Flat? Round? both round somewhat.

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Astronomers conduct experiments. You lose, again.

Oh really? What kind of experiments do Astronomers conduct?

Here, it's enough for your level.

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And also, every single Flat Earth Experiment is performed in conditions where the instruments recording the results over such an area would have a larger margin of error than the curvature of the Earth. Thus, we can say that every Flat Earth experiment is inconclusive, and your theory cannot be proved.

Why don't you explain for me, in detail, what kind of margin of error occurs in my experiment across the Monterey Bay. Here it is again in case you've missed it:

I live along the California Monterey Bay. It is a relatively long bay that sits next to the Pacific Ocean. The exact distance between the extremes of the Monterey Bay, Lovers Point in Pacific Grove and Lighthouse State Beach in Santa Cruz, is 33.4 statute miles. See this map.

On a very clear and chilly day it is possible to see Lighthouse Beach from Lovers Point and vice versa. With a good telescope, laying down on the stomach at the edge of the shore on the Lovers Point beach 20 inches above the sea level it is possible to see people at the waters edge on the adjacent beach 33 miles away near the lighthouse. The entire beach is visible down to the water splashing upon the shore. Upon looking into the telescope I can see children running in and out of the water, splashing and playing. I can see people sun bathing at the shore and teenagers merrily throwing Frisbees to one another. I can see runners jogging along the water's edge with their dogs. From my vantage point the entire beach is visible.

IF the earth is a globe, and is 24,900 English statute miles in circumference, the surface of all standing water must have a certain degree of convexity--every part must be an arc of a circle. From the summit of any such arc there will exist a curvature or declination of 8 inches in the first statute mile. In the second mile the fall will be 32 inches; in the third mile, 72 inches, or 6 feet, as shown in this chart. Ergo; looking at the opposite beach 30 miles away there should be a bulge of water over 600 feet tall blocking my view. There isn't.

Here's the math:



Suppose that the earth is a sphere with a radius of 3,963 miles. If you are at a point P on the earth's surface and move tangent to the surface a distance of 1 mile then you can form a right angled triangle as in the diagram.

Looking over a distance of 1 mile, we can use the theorem of Pythagoras:

a² = 3,963² + 1² = 15,705,370

and when we square root that figure we get a = 3,963.000126 miles

Thus your position is 3,963.000126 - 3,963 = 0.000126 miles above the surface of the earth.

0.000126 miles = 12 in * 5,280 ft * 0.000126 mi = 7.98 inches

Hence after one mile the earth drops approximately 8 inches.

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Ergo, looking across 30 miles the Pythagorean theorem becomes:

a² = 3963² +30² = 15,706,269

and when we square root that figure we get a = 3,963.113549 miles

Thus your position is 3,963.113549 - 3,963 = 0.113549 miles above the surface of the earth

0.113549 miles = 5,280 ft * 0.113549 mi = 599.53872 feet

Hence after 30 miles the earth drops approximately 600 feet.

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Whenever I have doubts about the shape of the earth I simply walk outside my home, down to the beach, and perform this simple test. The same result comes up over and over throughout the year under a plethora of different atmospheric conditions.

Here's more.

Man, this is pure entertaining.

Well, I can use your own logic to defeat your theory here. You said that all maps are innacurate, and the true shape of the Earth is unknown. So you don't know for sure that it is 33 miles. You're blindly believing.

And water is not perfectly flat, try downloading google earth and looking at the oceans.

And I see no recorded results, only your word for it and a diagram of a triangle.

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Here's more.

Man, this is pure entertaining.

Pure entertaining? How old are you? Do you know how to use proper English?

Stop posting. We already know what no controlled experiments with the cosmos are involved in Astronomy. It's all observation, as any child of five could realize. Agent Z has already realized this. I can give him credit for that. Reread this thread before posting further.

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Well, I can use your own logic to defeat your theory here. You said that all maps are innacurate, and the true shape of the Earth is unknown. So you don't know for sure that it is 33 miles. You're blindly believing.

I've traveled along the highway which wraps around the coast of my bay. I can look at my odometer. I'm fairly certain that the distance is correct.

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And water is not perfectly flat, try downloading google earth and looking at the oceans.

Google Earth is a 3D model. What are you telling me to look at here?

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And I see no recorded results, only your word for it and a diagram of a triangle.

As is all science; someone's word and math to support his claim. What's your point?

Round Earth Theory is draped in religious belief.

Here, Tom, more kids experiments for you.

No, I have not, I just gave up on you with astronomy, because I realised it would be a lost cause with you.

Why don't you just do what I say, and you'll find out. You are a scientist, right?

No, real science has a proof. Your experiment goes: I see it, so it must be true.

Your proof is inaccurate because the Earth is not a perfect sphere. Unless you look up the exact curvature of the Earth in that specific region, then reapply the figures, then rerun the test, your proof fails. (Remember those margins of errors I was talking about?)

This whole debate is stupid anyway. The lunar eclipse proves that the Earth is Round. Oh wait, I forgot

SHADOW OBJECT!!!!!1111!!!!!

Dude, your experiments are purely observational. There is no difference.

Well then, Tom, since you don't agree with observations, why do you agree the Earth is flat just by looking out of the window? That's an observation too, you know.

