121

Flat Earth Theory / Re: Total Eclipse July 02 2019

« on: May 16, 2019, 04:49:41 PM »

Wow... the Oppolzer northern hemisphere 1900-1918 eclipses map posted on FE wiki Eclipses, as a reference to promote FE, had a hidden twin map for the South Pole (Antarctic continent) as center.  That is fantastic.  So, FE now improved, it is double-sided.  See how Australia's shape is more real. I wonder which way UA pushes it.   

Calculating here how Sun and Moon, both at 4800km altitude and 48km in diameter could promote the strange total eclipse path on  11/13/2012, changing more than 30° of latitude in a matter of few hours, considering only a small 12° of circling difference (longitude) between Sun and Moon in 24 hours See, same altitude, can only promote a straight down vertical shadow, total or partial, never angled.    July/22/2028 will have more than 40 degrees of latitude change.  FE behavior is amazing.  Someone may say the Moon is way down below the Sun, with a chaotic circling path (we don't observe that in the real world), but both Sun and Moon being 48km in diameter the projected umbra shadow will never be wider then 48km, the minimum noted all times was never smaller than 120km. 

A lot of things don't add up. FErs scientists and high knowledge specialists need urgent to define and post the right numbers.

122

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?

123

Flat Earth Theory / Re: How does the Sun Create Energy on a Flat Earth?

« on: May 14, 2019, 03:16:06 PM »

Any spectrograph measurement, using diffraction gratings (>2000gr/mm), shows exactly the radiation solar spectrum, no matter if you think FE or RE, the Sun doesn't change.  The analysis points to Hydrogen fusion into Helium, releasing energy, no matter what.  So, FE Sun is a Hydrogen Fusion Reactor, but considering its tinny small size of only 30km in diameter, it is totally impossible to accumulate enough gas to promote such pressure and temperature necessary to ignite the fusion process. Not even considering that gravity doesn't exist in the FE world, any gas would disperse in vaccuum.   In the real universe, not even Jupiter could do it, with a diameter of 142984km, 4766 times larger diameter than the 30km FE Sun) composed with 90% hydrogen, mass equivalent to 318 Earths.  The smallest Red Dwarf star, is 80 times bigger than Jupiter, so it needs to pack another 79 Jupiters into the actual one for it to have a narrow chance to ignite fusion and become a star.  We are talking about 381000 times larger than the FE Sun.

Want to make a comparison?  Think about a 1/2" (12.7mm) small glass marble as being the FE Sun, now, 12.7mm x 381000 = 3.024 miles, that is the equivalent diameter of more than 650 city blocks put together to form a 3 miles diameter circular area (do you want me to post the calculation?), or the equivalent to 519,841,729 US school buses piled into a huge ball, and that is the smallest Red Dwarf known to be able to ignite.  Do you really think a tinny glass marble 1/2" in diameter will ignite fusion? 

US total yellow school buses in 2015/2016: 474194.  It would be necessary 1096 times the entire US school bus fleet to build such 3 miles diameter ball, just to make a small Red Dwarf ignite as a star, when compared to the size of FE Sun as a small 1/2" marble.   Think again.
https://files.schoolbusfleet.com/stats/SBFFB18StateByState.pdf
A regular US school bus is 2.6m wide, 13.7m long, 3.2m high, 114m³
A 3 miles diameter ball has a volume (V=1.33*PI*R*R*R) 5.92 E+10m³



Below the solar spectrograph, with the black absortion lines showing its radiation and gases composition.

124

Flat Earth Theory / Re: Scientific proof??

« on: May 13, 2019, 07:51:03 PM »

With so many online applications and software helping to locate the next visible satellite at evening time on your exact latitude/longitude, is there anyone in this world that never saw a cruising satellite reflecting solar light, yet???  Start looking up.

125

Flat Earth Theory / Re: How is it possible to see the sun rise or set?

« on: May 13, 2019, 07:32:23 PM »

Sorry Reer, you are wrong, about the Sun's altitude and viewing angle.

The FE statement for the Sun is 30 km in diameter, 3000 km in altitude.  I made no calculations whatsoever, but it seems FErs use this altitude because it is the only possibility to flat a sphere with a very far away Sun with parallel rays and have the same shadows based on the oblate spheroid model.  Also, the diameter is purely based on apparent size of view (angular size).

