The all-new conspiracy theory!

No, you have very memory. That is the one that tracks planes, and notes arrival and departure times as indicated by airports. Now, are you telling me that they're saying that the plane has arrived when the plane hasn't? Because if so, that would cause quite a few noticeable headaches, no? You go to pick up your friend from the airport, and he's still 8 hours away, eh?

Schedulers and passengers base their expectations on experience, not a model of the earth.  Or were you thinking the scheduler's and passengers were all doodling idle notes on manifold spaces and grassmannians while the planes fly?

And yes, jet streams dramatically alter real world flight times.

And yes, international flights are routinely delayed.  I've never thought to have a talk with the pilot about geodesics when my flight was delayed hours.

Schedulers and passengers base their expectations on experience, not a model of the earth.  Or were you thinking the scheduler's and passengers were all doodling idle notes on manifold spaces and grassmannians while the planes fly?

And yes, jet streams dramatically alter real world flight times.

And yes, international flights are routinely delayed.  I've never thought to have a talk with the pilot about geodesics when my flight was delayed hours.

On a Flat Earth you would have to have a (the location varies depending on how you want to arrange the continents)  flight path were every flight was delayed for like EIGHT HOURS.  Every single flight.  There's no way that the pilots that flew that route would ever notice, now is there.  By golly, the earth must be flat after all.

Manifold spaces and grassmannians? Uh...you don't need those.  You don't need any non-elementary math.  All you need is rough estimates of distances on the Earth and a rough idea of how fast your plane is going.  Which is readily available.  And division.  Or you could just measure fuel consumption.  Also, they're generally just called "manifolds."

Let me ask you a simple question:

Do you think that if New Zealand was twice as far away from Australia as the RE model says it is, than no one at all would notice?

There's no way that the pilots that flew that route would ever notice

Do pilots in the Southern Hemisphere have you on speed dial?

Quote

There's no way that the pilots that flew that route would ever notice

Do pilots in the Southern Hemisphere have you on speed dial?

Yes.  They do.

Its a wonder how none of the companies never noticed the fuel / distance consumption in southern hemisphere is threefold greater then in northern for no apparent reason at all... Or they did! Thats why there are so little flight paths in there!!

They just never came to research why it is so, not like it would save them $$$$ to know why and how...

In fact, planes from South America don't go to South Africa at all. There are very few international airports in South Africa, you see.

Oh good, a verifiable fact.
SA023 is a non stop flight from Sao Paulo to Jo'burg. It flies every day of the week
It has a flight time of approximately 11 hours.
How fast must it fly to do it in that time on your FE model?

As a starting point would you like to tell me how far it is in the FE model from Jo'burg to Sao Paulo.

I have not seen any detailed map posted here from which I can take an accurate measurement.
I could assume that your model plots the locations of cities on polar coordinates from the longitude and latitude and take your published disk diameter to calculate it.

btw is it just a coincidence that you claim the diameter of the world is the RE figure for equatorial circumference (24900 miles) or is there some empirical basis for this?
Could you also explain why the sun orbits at a height of exactly 3000 miles.
Is this just a 1 in a thousand coincidence, an arbitrary figure, or an approximation.
Again, what is the empirical basis for this figure?

Eratosthenes proved how far away the sun was with his famous experiment

Eratosthenes proved how far away the sun was with his famous experiment

Eratosthenes found the distance to the sun to be "σταδίων μυριάδας τετρακοσίας και οκτωκισμυρίας" (literally "of stadia myriads 400 and 80,000")

(...)

using the Greek, or Olympic, stadium of 185 metres, the figure of 804 million stadia that he quotes for the distance to the Sun comes to 149 million kilometres

So wheres the 3000 miles xO

You can confirm the 3000 miles yourself with some simple experiments.  I encourage you to try these: http://www.sacred-texts.com/earth/za/za23.htm

Actually, the 3000 miles is right there in the article you read:

Eratosthenes knew that on the summer solstice at local noon in the Ancient Egyptian city of Swenet (known in Greek as Syene) on the Tropic of Cancer, the sun would appear at the zenith, directly overhead. He also knew, from measurement, that in his hometown of Alexandria, the angle of elevation of the Sun would be 1/50 of a full circle (7°12') south of the zenith at the same time. Assuming that Alexandria was due north of Syene he concluded that the distance from Alexandria to Syene must be 1/50 of the total circumference of the Earth. His estimated distance between the cities was 5000 stadia (about 500 geographical or nautical miles). He rounded the result to a final value of 700 stadia per degree, which implies a circumference of 252,000 stadia. The exact size of the stadion he used is frequently argued. The common Attic stadion was about 185 m, which would imply a circumference of 46,620 km, i.e. 16.3% too large. However, if we assume that Eratosthenes used the "Egyptian stadion"[citation needed] of about 157.5 m, his measurement turns out to be 39,690 km, an error of less than 1%.[1]

You can confirm the 3000 miles yourself with some simple experiments.  I encourage you to try these: http://www.sacred-texts.com/earth/za/za23.htm

Actually, Rowbotham calculates the distance from the sun to earth to be 700 miles in that link.

