Proof of curvature

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Sliver

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Re: Proof of curvature
« Reply #90 on: May 30, 2010, 07:24:29 AM »
Hey, levee, how about just explaining why I could not see land while only about 50 miles off the coast of Maryland?  I should have been able to see Ocean City, but I couldn't.  The ocean was calm, the sky was clear, what kept me from seeing any of the buildings on land?
Those were some pretty pictures, levee.  But you ignored my question.

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flyingmonkey

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Re: Proof of curvature
« Reply #91 on: May 30, 2010, 07:33:26 AM »
If you actually looked at your own photos, you would clearly see that the more distant skyline becomes obstructed by water compared to the closer skyline.

Obviously you missed that bit.

AKA: Curvature.


I think, fm, that you have no idea what a round earth flying through space at some 107000 km/hr actually implies...without attractive gravity round earth theory falls flat on its nose...

It implies nothing, what it is is what it is.

All I know is that it works, and works fine because we are all here, debating about it.

Can you explain how we don't all fly off the Earth if it is a spinning disc?



Have some photos:


http://farm4.static.flickr.com/3122/3239550294_54321c861a.jpg
http://www.boatnerd.com/pictures/special/cgrportweller/torontofromstationPW-ns.jpg
http://upload.wikimedia.org/wikipedia/commons/3/33/Toronto_seen_across_lake_Ontario_from_Olcott_2.JPG
http://upload.wikimedia.org/wikipedia/en/e/e5/Toronto_from_LO.JPG
http://i82.photobucket.com/albums/j251/dawnd_01/Autumn%202007/winter%202007/IMGP8777ii.jpg
http://lh5.ggpht.com/_dwwIAAX3HDY/SvH6K92cJ5I/AAAAAAAAA6I/PYLKoieeKQk/s912/DSC02307.JPG
http://commondatastorage.googleapis.com/static.panoramio.com/photos/original/24715141.jpg

Just a few searches and look what I found.

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Lorddave

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Re: Proof of curvature
« Reply #92 on: May 30, 2010, 08:43:39 AM »
I'm still waiting to hear why Levee can see the Skydome but not the Harbourpoint III building in front of it.
You have been ignored for common interest of mankind.

I am a terrible person and I am a typical Blowhard Liberal for being wrong about Bom.

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sandokhan

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Re: Proof of curvature
« Reply #93 on: May 31, 2010, 03:38:46 AM »
You should not be waiting for me, since you have not done your homework.

http://www.tobuilt.ca/php/toneighbourhoods.php?sortfield=Buildings.%60BuildingName%60&sortby=ASC&search_fd8=Harbourfront

Harbour Point III is a 20-story building (if we take the maximum possible height, 260 feet, it still is well below the height of the Sky Dome (90 meters).

sliver and ld, how many times do you have to be told that when you bring a photo here, you must specify the following:

-place taken, a range of possible altitudes (at which the photographer was located), with additional photos, if needed, to support your opinion

-distance to the visual target

-height of visual target


Now, let me show that there is no curvature at all over lake Ontario. As always, we will use a 240 meter altitude for the photographer, even though the highest point outside Grimsby by some 2 km is located at 213 meters (Vinemount Ridge), and Beamer Falls Park can be found at an elevation of just 45 meters.

From 240 meters, on a round earth, we could barely see the signs of land from Toronto, over an immense curvature of 59 meters; but no curvature appears at all in these extraordinary photographs, just a flat surface of the lake.







A zoom from the same spot as in the first photograph, NO CURVATURE WHATSOEVER ALL THE WAY TO TORONTO:



(taken by Kerry-Ann Lecky Hepburn in 2007, a well known photographer from Grimsby)

http://www.flickr.com/photos/tundrabluephotography/312939439/


http://www.flickr.com/photos/suckamc/53037827/



NO CURVATURE OVER THE STRAIT OF GIBRALTAR, NOT ONE CENTIMETER:

THERE SHOULD BE A 3.31 METER CURVATURE OVER THE STRAIT OF GIBRALTAR, IN THE ROUND EARTH THEORY; even if we change the radius of the earth from 6300 to 6400 km, the curvature will vary from 3.30 to 3.35 meters. We should see an ascending slope starting from the shores of Spain, a midpoint curvature of 3.31 meters, and NOTHING BELOW 5 METERS FROM THE OTHER SIDE OF THE STRAIT.

