An inconvenient hemiplane.
« on: December 01, 2020, 06:46:35 PM »

Assuming the distance from the North Pole (NP) to the Equator (E) is 10,000 km whether speaking of FE or RE, and that the Equator to the southern edge (or South Pole) is an additional 10,000 km, we can easily calculate the area of each hemiplane using A = pi*r squared.

Where r = 10,000 km from NP to E, the area of the northern hemiplane = 314,159,265 kmsq.
Where r = 20,000 km from NP to SP (or southern edge) the area of the world = 1,256,637,061 kmsq (which, btw, is about 2.46 times the area in RE theory).
To calculate the size of the southern hemiplane, simply subtract the northern hemiplane from the total, which = 942,477,796 kmsq.

Therefore, the southern hemiplane is exactly 3 times the size of the northern hemiplane.

Does that agree with FE theory?

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Offline Pete Svarrior

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Re: An inconvenient hemiplane.
« Reply #1 on: December 01, 2020, 07:07:19 PM »
we can easily calculate the area of each hemiplane using A = pi*r squared
This relies on you substituting many unknowns within FET with your imagination. In conclusion:

Does that agree with FE theory?
Probably not.
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Re: An inconvenient hemiplane.
« Reply #2 on: December 01, 2020, 07:14:21 PM »
we can easily calculate the area of each hemiplane using A = pi*r squared
This relies on you substituting many unknowns within FET with your imagination. In conclusion:

Does that agree with FE theory?
Probably not.

No. Pick any number other than 10,000 km, and the math still holds. Going from the center of a circle half way to the edge will always define an area exactly 3 times less than the area defined by going from said half-way point to the edge. This is not RE vs. FE; this is simple geometry, and I want to know if FE theory recognizes this.
« Last Edit: December 02, 2020, 04:25:35 AM by stevecanuck »

Re: An inconvenient hemiplane.
« Reply #3 on: December 01, 2020, 10:30:40 PM »
But wait, there's more inconvenience to discuss:



The cities of Margaret River, Western Australia (34S, 115E) and Port Elizabeth, South Africa (34S, 25E) are a quarter of the way around the world from each other whether looking at a RE globe or a FE map. Of that there is no argument. What is in contention is the distance and the most direct line between them.

When I measure the RE distance on maps.google.com, it tells me they are 7,980 km apart, and that a straight-line between them runs as far south as 43 2/3 degrees, and lies about 600 km north of Kerguelen Islands.

FE distance and direction differ greatly. Given it is 10,000 km from the North Pole to the Equator, plus 34/90 of 10,000 km from the Equator to the south, the distance from the North Pole to each city is 10,000 + (34/90 x 10,000) = 13,778 km on lines that are 90 degrees apart.

Therefore, on a flat earth, a straight line between the 2 cities would be the hypotenuse of an equilateral right angle triangle with sides of 13,778 km. Therefore the distance between them would be 19,485 km, which is about 2.44 times farther in FET, and the line between them (the hypotenuse) would nearly touch the southern tip of India. This example can be duplicated for any two points on earth, and such a difference will occur every time.

Given that every shipping company in the world for the last 400+ years, and every airline in the world since their inception, have been successfully plotting course and distance based on RET, and that literally zero companies of any kind rely on knowing how far and which way it is from A to B by using FET, the only conclusion any rational person can come to is that the world is round.

To pile on further, the shape of every continent is known, their size is known, and the distance and direction from any point on earth to any other point is known. The only way they can exist as such is on a round earth.
« Last Edit: December 02, 2020, 04:30:29 AM by stevecanuck »

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Re: An inconvenient hemiplane.
« Reply #4 on: December 02, 2020, 06:24:20 AM »
No. Pick any number other than 10,000 km, and the math still holds.
Yes, that's how proportionality works, congratulations. How does that resolve your issue of simply making up the currently-unknown factors?

