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Offline rabinoz

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The dimensions of the Earth will not fit on a Flat Surface
« on: February 05, 2016, 03:04:26 AM »
This attempts to show that the accepted dimensions of the Earth simply will not fit on a flat surface.

It seems fairly well accepted(1) that the
distance from the equator to the north pole is (close to) 10,000 km and that the

equatorial circumference of the earth is (close to) 40,000 km(2).
So that the
(equatorial circumference) = 4 x (distance from the equator to the north pole).

But, on any round flat earth map(3) I have seen the distance from the "equator to the north pole" is the radius of the equatorial circle and we know that the
(circumference of equatorial circle) = 2 x π x (distance from the equator to the north pole) or 62,832 km

There is a big discrepancy here.  Put simply:
The accepted dimensions of the earth do not fit on a flat surface.
From what I can see the only "practical" object that these dimension can fit is a sphere! (4)


(1)One way to derive these dimensions is the original definitions of the Nautical Mile as one minute of either longitude or latitude.
So the equator to pole distance becomes 90° x 60' x 1Nm/min = 5,400 Nm or 10,000.8 km
and the circumference around the equator becomes 360° x 60' x 1Nm/min = 21,600 Nm or 40,003.2 km

(2)I know that the accepted figures are 40,075 km and 10,002 km resp, but I am using rounded numbers for simplicity.

(3)The Bi-Polar model has been suggested elsewhere, but I really do not see how that can help.

(4)Redefining π as 2 might help - bit that is a bit way out for me!.

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Offline Tom Bishop

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Re: The dimensions of the Earth will not fit on a Flat Surface
« Reply #1 on: February 05, 2016, 07:43:32 AM »
Please provide your evidence.

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Offline rabinoz

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Re: The dimensions of the Earth will not fit on a Flat Surface
« Reply #2 on: February 05, 2016, 08:59:56 AM »
Please provide your evidence.
Evidence of what?
No, I didn't pace out each degree from the equator to the north pole!
No, I didn't pace out each degree around the equator!

I do believe that you, yourself, have posted that a degree of latitude is 69.5 miles. (60.4 Nm, or 111.8 km - not different enough from my figures)
And, that the sun (or earth) rotates 15° each hour, making 360° all around the equator.
Now, if you are going to claim that a degree of longitude at the equator is not 60 nautical miles, take that up with the British Admiralty!
That was essentially the definition of the nautical mile.

In other words, I do believe that those figures are quite sound!

It does seem a little odd! I posted this back a little and got not the slightest interest!
I doubt that flat earth believers will rush at this. 
Before I put my foot too close to my mouth I would like to ask what are the accepted distances for:
Measurement
    Distance I would use
Equator to North Pole
    10,000 km
Circumference of Equator
   40,000 km
Rounding the distances to nice simple numbers would be nice, as high accuracy is not needed.
I believe I can justify these figures (or close to them) from previous writings of TFES or widely accepted data.

If you have different figures for the size of the earth I would like to see them.

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Offline Tom Bishop

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Re: The dimensions of the Earth will not fit on a Flat Surface
« Reply #3 on: March 18, 2016, 08:31:38 PM »
Please provide your evidence.
Evidence of what?
No, I didn't pace out each degree from the equator to the north pole!
No, I didn't pace out each degree around the equator!

Please provide your evidence for those distances, whether it was collected by yourself or others.

Re: The dimensions of the Earth will not fit on a Flat Surface
« Reply #4 on: March 18, 2016, 09:40:32 PM »
Who says how big a flat earth could be? Is there a limit?

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Offline nametaken

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Re: The dimensions of the Earth will not fit on a Flat Surface
« Reply #5 on: March 18, 2016, 10:11:57 PM »
Uhhh... Good point, except whenever I search I'm seeing that the best answer for the distance from NP to EQ is ~20k km, and EQ circumference is 40k km. If that's not the definition of 'radius', I don't know what is.

Granted, some quick searches do show WILD variance on the former's distance; I've seen from 7.5k km to 50k km in just the past few minutes. However, the first google result ("equator to north pole distance" I'm feeling lucky) states that the KM was actually designed on - whatever 10k of any measurement equaled said distance. So, meh. Looks like there isn't a consensus, but I just picked this up right as I saw this topic, so I don't really know.
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[H]ominem unius libri timeo ~Truth is stranger.

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Offline rabinoz

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Re: The dimensions of the Earth will not fit on a Flat Surface
« Reply #6 on: March 19, 2016, 06:38:15 AM »
Uhhh... Good point, except whenever I search I'm seeing that the best answer for the distance from NP to EQ is ~20k km, and EQ circumference is 40k km. If that's not the definition of 'radius', I don't know what is.