Here are the videos which prove, once and for all, now and forever, that there is no curvature, not one centimeter over the strait of Gibraltar:

Islamic History of Europe

#

Between 2:56 si 3:00 the author shows us the spanish beach and points towards the african coastline

Between 3:02 si 3:07 we can see clearly that there is no curvature all the way to Morocco; moreover, if we use the full screen option, we will see the waves splashing onto the opposing beach/shore...this is actually a closeup taken, again, from that beach...

Between 3:19 - 3:22, WE CAN SEE THE WAVES SPLASHING ONTO THE OPPOSING BEACH, EVEN WITH THE AUTHOR STANDING ON THE SPANISH SHORELINE, RIGHT NEXT TO THE STRAIT OF GIBRALTAR; on a round earth, we would see an ascending slope, with a midpoint curvature of 3.31 meters.

Between 3:43 si 3:45, the same thing, zero curvature...full screen option, the waves splashing onto the opposing beach/shore, WITH THE AUTHOR STADING RIGHT THERE ON THE SPANISH BEACH.

The Barbarians, here are the details, where we can see very clearly that there is no ascending slope, no midpoint curvature:

The Barbarians, hosted by Terry Jones

http://video.google.com/videoplay?docid=-811260411880444286&q=barbarians+terry+jones&total=22&start=10&num=10&so=0&type=search&plindex=1#

Between 38:28 - 38:35, we can see clearly ABSOLUTELY NO CURVATURE ALL THE WAY TO MOROCCO...the surface of the strait is completely flat...


So, there is no curvature whatsoever over the strait of Gibraltar, no matter the fairy tale you stubbornly want to believe in...

Here is a photograph taken right on the spanish beach, from the same place as that in the second video above:

http://www.flickr.com/photos/carlosromero/130948289/


SANDY HOOK - CONEY ISLAND

DISTANCE 7 MILES, 11.2 KM

CURVATURE 2.4 METERS

On a round earth, we should see a rising slope, with a midpoint visual obstacle of 2.4 meters, but there is no such thing in these photos taken right on the Sandy Hook beach:

http://www.flickr.com/photos/23956233@N04/2890814609/in/photostream/

http://www.flickr.com/photos/23956233@N04/2891651706/in/photostream/


NO CURVATURE OVER A DISTANCE OF 6000 KM LONDON - TUNGUSKA

The explosion at Tunguska (June 30, 1908, 7:15-7:20 am) took place at an elevation of 7 km. It was seen all the way from Irkutsk and Lake Baikal.

Lake Baikal is at a distance of some 600 km from the place of the explosion.



http://www.icr.org/research/index/researchp_sa_r05/

THE VISUAL OBSTACLE FOR A DISTANCE OF 600 KM IS 21.57 KM; NO WAY THAT AN EXPLOSION WHICH TOOK PLACE AT AN ALTITUDE OF 7 KM COULD BE SEEN FROM THAT DISTANCE.

Now, we will take a distance of just 6000 km between London and Tunguska.

Over this distance, WE HAVE A VISUAL OBSTACLE OF 4333 KM; ABSOLUTELY IMPOSSIBLE TO SEE ANYTHING WHICH IS LOCATED ON THE OTHER SIDE OF A GLOBE.

The explosion at Tunguska, on a round earth, should have been a local affair, restricted to an area of some 200 km x 200 km, nothing could be seen at 600 km, or at 6000 km (London).

Newspaper accounts from London:

http://www.bibliotecapleyades.net/ciencia/esp_ciencia_tunguska02.htm
http://www.nuforc.org/GNTungus.html

Now you must remember that the trajectory of the fireball which caused the explosion itself was observed for SOME 10 MINUTES (7:05 - 7:10) PRIOR TO THE EXPLOSION, HERE IS THE EXTRAORDINARY DESCRIPTION:

T.R. LeMaire, a science writer, continues this thought, by suggesting "The Tunguska blast's timing seems too fortuitous for an accident" (LeMaire 1980). He claims that a five-hour delay would make the target of destruction St. Petersburg, adding that a tiny change of course in space would have devastated populated areas of China or India.

