Photographic Evidence of Curvature Beyond the Visible Horizon

What? That photoshop job you did is full of colour gradients. Just like the dodgy boat picture. Its easy to see that you added the things in the sky, the buildings etc. A 'clean picture' looks like the aerial fell out the back of your black and white TV. Read the FAQ on the site.

Its so obvious that boat picture was a shoop.

Clean
Clean
Clean

Shopped
Shopped

I guess this one is also shopped
http://errorlevelanalysis.com/permalink/f5770d3/

And this seemingly obvious shopping doesn't turn up anything between the decks where you'd expect it.
http://errorlevelanalysis.com/permalink/40e25fc/

Yes, in the top one, they have added nice weather (including some fair weather fluffy clouds) and changed the lighting for the sea to match.

And yes, the second one is a complete abortion.

Here is one I found. You can tell its shopped by looking at what they did to her legs. Either that or she has an enormous boyfriend.
http://errorlevelanalysis.com/permalink/c120c31/

http://errorlevelanalysis.com/permalink/ff8adaf/
I took your non-shopped picture, and saved it as a low quality jpg.  Now it looks just like the one you thought was shopped.

Maybe it wasn't shopped after all.

Here is the other 8x/28x zoom photo with the two zooms compared side by side. (All I did was scale up the 8x and add lines and text.)

Notice that the waterline on both pictures is about equal. The extra zoom did nothing.

Quote from: Nolhekh on September 17, 2011, 05:32:40 PM

Quote from: three-dimensional-world on September 17, 2011, 05:27:01 PM

can you estimate the radius of the earth from these observations?

No declination or distance measurements were taken, so no.

So according to these pictures:

• The earth is (roughly) a sphere
• It's impossible to describe the size of this sphere

This is a completely unscientific approach.

I'm not proving the earth is a sphere, though with precise enough equipment I could.  What I am proving is curvature of the earth's surface.

There is an element of photoshopping in some of the OPs pictures.

A clean image should look like this.
http://errorlevelanalysis.com/permalink/212bc12/  or http://errorlevelanalysis.com/permalink/4037b7f/

An altered image looks like this
http://errorlevelanalysis.com/permalink/1b68cf2/

I just thought this relevant.

That's completely facetious and dishonst Thork.  The image's have been resaved in the process of converting from RAW, then later due to cropping.

If the images were cut and past jobs there would be more obvious evidence than what you presented.  In fact your analysis proves the images have not been photoshopped beyond what any normal process is conducted on a photograph in the workflow process.

If the images have been altered, then I cannot accept them on face value. Lord knows what has been done to them.

If the images have been altered, then I cannot accept them on face value. Lord knows what has been done to them.

I've got nothing to hide.  I will upload all untouched images shortly.  The images are converted to jpg from the camera's raw format using the "as shot" settings, only because flickr does not accept RAW images.  The entire set in RAW format is 980Mb.  If you want original RAW copies, pm me and I will email you copies of the spcific images you want to verify. 

Thork, not sure what you really know about photography, but your amatuerish assessment of my images via this tool (which incidently claims "If you are unsure how to interpret the results, please do not claim the results of this tool as proof of anything") does not say m uch I'm afraid.

Also, let's look at the assessment. All of these images were cropped and saved as jpg direct from Lightroom.  Others were further cropped for viewability by Faststone Maxview, which may explain the diffferences in analysis between the two container ship photos.  The image that you claim is "shopped" merely shows signs of being saved in a different program:

It is worth noting that edges and areas red in colour are often depicted as brighter in the ELA tests. This due to the way the photos are saved by various programs. It is not proof that image was manipulated.

The edges and areas that are brighter, are indeed red in the original picture.

btw, the "flickr user" is me, I am not trawling flickr for images to use; I took these images myself and am the only person who has handled them from start to finish.

If upon testing my originals, you still find the same results, then you are not reading the results correctly. 

http://errorlevelanalysis.com/permalink/ff8adaf/
I took your non-shopped picture, and saved it as a low quality jpg.  Now it looks just like the one you thought was shopped.

Maybe it wasn't shopped after all.

