  Anyway, the picture was taken from roughly 200 meters higher than Makulu, and your line is showing what?  The tops of the clouds?  Eye-level would have been center of frame had the camera been mounted on something to perfectly level it.  The curvature drop from Makalu to Kanchanjunga is 860m.  So you have a line of sight looking downward toward a peak 200m (230m including curvature drop) lower, and a taller background peak approx. 78 miles away that is  dropped from curvature by only 860m from the first peak.

Re: "Equator" problem
« Reply #439 on: November 18, 2014, 10:19:06 AM »
No.

So?

The view from the height of 8650 m from the ridge of Mt Everest.



On the right MAKALU (8463m - distance 12 miles)
In the back KANCHANJUNGA (8456m - distance 78 miles)

Dip for MAKALU : 29 meters
Dip for KANCHANJUNGA : 1 217 meters

HOW COME THAT KANCHANJUNGA IS AT EYE LEVEL AND ABOVE THE LEVEL OF MAKALU???

WHERE IS THE GROUND DROPPING FOR 1 217 METERS AT A DISTANCE OF KANCHANJUNGA?
WHY ARE YOU USING ALL CAPS? Do you think that makes your point more convincing? It doesn't. The best way to do that is to have a convincing point to start with.

Using your numbers, 8650 m for the photographer, 8463 m for Makalu, the photograph is taken from 187 m higher elevation than the summit of Makalu, so he has to be looking down toward the of top of Makalu from there, right? The top of that cloud deck seems about the same elevation (although this can fool you sometimes), and the top of Makalu is about even with the horizon formed by the cloud deck, so the horizon has to be below level. You're looking down at Makalu's summit, therefore you're looking down to the horizon, right?

Can you say where you got those elevations? I find 8,463 m for Makalu here but 8,586 m for Kangchenjunga here. How good is that 8650 m? ["29" beat me to this]

Anyway, using these numbers, since Kangchenjunga is 6.5 times as far away as Makalu (78 mi/12 mi = 6.5), then the straight sightline has dropped 1404 m below horizontal at Kangchenjunga's distance

(8463 m - 8650 m - 29 m) * 6.5 = (-187 m - 29 m) * 6.5
 = -216 m * 6.5
 = -1404 m (1404 m below level at distance)

If sag due to curvature is 1217 m at Kangchenjunga and its summit is 64 m lower than the photographer, then the sightline ought to intersect Kangchenjunga 123 m below its summit.

-1217 m - 64m = -1281 m (summit is 1281 m below level)
-1404 m -(-1281 m) = -123 m (sightline is 123 m below summit)

So the summit of  should appear slightly higher than Makalu, which it does. How much higher is sensitive to the elevation of the photographer, though. If the photograph was actually taken from 8670 m, that will cause the sightline to intersect  Kangchenjunga 110 m lower still.

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Now, you eagerly want to see this video, don't you: FLAT EARTH COMPASS CONFUSION :
How about running this topic to ground before moving on to something completely different.
"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 #440 on: November 18, 2014, 01:12:25 PM »
No.

So?

The view from the height of 8650 m from the ridge of Mt Everest.



On the right MAKALU (8463m - distance 12 miles)
In the back KANCHANJUNGA (8456m - distance 78 miles)

Dip for MAKALU : 29 meters
Dip for KANCHANJUNGA : 1 217 meters

HOW COME THAT KANCHANJUNGA IS AT EYE LEVEL AND ABOVE THE LEVEL OF MAKALU???

WHERE IS THE GROUND DROPPING FOR 1 217 METERS AT A DISTANCE OF KANCHANJUNGA?
WHY ARE YOU USING ALL CAPS? Do you think that makes your point more convincing? It doesn't. The best way to do that is to have a convincing point to start with.

Well, this is pretty exact imitation of above situation:





First bottle is 120 cm away from photographer.

Second bottle is 720 cm away from photographer.

