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Topics - spherical

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Flat Earth Theory / 100 proofs
« on: June 21, 2019, 06:51:57 PM »
I wish to start a conversation about some of the 100 proofs from the link below, text written year: 1885.
I have a list here I selected as good conversation points, 16, 33, 34, 37, 44, 49, 54, 55, 57 and 71.

But first, I want to ask if FEs agree with the 100 items posted on the file, and if not, which numbers should I eliminate?

Anyhow, I wish to start with numbers 33, 37 and 44, but pay attention to 44 first, and answer my question: If you fire 3 bullets, no, no bullets, just a baseball by hand, inside an airplane flying at 500 miles per hour, front to back then back to front, the last sideways. If you timed each ball to travel 1 meter, they would present different times?, meaning different speed?  Just remember, the airplane is the frame of reference. To confuse you a little bit more, remember that the "sideways" ball travel is in a real diagonal if observed from a stationary observer on the ground, but on board, it moves neatly sideways.  Try to calculate the 1m distance time/speed, from the point of view of a airplane passenger and from the observer on the ground.  Then read again the #44 below.

44) It is in evidence that, if a projectile be fired from a rapidly moving body in an opposite direction to that in which the body is going, it will fall short of the distance at which it would reach the ground if fired in the direction of motion. Now, since the Earth is said to move at the rate of nineteen miles in, a second of time, "from west to east," it would make all the difference imaginable if the gun were fired in an opposite direction. But, as, in practice, there is not the slightest difference, whichever way the thing may be done, we have a forcible overthrow of all fancies relative to the motion of the Earth, and a striking proof that the Earth is not a globe.

And no, the flight attendant does not walks the corridor in different speeds when going back or front in reference to passengers... that would be ridiculous to think.  In 1885 the world was different, ignorant, lacking tools, technology, instrumentation.  In 1880 only 19.5% of the population COULD READ. How this people could understand the world? They could not research, just believing on what the other ignorant say. And even the ones literate, what they read?

So, how many of the 100 can we discard now?

Flat Earth Theory / Investigating FE Jupiter
« on: June 05, 2019, 09:40:56 PM »
I found some old threads about Jupiter, but the discussion just bent to teachers, astronomy, etc, not concluding the discussion in a productive way.
My wish here, investigating FE Jupiter, is related to:

1.) Size
2.) Shape
3.) Altitude from FE
4.) Visible rotation
5.) Visible satellites passing in front and back
6.) How the Sun illuminate it and project its satellites shadow over its body
7.) Movement period over FE
8.) Why it changes visible size along the years
9.) Composition based on spectrometry

I wish the discussion stay focused on the subject.

Flat Earth Investigations / South Pole as the FE center?
« on: June 05, 2019, 08:31:11 PM »
I wish we could investigate why it needs to be the North Pole on the center of the FE ?

What evidences FErs have for such statement? Who defined that and based on what reason or proven evidence?

Why not the Antarctica in the center, and the Arctic as the ice wall all around? 

Flat Earth Theory / FE Wiki - Foucault Pendulum
« on: May 28, 2019, 04:03:29 PM »
On the link posted by the Wiki, about Airy's dismissing the Foucault Pendulum actually working, there is a continuation text below (emphasis mine), I think it must be part of the wiki as well.

Powell and Airy attempted to disabuse the public of doubts concerning the Foucault pendulum. On 9 May, Powell, a leading British popularizer of science, gave an address on the Foucault pendulum at a public demonstration of the experiment at the Royal Institution.
Listing the repetitions conducted by scientists in Britain and Europe, he noted that the "accordance of many of the results at different places within fair limits of error"
confirmed the validity of the experiment. Powell cautioned, nevertheless, that the "sources of error are numerous and not easy to be effectually guarded against." He reasoned that "these causes of error" affected "many of the public repetitions" whose results did not conform to theory.
Powell also noted that Airy had confirmed the experiment.
On 9 May, Airy had presented his results to the Royal Astronomical Society. Two months later, Airy observed in an address before the British Association for the Advancement of Science that the Foucault pendulum had "excited very great attention both in France and England" by "visibly proving, if proof were necessary," the earth's rotation.

Although now "certain" that "Foucault's theory is correct," Airy warned that "careful adjustments" were necessary.
"For want of these the experiment has sometimes failed."

Some free oscillation (non powered) pendulum experiments:

Search Youtube for large pendulums on museums and other places over several cities, Paris, Valencia, Houston, Portland, Norway, Chicago, Fermilab, San Francisco, Austin, St Andrews, Franklin's Institute...   It works so well, everybody wants one in theirs tall hall.

Flat Earth Theory / FE Wiki - Optics
« on: May 24, 2019, 08:28:23 PM »
"Horizon Limits with Refraction and Opacity
Horizon limits are easily explained by the fact that air is not transparent and refraction. As light travels through a denser medium, the object will appear to be smaller because light is refracted towards the normal. Furthermore, air is not transparent so it is not possible to see past a certain distance."

I am particular interested in someone that created such text above, to explain to me the science behind such (underlined) statement.
I can push an image photons through a denser medium, for instance a glass lens, and make the object appear bigger. Your wiki statement is incorrect.
Also, what is the meaning of "because the light is refracted towards the normal"? what is "normal"?
To finish, what is the "certain distance" in kilometers that it is impossible to see through because air is not transparent?
Where all this information came from? it is the wild guessing of somebody or it is multiple times science lab tested, duplicated, recorded and published?
Please provide source evidence.