Can we assume that the 'pilot' chose a cloudless day with excellent visibility from aloft to assure a safe drop? American Military strategy called for identical weather conditions; for a perfect strike on Hiroshima's industrial heart, the Enola Gay's bombardier was forbidden to release through a cloud cover: he had to see the target below. To maximize blast destruction, minimize radiation perils: the bomb was set to explode at a high altitude rather than against the ground. Similarly, the Siberian missile detonated high in the air, reducing or even eliminating fallout hazard (LeMaire 1980).

LeMaire maintains the "accident-explanation is untenable" because "the flaming object was being expertly navigated" using Lake Baikal as a reference point. Indeed, Lake Baikal is an ideal aerial navigation reference point being 400 miles long and about 35 miles wide. LeMaire's description of the course of the Tunguska object lends credence to the thought of expert navigation:

The body approached from the south, but when about 140 miles from the explosion point, while over Kezhma, it abruptly changed course to the east. Two hundred and fifty miles later, while above Preobrazhenka, it reversed its heading toward the west. It exploded above the taiga at 60degrees55' N, 101degrees57' E (LeMaire 1980).

THE TRAJECTORY ITSELF, PRIOR TO THE EXPLOSION, WAS SEEN ALL THE WAY FROM LONDON:

TO THE EDITOR OF THE TIMES.
Sir,--I should be interested in hearing whether others of your readers observed the strange light in the sky which was seen here last night by my sister and myself. I do not know when it first appeared; we saw it between 12 o'clock (midnight) and 12:15 a.m. It was in the northeast and of a bright flame-colour like the light of sunrise or sunset. The sky, for some distance above the light, which appeared to be on the horizon, was blue as in the daytime, with bands of light cloud of a pinkish colour floating across it at intervals. Only the brightest stars could be seen in any part of the sky, though it was an almost cloudless night. It was possible to read large print indoors, and the hands of the clock in my room were quite distinct. An hour later, at about 1:30 a.m., the room was quite light, as if it had been day; the light in the sky was then more dispersed and was a fainter yellow. The whole effect was that of a night in Norway at about this time of year. I am in the habit of watching the sky, and have noticed the amount of light indoors at different hours of the night several times in the last fortnight. I have never at any time seen anything the least like this in England, and it would be interesting if any one would explain the cause of so unusual a sight.
Yours faithfully,
Katharine Stephen.
Godmanchester, Huntingdon, July 1.?

More accounts:

A woman north of London wrote the London Times that on midnight of July 1st the sky glowed so brightly it was possible to read large print inside her house. A meteorological observer in England recounted on the nights of June 30th and July 1st:
A strong orange yellow light became visible in the north and northeast... causing an undue prolongation of twilight lasting to daybreak on July 1st...There was a complete absence of scintillation or flickering, and no tendency for the formation of streamers, or a luminous arch, characteristic of auroral phenomena... Twilight on both of these night was prolonged to daybreak, and there was no real darkness.
The report that most closely ties these strange cosmic happenings with Tesla's power transmission scheme is that while the sky was aglow with this eerie light it was possible to clearly see ships at sea for miles in the middle of the night.

To the Editor of the Times.
Sir,--Struck with the unusual brightness of the heavens, the band of golfers staying here strolled towards the links at 11 o'clock last evening in order that they might obtain an uninterrupted view of the phenomenon. Looking northwards across the sea they found that the sky had the appearance of a dying sunset of exquisite beauty. This not only lasted but actually grew both in extent and intensity till 2:30 this morning, when driving clouds from the East obliterated the gorgeous colouring. I myself was aroused from sleep at 1:15, and so strong was the light at this hour that I could read a book by it in my chamber quite comfortably. At 1:45 the whole sky, N. and N.-E., was a delicate salmon pink, and the birds began their matutinal song. No doubt others will have noticed this phenomenon, but as Brancaster holds an almost unique position in facing north to the sea, we who are staying here had the best possible view of it.
Yours faithfully,
Holcombe Ingleby.
Dormy House Club, Brancaster, July 1? (1908 )

Some people saw massive, silvery clouds and brilliant, colored sunsets on the horizon, whereas others witnessed luminescent skies at night. Londoners, for instance, could plainly read newsprint at midnight without artificial lights.