Quote from: Agent Z on October 28, 2007, 08:02:41 PM

Dude, your experiments are purely observational. There is no difference.

Have you read that copy-paste I made on how Rowbotham accounts for terrestrial refraction yet? Specialized instruments - hydrometers, barometers, thermometers - are used at multiple points along the experimentation route to rule out that the medium is not changing along the path of the experimentation area. This rules out light is being bent along the path.

Surveyor's instruments - plumb lines, theodolites, and high powered telescopes are used to ensure that the observer is looking across level paths.

Also, did you forget that I claim to have conducted the experiment at multiple times throughout the year during a variety of atmospheric condition. This repeated and controlled method rules out the likelihood of me seeing a chance optical illusion.

Now lets think about this criticially. There is a big difference looking across the surface of water here on earth to detect curvature and looking at a recurring shadow on a distant moon.

The Flat Earth experiments can be controlled with instruments to rule out terrestrial refraction, they can be conducted with water in long manmade containers, and they can be repeated over and over again at different locations upon the earth. This is essentially is a controlled experiment.

The phrase "look at the eclipse" is not a controlled experiment. That's an observation.

Why should I interpret the recurring shadow on the moon as the shadow of the earth as opposed to the shadow of any other celestial body?

There's a lot of celestial bodies up there, you know.

Yes, I read the Rowbotham experiment. It has no sources, no maths, basically no proof. It's the single word of one person. Just like your "repetitions" of this experiment across your bay. How do I know that you live in this particular place, and have performed this experiment. I don't. Until you provide proof that you do, it is invalid.

It's true that there are a lot of celestial bodies up there. What do you think causes a lunar eclipse? It has been widely recognised as the Earth since before Aristotle pointed out the fact that since the shadow is always spherical, then the Earth must be also.

Simple geometry can prove that the earth causes a lunar ecliipse. That's why it's common knowledge.

Aristotle pointed out the fact that since the shadow is always spherical, then the Earth must be also.

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Yes, I read the Rowbotham experiment. It has no sources, no maths, basically no proof. It's the single word of one person. Just like your "repetitions" of this experiment across your bay. How do I know that you live in this particular place, and have performed this experiment. I don't. Until you provide proof that you do, it is invalid.

Sure. You can go ahead and call me, Rowbotham, and every Flat Earth author liars. There's nothing stopping you from doing that. It just goes to show that our evidence is so powerful that it's the best rebuttal we can expect.

If you truly are curious about the shape of the earth, as you seem to be, then perhaps you should find a body of water near your location and preform the experiment yourself. Then you can rule out your "their all liars" hypothesis. I cannot prove that to you. You must prove it to yourself.

It's true that there are a lot of celestial bodies up there. What do you think causes a lunar eclipse? It has been widely recognised as the Earth since before Aristotle pointed out the fact that since the shadow is always spherical, then the Earth must be also.

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Simple geometry can prove that the earth causes a lunar ecliipse. That's why it's common knowledge.

Geometry does not prove that the shadow on the moon is an eclipse. The eclipse occurs a couple times a year, and so the Round Earth model has the moon rotating around the earth and aligning twice a year with the sun. All it takes to create a contradictory model is to place the moon above the Flat Earth and have some body align with the moon and sun every couple of years instead.

The Moon does not completely disappear as it passes through the umbra because of the refraction of sunlight by the Earth's atmosphere into the shadow cone; if the Earth had no atmosphere, the Moon would be completely dark during an eclipse. The red colouring arises because sunlight reaching the Moon must pass through a long and dense layer of the Earth's atmosphere, where it is scattered. Shorter wavelengths are more likely to be scattered by the small particles, and so by the time the light has passed through the atmosphere, the longer wavelengths dominate. This resulting light we perceive as red. This is the same effect that causes sunsets and sunrises to turn the sky a reddish colour; an alternative way of considering the problem is to realise that, as viewed from the Moon, the Sun would appear to be setting (or rising) behind the Earth.

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Aristotle pointed out the fact that since the shadow is always spherical, then the Earth must be also.

Wait. You're talking nonsense now. The shadow on the moon is not spherical. What makes you think that a SHADOW can be spherical?

Sure. You can go ahead and call me, Rowbotham, and every Flat Earth author liars. There's nothing stopping you from doing that. It just goes to show that our evidence is so powerful that it's the best rebuttal we can expect.

I'm not calling you a liar. But if your evidence is so convincing, where is it? You've provided all the math, and yet all you have to say is, "I have seen it." Why don't you take a still shot with your telescope and post it on the forum? That constitutes proof.

Some body? Where is the evidence for such a body?
I watched a lunar ecplise in March, it was quite spectacular. And strangely, I didn't see any body moving off after it finished.

I've just realised what your error is.

In this picture

You have line x as both a curve and a straight line in your proof. That's your error.

a² = 39632 +302 = 15,706,269

and when we square root that figure we get a = 3,963.113549 miles

Thus your position is 3,963.113549 - 3,963 = 0.113549 miles above the surface of the earth

0.113549 miles = 5,280 ft * 0.113549 mi = 599.53872 feet

Hence after 30 miles the earth drops approximately 600 feet.

Explain this, then:

Do you have any better math we can use to calculate the drop on a Round Earth model?