Flat Earth Theory / FE Wiki - Travelling East
« on: May 24, 2019, 08:10:02 PM »
On FE Wiki:
"As it happens, on a Round Earth you do not travel perfectly straight when traveling East or West either. Consider this thought experiment:
You are on a Round Earth standing 10 feet away from the North Pole. You are then directed to travel East and are instructed to continue to do so. What happens to your path? You end up traveling in a circle, and not in a straight line that you previously thought you would.
The exact same thing happens regardless of where are you on the Flat Earth. Your path will not be straight without you having to constantly change the direction you are traveling in reference to a compass

Except of course, if you are over the equatorial line. 
On RE you don't make any turn, you just walk straight to East. 
On FE you NEED to continuously make a left turn of 0.0057°/km (0°00'20"63/km) walking East.
That is a big difference.  So, the wiki is not correct and it is a misleading.

Also, even if you walk in a very straight line on RE, and even starting form the North Pole to ANY DIRECTION, you will end up in the same place you started after make a round trip around the globe. On FE you WILL end up hitting the ice wall, no matter where you start and the direction you take.

Flat Earth Theory / Wiki - Tom Bishop Experiment
« on: May 22, 2019, 10:02:30 PM »
On Tom Bishop Experiment at Wiki, he reports to have seem people playing at Lighthouse State Beach at Santa Cruz in Monterey Bay CA, from Lovers Point, 23 miles away, using a "good"  telescope, on a cold clear day.   Below a picture of the place he was, pointing directly to Lighthouse State Beach (Santa Cruz).  The picture is from Google Street View, with the maximum magnification it allows, perhaps 3 or 4 times.  Note the map at left and the compass at right, what help me to try to point to the Lighthouse State Beach.  I may be wrong with the exact location, tried my best.  Of course that camera is pretty bad, can't see anything on the other side of the bay, barely the mountains.

Can I ask you what brand and model of the telescope you used?
what aperture? eyepiece?
You said about chest on the ground, your telescope was not on tripod?
Do you have any pictures from the beach through the telescope?
It would be nice to have the pictures at the Wiki, don't you think?

The second picture is the opposite, from the Lighthouse State Beach directly to Lovers Point, at maximum magnification (3 or 4x).

Flat Earth Projects / Solar Eclipse Umbra and Penumbra sizes
« on: May 21, 2019, 05:56:51 PM »
As an interesting topic for improving general knowledge about light and shadow casting, I am starting here a discussion about how to calculate the Total Solar Eclipse Umbra and Penumbra sizes, angles of projection, etc, in order to calculate the proportional Sun and Moon sizes and respective distances.   As the umbra (the small disc of total shadow where you can not see the Sun during the eclipse) is a conical projection, it assumes a round shape if casting close to 90° angle to Earth's surface, or an ellipsoid when this angle decreases. 

There are 8 variable in this project.

1.) The Sun distance
2.) The Moon distance
3.) The Sun diameter
4.) The Moon diameter
5.) The Umbra diameter if a circle
6.) The Umbra dimensions (length and width) if an elipse
7.) The Penumbra dimensions
8.) The Shadow casting Angle on Earth's surface

The Umbra and Penumbra round shape can tell us the conical result of projection, with 5 and 7 we can calculate the proportions between 3 and 4   and between 1 and 2 above.   That is pure optics and shadow casting.

A) If the Sun is bigger than the Moon (what it is), it will projects a Moon shadow in a conical form, smaller the diameter as far it gets from the Moon, until the distance is such that the cast shadow reach the apex, a very narrow point.  If nothing intercepts this cast conical shadow, it just disappears in the space, nobody sees the cone.  If anything comes inside such dark cone, it will intercepts and shows the shadow.   The apex distance from the Moon is directly proportional to the sizes Sun & Moon and the distance between them.

B) If the Sun and the Moon has the same size, it will projects a Moon shadow in a conical form with infinite apex, known as "cylinder".  The shadow diameter would be the same as the Moon and will be projected at far distances without changing size.

C) If the Sun is smaller than the Moon, it will projects a Moon shadow with an inverted cone, apex on the Sun.  The cone would increase in diameter as it distances from the Moon, blocking solar light in vast area, in such way that the entire Andromeda galaxy could be in this inverted cone shadow.

Considering history and records, close to 90° of Sun's position where Umbra happens on total eclipse, the Umbra diameter was never smaller than 120km in diameter.   

Considering (A) above, the Moon must be way larger than 120km in diameter, the Sun must be way larger than the Moon and they are not close at all.  With the same measurement date for the Penumbra, one could calculate the size proportions between Sun & Moon and the proportional distances between them and Earth.