In London on the night of June 30th the air-glow illuminates the northern quadrant of the heavens so brightly that the Times can be read at midnight. In Antwerp the glare of what looks like a huge bonfire rises twenty degrees above the northern horizon, and the sweep second hands of stopwatches are clearly visible at one a.m. In Stockholm, photographers find they can take pictures out of doors without need of cumbersome flash apparatus at any time of night from June 30th to July 3rd.

THIS WOULD BE POSSIBLE ONLY ON A FLAT EARTH, GIVEN THE 4333 KM VISUAL OBSTACLE PRESENT on a round earth.


Let us go back to the Lake Michigan story; here are the main points:

'As twilight deepened, there were more and more lights.'

Bringing out a pair of binoculars, Kanis said he was able to make out the shape of some buildings.

'With the binoculars we could make out three different communities,' Kanis said.

According to one Coast Guard crewman, it is possible to see city lights across the lake at very specific times.

Currently a Coast Guard crewman stationed in Holland, Todd Reed has worked on the east side of Lake Michigan for 30 years and said he's been able to see lights across the lake at least a dozen times.

THE CURVATURE FOR 128 KM IS 321 METERS.

THE HOUSE OF THOSE RESIDENTS IS LOCATED RIGHT NEXT TO THE LAKE, BUT LET US INVESTIGATE VARIOUS ALTITUDES, FOR THE SAKE OF DISCUSSION.

h = 3 meters BD = 1163 METERS

h = 5 meters BD = 1129 METERS

h = 10 meters BD = 1068 METERS

The highest building in Milwaukee has a height of 183 meters, the difference from h = 5 meters in altitude being 946 meters, and those residents saw the buildings from THREE DIFFERENT COMMUNITIES, two of which have buildings whose heights measure way under 183 meters.

Therefore, the only way those buildings could be seen, given the 128 km distance, would be if the surface of Lake Michigan is completely flat.

The home of the Holland (MI) resident is located right next to the beach itself (Lakeshore Drive), therefore we can take an altitude of 5-10 meters for the deck of his residence, from where he saw the views.

And, National Service Service meteorologist J. Kowaleski said that on that Monday night the sky was clear.

With a visual obstacle of at least 1068 meters, there is NO WAY that the shapes of buildings from Milwaukee (and two other communities) could be seen from 128 km away.

One of those communities is Racine, Wisconsin, where the tallest building (County Court House) measures some 40 meters in height, so we can increase the visual obstacle by at least 140 meters (tallest building in Milwaukee = 183 meters).


Still one of the very best proofs that over a distance of 128 km, there is no curvature whatsoever, not one centimeter.
« Last Edit: May 31, 2010, 03:48:38 AM by levee »

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flyingmonkey

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Re: Proof of curvature
« Reply #94 on: May 31, 2010, 06:11:11 AM »
Now, where do you take into account atmospheric temperature and pressure?

I missed it.

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Lorddave

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Re: Proof of curvature
« Reply #95 on: May 31, 2010, 08:32:58 AM »
First, I'm sorry, I meant Harbourgh I, not III.  Got the buildings confused.  If you look at the picture I posted of the Toronto skyline from up close, that is larger (if by only 10-15m) than the dome.

Well I looked at the videos and I didn't see anything you said.
I saw the white lines along the horizon but it was uniform along the whole thing so I don't think they were waves.  Especially since they were 1 pixel in size.

I think I see the problem here though:
You don't know what infinity looks like.  You keep saying "I see no curvature what so ever" yet it's blatantly obvious.  There's so much you can't see that you should be able to, like the rest of the Ocean.  It should trail off from detail to non-detail until blurring into a giant blob of blue with the ground showing up blobs of green or brown squashed together.

It's the same principal as a hill.  If you look at a hill while you're on the hill it looks flat until it hits the visual horizon.  Assuming, of course, that you don't have any visual indications that you're on a hill.

http://kronski.com/wp-content/uploads/2007/02/walking-uphill-600x400.jpg

Perfect example.
See how it looks flat then dips but you can't see the dip, it just looks like a "horizon"?  That's what you see in all those videos.  You just think it's what a flat plane would look like. 

Again, to illustrate my point, you can't see the island airport IN FRONT of the CN tower.
You have been ignored for common interest of mankind.

I am a terrible person and I am a typical Blowhard Liberal for being wrong about Bom.