Then, based on your assumption, the Sun being over the ICE wall (worst case) and the observer being also over the 180° opposite ICE wall, the rectangle triangle would have a base of 20000 km and the vertical of 3000 km, what gives a (atan(3/20)) of 8.53 degrees.  This would be the lowest inclination (altitude) the Sun would appear anywhere over the FE for an observer.  Anywhere the observer or the Sun moves, the altitude will increase.   

The best possible analogy for what is 8.5° of altitude, is looking to your home front door from the curb across the street.   A regular door is about 80 inches tall, a regular city street is about 30 ft wide plus 15 ft from the curb to the door, total 45ft = 540 inches.  It would be atan(80/540) = 8.4°.

So, just walk to the curb across the street and look back to the top of your home front door, that is the lowest altitude the Sun would be anywhere over FE.

Now, thinking about apparent size.  If the FE Sun right over you will have "x" view diameter, and it is 3000 km of altitude, on that viewing experience Ice wall to Ice wall, the hypotenuse will be sqr(3000²+20000²) = 20223 km, the delta size = 3000/20223 = 0.15 or 15%.  Suppose the apparent Sun size right over you is around a US Quarter Coin, at that longest distance it will be the size of your shirt button.  That is big enough to be completely visible and shinning bright on the sky, mostly considering that (according to FE wiki) the Sun is a globe spinning, shinning in all directions, not only as a disc spotting light down, as it was said before.   Notice that according to this size and altitude, vanishing point does not make it disappear at all. It would looks like a street lamp at 150ft (50m) away.

So, where is the night sun?
 
 

126

Flat Earth Theory / Re: Theory/Model Request

« on: May 13, 2019, 04:00:25 PM »

Oh so are you assuming that the sun is just a big lamp or what?

Hmmm, I guess it is not a natural fusion floating device generating so much energy in a flimsy 30km diameter thing.  Even Saturn is not big and dense enough to ignite fusion.  If yes, we need to find out how it works so we could create few more on the earth's surface... can you imagine the free energy?   30km is my daily comute distance, pretty small.  It is not fission device, at such distance we would be all cooked by radiation.   

127

Flat Earth Theory / Re: Angle and Length of a pole's shadow

« on: May 13, 2019, 03:54:33 PM »

Curious as to why you chose 4pm instead of 3pm?  3pm would have a 45° elevation angle just like 9am does, but 4pm will not.

45° is already there at 9pm, why repeat?
Now, 4pm was chosen exactly for the little offset on the shadow and angle, as a "control experience" and being high enough in the sky to suffer minuscule refraction on the atmosphere.

I wonder if Tom Bishop would have time to help with the numbers.

128

Flat Earth Theory / Re: Crisp clear horizon line

« on: May 12, 2019, 04:37:48 PM »

I've performed the experiment as well. I printed out a ship that had a hull 1/8th of an inch tall and the hull disappeared once I got far enough away.

I just performed such experiment, printed a whole ship on a letter size paper.
Taped it to a card and nailed to the middle of a tree trunk.
Walk away until I barely could see the paper (80m+) - binoculars show the whole page and ship, perfectly visible, no missing hull.
Walk towards the tree until I could see the image on paper but could not identify (30m+) - binoculars show perfect image.
Walk towards the tree until I could recognize the ship on paper (10m+), could see it entirely, no missing hull.

Repeated the experience with the card and paper touching the ground.
The views and results exactly as above.
No missing hull at anytime.

Just remember that ships on ocean sink completely under the horizon based on distance, not only hull, and the visible part has great visibility.
Also, after part or the whole ship disappears, the use of binoculars or telescopes doesn't bring it back, what eliminates any relation to the human eye acuity or optical resolution.

I don't know how familiar are you with telescopes and binoculars, I am very much. Using a telescope I can see a small Moon's crater or the beautiful Jupiter's red spot and its natural satellites, impossible at naked eye, what means, it didn't "disappear", it still there and optical apparatus could be used to still it yet, not the case of the ship's hull or the entire ship disappearing under the horizon, because in that case the curved horizon just hid the object.

