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181
Flat Earth Theory / Re: Southern Celestial Rotation in the wiki
« on: July 16, 2020, 12:29:46 PM »
I didn’t mention South Africa because at the time I specified it will be daylight, the late morning. Readers might like to check this.

182
Flat Earth Theory / Re: Southern Celestial Rotation in the wiki
« on: July 16, 2020, 08:21:30 AM »
Which map do you mean? I find no agreement on what an FE map looks like, apart from the North Pole being in the centre. However, the logic is that north is pointing towards the north pole, so south points towards the rim of the known FE model of the world. If South America and Australia are in fact in opposite directions from the north pole then looking south from each country towards the Southern Cross is to look in opposite directions on the world. Therefore the same stars should not be visible at the same time from Australia and from South America and the wiki says as much in the question it both poses and answers.

 However, they are, according to reliable star charts. That’s the problem.

183
Flat Earth Theory / Re: Southern Celestial Rotation in the wiki
« on: July 16, 2020, 07:13:45 AM »
...But there is an even simpler way to check if we can see the same celestial body at the same time from Argentina and Australia. Just wait until the December solstice. See the Sun rising in Ushuaia and setting in Sydney, or even as far as Christchurch: https://www.timeanddate.com/worldclock/sunearth.html?iso=20201221T0852 . You don't even need to go there or find someone who lives there, a live webcam will do the trick.

No offence, and it’s a good idea at face value, but to see Crux you need darkness - twilight can swamp the sky enough to obscure the constellations after sunset and before sunrise. The times given take account of this, they should be sufficiently dark to give a good view. Further, although Crux is shown as you say in the sky, from Perth it dips partially below the horizon at times - the declination figure is likely for the centre of the constellation and not all its stars. You also need a very clear horizon to see a body at, say, 2 degrees above it.

184
Flat Earth Theory / Southern Celestial Rotation in the wiki
« on: July 15, 2020, 06:10:24 AM »
The wiki article Southern Celestial Rotation poses the question:–

Quote
Q. How can two people on opposite sides of the earth in Australia and South Ameirca (sic) both see the same South Pole Stars simultaneously?

and answers as follows:–

Quote
A. Since those areas are many hours apart from each other, when it is night or dusk for one area it is likely day or dawn for the other. It is questioned whether it is the case that those observers see the same stars simultaneously. Due to the time difference it may be that they see the stars alternately.

Which South Pole stars are meant is not specified. There is also a link in the wiki article to a video by Kyle Adams making the claim that viewers in Australia and South America don't see the Southern Cross at the same time.

At present it's summer north of the Equator and winter south of it, so the nights are longer in South America and Australia. The stargazing program Stellarium shows on this date, July 15th 2020, at 11:18 (UTC+01) the Southern Cross is visible from Ushaia, Argentina at about 25 degrees above the horizon in the south. At the same time, the Southern Cross is visible from Perth, Australia at about 60 degrees above the horizon in the south. The local times will be 07:18 (UTC-3) in Ushaia and 18:18 (UTC+8) in Perth. Sunset in Perth is 17:30 and sunrise in Ushaia is 9:47.

It looks like we have two possibilities: (a) Stellarium is wrong (b) the wiki is wrong. They can't both be right.

Have we FE members living in or near these two locations who can tell us if the Southern Cross is visible at the time stated?

185
Flat Earth Investigations / Southern Celestial Rotation in the wiki
« on: July 14, 2020, 09:30:20 PM »
The wiki poses the question:–
Quote
Q. How can two people on opposite sides of the earth in Australia and South Ameirca (sic) both see the same South Pole Stars simultaneously?

and answers as follows:–
Quote
A. Since those areas are many hours apart from each other, when it is night or dusk for one area it is likely day or dawn for the other. It is questioned whether it is the case that those observers see the same stars simultaneously. Due to the time difference it may be that they see the stars alternately.

At present it's summer north of the Equator and winter south of it, so the nights are longer in South America and Australia. The stargazing program Stellarium shows on this date, July 14th 2020, at 11:18 (UTC+01) the Southern Cross is visible from Ushaia, Argentina at about 25 degrees above the horizon in the south. At the same time, the Southern Cross is visible from Perth, Australia at about 60 degrees above the horizon in the south. The local times will be 07:18 (UTC-3) in Ushaia and 18:18 (UTC+8) in Perth.

There is also a link in the wiki article to a video by Kyle Adams making the same claim that viewers in Australia and South America don't see the Southern Cross at the same time. It looks like we have two possibilities: (a) Stellarium is wrong (b) the wiki is wrong. They can't both be right.

Have we FE members living in or near these two locations who can tell us if the Southern Cross is visible at the time stated?

186
Flat Earth Community / Re: ... and so easy to prove to yourselves.
« on: July 14, 2020, 05:10:35 PM »
People come to this forum for all sorts of reasons. Some to argue for FE, others to argue for RE. Others still for reasons difficult to fathom. Debates take all sorts of angles on these lines but you might find it instructive to read the 7-page thread another member gave you a link for in your other recent thread.

