**#60**Incomprehension of geometry/perspective

No, the horizon would not be noticeably curved at any altitude above sea-leavel. What does the horizon form on a sphere? Right, a circle. What does a circle look like when you observe it from the center? Right, a line. Now, let's take a hula hoop, put it around your head and look at it from the center. What do you see? Right, a LINE. Then you would say "hey, but I'm looking at the horizon from above!"

Ok. So let's say you're a tall guy, 2m, and you looked out at sea. How far could you see? Roundabout 5km according to this formula:

https://en.wikipedia.org/wiki/Horizon#Geometrical_model

Now, with a hula hoop measuring 70cm in diameter that gives you a visual radius uf 35cm. The horizon drop for 5km is roughly another 2m:

http://www.swisstopo.admin.ch/internet/swisstopo/de/home/topics/survey/faq/curvature.html

So to determine your correct observing position, all you have to do is calculate 35cm*4/5000 = 0.028 cm. That's a quarter millimetre. Can you raise your eye exactly 0.28mm above the center of the hula hoop and now tell me how curved it looks? It doesn't.

Then you say "what if I go higher, like 1 km?". Ok, let's try that. Problem is, the higher you go, the further expands your horizon. It will be at roughly 113 km for 1 km, total drop to the horizon (including altitude) is 2km. Calculating 35cm*2/113 = 0.3 cm. You can't see any curve with your bare eyes from there. That's why you have to get really, really high up in order to see that.

By the way, according to the flatearthers "law of perspective", which for some reason should create a "vanishing point" at some random distance, you would get a horizon forming a visual circle as well! Rising in altitude should create the exact same kind of effect, only with roughly half the drop because of missing curvature, but still, with a drop and a curve (due to rising altitude and the circular shape of the object). So what is that conjecture supposed to prove in the first place?

What you CAN see from certain places on earth is the dropping horizon. Here's a quick tip for your next holiday: Hawaii.

Get yourself a perfectly straight wooden plank, climb up on Mauna Kea on a clear day (4km high) and look out at sea. Just a little bit down from the observatories is also a popular place for hobby astronomers. Everybody can go there. When you place your plank perfectly parallel to ground level and look at it from each end, you will see the horizon drop down slightly from that line on both ends. If you happen to have a theodolite with you, you can even verify the little 2° of drop on each end which the spheric model predicts.

There are also a lot of photos from the summit of Manua Kea that show you the same thing. The following video is a beautiful example.

It's taken right next to the Gemini North Telescope (observers elevation roughly 4205m). The telescope that comes into view right next to the setting sun is Subaru Telescope. The top of its enclosure is 4182m high (4139+43).

http://subarutelescope.org/Introduction/telescope.html

With the distance between the telescope and the observer of roughly 750m we can calculate arctan(23/750)=1.76°. What does that mean? It means, if we see a far-away object dropping below the roof of Subaru Telescope, it must be (roughly) at least 1.76° below eye-level.

If the horizon wouldn't drop, it had to be visible ABOVE the roof of Subaru Telescope, as it stands lower than the observer's eyelevel, but the horizon has clearly dropped below, including clouds.

**#61**see #60

**#62**False claim

As said in #6, the Bedford Level experiment has been falsified multiple times, once it was done with proper controls to factor out refraction as source of error.

Fun fact, the first one to prove that the initial experiment was botched (and winning a bet by doing so), was Alfred Russel Wallace, not only a qualified surveyor, but also the often un-recognized co-creator of the theory of evolution by natural selection.

Both sides (John Hampden & A. R. Wallace) had agreed on an independent umpire (John Henry Walsh) to decide who won the bet. The umpire decided in favor of Wallace, as published in "The Field" March 26, 1870. Funny story, not only did Hampden not accept that, he even theatened to kill Wallace..

https://www.youtube.com/watch?v=RIZbh3g16-8#t=17m43s

Another even funnier fact: The footage that Mr. Dubay uses is a from video called "In search of the edge of the world" that was produced in 1990 as an educational program for children on how to tackle misleading information with critical thinking. The trap was set and he totally fell for it:

Here's the set-up of Wallace's experiment:

Here's what was seen through the telescope (up-down inverted), confirmed by both parties:

Here's a right-side-up-sketch of what they saw:

It is beyond my understanding how someone can see this, confirm that he sees it and then deny that it is a dead giveaway against a flat earth. For further reading, please check the following links:

http://www.cantab.net/users/michael.behrend/ebooks/PlaneTruth/pages/Chapter_02.html

http://blogs.scientificamerican.com/rosetta-stones/wallace-8217-s-woeful-wager-how-a-founder-of-modern-biology-got-suckered-by-flat-earthers/

https://en.wikipedia.org/wiki/Bedford_Level_experiment

**#63**Botched experiment

As said in #62, Rowbothams experiments were botched and his calculations did not include sources of error like atmospheric refraction; they are therefore absolutely useless. No valid experiment of today, with the necessary controls implemented, has been able to verify his findings.