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Sliver

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Re: Proof of curvature
« Reply #96 on: June 01, 2010, 07:31:15 AM »
sliver and ld, how many times do you have to be told that when you bring a photo here, you must specify the following:

-place taken, a range of possible altitudes (at which the photographer was located), with additional photos, if needed, to support your opinion

-distance to the visual target

-height of visual target

OK, but I didn't bring you a picture.  I asked you to explain why, when I was 50 miles off the coast of Ocean City, MD on a clear calm day, I could not see land.    If there is no curvature of the Earth, I should have been able to see Ocean City without any problems, right?  Please, explain it for me.

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sandokhan

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Re: Proof of curvature
« Reply #97 on: June 02, 2010, 01:56:47 AM »
sliver, the same thing applies to your observation. You must tell us at what altitude you were located, the distance itself (some kind of map at least, so that we know what you are talking about), and what visual target you had in mind. Were you using a pair of binoculars, a Canon 300, or just the naked eye? You must remember that the densities of both air and aether will modify the speed of light (which is a variable and not a constant), thus only under some special conditions will you (or the Holland residents) be able to see images of visual targets which are found at some 90 - 128 km distance.



Makes no difference. Harbour Point I, 98 meters in height appears in that photograph, just the top floors, as do ALL the other buildings from downtown Toronto. The island airport cannot appear in that picture, as you should know by now, given the height of its buildings; that picture captures details which are above 90 meters.

ld, you are joking of course. Do you UNDERSTAND what we are talking about here? Or perhaps not?

A round earth means CURVATURE: an ascending slope, a midpoint curvature, and a visual obstacle.

Now, here are videos themselves again (youtube is just the media outlet through which we can have access to these extraordinary videos; they were presented on the most prestigious cultural channels in England, France, Spain and much more...):

THERE SHOULD BE A 3.31 METER CURVATURE OVER THE STRAIT OF GIBRALTAR, IN THE ROUND EARTH THEORY; even if we change the radius of the earth from 6300 to 6400 km, the curvature will vary from 3.30 to 3.35 meters. We should see an ascending slope starting from the shores of Spain, a midpoint curvature of 3.31 meters, and NOTHING BELOW 5 METERS FROM THE OTHER SIDE OF THE STRAIT.

Here are the videos which prove, once and for all, now and forever, that there is no curvature, not one centimeter over the strait of Gibraltar:

Islamic History of Europe

#

Between 2:56 si 3:00 the author shows us the spanish beach and points towards the african coastline

Between 3:02 si 3:07 we can see clearly that there is no curvature all the way to Morocco; moreover, if we use the full screen option, we will see the waves splashing onto the opposing beach/shore...this is actually a closeup taken, again, from that beach...

Between 3:19 - 3:22, WE CAN SEE THE WAVES SPLASHING ONTO THE OPPOSING BEACH, EVEN WITH THE AUTHOR STANDING ON THE SPANISH SHORELINE, RIGHT NEXT TO THE STRAIT OF GIBRALTAR; on a round earth, we would see an ascending slope, with a midpoint curvature of 3.31 meters.

Between 3:43 si 3:45, the same thing, zero curvature...full screen option, the waves splashing onto the opposing beach/shore, WITH THE AUTHOR STADING RIGHT THERE ON THE SPANISH BEACH.


The Barbarians, here are the details, where we can see very clearly that there is no ascending slope, no midpoint curvature:

The Barbarians, hosted by Terry Jones

http://video.google.com/videoplay?docid=-811260411880444286&q=barbarians+terry+jones&total=22&start=10&num=10&so=0&type=search&plindex=1#

Between 38:28 - 38:35, we can see clearly ABSOLUTELY NO CURVATURE ALL THE WAY TO MOROCCO...the surface of the strait is completely flat...


So, there is no curvature whatsoever over the strait of Gibraltar, no matter the fairy tale you stubbornly want to believe in...

Here is a photograph taken right on the spanish beach, from the same place as that in the second video above:



A COMPLETEY FLAT SURFACE OF THE WATER FOR 13 KM, WITH NO CURVATURE WHATSOEVER.


ld, do you NOW understand what is going on? We have a very clear and definite visual target, WITH NO CURVATURE WHATSOEVER, there is no ascending slope, no midpoint curvature (3.3 meters); WE CAN SEE THE WAVES SPLASHING ONTO THE OPPOSING SHORELINE, WHICH IS IMPOSSIBLE ON A ROUND EARTH.

