The Flat Earth Society
Flat Earth Discussion Boards => Flat Earth Theory => Topic started by: Bobby Shafto on October 07, 2018, 02:26:11 PM
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Can you see all of the Cliffs of Dover from France? This post lifted from another topic claims yes. "No curvature whatsoever."
What do you think? Is this convincing evidence of earth's lack of rotundity?
The English Channel: 34 km distance from Cap Gris Nez to Dover, a curvature of some 22.4 meters on a round earth.
(https://image.ibb.co/hPHJxo/dover1.jpg)
(https://image.ibb.co/e1Daco/dover2.jpg)
The original webpages, as they were posted on flickr.com
The photographers located between Cap Blanc Nez and Cap Gris Nez: we will ascend to 30 meters.
(https://image.ibb.co/kf7qA8/doverbest2.jpg)
And now the photograph itself: no curvature whatsoever, all the way to the other shoreline, the Dover cliffs seen in their entirety (on a round earth, from 30 meters, we could not see anything under 16.5 meters from the other side), the ships are not part of an ascending/descending slope, no midpoint curvature of 22.4 meters:
(https://image.ibb.co/jBn7q8/doverbest.jpg)
Another photograph taken right on the beach of Cap Gris Nez: no curvature over a distance of 34 km:
(http://www.expedition360.com/journal/white_cliffs.jpg)
Dover cliffs:
(https://farm5.staticflickr.com/4070/4521816996_2971e62065.jpg)
(https://farm2.static.flickr.com/1051/4726849923_389dba2176.jpg)
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It states the curvature should be 22m, but the cliffs are over 100m high so you would expect to see most of them
https://en.m.wikipedia.org/wiki/White_Cliffs_of_Dover
The claim that you can see them all the way down to the waterline is very hard to verify from that photo.
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On a globe it all depends on the height of the observer. Cap Gris-Nez itself is said to be 30 meters above sea level. But a topographic map of Cap Gris-Nez shows that the direct surrounding can be up to 20 meters higher. Therefore a picture addressed to be taken from Cap Gris-Nez can easily be taken from an altitude of 50 meters:
(https://docs.google.com/drawings/d/e/2PACX-1vTo2pcBW2QnR_eOn1-HVbFQFbb2TuE80Z2ZBD332FqtwLoIiN_HVXf4u0TyfkeQg2ErBf524bUkJcyu/pub?w=577&h=556)
In that case the target hidden height is only 4,72 meters. The average elevation of the Dover cliffs is about 100 meters. This means that when the picture is taken from 50 meters altitude, only ± 5% of the cliffs are hidden by the curvature, and ± 95% will be clearly visible above the horizon.
(https://docs.google.com/drawings/d/e/2PACX-1vQe6_WyTInGuyYxtJxl-RyZyuMQhxFaPyb_bELlYkqBFyNHB8Bp2wSKC1QE2L8U0Cq8tU2uBFh-BrCM/pub?w=571&h=380)
Hard to tell the difference on a picture like this if we see 100% of the cliffs, or if we are missing the bottom 5%:
(https://docs.google.com/drawings/d/e/2PACX-1vR2FLJb2egdDc2R-OLjD9UmU4Q7xFKhrnunoNV7eg8nbwgIbj3l2D2G1g7PWMVeiUEDZRKHzDg2e9py/pub?w=608&h=396)
But many pictures of the Dover cliffs taken from France are not taken from Cap Gris-Nez, but from other locations nearby, like Cap Blanc-Nez, with an elevation of 134 meters:
(https://docs.google.com/drawings/d/e/2PACX-1vTxBesATkdXazBf8m2uLd-Mo8o0lYa-_49JfmmCx5a2X78SqVf_eEqDt0NJcw_PosTjIqhJYmWpLGHb/pub?w=557&h=253)
A recent picture of Cap Blanc-Nez, May 2018:
(https://docs.google.com/drawings/d/e/2PACX-1vRHwtrpmJ-oAuUzPKN31gbu1-nvPhGnQRLW-D7gl3CCFGfuyG1PHOI6D3hRnYB6Uwesrk9G1UYzFmDg/pub?w=604&h=388)
The earth curvature calculator shows us that from a distance of 33 km we need an altitude of 84 meters above sea level to reduce the target hidden height to zero:
(https://docs.google.com/drawings/d/e/2PACX-1vQHKnUIKMNfvETRpLXD3enJf6NNsDJuZmhMY2ecZD1VUX5k-QpjmvqsNwq2ty5PgMrvkaYERa7jP1BE/pub?w=602&h=379)
Conclusion, pictures from the Dover cliffs taken from the French coast do not prove the earth is flat. But the fact that those pictures are never taken from the beach but always from a higher altitude is proof for the curve.
(https://docs.google.com/drawings/d/e/2PACX-1vSwzapkP0v5nywwwZ136s4rAnXpirq5rGEW77RI_Xef9fYj66IpmTca0WXO-yT2X-aDa4zoyKhd1hXB/pub?w=602&h=383)
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Is this supposed to be a joke?
The fifth photograph was taken right on the Cap Gris Nez beach.
The photographers are right on the beach in the first and second images: out of the goodness of my heart, I accepted to ascend all the way to 30 m (you can see Cap Blanc Nez, the 130 m cliffs, in the background, the photographers are much closer to Cap Gris Nez).
No curvature of the water whatsoever: the boat is not part of either an ascending slope, a midpoint obstacle, or a descending slope.
The entirety of the Dover cliffs can be seen.
You wanna play games with me fighting for a meter of curvature?
I'll show you how it's done.
Let us increase the distance to 55 km (Grimsby - Toronto):
https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1616955#msg1616955
Zero curvature across a distance of 55 km, no 59 midpoint curvature whatsoever, the photographs include a ZOOM from the Niagara escarpment.
Remember, the RE called Ms. Kerry-Ann Lecky-Hepburn to confirm the photographs are real.
Toronto, more photographs:
https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1969076#msg1969076
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The photographers are right on the beach in the first and second images......
