# Is The Earth Round Or Flat?

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#### Secret User

• Flat Earth Believer
• 4233
##### Is The Earth Round Or Flat?
« on: December 29, 2008, 03:15:39 PM »
Is The Earth Round Or Flat?
The Mathematical Gazette Volume 16, No.221
Dec, 1932
By A.W King

Quote
SIR,-Has the above question any meaning ? If it is not possible for human beings to prove that the Earth is either round or flat, surely the question becomes meaningless. I give below reasons for thinking that we cannot answer the question one way or the other. Let us take a system of three unit vectors, e1, e2, e3, at right angles to each other and use spherical polar coordinates, viz. Φfor the co-latitude measured from e3, Θ for the meridian angle measured from e1, r for the radius vector. The differential vector dr of Euclidean 3-space using these coordinates is (1) dr=r(cos Φ cos Θ. e +cos Φ sin . e2 - sin Φ +. e3)dΦ + r ( - sinΦ) sin Θ. el + sin Φ ( cos Θ. e2) dO + (sin Φ cos Φ  + sin Φ) sin Θ. e2 + cos Φ. e3)dr.

Squaring (1) we get for the square of the line element (or ground form) (2) ds2=(dr)2 = r2 dΦ2 + r2 sin^2 Φ dΘ2 + dr^2.

Putting r=a in (1) we get for the differential vector of a sphere of radius a, in 3-space, (3) dr=a(cos Φ cos Θ. e1 +cos Φ ( sin Θ . e2 - sin Φ . e3) d +a (- sin Φ sin Θ. e^1 + sin Φ cos Θ. e2) d Θ,

with ground form (4) ds^2 =a^2 d Φ^2 +a^2 sin^2 Φ d Θ^2.

Next consider the non-Euclidean 3-space whose differential vector is, with Φ, Θ and r as parameters, (5) do- =r . e1 Φ + r sin Φ. e2 dΘ + e3. dr.

Squaring it, we get its ground form: (6) ds^2= r^2 dΦ^2 r^2 sin^2 4 dΘ^2 + dr^2.

Consider the Riemannian 2-pole elliptic plane with constant 1/a, lying in this non-Euclidean 3-space. It is obtained by putting r=a in (5). Its differential vector is (7) do-=a. e1 dΦ +a sin Φ) .e^2d Θ.

Its ground form is (8 ) ds^2=(do-)^2 =a^2 d Φa ^2 + a^2 sin^2 Φ dΘ^2.

By comparing their ground forms (2) and (6), we see that the Riemannian 3-space is " applicable " to Euclidean 3-space.

By comparing (4) and (8 ) we see that the Riemannian plane is " applicable" to the Euclidean sphere. Let us now suppose that two persons E and N move about the Earth in company with each other. Any measurements they may make will be the same, e.g. if they measure the sides and angles of a geodesic triangle, they will get the same relations connecting the sides and angles as given in spherical trigonometry. E chooses to interpret such measurements as proving that the surface is a sphere of radius a, lying in Euclidean 3-space. N chooses to interpret them as proving that the surface is the above-mentioned Riemann plane lying in the non-Euclidean 3-space (5). The geometries of these surfaces and spaces are the same. Therefore no possible experiment can decide between them. The proofs given in books on geography and astronomy beg the question by assuming our 3-space Euclidean. A corresponding argument applies to the case of a spheroid.
I vote for SecretUser as supreme overlord of TFES.

condemnant quod non intellegunt

#### Perfect Circle

• 734
• You are a pirate!
##### Re: Is The Earth Round Or Flat?
« Reply #1 on: December 29, 2008, 03:24:45 PM »
Correct me if I'm wrong, but wouldn't there be conflicting measurements in the southern hemisphere? Going in the direction of the ring of the geographical south pole on a flat earth yields an infinite possibility of angles in a non-Euclidean triangle. In the northern hemisphere, the angles and arc lengths would be the same on both models.
Like the sun, the stars are also expanding and contracting their diameter as they spin around the hub every six months.