Hence it is demonstrable that the distance of the sun over that part of the earth to which it is vertical is only 700 statute miles.

It should be done on the equinox..

Actually, the 3000 miles is right there in the article you read:

Quote from: Wikipedia

Eratosthenes knew that on the summer solstice at local noon in the Ancient Egyptian city of Swenet (known in Greek as Syene) on the Tropic of Cancer, the sun would appear at the zenith, directly overhead. He also knew, from measurement, that in his hometown of Alexandria, the angle of elevation of the Sun would be 1/50 of a full circle (7°12') south of the zenith at the same time. Assuming that Alexandria was due north of Syene he concluded that the distance from Alexandria to Syene must be 1/50 of the total circumference of the Earth. His estimated distance between the cities was 5000 stadia (about 500 geographical or nautical miles). He rounded the result to a final value of 700 stadia per degree, which implies a circumference of 252,000 stadia. The exact size of the stadion he used is frequently argued. The common Attic stadion was about 185 m, which would imply a circumference of 46,620 km, i.e. 16.3% too large. However, if we assume that Eratosthenes used the "Egyptian stadion"[citation needed] of about 157.5 m, his measurement turns out to be 39,690 km, an error of less than 1%.[1]

He's calculating earth's cimcumference there, not the distance to sun... And i still cannot find 3000 miles in this quotation. Be a bit more specific please?

He's calculating earth's cimcumference there, not the distance to sun... And i still cannot find 3000 miles in this quotation. Be a bit more specific please?

That is only because he incorrectly assumed the Earth was round.

You can confirm the 3000 miles yourself with some simple experiments.  I encourage you to try these: http://www.sacred-texts.com/earth/za/za23.htm

OK, so the sun is 3000 miles above the earth on the basis of an experiment that concludes it is 700 miles up.
This works for you does it?

He's calculating earth's cimcumference there, not the distance to sun... And i still cannot find 3000 miles in this quotation. Be a bit more specific please?

Here is Tom Bishop's boilerplate explanation:

On March 21-22 the sun is directly overhead at the equator and appears 45 degrees above the horizon at 45 degrees north and south latitude. As the angle of sun above the earth at the equator is 90 degrees while it is 45 degrees at 45 degrees north or south latitude, it follows that the angle at the sun between the vertical from the horizon and the line from the observers at 45 degrees north and south must also be 45 degrees. The result is two right angled triangles with legs of equal length. The distance between the equator and the points at 45 degrees north or south is approximately 3,000 miles. Ergo, the sun would be an equal distance above the equator.

However, if you choose different latitudes for the measurements, then the distance to the sun changes too.  He never really does explain why that happens.

Quote from: lived_eht_asan on May 15, 2008, 09:37:11 AM

You can confirm the 3000 miles yourself with some simple experiments.  I encourage you to try these: http://www.sacred-texts.com/earth/za/za23.htm

OK, so the sun is 3000 miles above the earth on the basis of an experiment that concludes it is 700 miles up.
This works for you does it?

Sometimes people will cite a reference without bothering to see if it actually supports their claim.

The point is, FET has already demonstrated empircal methods to find the distance to the sun, which you never bothered to research.  Don't worry about the exact figure until you accept that it is nowhere near as far as you think it is.

Eratosthenes figures are likely where the 3000 figure first came from.  Assume the world is flat, and the sun is not an infinite distance away- his numbers produce a triangle with a 7°12' angle and the short end 5000 stadia.  That puts the sun about 3000 miles up.

Here is Tom Bishop's boilerplate explanation:

On March 21-22 the sun is directly overhead at the equator and appears 45 degrees above the horizon at 45 degrees north and south latitude. As the angle of sun above the earth at the equator is 90 degrees while it is 45 degrees at 45 degrees north or south latitude, it follows that the angle at the sun between the vertical from the horizon and the line from the observers at 45 degrees north and south must also be 45 degrees. The result is two right angled triangles with legs of equal length. The distance between the equator and the points at 45 degrees north or south is approximately 3,000 miles. Ergo, the sun would be an equal distance above the equator.