Granted, some quick searches do show WILD variance on the former's distance; I've seen from 7.5k km to 50k km in just the past few minutes. However, the first google result ("equator to north pole distance" I'm feeling lucky) states that the KM was actually designed on - whatever 10k of any measurement equaled said distance. So, meh. Looks like there isn't a consensus, but I just picked this up right as I saw this topic, so I don't really know.
Notwithstanding that
Quote
Best Answer:  North pole to equator= 12429.91 miles
or
20004 kilometers
Or 12.8 weeks in a kayak ; )
There is not much disagreement about the distance from the equator to the north pole being close to 10,000 km.
  • That is how Napoleon defined the metre.

  • The Wiki says so!
    Quote
    Latitude
    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
    Knowing that as you recede North or South from the equator at equinox, the sun will descend at a pace of one degree per 69.5 miles, we can even derive our distance from the equator based upon the position of the sun in the sky.
    Since there are 90° of latitude from the equator to the north pole, the distance must be 90 x 69.5 = 6,250 miles or 10,058 km, close enough for this application. Mind you Napoleon was closer! 

  • Modern value about 10,002 km from on http://www.transnav.eu/   
    "International Journal on Marine Navigation and Safety of Sea Transportation,  Volume 7, Number 2, June 2013"
The equatorial circumference is where the disagreement will come.
« Last Edit: March 19, 2016, 08:04:24 AM by rabinoz »

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Offline nametaken

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Re: The dimensions of the Earth will not fit on a Flat Surface
« Reply #7 on: March 19, 2016, 07:48:15 AM »
Yeah saw that meager yahoo questions source getting thwarted a mile off.

Not so bad for the first egg I'll be scraping off my face here. I got nothing on this for now. You did inspire my interest however, so I'm not conceding defeat, just reading through these right now. I mean, I'm not one to go against Napoleon, but the wiki here serves mainly as a deterrent to the non-zetetic minded who stumble upon it; I think we all know that. At least, I personally cannot see any true FE'er advocate wholeheartedly believe anything there (I am by no means condemning it; that deterrent has proved a great firewall since decades before I was born).

tl;dr you win for now as far as I'm concerned. But I know how to turn ice into fire.
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Offline rabinoz

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Re: The dimensions of the Earth will not fit on a Flat Surface
« Reply #8 on: March 19, 2016, 08:04:50 AM »
Yeah saw that meager yahoo questions source getting thwarted a mile off.
Not so bad for the first egg I'll be scraping off my face here.
tl;dr you win for now as far as I'm concerned. But I know how to turn ice into fire.
Don't worry about the egg, that was largely that silly Yahoo answer's fault. I should not have put all that in to answer your post. It was really aimed at Tom Bishop, not you. So I am going to delete most of that reply to you and try to put a "more compact" version into an answer to Tom Bishop's "evidence" post.

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Offline rabinoz

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Re: The dimensions of the Earth will not fit on a Flat Surface
« Reply #9 on: March 19, 2016, 12:48:31 PM »
Please provide your evidence.
I apologise for the length, but you wanted evidence!

There is not much disagreement about the distance from the equator to the north pole being close to 10,000 km.
  • That is how Napoleon defined the metre.
  • The Wiki says so!
    Quote
    Latitude
    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
    Knowing that as you recede North or South from the equator at equinox, the sun will descend at a pace of one degree per 69.5 miles, we can even derive our distance from the equator based upon the position of the sun in the sky.
    Since there are 90° of latitude from the equator to the north pole, the distance must be 90 x 69.5 = 6,250 miles or 10,058 km, close enough for this application. Mind you Napoleon was closer! 
The equatorial circumference is where there may be a dispute.
The accepted (globe) distance is 40,075 km.

A value very close to this this has been accepted for over 1,000 years. Of the medieval Persian Abu Rayhan al-Biruni (973–1048) it is said:
Quote
Al-Biruni
Important contributions to geodesy and geography were also made by Biruni. He introduced techniques to measure the earth and distances on it using triangulation. He found the radius of the earth to be 6339.6 km.
This gives an equatorial circumference of 39,833 km - not far from the current 40,075 km. Note that his methods were some of the earliest examples of Geodetic-Surveying. Al-Biruni is regarded as "the father of Geodetic-Surveying".