Also the pinned thread at the top of this forum.

187
Noted. No whine intended, I’ll wait patiently instead.

188
Flat Earth Investigations / Re: shoemaker-levi question.
« on: July 14, 2020, 05:28:14 AM »
That’s not how it works around here. You have to prove these observations were made, in the manner you describe, and that no CGI or skullduggery is involved. Good luck with that, you’ll find the bar fairly high.

189
Flat Earth Community / Re: ... and so easy to prove to yourselves.
« on: July 14, 2020, 05:14:49 AM »
You do know photo/video evidence is not trusted? That distortion is always expected, leading to images of a concave Earth, a convex ocean, a moon which isn’t at the place in the sky shown? And what would a laser bounced off reflectors on the moon do, assuming there are reflectors there, which is not believed?

190
Have you tried the experiment yourself? The purpose of the photo in my post is to illustrate what to do in the interests of finding out ...

No answer yet, so I presume the OP is outside with a golf or ping pong ball trying it for him/herself. It’s a particularly good time to try it, the moon is at last quarter right now so the terminator line is especially easy to see.

Let us know how you get on.

191
Have you tried the experiment yourself? The purpose of the photo in my post is to illustrate what to do in the interests of finding out for myself without using wideangle photographs (with distortion) or panoramic compositions stitched together in Photoshop which show much wider fields of view than the human eye can possibly see (also with distortion). Perspective has nothing to do with it.

If the illusion is caused by EA, the light striking the ping pong ball just 7 feet above Earth's surface would show a much different terminator line angle than that on the moon some thousands of miles above the Earth, but until some more homework is done on that equation in the Wiki, no calculation can be done on the difference to be expected.

192
The explanation for the Moon Tilt Illusion in RE is an effect of perspective.

When viewing the Moon Tilt Illusion, the Moon will often be tilted upwards:





RE Theorists explain this as result of a perspective effect (my emphasis)


I've searched for RE explanations of how extreme perspective changes explain the Moon Tilt Illusion, but found none so far. If the OP can supply some examples (other than copypasta from the wiki) I'd be obliged.

Regarding the apparent illusion, and if I read the FE explanation correctly, this is accounted for by Electromagnetic Acceleration affecting the light from Sun and Moon as we see them from Earth's surface. This implies to me that near dawn with a waning gibbous moon, or near sunset with a waxing gibbous moon, the light from the sun (just above the horizon) is most affected by EA while the light from the moon (fairly high in the sky) is much less affected by EA, thus accounting for the apparent anomaly where the moon's terminator doesn't appear to accurately indicate the sun's position. But is the anomaly real?

Every schoolboy should know the following experiment, and if you haven't tried it I urge you to do it. On such a morning described, go outside and hold a golf ball, ping-pong ball or similar in line with the moon and compare the shadow on the ball with that on the moon's face. The two will match.

Quote
The Ping Pong Perspective

Holding a white ball at arm’s length in the direction of the Moon shows how lunar phases depend on where the Moon is in the sky with respect to the Sun. S&T: J. Kelly Beatty



The Moon's phases are actually related to orbital motion, and there's a simple and fun observation that shows how they're connected. All you'll need is a Ping-Pong ball to simulate the Moon—actually, any small, white sphere would work. Then head outside about an hour before sunset, or around the time of a first-quarter Moon. Find the Moon in the southern part of the sky, then hold the ball up at arm's length right beside it.

You'll see that the ball shows exactly the same phase as the Moon. The Sun illuminates both the ball and the Moon from the same direction, and you see them as partly sunlit and partly in shadow, their bright and dark portions mimicking each other perfectly. If the weather stays clear, you can repeat this observation on the next several afternoons. Each day the Moon's orbital motion has carried it farther east, and the sunlit portion of its disk has grown larger. If you hold your ball up near the Moon, you'll see that its “phase” has thickened too.

To sneak a preview of the Moon's appearance in the days to come, simply move the ball farther east. And if you move it all the way over so your arm points low in the eastern sky, the side of the ball that's facing you will be almost completely illuminated — nearly a “Full Ball,” so to speak. And, sure enough, a day or two before full Moon, the Moon hangs low in the eastern sky just before sunset and is almost completely illuminated.

https://skyandtelescope.org/astronomy-resources/what-are-the-phases-of-the-moon/

193
Flat Earth Media / Re: International Shipping Agent
« on: July 11, 2020, 09:20:00 AM »
It also doesn’t help that a “great circle” route doesn’t go from Oakland to Guam via Hawaii, it passes well to the north of Hawaii.

http://www.gcmap.com/mapui?P=oak-gum

194
Flat Earth Media / Re: International Shipping Agent
« on: July 08, 2020, 08:31:22 PM »
For an overview of shipping routes across the world just 8 years ago, the interactive map at https://www.shipmap.org/ is fascinating. Once the voiceover has finished, you can play with the settings to display routes by container as well as other ships. There's very little of the world's oceans not covered by shipping routes, despite what the shipping agent has to say.