Exactly. Because Thork can't read the analysis correctly, he is assuming the dark sky is a sign of photoshopping.  Yet in the analysis it clearly says "they will stand out as a different colour", not as a large area of black, which appears instead to point to high jpg compression on areas of mostly one colour.  The large black area doesn't correspond to the example given by the author of the tool.

Hurriedly, I'll type another brief message. I hope to give the subject the time it deserves later.

Quote from: pitdroidtech on September 18, 2011, 03:44:53 AM

The refracted image, along with the fact that slightly more of the view is visible in the telescopic shots, is consistant with my findings.  See below for images with white lines drawn through the fold line of the refraction. 

I wholly agree with you, in fact, I had thought about including these lines, but it was already late and I slouched. The problem with your interpretation is that the white line drawn on the inferior mirage represents the true horizon. In an inferior mirage the mirage is mirrored below the horizon. I'm sure we can agree with this.
Again, the hill and buildings thereon are restored by magnification. If everything under the white line is projected below the horizon on water, the white line represents the hill of water that rotundity insists is there. The hill of water should not recede by magnification. I will approach this again later tonight, if you wish, but I think I've demonstrated the point sufficiently for everyone to grasp at the moment.

0m 8x

Inferior Mirage by max_wedge, on Flickr

0m 28x

Inferior Mirage by max_wedge, on Flickr

Quote

  Nevertheless, when all the images are taken together, it's quite clear that more is visible at higher elevations.  Indeed, even if the telescopic view HAD restored part of the view, the non-telescopic elevated view restored FAR MORE of the view:

Of course they do because by raising your elevation you are allowing your eye-line to recede farther before it shrinks. This is covered in ENaG, and I don't have time to post more deeply at the moment.

Quote

In a Flat Earth Scenario, increased elevation can not restore any part of the image, since there is no hill of curvature to obsure anything in the first place.

This is not true, and the misunderstanding is because you are misrepresenting true perspective. A simple observation on a locally level surface (basketball court?) will demonstrate this for you. Place a small coin upright at the edge of he court and place your eye as close to the surface at the other end of the court. You will be unable to discern the entirety of the coin until you raise your eye level, but this has nothing to do with a hill of concrete in between you and the coin, and everything to do with your ability to discern objects at distance.

Quote

Also please note, I will gladly do more such experiments, and also hope to do some on a day were there is no mirage evident.

This would be splendid. Thank you again. I will also address some of the other photos you have taken when I have the time, but this series of pictures showed the effect for markedly than the others.

We may have to agree to disagree, but I cannot accept Rowbotham's theory on perspective.  He provides no explanation and his drawings are non-sensical.  There are no lines of sight or reflection/refraction used to demonstrate exactly how his perspective is supposed to work.  It's easy to verify the laws of optics with regards to reflection, refraction, deflection and defraction of light rays.  Rowbotham's perspective on the other hand can not even be represented mathematically (using geometry).

The example you give of the coin is odd and does not tally with my experience.  I can easily see the whole coin at that distance, certainly the bottom half of it is not obscured, and standing up and viewing the coin does not make it any clearer.

But also note what we are talking about here, is not "clarity" but the actual blocking from view of the lower part of something.  Even if the coin was not clear from the ground, but more clear from the standing height, it would not be ab accurate analogy of what my photos are showing.

My photos show clearly, that as elevation is increased more of the image comes within view, a part of the view that was not visible at all at the lower elevation.  The clarity of the images at the different elevations is the same, but more of the lower segment of the view is visible at the higher elevation.

I wholly agree with you, in fact, I had thought about including these lines, but it was already late and I slouched. The problem with your interpretation is that the white line drawn on the inferior mirage represents the true horizon. In an inferior mirage the mirage is mirrored below the horizon. I'm sure we can agree with this.
Again, the hill and buildings thereon are restored by magnification. If everything under the white line is projected below the horizon on water, the white line represents the hill of water that rotundity insists is there. The hill of water should not recede by magnification. I will approach this again later tonight, if you wish, but I think I've demonstrated the point sufficiently for everyone to grasp at the moment.

Please accept my appologies for rather clumsily failing to mark my photos accurately with elevation YET AGAIN(!), but for further reference, all 28x photos that I have posted to this point, where actually taken at 1.5m and not sea level.

