Earths circumference (at the equator)

well if you can easily measure one point, then easily measure several points, fly a plane over these points and get a rough estimate of the distance. I mean this has been done, but for some reason its not valid, apparently.

Its just stupid. If you have a  shape, you measure it, it is 2d and has 3 sides, each side is the same length, then you can make a pretty confident guess its going to be an triangle, its not going to be a pyrimid. If you measure the world, the measurments match those of a sphere, then its not going to be a circle.

well if you can easily measure one point, then easily measure several points, fly a plane over these points and get a rough estimate of the distance. I mean this has been done, but for some reason its not valid, apparently.

Have you ever been in a plane?  A measurement has to be conducted by some type of standard or reference otherwise it will not be accurate.  A plane does not come even remotely close to a perfectly straight line.

hence i said rough estimate. Which is all you need with this kind of difference in distances.

Anyway whered roboman go?

ok, well due to lack of a valid number from any FE'er for the equatorial circumference, or an explanation as to where any of their measurments came from, or why the RE measurments are invalid. We can conclude that the measurments made of the earth fit with a spherical shape, and not with a flat shape, and until FET has any valid measurments which point to the earth being flat, the earth shall remain a sphere.

Sorry what? I really dont follow. How is the equator an arbitrary circle in FE??? It still sepereates the northern and southern hemispheres, well northern and southern "parts"
in FE i guess.

The Equator in FET is no more significant than any other line of latitude. The hubward circle and rimward annulus do not have equal areas, so the most that can be said for the Equator is that it is the locus of all points which bisect some meridian.

Also your point about their "significance" is completely insignificant (irony yes lol). Because its about which measument was made correctly in FE opinion. For some reason they decided that the RE measurment for the eqautorial circumference should be the measurment for the diameter of their earth. Or they did this by accident and they were meant to decide that the RE measurment for the circumference around the poles was measured correctly, and that our circumference was measured incorrectly. I want to know where the hell this decision came from. Why do they believe any of the measurments?? Why just one?

I don't know where that particular decision came from. You'd have to ask whoever wrote the FAQ; I don't know who did.

Well if you can measure one point between two lines of longitude then you can just extrapolate from that to get your full circumference can you not?

How is it any more challenging to make this measurment, then the one between the poles.

If you can show me a technique for extrapolating from a single data point, I would be delighted to read of it.

So tell me how far is from Oslo to Stockholm? What do you get when you calculate it by RE's Great Circle method? What do you get when you get using FE?

Here are my answers?
Actual: 417 kilometers or 225 nautical miles. Reference: http://www.convertunits.com/distance/from/Oslo,+Norway/to/Stockholm,+Sweden
RET Great Circle:
 Olso: 59  57 N  10  42 E, Reference: http://www.infoplease.com/ipa/A0001769.html
 Stockholm: 59  17 N  18  3 E, Ibid
 419km Reference: Google Earth 5.0, Ruler Tool
FET
 We need to follow RS's (inane) comment that he can calculate the distance around the Equator in FE by simple geometry. Let's apply it to this problem too!

First let's calculate the distance around the FE on the 59th N.
 First the radius. The 59th N is 31 of 180, so the radius of the 59th is (the diameter of the FE)/2 * 31/180 = (24900 miles)/2 * 31/180 = 2,144.16667 miles.
 Second the length around the 59th is that radius * pi =  6,736 miles
 Third the distance from 10⁰ 42' to 18⁰ 3' = (18⁰ 3') - (10⁰ 42'))/360 * that radius = 7.35/360 * 6,736 miles = 137.5 miles = 221 km.
 Fourth we can get a great upper bound by using the taxi cab metric and adding in the FE distance north to south and using this more southern latitude. The north-south distance would be (40'/360⁰)(diameter of the FE)/2 = 23 miles = 37 km. So the upper bound is 258 km.

So we have a conclusive, real-world demonstration that RE's prediction of 419km is better than FE's prediction of less than 258km against the actual of 417km.