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#### Robbyj

• Flat Earth Editor
• 5459
##### Re: Is The Earth Round Or Flat?
« Reply #2 on: January 26, 2009, 03:03:51 PM »
It's round. Next topic

Still contributing to the discussion immensely, I see.

In the northern hemisphere, the angles and arc lengths would be the same on both models.

Not really.  If you assign the radius in fe in accordance to the arc length between latitudes on re even the northern hemisphere is off.
« Last Edit: January 27, 2009, 05:48:02 AM by Robbyj »
Why justify an illegitimate attack with a legitimate response?

• Official Member
• 35374
• Former President of Iraq
##### Re: Is The Earth Round Or Flat?
« Reply #3 on: February 01, 2009, 06:44:12 PM »
FLAT

NOW FUCK OFF

How productive.

Is The Earth Round Or Flat?
The Mathematical Gazette Volume 16, No.221
Dec, 1932
By A.W King

Quote
SIR,-Has the above question any meaning ? If it is not possible for human beings to prove that the Earth is either round or flat, surely the question becomes meaningless. I give below reasons for thinking that we cannot answer the question one way or the other. Let us take a system of three unit vectors, e1, e2, e3, at right angles to each other and use spherical polar coordinates, viz. Φfor the co-latitude measured from e3, Θ for the meridian angle measured from e1, r for the radius vector. The differential vector dr of Euclidean 3-space using these coordinates is (1) dr=r(cos Φ cos Θ. e +cos Φ sin . e2 - sin Φ +. e3)dΦ + r ( - sinΦ) sin Θ. el + sin Φ ( cos Θ. e2) dO + (sin Φ cos Φ  + sin Φ) sin Θ. e2 + cos Φ. e3)dr.

Squaring (1) we get for the square of the line element (or ground form) (2) ds2=(dr)2 = r2 dΦ2 + r2 sin^2 Φ dΘ2 + dr^2.

Putting r=a in (1) we get for the differential vector of a sphere of radius a, in 3-space, (3) dr=a(cos Φ cos Θ. e1 +cos Φ ( sin Θ . e2 - sin Φ . e3) d +a (- sin Φ sin Θ. e^1 + sin Φ cos Θ. e2) d Θ,

with ground form (4) ds^2 =a^2 d Φ^2 +a^2 sin^2 Φ d Θ^2.

Next consider the non-Euclidean 3-space whose differential vector is, with Φ, Θ and r as parameters, (5) do- =r . e1 Φ + r sin Φ. e2 dΘ + e3. dr.

Squaring it, we get its ground form: (6) ds^2= r^2 dΦ^2 r^2 sin^2 4 dΘ^2 + dr^2.

Consider the Riemannian 2-pole elliptic plane with constant 1/a, lying in this non-Euclidean 3-space. It is obtained by putting r=a in (5). Its differential vector is (7) do-=a. e1 dΦ +a sin Φ) .e^2d Θ.

Its ground form is (8 ) ds^2=(do-)^2 =a^2 d Φa ^2 + a^2 sin^2 Φ dΘ^2.

By comparing their ground forms (2) and (6), we see that the Riemannian 3-space is " applicable " to Euclidean 3-space.

By comparing (4) and (8 ) we see that the Riemannian plane is " applicable" to the Euclidean sphere. Let us now suppose that two persons E and N move about the Earth in company with each other. Any measurements they may make will be the same, e.g. if they measure the sides and angles of a geodesic triangle, they will get the same relations connecting the sides and angles as given in spherical trigonometry. E chooses to interpret such measurements as proving that the surface is a sphere of radius a, lying in Euclidean 3-space. N chooses to interpret them as proving that the surface is the above-mentioned Riemann plane lying in the non-Euclidean 3-space (5). The geometries of these surfaces and spaces are the same. Therefore no possible experiment can decide between them. The proofs given in books on geography and astronomy beg the question by assuming our 3-space Euclidean. A corresponding argument applies to the case of a spheroid.

I'm a little confused here.  Does this conclusion support RET or FET?