Considering (B), both Moon and Sun must have 120km in diameter in order to cast a same size shadow.  The problem here is that the Umbra diameter changes even at 90° (round Umbra), in different eclipses according to factors, one is the actual distance Moon-Earth at the moment of eclipse, another is the angle of shadow casting over Earth surface.  Considering only a closer to 90° Sun & Moon position, then both must have 120km in diameter.  The killing factor here is that on this situation the only way to have Penumbra is when the same size Sun is far away from the Moon, it can be calculated, measured, simulated on any optical lab or kitchen table.   Grab your flashlight, put it face down over a paper, using the flashlight face draw a circle with a pencil, cut this disc of paper (lets call it "blocker"), now projects the flashlight light beam 90° against the wall and use the blocker to block the light, change distances Flashlight-Wall, Flashlight-Blocker and see what happens.  No matter the distance from the wall, the blocker shadow diameter will be almost constant.   Now cut another blocker half the diameter of the first one and repeat the experience, you will notice the conical shadow projected on the wall, if the flashlight and the smaller blocker is away from the wall, there will be no casting shadow visible defined, as (A).   With the blocker same size of flashlight, the Umbra never changes size, and it changes in real life, this option (B) is eliminated.

Considering (C), Cut a blocker twice the diameter of the flashlight and play with it against the wall... it will be impossible, the cast shadow will always be larger than the blocker or the flashlight.  If flashlight and blocker are very close, there will be no light projected to the wall at all, no Penumbra. So, this option (C) is eliminated.  The only way to have a 120km round shadow Umbra (90°) and larger Penumbra for a 48km diameter Sun is for the Moon to be way from the Sun and larger.  It is easy to calculate.  With the FE Sun at altitude of 4800km and 48km in diameter, to project a Umbra shadow of 120km in diameter, the blocker (Moon) can be calculated using this formula: 

SD = Sun Diameter
MD = Moon Diameter
SPD = Shadow Projected (on Earth) Diameter
H1 = Distance Sun to Cone Apex (behind/above the Sun)
H2 = Distance Sun to Moon
H3 = Distance Earth to Moon
H = H1+H2+H3

First you need to find H1

H2+H3 = 10
SD = 2
SPD = 8

2 = 8 * H1 / (10+H1)
2 * (10+H1) = 8 * H1
8 * H1 / 2 = 10+H1
4 * H1 - H1 = 10
H1 = 10 / 3
H1 = 3.333

Then we have the total cone H as 13.333.

Now you can calculate Moon Diameter (MD) for a certain distance (H3)

MD = SPD * (H2+H1) / H

Suppose all the numbers above the same, and H2 (Sun to Moon Distance) is 3.
MD = 8 * (3+3.333) / 13.333
MD =3.8
So, if the Sun (diameter 2) is 10 from Earth, Moon is 7 from Earth, Shadow is 8, Moon diameter must be 3.8, almost double the Sun.

If H2 = 1, MD = 8 * (1+3.333) / 13.333
MD = 2.6, still bigger than the Sun even being very much closer to the Sun.

So, for a Sun = 48km (SD=48), H2+H3 = 4800km, SPD = 120km
48 = 120 * H1 / (4800+H1)
48 * (4800+H1) = 120 * H1
120 * H1 / 48 = 4800+H1
2.5 * H1 - H1 = 4800
H1 = 4800 / 1.5 = 3200km

Now, calculate diameter of the Moon, in kilometers, by distance from Sun:

MD = 120 * (H2+3200) / 8000
MD = 0.015 * (H2 + 3200)
MD = H2*0.015 + 48

See, no matter the distance of the FE Moon from the Sun, it starts with at least 48km in diameter if very near the Sun, or 120km if very near to Earth. The Moon should increase 15 meters in diameter for every kilometer from the Sun.  The FE wiki says Sun and Moon are at the same altitude and size, in that case, a total solar eclipse would obscure the whole Earth. On that Nodal calculation of the Moon performing an angled path under and over the Sun, it becomes clear that the FE Moon would be not at the same altitude as the Sun during a total eclipse, it will be lower, so, to projects a 120km Umbra, the Moon needs to be bigger than the Sun.  If FErs publish the numbers, angle of inclination, altitude, etc, one could even calculate the Moon diameter based on the 120km Umbra and Penumbra.   It will be interesting to see those FE numbers, mostly to produce the very large penumbra that can cover diferentes continents at once.

Then you need to use the Penumbra numbers to calculate how far is the Sun from the Moon.  For example, at the maximum point of the Great American Eclipse in August 2017, the penumbral shadow spanned from Panama all the way to Greenland, covering North America, Central America, and large parts of the northern polar regions.  For such gigantic Penumbra, only option (A) is possible.  Moon is way bigger than 120km in diameter, very far away from Earth and the Sun is way bigger than the Moon, 400 times farther than the Moon.

Interesting is that if you use the RE numbers, Umbra and Penumbra fits perfectly on the solar eclipses numbers on record, calculating with Sun and Moon distances, angles of projections and ellipsoid sizes if lower than 90° projection, and lines of nodes.  The average Umbra conical shadow distance is no more than 380,000 km.

I wonder if FEs could provide some simple calculations to justify their own numbers for Sun/Moon altitude and lines of nodes, to create the same Umbra/Penumbra results.

Suppose FE Moon is farther from the North Pole than the Sun, shadow lower than 90°, and by some way it projects a shadow further yet, the ellipsoid cast shadow would be wider in the farther side from the North Pole, and narrow on the closer side.  Also, the farther side on FE will be fully projected on the ground, with good contrast, while on the RE the further side disappears due planet curvature.  The same would happen if Sun/Moon aren't well aligned to 90° and over a little different longitude.