I doubt that FE supporters will accept even this evidence, so I have calculated my estimate of the circumference from a couple days travel, on mainly west to east journeys on an "almost round the block trip".
Of course the road (even across the Nullabor) is not quite straight. So I actually used the Garmin GPS Map for the distance.
Just in case that naughty NASA has been tricking us with the GPS readings I compared used the car's odometer with the map and GPS (the Landcruiser Prado odometer is almost exact). From the point-to-point distances I worked the west-east component of distance, the longitude difference and the average latitude of each of two journeys.
From these figures I can calculate the km/degree at the latitude of that journey and hence the (circumference at that latitude) = 360 x (km/degree).

If these figures are accepted, we now have to work out what the equatorial circumference.
On the globe that is easy (at least to a good appoximation), where the (equatorial circumference) = (circumference at that latitude)/cos(latitude).

But, for the flat earth we have the problem that no-one seems certain of the accepted map! I will take it as the map on the right on which we should be able to calculate the (equatorial circumference) = (circumference at that latitude)*90/(90-latitude),
since on this map the meridians of longitude are straight lines radiating from the north pole.

The most widely accepted map model of a flat earth.
I was going to put the detail of calculations in, but it got too large! In summary:
Origin and Destination
"Long Diff" 
"at Lat" 
"km/deg" 
"Circ at Lat" 
"Circ at Equ Globe" 
"Circ at Equ Flat" 
Balladonia (Western Australia) to Eucla (Western Australia) 
5.264° 
-32.01° 
94.5   
34,021 km   
40,123 km   
31,302 km   
Eucla(Western Australia) to Penong(South Australia) 
4.125° 
-31.80° 
94.7   
34,087 km   
40,108 km   
32,055 km   
The circumference at the equator for the globe reasonably quite well with the "accepted value" of 40,075 km.
The value, however, for the flat earth does not help the flat earth case at all. It is simply based on measured distances, scaled by the ratio of the radiating meridian spacing.

It should be stressed here that the actual distances used were from a (GPS) map, but they were checked against driving distances between the same locations. The driving distances did come out a little larger (eg for the first case 532 km on road, 503 km direct).
It would be good if others could do similar measurements at other locations. We have quite a few long E-W roads, but I imagine USA has similar of better in Texas, Arizona of Nevada.

In summary I contend that the Equator to North Pole is indeed close to 10,000 km and
the Equatorial Circumference is indeed close to 40,000 km.

Now, if anyone disagrees, please present your evidence!

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Offline rabinoz

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Re: The dimensions of the Earth will not fit on a Flat Surface
« Reply #10 on: March 19, 2016, 12:54:46 PM »
Who says how big a flat earth could be? Is there a limit?
No-one is limiting the size of the "flat earth". I am only asking two things:
The distance from the equator to the north pole and distance around the equator of the real earth, whatever shape that should turn out to be.
They are values well known to any navigator and should be quite measurable. If you postulate an "infinite flat earth" outside the "ice-wall", well that's up to you!

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Offline nametaken

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Re: The dimensions of the Earth will not fit on a Flat Surface
« Reply #11 on: March 19, 2016, 06:29:44 PM »
This is how this topic makes me feel; like I'm starting my Spring Break off with Insanity.

Going [further] off topic now. I have a "collection of Flat Earth Talking Points" index I'm making (already posted an alpha, but it needs a lot of work); would you be on board with me including this 10k vs 40k piece? I can cross reference this post if you want (in my alpha version, I already mentioned Geodetic Surveying once, but not Al-Biruni). Even if the distances are disproved, it still qualifies, as FETP is an index.

Edit: walked away and thought about it a bit. It obviously all comes down to the actual equator circumference, seeing as the Kilometer by definition was BORN by measuring from the North pole to the Equator. Going all the way back to one of the first major Geodetic missions, I haven't found one that actually circumnavigated the equator; most seem to make mathematical inferences from scientific observations instead.
« Last Edit: March 19, 2016, 07:40:20 PM by nametaken »
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Offline rabinoz

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Re: The dimensions of the Earth will not fit on a Flat Surface
« Reply #12 on: March 30, 2016, 11:12:55 PM »
This is how this topic makes me feel; like I'm starting my Spring Break off with Insanity.

Going [further] off topic now. I have a "collection of Flat Earth Talking Points" index I'm making (already posted an alpha, but it needs a lot of work); would you be on board with me including this 10k vs 40k piece? I can cross reference this post if you want (in my alpha version, I already mentioned Geodetic Surveying once, but not Al-Biruni). Even if the distances are disproved, it still qualifies, as FETP is an index.

Edit: walked away and thought about it a bit. It obviously all comes down to the actual equator circumference, seeing as the Kilometer by definition was BORN by measuring from the North pole to the Equator. Going all the way back to one of the first major Geodetic missions, I haven't found one that actually circumnavigated the equator; most seem to make mathematical inferences from scientific observations instead.
Well, how would you measure you distance precisely even if you did exactly "circumnavigate the equator".