In addition to the sums being wrong at different latitude, the equatorial observer could mark the track of the sun on the ground over the period of two hours before, to two hours after midday.

Ok, FEers; what would the path be like?

The point is, FET has already demonstrated empircal methods to find the distance to the sun, which you never bothered to research.  Don't worry about the exact figure until you accept that it is nowhere near as far as you think it is.

Eratosthenes figures are likely where the 3000 figure first came from.  Assume the world is flat, and the sun is not an infinite distance away- his numbers produce a triangle with a 7°12' angle and the short end 5000 stadia.  That puts the sun about 3000 miles up.

And these empirical methods give contradictory results. Yay!

Do NOT assume the earth is flat, it is something we are trying to PROVE.

Eratosthenes figures are likely where the 3000 figure first came from.  Assume the world is flat, and the sun is not an infinite distance away- his numbers produce a triangle with a 7°12' angle and the short end 5000 stadia.  That puts the sun about 3000 miles up.

Did you do maths at school?

5000 stadia = 785Km. (or possibly 925km)

392.5/tan 3.56 = 6308km.

So even if FE were right this would put the sun 3919 miles up (over 4000 with the longer stadia)

Is this what passes for good science to FEers.

Quote from: lived_eht_asan on May 15, 2008, 11:22:35 AM

Eratosthenes figures are likely where the 3000 figure first came from.  Assume the world is flat, and the sun is not an infinite distance away- his numbers produce a triangle with a 7°12' angle and the short end 5000 stadia.  That puts the sun about 3000 miles up.

Did you do maths at school?

5000 stadia = 785Km. (or possibly 925km)

392.5/tan 3.56 = 6308km.

So even if FE were right this would put the sun 3919 miles up (over 4000 with the longer stadia)

Is this what passes for good science to FEers.

Sadly, yes.

Please don't talk about math when you do this: 7°12' / 2 = 3.56°.  That is just plain retarded.


3000 is not my calculation, its an approximation accepted by FES, and the relative distance is supported by the observations of your own eyes when you look out the window.

It is most certainly not millions of km's, as you claim.

and the 3000 would vary from season to season as it approaches and recedes from the earth. Also, if you are calculating the distance to the observer, you may get a figure higher than this, as the 3000 miles is the height above the earth directly above it (ie. above the equator on equinox).

Oh no? What about the extremely entertaining parallax argument I made before. Go dig it up, its somewhere around, and no-one answered it...

And still, no-one has an argument for my point. Allow me to put a seed of doubt in your heads. If the distance was longer, and airports only load up just enough fuel for the journey(ironically, to save fuel due to weight), and the distance is up to 5 times longer, how on Earth do planes make it? And about your argument about times being based on experience rather than actual distance, how is it that a plane with a cruising speed of less Mach 1 is able to either travel further than the distance time ratio would allow on a FE, or break the speed of sound without anybody noticing?

Please don't talk about math when you do this: 7°12' / 2 = 3.56°.  That is just plain retarded.

Even though, the difference between tan 3.56 and tan 3.60 is around 1%, that doesnt make up for the difference

Please don't talk about math when you do this: 7°12' / 2 = 3.56°.  That is just plain retarded.


3000 is not my calculation, its an approximation accepted by FES, and the relative distance is supported by the observations of your own eyes when you look out the window.

It is most certainly not millions of km's, as you claim.

Ok, in haste I mis-read the angle. It does not significantly change the point though.

By repeating the figure of 3000 miles you have implicitly accepted it, as you have again above.

So just what observation can you make when you look out of a window to support this?

Essentially if you apply a binomial expansion using the congruency shown by the cosine of the lengths, then we can draw the conclusion that the angle of elevation will be in the region of 180 degrees, which is impossible as this means the sun would remain over the same point. However if you look at the reciprical and take into account the nature of Handels left hand rule, then we can deviate through calculus that the sun spins at 66426.228(3.s.f) meters per second, if this was the case, then it would be constantly be producing sonicbooms, which surely we would hear, i reckon your figures are way off!!!

Essentially if you apply a binomial expansion using the congruency shown by the cosine of the lengths, then we can draw the conclusion that the angle of elevation will be in the region of 180 degrees, which is impossible as this means the sun would remain over the same point. However if you look at the reciprical and take into account the nature of Handels left hand rule, then we can deviate through calculus that the sun spins at 66426.228(3.s.f) meters per second, if this was the case, then it would be constantly be producing sonicbooms, which surely we would hear, i reckon your figures are way off!!!

Narcberry, is that you?