Records show (and it is possible to calculate) the Umbra speed is around 1700km/h Eastward, the total eclipse umbra lasts no more than 7 1/2 minutes at a specific point.  A Concorde flight was able to fly under Umbra at Mach 2.5 on June 1973 for more than 74 minutes.

MIR image shadow from the Solar Eclipse from Aug/11/1999:

From ISS on March 29, 2006:

I invite everyone to participate in this exercise of numbers.

Flat Earth Theory / Total Eclipse July 02 2019
« on: May 14, 2019, 05:48:11 PM »
Can any FEr demonstrate how the Total Solar Eclipse of July 2nd 2019 is predicted under FE map and conditions?
How the FE Moon comes under the FE Sun on that particular path?

Flat Earth Theory / Angle and Length of a pole's shadow
« on: May 09, 2019, 05:54:23 PM »
I am trying to crunch numbers for two shadows, but for some reason it became complex:

What: A vertical pole 10 meters high with an arrow on top
Location: Close to Quito Equador (Lat: 0°, Long: 78°W), very good flat ground.
Date: March 20 2019 or 2020 (Equinox)
Shadow Time (#1): 09:00h (9am) local time, no Daylight Savings Time in Quito after 1993.
Shadow Time (#2): 16:00h (4pm) local time.

Based on FE map, I am interested to calculate the pole's shadows (#1 and #2) length on
the ground and the angle of arrow projection from the North Pole in degrees.

I'm having some difficulties with this math.

Some volunteers please? One decimal digit will be enough.
Tom Bishop's numbers would be nicely welcome.

Answer #1:  Length _____m,  Angle _____°
Answer #2:  Length _____m,  Angle _____°

Flat Earth Theory / Red Moon
« on: May 08, 2019, 06:02:52 PM »
First let me apologize, I read the FEwiki about Red Moon, but my limited capacity of understanding did not grasp exactly what was said there.
Based on that, I humble request further explanations.

The FE wiki:

"The Lunar Eclipse is red because the light of the sun is shining through the edges of the Shadow Object which passes between the sun and moon during a Lunar Eclipse. The red tint occurs because the outer layers of the Shadow Object are not sufficiently dense. The Sun's light is powerful enough to shine through the outer layers of the Shadow Object, just as a flashlight is powerful enough to shine through your hand when you put it right up against your palm."

My doubts is in reference of physical positioning. 
See, I understand the explanation about a shadow object being positioned between the Sun and the Moon during a Lunar Eclipse.   

Also, the red tint is caused by the outer layer of the shadow object not being sufficient dense. 
So, it is like a semi-transparent shadow being projected?  But in what color? Red?  it means the shadow object is red in essence? like a red acrylic under the sun, projects a redish shadow on floor? Is that it? 

But what really made me confused is, if the physical positioning; first the Sun high above all, then the Shadow Object in the middle, then the Moon lower of the three, then we down here, everything aligned in a straight line.  How can we see the Moon illuminated by the Sun, if the Sun is above the Moon?   Okay, I guess the Moon is transparent. 

But still, why on all Moon Eclipses, caused by this shadow object, we don't see its redish outer layer right on the edge between light and dark on the eclipsed Moon everytime?  and worse, why when this shadow object just crosses in front of the Sun, without Moon involved, we don't see its dark projection on Earth's anywhere, with the redish ring around it? 

My big lack of understanding is related to the real time positioning.
See, all Moon Eclipses happen when the Sun is not on the sky.  All of them happen after sunset, during the night or before sunrise.  So, how the Sun higher than everything else, over the other side of the flat earth disc (it is night here), can project its light all over the north pole (I thought the sun projects a limited size downwards spot), the solar light then reach and illuminate the Moon (over us in this side of the disc), but now the Shadow Object that creates the eclipse is where? above the Moon? below the Moon?, between the Sun and Moon or between Moon and us?  Are the Moon and the Shadow Object (with a semi-transparent redish outer layer) thin discs or spherical object?  If the Moon is a flat disc as the earth, how can I see it round in the sky in the East or West and not an elipse due the angle of view?  If it is spherical, how it is fully transparent? 

What could really help me to calculate and reach conclusions is to know finally what are the altitudes and sizes of the Moon, Sun and the shadow object.

I am really confused.

Flat Earth Theory / Polaris & Alpha Crucis visibility
« on: May 04, 2019, 08:31:49 PM »
I search on wiki and FAQs and found none about this.

First the numbers, according to FE available maps:
Distance from North Pole to Equator: 10,000 km
Distance from North Pole to Australia or South of Argentina: 15,000 km
Distance from North Pole to South Pole Ice Wall:  20,000 km.

Then, the Control experiment:
Any northern country, Canada, USA, Iceland, Russia, Norway, can easily see Polaris few degrees out of vertical on North Pole, almost simultaneously during winter time, January, since the darkness of the sky allows it.  That is an unquestionable fact, accepted by FE and RE.

Then, the Text experiment:
Any southern country, south of Argentina, South Africa, Australia, New Zealand, can easily see Alpha Crucis (the bright absolute -4.14 magnitude star on the Crux Constellation), almost simultaneously during winter time, July, since the darkness of the sky allows it.  That is a fact for millions that live on the very southern hemisphere, they can even take pictures with date/time to show it.