Nope, the vegetation around their feet doesn't grow on beaches, the sharp edge and the size of the waves below clearly shows they are standing on a cliff. Altitude unknown, but if close to Cap Gris-Nez it can easily be 50 meters above sea level:
(https://docs.google.com/drawings/d/e/2PACX-1vTo2pcBW2QnR_eOn1-HVbFQFbb2TuE80Z2ZBD332FqtwLoIiN_HVXf4u0TyfkeQg2ErBf524bUkJcyu/pub?w=577&h=556)
The entirety of the Dover cliffs can be seen.
That picture cannot be used to make such a claim, because we do not have a reference of another picture showing those same cliffs clearly top to bottom, and without such a reference nobody can tell what we see; the entirety of the cliffs or just 84% of the upper part.
Just like we should compare this picture of the Toronto skyline:
(https://docs.google.com/drawings/d/e/2PACX-1vQrcCmC1SoFAlP80PxaVVlUcze1MHcY2jkOI3ygqtNh8uqkn72pqKNhRQTN8eikGOV6whPPM0qK3r_6/pub?w=731&h=483)
With this one,
(https://docs.google.com/drawings/d/e/2PACX-1vTT33ZsZ1glVePjcI3Q6F3sHMfM4ZqEDUCBC5wTn3OG3f2GOSiF8_F-3Z_9STlDwcX9KBbB7NRJ2H1C/pub?w=732&h=483)
to see if the first picture is showing the entirety of the Toronto skyline or if a part of it is hidden behind a curved surface of water:
No curvature of the water whatsoever: the boat is not part of either an ascending slope, a midpoint obstacle, or a descending slope.
Did someone ever mention the word "perspective" to you? Probably not.
The curve you are trying to see is calculated by "8 inch/miles²" Do you know how big 8 inch is when seen over a distance of 1 mile? You need very sharp eyes to see an object of 8 inch over a distance of 1 mile. From that distance that 8 inch is seen within an angle of 0.086 degrees of an arc, which is equivalent of 0.5 inch watched over a distance of 30 feet'
Let's say the boat in your picture is somewhere in the middle between France and England, 10 miles away from the observer. then the drop of curvature will be 800 inch, or 67 feet. Over a distance of 10 miles 67 feet is still only 0.86 degrees of an arc. Now the problem is that with the naked eye it is hard to tell if a horizontal line 10 miles away is exactly on our eye level or just 0.86 degrees of an arc below eye level. Therefore with the naked eye we can not see the curvature of the globe as a descending slope, because perspective is hiding that slope for us. For our natural observation it is impossible to tell if we're watching a flat surface, or a surface that's curving down from our point of view with only 8 inch/miles².
To tell if a body of water is flat or curving down with 8 inch/miles² we need the help of precision instruments like theodolites, or a trick like watching a boat on the horizon from two different levels of observation, like on this picture: same boat, same distance but two pictures taken 7 seconds and 4 feet apart from each other, showing that the first picture of the boat is more obstructed by a bulge of curved water than the second one. That's how we see the curvature.
(https://docs.google.com/drawings/d/e/2PACX-1vQ-ZQlQ3Hg_SCIDXezc5z9RBInOWT1CSFHbNTCrApSD1poqQPBH_qoW1u4KmiIyjMBs1RVN74h6S9Zl/pub?w=959&h=633)
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The claim that you can see them all the way down to the waterline is very hard to verify from that photo.
I felt the same thing. Here's the photo.
(http://oi67.tinypic.com/29xfyno.jpg)
That section of the cliffs are as much as 400' high. Distance is at least 21 miles.
For comparison, here's a recent photo I took of an island rising 400' above sea level from a distance of 17 miles. Can you tell if the island is or isn't fully visible?
(http://oi65.tinypic.com/331pp5e.jpg)
Mine is most definitely taken from about 3' above the tidal water level. If one can tell that the full extent of the Cliffs of Dover is visible, then the same should be true of my photo too.
But look again from a view of 400'
(http://oi64.tinypic.com/2qlx1c0.jpg)
Compare:
(http://oi64.tinypic.com/2l9jvrt.jpg)
I don't believe that Dover image was taken from sea level. I think it was taken from around 90-100' in elevation:
(http://oi66.tinypic.com/166ez50.jpg)
The inset box is the section of the Cliffs at Samfire Hoe, which is an area built up at the base of the cliffs up to 60' above sea level. The train tracks there enter the Shakespeare Tunnell. The cliffs above rise to 430'.
From a viewing elevation of 100' you could expect 30-60' MSL down to the water's edge of Samphire Hoe to be hidden depending on the refractive index. You might even get looming on occasion and actually be able to see the full extent...from 100 ft viewing height. I marked the no/low refraction (sinking) and standard refraction elevations on the Google image with red and yellow lines respectively. I didn't mark where high refractive index (looming) would be since that could be to the water's edge.
Back to the WStreet photo, 33' amounts to around 3 pixels, and double that is 6 pixels, of course. Can someone claim to be able to see if 3 (or 6) pixels at that resolution are missing or fully visible? Color me skeptical. I can definitely make out the upper elevation of Samphire Hoe, but there's not enough there to account for the fully angular height of 60 feet that would be true if "full extent" was visible. Of course, it could be compressed by a different refractive index close to the water level.
But who knows? I sure wouldn't present that photo as evidence one way or the other. You can't tell. If it was taken from 100', then I think it shows what one would expect on a globe, with about 30' or so hidden, which is probably too small an amount to really tell at that resolution.
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As I said several times before, what the RE really want is a spherical earth with no curvature.
The curve you are trying to see is calculated by "8 inch/miles²" Do you know how big 8 inch is when seen over a distance of 1 mile? You need very sharp eyes to see an object of 8 inch over a distance of 1 mile. From that distance that 8 inch is seen within an angle of 0.086 degrees of an arc, which is equivalent of 0.5 inch watched over a distance of 30 feet'
This is priceless.
Here are the CORRECT formulas:
CURVATURE
C = R(1 - cos[s/(2R)]) - angle measured in radians
R = 6378,164 km
s = distance
VISUAL OBSTACLE
(http://i113.photobucket.com/albums/n206/dharanis1/Capture_zpswhoewt2o.jpg)
BD = (R + h)/{[2Rh + h2]1/2(sin s/R)(1/R) + cos s/R} - R
BD = visual obstacle
h = altitude of observer
The MIDPOINT OBSTACLE for a distance of 34 km is 22.4 meters.