All other details are accurate.

So the remaining issue is the question of Rowbotham's perspective, which needs to be resolved before we can assume from my photos that the surface of the earth is curved.

However, I hope you can agree, my photos do tend to show that elevation is the factor that makes more of the view visible, and that objects do atleast appear to fall below the horizon and that viewing through a telescope does not minimise this affect.

That said, and as said before I will continue to conduct these experiments, since the collection of repeat data is what allows us to eliminate chance environmental conditions as the cause of our apparent findings.

The photographs posted on this thread can be explained very easily; no need to accuse somebody of photoshopping anything...the lower portion of the ships disappear very simply because the camera could not capture all the details at that distance; with a reflector telescope we would recover the whole image...let me explain what is going on...




BD = (R + h)/{RAD[2Rh + h^2](sin s/R)(1/R) + cos s/R} - R

RAD = SQUARE ROOT OF []

R = 6378.164 km

h = AE = height of observer/photographer

s = distance at the surface, for example 34 km between England and France across the English Channel

BD = height of observable visual target on a round earth


Data for St. Catharines, Lake Ontario, distance to Toronto, 50 km:

2 meters (observer) - 158 meters (visual obstacle)

3 - 150.5

5 - 138

10 - 117.5


http://ireland.iol.ie/~geniet/eng/refract.htm#Terrestrial (online terrestrial refraction calculation)



http://www.flickr.com/photos/j-a-x/129240474/sizes/l/in/photostream/



Height of Sky Dome: 90 meters, the building itself can be seen without any terrestrial refraction in the photograph, but we will include 10 meters, for the sake of the discussion; that is, the influence of the refraction will be some 10 meters...


Two other photographs, taken right there, on the same beach:

http://valdodge.com/wp-content/uploads/2007/08/pirate-ship-5137.jpg
http://valdodge.com/wp-content/uploads/2007/08/mirage-across-the-lake-5112.jpg





The altitude of the photographer can be easily estimated to be at or around 10 meters (if we would ascend to some 20 meters, that would mean that we are on top of a five-story building; certainly not the case here, as we can see from the photographs themselves; I would estimate some 5 meters, but we will go to 10 meters).

On a round earth, taking refraction into account, and ascending to some 10 meters, it would still be impossible to see the rooftop of the Sky Dome.


Now, the fact that the lower portion of a building/ship cannot be seen in some photographs is a result of the quality of the camera used:


http://www.flickr.com/photos/j-a-x/150629243/ (CN Tower barely visible)

http://www.flickr.com/photos/j-a-x/83867796/ (with a better camera, more details become visible)

http://www.flickr.com/photos/j-a-x/129240474/sizes/l/in/photostream/ (and the rooftop of the Sky Dome very visible, completely impossible on a round earth)


Now let us really take care of business.

As always, for Grimsby, we will ascend to some 240 meters (even though the highest altitude there is Vinemount Ridge, 213 meters); from that height, we could barely see the first signs of the beach from Toronto, distance 55 km, curvature of 59 meters.

In the following photographs, there is no curvature whatsoever, not a single centimeter, no 59 meters curvature in sight.

http://www.weatherandsky.com/Mirages/Mirages.html





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




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

NO CURVATURE ACROSS THE ENGLISH CHANNEL:

The original webpages as they appeared on flickr.com about four years ago:




The photographers are located right on the Cap Gris Nez beach, at an altitude of about 2-3 meters...the small rectangle in the photo is Cap Blanc Nez:



SHIPSPOTTING ON CAP GRIZ NEZ, ZERO CURVATURE ALL THE WAY TO ENGLAND, WHITE CLIFFS, DOVER:



No curvature whatsoever, a completely flat surface of the English Channel


white cliffs dover


Another photograph taken from Cap Gris Nez:

http://www.expedition360.com/journal/archives/2007/09/



To meet the requirements of the RE, here are the numbers for different altitudes (we will go all the way to 20 meters, that is, the height of a five-story building):


h = 3 m BD = 60.6
h = 5 m BD = 53
h = 10m BD = 40.4
h = 20m BD = 25.5

That is, from an altitude of 20 meters, we would not see anything below 25.5 from the other side; the White Cliffs are in full view...