You also wrote (about gravity): It implies nothing, what it is is what it is.

All I know is that it works, and works fine because we are all here, debating about it.

Can you explain how we don't all fly off the Earth if it is a spinning disc?


Again, please read up the facts: there is no such thing as attractive gravity, ample proofs from each and every branch of physics:

http://theflatearthsociety.net/talk/viewtopic.php?f=7&t=1183&start=15#p35541
http://theflatearthsociety.net/talk/viewtopic.php?f=7&t=1183&start=15#p35542

What spinning disc? The Earth is completely STATIONARY, and not moving anywhere.


I wrote earlier: Here is the most complex formula for terrestrial/atmospheric refraction, you CANNOT see the roof top of the Sky Dome from that distance, no matter what you say...

http://ireland.iol.ie/~geniet/eng/refract.htm#Terrestrial


« Last Edit: June 02, 2010, 02:01:48 AM by levee »

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Lorddave

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Re: Proof of curvature
« Reply #98 on: June 02, 2010, 12:31:16 PM »
*very long post*


Umm... NONE of the buildings are visible.  I'm not sure what you're seeing but it's not there.  Perhaps if you grabbed a photo and circled where you see the HarbourghPoint buildings.

And again, what upward slope?  it's a curve and everything infront of you is going DOWN! (from your point of view)
You have been ignored for common interest of mankind.

I am a terrible person and I am a typical Blowhard Liberal for being wrong about Bom.

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Sliver

  • 557
Re: Proof of curvature
« Reply #99 on: June 02, 2010, 07:01:00 PM »
sliver, the same thing applies to your observation. You must tell us at what altitude you were located, the distance itself (some kind of map at least, so that we know what you are talking about), and what visual target you had in mind. Were you using a pair of binoculars, a Canon 300, or just the naked eye? You must remember that the densities of both air and aether will modify the speed of light (which is a variable and not a constant), thus only under some special conditions will you (or the Holland residents) be able to see images of visual targets which are found at some 90 - 128 km distance.
I was at sea level, as I stated.  I was roughly 50 miles east of Ocean City, MD, as I stated.  I was looking toward shore with my naked eyes.

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markjo

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Re: Proof of curvature
« Reply #100 on: June 02, 2010, 07:38:08 PM »
You must remember that the densities of both air and aether will modify the speed of light (which is a variable and not a constant), thus only under some special conditions will you (or the Holland residents) be able to see images of visual targets which are found at some 90 - 128 km distance.

Special conditions like the ones that cause atmospheric refractive phenomena such as looming?
http://einhornpress.com/loominglightrefraction.aspx
http://scubageek.com/articles/wwwlom.html
http://en.wikipedia.org/wiki/Looming_and_similar_refraction_phenomena
Science is what happens when preconception meets verification.
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Besides, perhaps FET is a conspiracy too.
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It is just the way it is, you understanding it doesn't concern me.

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sandokhan

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Re: Proof of curvature
« Reply #101 on: June 04, 2010, 01:43:09 AM »
The thermal inversion effect is taken into consideration in the formula I posted earlier:

http://ireland.iol.ie/~geniet/eng/refract.htm#Terrestrial

A very strong inversion is necessary before the superior mirage is noticeable. There is no analogue to the heated surface and sharp temperature gradient above it in this case, so the superior mirage with crossing of rays and reflection is seldom observed. The superior mirage is also called the polar mirage, because it is most frequently observed in that region, over a very cold surface.

Even by using your formula (http://einhornpress.com/loominglightrefraction.aspx ) it is absolutely impossible to see the County Court House from Racine, Wisconsin, over a 128 km distance; even the tallest building in Milwaukee, 183 meters, could not be seen at all from Lakeshore Rd. in Holland.

And remember, thermal inversion takes place mostly in the polar regions...

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markjo

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Re: Proof of curvature
« Reply #102 on: June 04, 2010, 06:54:47 AM »
A very strong inversion is necessary before the superior mirage is noticeable. There is no analogue to the heated surface and sharp temperature gradient above it in this case, so the superior mirage with crossing of rays and reflection is seldom observed.