Suppose you are standing on a cliff 100 ft in height in Cap Gris Nez. On a spherical earth you will observe this: an ascending slope, a huge midpoint bulge/visual obstacle of 22.4 meters, and a descending slope all the way to the other shoreline.
None of these features are being observed in this fantastic photograph:
(https://image.ibb.co/jBn7q8/doverbest.jpg)
None of these features are being observed in these photographs taken from the Niagara escarpment:
(https://image.ibb.co/kJp4no/torontoboat.jpg)
(https://i4.photobucket.com/albums/y108/RichardF/IMG_0078-2.jpg)
(https://i4.photobucket.com/albums/y108/RichardF/IMG_0086.jpg)
No ascending slope, no huge 59 meter midpoint obstacle, no descending slope.
HOW would the water stay curved on a spherical earth? No one can explain how two gravitons attract each other.
How would a graviton released by the iron/nickel core interact with a graviton emitted by lake Ontario?
Does it make any sense at all? And if not, how can anyone believe this nonsense?
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Let us increase the distance to 55 km (Grimsby - Toronto):
https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1616955#msg1616955
Zero curvature across a distance of 55 km, no 59 midpoint curvature whatsoever, the photographs include a ZOOM from the Niagara escarpment.
Remember, the RE called Ms. Kerry-Ann Lecky-Hepburn to confirm the photographs are real.
Toronto, more photographs:
https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1969076#msg1969076
As expected on a globe earth:
(http://oi64.tinypic.com/2cckr4l.jpg)
(http://oi65.tinypic.com/22f6ef.jpg)
Compare with what Toronto skyline would look like from that elevation at that distance if the earth was flat (and horizon really DID rise to eye level).
(http://oi65.tinypic.com/2lb2ohc.jpg)
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Here's what the curve and "bulge" between the Beamer Conservation Overlook near Grimbsy to Toronto look to scale:
(http://oi63.tinypic.com/2q0uiy9.jpg)
(If image is broken or width is messing up the page display, here's a direct link (https://drive.google.com/file/d/1aeeLD5iunvJGocooBxYkwHjs3V2dauql/view?usp=sharing).)
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As expected on a globe earth:
No.
On a globe earth, for a distance of 55 km, you will have a 59 meter midpoint obstacle.
And both an ascending slope to that 59 meter curvature midpoint, and a descending slope all the way to the other shoreline.
What you seem to preach is a globe earth WITHOUT CURVATURE.
For a distance of 55 km, on a globe earth, you will have a 59 meter midpoint visual obstacle. It is as simple as this.
http://oi63.tinypic.com/2q0uiy9.jpg
Completely misleading!
What you are comparing in that image is the 55 km distance to the 59 meter curvature: a catastrophic analysis.
What you have to compare is the 59 meter curvature to the buildings/skyline of Toronto!
That is, if you are on the Niagara escarpment, what you will see on a spherical earth is an ascending slope, a midpoint curvature of 59 meters which would be immediately seen by comparison with the buildings from Toronto, and a descending slope.
NO such features are being observed in the pics from Grimsby.
Let us visit the strait of Gibraltar.
No curvature across the strait of Gibraltar, no ascending slope, no midpoint 3.5 meter visual obstacle, a perfectly flat surface of the water all the way to Africa:
http://www.dailymotion.com/video/x42v7ip
38:28 to 38:35
(http://image.ibb.co/n3bHw6/gib.jpg)
From the same spot, a splendid photograph:
(https://farm1.static.flickr.com/55/130948289_44854d63fa_b.jpg)
No curvature whatsoever, just like the image in the video itself.
You have some explaining to do.
HOW would the water stay curved on a spherical earth? No one can explain how two gravitons attract each other.
How would a graviton released by the iron/nickel core interact with a graviton emitted by lake Ontario?
Does it make any sense at all? And if not, how can anyone believe this nonsense?
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I dare anyone to say they can visually detect an "ascending slope" of 146 feet over a longitudinal view of 16 miles.
You can't. Expecting to detect the "bulge" and thinking that not seeing it means the surface is flat is naive. The horizon is where the curve reveals itself.
Example:
Point Loma (San Diego) to Isla Coronados Middle Grounds (Mexico) is ~ 17 miles.
Here are four views on the same day, taken one right after the other, but from ascending viewing heights and their associated horizon points (with standard refraction):
Observation 1: 3 feet
Observation 2: 30 feet
Observation 3: 60 feet
Observation 4: 400 feet (not depicted)
(http://oi66.tinypic.com/210mtdi.jpg)
The "bulge" is the same for all (48' at the midway point of 8 1/2 miles) but, even once the horizon point is beyond that bulge point, you can't detect the crest of a 48' gradual slope 8.5 miles in the distance. If there were gridlines on the earth, then maybe. But such a feature is lost to perspective and depth perception.
But the horizon? That's plain:
(https://i.imgur.com/68HnvSz.jpg)
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I dare anyone to say they can visually detect an "ascending slope" of 146 feet over a longitudinal view of 16 miles.
Sure you can.
You have to compare the curvature to the visual target, NOT to the length of the distance.
Grimsby - Toronto 55 km
A midpoint curvature of 59 meters
To borrow a line from the famous Wendy commercial:
WHERE'S THE CURVATURE?
(https://image.ibb.co/ndVDxo/lakeontario53_zps743773f9.jpg)
No 59 meter curvature at all.
(https://image.ibb.co/eRE3V8/Toronto_Day.jpg)
No ascending slope, no midpoint visual obstacle of 59 meters, no curvature whatsoever.
From the very same spot, Ms. Lecky-Hepburn used a reflector telescope for this zoom:
(https://image.ibb.co/fQjnq8/thor2h.jpg)
Everything in plain sight: no curvature whatsoever.
You still have some explaining to do.
HOW would the water stay curved on a spherical earth? No one can explain how two gravitons attract each other.
How would a graviton released by the iron/nickel core interact with a graviton emitted by lake Ontario?
Does it make any sense at all? And if not, how can anyone believe this nonsense?
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You still have some explaining to do.
HOW would the water stay curved on a spherical earth? No one can explain how two gravitons attract each other.