Let us go to Lake Michigan...

From Holland Michigan, across the Lake Michigan, lights of three different communities were seen (one of them Milwaukee), across a distance of 128 km.

http://nl.newsbank.com/nl-search/we/Archives?p_action=keyword&s_search_type=keyword&p_product=HSHH&p_theme=gatehouse (on the archive webpage, May 28, 2003, Oh Say Can You See article)





'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

h = 20 meters BD = 984 METERS

h = 50 meters BD = 827.6 METERS

h = 100 meters BD = 667.6 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 (you can also use the above formula on atmospheric refraction to see how impossible it is to see shapes of buildings over a 128 km distance).

THE TALLEST BUILDING IN RACINE IS THE COUNTY COURTHOUSE, 40 METERS; IT WOULD BE ABSOLUTELY IMPOSSIBLE TO SEE THIS COURTHOUSE FROM 128 KM DISTANCE, FROM HOLLAND.

with a reflector telescope we would recover the whole image...let me explain what is going on...

In just one phrase levee has shown us that he does not have even a hint of knowledge on Optics.

Reflector telescopes and Refractor telescopes produce about the same quality of images for a given aperture. The only real difference is chromatic aberration, which is a lot less in current telescopes due to the advance in optical glass and lens design.

If we were having chromatic aberration problems, colored rainbows would appear around bright objects, reducing overall resolution. But in these photographs we do not see the first hint of chromatic aberration.

And then, if we did post photographs made with a reflector telescope, you would declare that silver coated reflectors are not good enough, that some other telescope design is needed to make the ship visible again. And then you would start with the camera, or anything else. That is called special pleading.

PS., levee does not show one photograph in which the photographer explains the circumstances in which he took the photos. Now that he has embarked in Rowbotham's idea of different lenses recovering different parts of the objects, we can tell him to give us the information about the lenses and other circumstances or stop showing the photos he got from the Internet or other threads.

As always, for Grimsby, we will ascend to some 240 meters (even though the highest altitude there is Vinemount Ridge, 213 meters); from that height, we could barely see the first signs of the beach from Toronto, distance 55 km, curvature of 59 meters.

In the following photographs, there is no curvature whatsoever, not a single centimeter, no 59 meters curvature in sight.

http://www.weatherandsky.com/Mirages/Mirages.html





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




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

Actually, by my RE calculations, at 240 metres, your horizon would be 55.3 kilometers away.  The CN Tower, is only 53 kilometers away from Grimsby, making the beach of Toronto well within the visible range of the Vinemount Ridge.  No curvature should be visible in front of Toronto at this height, and thanks to your posted images, we can confirm that this is the case.

Another thing to note, is that with a Flat Earth, the horizon should represent the eye level.  If the observer's altitude is 240m, then the horizon should cut through every object in the distance at that level.  The CN Tower is about 550 metres tall.  At 240m we should see the horizon cut through the CN tower just below halfway to its top.  Instead, the entire city seems to be sitting on top of the horizon.  This should be impossible even for rowbotham's perspective as the lower parts of the buildings should disappear rather than rise above the horizon.  However, with Toronto at a distance of roughly 55 km, and with round earth geometry predicting a 55 km distant horizon, these photographs are completely in line with round earth predictions.

Levee, before you go posting the english channel more,
[quote title=http://www.gitesduventus.com/index.php?option=com_content&view=article&id=7&Itemid=8&lang=en] The cliffs of the Cap are the closest point of France to England - 34 km (20 miles) from their English counterparts at Dover. Smothered in sea pinks and thrift, the cliffs are a perfect vantage point to see hundreds of ships from oil tankers to little fishing trawlers plying the waters below. On a clear day, the emblematic white cliffs of Dover on the English shore can be seen. Cap Gris Nez (height 45 m.) is a well-known migration hot-spot and sea watching site.
[/quote]

Yeah, so the horizon is 23 km away., and the cliffs are 34 km away.
The white cliffs of dover can be over 100 meters tall.

34-23=11 km

sqrt(11^2+6300^2)-6300)*1000 = 9.6 meters should be hidden.
You can't tell if 10% of the cliffs are invisible from the photos

Also, Nolhekh is right about 240 metres. Your math is incorrect, and the ontario pictures agree with a round earth, and again show proper curvature.