And I'm willing to wager that "the County Court House from Racine, Wisconsin, over a 128 km distance" is seldom observed as well.
Science is what happens when preconception meets verification.
Quote from: Robosteve
Besides, perhaps FET is a conspiracy too.
Quote from: bullhorn
It is just the way it is, you understanding it doesn't concern me.

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Lorddave

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Re: Proof of curvature
« Reply #103 on: June 04, 2010, 01:56:55 PM »
A very strong inversion is necessary before the superior mirage is noticeable. There is no analogue to the heated surface and sharp temperature gradient above it in this case, so the superior mirage with crossing of rays and reflection is seldom observed.

And I'm willing to wager that "the County Court House from Racine, Wisconsin, over a 128 km distance" is seldom observed as well.

Which is odd since, according to Flat Earth Theory, it should be seldom NOT observed.
You have been ignored for common interest of mankind.

I am a terrible person and I am a typical Blowhard Liberal for being wrong about Bom.

Re: Proof of curvature
« Reply #104 on: September 29, 2011, 02:52:45 AM »
The thermal inversion effect is taken into consideration in the formula I posted earlier:

http://ireland.iol.ie/~geniet/eng/refract.htm#Terrestrial

Actually, the page you linked actually specifically states that the thermal inversion affect is not taken into account in those calculations:
Quote
Remember that the above does not include atmospheric conditions like convective boundary layers and inversions.

A very strong inversion is necessary before the superior mirage is noticeable. There is no analogue to the heated surface and sharp temperature gradient above it in this case, so the superior mirage with crossing of rays and reflection is seldom observed. The superior mirage is also called the polar mirage, because it is most frequently observed in that region, over a very cold surface.
seldom observed ie: perhaps 12 times in 30 years?  Sound pretty seldom to me.  In any case the mirage they are talking about, with a crossing or rays and reflection - is where an upside down image of something is reflected on top of itself.  This is not what happened in Racine.

Even by using your formula (http://einhornpress.com/loominglightrefraction.aspx ) it is absolutely impossible to see the County Court House from Racine, Wisconsin, over a 128 km distance; even the tallest building in Milwaukee, 183 meters, could not be seen at all from Lakeshore Rd. in Holland.
That formula does not account for refraction - it is simply the line of sight for a given distance due to curvature.


And remember, thermal inversion takes place mostly in the polar regions...
Mostly, but not only.  And as stated, the type of superior mirage that occurs mostly in polar regions is not the kind of superior mirage that was seen at Racine.
First human spacewalker, Cosmonaut Alexei Leonov: “Lifting my head I could see the curvature of the Earth's horizon. ’So the world really is round,’ I said softly to myself, as if the words came from somewhere deep in my soul. "

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sandokhan

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Re: Proof of curvature
« Reply #105 on: September 29, 2011, 03:09:53 AM »
A question for the moderators: why was an answer provided here, when the discussion went on at another thread? The quotes provided above are from another thread...

http://www.theflatearthsociety.org/forum/index.php?topic=50707.msg1249773#msg1249773

pitrheumatoidtech, you think or imagine that by changing the place, your luck will change also?

YOU HAVE NOT PROVIDED ANY FORMULA TO BACK YOUR WILD CLAIMS!

LOOMING CANNOT CHANGE THE FACTS: there is a 984 meter visual obstacle over a distance of 128 km, from Holland to Racine.

WHY do you keep bullshitting everybody here?

And as stated, the type of superior mirage that occurs mostly in polar regions is not the kind of superior mirage that was seen at Racine.

WHERE IS YOUR PROOF, THE PRECISE FORMULA?


There is no formula whatsoever which will explain the fact that a 40 meter building, can be seen from a 128 km distance.


Just as in the case of attractive gravity, YOU HAVE NO SCIENTIFIC DATA AT YOUR DISPOSAL, AND YOU ARE MAKING UP FACTS AS YOU GO ALONG, THINKING IT WILL WORK. IT WON'T. Not with me.

Please research your bullshit claims very carefully, using any formula you can find; it won't help you, not at all.

You cannot see a 40 meter building, given the 984 meter visual obstacle.



EDIT


You keep telling me to provide formulas, then when I show how the refractive index of an inversion later is calculated, you completely ignore the information and tell me yet again that I haven't shown the formula.

