Flat? Round? both round somewhat.

Quote

Explain this, then:

We already have. It's a perspective effect which occurs when you look at distant bodies with the naked eye.

Using a proper equipment with proper optical (not digital) magnification reverses this effect and brings the scene back into view. This is a proof for a Flat Earth.

It is because of the Earth's curvature that causes the CN Tower to look "dropped" into the ocean, as you move further away from it.

Because of the curvature, the bottom half of the CN Tower seems to disappear. Therefore:

Which looks like this:

With an "optical magnification":

Even with magnification or telescope, the result will still be the same. You will still not able to see the bottom half of the CN Tower due to the curvature. However, if you want to discredit this, try to get a spot where the view is exactly like Gulliver's picture, get a "optical magnification" telescope, and see the bottom half of the Tower. Bring the picture here: it must show that you can see the bottom half by using the magnification, not by walking to the tower and taking a picture as to claim you've shown it's possible to see it.

This is why I've said "even if you have a telescope able to see from Earth to Mars, you still can't see from New York to London with it, assuming the Earth is flat." Since you will just say something like "distortion" or "discoloration" due to the atmospheric effect, why don't you do this experiment under an exceptionally clear day in Manhattan. With a telescope that powerful, it should be easy for you to see London even with atmospheric discoloration or distortion.

However, I will not be surprised if you can actually try to counter this because, as demonstrated before, you like to make stuffs up.

Quote

Explain this, then:

We already have. It's a perspective effect which occurs when you look at distant bodies with the naked eye.

Using a proper equipment with proper optical (not digital) magnification reverses this effect and brings the scene back into view. This is a proof for a Flat Earth. This effect is part of why reason why the ancients mistakenly concluded that the earth is a globe.

See this post for more information and personal testimonies: http://theflatearthsociety.org/forum/index.php?topic=17435.msg301724#msg301724

No you haven't, because the sinking ship effect is a pile of bullshit.

Quote from: Jack. on October 28, 2007, 06:25:07 PM

Here, Tom, more kids experiments for you.

Astronomers simply look into their telescopes, observe, and interpret.

Again, please think before you post.

tom, i believe you've made yourself into a bit of a hippocrate there. but to refresh my memory, what is it that you were doing on the bay?

OK, I have the proof. I'll be using an image I drew myself for reference, as Tom's Image didn't highlight the correct areas to solve the problem.



The goal here is to find the length a, which is the vertical difference between A and B.

Let's use the above model, where O is the centre of the Earth, A is our standing position, and B is our viewing position. The diagram shows us that (assuming that the earth is a perfect sphere for this example, otherwise it becomes far too complicated,) the radius r is the same along the line OA as OB. Since we know the length of the arc AB (which I just realised I forgot to label, let's just call it l) we can find the angle θ, (in degrees, it would be ^l/_2πr x 360.)

Now, let's assume the position of the Earth is at (0,0). (Note that for this example we do not need to worry about moving in the z direction, because the example assumes no movement in this direction, and thus can be ignored.) Now using simple geometry, We can define the point A (0,r) and B (rsinθ, rcosθ).

Now we know that the length a is the difference in height between A and B. So by defining the coordinates, we get, a = r - rcosθ.

And this is the difference in height between any 2 points on an arc of known length.

(Please note that I have not yet tried the numbers given in the example so I don't yet know if this will help my argument or not. I did this deliberately so my proof would not be biased in any way, thus the use of unassigned letters to show the answer.)

I will apply the numbers from your 30 mile example and post it in a new post.

Plus, a further note, I would like someone to check my reasoning with this proof before I apply it to the numbers given by the example, since if it is not accurate, it is a waste of time.

From Gulliver:

"Here's my math:
The diameter of the Earth is approximately: 7,925 miles
The distance to the object: d
The point of the Observer: A
The center of the Earth: B
The point of the object: C
The point level to the observer directly over C: D
The angle DAB is a right angle.
AB is one side of a right triangle, the radius of the Earth.
The length of AB is: 3,960 miles approximately
AC is equal AB
The distance for CD is the sought value: x
ACD is a straight line.
ACD is the hypotenuse of the right triangle ABD.
By the Pythagorean theorem:
d2  + AB2 = (AC+x)2
==> AC+x= sqrt(d2  + AB2)
==> x = sqrt(d2  + AB2) - AC
d (miles)   x (feet)
1             0.67
2             2.67
3             6.00
4           10.67
5           16.67
6           24.00  <===
7           32.67
8           42.67
9           54.00
10          66.67"

I think my method is better.