How would a graviton released by the iron/nickel core interact with a graviton emitted by lake Ontario?
Does it make any sense at all? And if not, how can anyone believe this nonsense?
No, I don't have to explain that. I'm just observing curvature. How and/or why it exists is a different concern.
Curve is evidenced by the horizon; not by trying to detect "ascending/descending slope" of "the bulge." Curvature is detected by horizon.
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From the very same spot, Ms. Lecky-Hepburn used a reflector telescope for this zoom:
(https://image.ibb.co/fQjnq8/thor2h.jpg)
Everything in plain sight: no curvature whatsoever.
The horizon is at or further than the "everything in plain sight" in that photo, so of course curve isn't detectable.
Actually, it is if you know where "eye level" is and do a little analysis of the "dip." But just looking out over the vista? You're not going to see what you think you should see. The degree of slope is too slight. That inability to detect the slope isn't indicative of "no curvature."
I know you've been harping on this for a long time. But you're looking for curve in the wrong way. The way and what you're looking for, sure, you're never going to see it. Because it isn't manifest in the way you say it should be.
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No, I don't have to explain that
But you have to: otherwise, the readers will think that the water stays curved by pure magic.
How does a graviton emitted by the iron/nickel core interact with a graviton released by lake Ontario?
I'm just observing curvature.
Not in the photographs I provided: no curvature at all is being observed.
The degree of slope is too slight. That inability to detect the slope isn't indicative of "no curvature."
You could not see anything at all from the shoreline of Toronton given the 59 meter curvature. The zoom shows in plain sight each and every detail: not possible on a globe earth.
Curvature is detected by horizon.
Right.
Let us increase the distance to 5,200 km.
(https://i739.photobucket.com/albums/xx38/jorroa5990/Tunguskadistance_zps4429f436.png)
(https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcTXcyUgDO5cacT2pNraN0OSkpteUdUkOZ5lYNyWFafLr5OIPAhSrg)
The Tunguska event was observed INSTANTANEOUSLY from London, Antwerp, Stockholm, Berlin.
https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1676400#msg1676400 (six consecutive messages)
It could not have been caused by either a comet, or an asteroid, or a meteorite.
The precise geomagnetic pulses were observed THREE DAYS BEFORE THE EVENT:
(https://image.ibb.co/i4nX8y/tgr1.jpg)
(https://image.ibb.co/h8hFTy/tgr2.jpg)
(https://image.ibb.co/cjwLvd/tgr3.jpg)
(https://image.ibb.co/kAeOFd/tgr4.jpg)
TWO OBJECTS CAUSED THE TUNGUKSA EXPLOSION:
https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1995026#msg1995026
The initial map of the trajectory:
(https://image.ibb.co/jRMzZS/tunguska.jpg)
The final map: two trajectories, whose paths were modified in mid-air, no natural object is capable of such a feat.
(https://image.ibb.co/fpJJTn/tung03.jpg)
https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1995521#msg1995521
(https://image.ibb.co/bGhe3y/tung4.jpg)
It was found that the pattern of ytterbium’s distribution at Tunguska follows the projection of the “southern” TSB’s path on the ground. Similar shapes have been formed at Tunguska for the surface distribution of lanthanum, lead, silver and manganese (Zhuravlev & Demin, 1976). Only these five elements have patterns of distribution in Tunguska soils and peats that follow the projection of the TSB path on the ground, and only ytterbium follows this path strongly enough to be considered as the most likely main ingredient of the TSB substance.
This is an amazing outcome, one should note. This soft silvery-white rareearth metal, discovered in 1878, is now used mainly for improving the hardness of stainless steel, as well as in making high-power lasers. Definitely, if the chief chemical component of the TSB was ytterbium it hardly could have been a natural space body.
If the light from the Sun could not reach London due to curvature and/or any light reflection phenomena, then certainly NO LIGHT from an explosion which occurred at some 7 km altitude in the atmosphere could have been seen at all, at the same time, on a spherical earth.
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As I said several times before, what the RE really want is a spherical earth with no curvature.
Weirdly, I often notice that FE wants a flat earth but with curvature.
A lot of the "long distance photos" arguments I see as evidence for a flat earth are along the lines of "Well, the curve calculator says that 'x' should be hidden but only 'y' is hidden!"
OK, well maybe that's because of refraction or a mistake in your calculations of distance or viewer height, but if the earth is flat why is 'y' hidden? Surely you should be able to see all of the distant objects?
Suppose you are standing on a cliff 100 ft in height in Cap Gris Nez. On a spherical earth you will observe this: an ascending slope, a huge midpoint bulge/visual obstacle of 22.4 meters, and a descending slope all the way to the other shoreline.
No, silly! Because you are effectively standing on "top" of the earth. The earth slopes away from you in all directions, just very gradually.
If you're standing on top of a ball why would there be any upwards slope?
On a globe earth you see the top of these two diagrams. The horizon is you looking over the curve of the earth, represented by the red line.
(https://image.ibb.co/cuLRVx/Horizon.jpg)
This is why you see further at altitude, you can see further over the curve of the earth.
The slope away from you is far too subtle to see though.
On a flat earth the horizon would be the limit of your vision, bottom diagram.
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On a spherical earth you will observe this: an ascending slope, a huge midpoint bulge/visual obstacle of 22.4 meters, and a descending slope all the way to the other shoreline.
That is not what you will observe on a spherical earth, because this is not the curve, this is a "straw man"
What is a straw man?
(https://docs.google.com/drawings/d/e/2PACX-1vTZGRzpSdXVBo2gxNJ5XqwIfMBuIdDoPKv7ZoaBEDdYHBk2q8wYiuWHUnlVbIEneL_R9f1dwueZ5uAV/pub?w=463&h=285)
What FE'ers do who claim they can't see the curve is they first create a "straw man" of the curve; a totally false image in their prepossessed anti-globular mind about how a curved surface of the earth would be visible. Then they look around if they can see this caricature of a curve. Of course they can't see it, because they are not looking for the curve itself, but for a misconception about that curvature. And because such a misconception does not exist in the real world, they incorrectly conclude that there is no curve, while that curve is just in front of their nose if they are watching the horizon on a clear day.
That's what I call: "A straw man attack on Mother Earth spherical shape."