As for your account of the coast guard
"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."

A dozen times in 30 years? On a flat earth, that should happen way more often. Sometimes, the air does funny things, and mirages occur, like:

I tried this.  One thing I notice is that the penny, because it is flat, diminishes in vertical apparent size as you go from viewing it from slightly above to viewing it edge on.  This makes a penny a poor comparison to a building.  Another thing I notice, is that I can still see the floor where the penny was.  There's no actual "horizon" blocking my view.

You are simulating an object very distant. As you look at a building from a mostly vertical angle the "apparent size" decreases as well. You can still see the floor where the penny was, just as you can still see the sea. There is no actual "horizon" blocking your view. But again, I appreciate the fact you actually performed this experiment instead of accepting "axioms" as fact.

We may have to agree to disagree, but I cannot accept Rowbotham's theory on perspective.  He provides no explanation and his drawings are non-sensical. "

Actually, they do make sense, but because they are not what we are taught, it takes a few minutes to grasp the significant points. I had trouble with it as well, when I first read it.

Rowbotham's perspective on the other hand can not even be represented mathematically (using geometry).

I'm sure they could be, but it would take someone brighter than I to do the math, I am sure.

The example you give of the coin is odd and does not tally with my experience. 

Please actually try this at a local basket ball court. Anyone can demonstrate this with a minimum of effort, and it was done faithfully above by another poster.

I can easily see the whole coin at that distance, certainly the bottom half of it is not obscured, and standing up and viewing the coin does not make it any clearer.

The question is certainly not of clarity, but of ability to resolve the coin in any way.

My photos show clearly, that as elevation is increased more of the image comes within view, a part of the view that was not visible at all at the lower elevation. 

Correct, just as Rowbotham predicts. You are raising your eye's level extending the horizon.

I can't really add to what's already been said.  Trig, Momentia and Nolhekh covered it perfectly well.  The so called flat earth evidence posted by levee, not even the work of the levee themselves but random pics gathered from the net with no comparative views and sketchy details, really doesn't stand up against my findings.  Might I say too, that my findings aren't definitive.  I need to do more work to build up a comprehensive study of photos of ships on and past the horizon in different conditions to present an unarguable position.  Yet my photos are already far more powerful a visual proof than levee's random net gatherings.

Levee:

The photographs posted on this thread can be explained very easily; no need to accuse somebody of photoshopping anything...the lower portion of the ships disappear very simply because the camera could not capture all the details at that distance; with a reflector telescope we would recover the whole image...let me explain what is going on...

Thanks for your support, I certainly haven't photoshopped anything, but I'm not worried by claims that I have since this evidence can be repeated by any flat earther anytime they want and posted on this thread to counter my evidence.  If flat earther's can't produce experimental evidence to counter my findings, with atleast some details of elevation, distance and so on and comparative views, then their claims of photoshopping just look like insecure blustering (or the trolling that it is).

Regarding the lower portion of the ship disappearing due to the camera not being capable of resolving the detail; this is patently absurd.  The detail on the upper sections of the ships is quite clear.  What you are proposing is that the details of the lower portions of the hull would require more powerful magnification to resolve than the upper portion.  This isn't going to be achieved with a reflector telescope. You need a tool from Harry Potter's magic box to achieve that.

It's time for the FE'ers to cut the mustard regarding Rowbotham's supposed working of perspective.  They need to propose just how such a fanciful notion can occur. Quoting ENag isn't enough.  eNAg doesn't provide any explanation that involves a geometric model that shows how rays of light are reflected or refracted to cause the illusion of the ship disappearing from the bottom up.  Quoting eNAG when talking about perspective can only be interpreted as trolling from now on.  Come on FE'ers, surely one of you has enough mathematical ability to propose a model that accounts for this affect of perspective that the FE theory so strongly relies on?  Free yourself from eNaG I say!

Please accept my appologies for rather clumsily failing to mark my photos accurately with elevation YET AGAIN(!), but for further reference, all 28x photos that I have posted to this point, where actually taken at 1.5m and not sea level.

I had assumed the pictures to be taken at some distance above actual sea level or nothing would effectively seen, even with a calm sea.