Please learn some elementary geometry. You have made two errors which have caused your FE measurement to be exactly half of what it should be.

well no, they wouldnt be the same distance. This is the FUCKING PROBLEM. Wow, maybe this is why you dont get it. The distance around a sphere from the equator to the north pole is not the same as the distance along a straight line, on a flat earth form the equator to the north pole.

or you can think of it as, the distance on RE from the equator to the north pole along the surface, is not the same as half the diameter of the earth (which is where the north pole would lie if the earth was flat, in the centre of the RE).

You are assuming, without reason to do so, that the RE Equator and the FE Equator are the same length. Why?

I predict the RS will be just as silent as he is about the math behind "bendy light" that he promised to deliver by mid-April. I predict TB will claim, if he shows up, that the math is just a ruse.

I have been busy with school recently. I will get to the specifics of the EA theory when I have time; unfortunately, I must place my other commitments first at the present time.

Btw steve, you are also yet to tell me if u actually believe the world is flat. (not if you are a "true believer in FET")

I don't know what the difference is supposed to be, but yes I do believe the Earth to be flat.

none of what you just said changes my previous statement. But ill reply to what you said when i have time, as im going to sleep now.

But ill reply to what you said when i have time, as im going to sleep now.

Yet, nearly fifteen minutes later, you are still online.

The Equator in FET is no more significant than any other line of latitude.

LOL@fail. Try learning.

http://en.wikipedia.org/wiki/Equator

The equator is one of the five main circles of latitude that are based on the relationship between the Earth's axis of rotation and the plane of the Earth's orbit around the sun. It is the only line of latitude which is also a great circle. The imaginary circle obtained when the Earth's equator is projected onto the sky is called the celestial equator.

The Sun, in its seasonal movement through the sky, passes directly over the equator twice each year, on the March and September equinoxes. At the equator, the rays of the sun are perpendicular to the surface of the earth on these dates.

Places on the equator experience the quickest rates of sunrise and sunset in the world. Such places also have a theoretical constant 12 hours of day and night throughout the year (in practice there are variations of a few minutes due to the effects of atmospheric refraction and because sunrise and sunset are measured from the time the edge of the Sun's disc is on the horizon, rather than its centre). North or south of the equator day length increasingly varies with latitude and the seasons.

Quote

The equator is one of the five main circles of latitude that are based on the relationship between the Earth's axis of rotation and the plane of the Earth's orbit around the sun. It is the only line of latitude which is also a great circle. The imaginary circle obtained when the Earth's equator is projected onto the sky is called the celestial equator.

The Sun, in its seasonal movement through the sky, passes directly over the equator twice each year, on the March and September equinoxes. At the equator, the rays of the sun are perpendicular to the surface of the earth on these dates.

Places on the equator experience the quickest rates of sunrise and sunset in the world. Such places also have a theoretical constant 12 hours of day and night throughout the year (in practice there are variations of a few minutes due to the effects of atmospheric refraction and because sunrise and sunset are measured from the time the edge of the Sun's disc is on the horizon, rather than its centre). North or south of the equator day length increasingly varies with latitude and the seasons.

The first sentence in your quote speaks of the Earth's axis of rotation and its orbit around the Sun, neither of which is applicable to FET.

The first sentence in your quote speaks of the Earth's axis of rotation and its orbit around the Sun, neither of which is applicable to FET.

And yet points on the equator do recieve 12 hours (ish) of daylight followed by 12 of night. And yet the sun passes directly over the equator twice a year, and yet the equator experiences the quickest sunrise and sunset. And yet the equator is the equidistant point between the north and south pole.

All of these are simple facts about the equator which apply to both FE and RE.

Seriously try learning a little about the subject before you hit the post button. It really would make your posts more interesting.

...

Quote from: MayTheBetterModelWin on May 12, 2009, 05:17:34 AM

So tell me how far is from Oslo to Stockholm? What do you get when you calculate it by RE's Great Circle method? What do you get when you get using FE?