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The previous two messages should be moved to CN.
They don't belong anywhere else.
Imagine having to explain basic geometry to the stubborn RE.
You are standing right on the Cap Gris Nez. Distance to the Dover cliffs, 34 km. Spherical earth hypothesis.
Unite the two beaches (Cap Gris Nez and Dover) by a straight line. The midpoint curvature/basic visual obstacle will measure 22.4 meters (greatest distance from the arclength to the straight line).
You compare the midpoint curvature to the visual target (which I always include, be it the Toronto skyline, or the rock of Gibraltar, or the Dover cliffs), AND NOT to the length of the distance.
Even if one ascends to 45 meters in altitude (Cap Gris Nez cliffs), you still are going to have to deal with a 22.4 meter midpoint bulge, an ascending slope, and a descending slope.
That midpoint curvature is totally missing from the photographs taken in France, Grimsby, Gibraltar.
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That midpoint curvature is totally missing from the photographs taken in France, Grimsby, Gibraltar.
There is no midpoint curvature, my diagram shows why.
If you want to provide your own diagram explaining my mistake then please do, just repeating yourself does nothing to advance this discussion.
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Your diagrams are comparing the arclength distance to the value of the midpoint curvature.
The midpoint curvature figure (English Channel 22.4 meters, lake Ontario 59 meters, strait of Gibraltar 3.5 meters) has to be compared to the heights of the visual targets (SkyDome 86 meters, Dover Cliffs 100 meters, rock of Gibraltar), that is why I always include the visual target.
Then, your globe earth hypothesis falls apart.
Please explain to your readers how two gravitons attract each other.
You need attractive gravity to explain the supposed curved line of the body of water.
How does it work?
How does a graviton emitted by the iron/nickel core interact with a graviton released by lake Ontario?
If you cannot explain this, you are going to have to admit you are relying on pure magic as an explanation.
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Your diagrams are comparing the arclength distance to the value of the midpoint curvature.
They're not comparing anything. You seem to be asserting that there is a big hill of water rising up in front of you.
Actually, you're looking slightly down (horizon dip) over the hill of water.
The line of sight is the red line (I did this for the boat/laser experiment)
(https://image.ibb.co/f7hvrn/Laser_Again.jpg)
But note that the curve is massively exaggerated in this diagram, you wouldn't be able to perceive it. It would block the view of anything beyond the curve but the amount blocked would depend on viewer height and distance of the object.
Bobby gave a more accurate "to scale" diagram above, you can see how slight the curve is in real life.
You need attractive gravity to explain the supposed curved line of the body of water.
How does it work?
Why do I need to understand how it works in order to observe that it exists? Every time I see something fall I observe it exists.
The Cavendish experiment demonstrates it is attraction between objects, not the earth pushing upwards.
Do I have to understand how magnetism works to notice that magnets attract (or repel)
All I need to do is observe that it occurs. I believe the latest model is to do with curved space-time, gravitons are theoretical, but the mechanism is irrelevant.
If you cannot explain this, you are going to have to admit you are relying on pure magic as an explanation.
Says the man who talks about "ether whirlpools" a lot...
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Even if one ascends to 45 meters in altitude (Cap Gris Nez cliffs), you still are going to have to deal with a 22.4 meter midpoint bulge, an ascending slope, and a descending slope.
That midpoint curvature is totally missing from the photographs taken in France, Grimsby, Gibraltar.
You still cling to your straw man as if losing your straw man would be losing your life.
(https://docs.google.com/drawings/d/e/2PACX-1vSVoPJ8NHWxhbxD7SkwMqd-VPTK8ArRdMSGvDj8p0JpSx1X-6adzy3lsVfi7afk3YxeCC9U5KFV6HNH/pub?w=641&h=457)
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The Cavendish experiment demonstrates it is attraction between objects, not the earth pushing upwards.
You must be dreaming: the Cavendish experiment was DEBUNKED and proven to be false by Dr. Steve Lamoreaux (Yale University) in one of the most famous experiments of the past few decades.
LAMOREAUX EXPERIMENT
Steve Lamoreaux (Yale University): proof of the existence of negative energy (zero point vacuum energy - that is, subquark strings/telluric currents/magnetic monopoles double torsion strings):
https://www.youtube.com/watch?v=igpbED2xOKM
(starts at 7:50 - Dr. Lamoreaux explains the pushing gravity experiment)
Steve reasoned that if he created a narrow-enough region of empty space like the area between the two ships, then some of the shimmering zero-point energy would not fit inside it.
The energy of empty space outside the narrow region would be stronger and force it to shrink.
That force would be the signature of negative energy, and Steve set out to create it in his lab.
It was an idea that would consume him for more than a decade.
Inside this vacuum chamber are two small metal plates sitting less than the width of a human hair apart from one another.
To get them that close and not touch, the metal has to be perfectly flat, down almost to the atomic level.
The zero-point fluctuations of free space won't fit between those plates, as well, so when you bring these two plates together, there are fewer fluctuations between the plates than there are outside the plates.
The force builds up, and it actually gets stronger and stronger as the plates get closer together, and that force we refer to as arising from negative energy.
The zero-point energy fluctuations outside the plates are stronger than those between, so pressure from the outside pushes them together.
Or think of it another way.
The negative energy between the plates expands space around it.
Steve's years of meticulous labor have made him the first person on Earth to have measured a force produced by negative energy.
THE PLATES ARE PUSHED BY THE OUTSIDE PRESSURE OF THE TELLURIC WAVES: NO ATTRACTIVE GRAVITY.
EXPERIMENT PERFORMED IN VACUUM BY DR. STEVE LAMOREAUX.