8586 - 8463 = 123
8650 - 8462 = 188

Height of the first bottle = 30 cm
Hight of the second bottle = 31 cm

8586 / 123 = 69,8
30 / 69,8 = 0,42

8650 / 188 = 46
30 / 46 = 0,65

0,42 + 0,65 = 1,07

31 - 30 = 1 which is approximately 1,07 (compensation for the height-differences among  all three mountain's peaks)

Above imitation has been made on a flat surface. If the Earth was round Kanchenjunga's peak would be at the level of lowest red line in above picture or at the level of red dash (right above the lowest red line) at best.

According to what we see in our picture that was taken from Mt Everest, Kanchenjunga is too far above Makalu's (peak) line, but the reason for that can't be convexity of the Earth, quite opposite than that...

Alpha2Omega, you have to make some experiments, that would be of much more help comparing it with your mumbo jumbo seductive math calculations..
« Last Edit: November 19, 2014, 04:18:13 AM by cikljamas »
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ausGeoff

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Re: "Equator" problem
« Reply #441 on: November 18, 2014, 03:21:21 PM »



Well, this is pretty exact imitation of above situation:



First bottle is 120 cm away from photographer.

Second bottle is 720 cm away from photographer.
Are you serious?  A couple of bottles in your study represents massive mountains in the Himalayas?  Well, if you reckon so.

I'm guessing—that like a lot of flat earthers—you incorrectly interpret the effects of perspective in the real world.  You've effectively "cheated" by positioning the height of your camera above the floor to suit your desired perspective of the bottles—and the convenient red lines you've drawn.  So your bottles image doesn't really offer much evidence—although I can see what you're trying to get at.

Can you lower your camera's position by 15cm vertically and post the image?  And maintain the relative 120cm and 720cm distances too.

But I do thank you—in this case—for at least taking the time and effort to post these pics—it's a lot more than most other flat earthers do in an effort to clarify their claims.    :)

Re: "Equator" problem
« Reply #442 on: November 18, 2014, 08:11:45 PM »
Well, this is pretty exact imitation of above situation:




Interesting exercise. Let me see if I can follow what you're doing here - bear with me, please (I'll use the comma as decimal separator for consistency here - I'm not used to it, so may goof it up some).

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First bottle is 120 cm away from photographer.
Second bottle is 720 cm away from photographer.

12 mi : 78 mi would be equivalent to 120 cm : 780 cm
Why is the distant bottle 720 cm instead of 780 cm? If there's not enough room, moving the near bottle to 111 cm would preserve the 6,5 : 1 ratio of distances. This is because 720 cm/6,5  = 110,8 cm.
Alternatively, you could place them at 100 and 650 cm if 1 m isn't too close to your camera.

8586 - 8463 = 123   Height difference between Kangchenjunga and Makalu, right?
8650 - 8462 = 188   Height difference between viewpoint and Makalu, right? Why 8462 instead of 8463?

Height of the first bottle = 30 cm
Height of the second bottle = 31 cm

8586 / 123 = 69,8
30 / 69,8 = 0,42  Excess height of Kangchenjunga over Makalu scaled to a 30 cm 'Kangchenjunga'?

8650 / 188 = 46
30 / 46 = 0,65  Excess height of viewpoint above Makalu scaled to a 30 cm 'viewpoint'?

This isn't right. You need to keep a fixed scale factor for all the vertical displacements or your model isn't meaningful.

0,42 + 0,65 = 1,07  What is this? Your camera should be  somewhat above 30-cm MakaluBottle because the viewpoint is higher, and KangchenjungaBottle should be slightly taller than MakaluBottle. Adding these numbers doesn't have any physical significance.