I already wrote about that, the problem to use ships below the horizon, is that the reference is always a complete bad video, fuzzy, lots of mist and moisture in the middle, stabilization, etc.  Those videos are the horror in full extend for anyone involved with expensive optics.   When I see those videos I wonder what a heck this people are using? holding the camera by hand? no steady tripod? solar filter oil all over the lens? no image filters? no post-processing software to clean up the mist?   Even my first home made telescope decades ago could produce a crisp and better image.  The chosen video is on purpose, to create more discussion than answers.

Below some videos with better visibility, but lacking knowledge of "controlled experience", no "mirage", "refraction" or "optical resolution", we can still see the very narrow masts and tops all the way.
https://www.youtube.com/watch?v=sNGorPyX9OQ

The following is a little bit better, using calculation and flashing lights from the target, what makes it easier for the geometry.
https://www.youtube.com/watch?v=GrihjP5tTTM

A little more good technology used in the next one.
https://www.youtube.com/watch?v=QVa2UmgdTM4

The following uses a control reference setup (over the hill recording), great video.
https://www.youtube.com/watch?v=RX2m_NfrJD4

Another great experiment, you can see the bottom of the building didn't just vanish due optical resolution, mirage, refraction.
https://www.youtube.com/watch?v=MoK2BKj7QYk

129

Flat Earth Theory / Re: Angle and Length of a pole's shadow

« on: May 12, 2019, 02:52:04 PM »

Read about the equinox here: https://wiki.tfes.org/Equinox

Sorry Tom, found no answer on the link to help me calculate the angles requested.

No matter very tinny refraction of the sun's light on the atmosphere, an observer in Quito will see the 9am Sun on Mar20/21 pretty close to altitude 45° at azimute 90° (totally East).

Can you pretty please dedicate few minutes and calculate the angles and shadows sizes for me?

Oh, I just realize the 9am shadow length, because the Sun is at 45° it forms an isosceles right triangle, and the shadow will be exactly the same as the height of the pole, 10m.   Am I correct Tom?

130

Flat Earth Theory / Re: Angle and Length of a pole's shadow

« on: May 11, 2019, 04:41:16 PM »

A modest contribution to help Tom Bishop, I made the drawing below, based on a FE map.
I am almost able to calculate the angles, just waiting Tom numbers for confirmation.
The shadow lengths on the ground are more difficult, due FE perspective and vanishing points.

131

Flat Earth Investigations / Re: Round Earth proof - comments?

« on: May 11, 2019, 03:57:43 PM »

I post a similar but easier experience, a single pole on Quito Equator, and kindly request Tom Bishop to answer.
I know he is very busy, but I am sure he will be glad to answer.

https://forum.tfes.org/index.php?topic=14748.0

132

Flat Earth Investigations / Re: Flat Earth Map

« on: May 11, 2019, 03:48:55 PM »

Pull up Rowbotham's map. Australia is its normal size there. The map you usually see online is a globe projection.

Tom, can you please be dear and post here two easy things, according to FE map:

1) The direct physical distance from Perth to Sidney in Australia
2) The direct physical distance from Perth (AU) to Cape Town in Africa

The numbers don't need to be very precise, any 10 km error is acceptable.

133

Flat Earth Media / Re: Flat Earth - No Boat Went Over The Horizon

« on: May 11, 2019, 03:36:12 PM »

This kind of absurd observation under no control whatsoever is the same as measuring a patient blood pressure and pulse rate during a sky diving, and assume the patient is ill due the rapid changes in the numbers.   I really don't know why people continues to assume observing a boat going down the horizon, under terrible observation will be the answer for all the questions.  On the video above there is no way to say the experience has any control, so it is totally invalid for any statements both from REs or FEs... it is ridiculous to pursue it otherwise.    One could say, it is the same as measuring the size and weight of the chicken while it runs like crazy all over the yard.  Situation like that is just to promove endless debating without any scientific repeatable results and answer, it seems it was chosen on purpose for that end.

134

Flat Earth Theory / Re: The IceWall NASA lies and their is an IceWall at the end of the earth

« on: May 11, 2019, 03:23:18 PM »

Last picture of my previous post disappeared.
Here goes again, the map of Antarctica with the locations of the stations from all over the world.