However, I hope you can agree, my photos do tend to show that elevation is the factor that makes more of the view visible, and that objects do atleast appear to fall below the horizon and that viewing through a telescope does not minimise this affect.

I agree with your first point, but not the second. Buildings that were not present at 8x magnification were present at 28x. This is impossible to assign to the inferior mirage. Magnification should not move the visible horizon or affect the sight lines. The light will take the same path from hill/buildings to lens regardless of magnification. Clearly, restoration was present in your photographs.
Some small amount of restoration was also seen in the pictures of the apartments, but because of their relative nearness, it was much less prominent (near negligible really). Ideally, we could have a picture of a stationary ship at the horizon. Unfortunately, lightships are no longer en vogue. I noticed in at least one of your pictures the horizon is farther than the ship, this makes looking for restoration useless (this is by no means criticism, as I said, I am very pleased that you made the results of your effort available to us).

That said, and as said before I will continue to conduct these experiments, since the collection of repeat data is what allows us to eliminate chance environmental conditions as the cause of our apparent findings.

This would be ideal. The more viewings of the same sights, should give us the more reliable data to base our findings on.

Correct, just as Rowbotham predicts. You are raising your eye's level extending the horizon.

Regarding the math, it's already been done.  The study of optics is a well proven and demonstrated body of mathematics.  It simply does not account for Rowbotham's hypothesis.

As to the increasing view of the horizon upon elevation; Actually no, but in practice, the eye is easier able to make out the details of the tops of objects near the viewer than at low elevations.  So if for example the coin were laid flat, then as you stand up, more details of the coin's face will become discernable.  This is not the same as extending your view of the horizon.  You can still see all the way to the horizon from a low altitude, but you can only see the vertical component of the object ie: you can not see any detail of the top of the object.  Even if the object is lower than oneself, you still cannot see the details because the angle of incidence is so low that the details are squashed into a narrow band which the human eye can not resolve.  But you can see the vertical face of the object clearly.

Don't worry I will conduct your experiment, this thread is very much about experimentaly repeatable scientific findings.  There has been enough of people claiming "read ENaG" without being prepared to go and see for themselves how things work.  They are giving Zeteticism a bad name.

As to the increasing view of the horizon upon elevation; Actually no, but in practice, the eye is easier able to make out the details of the tops of objects near the viewer than at low elevations.  So if for example the coin were laid flat, then as you stand up, more details of the coin's face will become discernable. 

I am not at all concerned with the face of the coin unless it is upright. The issue is your eyes ability to resolve the coin or any other object at distance.

Don't worry I will conduct your experiment, this thread is very much about experimentaly repeatable scientific findings.  There has been enough of people claiming "read ENaG" without being prepared to go and see for themselves how things work.  They are giving Zeteticism a bad name.

I think if you read ENaG, and then perhaps re-read ENaG, his laws make sense. It took more than one reading for me to sort them out as well. Perhaps I will try to make new diagrams and take some pictures of my own to set them out. I wish I lived closer to the sea.

I think if you read ENaG, and then perhaps re-read ENaG, his laws make sense. It took more than one reading for me to sort them out as well. Perhaps I will try to make new diagrams and take some pictures of my own to set them out. I wish I lived closer to the sea.

I have read it and reread it to ascertain just how Rowbotham has drawn his conclusions.  There are base assumptions that he makes that underly his drawings.  There is no attempt made to qualify these base assumptions.

Regarding the coin, standing on it's end, perhaps you should take some photos of this experiment yourself.  A part of me is inclined not to do the experiment, on the basis that you may simply be trolling and laughing at me as I go about conducting experiments that you yourself haven't actually done! 

But tbh, I enjoy debate and I enjoy photography and this gives me something different to do with my (shortage of) spare time.

Quote from: Nolhekh on September 18, 2011, 11:14:29 AM

I tried this.  One thing I notice is that the penny, because it is flat, diminishes in vertical apparent size as you go from viewing it from slightly above to viewing it edge on.  This makes a penny a poor comparison to a building.  Another thing I notice, is that I can still see the floor where the penny was.  There's no actual "horizon" blocking my view.

You are simulating an object very distant. As you look at a building from a mostly vertical angle the "apparent size" decreases as well.

