Fatally flawed FETs

I went to new jersey and looked east over the coast. I didn't see any land across the atlantic at all!

preco, here are the facts of life for you:

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 also find it 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.

NO curvature across lake Ontario:

http://www.theflatearthsociety.org/forum/index.php?topic=29694.msg722674#msg722674

http://www.theflatearthsociety.org/forum/index.php?topic=29694.msg719893#msg719893

http://www.theflatearthsociety.org/forum/index.php?topic=22317.msg718436#msg718436


NO curvature from Port Credit to Toronto:

http://www.theflatearthsociety.org/forum/index.php?topic=29694.msg722741#msg722741


NO curvature from Etobicoke to Toronto:

http://theflatearthsociety.net/talk/viewtopic.php?f=7&t=1090


And the very best proof, no curvature over distances of 600-1000 km (Lake Baikal/Irkutsk) and over a distance of 7000 km (London - Tunguska; curvature of 936 km, with a visual obstacle of 7436 km):

http://theflatearthsociety.net/talk/viewtopic.php?f=7&t=1142


And you know already what is going over the Strait of Gibraltar, over the English Channel...

I knew it! 16 days of nothing from the FEers & then this clown posts after I sign off.

God, where do you start with that impressive load of crap?

Well I guess the glaringly obvious first point is, it still doesn't address the op. You should learn to read before you learn to copy & paste!

Probably just as obvious is the fact that FES doesn't permit photos as evidence. To accept your photos means you have to accept photos from planes showing curvature.

All that aside, those photos don't show anything as there is no information on the photographer's height above sea level compared to the subject. Unless the camera lens is at water level (which it clearly isn't), allowing for wave height (which isn't much in those photos), there is no point even looking at your figures.

And to claim as you do with one of photos that you can distinguish that there is not even a 1cm drop is just laughable.


Now if there are really no serious attempts at addressing the FATAL FLAWS OF FE !!!!!!! I will again take my leave.

Not to mention that one of the photos, (http://farm4.static.flickr.com/3089/2379255560_d357df6305_o.jpg), shows a very intense fata morgana along the horizon, which means that, regardless of your FET/RET position, there is obviously some kind of atmospheric refraction occurring which hasn't been accounted for.

Next, the majority of experiments performed by Rowbotham were to try & detect curvature over a distance of approx 6 - 10 miles which is obviously well within the visual limits of any horizon.

The math for a round earth seems to say there would be a visual drop over that space.

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the math says that for a round earth there would be a 6 ft drop over 6 miles of the horizon.  if you can visually judge that, id be very impressed.

the math says that for a round earth there would be a 6 ft drop over 6 miles of the horizon.  if you can visually judge that, id be very impressed.

Oh but you can dazed...unless your one of them that denies everything just for the heck of it.   
6 miles is actually very limited distance if you were to overlook the horizon (ex. the atlantic or pacific) and the curvature is easily seen. However as one debater stated; would the curvature flip if we were to stare at it upside down?