Antarctica Live Cams:
https://www.usap.gov/videoClipsAndMaps/mcmwebcam.cfm
https://www.usap.gov/videoclipsandmaps/spwebcam.cfm
http://www.antarctica.gov.au/webcams
https://www.geocam.ru/en/in/antarctica/

135

Flat Earth Theory / Re: The IceWall NASA lies and their is an IceWall at the end of the earth

« on: May 10, 2019, 08:15:13 PM »

Antarctica is not an ICE WALL, it is just a regular beautiful continent, mountains, vast land, covered in snow and ice.  There are several research groups and facilities there, zillions of pictures all over the internet.   Let me ask you this:  Do you ever went to Africa?  if not, how do you know it exists?

136

Flat Earth Investigations / Re: Flat Earth Map

« on: May 10, 2019, 07:58:21 PM »

Agreed, southern hemisphere FE map fails tremendously.  The inventors of FE thought people from southern hemisphere are native dumb people that can not read or write, and would never contest such very wrong map and statements.  They just forget the 11% world population down there are not monkeys, there are plenty of universities, very large cities, heavy industry, research centers, scientists.  Antarctica is packed with research groups from all over the world.  Everything related to the Southern Hemisphere, South Pole and Antarctica literally kills any FE statements.  They also count with people from North Hemisphere never traveled overseas, never stepped inside an airplane, can't use a calculator, can't do trigonometry, never studied physics (never studied anything), never gazed the universe through a telescope.  The FE southern hemisphere map is a literal attack to human intelligence.

137

Flat Earth Projects / Re: Wiki - Sun

« on: May 10, 2019, 04:27:25 PM »

Just some numbers to help the calculations:
The solar energy perpendicular that hits Earth's ground is 1kW/m².
It is also calculated the average energy all lands receive under direct sunlight, due spherical inclination, to be around 300W/m².
The FE has no steep angles like a sphere has, so lets increase this power to 500W/m² on FE as a logical guess.

If you calculate the average FE area of the FE disc that actually receives solar radiation, you can calculate the solar energy radiated from that particular steradian of the Sun, extrapolate for the whole solar sphere radiance energy. 

If the covered FE area is "A m²" and the average energy measured on the ground is 500W/m² then the energy necessary to cover such area is "A x 500W".  Considering the Sun a sphere radiating energy all around, you can calculate the exact angle from the FE Sun that covers the stated FE area. 

As this area is larger than the 30km (diameter of the FE Sun), more than half of the sphere will be projecting radiation down to the FE area, but on the steep angles the radiation is angled and reduced, so, to facilitate the calculations, lets assume just 1/3 of the FE Sun's "visible disc" is illuminating the FE area. It would represent 1/6 from the whole sphere producing energy hits FE.

I am being good here, since the Isosceles Triangle formed between FE diameter (40000km) and Sun's height (3000km) represents the illumination "cone"  that hits FE with 162° of aperture.

It means we could assume the FE Sun produces 6 times more energy that what FE receives, so the FE Sun could produces "6 x A x 500W".

Due the extreme difficult to understand and calculate the illuminated area over FE, considering now, May 10 Central Florida, I can see the Sun from 6:38am to 8:06pm, it is 13h 28min of solar exposure.  If 12 hours of sun represents 180° of solar coverage over the equator line (15°/h) with an average power of 500W/m², then 13:28h would represent 193° of a circular line around FE North Pole.  That great semicircle means the Sun is illuminating the FE from East to West, reaching the FE Ice Wall, and obviously the FE North Pole. 

As only FErs can calculate that, I will only guess a possible 40% of the total FE disc being illuminated by the solar light.

As the distance from North Pole to equator is 10,000 km, the FE disc radius would be 20,000km, what means a total area of (PI*R²) 1256636000 km² or 1.25 E+9 km² or 1.25 E+15 m².   So, 40% of that is 5 E+14 m², applying the formula 6 x A x 500W, result in total Sun's energy radiated as 6 x 5 E+14 x 500W = 15000 E+14 = 1.5 E+18 W , or, 1500 PetaWatt.

