In my opinion, this photograph represents one of the most extraordinary proofs, this time taking into account a portion of land, that the surface of the earth is flat.
Now, to meet even Ben's possible criticism, I will increase the altitude of the photographer all the way to 6000 FEET (1828 METERS).
First, let us suppose that the photographer was flying in a helicopter, at 3000 FEET (about 900 meters). Even then, we could hardly see the first signs of land from Montreal, the curvature itself measuring some 268 meters in height. No such thing present there, in the photograph itself.
And now, let us imagine that the photographer was in a plane, flying at some 6000 feet (1828 meters). Since the curvature itself represents 1/7 of the altitude of the photographer, we could easily discern it in that photograph, especially given the fact that we have a definite visual target, Montreal, whose tallest building measures some 225 meters. That is, we should see a midpoint curvature of 268 meters, exceeding the visual target by some 40 meters, and being 1/7 of the height at which the photographer finds himself.
But no such thing happens, we can see a very plain field, all the way to Montreal, with no midpoint curvature of 268 meters. This in my opinion represents a fantastic proof that between Burlington (VT) - Montreal, the surface of the earth is flat.
Let us go over to lake Michigan, over a distance of some 128 km.
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=doc&p_docid=122D5519C959F390&p_docnum=1&p_theme=gatehouse&s_site=HSHH&p_product=HSHH(you can find the article on their site, 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.
If we resort to the same terrestrial/atmospheric refraction formula, it will inform us of the impossibility of seeing any details from Milwaukee given the 984 meter visual obstacle.
For those who like optical reflection, they should know that the best such phenomenon known today, is the Ice Blink occurrence, where a diffuse, unclear view of the ice sheet is reflected on clouds which travel at a low altitude. There is no way to see buildings reflected in clouds, given the visual obstacle of some 984 meters, not to mention the fact that the shape of the buildings themselves were seen from Holland.
What other proof could the round earth supporters be looking for? No curvature whatsoever, over a distance of 128 km.
As for the UA accelerator, there remain the same objections I raised earlier...
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