The test conclusion questions:
1. How come (see the picture below for reference illustration) when Alpha Crucis is visible for the Argentina because it will very high in the sky over the South Pole for them, and obviously on top of them on FE map, the people from Australia also can see it, even when the star is over the ice wall in the opposite side of the FE disc, 35000 km (land distance) away, and they can not see Polaris that is less than half way the distance, 15000km (land distance)?

2. How come in RE the Australia and Argentina see the Crux Constellation rotated in the sky, the same way Canada and China see Ursa Major also rotated?  It is understandable that the Canadians and Chinese are facing each other when looking to Ursa Major, but based on FE map, Argentinians will be facing the ice wall to see Alpha Crucis, but the Australians will be facing all over the FE to see the same star, and would see in the constellation in the same angle as the Argentinians.

3. On the FE map, the visible measurable distance between the stars (Alpha, Beta, Gama and Epsilon Crucis) in the Crux Constellation would be different when measured form Argentina, Brazil, Uruguay, Colombia or Australia, due different distances from them to the stars that are just few thousands kilometers up. In RE they are exactly the same with great precision.

4. In the FE, stars and the whole sky make a full rotation over FE once a day.  Then why the Crux Constellation appears almost at the same position in the South, rotating pivoted to a common point close to it, during the period of 24 hours?  As a fast rotating carousel, we should have several different stars occupying the same place of the Crux, but it is not what is being observed, that is a similar rotation as the North Pole sky, in opposite direction.


Additional data:
According to FE maximum visibility by perspective, you can not see further than North Pole - Equator distance, 10,000 km.  Even so, a person in Equator can see the Sun for 12 consecutive hours, what makes 90+90° on FE map's equator line, such line of view is a land distance of 14100 km before sunset (hypotenuse of an isosceles triangle), not even considering the Sun should be 41% smaller at sunset than noon time.  Based on that, people from Rio de Janeiro (latitude 22.9°S), 12971 km from the North Pole, should be able to see Polaris easily, but they are unable to do so.  Also, FE perspective doesn't change angles by object brightness.  Feel free to replace the Sun with any other star on the sky's Equator, visible for 12 hours, same brightness as Polaris.

If this subject was extensively debated previously in the forum with non productive results, please inform and point, so I can remove this post.


Flat Earth Theory / Timezone Northern and Southern Hemisphere
« on: April 26, 2019, 09:55:59 PM »
Can you please, inform what is the distance in kilometers or miles between time zones on the latitude 28.5°N (Orlando FL) and 28.5°S (Rio Grande do Sul Brazil) ?   Considering that all 24 time zones lines concentrate on the North Pole, on the flat earth design they open up like an umbrela or slices of a pizza all over the flat map,  right?  So, as your South pole is the border ice wall, it is huge, it will be 80 thousand kilometers circumference of your ice wall, sliced in 24 lines, equal to 3333 km between each time zone.   Is that correct?   

Well, lets calculate.  On the oblate spheroid planet, the equator line is at latitude zero, North pole is latitude 90°, so Orlando at 28.5°N will have a circumference of 40000 * sin(90-28.5), or 40000 * sin(61.5) = 35152km, divided by 24 = 1464km between each time zone.  Now, we know it is not exactly like that because it is oblate, but lets consider spherical.   Now, The 28.5°S would be the the same, 1464km between each time zone.

Now, considering your flat earth map, for the 28.5°S, it will not use sine, it will be a flat proportion, just considering radius and PI.   So, again, equator has 40000 km in the equator circumference, 28.5°N, Orlando FL, will be  40000 * (90-28.5) / 90 = 22888km / 24 = 953 km between each time zone.   Now, for the 28.5°S on your flat earth map, it will be 40000 * (1 + ((90-28.5) / 90)) = 67333/24 = 2805 km between each time zone, that is preposterous ridiculous.   If you once traveled to the Southern Hemisphere, you know that is totally not true.   

Look below the world time-zones, there are 5 hours zones between Africa and Brazil, and only 7700 km between South Africa and Brasilia. This information (distance) is not a straight parallel line, this is why it end up using 1540km per time zone.    The point is that according to your flat earth map, it would take 2805 km for only ONE timezone... those 5 time-zones would take double distance (14027km) than what is measured, that is horrendous.

Now lets compare Australia to USA.
They have almost the same horizontal (longitude) size (New York to SFrancisco almost the same as Pert to Sidney), see the picture below.  New York is at latitude 40°N.   As USA, Australia also has almost 3 hours of timezone differences, see the first picture and uses 40° of longitude, or 40/360 of the world circumference, according to last picture.

Australia is at 35°S Latitude, and according to the flat earth map, the circumference on the parallel of Sidney is [40000 * (1+ ((90-35) / 90))] = 64444km / 24h = 2685 km per time zone, multiplied by lets say 2.7 = 7250km.  The measured distance between Pert to Sidney is 3291km by air, or even zigzagging by land it is no more than 3930km, way below the calculated on flat earth map (7250km).

Last proven number:  40° of longitude (Australia) correspond to 40/360=0.11111 of the world's circumference, what is 24 * 0.1111 = 2.6666 hours, not bad for my guess of 2.7 time-zones above.