Here are my answers?
Actual: 417 kilometers or 225 nautical miles. Reference: http://www.convertunits.com/distance/from/Oslo,+Norway/to/Stockholm,+Sweden
RET Great Circle:
 Olso: 59  57 N  10  42 E, Reference: http://www.infoplease.com/ipa/A0001769.html
 Stockholm: 59  17 N  18  3 E, Ibid
 419km Reference: Google Earth 5.0, Ruler Tool
FET
 We need to follow RS's (inane) comment that he can calculate the distance around the Equator in FE by simple geometry. Let's apply it to this problem too!

First let's calculate the distance around the FE on the 59th N.
 First the radius. The 59th N is 31 of 180, so the radius of the 59th is (the diameter of the FE)/2 * 31/180 = (24900 miles)/2 * 31/180 = 2,144.16667 miles.
 Second the length around the 59th is that radius * pi =  6,736 miles
 Third the distance from 10⁰ 42' to 18⁰ 3' = (18⁰ 3') - (10⁰ 42'))/360 * that radius = 7.35/360 * 6,736 miles = 137.5 miles = 221 km.
 Fourth we can get a great upper bound by using the taxi cab metric and adding in the FE distance north to south and using this more southern latitude. The north-south distance would be (40'/360⁰)(diameter of the FE)/2 = 23 miles = 37 km. So the upper bound is 258 km.

So we have a conclusive, real-world demonstration that RE's prediction of 419km is better than FE's prediction of less than 258km against the actual of 417km.

Please learn some elementary geometry. You have made two errors which have caused your FE measurement to be exactly half of what it should be.
...

Do you have any evidence for your outlandish claim?

...

Quote from: MayTheBetterModelWin on May 12, 2009, 06:20:04 AM

I predict the RS will be just as silent as he is about the math behind "bendy light" that he promised to deliver by mid-April. I predict TB will claim, if he shows up, that the math is just a ruse.

I have been busy with school recently. I will get to the specifics of the EA theory when I have time; unfortunately, I must place my other commitments first at the present time.

...

Why is it that you can find the time to post here on other topics while missing your commitment on "bendy light"? You provide us with clear evidence that you have the time. We can only surmise that you're just not capable of providing the specifics.

Quote from: Colossal Death Robot on May 13, 2009, 03:39:09 AM

But ill reply to what you said when i have time, as im going to sleep now.

Yet, nearly fifteen minutes later, you are still online.

well if your really interested, i was on my gf's laptop. I posted this, closed the browser, went and brushed my teeth and checked e-mails etc on my comp for 5mins in my room, then went back 2 my gf's room, said goodbye to the person i was talking to on msn, then went to bed. Suffice as an explanation? aka i am going to sleep was faster to type. So pls, keep ur lame comments 2 urself.

Im only posting now, as i got up for an 8am surgery lab, which i realised was at 8.30 after i got up, so im killing time.

btw gogo point out his "elementary" geometry mistake

And yet the equator is the equidistant point between the north and south pole.

The Equator is not a point. I think you are the one that needs education.

Quote from: Robosteve on May 13, 2009, 03:17:23 AM

...

Quote from: MayTheBetterModelWin on May 12, 2009, 05:17:34 AM

So tell me how far is from Oslo to Stockholm? What do you get when you calculate it by RE's Great Circle method? What do you get when you get using FE?

Here are my answers?
Actual: 417 kilometers or 225 nautical miles. Reference: http://www.convertunits.com/distance/from/Oslo,+Norway/to/Stockholm,+Sweden
RET Great Circle:
 Olso: 59  57 N  10  42 E, Reference: http://www.infoplease.com/ipa/A0001769.html
 Stockholm: 59  17 N  18  3 E, Ibid
 419km Reference: Google Earth 5.0, Ruler Tool
FET
 We need to follow RS's (inane) comment that he can calculate the distance around the Equator in FE by simple geometry. Let's apply it to this problem too!