The catastrophic, delirious, unimaginable, stupendous, monstruous errors committed by Cavendish:
https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg805751#msg805751
NIPHER EXPERIMENT
The relationship between gravitation and the electric field was first observed experimentally by Dr. Francis Nipher. Nipher's conclusion was that sheilded electrostatic fields directly influence the action of gravitation. He further concluded that gravitation and electrical fields are absolutely linked.
http://www.rexresearch.com/nipher/nipher1.htm
The relationship between gravitation and the electric field was first observed experimentally by Dr. Francis Nipher. Dr. Francis Nipher conducted extensive experiments during 1918, on a modified Cavendish experiment. He reproduced the classical arrangements for the experiment, where gravitational attraction could be measured between free-swinging masses, and a large fixed central mass. Dr. Nipher modified the Cavendish experiment by applying a large electrical field to the large central mass, which was sheilded inside a Faraday cage. When electrostatic charge was applied to the large fixed mass, the free-swinging masses exhibited a reduced attraction to the central mass, when the central mass was only slightly charged. As the electric field strength was increased, there arose a voltage threshold which resulted in no attraction at all between the fixed mass and the free-swinging masses. Increasing the potential applied to the central mass beyond that threshold, resulted in the free-swinging masses being repelled (!) from the fixed central mass. Nipher's conclusion was that sheilded electrostatic fields directly influence the action of gravitation. He further concluded that gravitation and electrical fields are absolutely linked.
Do I have to understand how magnetism works to notice that magnets attract (or repel)
You better understand how ether magnetism works.
Here are the REAL TIME experiments and photographs performed by SPINTRONICS:
https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg759332#msg759332
I believe the latest model is to do with curved space-time, gravitons are theoretical, but the mechanism is irrelevant.
What?
You must be very uninformed.
TOTAL DEBUNKING OF TGR/TSR:
https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg769750#msg769750
Your diagrams/pseudo-explanations amount to nothing at all.
You are still comparing a midpoint curvature figure with the total distance.
You need to learn basic geometry.
You are standing right on the Cap Gris Nez. Distance to the Dover cliffs, 34 km. Spherical earth hypothesis.
Unite the two beaches (Cap Gris Nez and Dover) by a straight line. The midpoint curvature/basic visual obstacle will measure 22.4 meters (greatest distance from the arclength to the straight line).
You compare the midpoint curvature to the visual target (which I always include, be it the Toronto skyline, or the rock of Gibraltar, or the Dover cliffs), AND NOT to the length of the distance.
Even if one ascends to 45 meters in altitude (Cap Gris Nez cliffs), you still are going to have to deal with a 22.4 meter midpoint bulge, an ascending slope, and a descending slope.
That midpoint curvature is totally missing from the photographs taken in France, Grimsby, Gibraltar.
You do not compare the 55,000 meters with the 59 meter curvature: you must compare the 59 meter curvature with the height of the SkyDome, 86 meters.
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You are standing right on the Cap Gris Nez. Distance to the Dover cliffs, 34 km. Spherical earth hypothesis.
The spherical earth theory includes a gravitational force curving water around the centre of the globe, and causes the observer to realize that "down" is always in the direction of the centre of the globe.
In this frame of reference from the point of view of the observer, the surface of the earth is always curving downwards, from the place he is standing towards his horizon and beyond.
If there would be an ascending slope of curvature between the observer and the horizon on a spherical earth, the horizon would be above eye level, which on a sphere is utter nonsense.
If you really believe that on a globe there would be ascending slopes of curvature, than you seriously confuse Globe earth theory with Concave earth theory.
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The spherical earth theory includes a gravitational force curving water around the centre of the globe
How can it be a "theory" if you cannot as much as explain how two gravitons attract each other?
What you have is a mere hypothesis, being ripped into shreds by the Biefeld-Brown effect, experiments carried out in full vacuum: no such thing as attractive gravity.
The gravitational field consists of gravitons.
Please explain to your readers how two gravitons attract each other.
I have provided both the theoretical proof of ether waves (Whittaker's 1903 and 1904 papers) and the experimental proofs (Galaev ether drift experiments) of their existence.
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Your diagrams/pseudo-explanations amount to nothing at all.
You are still comparing a midpoint curvature figure with the total distance.
You need to learn basic geometry.
You are standing right on the Cap Gris Nez. Distance to the Dover cliffs, 34 km. Spherical earth hypothesis.
Unite the two beaches (Cap Gris Nez and Dover) by a straight line. The midpoint curvature/basic visual obstacle will measure 22.4 meters (greatest distance from the arclength to the straight line).
(http://oi67.tinypic.com/ou4i1k.jpg)
Using your diagram from earlier, I've added (in red) what you believe is important: the chord AB from the French coast line to the English cliffs. Point F is the midpoint, and there is a distance between that and the midpoint of the arc between those same two points - labeled G.
That distance, commonly referred to as the "bulge" is, as you say, about 22m across an arc distance of 34,000m in the case of Cap Gris Nez to Dover Cliffs.
In the Beamer Conservation Area to Toronto scenario, it's about 45m across an arc distance of over 52,000m.
The graphic above is great for explaining the geometry, but it's misleading in illustrating the reality of the height of the bulge in relation to the distance across the arc.
Work out the geometry and the difference between the arc distance and the chord distance is half a foot over Lake Ontario; less than that across the English Channel.
(http://oi63.tinypic.com/2q0uiy9.jpg)
This is as close to a scale image as I can get of what that arc/chord looks like relative to the distance. That "bulge" barely registers. More importantly, that "bulge" doesn't present as a visual obstacle to anything. It's not an obstacle. Horizon is the obstacle. And the scenario of Grimbsy-Toronto, the horizon is at or beyond the target. In the scenario of France-Dover, the horizon is beyond the midpoint and it, not "bulge" is what's causing any visual obscuring (which is to little to resolve in the photo provided).
You compare the midpoint curvature to the visual target (which I always include, be it the Toronto skyline, or the rock of Gibraltar, or the Dover cliffs), AND NOT to the length of the distance.
Even if one ascends to 45 meters in altitude (Cap Gris Nez cliffs), you still are going to have to deal with a 22.4 meter midpoint bulge, an ascending slope, and a descending slope.
That midpoint curvature is totally missing from the photographs taken in France, Grimsby, Gibraltar.
You do not compare the 55,000 meters with the 59 meter curvature: you must compare the 59 meter curvature with the height of the SkyDome, 86 meters.
You NEVER have to "deal with" the midpoint bulge. It's immaterial. Your view cannot ever be obstructed by the bulge (except when the horizon happens to coincide with it). You are trying to see something that you can't see and that doesn't matter anyway.
Point C on the line tangent to arc AB is what matters. That's the visual obstruction. Not G.