I think flat earth believers are aching hard to answer and contest those simple proved numbers.  Grab your calculator and lets have a tea.

Flat Earth Theory / GPS constellation of satellites
« on: April 26, 2019, 08:46:37 PM »
Would you care to explain please, how GPS, satellite phones and communication in general work in high-seas or in the deep deserts, in a flat earth model?  and please, don't attack my intelligence writing "balloons" or something like that.   If your high altitude "dome" could reflect radio waves, why can't we have waves reflected at any transmitted angle, and only works when transmitter and receiver points exactly where geo satellites are, within narrow angle?

Flat Earth Theory / Clock the Sky
« on: April 26, 2019, 08:17:19 PM »
You can adjust your clock by the movement of things on the sky.
The time a star or the Sun takes to move 15 degrees is exactly 60 minutes, or, 4 minutes per degree, EXACTLY.
It doesn't matter if right after rising in the East, at the top of the sky or right before setting in the west, the time and angle distribution is the same.

This means, the apparent movement of things in the sky is perfectly linear, and that have a simple explanation, the sky is still, our planet is what rotates in a very steady and constant speed exposing us to the universe.

You can NOT have this linearity if dealing with two parallel planes moving one against another, this is exactly what the flat earth model shows, a flat earth disc and a flat sky disc rotating overhead.  When parallel, your angle of view changes constantly, along with the apparent size and time per distance. 

The two images below, the left from the North pole and all the stars rotating around it CCW, the other from the South pole, all stars rotating around it in CW... that for itself kills the flat earth theory.

I wonder when some flat earth believer would contest and explain how this work in their flat world.

Flat Earth Theory / Equator line walk
« on: April 26, 2019, 05:41:52 PM »
On the oblate spheroid planet, if you walks over the equator you do it in a perfect straight line, zero degrees for any side from that line, never deviate even a tenth of degree, and after 40 thousand km you will find yourself back to your starting point. 

On the flat earth model, your walk over the equator is not straight, during the 40 thousand km walk you make a whole 360° turn.  If you are walking west, you will make a long right turn to northwest all the time, it will be a 40 thousand km walk, 360°, 111 km per degree, that would be pretty much noticeable and measurable.   


On the oblate spheroid planet, you can sit down and watch the movement of the stars, anything that raise at the perfect East will set perfectly at West. Anything that raises from Northeast will cross the northern sky, never coming over you, and sets on Northwest.  Anything that raises from Southeast, will cross the southern sky, never coming over you, and setting on Southwest.

On the flat earth model, everything on the sky makes a complete 180° turn over you during a 12 hous period of time, nothing will raise or set, things would appear (?) from the Northeast view, make a clockwise 90° turn until it gets right over you, then makes another clockwise 90° turn and disappear (?) over the Northwest, the whole sky is rotating in a circle over you, right?


Go outside right after sunset, with a 12" or 30cm ruler, fully extend your arm with the ruler in hand, measure the distance between any two bright stars close to the eastern horizon in the sky that would be around 2 to 5 cm apart.  Make note of such measurement.  Come back 5 hours later, the same stars will be closer to the top of the sky, make another measurement in the same way, return 5 hours later and measure them almost setting in the western sky horizon.   You will have a very important and nice surprise.  All the measurements will be exactly the same, to the microns of measurement.   I know this experience could not make any sense to most flat earth believers, but it should do.    I will explain why.   In your flat earth model, the sky is only 3000 miles high and flat, rotating over us.  Go to a straight street, put two soda cans 1 meter/yard apart on the walkway, go to the walkway across the street, walks away 50 steps, turn back and walks 100 steps, even being on the other side of the street you will be approaching the soda cans in the middle of the walk.  Repeat the ruler measurement with fully extended arm, while you walks, you will notice that the measurement would show a larger number when you will be closer to the cans, right across the street, 90° angle, and smaller when you are at smaller angles and far from the cans.

What conclusion you have with this cans experience?  That while you moves on the street, or the street moves on you, objects distances will change while in a lower angle from you, and that distance will be largest while it is right on top of you.

From the 50 steps to one direction and other on the other side of the street and the soda cans, probably you made a 176° of a sweep, triangle of 20 m of height by 100 m of base, your observation distance changed from 20m (across the street) to 54m at 50m away.  Your observation of the flat earth sky during horizon to horizon would be 20 thousand km walk (triangle base) to 4.8km high,  it will form a triangle with a sweep of  128°, your distance from your sky will be from 4.8km (right on top, 12pm) to 10.1km (at 6am or 6pm), that is more than double the distance, the object size in the sky will be completely different in size.  The size formula for different distances is pretty simple, just a ratio.  If you see the sun at 12pm as (lets say) 20mm on the sky, at 6pm it will appear as 20 * 4.8 / 11.1 = 8.6mm.  But it does not, right?  Measure your flat earth sun at noon and at 6pm right before sunset... will see it has exactly the same size.   The only explanation is that it is tremendously far away, millions of times further than just 4.8km.