I just realised I got my coordinates for B back to front. Will fix accordingly as it destroys the proof.

applying my proof gives a = 597.45 feet.

So Tom Bishops method was almost correct but the method would get worse and worse the larger the arc.

Which is why I said it is accurate as long as a/R<<1.

My model of telescope does not have a camera mount. But if you would like to see something photographic which demonstrates that the surface of water does not curve in the way predicted by the Round Earth model we just need to search the internet for telescopic images across bodies of water.

See this website for example: http://www.weatherandsky.com/Mirages/Mirages.html

it's time to cut the shit tom and quit lying. a child can tell by looking at the lake's surface that the image was taken from some height. the very web page you direct to shows it. here's another picture from the same page:



huh, what is that you ask? why, the tops of trees. please explain how the tops of trees got to be at ground level by the lake? and yet you state otherwise. if you lie about this, why should we believe anything you say?

applying my proof gives a = 597.45 feet.

So Tom Bishops method was almost correct but the method would get worse and worse the larger the arc.

neither one of you are taking atmospheric refraction into account for apparent curvature. our engineers had to take this into account in afghanistan for backup relays, or guess what: they wouldn't work. bishop's posting of some guy's fe calculation of refraction is laughably simplistic and wrong. for starters, it assumes a flat earth and gets worse from there. first link below has the actual formulas (scroll past astronomical refraction...or read it also to debunk bishops "sun is a spotlight because of snell's law" nonsense). all verifiable with other sources, online and off.

http://www.iol.ie/~geniet/eng/refract.htm
http://mintaka.sdsu.edu/GF/mirages/mirintro.html
http://www.connect802.com/antenna_c_main.php

if our atmosphere were more dense - probably still in the range of long-term survivability - the world would seem concave from the surface...at least until a certain altitude. dog only knows what bishop would believe by looking out his window then.

neither one of you are taking atmospheric refraction into account for apparent curvature.

Dr. Samuel Birley Rowbotham accounts for terrestrial refraction in Experiment 9 of Earth Not a Globe:

...

The only modification which can be made in the above calculations is the allowance for refraction, which is generally considered by surveyors to amount to one-twelfth the altitude. of the object observed. If we make this allowance, it will reduce the various quotients so little that the whole will be substantially the same. Take the last case as an instance. The altitude of the light on Cape Bonavista, Newfoundland, is 150 feet, which, divided by 12, gives 13 feet as the amount to be deducted from 491 feet, making instead 478 feet, as the degree of declination.

Many have urged that refraction would account for much of the elevation of objects seen at the distance of several miles. Indeed, attempts have been made to show that the large flag at the end of six miles of the Bedford Canal (Experiment 1, fig. 2, p. 13) has been brought into the line of sight entirely by refraction. That the line of sight was not a right line, but curved over the convex surface of the water; and the well-known appearance of an object in a basin of water, has been referred to in illustration. A very little reflection, however, will show that the cases are not parallel; for instance, if the object (a shilling or other coin) is placed in a basin without water there is no refraction. Being surrounded with atmospheric air only, and the observer being in the same medium, there is no bending or refraction of the eye line. Nor would there be any refraction if the object and the observer were both surrounded with water. Refraction can only exist when the medium surrounding the observer is different to that in which the object is placed. As long as the shilling in the basin is surrounded with air, and the observer is in the same air, there is no refraction; but whilst the observer remains in the air, and the shilling is placed in water, refraction exists. This illustration does not apply to the experiments made on the Bedford Canal, because the flag and the boats were in the same medium as the observer--both were in the air. To make the cases parallel, the flag or the boat should have been in the water, and the observer in the air; as it was not so, the illustration fails. There is no doubt, however, that it is possible for the atmosphere to have different temperature and density at two stations six miles apart; and some degree of refraction would thence result; but on several occasions the following steps were taken to ascertain whether any such differences existed. Two barometers, two thermometers, and two hygrometers, were obtained, each two being of the same make, and reading exactly alike. On a given day, at twelve o'clock, all the instruments were carefully examined, and both of each kind were found to stand at the same point or figure: the two, barometers showed the same density; the two thermometers the same temperature; and the two hygrometers the same degree of moisture in the air. One of each kind was then taken to the opposite station, and at three o'clock each instrument was carefully examined, and the readings recorded, and the observation to the flag, &c., then immediately taken. In a short time afterwards the two sets of observers met each other about midway on the northern bank of the canal, when the notes were compared, and found to be precisely alike--the temperature, density, and moisture of the air did not differ at the two stations at the time the experiment with the telescope and flag-staff was made. Hence it was concluded that refraction had not played any part in the observation, and could not be allowed for, nor permitted to influence, in any way whatever, the general result.