(http://oi67.tinypic.com/ou4i1k.jpg)
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In the Beamer Conservation Area to Toronto scenario, it's about 45m across an arc distance of over 52,000m.
What?
It is 59 meters, not one meter less.
Work out the geometry and the difference between the arc distance and the chord distance is half a foot over Lake Ontario; less than that across the English Channel.
We are not comparing ANYTHING to either the arc distance or the chord distance.
We are comparing the midpoint curvature with the visual target.
That "bulge" barely registers. More importantly, that "bulge" doesn't present as a visual obstacle to anything. It's not an obstacle. Horizon is the obstacle.
We are NOT dealing with the horizon. Each photograph has a definite visual target. That is what you have to deal with.
It's immaterial.
What?
Then we are done here: you have admitted that the midpoint bulge does not exist, it is "immaterial".
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Also, as AllAroundtheWorld has pointed out, you and your objective are not leaning toward parallel with the midpoint. They are tilted away from each other by the degree of arc. So, if you are at point A but viewing from a height above A (point E), your level is perpendicular to the gravity vector, and a "dip" angle is formed between it and the line tangent at point H.
(http://oi64.tinypic.com/dqpyc2.jpg)
You're interpreting the scenario like the thick red lines added to the graphic originating from points A and B, expecting that "hill" of curvature to ascend and descend between the viewer and the objective. But it's not a "hill". It's a slope away from horizontal. The highpoint of the bulge matters not at all to the line of sight from E. The horizon at point C is what matters, and that can be before, at or after the midpoint bulge of G.
Horizon is what manifests curvature. Not "bulge." Proclaiming bulge height numbers that are tiny fractions of the distance across which they rise and descend is meaningless. You can't perceive it, and it doesn't obstruct visually. Horizon does.
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The spherical earth theory includes a gravitational force curving water around the centre of the globe
How can it be a "theory" if you cannot as much as explain how two gravitons attract each other?
What you have is a mere hypothesis, being ripped into shreds by the Biefeld-Brown effect, experiments carried out in full vacuum: no such thing as attractive gravity.
The gravitational field consists of gravitons.
Please explain to your readers how two gravitons attract each other.
I have provided both the theoretical proof of ether waves (Whittaker's 1903 and 1904 papers) and the experimental proofs (Galaev ether drift experiments) of their existence.
This topic is about observing the Dover cliffs from the French coast, not about explaining gravity. You are now replacing your indefensible straw man with a red herring:
(https://docs.google.com/drawings/d/e/2PACX-1vRBVrA0tGMrLUOtoDo-oJME2VjJfsfCnrR2qO0fc60opOFnTTAmv1qkNm2sweoqA9DpvicRRbctRAHA/pub?w=497&h=298)
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In the Beamer Conservation Area to Toronto scenario, it's about 45m across an arc distance of over 52,000m.
What?
It is 59 meters, not one meter less.
It doesn't matter, but you're right. Physically, it's 59m. If accounting for standard atmospheric refraction which has the visual effect of increasing the radius of the earth, IF you could actually perceive this "hill" of water, it would appear to be 45m and not 59m.
But, if you want to be pedantic about it, yes. You are right. It is physically 59m between chord and arc length on the globe earth. Not that it matters since you can't see it either way.
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In the Beamer Conservation Area to Toronto scenario, it's about 45m across an arc distance of over 52,000m.
What?
It is 59 meters, not one meter less.
Work out the geometry and the difference between the arc distance and the chord distance is half a foot over Lake Ontario; less than that across the English Channel.
We are not comparing ANYTHING to either the arc distance or the chord distance.
We are comparing the midpoint curvature with the visual target.
We are not. You are, for some odd reason. There is no material reason to compare "midpoint curvature" to visual target. The 59m you avidly corrected me about is the height difference between chord and arc. It is tiny compared to their distances. You can't see it. It doesn't get in the visual way of the target. You're wrong to think it should be detectable as evidence of curvature.
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not about explaining gravity.
Unless you can explain how two gravitons attract each other, you are believing a fairy tale.
Certainly you want the level of lake Ontario to be curved: if you cannot as much as explain how that takes place, it means the globe earth hypothesis is totally wrong.
Proclaiming bulge height numbers that are tiny fractions of the distance across which they rise and descend is meaningless.
Not me: YOU.
YOU have compared the midpoing curvature to the distance across which they rise, NOT ME.
I have done no such thing.
I am comparing the midpoint curvature to the visual target.
Join points A and B on the respective beaches: point C is the midpoint curvature.
That is what you have to deal with: even if you ascend to a point higher than C (let's say 45 meters in Cap Gris Nez), and you have a visual target, the ascending slope, the midpoint curvature (22.4 meters) and the desdending slope are still there on a globe earth.
It is just that they are totally missing in the photographs: the ship is not part of any slopes, just a perfectly straight surface of the English Channel.
If accounting for standard atmospheric refraction
What?
My photographs include no refraction effects, I always take care of this aspect.
You can't see it.
If "you can't see it" and if it's "immaterial" then the earth is flat.
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That "bulge" barely registers. More importantly, that "bulge" doesn't present as a visual obstacle to anything. It's not an obstacle. Horizon is the obstacle.
We are NOT dealing with the horizon. Each photograph has a definite visual target. That is what you have to deal with.
We are not dealing with horizon in the Toronto image. That's why you can see everything. Bulge isn't an obstacle. Horizon is. The horizon point in the Toronto picture is at the distance of the Toronto skyline. Thus, no surprise that you can see all of Toronto skyline.
In the Dover Cliffs scenario, you ARE dealing with horizon. But the image you provided does not offer the resolution necessary to affirm its impact on visuals. Whatever is hidden (or not hidden) of the cliffs (like the Samphire Hoe), it's due to horizon, and not bulge.
It's immaterial.
What?
Then we are done here: you have admitted that the midpoint bulge does not exist, it is "immaterial".
We are done here. Me saying it is immaterial doesn't mean that it doesn't exist. I (we) provided you with sincere critique of your reasoning. You're playing games to preserve a long-held assertion that is flawed. You delve into more difficult, deep and arcane concepts than this, but this simply geometric difference between "bulge" and "horizon" eludes you and you insist that this "midpoint curvature" must be detectable as an obstruction. Sorry, but if you can't see how that is in error, then I have to wonder about the rest of your FAQ.