Except that our experiment involves things disappearing when viewed from a horizontal angle, which a building does not do.  You can simulate a building better with a lego brick.  the 2 x 2 kind, which has the same width as a penny.  Preferably red, so that you can be sure if it's really diminishing beyond our minimum ocular resolution.

Quote from: pitdroidtech on September 18, 2011, 04:45:09 PM

We may have to agree to disagree, but I cannot accept Rowbotham's theory on perspective.  He provides no explanation and his drawings are non-sensical. "

Actually, they do make sense, but because they are not what we are taught, it takes a few minutes to grasp the significant points. I had trouble with it as well, when I first read it.

I developed my own understanding of perspective through hours of mathematical study and revelation to gain an advantage over other students.  So, while I was still tought, my lack of acceptance of Rowbotham's explanations has nothing to do with my education.  My own spatial sense tells me right away when something he claims is mathematically out of place.

Quote

Rowbotham's perspective on the other hand can not even be represented mathematically (using geometry).

I'm sure they could be, but it would take someone brighter than I to do the math, I am sure.

So far I have been unsuccessful.  Some of his claims are in direct conflict with trigonometry, which is a major component of my mathematical understanding of perspective.

My photos show clearly, that as elevation is increased more of the image comes within view, a part of the view that was not visible at all at the lower elevation. 

Correct, just as Rowbotham predicts. You are raising your eye's level extending the horizon.
[/quote]
This extending of the horizon in the context of a flat earth is, to my knowledge, mathematically impossible, but is possible on a round earth, and thanks to Levee's post and given altitude I was able to mathematically determine that his photo was precisely in line with the prediction I calculated using RET.

I think if you read ENaG, and then perhaps re-read ENaG, his laws make sense. It took more than one reading for me to sort them out as well. Perhaps I will try to make new diagrams and take some pictures of my own to set them out. I wish I lived closer to the sea.

Don't worry, I'm fairly familiar with ENaG.

I posted a similar argument to the one below before, but never got a response. My argument will use ship 2 photos:

If:

T1) If [there is nothing obscuring an object from view] then:
[Iff the angular diameter of an object is less than 1 arc minute (2.909*10^-4 radians), that object is invisible.]
(This is Rowbotham's theorem, mathematically.)

T2) The angular diameter of an object is about the object's height in radians divided by the distance to the object times the magnification of the lens used. (AD = (height)*(magnification)/(distance) )

Then:

L1) Let: the distance to the ship be 25 km.
h be the height of the part of the hull that is invisible.
magnification of the lens = 28.

L2) By T2, an object viewed through a 28x lens at 25 km will have an angular diameter of 1 arc second or greater iff the object's height, y, satisfies:
y ≥ (distance)*(AD)/(magnification) = 25000*2.909*10^-4 / 28
y ≥ 0.26 m.
So, anything bigger than 0.26 m across at 25 km is greater than one arc-minute.

L3) h ≥ .26 m by inspection of the photo taken from a high position. By L2, The hidden hull has an angular diameter of 1 arc minute or greater. (Indeed, much greater)

L4) By L3 and T1, since the hull is invisible and the hull has an angular diameter of more than one arc minute, it must be obscured by something. This is zetetically observed to be water.

L5) By L4, the water demonstrates curvature, which is indicative of an RE.


That is all. If you wish to argue, please point to the line you believe is in error, and correct it.

Actually, by my RE calculations, at 240 metres, your horizon would be 55.3 kilometers away.  The CN Tower, is only 53 kilometers away from Grimsby, making the beach of Toronto well within the visible range of the Vinemount Ridge.  No curvature should be visible in front of Toronto at this height, and thanks to your posted images, we can confirm that this is the case.

Do you understand where you are, and what is being debated here?

We have a distance of 55 km (as I said, we do ascend to 240 meters), and a curvature of 59 meters; that is, an ascending slope, a midpoint curvature of 59 meters, which does not exist. Of course that curvature should be visible, there would not be a way to avoid seeing it; what you want is a round earth with no curvature...those photographs prove clearly that there is no curvature across Lake Ontario.