31 - 30 = 1 which is approximately 1,07 (compensation for the height-differences among  all three mountain's peaks)

If you think about it in a physical sense, Makalu is the lowest of these three elevations at 8463 m, so it should be the lowest of your model (make it 30 cm here). Kangchenjunga is next higher at 8586 m, and your camera, representing the observer, highest at 8650 m. If you can get two 30-cm tall bottles, here's what I'd do:

Use 8463 m = S * 30 cm, where S is a scaling factor to force Makalu to be a scaled 30 cm.

S = 8463 m / 30 cm = 282,1 m/cm
which means that in your model, each cm represents 282,1 m in the real world.

Viewpoint:  8650 m / S = 30,66 cm
Makalu: 8463 / S = 30 cm (sanity check)
Kangchenjunga: 8586 / S = 30,44 cm


Above imitation has been made on a flat surface. If the Earth was round Kanchenjunga's peak would be at the level of lowest red line in above picture or at the level of red dash (right above the lowest red line) at best.

According to what we see in our picture that was taken from Mt Everest, Kanchenjunga is too far above Makalu's (peak) line, but the reason for that can't be convexity of the Earth, quite opposite than that...

For this to be a meaningful model, your camera's optical axis should have been 30,65 cm above datum (the floor) and level. Not having the axis level may or may not have much effect, but it's best to eliminate variables. From your picture, I don't think the camera was at the correct height, and, at 31 cm, KanchenjungaBottle is a little too tall (should be 30,44 cm).

Can you repeat the experiment using two 30-cm bottles for the mountains and shim KanchenjungaBottle with about 4,4 mm of paper, or wood, or something, or maybe cap MakaluBottle snugly, and unscrew KanchenjungaBottle's cap to make it the requisite 4,4 mm higher? Get a stubby tripod for your camera and set the camera pointing level, with the center of its lens 30,65 cm above datum (floor) and retake the shot.


Next, if you're feeling adventurous, try this:

Since the curvature of the earth puts the datum at our viewpoint 1217 m above datum at Kanchenjunga, raise the camera by 1217 m / (282,1 m/cm) = 4.31 cm, from 30,66 cm to a height of 34,97 cm at the optical axis, and raise MakaluBottle  the equivalent 1217 m - 29 m = 1188 m, or  1188 m / (282,1 m/cm) = 4,21 cm. KanchenjungaBottle is left as it was. Take another picture. If your set-up is accurate (and no math mistakes here), I expect to see MakaluBottle below level and KangchenjungaBottle slightly higher than MakaluBottle, but by less than the flat model - which only makes sense.

Alpha2Omega, you have to make some experiments, that would be of much more help comparing it with your mumbo jumbo seductive math calculations..
If your model is accurately scaled it can be informative, but math, if used correctly, can also, and is much easier to work with if you're familiar with it. If either is incorrectly set up or used, the results will most likely be misleading.

Seductive...  ooohh! No one's ever called stuff like this from me seductive before; elegant, maybe (very rarely), seductive, no! Thanks, I think.
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

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cikljamas

  • 2432
  • Ex nihilo nihil fit
Re: "Equator" problem
« Reply #443 on: November 19, 2014, 03:51:30 AM »
My math:

Quote
8586 - 8463 = 123
8650 - 8462 = 188

Height of the first bottle = 30 cm
Hight of the second bottle = 31 cm

8586 / 123 = 69,8
30 / 69,8 = 0,42

8650 / 188 = 46
30 / 46 = 0,65

0,42 + 0,65 = 1,07

Your math:

Quote
Viewpoint:  8650 m / S = 30,66 cm
Makalu: 8463 / S = 30 cm (sanity check)
Kangchenjunga: 8586 / S = 30,44 cm

We get basicaly the same result!
Quote
For this to be a meaningful model, your camera's optical axis should have been 30,65 cm above datum (the floor) and level. Not having the axis level may or may not have much effect, but it's best to eliminate variables. From your picture, I don't think the camera was at the correct height, and, at 31 cm, KanchenjungaBottle is a little too tall (should be 30,44 cm).