So, why the stars distances are all the same while they cross all your sky?  Why they don't change distances from horizon to top of the sky?   If you learn any geometry at middle school, you did learn that a 1 meter diameter rotating drum or circle over the ground, the distance from any point on the surface to ground will be a variance from zero to 1 meter, easily measured.  If this same drum is lift 10 km up in the sky, now the variation of distance to the ground from any point on the surface of the drum will be a variance from 10000m to 10001m, almost impossible to measure or to notice.

That is the same with stars, they are so far away, several trillions of miles away or much much further, the 6300km difference in distance to the stars of 6 hours Earth rotating 90° will add so little to the total distance, that makes the stars to be at the exact same distance among them.  It is the same as in the soda can experience, if you move yourself not just to the other side of the street, but 10 blocks away, and look and measure the soda cans distance through a binoculars, they will be measured the same distance, no matter if you walk 200 yards parallel to them.  This is the same effect when traveling by car in the highway, things closer to the road start appearing small at distance, become huge when closer to you, pass fast, reduce size as it distances away.  But far away trees and mountains appear to be moving with you at the same size.   

When you say the flat earth sky, sun, moon, planets and stars are only 3000 miles (4800 km) high, that is too low for them to keep the same distances when moving all over you through the 20 thousand km half circle of the flat earth equator sky, that is impossible. It is the same as the soda cans across the street, you WILL measure a large difference in distances.   Just go outside and measure, you will find none, what proves the spheroid planet.

Oh, too much calculations, numbers and angles for you?  Then, sorry, you should not be involved in trying to understand the ins and outs of the scientific world, we call it geometry and astrophysics.  I would recommend other areas of interest for you.

Flat Earth Theory / Solar Light Composition
« on: April 25, 2019, 10:29:10 PM »
The flat earth theory states the Earth's equator line is 40000 km in circumference, the diameter of the equator is then 12738 km, and as the equator is a center line between the North pole and the "ice wall" South pole, then the total diameter of the flat earth is 25477 km, making the ice wall of a circumference of 80038 km.  Clear?  Okay, the theory also states the Sun has the same diameter of the Moon, around 48 km (30 miles), and both are at the altitude of 4800 km (3000 miles).

Okay, based on what we can analyse and decompose the light received from the sun, we can even detect what temperature and what gases are being converted there, Hydrogen to Helium, we can simulate the fusion in computers, even calculate the amount of energy it needs to produce and based on its size, we can calculate how long it is doing that, and how long it probably will continue doing that.   But this is based on our normal universe we know, that follows the same laws of physics.

Now, based on flat earth theory, how the 30 miles across disc sun generates radiation? how long it is being doing it?  We can measure the sun light power that hits earth surface and measured around 1kW/m², that is a fact.  We can add all the energy from the sun that hits planet earth at a single time or second, that is huge and most of it is invisible infrared.  Even so, as the Sun radiates all around it, and Earth is very far away from it, we receive just a tinny bitty small part of the total radiation emitted by the Sun.

Lets calculate it using the flat earth theory numbers, and think that all the Sun's radiation is "coned" down to the flat earth.

Radius of flat earth: 25477 km.  Area of a disc is = Radius² x PI = 2,038,216,561 km², or 2 x 10^9 km².  Clear?
Each km² is a 1000 x 1000 m, so it is 1 million m², or 1 x 10^6 m².
Now, converting 2x10^9km² to m² = 2x10^9 x 1x10^6 = 2x10^15m².  Still clear? Math correct, right?
Now consider that your flat earth theory 33% of earth area actually "coned" illuminated by the sun at any given time, divide this big number by 3, ending up with 6x10^14m², now multiply it by 1kW or 1000 Joules, that the sun pours of energy on the illuminated area, 6x10^14m² x 1x10^3W = 6x10^17W, that is beautiful, simple calculations.
Now, do you know how much energy all our hydroelectric plants, nuclear, coal, oil, wind, etc, on the whole planet generates at any given time?
In 2013 everything together produced 5.67x10^20 Joules, that is during a whole year, that has 31536000 seconds, or 3.1x10^7, so divide the 5.67x10^20 by 3.1x10^7 = 1.82x10^13 Watts per second, or Joules.   So, the sun, your little disc of only 30 miles in diameter, 3000 miles up, can generate 4 levels of magnitude more energy that ALL our generation on Earth?   

The question is HOW a 30 miles flat disc can generate 33 THOUSAND times more energy than ALL our energy production on the planet, and please, answer that with chemical/scientific explanations, so we could create an artificial Sun and solve all our energy problems.  See, according to this, an artificial SUN using the same flat earth magic energy generation, to fulfill ALL our energy needs could generate just double of what we are producing now, to do that, it could have an area 16500 times smaller than the 30 miles diameter flat earth sun. 

Area = Radius² x PI.   A 30 miles (48km) diameter disc has an area of 1.8 x 10^9 m², dividing it by 16500 = 1.1 x 10^5 m² = radius of 186 meters only...  Come on, this is smaller than two city block, LETS DO IT RIGHT NOW !!!!    Flat earth theory found the solution for our energy needs.  Perhaps a small 8 inch disc made of the same flat earth sun material, can fuel an electric car forever, can you imagine? no more fossil fuel engines? and better, using the fantastic levitation techniques of the flat earth sun, we can make cars flying without any propulsion.  Fantastic.

But first, please explain how your flat earth sun works, in technical details.