All of these measurements are done under the assumption that:
  • On the Flat Earth, the sun rotates (above the equator at the equinoxes) at a precisely uniform rate. I don't think there is too much argument here - it is basically how time has been measured for millenia (much more precise methods now).
  • On the Globe, the earth rotates at a precisely uniform rate, again I don't think there is too much argument.
then the one or more degrees of longitude is measured (Geodetic Surveyors again) and scaled up to 360° for the full circumference.

A relevant Flat Earth reference is
Quote from: the Wiki
See Finding your Latitude and Longitude
Latitude
To locate your latitude on the flat earth, it's important to know the following fact: The degrees of the earth's latitude are based upon the angle of the sun in the sky at noon equinox.
That's why 0° N/S sits on the equator where the sun is directly overhead, and why 90° N/S sits at the poles where the sun is at a right angle to the observer. At 45 North or South from the equator, the sun will sit at an angle 45° in the sky. The angle of the sun past zenith is our latitude.
Knowing that as you recede North or South from the equator at equinox, the sun will descend at a pace of one degree per 69.5 miles, we can even derive our distance from the equator based upon the position of the sun in the sky.
Longitude
To find your longitude you just need to know how many hours apart you are from Greenwich, UK and a vertical stick to know when the sun is at its zenith over your present location.

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Offline Venus

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Re: The dimensions of the Earth will not fit on a Flat Surface
« Reply #13 on: April 06, 2016, 09:18:52 AM »
Uhhh... Good point, except whenever I search I'm seeing that the best answer for the distance from NP to EQ is ~20k km, and EQ circumference is 40k km. If that's not the definition of 'radius', I don't know what is.

Granted, some quick searches do show WILD variance on the former's distance; I've seen from 7.5k km to 50k km in just the past few minutes. However, the first google result ("equator to north pole distance" I'm feeling lucky) states that the KM was actually designed on - whatever 10k of any measurement equaled said distance. So, meh. Looks like there isn't a consensus, but I just picked this up right as I saw this topic, so I don't really know.

If you google "circumference of the earth through the poles" you get this website ... http://www.space.com/17638-how-big-is-earth.html

It says the circumference at the equator is 40,075 km (24,902 miles), whereas the circumference through the poles is 40,008 km (24,860 miles)

So rounding off to 40,000 km for the circumference through the equator, and 10,000km for the distance from the equator to the North Pole is totally acceptable.

From memory the original post contained nothing about the radius of the earth, which would be the distance from either one of the poles or from the equator, directly to the centre of the earth, which is about 6,370 km

Because I live on the 'bottom' of a spinning spherical earth ...
*I cannot see Polaris, but I can see the Southern Cross
*When I look at the stars they appear to rotate clockwise, not anti-clockwise
*I see the moon 'upside down'
I've travelled to the Northern Hemisphere numerous times ... and seen how different the stars and the moon are 'up' there!
Come on down and check it out FE believers... !!

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Offline rabinoz

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Re: The dimensions of the Earth will not fit on a Flat Surface
« Reply #14 on: April 06, 2016, 10:18:34 AM »
Uhhh... Good point, except whenever I search I'm seeing that the best answer for the distance from NP to EQ is ~20k km, and EQ circumference is 40k km. If that's not the definition of 'radius', I don't know what is.

Granted, some quick searches do show WILD variance on the former's distance; I've seen from 7.5k km to 50k km in just the past few minutes. However, the first google result ("equator to north pole distance" I'm feeling lucky) states that the KM was actually designed on - whatever 10k of any measurement equaled said distance. So, meh. Looks like there isn't a consensus, but I just picked this up right as I saw this topic, so I don't really know.

If you google "circumference of the earth through the poles" you get this website ... http://www.space.com/17638-how-big-is-earth.html

It says the circumference at the equator is 40,075 km (24,902 miles), whereas the circumference through the poles is 40,008 km (24,860 miles)

So rounding off to 40,000 km for the circumference through the equator, and 10,000km for the distance from the equator to the North Pole is totally acceptable.

From memory the original post contained nothing about the radius of the earth, which would be the distance from either one of the poles or from the equator, directly to the centre of the earth, which is about 6,370 km
I simply cannot seem to provoke a reasonable debate on what the actual circumference through the equator is.
Of course I agree with your figures which makes the North Polar Equidistant Azimuthal map quite impossible.
Living in Australia is a great incentive to getting rid of that map!