Your effort is greatly appreciated...but, you only posted a formula for the determination of the ray curvature...and you know very well what we need here, don't you? As usual, I have to do the research for the RE...

HERE IS THE ONLINE CALCULATION OF THE LOOMING/THERMAL INVERSION EFFECTS ON CURVATURE:

http://mintaka.sdsu.edu/GF/explain/atmos_refr/altitudes.html

We need to specify the following: temperature/height at the place of observation, temperature/height of the visual target, distance, and most important of all the STANDARD OR ARBITRARY LAPSE RATE (an arbitrary lapse rate corresponds to taking into account the looming/thermal inversion effects).

How to compute the arbitrary lapse rate is described in the paragraphs preceding the formula, and also here (you missed these things, didn't you?):

http://mintaka.sdsu.edu/GF/explain/atmos_refr/bending.html


Let us give it a try with a distance of 50 km, temperature of 20C, observer at 2 meters, and visual target at 180 meters.

We will get a POZITIVE VALUE for the angle in question, that is, the visual target can be seen/observed.

Let us now change the height of the visual target to 20 meters, something we know that it could not be seen on a round earth. Also, we will use an arbitrary lapse rate.

We will get a NEGATIVE VALUE for the angle, that is, the visual target could not be seen.


For Lake Michigan, we will use multiple sets of numbers, to satisfy the requirements of the RE. Temperatures at Holland of 5, 10, 20C (remember, it was Memorial Day 2003, we could expect some 10-15C there), and at Milwaukee/Racine of 5, 10, 20C; height of the resident at some 20 meters (and I am being very generous here; actually we could take some 5-10 meters, but I will look the other way), height of visual target: first 40 meters, then 183 meters.

We will use, of course, ONLY AN ARBITRARY LAPSE RATE, to take into account the full LOOMING/THERMAL INVERSION/DUCT FORMATION EFFECTS.


NO MATTER HOW THE FORMULA IS USED WE GET A NEGATIVE VALUE/ANSWER: that is, the visual target CANNOT BE SEEN.

This was to be expected of course...as I have clearly described the facts.

The final word and conclusion: there is no curvature over Lake Michigan, across a distance of 128 km.
« Last Edit: October 04, 2011, 01:52:22 AM by levee »

Re: Proof of curvature
« Reply #106 on: September 29, 2011, 06:34:14 AM »
A question for the moderators: why was an answer provided here, when the discussion went on at another thread? The quotes provided above are from another thread...

http://www.theflatearthsociety.org/forum/index.php?topic=50707.msg1249773#msg1249773

pitrheumatoidtech, you think or imagine that by changing the place, your luck will change also?

YOU HAVE NOT PROVIDED ANY FORMULA TO BACK YOUR WILD CLAIMS!

LOOMING CANNOT CHANGE THE FACTS: there is a 984 meter visual obstacle over a distance of 128 km, from Holland to Racine.

WHY do you keep bullshitting everybody here?

And as stated, the type of superior mirage that occurs mostly in polar regions is not the kind of superior mirage that was seen at Racine.

WHERE IS YOUR PROOF, THE PRECISE FORMULA?


There is no formula whatsoever which will explain the fact that a 40 meter building, can be seen from a 128 km distance.


Just as in the case of attractive gravity, YOU HAVE NO SCIENTIFIC DATA AT YOUR DISPOSAL, AND YOU ARE MAKING UP FACTS AS YOU GO ALONG, THINKING IT WILL WORK. IT WON'T. Not with me.

Please research your bullshit claims very carefully, using any formula you can find; it won't help you, not at all.

You cannot see a 40 meter building, given the 984 meter visual obstacle.
I provided an answer here, because YOU linked to this thread.  So cop it sweet buddy.

I also gave you a very precise explanation of looming and how it can travel parallel to the earth's curved surface. I suggest you read it and respond without copying and pasting the same irrelevant info YET again.

Despite your very best efforts to provoke me into name calling, by calling me names, it isn't going to happen  8)
First human spacewalker, Cosmonaut Alexei Leonov: “Lifting my head I could see the curvature of the Earth's horizon. ’So the world really is round,’ I said softly to myself, as if the words came from somewhere deep in my soul. "