First let's calculate the distance around the FE on the 59th N.
 First the radius. The 59th N is 31 of 180, so the radius of the 59th is (the diameter of the FE)/2 * 31/180 = (24900 miles)/2 * 31/180 = 2,144.16667 miles.
 Second the length around the 59th is that radius * pi =  6,736 miles
 Third the distance from 10⁰ 42' to 18⁰ 3' = (18⁰ 3') - (10⁰ 42'))/360 * that radius = 7.35/360 * 6,736 miles = 137.5 miles = 221 km.
 Fourth we can get a great upper bound by using the taxi cab metric and adding in the FE distance north to south and using this more southern latitude. The north-south distance would be (40'/360⁰)(diameter of the FE)/2 = 23 miles = 37 km. So the upper bound is 258 km.

So we have a conclusive, real-world demonstration that RE's prediction of 419km is better than FE's prediction of less than 258km against the actual of 417km.

Please learn some elementary geometry. You have made two errors which have caused your FE measurement to be exactly half of what it should be.
...

Do you have any evidence for your outlandish claim?

Erroneous parts bolded.

Why is it that you can find the time to post here on other topics while missing your commitment on "bendy light"? You provide us with clear evidence that you have the time. We can only surmise that you're just not capable of providing the specifics.

In case you haven't noticed, I haven't been posting very much lately. I have a significant quantity of assessable tasks coming up as it is nearing the end of the semester. When I have the time, I will provide you with the specifics regarding the EA theory; until then, use Euclid's empirical formula if you want to perform any calculations regarding bendy light.

Quote from: MayTheBetterModelWin on May 13, 2009, 05:21:34 AM

Second the length around the 59th is that radius * pi =  6,736 miles

Erroneous parts bolded.

Give him a break, he's only off by over 6000 miles.

The equator is a point on any latitude.

Wow.

Hopefully now you can see that the equator is "significant"?

Hopefully now you can see that the equator is "significant"?

All I see is that you are clueless when it comes to geography.

Quote from: dr.spock on May 14, 2009, 04:25:10 AM

Hopefully now you can see that the equator is "significant"?

All I see is that you are clueless when it comes to geography.

Says the boy that posted this:

The Equator in FET is no more significant than any other line of latitude.

Fucking epic.

i think hes pointing out you meant to say longitude. They are easy 2 confuse

i think hes pointing out you meant to say longitude. They are easy 2 confuse

I know. That's about all he's capable of doing. I really hope there's a job out there for him which requires these kinds of troll skills.

Says the boy that posted this:

Quote from: Robosteve on May 13, 2009, 03:17:23 AM

The Equator in FET is no more significant than any other line of latitude.

Fucking epic.

You have yet to show otherwise.

Quote from: dr.spock on May 14, 2009, 05:01:12 AM

Says the boy that posted this:

Quote from: Robosteve on May 13, 2009, 03:17:23 AM

The Equator in FET is no more significant than any other line of latitude.

Fucking epic.

You have yet to show otherwise.

That's what makes you the extra special retard that you are. I already have.

Now read. Lurk moar. Maybe molest a family member and tell them its a secret game.

That's what makes you the extra special retard that you are. I already have.

No you haven't. You've only stated truths regarding the Celestial Equator.

please give us the FE equator distance.

MEASURE THE EARTH. UNTIL YOU DO, THE MEASURMENTS ARE THAT OF A SPHERE.

MEASURE THE EARTH. UNTIL YOU DO, THE MEASURMENTS ARE THAT OF A SPHERE.

Cruise control is only really useful when you're on the highway.

Always best to measure girth at the middle, where it's thickest.

Quote from: Colossal Death Robot on May 15, 2009, 05:28:44 AM

MEASURE THE EARTH. UNTIL YOU DO, THE MEASURMENTS ARE THAT OF A SPHERE.

Cruise control is only really useful when you're on the highway.

Surely these (i think) fictional FE scientists have enough money, resources and determination to make two measurments of the earth and see if they are consistent with a circle.

surely its the most logical way to determine the shape of something too big for us to just observe. Just measure it, and see what shape the measurments are consistent with. The measurments are consistent with a sphere. Unless the FE'ers make some measurments that say otherwise, then the world will remain a sphere until that day.