In may, the author delivered a course of lectures in the Mechanics' Institute, and afterwards at the Rotunda, in Dublin, when great interest was manifested by large audiences; and he was challenged to a repetition of some of his experiments--to be carried out in the neighbourhood. Among others, the following was made, across the Bay of Dublin. On the pier, at Kingstown Harbour, a good theodolite was fixed, at a given altitude, and directed to a flag which, earlier in the day, had been fixed at the base of the Hill of Howth, on the northern side of the bay. An observation was made at a given hour, and arrangements had been made for thermometers, barometers, and hygrometers--two of each--which had been previously compared, to be read simultaneously, one at each station. On the persons in charge of the instruments afterwards meeting, and comparing notes, it was found that the temperature, pressure, and moisture of the air had been alike at the two points, at the time the observation was made from Kingstown Pier. It had also been found by the observers that the point observed on the Hill of Howth had precisely the same altitude as that of the theodolite on the pier, and that, therefore, there was no curvature or convexity in the water across Dublin Bay. It was, of course, inadmissible that the similarity of altitude at the two places was the result of refraction, because there was no difference in the condition of the atmosphere at the moment of observation.

Your rebuttal?

There is no experimental evidence for such a body because Astronomy, even Flat Earth Astronomy, is observational and interpretational only.

The current Flat Earth hypothesis of the Shadow Object places a body between the moon and sun during an eclipse, not the moon and earth. The shadow object simply casts a shadow upon the moon, it does not obscure it directly.

You're not going to be able to see the shadow of the Shadow Object obscure the stars because the shadow object moves in circular patterns with the swirling multiple system which moves at one rotation per twenty four hours. The Shadow object is sitting somewhere in the whirling stellar system which rotates above our heads, moving as it moves, circling the hub of the earth at one rotation per 24 hours.

The existence of this body is visible only when the Sun, Shadow Object, and Moon perfectly align, manifesting as a shadow upon the moon.

why is it then that we only see lunar eclipses during a full moon?

mere coincidence?

...which of course begs the question of how the moon even appears in the same phase to half the world all at the same time anyway, but that's a whole 'nother "optical illusion" to untangle...

Quote from: Agent Z on October 29, 2007, 08:31:47 AM

applying my proof gives a = 597.45 feet.

So Tom Bishops method was almost correct but the method would get worse and worse the larger the arc.

neither one of you are taking atmospheric refraction into account for apparent curvature. our engineers had to take this into account in afghanistan for backup relays, or guess what: they wouldn't work. bishop's posting of some guy's fe calculation of refraction is laughably simplistic and wrong. for starters, it assumes a flat earth and gets worse from there. first link below has the actual formulas (scroll past astronomical refraction...or read it also to debunk bishops "sun is a spotlight because of snell's law" nonsense). all verifiable with other sources, online and off.

http://www.iol.ie/~geniet/eng/refract.htm
http://mintaka.sdsu.edu/GF/mirages/mirintro.html
http://www.connect802.com/antenna_c_main.php

if our atmosphere were more dense - probably still in the range of long-term survivability - the world would seem concave from the surface...at least until a certain altitude. dog only knows what bishop would believe by looking out his window then.

I didn't need to apply refraction. I was merely providing a more accurate model.

Which is why I said it is accurate as long as a/R<<1.

I missed that quote, chief.