However, with Toronto at a distance of roughly 55 km, and with round earth geometry predicting a 55 km distant horizon, these photographs are completely in line with round earth predictions.

No they are not...there is no ascending slope, no midpoint curvature of 59 meters...the surface of Lake Ontario is perfectly flat...that what can be seen quite clearly...

As for your account of the coast guard
"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."

A dozen times in 30 years? On a flat earth, that should happen way more often. Sometimes, the air does funny things, and mirages occur, like:

On a round earth you cannot see the County Courthouse of Racine, or buildings from Milwaukee, PERIOD. It does not matter how many times it was seen; they are not looking at the other shoreline all the time...

The visual obstacle is over 1000 meters...you cannot use terrestrial refraction, or optical reflection to explain your way out of this...the reason those buildings can be seen is that the surface of Lake Michigan is perfectly flat.

Yeah, so the horizon is 23 km away., and the cliffs are 34 km away.
The white cliffs of dover can be over 100 meters tall.

34-23=11 km

sqrt(11^2+6300^2)-6300)*1000 = 9.6 meters should be hidden.
You can't tell if 10% of the cliffs are invisible from the photos

You are dreaming. From those

Actually, by my RE calculations, at 240 metres, your horizon would be 55.3 kilometers away.  The CN Tower, is only 53 kilometers away from Grimsby, making the beach of Toronto well within the visible range of the Vinemount Ridge.  No curvature should be visible in front of Toronto at this height, and thanks to your posted images, we can confirm that this is the case.

Do you understand where you are, and what is being debated here?

We have a distance of 55 km (as I said, we do ascend to 240 meters), and a curvature of 59 meters; that is, an ascending slope, a midpoint curvature of 59 meters, which does not exist. Of course that curvature should be visible, there would not be a way to avoid seeing it; what you want is a round earth with no curvature...those photographs prove clearly that there is no curvature across Lake Ontario.

However, with Toronto at a distance of roughly 55 km, and with round earth geometry predicting a 55 km distant horizon, these photographs are completely in line with round earth predictions.

No they are not...there is no ascending slope, no midpoint curvature of 59 meters...the surface of Lake Ontario is perfectly flat...that what can be seen quite clearly...

As for your account of the coast guard
"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."

A dozen times in 30 years? On a flat earth, that should happen way more often. Sometimes, the air does funny things, and mirages occur, like:

On a round earth you cannot see the County Courthouse of Racine, or buildings from Milwaukee, PERIOD. It does not matter how many times it was seen; they are not looking at the other shoreline all the time...

The visual obstacle is over 1000 meters...you cannot use terrestrial refraction, or optical reflexion to explain your way out of this...the reason those buildings can be seen is that the surface of Lake Michigan is perfectly flat.

Yeah, so the horizon is 23 km away., and the cliffs are 34 km away.
The white cliffs of dover can be over 100 meters tall.

34-23=11 km

sqrt(11^2+6300^2)-6300)*1000 = 9.6 meters should be hidden.
You can't tell if 10% of the cliffs are invisible from the photos

You are dreaming. From those 2 meters right there on the beach of Cap Gris Nez, you CANNOT SEE ANYTHING BELOW 65 METERS, THAT IS WHY I POSTED THE PRECISE FORMULAS. The full view of the White Cliffs of Dover are in sight, no curvature whatsoever across the English Channel, no ascending, no descending slope.




The so called flat earth evidence posted by levee, not even the work of the levee themselves but random pics gathered from the net with no comparative views and sketchy details, really doesn't stand up against my findings.

Your photographs prove nothing about a flat or round earth...you state: 15 to 25 km...and a altitude of 12 meters, or even lower than that...the pictures show that your camera did not capture the entire details, that is, the lower portion of the ships, as can be seen quite clearly from the messages you posted.

My photographs show that there is no curvature across Lake Ontario and the English Channel.





NO ASCENDING SLOPE, NO MIDPOINT CURVATURE OF 59 METERS...NO CURVATURE WHATSOEVER.


And you cannot explain this one either:

http://www.flickr.com/photos/j-a-x/129240474/sizes/l/in/photostream/



The rooftop of the Sky Dome visible from the St. Catharines beach; I already posted the numbers...you cannot see anything below at least 120 meters from St. Catharines...