See, I have a small scientific calculator TI-36X and can do wonders with it.   Some people have tremendous difficult to understand TAN-1 or e-based Logs function, it is not their fault, some animals have no competent cognitive capacity, that is their nature.

Flat Earth Theory / Size of planets
« on: April 25, 2019, 10:06:16 PM »
The flat earth theory states the Earth's equator line is 40000 km in circumference, the diameter of the equator is then 12738 km, and as the equator is a center line between the North pole and the "ice wall" South pole, then the total diameter of the flat earth is 25477 km, making the ice wall of a circumference of 80038 km.  Clear?  Okay, the theory also states the Sun has the same diameter of the Moon, around 48 km (30 miles), and both are at the altitude of 4800 km (3000 miles).  Now, what is the size and altitude of Jupiter, Saturn, Mars, Venus, Mercury?  What about the size of the Jupiter natural satellites, Europa, Ganymede, Io, Callisto?   Also, how can we actual visually see Jupiter's satellites move around Jupiter after few hours, it proves they are orbiting Jupiter.  If you never did it, just grab any average telescope, set with an ocular no less than 25mm and enjoy the view, make drawings of pictures about the moons and compare 2 hours later, or next night.  You can actually see the moons go behind Jupiter and then cross in front of it, proving the orbital plane.  How this "orbital" can happens and how big are those moons in the flat earth theory design?  On the several pictures collected animated GIF below, see the moons disappearing on the Jupiter shadow on the left side, fir Io then Europa, the closer ones. 

I am very, very curious about the sizes of such planets and moons, please someone answer.

Flat Earth Theory / Magnetic Poles
« on: April 25, 2019, 09:20:52 PM »
Can you please explain how flat earth model has a somehow tick disc of dirt and rocks, with strong magnetic poles, being the North in the center and the South all around such disk?   Even a magnet in form of a donut or torus can not present such property.  It is impossible, it will be an external monopole.  If you lean a little bit about magnetic lines fields of force, you will realize each magnetic line leaves one pole and travels in a circular form and arrives at the other pole, so all lines are equivalent.  Imagine lines of force leaving your flat earth south pole, all around the ice wall, billions of lines, and arriving to a single point in the North pole, the magnetic force there would be fantastic and impossible strong.   That is not the case, se can measure the magnetic force properties on the South pole and on the North magnetic pole and find them exactly the same.   Care to explain this both discrepancies when considering the flat earth model?  It would be a complete havoc, but it is not.    The third point is, I am really curious to read about your explanations about how the flat earth generates a movable magnetic force, if not as a huge ball of melting rocks and iron rotating inside your planet, generating magnanimous electric current and magnetic fields.  Remember, the magnetic poles move in time, this is only possible with the spherical model.  Please explain.

Flat Earth Theory / Moon view angle.
« on: April 25, 2019, 09:09:59 PM »
In the flat earth model, the Moon is not too distant only 4800 km up, suppose it is right on top, azimuth, midnight, full moon.  It shows some features, craters, mountains, etc. A person 80° East, shall also see the Moon at 80° to the West almost setting, it will be almost 5:20 at morning for him.  Another person at 80° West, will see the same Moon at 80° East, it recently raised, it will be around 6:40 afternoon for him.   As this two persons are seeing the Moon at the same time, but with 160° distant one to another, the different angle of view would allow them to see different features, mountains and craters of the Moon, considering the Moon is also spherical like the video below, Lunar libration, showing a composition of 28 days.  But no, they both see the exactly same image of the Moon, because it is much far away than the flat earth model, impossible to see the "sides" of the Moon.  In flath earth model, the Moon is 30 miles in diameter and 3000 miles distant, just 100 times its diameter. Considering the Equator line (even in the flat earth model) is 40 thousand km in circumference, those two persons 160° apart will have an arc of 17700 km and a cord of 12500 km between them, what makes a triangle to the Moon of a base of 12500 km and a vertical of 4800 km, each one would see a side of the Moon by 21°, and that is pure math and geometry.  It means, one would see 138° of the Moon face from the other, but not, they see the same face and far stars pretty close to the Moon direction in the same position for each other what proves the noon parallax error.  The Moon is not flat, see the image below or grab the nearest binocular or telescope and confirm. I have both at home, tired to see craters and mountains of the Moon in detail.  With little understanding of mid school geometry you can explain the whole spherical system. Care to explain the flat one?   

This example is proportional and the same as two persons, one in each side of the 20m wide road, watching a car at 80m away, of course the person at the left side of the road will be able to see the left door, while the person at the right side will be able to see the right door, and not the other way around.  This doesn't happen in the above example of the Moon, the proportional distances are the same.

Also, as the flath earth equator diameter is 12700 km, if the two persons were 180° one to another, and if the Moon is straight up azimuth over the first, 4800km up, the triangle formed over the flat earth model will be 4800 up, 12700 ground and a hypotenuse of Sqr(4800²+12700²) = 13576km, what makes an angle of 20.7° for the other person, and we know we see the Moon disappear under the horizon, we can see it with bare eyes.  This alone explains the spherical model.  Care to explain why we don't see the Moon all the time in the sky, from 20° to 89° constantly? 

And yes, we can see the Moon during the day, if the angle against the Sun allows to illuminate it.

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