This is how i have done it (i have used third bottle as a referent point):


Quote
Next, if you're feeling adventurous, try this:

Since the curvature of the earth puts the datum at our viewpoint 1217 m above datum at Kanchenjunga, raise the camera by 1217 m / (282,1 m/cm) = 4.31 cm, from 30,66 cm to a height of 34,97 cm at the optical axis, and raise MakaluBottle  the equivalent 1217 m - 29 m = 1188 m, or  1188 m / (282,1 m/cm) = 4,21 cm. KanchenjungaBottle is left as it was. Take another picture. If your set-up is accurate (and no math mistakes here), I expect to see MakaluBottle below level and KangchenjungaBottle slightly higher than MakaluBottle, but by less than the flat model - which only makes sense.

Well, this was really hard, i had to make a lot (i mean a lot) of shots to accomplish this task to be as close to reality as is possible under these conditions...The hardest part was to perfectly leveled the horizontal line. If you go wrong just for one mm or so, while leveling the camera, the result you get is far off from what it has to be...

But finally, this is what we have got after a lot of shots:



Round Earth version on the left...

Flat Earth version on the right...

So, you were right saying:

Quote
I expect to see MakaluBottle below level and KangchenjungaBottle slightly higher than MakaluBottle, but by less than the flat model - which only makes sense

Now, we have to see once more how it looks in the reality:



What do you think?

Zoomed Kanchenjunga (amazing, isn't it?):
« Last Edit: November 19, 2014, 04:53:39 AM by cikljamas »
"I can't breathe" George Floyd RIP

Re: "Equator" problem
« Reply #444 on: November 19, 2014, 07:49:25 AM »
Cool! Good work.

My math:

Quote
8586 - 8463 = 123
8650 - 8462 = 188

Height of the first bottle = 30 cm
Hight of the second bottle = 31 cm

8586 / 123 = 69,8
30 / 69,8 = 0,42

8650 / 188 = 46
30 / 46 = 0,65

0,42 + 0,65 = 1,07

Your math:

Quote
Viewpoint:  8650 m / S = 30,66 cm
Makalu: 8463 / S = 30 cm (sanity check)
Kangchenjunga: 8586 / S = 30,44 cm

We get basicaly the same result!
They are quite close, insignificant here, really. I was looking for why I was getting slightly different values - they should have been exactly the same if we were doing the same math - and found the changed scaling factor.

Quote
Quote
For this to be a meaningful model, your camera's optical axis should have been 30,65 cm above datum (the floor) and level. Not having the axis level may or may not have much effect, but it's best to eliminate variables. From your picture, I don't think the camera was at the correct height, and, at 31 cm, KanchenjungaBottle is a little too tall (should be 30,44 cm).

This is how i have done it (i have used third bottle as a referent point):


Quote
Next, if you're feeling adventurous, try this:

Since the curvature of the earth puts the datum at our viewpoint 1217 m above datum at Kanchenjunga, raise the camera by 1217 m / (282,1 m/cm) = 4.31 cm, from 30,66 cm to a height of 34,97 cm at the optical axis, and raise MakaluBottle  the equivalent 1217 m - 29 m = 1188 m, or  1188 m / (282,1 m/cm) = 4,21 cm. KanchenjungaBottle is left as it was. Take another picture. If your set-up is accurate (and no math mistakes here), I expect to see MakaluBottle below level and KangchenjungaBottle slightly higher than MakaluBottle, but by less than the flat model - which only makes sense.

Well, this was really hard, i had to make a lot (i mean a lot) of shots to accomplish this task to be as close to reality as is possible under these conditions...The hardest part was to perfectly leveled the horizontal line. If you go wrong just for one mm or so, while leveling the camera, the result you get is far off from what it has to be...
This is why I recommended something like a mini-tripod; it would have made set-up and adjustment easier and more consistent. You gotta work with what you have, though.