Based on the same calculations, A x 500W would be the energy FE receives directly from the Sun, 5 E+14 x 500W = 2500 E+14 = 2.5 E+17 or 250 PetaWatts.

On RE measurements, we know the Sun pours 174 PetaWatts directly over our planet, 30% of it is reflected back to the universe, the 70% rest is absorbed by clouds, oceans and land.  It means around 122 PetaWatts (122 E+15 W).   With all the guessing, my calculations above results in 250 PW, what is just double the actual RE calculations.

See, to adjust the solar radiation over FE from 250PW to 122PW, we need to reduce the land area illuminated by the Sun, from 40% to 20% in the calculations above.  But with only 20% of area, it would be impossible for the Sun to cover 193° over the circle at 38°N, I would NOT be seeing the Sun raise at 6:38am and set at 8:06pm on Central Florida on May 10, the day light would short than that.

Somebody must explain this to me.

 

138

Flat Earth Theory / Re: Gravity is not constant so why is acceleration?

« on: May 10, 2019, 02:30:49 PM »

I am dying to measure the precise length of a spring, like the one in the picture, in the vacuum, at sea level (1 ATM, 14.7 PSI) and even 10m underwater (2 ATM, 200kPa), considering the spring metal at the same temperature in each measurement.    Considering most of weight scales use a simple spring to counteract the gravity acceleration, air pressure will not interfere with measurement.  The electronic ones use strain gage resistive sensors and measurement electronics on a tick piece of metal, second image, measuring the microns of metal deformation, what acts exactly like a spring but much better resolution and precision.

139

Flat Earth Theory / Re: Crisp clear horizon line

« on: May 09, 2019, 09:57:38 PM »

Lake Pontchartrain transmission line, 15 miles, at 8" /mile² is 1800" (45.72m) down the horizon curvature.

140

Flat Earth Theory / Re: Crisp clear horizon line

« on: May 09, 2019, 09:28:03 PM »

I wonder if we can have real pictures demonstrating figure 73 above.
Any hand drawing objects used as proof of scientific research is at least childish.

I understand pretty well "optical resolution", I have several telescopes and deal with for deep sky observation.
Optical resolution has nothing to do with an artifact disappear on your vision, it is related to "resolving details" by Rayleigh criterion.
On your figure 73 item C, the center white dot would become fuzzy and edges out of focus, not disappearing like that.
When the wavelength becomes larger than the viewing angle, it just mix to each other, not disappears. 

This is a serious issue on microscopy since the observed micro elements can be out of resolution, needing much better and expensive optics.  It is not by chance that the electronic industry was improving its light wavelength projection every time they reduce silicon wafer width traces and slits. Decades ago they used regular visible light, as it become narrow and narrow, the slit definition in the films went way over the optical resolution of the light, they needed to migrate to blue, UV and even RX for the wavelength to be able to go through the film image density of details.

Also, lighter details tend to overcome the wavelength of less lighter artifacts, so, sorry, the white dot would even spread its fuzzy over its edge becoming a little bit larger.  This is why on sniper training they use while dots on center target, so the sniper would see the center target even a great distances. Also, snipers are trained to seek and focus on lighter artifact on the target for effect, even if smaller.
The same for your figure 74-C, that is not how it works, it will become fuzzy, but no magical disappearance.

In real life, bottom of several examples are normally in the shadow of itself, less brightness, tend to be more difficult to see.  The bottom of a car at distance tend to be confused with the road itself, but they don't disappear, you can see them, measure them, even zoom and nicely photograph them.  The bottom of a ship over the sea at distance just don't magically disappear by optical resolution, in that case is just below the horizon, you can zoom, telescope, whatever you want, the bottom will never to be seen, you can even see in details the windows and even rivets at the water level, what means resolution is pretty great.  And above all, if you lift yourself by helicopter, not changing your distance to the chip, you still be able to see the bottom of the chip, since you changed the horizon level.  Neat, isn't it?

Below, the fuzzy image (a) and (b) shows that effect, and better, the radiance of (b) didn't change from (a), since both still present in the same fuzzy image.  They don't disappear.

Sorry, no cigar this time.




Rayleigh criterion:



Further reading:
https://www.olympus-lifescience.com/en/microscope-resource/primer/digitalimaging/deconvolution/deconresolution/