Frankly, I admire that you have a passionate and individualistic approach to the flat earth. You are critical of TFES FAQ and some of its "FET" statements. I like that. (Just as I appreciate JRowe's individual and unique Dual Earth Theory, even though I also disagree with it.) However, your salesmanship leaves a lot to be desired. Given your seeming grasp on more complex concepts, I didn't expect such resistance (and belligerence) to criticism of your analysis of photos and curvature.
If we're at loggerheads on this simple concept, good luck to you. I'm going to keep preaching how flat earthers get the globe wrong. You keep doing your thing.
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You are now replacing your indefensible straw man with a red herring
Hi Humble. While I'm sure your remarks on logical fallacies are nothing short of exquisite, they absolutely do not belong to this thread. If you want to talk about how much you hate sandokhan's logic, do so in AR.
I would also point out that you were the one to revive the discussion of gravitational forces - it's in poor form for you to then complain that your statements are being discussed.
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I'm going to keep preaching how flat earthers get the globe wrong.
Case in point:
If accounting for standard atmospheric refraction
What?
My photographs include no refraction effects, I always take care of this aspect.
If it's a photograph taken through an atmoSPHERE, there are refraction effects.
The effect of an atmoSPHERE isn't only distortions like mirage, or stooping, Fata Morgana, shimmering, chromatic prism artifacts or any number of other phenomena. In a perfectly stable atmosphere with a standard temperature/density lapse rate, light will still bend without creating distortion, altering the apparent vertical position of distant objects compared to a no-atmosphere sighting. It can't be helped. Only if there was absolutely no gradient, or the angle of incidence is steep enough, does atmospheric refraction have no (or little) effect.
"Globebusters" are always discarding atmospheric refraction. On a flat earth, with an atmoPLANE, you could do that (I suppose.) On a globe earth with a conforming atmoSPHERE, you can't. The lack of apparent distortion does not mean there is no effect due to atmospheric refraction.
A month or so ago I started up a topic on atmospheric refraction, hoping to address this very common misconception. When looking across any expanse and trying to ascertain curve or flat based on what you can or can't see vs what one should or shouldn't be able to see if the earth surface is convex, you can't just toss out the effect of atmospheric refraction because you can't see any distortion. Refraction will alter light paths and sight lines without distortion.
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Join points A and B on the respective beaches: point C is the midpoint curvature.
That is what you have to deal with: even if you ascend to a point higher than C (let's say 45 meters in Cap Gris Nez), and you have a visual target, the ascending slope, the midpoint curvature (22.4 meters) and the desdending slope are still there on a globe earth.
That is what we deal with, but what you completely overlook is that on a globe that the line that joins points A and B on the respective beaches is not a horizontal line parallel to the observers eye level. Due to the curvature the other beach is always on a lower position (from the point of view of the observer) than the position of the observer, and therefore that line connecting those beaches is a line going down from the point of view of the observer. And because of that, on a globe the observer is always on a higher point than C, even if he/she is standing at sea level ->
(https://docs.google.com/drawings/d/e/2PACX-1vSJPqhMsKrGzx6AI7P3745x3GJXuJbDjGOo2RS-0OYqx3CqGCbqAeTd62ttZ0cawsUXf_NvmXydaGMW/pub?w=442&h=452)
That's why on a globe there is no such thing as an "ascending slope" between the observer and his horizon, only a descending slope.
It is just that they are totally missing in the photographs: the ship is not part of any slopes, just a perfectly straight surface of the English Channel.
There is a descending slope between you and the ship. The reason this slope is not visible with the naked eye or on a photograph is the result of perspective, as I pointed out earlier in tis thread:
Let's say the boat in your picture is somewhere in the middle between France and England, 10 miles away from the observer. then the drop of curvature will be 800 inch, or 67 feet. Over a distance of 10 miles 67 feet is still only 0.86 degrees of an arc. Now the problem is that with the naked eye it is hard to tell if a horizontal line 10 miles away is exactly on our eye level or just 0.86 degrees of an arc below eye level. Therefore with the naked eye we can not see the curvature of the globe as a descending slope, because perspective is hiding that slope for us. For our natural observation it is impossible to tell if we're watching a flat surface, or a surface that's curving down from our point of view with only 8 inch/miles².
To tell if a body of water is flat or curving down with 8 inch/miles² we need the help of precision instruments like theodolites........
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PS:
If you want to talk about how much you hate sandokhan's logic....
I do not hate sandokhan's logic, in contrary, I love it because he is willing to put time and effort in substantiating his point of view. That's why he is my favorite flat earther.
What I do not love is people making all kinds of bold claims and wild accusations without any substantiation.
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Unless you can explain how two gravitons attract each other, you are believing a fairy tale.
Again, I don't need to understand the mechanism behind how something works to observe that it does.
Did rainbows only start appearing in the sky after people understood how the light travelled through raindrops to cause the effect?
I'll ignore your word soup about the Cavendish Experiment, suffice to say that when I Googled "Cavendish Experiment Debunked" most of the results came, predictably, with flat earth websites. In the real world it is a well accepted and repeatable experiment.
https://www.quora.com/Is-the-Cavendish-experiment-a-pseudoscience-hoax
That is what you have to deal with: even if you ascend to a point higher than C (let's say 45 meters in Cap Gris Nez), and you have a visual target, the ascending slope, the midpoint curvature (22.4 meters) and the desending slope are still there on a globe earth.
Why do you keep on talking about ascending slopes? There is no ascending slope, just a descending one. The actual situation is this:
(https://image.ibb.co/k1VLg9/Descending_Slope.jpg)
Your line of sight is the red line. The horizon is where you see the edge of the earth, H. Any object nearer to you than that, such as a boat at A, should be fully visible - waves or other obstacles permitting. Any object further away than that such as the cliffs at B would be partly occluded by the curve of the earth. How much depends on your viewer height and refraction would be a factor. It just isn't clear from the picture of the channel whether the cliffs are occluded or by how much.
You have been shown plenty of photos in this thread of distant shorelines clearly occluded.
If you believe my diagram to be flawed somehow then please produce your own explaining why.