But wait, there's more inconvenience to discuss
Nah, we'll deal with your failures one at a time. There will be no need to discuss further consequences of your incorrect assumptions once you've corrected those.
« Last Edit: December 02, 2020, 06:28:14 AM by Pete Svarrior »
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Re: An inconvenient hemiplane.
« Reply #5 on: December 02, 2020, 06:57:10 AM »
The Wiki gives the following information under “Eratosthenes on diameter”:-

Quote
A circle with a diameter of 25,000 miles across is simply the area of land which the light of the sun affects, and represents the area of our known world.

25,000 miles is 40233.6km

The kilometre was originally defined as one ten thousandth of the meridional distance (through Paris) from north pole to the equator. Presumably that distance was known when the definition was declared by the French - pretty silly if they didn’t.
« Last Edit: December 02, 2020, 07:30:50 AM by Longtitube »
Once again - you assume that the centre of the video is the centre of the camera's frame. We know that this isn't the case.

Re: An inconvenient hemiplane.
« Reply #6 on: December 02, 2020, 03:10:57 PM »
No. Pick any number other than 10,000 km, and the math still holds.
Yes, that's how proportionality works, congratulations. How does that resolve your issue of simply making up the currently-unknown factors?

But wait, there's more inconvenience to discuss
Nah, we'll deal with your failures one at a time. There will be no need to discuss further consequences of your incorrect assumptions once you've corrected those.

The question still stands. Does FET recognize that the southern hemiplane is 3 times the size of the northern hemiplane. Again, this is simple geometry. We could be taking about a dinner plate or a circular throw rug and the question is the same. The actual measurements need not be known.

Same with the Margaret River to Port Elizabeth question. Actual numbers are not needed, just relative distance. Does FET believe they are 2.44 times farther apart than, and in a different direction from each other than RET? Your method of denial causes me to reiterate that this is purely a geometric relationship that does not require actual measurements.

Also, the mocking tone you take is unbecoming the moderation of a serious forum.

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Re: An inconvenient hemiplane.
« Reply #7 on: December 02, 2020, 04:56:40 PM »
The question still stands.
Well, it doesn't. And there is no "still" about it - you have yet to render your question complete and coherent. In other words, it never stood to begin with.

You're asking a question based on your assumptions. I'm telling you that the factors you substituted with your imagination are unknown within FET.

You would have known this much if you followed the basic rules of engagement and read one of the bajillion threads about distance measurements that already exist on this forum. But you didn't.

Also, the mocking tone you take is unbecoming the moderation of a serious forum.
If you choose to read my statements as mockery, that's on you. You're not being mocked (yet) - I'm merely pointing out your errors to you. By following your logic, one could interpret your repeated use of "simple geometry" as patronising, same goes for demanding that we "recognise" your opinions; and yet I have a feeling that you're simply stating what you consider to be fact.

Similarly, your declaration that your assumptions are "inconvenient" to us carries a certain aura of smugness (doubly so if we assume you're referring to Al Gore's propaganda piece with a similar title). It is most indubitably unbecoming of a prestigious member of this website on the Internet, a position I'm sure you'll agree carries much pizzazz. (In case of any doubt: that was me mocking your idea that a janitor on a relatively small forum should aspire to some unspecified high standards.)

Your unwillingness to resolve the holes in your reasoning, and your attempts at distracting from the conversation by tone-policing are duly noted, though.
« Last Edit: December 02, 2020, 05:04:29 PM by Pete Svarrior »
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Re: An inconvenient hemiplane.
« Reply #8 on: December 05, 2020, 10:15:40 PM »

I'm willing to try this again if you're interested. First, I would like to apologize for anything I posted that sounded smug or condescending, including the title of this thread. I'm also going to stay away from exact measurements because they're a distraction. My question can be expressed purely in terms of relative size.

For a circle, the area (A = pi times radius squared), starting from the middle and going to the halfway point of the radius, describes an 'inner circle' that is one quarter the size of the whole circle. So, subtracting the inner circle from the whole circle shows that the outer circle is three times the size of the inner circle (the area of a doughnut with these dimensions is three times the size of the hole).

Therefore, in FE terms, the southern hemiplane would have three times the surface area of the northern hemiplane, and my question is whether that is claimed to be so in FET.