**#64**Incomprehension of geometry

Again, as stated in #60, the horizon on a sphere forms a circle of equal relative altitude to your point of observation. When you look at it from low altitudes, you see nothing but a line and any ship going along the horizon doesn't have to go up or down (not even by a nanometer) as seen from your point of view.

**#65**Botched experiment

1) Every sailor knows that the speed of a vessel and its true speed over ground are entirely different things. Due to currents they can have completely different values. Mr Rowbotham did not factor this into his calculations.

2) Atmospheric refraction is a big factor and can make objects visible that geometrically are far beyond the horizon, especially when you look out on the sea on a sunny day. Again, this was not factored in.

3) Assuming a mast-height of 80 feet (roundabout 25m) and an observing position at 6 feet of elevation (roundabout 2m), you should be able to see the mast-head up until at least 23km away geometrically.

http://www.ringbell.co.uk/info/hdist.htm

Therefore, factoring in refraction and true speed of the vessel, a steamboat sailing away for four hours can be well within your visible range.

**#66**Botched experiments

Again, and again: Ask any qualified surveyor. These people don't have to put quotation marks around the word "theodolite" because they use it everyday, they know how it works and how to conduct geodesic measurements.

Any surveyor will tell you that and explain to you why Mr Rowbothams experiments involved phony calculations (#65), ignored geometrical basics (#64), did not factor in sources of error like refraction (#62-65), are non-verifiable using today's standards and therefore absolutely worthless from a scientific point of view.

**#67**Incomprehension of the model

Great Orme's elevation is 207m

https://en.wikipedia.org/wiki/Great_Orme

Only looking at google maps, one waterfront road in Douglas Harbour has an elevation of 70m. I'm sure if you looked at a topographical map, you could find an even higher point of observation, but let's stick with this one.

The drop due to curvature over 60 miles (97km) is roughly 740m:

http://www.swisstopo.admin.ch/internet/swisstopo/de/home/topics/survey/faq/curvature.html

Refraction coefficients of more than 0.13 are not only entirely possible (even up to+16), but they are also very likely especially for sunny days over open water:

http://onlinelibrary.wiley.com/doi/10.1029/2010JD014067/full

Even just a value of 0.18 will reduce your visible drop by 130m for this given distance. Let's add these 130m to one of the two locations. You have one with 207m, its sight line meets the horizon at 51.4km:

http://www.ringbell.co.uk/info/hdist.htm

The other one with (refraction-induced virtual) 70+130 = 200m meets the horizon at 50.5km. That gives us a sum of 101.9km. If the distance between these two locations is 97km that means that THEY CAN SEE EACH OTHER under clear conditions, it doesn't even have to be a hot summer day.

**#68**Phony numbers

Distance from Apple Pie Hill to the Philadelphia skyline is 52 km (32.3 and not 40 miles).

The highest point on Apple Pie Hill is the firetower. It stands 18m above Apple Pie Hill's elevation of 62m, making it 80m total (262 and not 205 feet).

https://en.wikipedia.org/wiki/Apple_Pie_Hill

Strangely enough, this 360° panoramic picture (available at the above Wikipedia link for higher res) around the tower neither shows the same features as Mr Dubays picture (river & open area in the foreground), nor can you "clearly" see the Philly skyline.

Let's do the math anyways: Comcast Tower in Philadelphia is 297m high, with ground level elevation of 12m thats 309m total.

https://en.wikipedia.org/wiki/List_of_tallest_buildings_in_Philadelphia

That alone gives you a sea-level horizon at 62.8km.

http://www.ringbell.co.uk/info/hdist.htm

That's it. You don't even need the firetower on Apple Pie Hill. You don't need refraction. No more math needed. The tallest buildings in Philadelphia must be visible from Apple Pie Hill under clear conditions.

**#69**False claim + phony numers

I measured the distance from Bear Mountain to Manhattan and it's 40 miles, not 60. The distance to Manhattan Island is between 60 and 65 km. Let's take the 63 km to Empire State Building. You can calculate the distance to the horizon dependent on your height of observation using this more accurate formula:

https://en.wikipedia.org/wiki/Horizon#Geometrical_model

If you don't want to do that, you can just punch in your elevation on this website and get the same result:

http://www.ringbell.co.uk/info/hdist.htm

The result is 70.8 km for 393 m height (Bear Mountain). Even if you don't factor in atmospheric refraction or the height of the buildings and even if you don't factor in the 14m ground elevation at Empire State Building according to this topographical map

http://en-us.topographic-map.com/places/New-York-168850/

.. it's a geometrical fact that you have to be able to see Manhattan from Bear Mountain.

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