(caution: Mercator-like projection of Round Earth model used)

195
Flat Earth Projects / Re: UA Circular Motion Theory
« on: July 07, 2020, 08:58:48 PM »
That's really weird on the small laboratory scale. On a smallish scale, dropping an object would show oddities too – it wouldn't fall vertically. It might be interesting to find out at what scale that wouldn't be measurable.

196
Flat Earth Projects / Re: UA Circular Motion Theory
« on: July 06, 2020, 09:02:38 PM »
There is a centrifugal force calculator where you can play around with the fields and get some possible values for a centrifugal acceleration of 1g.

https://www.omnicalculator.com/physics/centrifugal-force

(Mass field can be any value and won't affect acceleration or velocity fields)

Centrifugal acceleration: 1g
Tangential velocity: 100,000 mph
Radius: 126,626 miles

Centrifugal acceleration: 1g
Tangential velocity: 250,000 mph
Radius: 791,413 miles ..................

...

...




That calculator is a good find! I ran your figures through it myself to see how long each rotation would take:–

Tangential velocity (Tv) 100,000mph    Revolutions per year (Rpy) 1101  which is about 3 a day!
Tv 250,000mph    Rpy 440.45
Tv 500,000mph    Rpy 220.2
Tv 1,000,000mph      Rpy 110.1
Tv 2,000,000mph      Rpy 55  (6.64 days per revolution)
Tv 20,000,000mph    Rpy 5.5 (66.4 days per revolution)
Tv 200,000,000mph  Rpy 0.55 (1.88 years per revolution)

I think 1.88 years per revolution is still a bit quick to not notice. However, the last calculation is:–

Tv 671,000,000mph which gives 0.164 revolutions per year, or about 6 years per revolution

– but this is an upper limit to the period of revolution, because it's a fraction short of the speed of light in a vacuum.

I hope I haven't made any glaring arithmetical error here...

197
Flat Earth Media / Re: International Shipping Agent
« on: July 06, 2020, 04:16:49 PM »
Bizarre. Makes me wonder why sailing ships took the routes they did from Europe to Australia and back, eg “The Last Grain Race” by Eric Newby. Would have made more sense to head north on the way back, skirt the Philippines and Japan, head along the west coast of N and S America before rounding the Horn and back up the Atlantic. Might have been safer too, the storm descriptions from these voyages were hair-raising.

198
Flat Earth Theory / Re: Electromagnetic Acceleration calculation
« on: July 05, 2020, 08:19:22 PM »

199
Flat Earth Theory / Electromagnetic Acceleration calculation examples
« on: July 05, 2020, 08:17:10 PM »
Here's three graphs of the result for β equal to 1, 20 and 0.2   disclaimer: I don't know the actual value of β






200
Flat Earth Theory / Electromagnetic Acceleration calculation
« on: July 05, 2020, 07:57:39 PM »
Hello, I'm new here so please forgive if I haven't got things straight. Been reading the Wiki as recommended and I'm intrigued by the Electromagnetic Acceleration equation.
Quote
This is a limit of a more complex (and not yet final) expression as x approaches infinity, so this will only work when y is much greater than x - that is to say, when the vertical distance traveled is much greater than the horizontal distance traveled. Put another way, its accuracy will improve the closer the light ray is to vertical. Therefore, while it is not valid for short-range experiments, it can give an idea of how much sunlight would bend on its way to the Earth, for instance.



Where (0,0) is understood to be the point at which the light ray is horizontal (that is, the derivative of this function is zero).

Definition of terms:

x, y - co-ordinates in the plane of the light ray, where y is increasing in the direction of fastest decreasing Dark Energy potential, and x is increasing in the direction of the component of propagation of the ray which is perpendicular to y.

c - the speed of light in a vacuum.

β - the Bishop constant, which defines the magnitude of the acceleration on a horizontal light ray due to Dark Energy. When the theory is complete, attempts will be made to measure this experimentally.

It is believed that the bending of light does not simulate the rate of globe earth curvature. Instead, the bending occurs more gradually over a greater distance.


I hope I'm understanding this correctly and have built an Excel spreadsheet to calculate values of vertical position of light from a starting position on the ground where the light starts horizontal, eg at sunset. The worksheet calculates vertical position up to 20,000 kilometres (horizontally) from the starting point for any value of β you care to input (the default value of β is 1). The worksheet displays results on a graph of y in metres and x in kilometres, although x is converted to metres for the calculation.

Hopefully people will find it useful!

NB I've checked the calculation by hand with repeated examples, but it's always possible I've missed something. The worksheet is locked so only the β value can be changed when first using it. However, should you want to edit anything, the lock is not password protected, so you can unlock it if you want.
The Scratchpad below the graph lets you input any two values of x you like and it will calculate y for the β value specified.

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