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Messages - Wuz Ranger

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Flat Earth Debate / Re: What this site is REALLY about
« on: August 25, 2008, 12:32:47 PM »
You have made a very well endowed point.  I've often tried to come up with a conclusion as towards why people think the earth is flat, I've only come up with one solution.  In grade school, the teacher showed them the map of the US......thus thinking the world is flat. 

Great post though!!!!

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This question was asked and never answered.  I do have an additional question to add.  All this FE crap is based on what you see must be real right?  So who can draw a map of FE?  RE maps have been in existence for hundreds, some might speculate thousands, of years.  All the early explorers who mapped the earth were all wrong?

C-Ray you are on to something my man. http://en.wikipedia.org/wiki/Spherical_Earth

 There are several reasonable ways to approximate Earth's shape as a sphere. Each preserves a different feature of the true Earth in order to compute the radius of the spherical model. All examples in this section assume the WGS 84 datum, with an equatorial radius "a" of 6,378.137 km and a polar radius "b" of 6,356.752 km. A sphere being a gross approximation of the spheroid, which itself is an approximation of the geoid, units are given here in kilometers rather than the millimeter resolution appropriate for geodesy.

    * Preserve the equatorial circumference. This is simplest, being a sphere with circumference identical to the equatorial circumference of the real Earth. Since the circumference is the same, so is the radius, at 6,378.137 km.
    * Preserve the lengths of meridians. This requires an elliptic integral to find, given the polar and equatorial radii: \frac{2a}{\pi}\int_{0}^{\frac{\pi}{2}}\sqrt{\cos^2\phi + \frac{b^2}{a^2}\sin^2 \phi}\,d\phi. A sphere preserving the lengths of meridians has a rectifying radius of 6,367.449 km. This can be approximated using the elliptical quadratic mean: \sqrt{\frac{a^2+b^2}{2}}\,\!, about 6,367.454 km.
    * Preserve the average circumference. As there are different ways to define an ellipsoid's average circumference (radius vs. arcradius/radius of curvature; elliptically fixed vs. ellipsoidally "fluid"; different integration intervals for quadrant-based geodetic circumferences), there is no definitive, "absolute average circumference". The ellipsoidal quadratic mean is one simple model: \sqrt{\frac{3a^2+b^2}{4}}\,\!, giving a spherical radius of 6,372.798 km.
    * Preserve the surface area of the real Earth. This gives the authalic radius: \sqrt{\frac{a^2+\frac{ab^2}{\sqrt{a^2-b^2}}\ln{(\frac{a+\sqrt{a^2-b^2}}b)}}{2}}\,\!, or 6,371.007 km.
    * Preserve the volume of the real Earth. This volumetric radius is computed as: \sqrt[3]{a^2b}, or 6,371.001 km.

Note that the authalic and volumetric spheres have radii that differ by less than 7 meters, yet both preserve important properties. Hence both, and occasionally an average of the two, are used.

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I was looking at my LES and i never saw this but there was a wierd $50 tax... I'm FE now.




LOL Nice...you should have made the text the same size and font though....LOL

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Flat Earth Q&A / Re: FAQ Question
« on: August 20, 2008, 08:46:39 AM »
Wow I can see that I started something quiet bad here.  My intentions were to not make war but rather find out real answers to my questions.  I ask these questions because I don't know the answer to them, like one poster mentioned before, you cannot perform a search on this site to get a real answer to save your life. 

Ok here is another question (I hope this one doesn't cause for harsh words)

FE'rs say that NASA is a conspiracy theory.  What about the private investors who are trying to make space travel possible?  Are they part of the conspiracy theory? Why do we have time zones? Why would it be dark in Boston yet be sunny in L.A?  The FAQ site doesn't answer this?


Thanks,

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Flat Earth Q&A / FAQ Question
« on: August 19, 2008, 12:37:27 PM »
Good Afternoon everyone: Below is from the FAQ:

Q: "If the world was really flat, what would happen if you jump off the disc's edge?"

A: You would enter an inertial reference frame, moving at a constant velocity in the direction the Earth was moving before you jumped. The Earth would continue accelerating upwards past you at a rate of 1g, so it would appear to you that you were falling into space.

How would anybody really know this.  I mean if the government is stopping us from reaching the wall, how does anybody really know the answer.  Besides, if you did make it, would you still be alive to tell about it or even communicate your findings?

Just a question thanks

Ranger

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