According to the flat-heads, those pictures were taken with special fisheye-type lenses that distort the natural flatness of the earth to distort it to look like a sphere.
The stupid, it hurts.
This one is awesome:
Where does he think the 1000 mph sideways motion of the balloon is supposed to go, when then balloon rises from the Earth? If you’re in a moving car and toss a ball in the air, it doesn’t instantly smash into the back window, does it? And that’s because the ball is moving at the same forward speed as the car.
And at least a third of these are ignoring the fact that, according to the spherical earth theory, the force of gravity is normal to the surface of the earth. Therefore, trains don’t have to ascend and descend a 3 mile hump when travelling from London to Liverpool. Yes, it turns out gravity always pulls “down”, but down is toward the center of mass of the sphere, not down according to some external point. If that were the case then, you know, people in Australia would fall off, all the water on Earth would have poured off the equator, and so on.
More specifically, it depends on the spatial gradient of density perpendicular to the direction of travel of the light.
Atmospheric refraction also affects sound propagation. Depending on the atmospheric temperature profile on any given day, I may or may not be able to clearly hear traffic noise from the highway that’s a mile from my house.
Long but very interesting PDF article about acoustic refraction in the atmosphere.
Anyway, back to optics…atmospheric refraction can certainly explain the oddly quantified results presented in the OP.
Not only that, but balloons are suspended in the atmosphere, which is viscously coupled to the surface of the Earth. If it wasn’t, we’d have 1,000mph winds all the time…
Heinrich Hertz showed that electromagnetic waves are also refracted by the atmosphere (specifically radio waves), thus demonstrating that light is indeed a form of electromagnetism just like James Maxwell predicted.
I get a kick out of reasoning like this. Since the distances that the lights can be seen is more than their calculations show the Earth must be a larger globe than is commonly believed. But since the lighthouses aren’t seen above the horizon out to any given distance then, in fact, the Earth is spherical: we’re just discussing the value of the diameter.
For fun:
The rule of thumb (from Bowditch) I always used was “square root of height = distance in nautical miles” with an error of about 6 percent. The more accurate rule is "square root of height * 1.32 = distance in statute miles. That adds in refractive effects to a decent degree. The conspiracy math isn’t hugely off (like an order of magnitude) but still ignores important effects.
As Mijin noted, the very first “proof” falls, um, flat. The writer claims that you can look at the horizon and see that it is perfectly flat (mountains notwithstanding, I suppose). I can stand on the beach on a clear day, look at the ocean and clearly see that the horizon is not perfectly flat. The curvature is pretty subtle, but it is observable to the naked eye.
The stack of turtles underneath, though, I cannot account for them.
Actually, on a smooth sphere the horizon is perfectly flat. If you were to draw a line along the “horizon” on the surface of this sphere, you would get a circle with you at the center and a radius given by the “rules of thumb” that MonkeyMensch cited.
I get that modern flat-earthers have to deal with the problem that if you’re in Seattle at local noon and call your friend in London, you find that the sun has already set there. Except how is that supposed to work exactly? The sun can’t ever go below the horizon on every point on the surface of the flat earth. It has to be possible to see the sun in Seattle, yet the sun is below the horizon in London. Except that isn’t possible on a flat Earth. No matter what pattern the sun describes as it floats in circles a few thousand miles above the flat Earth, you’d be able to see it from anywhere on Earth.
(Missed the edit window) Assuming that your eye level above the surface is much less than the diameter of the earth. If you go high enough you can see curvature…
Tidal level at the point of observation could have a significant influence on visibility of such phenomenon as well by moving the observer up/down 2-3 meters.
Thanks for that, scr4, even though my brain got rather refracted after a few paragraphs.
So, I’m not sure if atmospheric refraction covers what I am about to write. The concept of seeing a faraway light is often not a straight line-of-sight matter.
When I drive from Indianapolis to Anderson, there’s a small airport a few miles off to my left. I’ve never been there, but I can see the rotating beacon’s light swinging across the sky, even though I’ve never seen the beacon itself through the thousands of trees in the way.
Likewise, anybody driving at night can see a city’s glow in the sky miles before seeing a glimpse of the city.
If you’re on the deck of a freighter bobbing out in the ocean (60 ft. above sea level, by the way,) you’ll see the scattered light of a lighthouse miles before you can see the light itself.
Tides in the open ocean aren’t that high. They’re something like a couple feet. The 2-3 meters they get along the coasts is due to piling up.
The atmosphere is not completely clear. What you are seeing are all the little bits of dust and water droplets being lit by the lights.
No one answered this directly, so I brushed off an old Taylor’s series to discover that, with usual assumptions, distance to the horizon is simply the geometric mean of height and Earth’s diameter.
Assume that the Earth is a perfect sphere and much larger than the observer. Ignore the effect of refraction on apparent horizon.
cos x = r / (r + h) = 1 - x[sup]2[/sup]/2
where r is the Earth’s radius; h the distance from a point (e.g. your eyeballs) to the Earth’s surface; x is the tiny angle in the triangle defined by your eyeball, the horizon and Earth’s center; and the third term is the Taylor’s series for cosine when x is tiny.
Rearranging terms yields x = √(2h/(r+h)) but we will ignore the tiny h in the denominator. To convert angle x to distance d, when h is tiny, simply multiply by r and get d = √(h(2r))
Distance to the horizon is the geometric mean of eyeball height h and Earth’s diameter 2r.
The Earth’s diameter is 12.74 (3.57[sup]2[/sup]) megameters, so distance to horizon is given in kilometers as 3.57 times the square root of height in meters. With eyeballs at 2 meters, the horizon is at 7.14 kilometers.
Lets translate this into metric to get “99 meters above sea level … can be seen as far as 96.6 kilometers away … an impossible 632 meters below the horizon.”
If you are 632 meters above the base of the Statue of Liberty, the horizon is 3.57 * √632 = 89.7 kilometers distant. Add this to the 7.14 kilometers to get close agreement with the 96.6 kilometer claim. So the claim is correct! …
…*** if the effects of refraction are ignored.***
Disclaimer: Use my very rusty trig and arithmetic at your own risk. No animals were tortured during its preparation.
nm
Mercy … what have you been doing to the plants then?
I am having a hell of a time charting out a sixty-mile line of sight that you could use to see the Statue of Liberty. New York harbor has a lot of stuff built up around it on all sides.
[QUOTE=septimus]
distance to horizon is given in kilometers as 3.57 times the square root of height in meters.
[/QUOTE]
I assumed OP’s numbers were correct for refraction, and answering Mr. Pearse’s question for the non-refraction case, i.e to verify the arithmetic for “impossible 2,074 feet below the horizon” . Without refraction Miss Liberty’s torch would appear at the horizon sea-level at distance 3.57 * sqrt(99) = 35.3 kilometers.
And in addition to the problem Dr. Fidelius points out, there may be some risk that David Copperfield has disappeared the Lady.
As I pointed out in another thread we are torturing plants here, but for reasons unrelated to this thread. My wife laboriously watered dozens of baby banana trees, timing it so the monsoon rains would take over at just the right time.
But the rains haven’t come. And what water there is is too warm for plants. Dozens of our banana trees are dying, along with much other vegetation.
There is another thread where I am trying to torture animals, especially myself. But only Homo sapiens is targeted — no innocent animal is being tortured there.
And if you were on the crow’s nest of a ship 30 meters above the water, that would add 3.57*sqrt(30) = 19.6km, so that you would see the tip of the torch about 55km away…