Quote
But finally, this is what we have got after a lot of shots:



Round Earth version on the left...

Flat Earth version on the right...

So, you were right saying:

Quote
I expect to see MakaluBottle below level and KangchenjungaBottle slightly higher than MakaluBottle, but by less than the flat model - which only makes sense

Now, we have to see once more how it looks in the reality:



What do you think?

Zoomed Kanchenjunga (amazing, isn't it?):


Interesting experiment. Your wife must have thought you were nuts.

I like that last picture. 
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

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cikljamas

  • 2432
  • Ex nihilo nihil fit
Re: "Equator" problem
« Reply #445 on: November 19, 2014, 08:03:47 AM »
Cool! Good work.

Interesting experiment. Your wife must have thought you were nuts.

I like that last picture.

Thanks!

She had become aware of it long time ago...  ;D

Well, after all effort that i have taken, now i noticed that i have placed my camera one whole cm above the appropriate place, so i had to repeat this strenuous procedure of shooting one more time, but at least now i can say that i have finished it finally:



So, what can we conclude out of all this?

edit:

It seems that this photograph has been taken from the same spot:



This one could be decisive in our case:
« Last Edit: November 19, 2014, 09:18:12 AM by cikljamas »
"I can't breathe" George Floyd RIP

Re: "Equator" problem
« Reply #446 on: November 19, 2014, 10:14:51 AM »
This one could be decisive in our case:

By lining up features of the two peaks, it looks to have been taken from about 8750m.

Re: "Equator" problem
« Reply #447 on: November 19, 2014, 10:35:47 AM »
This is how i have done it (i have used third bottle as a referent point):

I see you are using a Samsung WB150F (correct me if I'm wrong).  According to the specs, you have manual control over exposure up to 16" and ISO settings.

If you are still in doubt, or still can't understand, how one can get sharp pictures of star trails in addition to sharply focused landscape since Earth is rotating, and as you claimed back on page 4 that taking a picture from a moving object would result in blurriness, here's an experiment.

Set that camera for an exposure of several seconds, anchor it securly in a vehicle so that it's looking out the windshield with some of the interior or some exterior part is framed in the shot too, drive down a street with street lamps at night, and take a picture.  The longer the exposure the better.

Re: "Equator" problem
« Reply #448 on: November 19, 2014, 01:55:37 PM »
Cool! Good work.

Interesting experiment. Your wife must have thought you were nuts.

I like that last picture.

Thanks!

She had become aware of it long time ago...  ;D

Well, after all effort that i have taken, now i noticed that i have placed my camera one whole cm above the appropriate place, so i had to repeat this strenuous procedure of shooting one more time, but at least now i can say that i have finished it finally:



So, what can we conclude out of all this?

edit:

It seems that this photograph has been taken from the same spot:



This one could be decisive in our case:


I think there is no way to deny that the surface is flat based on the photos. In fact, imagining curvature where there is none is a sign of some psychological disorder, I am guessing. Mostly provoked by the desire to maintain the fantasy world which immediately starts to collapse when confronted with real-world observations. It hurts, so people prefer to ignore it.

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ausGeoff

  • 6091
Re: "Equator" problem
« Reply #449 on: November 19, 2014, 02:40:38 PM »
In fact, imagining curvature where there is none is a sign of some psychological disorder, I am guessing.
I'm pleased to see you admit that this nonsensical opinion is nothing more than a "guess", as any psychologist would unhesitatingly disagree with it.

Quote
Mostly provoked by the desire to maintain the fantasy world which immediately starts to collapse when confronted with real-world observations. It hurts, so people prefer to ignore it.
A belief in the flat earth notion is more than likely to indicate the maintenance of a fantasy world by such believers.  The scientific status quo is the spherical earth model.

Thus far, there is zero empirical evidence supporting the notion of a flat earth.  No working model; no agreed-upon map.  No real world observations. No photographic images.  Nothing.  Period.